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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
(7-1)
The rate of change of momentum is equal to the
net force:
(7-2)
This can be shown using Newton’s second law.
Example 7-1
For 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?
v 2 = 55 m/s, v1 = 0, t = 0.004 s
p mv 2 - mv 1 (0.060 kg)(55 m/s) - 0
F=
=
=
= 800 N
t
t
0.004 s
To lift a 60 kg person would require a force of
mg = (60 kg)(9.80 m/s 2 ) = 600 N.

Example 7-2
Water 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?
p initial = mv = (1.5 kg)(20 m/s) = 30 kg  m/s, p final = 0
p p final - p initial 0 - 30 kg  m/s
=
=
= -30 N
t
t
1.0 s
The minus sign indicates that the force on the water
F=
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 Momentum
During 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 7-3
A 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?
p initial = mAv A +mBv B = mA v A , p final = (mA +mB )v'
p initial = p final
mAv A = (mA +mB )v'
mA
(10000. kg)
v' =
vA =
(24.0 m/s) =12.0 m/s
mA +mB
(10000. kg +10000. kg)
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 7-4
Calculate the recoil velocity of a 5.0 kg rifle that shoots a 0.020 kg
bullet at a speed of 620 m/s.
momemtum before = momentum after
mBv B +mR v R = mBv'B +mR v'R
0 +0 = mBv'B +mR v'R
mBv'B
(0.020 kg)(620 m/s)
v === -2.5 m/s
mR
(5.0 kg)
'
R
7-3 Collisions and Impulse
During a collision, objects
are deformed due to the
large forces involved.
Since
write
, we can
(7-5)
The definition of impulse:
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 Impulse
The 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.
(a) We need to know the velocity of the person
just before striking the ground.
v 2 = v 20 + 2a(y - y 0 )
v = 2g(y - y 0 ) = 2(9.8 m/s 2 )(3.0 m) = 7.7 m/s
Ft = p = mv = (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)
7.7 m/s + 0 m/s
= 3.9 m/s
2
d 1.0x10 -2 m
Collision lasts for : t = =
= 2.6x10 -3 s
v
3.9 m/s
540 N  s
Ft = 540 N  s  F =
= 2.1x10 5 N
-3
2.6x10 s
F = Fy = Fgrd - mg  Fgrd = F + mg = 2.1x10 5 N +(70 kg)(9.80 m/s 2 ) = 2.1x10 5 N
(b) Average speed during collision
: v=
d 0.50 m
=
= 0.13 s
v 3.9 m/s
540 N  s
F=
= 4.2x10 3 N
0.13 s
Fgrd = F + mg = 4.2x10 3 N + (70 kg)(9.80 m/s 2 ) = 4.9x10 3 N
(c) t =
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 7-7
Billiard 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?
Conservation of momentum gives
:
mv = mv 'A +mv 'B  v = v'A + v'B
Conservation of kinetic energy gives
:
v A - v B = v'B - v'A  v = v'B - v'A
So we can set our 2 equations equal :
v'B - v'A = v'A + v'B  0 = 2v'A  v'A = 0
v = v'B - v'A so v'B = v

Example 7-8
A 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.
mp v p +0 = mp v'p +mHev'He
v p - 0 = v'He - v'p  v'p = v'He - v p
We plug in for v
'
p
in our first equation :
mp v p = mp v'He - mp v p +mHev'He
'
He
v
2mp v p
2(1.01 u)(3.60x10 4 m/s)
=
=
=1.45x10 4 m/s
mp +mHe
1.01 u + 4.00 u
v'p = v'He - v p =1.45x10 4 m/s - 3.60x10 4 m/s = -2.15x10 4 m/s
