Momentum and Impulse

Physics 6A

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Momentumany moving object will have both momentum and kinetic energy

We use a lowercase p for momentum – here is the formula

vmprr⋅=

Notice that this is a vector, so you will have to break into components when adding up the momenta of several objects.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Momentumany moving object will have both momentum and kinetic energy

We use a lowercase p for momentum – here is the formula

vmprr⋅=

Notice that this is a vector, so you will have to break into components when adding up the momenta of several objects.

Impulseif the momentum of an object changes, then either its mass or its velocity must have

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For Campus Learning Assistance Services at UCSB

if the momentum of an object changes, then either its mass or its velocity must have changed. In most cases, we will consider the mass to be constant, so a change in momentum will mean the object accelerated (velocity changed).

So there must have been a force applied to it. This gives us a formula:

tFpJ avg ∆⋅=∆=rrr

This is just Newton’s 2nd Law – can you see why?

Momentumany moving object will have both momentum and kinetic energy

We use a lowercase p for momentum – here is the formula

vmprr⋅=

Notice that this is a vector, so you will have to break into components when adding up the momenta of several objects.

Impulseif the momentum of an object changes, then either its mass or its velocity must have

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

if the momentum of an object changes, then either its mass or its velocity must have changed. In most cases, we will consider the mass to be constant, so a change in momentum will mean the object accelerated (velocity changed).

So there must have been a force applied to it. This gives us a formula:

tFpJ avg ∆⋅=∆=rrr

This is just Newton’s 2nd Law – can you see why?

Solve this for Favg to get amt

vmtp

Favgr

rrr=

∆

∆⋅=

∆

∆=

Typical Example:

A golf ball (initially at rest) is struck by a club, and the ball is given a velocity of 50 m/s.

If the mass of the ball is 46 grams and the club is in contact with the ball for 5 ms, what is the average force applied to the ball by the club?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Typical Example:

A golf ball (initially at rest) is struck by a club, and the ball is given a velocity of 50 m/s.

If the mass of the ball is 46 grams and the club is in contact with the ball for 5 ms, what is the average force applied to the ball by the club?

We can find the impulse (change in momentum) directly in this problem:

Take a minute and compute it for yourself…

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Typical Example:

A golf ball (initially at rest) is struck by a club, and the ball is given a velocity of 50 m/s.

If the mass of the ball is 46 grams and the club is in contact with the ball for 5 ms, what is the average force applied to the ball by the club?

We can find the impulse (change in momentum) directly in this problem:

The ball starts at rest, so the initial momentum is 0.

After the club hits it, the ball has momentum p =(0.046kg)(50m/s)=2.3kg-m/s

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For Campus Learning Assistance Services at UCSB

After the club hits it, the ball has momentum pfinal=(0.046kg)(50m/s)=2.3kg-m/s

So the impulse is ∆p=2.3kg-m/s

Typical Example:

A golf ball (initially at rest) is struck by a club, and the ball is given a velocity of 50 m/s.

If the mass of the ball is 46 grams and the club is in contact with the ball for 5 ms, what is the average force applied to the ball by the club?

We can find the impulse (change in momentum) directly in this problem:

The ball starts at rest, so the initial momentum is 0.

After the club hits it, the ball has momentum p =(0.046kg)(50m/s)=2.3kg-m/s

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For Campus Learning Assistance Services at UCSB

After the club hits it, the ball has momentum pfinal=(0.046kg)(50m/s)=2.3kg-m/s

So the impulse is ∆p=2.3kg-m/s

Now we can divide by ∆t to find Favg=460N

- Impulses always occur as action-reaction pairs- Longer impact time means smaller force, and vice-versa“Real-Life” Examples:Automobile safety:DashboardsSeatbelts and AirbagsCrumple zones

Product packagingStyrofoam spacersEgg cartons

SportsBaseball: Padded catcher’s mitt

Notes on Impulse

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Baseball: Padded catcher’s mittGolf: Deformation of ball during brief impactBoxing gloves: Padding means smaller impact to the hands (and face, etc.)Cycling: Always wear your helmet!

Sleep technologyHead on a pillow more comfortable than head on a block of wood.

Conservation of MomentumAs long as there are no outside forces, the total momentum of a system of objects will not change.

We will use this idea mainly in collision problems (but it is much more fundamental, and always applies)

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Conservation of MomentumAs long as there are no outside forces, the total momentum of a system of objects will not change.

We will use this idea mainly in collision problems (but it is much more fundamental, and always applies)

Collisions (2 main categories)Elastic Collisions – KE is conserved (no dissipative forces)

A good approximation for things like billiard balls

Your textbook has shortcut formulas for head-on elastic collisions (page 244).

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Inelastic Collisions – some KE is lost due to dissipative forces during impact

Special Case: When the objects stick together, the collision is perfectly inelastic. This will be the easiest type of problem to solve, because there is only 1 object after the collision.

Momentum is conserved in all collisions.

Use conservation of momentum first in all collision problems.

Use these formulas if you want to avoid some possibly difficult algebra

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=?? v=3m/s

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BEFORE AFTER

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=?? v=3m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

BEFORE AFTER

This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.

Write down the initial and final momentum for the system, then set them equal.

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=?? v=3m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

BEFORE AFTER

This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.

Write down the initial and final momentum for the system, then set them equal.

)v)(kg50()5)(kg100()3)(kg50kg100(

pp

boysm

sm

if

+=+

=

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=?? v=3m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

BEFORE AFTER

This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.

