Physics 11 Assignment KEY
Chapter 7: Momentum & Collisions
1. Define the following terms:








Momentum - the product of an object’s mass and its velocity
Impulse - the product of the force exerted on an object and the time interval over which the
force acts
Impulse-momentum theorem - the impulse is equal to the change in momentum of an object
involved in an interaction
Closed system – a system where no objects enter or leave
Isolated system – a system upon which no external forces act
Elastic collisions – collisions where both momentum and kinetic energy are conserved
Inelastic collisions – collisions where only momentum is conserved
Law of conservation of momentum – in any closed, isolated system when two or more
objects interact, the total momenta of all objects before the interaction equals the total
momenta of all objects after the interaction
2. We claim that momentum is conserved. Yet most moving objects eventually slow down and stop.
Explain.
Momentum is conserved if no outside forces are acting on an object. In our everyday
experiences outside forces such as friction and air resistance are present to oppose the
motion of most moving objects, which in time will slow them down and/or stop them. If an
“isolated” system were to be broadened to include friction and air resistance, i.e., all of
earth, then momentum would be conserved.
3. When a person jumps from a tree to the ground, what happens to the momentum of the person
upon striking the ground? Explain.
Consider this problem as a very light object hitting and sticking to a very heavy object. The
large object – small object combination (Earth + jumper) would have some momentum after
the collision, but due to the very large mass of the Earth, the velocity of the combination is
so small that it is not measurable. Thus the jumper lands on the Earth, and nothing more
happens.
4. A person carrying a heavy backpack is stranded in the middle of a frozen, frictionless lake. Her
legs are frozen stiff and cannot be used to walk or crawl. What could the person do to save herself
and move closer to the shore?
If the person would have faced away from the shore and thrown the backpack directly away
from the shore, she would have acquired a velocity towards the shore by conservation of
momentum. Since the ice is frictionless, she would slide all the way to the shore.
5. Today’s automobiles are designed with ‘crumple zones’ that collapse upon impact. How do these
crumple zones prevent serious injury, compared to older, more rigid vehicle designs?
“Crumple zones” are similar to air bags in that they increase the time of interaction during
a collision. Since the change in momentum of the collision is constant regardless of
whether or not ‘crumple zones’ are used, according to the impulse-momentum theorem, a
lower average force is required F  t  p .
6. Is it possible for a body to receive a larger impulse from a small force than from a large force?
The impulse is the product of the force and the time duration that the force is applied. So
the impulse from a small force applied over a long time can be larger than the impulse
applied by a large force over a small time.
7. A light body and a heavy body have the same kinetic energy. Which has the greater momentum?
Explain.
The momentum of an object can be expressed in terms of its kinetic energy, as follows:


1

p  mv  m 2 v 2  m mv 2  2m mv 2   2mEk 
2

Thus if two objects have the same kinetic energy, then the one with more mass has the
greater momentum.
8. What is the magnitude of momentum of a 22-g sparrow flying with a speed of 8.1 m/s?
p = mv = (.022 kg)(8.1 m/s) = 0.18 kg  m/s
9. A child in a boat throws a 5.40-kg package out horizontally with a speed of 10.0 m/s. Calculate the
velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg
and that of the boat is 55.0 kg.
During the throwing, we use momentum conservation law for the one-dimensional motion:
(m boat + m child  m package )  v = (m boat + m child )  v boat  child  m package  v package
0 = (m boat + m child )  v boat  child  m package  v package
- (m boat + m child )  v boat  child = m package  v package
v boat  child =
v boat  child =
m package  v package
- (m boat + m child )
(5.40 kg)  (10.0 m/s)
 0.67 m/s (opposite to the direction of the package
- (55.0 kg + 26.0 kg )
10. Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of
1300 kg/s with a speed of 40,000 m/s (at the moment of takeoff).
The force on the gas can be found from its change in momentum. The force on the rocket
is equal in magnitude and opposite in direction to the force on the gas (Newton’s 3rd law).
F  t  p  mv
F 
mv mv  m 

