CHAPTER 7: Linear Momentum
Answers to Questions
1.
For momentum to be conserved, the system under analysis must be “closed” – not have any
forces on it from outside the system. A coasting car has air friction and road friction on it,
for example, which are “outside” forces and thus reduce the momentum of the car. If the
ground and the air were considered part of the system, and their velocities analyzed, then the
momentum of the entire system would be conserved, but not necessarily the momentum of
any single component, like the car.
7.
“Crumple zones” are similar to air bags in that they increase the time of interaction during a
collision, and therefore lower the average force required for the change in momentum that
the car undergoes in the collision.
10.
The momentum of an object can be expressed in terms of its kinetic energy, as follows.


p  mv  m2 v 2  m mv 2  2m

1
2

mv 2  2mKE .
Thus if two objects have the same kinetic energy, then the one with more mass has the
greater momentum.
Solutions to Problems
12. Consider the motion in one dimension with the positive direction being the direction of
motion of the bullet. Let “A” represent the bullet, and “B” represent the block. Since there
is no net force outside of the block-bullet system (like frictions with the table), the
momentum of the block and bullet combination is conserved. Note that vB  0 .
pinitial  pfinal  mAv A  mB vB  mAvA  mB vB 
vB 
mA  vA  vA 
mB

 0.023 kg  230 m s  170 m s 
2.0 kg
 0.69 m s
15.
(a)
The impulse is the change in momentum. The direction of travel of the struck
ball is the
positive direction.
p  mv   4.5  102 kg   45 m s  0   2.0 kg m s
(b)
The average force is the impulse divided by the interaction time.
p 2.0 kg m s
F

 5.8  10 2 N
3
t
3.5  10 s
22. Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have
vA  3.30 m s and vB  0 . Use Eq. 7-7 to obtain a relationship between the velocities.
vA  vB    vA  vB   vB  vA  vA
Substitute this relationship into the momentum conservation equation for the collision.
mA vA  mB vB  mA vA  mB vB  mA vA  mA vA  mB  vA  vA  
0.220 kg
 mA  mB 
vA 
 3.30 m s   1.10 m s  east 
0.660 kg
 mA  mB 
vB  vA  vA  3.30 m s  1.10 m s  4.40 m s  east 
vA 
35 Use conservation of momentum in one dimension. Call the direction of the sports car’s
velocity the positive x direction. Let A represent the sports car, and B represent the SUV.
We have vB  0 and vA  vB . Solve for vA .
pinitial  pfinal  mA vA  0   mA  mB  vA  vA 
mA  mB
vA
mA
The kinetic energy that the cars have immediately after the collision is lost due to negative
work done by friction. The work done by friction can also be calculated using the definition
of work. We assume the cars are on a level surface, so that the normal force is equal to the
weight. The distance the cars slide forward is x . Equate the two expressions for the work
done by friction, solve for vA , and use that to find vA .
Wfr   KEfinal  KEinitial after
collision
 0  12  mA  mB  vA2
Wfr  Ffr x cos180o    k  mA  mB  g x
 12  mA  mB  vA2    k  mA  mB  g x  vA 
vA 
mA  mB
mA
vA 
mA  mB
2  k g x 
mA
2  k g x
920 kg  2300 kg
920 kg

2  0.80  9.8 m s 2
  2.8 m 
 23.191m s  23 m s
40. Use this diagram for the momenta after the decay. Since there was no
momentum before the decay, the three momenta shown must add to 0 in both
the x and y directions.
 pnucleus  x  pneutrino
 pnucleus  y  pelectron
pnucleus 

  tan 1
 pnucleus  x   pnucleus  y
2
 5.40 10
 pnucleus  y
 pnucleus  x
23
2

 pneutrino    pelectron 
2
2
   9.30 10 kg m s   1.08 10 kg m s
 9.30 10 kg m s   59.9
p 
 tan
p 
 5.40 10 kg m s 
kg m s
2
2
23
22
23
 tan 1
p nucleus
pneutrino
electron
1
o
23
neutrino
The second nucleus’ momentum is 150o from the momentum of the electron.

pelectron
43. Call the final direction of the joined objects the positive x axis. A diagram of
the collision is shown. Momentum will be conserved in both the x and y
directions. Note that vA  vB  v and v  v 3 .
py :
 mv sin 1  mv sin  2  0  sin 1  sin  2  1   2
px : mv cos 1  mv cos  2   2m  v 3  cos 1  cos  2 
cos 1  cos  2  2 cos 1 
2
3
2
3
mv A
1
2
mv B
 1  cos 1 13  70.5o   2
1   2  141o
47. Choose the carbon atom as the origin of coordinates.
10
mC xC  mO xO 12 u  0   16 u  1.13 10 m
xCM 

 6.5  1011 m from the C
mC  mO
12 u  16 u
atom.


J7: What average force does a 2.0 kg ball travelling at 20 m/s exert on a wall if it
rebounds at 10 m/s? Assume normal incidence, and that the collision lasts 0.001 seconds.