Physics 30 Unit 1 – Momentum and Impulse To accompany Pearson Physics.

Physics 30

Unit 1 – Momentum and Impulse

To accompany Pearson Physics

Momentum

• “quantity of motion” - Newton• is the product of mass and velocity• Is a vector,

• has units of m

kgs

p m v

Momentum

• Newton’s 2nd Law, , can be rewritten as

as your text shows on page 450. We’ll use this later to determine a quantity called impulse

• Try Practice Problem 1 page 451

F ma

net

pF

t

Momentum

• m = 65 kg + 535 kg = 600 kg

3600 11.5 / = 6.9 10m

p m v kg m s kgs

Since velocity is a vector, direction must be given along with the numerical answer as shown on page 451

Practice Problem 1, page 451:

Momentum

• Momentum “varies directly as” or “is directly proportional to” both

Study Example 9.2 page 452

• Try Practice Problem 1 page 452

and m v

Momentum

• Practice Problem 1, page 452

3 34 2

3 3 9 94 2 8 8

8.25 W

8.25 9.28 W

mp m v kg m m v v

sm m

p m v m v m v kg kgs s

Momentum

• Do questions 1 – 3, 5, 6, 8, 9, 11 page 453

Momentum and Impulse

• Quick Lab 9.2 page 455 (discuss or do and discuss)

• Earlier you saw that Newton’s 2nd Law could be rewritten as:

net

pF

tThis equation can be rewritten as:

netp F t


• The change in momentum is equal to the net force times the time for the change,

netF t

This quantity is called IMPULSE and is equal to the change in momentum


• Since , netp F t

netm v F t

For a given change in momentum (or change in velocity if mass is constant), a fast change will mean a large force while a slower change will mean a smaller force


Practical applications?

• cushioned soles on runners

• airbags

• crumple zones in cars

• others?

Additional examples p. 463 - 465


• Units of impulse:

• Units of momentum:

Since impulse = change in momentum, these units must be same. Your text shows this on page 457.

N s

m

kgs


• Direction of Impulse???

Same as the direction of the change of momentum

If you are travelling west in a car at constant speed, and hit the brakes,

If you are travelling west in a car at constant speed, and step on the gas,

Impulse is east

Impulse is west


• Try Practice Problems 1 and 2 page 458


• Practice Problem 1, p. 458

a) Impulse 200 3.64 728N s N s S

b) Impulse 728 1100

728 0.662 /

1100

p N s k vg

N sm s S

kgv


• Practice Problem 2, p. 458

Since velocity is a vector, it is necessary to take direction into account when determining change in velocity

400 1.80 / 0.200 / 4.20

190 forward

net

net

net

m v t

kg m s m s

F N

Fs

F


• When the net force is not constant, the impulse can be calculated by finding the area under an F vs t graph

• Your textbook has a number of examples on pages 459 – 461

• Discuss golf ball graph page 462

Momentum

• Review Example 9.4, page 462

• Do Practice Problem 1, page 462



c)

impulse

0.650 4.4

4.46.8 /

0.650

m

kg N s

N sg

vk

v

m

v

s

force vs time for a basketball shot

0

5

10

15

20

25

0 0.1 0.2 0.3 0.4

time (s)

forc

e (N

)

a)

b) Impulse = total area = A1+A2

1 12 20.15 22 0.25 22

4.4

s N s N

N s

A1 A2


• Do questions 1, 2, 4, 7, 8, 9b, 10, page 467

Momentum: 1d Collisions

• Note that in collisions friction is present as a constant force before, during, and after the collision

• Since collision times are relatively short, friction can be ignored in collision questions provided you are using only momentum of the objects immediately before or after the collision

• This will always be the case for you

Momentum

• Lab 9.5 page 471

Momentum : 1d Collisions

• Law of Conservation of Momentum:

When no external net force acts on a system, the momentum of the system remains constant

i fsys sysp p

• Review pages 474 – 479including examples 9.5, 9.6, 9.7, and 9.8


• Try Practice Problem 1 on page 476, Practice Problem 2 on page 477, Practice Problem 1 on page 478, and Practice Problem 1 on page 479

• Types of interactions: explosions including:????hit and stickhit stationary and bounce backhit moving and bounce back



The negative sign on the velocity means it is opposite in direction to the astronaut, that is, towards the astronautFinal expression of answer???