  v  1300 kg/s (4.0 x 10 4 m/s)  5.2 x 10 7 N downward
t
t
 t 
11. A 9500-kg boxcar traveling at 16 m/s strikes a second car (which is initially at rest). The two stick
together and move off with a speed of 6.0 m/s. What is the mass of the second car?
For the one-dimensional motion, we take the direction of the first car for the positive
direction. For this perfectly inelastic collision, we use law of momentum conservation:
m boxcar #1  v boxcar #1  m boxcar #2  v boxcar #2 = (m boxcar #1 + m boxcar #2 )  v boxcars #1 & #2
m boxcar #1  v boxcar #1  0 = (m boxcar #1 + m boxcar #2 )  v boxcars #1 & #2
m boxcar #1  v boxcar #1
= m boxcar #1 + m boxcar #2
v boxcars #1 & #2
(9500 kg)  (16 m/s)
= m boxcar #1 + m boxcar #2
6.0 m/s
25333 kg = m boxcar #1 + m boxcar #2
25333 kg - 9500 kg = m boxcar #2
1.58  10 4 kg = m boxcar #2
12. A 15-g bullet strikes and becomes embedded in a 1.10-kg block of wood placed on a horizontal
surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is
0.25, and the impact drives the block a distance of 9.5 m before it comes to a rest, what was the
muzzle speed of the bullet?
On the horizontal surface after the collision, the normal force is FN = (m + M)•g. We find the
common speed of the block and bullet immediately after the embedding by using the workenergy principle for the sliding motion:
W friction  E k
1
( M  m)v 2
2
1
 k ( M  m)  g  d  ( M  m)v 2
2
1 2
 k  g  d  v
2
Ff  d 
20.25(9.81 m/s 2 )(9.5 m)   v
6.83 m/s  v
For the collision, we use law of conservation of momentum:
m  v  M  V = (M  m)  v M  m
m  v  0 = (M  m)  v M  m
(M  m)  v M  m
=v
m
(1.10 kg  0.015 kg )  (6.83 m/s )
=v
0.015 kg
507 m/s = v
13. A 13-g bullet traveling 230 m/s penetrates a 2.0-kg block of wood and emerges going 170 m/s. If
the block is stationary on a frictionless surface when hit, how fast does it move after the bullet
emerges?
Law of Conservation of Momentum gives us:
m1  v1  m 2  v 2 = m1  v1 '  m 2  v 2 '
v2 '
m1  v1  m 2  v 2  m1  v1 '
m2
(0.013 kg)  (230 m/s )  0  (0.013 kg)  (170 m/s )
2.0 kg
v 2 '  0.39 m/s
v2 '
14. A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact
with the ball for 5.0 x 10-3 s. Find (a) the impulse imparted to the golf ball, and (b) the average force
exerted on the ball by the golf club.
We find the impulse on the ball from:
(a)
F  t  p  mv  0.045 kg(45 m/s)  2.03 N  s
(b)
p
2.03 N  s
F

 405 N
t 5.0  10 -3 s
15. Suppose the force acting on a tennis ball (mass 0.060 kg)
points in the +x direction and is given by the graph below. Use
graphical methods to estimate (a) the total impulse given the
ball, and (b) the velocity of the ball after being struck,
assuming the ball is being served so it is nearly at rest initially.
(a) The impulse is the area under the F vs. t curve. The
value of each block on the graph is:
1 block = (50 N)(0.01 s) = 0.50 N•s
We estimate there are 10 blocks under the curve, so
the impulse is:
FΔt = (10 blocks)(0.50 N•s/block) ~ 5.0 N•s
(b) We find velocity from:
F  t  p  mv
F  t 5.0 N  s
v 

 83.3 m/s
m
0.06 kg
16. A 0.25-kg billiard ball (m1) moving at 5.00 m/s collides head on with a stationary 0.80-kg steel ball
(m2). If the billiard ball bounces directly backwards at 2.62 m/s, was the collision elastic?
Use law of conservation of momentum to determine velocity of steel ball after collision:
m1  v1  m 2  v 2 = m1  v1 '  m 2  v 2 '
v2 '
m1  v1  m 2  v 2  m1  v1 '
m2
(0.25 kg)  (5.00 m/s )  (0.80 kg)  (0 m/s )  (0.25 kg)  (-2.62 m/s )
0.80 kg
v 2 '  2.38125 m/s
To determine whether or not the collision is elastic, one must determine if kinetic energy is
conserved:
v2 '
E k1  E k2 = E k1 ' E k2 '
1
1
1
2
2
2
m1 v1   0 = m1 v1 '  m 2 v 2 '
2
2
2
1
1
1
(0.25 kg)  (5.00 m/s ) 2  0 = (0.25 kg)  (-2.62 m/s ) 2  (0.80 kg)  (2.38 m/s ) 2
2
2
2
3.126190625 J  3.126190625 J
Therefore, the kinetic energy is conserved and the collision is elastic.
17. An explosion breaks an object into tow pieces, one of which has 1.5 times the mass of the other.
If 7500 J were released in the explosion, how much kinetic energy did each piece acquire?
Use law of conservation of momentum to determine relative velocities of the pieces:
m1  v1  m 2  v 2 = m1  v1 '  m 2  v 2 '
0 = m1  v1 '  m 2  v 2 '
0 = m1  v1 '  1.5m1  v 2 '
v1 ' 
 1.5m1  v 2 '
m1
v1 '  1.5v 2 '
The kinetic energy of each piece is:
1
2
m 2 v 2 
2
1
1m 
1
1
2
2
2
E k1  m1 v1    2  - 1.5v 2 '  1.5 m 2 v 2 '  E k2
2
2  1.5 
2
2
E k2 
The energy supplied by the explosion produces the kinetic energy:
E total  E k1  E k2  2.5E k2
7500 J  2.5E k2
3000 J  E k2
4500 J  E k1