Chapter 7

Linear Momentum

Units of Chapter 7

•Momentum and Its Relation to Force

•Conservation of Momentum

•Collisions and Impulse

•Conservation of Energy and Momentum in Collisions

•Elastic Collisions in One Dimension•Inelastic Collisions

•Collisions in Two or Three Dimensions

•Center of Mass (CM)

•CM for the Human Body

•Center of Mass and Translational Motion

7-1 Momentum and Its Relation to Force

Momentum is a vector symbolized by the symbol p, and is defined as

The rate of change of momentum is equal to the net force:

This can be shown using Newton’s second law.

(7-1)

(7-2)

Example 7-1For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms, estimate the average force on the ball. Would this force be large enough to lift a 60 kg person?

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v2 = 55 m/s, v1 = 0, Δt = 0.004 s

F =Δp

Δt=

mv2 - mv1

Δt=

(0.060 kg)(55 m/s) - 0

0.004 s= 800 N

To lift a 60 kg person would require a force of

mg = (60 kg)(9.80 m/s2) = 600 N.

Example 7-2Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it. (That is, we ignore any splashing back.) What is the force exerted by the water on the car?

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pinitial = mv = (1.5 kg)(20 m/s) = 30 kg ⋅m/s,pfinal = 0

F =Δp

Δt=

pfinal - pinitial

Δt=

0 - 30 kg ⋅m/s

1.0 s= -30 N

The minus sign indicates that the force on the water

is opposite to the water's original velocity. The car

exerts a force of 30 N to stop the water, so the water

exerts a forcec of 30 N on the car (Newton's 3rd law).

7-2 Conservation of MomentumDuring a collision, measurements show that the total momentum does not change:

(7-3)

7-2 Conservation of Momentum

More formally, the law of conservation of momentum states:

The total momentum of an isolated system of objects remains constant.

Example 7-3A 10,000. kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed just afterward?

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pinitial = mAvA +mBvB = mAvA, pfinal = (mA +mB)v'

pinitial = pfinal

mAvA = (mA +mB)v'

v'=mA

mA +mB

vA =(10000. kg)

(10000. kg +10000. kg)(24.0 m/s) =12.0 m/s

7-2 Conservation of Momentum

Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.

Example 7-4Calculate the recoil velocity of a 5.0 kg rifle that shoots a 0.020 kg bullet at a speed of 620 m/s.

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momemtum before = momentum after

mBvB +mRvR = mBvB' +mRvR

'

0 +0 = mBvB' +mRvR

'

vR' = -

mBvB'

mR

= -(0.020 kg)(620 m/s)

(5.0 kg)= -2.5 m/s

7-3 Collisions and Impulse

During a collision, objects are deformed due to the large forces involved.

Since , we can

write

The definition of impulse:

(7-5)

7-3 Collisions and Impulse

Since the time of the collision is very short, we need not worry about the exact time dependence of the force, and can use the average force.

7-3 Collisions and ImpulseThe impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time.

This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

Example 7-6

(a) Calculate the impulse experienced when a 70. Kg person lands on firm ground after jumping from a height of 3.0 m. (b) Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged, and again (c) with bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm.

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(a) We need to know the velocity of the person

just before striking the ground.

v2 = v02 +2a(y - y0)

v = 2g(y - y0) = 2(9.8 m/s2)(3.0 m) = 7.7 m/s

F Δt = Δp = mΔv = (70 kg)(0 - 7.7 m/s) = -540 N ⋅s

The minus sign means the force acts upwards, as

it is opposed to the original momentum (downward).

Example 7-6 (cont’d)

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(b) Average speed during collision : v =7.7 m/s +0 m/s

2= 3.9 m/s

Collision lasts for : Δt =d

v=

1.0x10-2 m

3.9 m/s= 2.6x10-3 s

F Δt = 540 N ⋅s → F =540 N ⋅s

2.6x10-3 s= 2.1x105 N

F = ΣFy = Fgrd - mg → Fgrd = F +mg = 2.1x105 N +(70 kg)(9.80 m/s2) = 2.1x105 N

(c) Δt =d

v=

0.50 m

3.9 m/s= 0.13 s

F =540 N ⋅s

0.13 s= 4.2x103 N

Fgrd = F +mg = 4.2x103 N +(70 kg)(9.80 m/s2) = 4.9x103 N

7-4 Conservation of Energy and Momentum in Collisions

Momentum is conserved in all collisions.

Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic.

7-5 Elastic Collisions in One Dimension

Here we have two objects colliding elastically. We know the masses and the initial speeds.

Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds.

Example 7-7Billiard ball A of mass m moving with speed v collides head-on with ball B of equal mass at rest (vB=0). What are the speeds of the two balls after the collision, assuming it is elastic?

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Conservation of momentum gives :

mv = mvA' +mvB

' → v = vA' + vB

'

Conservation of kinetic energy gives :

vA - vB = vB' - vA

' → v = vB' - vA

'

So we can set our 2 equations equal :

vB' - vA

' = vA' + vB

' → 0 = 2vA' → vA

' = 0

v = vB' - vA

' so vB' = v

Example 7-8A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60x104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe=4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space.

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mpvp +0 = mpvp' +mHevHe

'

vp - 0 = vHe' - vp

' → vp' = vHe

' - vp

We plug in for v p' in our first equation :

mpvp = mpvHe' - mpvp +mHevHe

'

vHe' =

2mpvp

mp +mHe

=2(1.01 u)(3.60x104 m/s)

1.01 u + 4.00 u=1.45x104 m/s

vp' = vHe

' - vp =1.45x104 m/s - 3.60x104 m/s = -2.15x104 m/s