0

0

0 110 0.80 / 4000

110 0.80 /0.022 /

4000

i f

ss

s

sys sys

ast ss

ast ast ss

s

ss

v

v

p p

p p

m v m

kg m s kg

kg m sm s

kgv

Momentum:1d Collisions


2 2

2

1 1 2 1 1,2

32

1050 2.65 / 0 1050 0.78 /

1050 2.65 / 1050 0.78 /2.5 10

0.78 /

i f

car car

car

c

sys sys

car car car car c

ar

ar

p p

m v v m v

kg m s kg m s

kg m s kg

m m

m

m sm s

m kg

Since initial and final velocities are in the same direction, there is no need to assign positive and negative values


• Practice Problem 1 page 478

• 1.2 m/s [ W ]

• Note the use of positive and negative on the velocities: + right, [E], up, [N]

- left, [ W ], down, [S]

0.25 2.0 / 0 0.25 0.79 / 0.58

0.25 2.0 / 0.25 0.79 /1.2 /

0.58

i fsys sys

vb vb bb bb vb vb bb

bb

bb

bb

p p

m v m v m v m

kg m s kg m s kg

kg m s k

v

v

g m sk

v m sg



• 0.62 m/s [E]

72 1.6 / 87 1.4 / 72 0.84 / 87

72 1.6 / 87 1.4 / 72 0.84 /0.62 /

87

i fsys sys

sksb sb sk sk s

sk

s

k

k

b sb s

p p

m v m v m v m

kg m s kg m s k

v

v

v

g m s kg

kg m s kg m s kg m sm s

kg

Momentum : Elastic and Inelastic Collisions

• Elastic collisions: Ek conserved

• Since Ek is a scalar direction of v is unimportant

• No true elastic collisions in the macroscopic world:

Collision between: % Elastic

tennis ball and tile floor 66%

golf ball and tile floor 87%

basketball and tile floor 74%


• Inelastic collisions: Ek not conserved

• Macroscopic collisions are always inelastic to some degree

• 100% inelastic – both objects are deformed and stick together

• Collisions between vehicles always involve deformation even when the vehicles bounce off each other

• Review examples 9.9 and 9.10• Do Practice Problem 1 on each of pages

484 and 485


• Practice Problem 1, page 484• Energy conserved in pendulum swing:

• Momentum conserved in collision of bullet and block

0.00259 0 1.00 0.00259 1.01 /

1.00 0.00259 1.01 /391 /

0.00259

bullet block block bullet block bullet blbullet

bullet

bulle

ock

t

m m v m m v

kg kg kg m s

kg kg m sm s

kg

v

v

v

2122

2

0.00259 1.00 0.00259 1.00 9.81 0.0520

2 9.81 0.0520 1.01 /

bullet block

bullet block

mkg kg kg kg ms

m m

v

v m ss



212

212

Initial:

0.012 14 / 0 1.1

Final:

0.012 0.200 0.78 / 0.064

1.1 0.064 1.1

kdart k block glider

k dart block gli

k lost

der

E E kg m s J

E kg kg m s

J JE

J

J


• Do Check and Reflect Questions, page 486

1, 2a, 3, 6, 8, 10


• Real collisions are most often 3 dimensional, but the techniques to solve them are the same as those for 2 dimensional

• Because momentum is a vector, vector analysis is required:


• Look at the information available on the Physics 30 Data Sheets:

• I recommend, that like in Physics 20, you measure all angles with respect to the positive x-axis


• If you do this, x and y components of any vector R will be found by:

• The angle between the x axis positive or negative and the vector will always be determined by:

cos sinx yR R R R

1tan y

x

R

R


• You can save work by drawing a good vector diagram and using the law of sines and law of cosines to solve even non-right angle triangles

• I don’t recommend this• Review example 9.12 – I’ll redo it

here


• First change 12.0°[E of N] to 78.0° [N of E] (standard position)

• Since both masses are the same we can ignore mass in the calculation – work with velocities only

vector x-component y-component

0 1.20 m/s

0 0

? ?