Write down the initial and final momentum for the system, then set them equal.

sm

boy

boysm

sm

if

1v

)v)(kg50()5)(kg100()3)(kg50kg100(

pp

−=

+=+

=

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=?? v=3m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

BEFORE AFTER

This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.

Write down the initial and final momentum for the system, then set them equal.

sm

boy

boysm

sm

if

1v

)v)(kg50()5)(kg100()3)(kg50kg100(

pp

−=

+=+

=

Why is the answer negative? What does that mean?

Example: A 100kg man and a 50kg boy are at a father-son ice skating jamboree. The man skates toward his son at a speed of 5 m/s, and picks the boy up and puts him on his shoulders. If their final speed is 3 m/s, find the initial velocity of the boy.

v=5m/svboy=-1m/s v=3m/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

BEFORE AFTER

This is a perfectly inelastic collision (they stick together). So we only need conservation of momentum.

Write down the initial and final momentum for the system, then set them equal.

sm

boy

boysm

sm

if

1v

)v)(kg50()5)(kg100()3)(kg50kg100(

pp

−=

+=+

=

The boy is initially moving toward his father.

Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final velocity of each ball. Assume the collision is elastic.

10 m/s

BEFORE

V=??

AFTER

V=??

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For Campus Learning Assistance Services at UCSB

Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final velocity of each ball. Assume the collision is elastic.

10 m/s

BEFORE

V=??

AFTER

V=??

Since the collision is elastic we can use the formulas on page 244 of your book:

ba vmm

v ⋅−

=a v

m2v ⋅=

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For Campus Learning Assistance Services at UCSB

i,aba

baf,a v

mmmm

v ⋅+

−= i,a

ba

af,b v

mmm2

v ⋅+

=

Example: A golf ball (45 grams) rolls toward a billiard ball (135 grams). The initial speed of the golf ball is 10 m/s, and the billiard ball is initially at rest. Find the final velocity of each ball. Assume the collision is elastic.

10 m/s

BEFORE

Vb=??

AFTER

Va=??

Since the collision is elastic we can use the formulas on page 244 of your book:

ba vmm

v ⋅−

=a v

m2v ⋅=

ab a

b

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For Campus Learning Assistance Services at UCSB

sm

f,a

sm

f,a

i,aba

baf,a

5v

10kg135.0kg045.0kg135.0kg045.0

v

vmmmm

v

−=

⋅+

−=

⋅+

−=

sm

f,b

sm

f,b

i,aba

af,b

5v

10kg135.0kg045.0

)kg045.0(2v

vmm

v

+=

⋅+

=

⋅+

=

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?

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For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

We should start by writing down the components of each player’s initial velocity:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Next we can write down the formula for conservation of momentum in each direction:

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Conservation of momentum in each direction:

x,fsm

sm v)kg170()0()kg80()6.4()kg90(

directionx

⋅=⋅+⋅

−

y,fsm

sm v)kg170()4()kg80()9.3()kg90(

directiony

⋅=⋅+⋅

−

Total mass = 90kg + 80kg

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Conservation of momentum in each direction:

sm

x,f

x,fsm

sm

4.2v

v)kg170()0()kg80()6.4()kg90(

directionx

=⇒

⋅=⋅+⋅

−

sm

y,f

y,fsm

sm

9.3v

v)kg170()4()kg80()9.3()kg90(

directiony

=⇒

⋅=⋅+⋅

−

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Conservation of momentum in each direction:

sm

x,f

x,fsm

sm

4.2v

v)kg170()0()kg80()6.4()kg90(

directionx

=⇒

⋅=⋅+⋅

−

sm

y,f

y,fsm

sm

9.3v

v)kg170()4()kg80()9.3()kg90(

directiony

=⇒

⋅=⋅+⋅

−

Combine the components to find the magnitude and direction of the final velocity:

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Conservation of momentum in each direction:

sm

x,f

x,fsm

sm

4.2v

v)kg170()0()kg80()6.4()kg90(

directionx

=⇒

⋅=⋅+⋅

−

sm

y,f

y,fsm

sm

9.3v

v)kg170()4()kg80()9.3()kg90(

directiony

=⇒

⋅=⋅+⋅

−

Combine the components to find the magnitude and direction of the final velocity:

sm2

sm2

sm

f 6.4)9.3()4.2(v =+=

Example: This one will be 2-dimensional. Two hockey players are initially skating as shown in the figure. They collide, and stick together. Find their final velocity (magnitude and direction).

Player B

40°

vfinal=?This is a perfectly inelastic collision, so we should use conservation of momentum. We need to have separate formulas for x- and for y-directions

Components of each player’s initial velocity:

sm

sm

y,a

sm

sm

x,a

9.3)40sin()6(v

6.4)40cos()6(v

=⋅=

=⋅=

o

o

sm

y,b

x,b

4v

0v

=

=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Mass 80kg

v=4 m/s

Player A

Mass 90kg

v=6 m/s

Conservation of momentum in each direction:

sm

x,f

x,fsm

sm

4.2v

v)kg170()0()kg80()6.4()kg90(

directionx

=⇒

⋅=⋅+⋅

−

sm

y,f

y,fsm

sm

9.3v

v)kg170()4()kg80()9.3()kg90(

directiony

=⇒

⋅=⋅+⋅

−

Combine the components to find the magnitude and direction of the final velocity:

sm2

sm2

sm

f 6.4)9.3()4.2(v =+=

o584.29.3

)tan( =θ⇒=θ