fisica generale - alan giambattista, betty mccarty richardson copyright © 2008 – the mcgraw-hill...

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

Chapter 7: Linear Momentum

•Definition of Momentum

•Impulse

•Conservation of Momentum

•Center of Mass

•Motion of the Center of Mass

•Collisions (1d, 2d; elastic, inelastic)

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§7.1 Momentum

Consider two interacting bodies:

F21 F12

If we know the net force on each body then tm

t netFav

The velocity change for each mass will be different if the masses are different.

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tm 2111 Fv

Rewrite the previous result for each body as:

ttm 211222 FFv

Combine the two results:

ifif mm

mm

222111

2211

vvvv

vv

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The quantity (mv) is called momentum (p).

p = mv and is a vector.

The unit of momentum is kg m/s; there is no derived unit for momentum.

5



From slide (3):

21

2211

pp

vv

mm

The change in momentum of the two bodies is “equal and opposite”. Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other.

6



§7.2 Impulse

Definition of impulse: t Fp

One can also define an average impulse when the force is variable.

7



Example (text conceptual question 7.2): A force of 30 N is applied for 5 sec to each of two bodies of different masses.

30 N

m1 or m2

Take m1<m2

(a) Which mass has the greater momentum change?

t FpSince the same force is applied to each mass for the same interval, p is the same for both masses.

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(b) Which mass has the greatest velocity change?

Example continued:

m

pv

Since both masses have the same p, the smaller mass (mass 1) will have the larger change in velocity.

(c) Which mass has the greatest acceleration?

t

v

aSince av the mass with the greater velocity change will have the greatest acceleration.

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Example (text problem 7.9): An object of mass 3.0 kg is allowed to fall from rest under the force of gravity for 3.4 seconds. What is the change in momentum? Ignore air resistance.

Want p =mv.

(downward) m/s kg 100p

m/sec 3.33

vm

tgv

tav

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Example (text problem 7.10): What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds? Assume frictionless ice.

N 5.7

s 0.20

m/s 0.3kg 0.50

av

av

av

F

t

m

t

t

vpF

Fp

The force will be in the direction of motion.

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§7.3 Conservation of Momentum

v1i

v2i

m1 m2

m1>m2

m1 m2

A short time later the masses collide.

What happens?

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During the interaction:

N1

w1

F21

N2

w2

F12

x

y

1121

11 0

amFF

wNF

x

y

2212

22 0

amFF

wNF

x

y

There is no net external force on either mass.

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The forces F12 and F21 are internal forces. This means that:

ffii

ifif

2121

2211

21

pppp

pppp

pp

In other words, pi = pf. That is, momentum is conserved. This statement is valid during the interaction only.

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Example (text problem 7.18): A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet leaves the barrel?

As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved.

m/s 1.82m/s 820kg 4.5

kg 01.0

0

br

br

rrbb

fi

vm

mv

vmvm

pp

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§7.4 Center of Mass

The center of mass (CM) is the point representing the mean (average) position of the matter in a body. This point need not be located within the body.

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The center of mass (of a two body system) is found from:

21

2211

mm

xmxmxcm

This is a “weighted” average of the positions of the particles that compose a body. (A larger mass is more important.)

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In 3-dimensions write

ii

iii

cm m

m rr where

iimmtotal

The components of rcm are:

ii

iii

cm m

xmx

ii

iii

cm m

ymy

ii

iii

cm m

zmz

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Example (text problem 7.27): Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of10.0 grams. Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.0 cm, 5.0 cm)?

x

y

A

x

ba

bb

ba

bbaacm mm

xm

mm

xmxmx

ba

bb

ba

bbaacm mm

ym

mm

ymymy

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cm 8

cm 2g30g 10

g 10

b

bcm

x

xx

cm 20

cm 5g 10g 30

g 10

b

bcm

y

yy

Example continued:

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Example (text problem 7.30): The positions of three particles are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (-1.0 m, -2.0 m). The masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively. What is the location of the center of mass?

x

y

3

2

1

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m 92.1

kg 364

m 1kg 3m 2kg 6m 4kg 4321

332211cm

mmm

xmxmxmx

m 38.1

kg 364

m 2kg 3m 4kg 6m 0kg 4321

332211cm

mmm

ymymymy

Example continued:

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§7.5 Motion of the Center of Mass

For an extended body, it can be shown that p = mvcm.