AvBvAv

Bv

1.17 / cos78.0

0.243 /

m s

m s

1.17 / sin78.0

1.14 /

m s

m s


0 0.243 /

0.243 /

B x

B x

B

A x

x

xv v

m v

v

s

m

v

s

1.20 / 1.14 /

1.20 / 1.14 / 0.0556 /

B y

B y

y

B y

A yv v

m s m s

m s m s

v

v

mv s

We now have everything needed to find ! Bv


2 22 2

0.243 / 0.0556 /

0.250 /

BB x B y

B

v v m s

v

m s

m

v

s

1 1 0.0556 /tan tan 12.9

0.243 /y

x

R m sR m s

Since x is negative and y is positive the resultant is in the 2nd quadrant – therefore 12.9° [N of W ] or 167°


• Try Practice Problem 1, page 491


vector x-component Σx y-component Σy

0189

kg•m/s

0

0

189 kg•m/s 0

? ?

centp

defp

defp

centp

104 1.82 /

189 /

kg m s

kgm s

97.0 0.940 /

91.2 /

kg m s

kgm s

91.2 / cos341.5

86.5 /

kgm s

kgm s

91.2 / sin341.5

28.9 /

kgm s

kgm s


189 / 86.5 /

189 / 86.5 / 103 /def x

def x

x cent x def x

p

p p

kgm s k

p

gm s

kgm s kgm s kgm s

0 28.9 /

0 28.9 / 28.9 /def y

def

y cent y de

y

f y

p

p p

kgm s

kgm s kgmp s

2 2

2 2103 / 28.9 /

107 /

107 /1.03 /

104

def

def

de

def

def x de

f

f yp p

kgm s kgm s

kgm s

kgm sm s

kgv

p

p

p


• Practice Problem 1, page 491, continued

1 1 28.9 /tan tan 15.7

103 /y

x

R kgm sR kgm s

Since both x and y are positive, this is a 1st quadrant angle:

[15.7° N of E]


• Practice Problem 2, page 492vector x-component Σx y-component Σy

-150 kg•m/s

-150

kg•m/s

0-454

kg•m/s

0 -454 kg•m/s

? ?

centp

defp

pairp

108 4.20 /

454 /

kg m s

kgm s

100 1.50 /

150 /

kg m s

kgm s

2 2

2 2150 / 454 / 478 /

paipair

pa

r x pair y

ir

p p

kgm s kgm s kgp m

p

s


• Since x and y are both negative this is in a 3rd quadrant angle

71.7° [S of W ]

1 1

478 /2.30 /

100 108

454 /tan tan 71.7

150 /

a

y

p

x

ir

kgm sm s

kg kg

R kgm sR kgm s

v


• Review Example 9.14, page 494• Try Practice Problem 1, page 494



• Mass of missing 3rd fragment =

0.058 kg – 0.018 kg – 0.021 kg = 0.019 kg


Momentum of piece # x-component y-component

0 0

0 0.043 kg•m/s

0.034 kg•m/s 0

? ?

Practice Problem 2, page 494

1 0.018 2.40 /

0.043 /

p kg m s

kgm s

2 0.021 1.60 /

0.034 /

p kg m s

kgm s

1 2 3 0p

3 ?p

3

3

1

3

2

0 0 0.034 /

0.034 /

x x

x

x x

x

p p

kgm s

kg

p

p

mp s

31 2

3

3

0 0.043 / 0

0.043 /

y y yy

y

y

p p

kgm s

kg

p

p m

p

s


3

0.055 /2.9 /

0.019kgm s

mkg

v s

mass of 3rd piece

1 1 0.043 /tan tan 52.1

0.034 /y

x

R kgm sR kgm s

Since both x and y components are negative the angle is in the 3rd quadrant:

52.1° [S of W ]


• 2d collisions can likewise be elastic or inelastic; the test is to check for conservation of kinetic energy





• Since Ek is a scalar, it is not necessary to consider direction of velocity (nice)

• Ek puck:

• Ek goalie:

• Total Ek before:

212 0.168 45.0 / 170k kg m sE J

212 82.0 0.200 / 1.64k kg m sE J

170 1.64 172k J JE J


• Ek goalie and puck:

• Inelastic collision → percent elastic:

212 82.0 0.168 0.192 / 1.51k kg kg m sE J

1.51

100 10%elasticit 0 0.882%17

y2

k before

k after

E JE J


• Subatomic particles are the only “objects” that can have true 100% elastic collisions

• Do Check and Reflect, page 499,

questions 1, 5, 6, 7, 8, 10

Physics 30 Unit 1 – Momentum and Impulse To accompany Pearson Physics.

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Transcript of Physics 30 Unit 1 – Momentum and Impulse To accompany Pearson Physics.