This Week MomentumMondaybull What is momentum bull p = mv

Tuesdaybull How does momentum

change (impulse)bull I = mΔv = fΔt

Wednesdaybull Quizbull Conservation of

momentum bull pinitial=pfinal

Thursdaybull Elastic and inelastic collisions

Fridaybull Quizbull Momentum conservation in 2D

Monday bull Problem Solvingbull Integrating our knowledge

Today What is Momentum

Momentum in the Vernacular

bull In everyday experience momentum is the amount ldquounfrdquo an object has

So what factors affect the momentum of an object

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

A car will certainly hurt more whyBecause it is more massive (more mass)

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Today What is Momentum

Momentum in the Vernacular

bull In everyday experience momentum is the amount ldquounfrdquo an object has

So what factors affect the momentum of an object

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

A car will certainly hurt more whyBecause it is more massive (more mass)

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum in the Vernacular

bull In everyday experience momentum is the amount ldquounfrdquo an object has

So what factors affect the momentum of an object

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

A car will certainly hurt more whyBecause it is more massive (more mass)

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

A car will certainly hurt more whyBecause it is more massive (more mass)

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

What affects Momentum

Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph

A car will certainly hurt more whyBecause it is more massive (more mass)

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

What affects Momentum

Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph

The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum Defined

Momentum is the product of mass and velocity

This is normally written p = m x vBolded letters denote vectors

What are the units of momentump = m x vm kg v ms

p kg bull ms kilogram meters per second

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

p=mv

Which has more momentum my car or me

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

p=mv

Which has more momentum my car or me

vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Homework

bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice

on momentum

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Todayrsquos Schedule (Jan 34)

bull How does momentum change (impulse)bull I = mΔv = fΔt

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

How Does Momentum of an Object Change

p=mvConsider Δp=ΔmvWhat does this mean

Why is this not a change in momentum of the object

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

How Does Momentum of an Object Change

p=mvConsider Δp=mΔvWhat does this mean

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

How Does Momentum of an Object Change

p=mvConsider Δp=mΔv

This means that velocity is changing Unlike Δm Δv does not imply that the object is

falling apart or clumping together

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Introducing Impulse

bull Δp is know as impulse (I)

bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction

but I changed it)bull I had a sudden impulse to ldquordquo

(I was suddenly did something that was not planned)

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Δp = I = mΔv

A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms

What is the impulse the bowling pins provide to a bowling ball

I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Other Ways to Find Impulse

I = mΔv

Remember a = ΔvΔt therefore Δv = aΔt

Substituting in we getI = maΔt

Remember F=maSubstituting in we get

I=FΔt

I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

I = Δp = mΔv = FΔt

I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal

A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Bouncy ball Clay ball

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Homework

bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-

Momentum Change Theoremrdquo (link on website)

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Todayrsquos Schedule (Jan 4)

bull Quiz on What is momentumbull Conservation of momentum

pinitial=pfinal

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Two-Minute Problem

A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision

A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Conservation of Momentum

Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2

When I push on the desk it pushes back on me with equal force in the opposite direction

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Steel ball demoI = Δp = mΔv = FΔt

Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Steel ball demo

Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1

F12 = -F21

Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1

Δt12 = Δt21

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Derivation of Conservation of Momentum

Recall from last class that I = Δp = mΔv = FΔt

F1 = -F2

Δt1 = Δt2= Δt

I1= F1 ΔtI2 = F2 Δt = - F1 Δt

I1=-I2

In any interaction momentum gain of one object is equal to the loss of momentum from another

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Total Momentum of a System is conserved

If there are no outside forces acting on an system the momentum of that system remains constant it is conserved

This is the property momentum conservation

ptotal = Σp = p1 + p2 + p3 + hellip

ptotalinitial = ptotalfinal

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum Conservation

A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct

A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip

ptotailnitial = ptotalfinal

The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision

A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal

Time to get tricky you may need a calculator

A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision

A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Homework

bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I

expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom

ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Todayrsquos Schedule (Jan 5)

bull Summary of Equationsbull Schedule for the next week and a half

- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum

equation uses

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

p=mv

What is the momentum of a bee that weighs 10 grams and flies at 2 ms

10g=01kg p=mv=01x2= 02kgms

How does that compare to a tortoise that weighs 1kg and moves at 05ms

p=mv=1x05= 05kgms

The tortoise has more momentum

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The

football (300g) is kicked at 25ms What was the impulse

300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms

The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact

I = FΔt

75 = F 1

F= 75 N

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)

What is the impulse that the space shuttle provides to him

What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

ptotailnitial = ptotalfinal

ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f

ptotailnitial = ptotalfinal

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

No Official Homework

We will be working on more difficult problems from here on out

If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems

Review kinematics if the projectile motion was difficult

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Egg Throw

(bring a jacket)

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Other Similar Impulse Applications

bull Climbing ropesbull Airbagsbull Circus Netsbull Diving

GolfBaseball following through on your swings

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Sample Test Question

Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision

(a) Calculate the speed of Ball B immediately after the collision

(b) Calculate the horizontal displacement d

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one

The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall

The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

2 Dimensional MomentumWhich of the following are possible final momenta for the case below

A

B

C

D

E

F

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Full 2D Problem on Board

Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis

What is the velocity (direction and speed) of Ball 2

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Tricky Extension

In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision

(d) Calculate the y-component of incident ball As momentum immediately after the collision

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Tricky Extension

(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Tricky Extension

(d) Calculate the y-component of incident ball As momentum immediately after the collision

We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of

your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other

Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data

bull Do any one of the three worksheet problems posted on the website

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Practice Test Questions

RocketTurbine

Try out the problems in the worksheet section

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

RocketsCan we use momentum conservation

Yes there are no outside forces on our system since we are ignoring gravity

Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms

Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

Which sort of problem is this

A Total momentum

B Impulse

C Conservation of momentum

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63

Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water

B Impulse

We are dealing with change in momentum and forces over a period of time

Try solving now on your own

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50
• Slide 51
• Slide 52
• Slide 53
• Slide 54
• Slide 55
• Slide 56
• Slide 57
• Slide 58
• Slide 59
• Slide 60
• Slide 61
• Slide 62
• Slide 63