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CT11-1
A ball bounces off the floor as shown. The direction of the impulse of the ball,
...
A:
straight up 
B:
straight down 
C:
to the right 
D:
to the left 

 p , is
Answer: straight up.
  



p  p f  p i , p f  p i  p. Also, since
 
p  Favg t , the direction of the force from the floor is the same as the direction of the

impulse  p
pf
p
pi
CT11-2
Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart
for 3 s, and then the other for the same length of time, exerting equal force on each, the
momentum of the light cart is..
A: four times
D: one-half
B: twice
C: equal to
E: one-quarter
the momentum of the heavy cart.


Answer: the momenta are equal. p  Favg t so if we apply the same force for the same
time, we will get the same increase in momentum. Applying the same force to the heavier
cart will make it go more slowly, but it has more mass, so it will get the same
momentum.
CT11-3
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same
momentum, and you exert the same force to stop each. How do the time intervals to stop
them compare?
A: It takes less time to stop the ping-pong ball.
B: Both take the same time.
C: It takes more time to stop the ping-pong ball.
Correct answer is both take the same time. Both have the same momentum (the bowling
ball must
be going very slow or else the ping-pong ball is going really fast) and
 
p  Favg t . If we apply the same force to each, it will take the same time to stop both.
CT11-4
A big ball, mass M=10m, speed v, strikes a small ball, mass m, at rest. Could the
following occur: The big ball comes to a complete stop and the small ball takes off with
speed 10v?
10
m
v
m
m
10v
M
A: Yes, this can occur.
B: No, it cannot occur because it would violate momentum conservation
C: No, it cannot occur because it would violate conservation of energy.
Answer: It cannot occur because it would violate conservation of energy. Although
momentum is conserved in this collision,
Ptot,i = (10m)v = Ptot,f = m(10v),
Energy is not conserved, since the KE after is 10 times bigger than the KE before. If any
heat were generated in the collision, energy conservation would be violated even more.
KEi = (1/2)(10m)v2  KEf= (1/2)m(10v)2 =(1/2)m(100)v2
In real life, it would not be possible for the big mass (10m) to come to a complete stop as
a result of the collision.
CT11-5 Ball 1 strikes stationary Ball 2 in 1D elastic collision. The initial momentum of
p1i
p2f


p1i , and the final momentum of Ball 2, p 2 f , are shown on the graph.

What is the magnitude of the x-component of p 1f ?
Ball 1,
A: 0
B: 1
C: 2
D: 3
E: None of these
Answer: Since this is a 1D collision, I can write ptot = p1i = p1f + p2f (since p2i = 0). So
p1f = p1i - p2f = 4 - 5 = -1. The magnitude of p1f is +1.
CT11-5 Ball 1 strikes stationary Ball 2 in 2D elastic collision. The initial momentum of


Ball 1, p1i , and the final momentum of Ball 2, p 2 f , are shown on the graph.

What is the x-component of p 1f ?
A: 0
B: 1
C: 2
D: 3
p2f
p1i
E: None of these
Answer: 1 If there are no external forces on the system of 2 masses, then the total vector
momentum must remain constant. We must have ptot  p1i  p1f  p2f
p2f
p1f
p1i