From this it follows that Fext = macm.

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Example (text problem 7.35): Body A has a mass of 3.0 kg and vx = +14.0 m/s. Body B has a mass of 4.0 kg and has vy = -7.0 m/s. What is the velocity of the center of mass of the two bodies?

Consider a body made up of many different masses each with a mass mi.

The position of each mass is ri and the displacement of each mass is ri = vit.

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For the center of mass:

ii

iii

m

mt

rvr cmcm

Solving for the velocity of the center of mass:

ii

iii

ii

i

ii

m

m

mt

m vr

vcm

Or in component form:

ii

ixii

m

vm ,

xcm,v

ii

iyii

m

vm ,

ycm,v

Example continued:

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Applying the previous formulas to the example,

m/s 6

kg 43

m/s 0kg 4m/s 14kg 3

,,,

ba

xbbxaaxcm mm

vmvmv

m/s 4

kg 43

m/s 7kg 4m/s 0kg 3

,,,

ba

ybbyaaycm mm

vmvmv

Example continued:

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§§ 7.6 & 7.7 Collisions in One and Two Dimension

When there are no external forces present, the momentum of a system will remain unchanged. (pi = pf)

If the kinetic energy before and after an interaction is the same, the “collision” is said to be perfectly elastic. If the kinetic energy changes, the collision is inelastic.

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If, after a collision, the bodies remain stuck together, the loss of kinetic energy is a maximum. This type of collision is called perfectly inelastic.

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Example (text problem 7.41): In a railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.0m. The two cars couple together upon collision.

(a) What is the speed of the two cars after the collision?

m/s 2.0

0

121

1

212111

2121

vmm

mv

vmmvmvmvm

pppp

pp

ffii

fi

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(b) Suppose instead that both cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.0 m/s.

m/s 25.0

00

12

12

2211

2121

ii

ii

ffii

fi

vm

mv

vmvm

pppp

pp

Example continued:

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Example (text problem 7.49): A projectile of 1.0 kg mass approaches a stationary body of 5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse direction along the same line with a speed of 5.0 m/s. What is the speed of the 5.0 kg mass after the collision?

m/s0.3m/s 5.0m/s 10kg 0.5

kg 0.1

0

112

12

221111

2121

fif

ffi

ffii

fi

vvm

mv

vmvmvm

pppp

pp

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Example (text problem 7.58): Body A of mass M has an original velocity of 6.0 m/s in the +x-direction toward a stationary body (B) of the same mass. After the collision, body A has vx=+1.0 m/s and vy=+2.0 m/s. What is the magnitude of body B’s velocity after the collision?

A

B

vAi

Initial

B

AFinal

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fxfxix

fxfxixix

fxix

vmvmvm

pppp

pp

221111

2121

0

fyfy

fyfyiyiy

fyiy

vmvm

pppp

pp

2211

2121

00

x momentum:

Solve for v2fx: Solve for v2fy:

m/s 00.5

11

2

11112

fxix

fxixfx

vv

m

vmvmv

y momentum:

m/s 00.2

12

112

fyfy

fy vm

vmv

m/s 40.522

22

2 fxfyf vvvThe mag. of v2 is

Example continued:

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Summary

•Definition of Momentum

•Impulse

•Center of Mass

•Conservation of Momentum

fisica generale - alan giambattista, betty mccarty richardson copyright © 2008 – the mcgraw-hill...

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x slide

inelastic slide

center of mass collisions

unit of momentum

larger mass

total momentum

body system