Download Unit 6 Review Key

Impulse-Momentum Review Sheet
(All units of momentum in the momentum conservation diagrams are kgsm .)
1. A ball of mass 3.0 kg, moving at 2 m/s eastward, strikes head-on a ball of mass 1.0 kg that is
moving at 2 m/s westward. The balls stick together after the impact. Complete the momentum
conservation diagram. What is the magnitude and direction of the velocity of the combined mass
after the collision?
event: Completely inelastic collision where two moving objects collide and stick together.
initial
object/mass/velocity
final
object/mass/velocity
3.0 kg at 2.0 m/s
4.0 kg at ?
m/sm/s
1.0 kg at -2.0 m/s
+
–
E
0
+
4
0
4
-4
–
m3iv3i  m1iv1i  (m3  m1)v f
3.0kg(2.0 ms )  1.0kg( 2.0 ms )  (3.0kg 1.0kg)v f  v f  1.0 ms eastward
2. One way of measuring the muzzle velocity of a bullet is to fire it horizontally into a massive block
of wood placed on a cart. Assuming no friction, we then measure the velocity with which the wood
containing the bullet and cart begin to move. In one experiment the bullet had a mass of 7.5 g and
the wood and its cart had a mass of 5.0 kg. After the shot, the cart, wood, and bullet moved at a
constant speed, traveling 2.4 m in 4.0 s. From these data, complete a momentum conservation
diagram and determine the original speed of the bullet.
event:
Completely inelastic collision where moving object collides with stationary
object and they stick together.
initial
object/mass/velocity
final
object/mass/velocity
5.0 kg at 0 m/s
5.075 kg at ?
0.0075 kg at ? m/s
+
–
E
0
3.0
0
3.0
–
+
m0.05iv 0.05i  0  (m20  m0.05 )v f
vf 
2.4m
 0.60 ms  (0.0075kg)v 0.05i  0  (5.0kg  0.0075kg)0.60 ms  v 0.05i  400 ms
4.0s
©Modeling Instruction - AMTA 2013
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U9 Momentum - review v3.1
3. A raft of mass 180 kg carries two swimmers of mass 50. kg and 80. kg. The raft is initially floating
at rest. The two swimmers simultaneously dive off opposite ends of the raft, each with a horizontal
velocity of 3 m/s. Complete the momentum conservation diagram and determine the final velocity
and direction the raft.
event: One stationary body with 3 parts, 2 of the parts push off the 3 rd.
initial
object/mass/velocity
final
object/mass/velocity
50 kg at 0 m/s
50 kg at -3.0 m/s
80 kg at 0 m/s
80 kg at 3.0 m/s
180 kg at 0 m/s
180 kg at ? m/s
+
–
0  0  0  m180f v180f  m50f v50f  m80f v80f
0  (180kg )v180f  50kg(3.0 ms )  80kg(3.0 ms )  v180f 
0
120
0
-120
–
+
150 kgsm  240 kgsm
 0.50 ms
180kg
4. Comment on the advisability of attempting to jump from a rowboat to a dock that seems just within
jumping distance.
You push back on the boat so you can jump forward. Since there is so little friction to hold the
boat in place the boat will move backward as you push.
5. a. Why is it difficult for a fire-fighter to hold a hose that ejects large amounts of high-speed water?
The hose is pushing the water forward therefore the water pushes the hose backwards. The
fire-fighter is holding onto the hose being pushed backwards.
b. Calculate the force needed to hold a 6 cm diameter fire hose in place when the water flow rate
is 110 m3/hour. (density of water: 1000 kg/m3)
3
kg
1hr
m
1hr
m
110 mhr (1000 mkg3 ) 3600
s  30.6 s  110 hr ( 3600 s )  0.0306 s 
3
3
3
0.0306 ms
 10.8 ms
2
 (.03m)
F  30.6 kgs (10.8 ms )  330N
©Modeling Instruction - AMTA 2013
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U9 Momentum - review v3.1
6. A sailboat is stalled on a lake on a windless day. The skipper's only piece of auxiliary equipment is
a large fan, which can either be set up to blow air into the sail or the fan can be pointed the off of
the back of the boat with the sail taken down. Will either technique get the boat to shore? Which
technique would work best? Explain your answer in terms of impulse and momentum.
The fan will push the air backward and the air
will push the fan (and the boat) forward. The
boat will move forward much faster than the
previous method.
The fan pushes air forward and the air pushes
the fan (and the boat) backward. The air hits
the sail and pushes the sail (and the boat)
forward. One would think that the boat would
not move because those two forces on the boat
would cancel out. However, the air will
bounce off the sail giving it a greater impulse
than it gave back on the fan. The boat will
move very slowly forward.
A practical application related to question 6.
7. a. When a bug smacks into your windshield, which gets hit harder, the bug or the windshield?
Justify your answer.
The bug hits the window with the same force as the window hits the bug. The two forces are a
3rd Law pair so they must be equal.
b. Which will experience the greater change in momentum? Justify your answer.
Both objects experience the same change in momentum. The forces are equal, the time of
interaction is equal, therefore the impulses are equal. The numerical value of the impulse is
always equal to the change in momentum.
c. Which will experience the greater acceleration? Justify your answer.
F
The bug will have the greatest acceleration. a  net and both objects have the same net force
m
therefore the bug, having less mass, will have greater acceleration.
©Modeling Instruction - AMTA 2013
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U9 Momentum - review v3.1
8. Accident investigators come upon the following scene involving a collision between two cars of
similar mass. Both drivers had intended to travel through the intersection without turning.
Driver A says that after stopping at the stop sign and
then proceeding, car B ran the stop sign so there was
no time to stop before hitting the side of car B.
B
A
Driver B says that after stopping at the stop sign and
then proceeding, car A came out of nowhere,
evidently running the stop sign and T-boning car B.
car A’s path
stop
sign
stop
sign
car
B’s path
a. Which car ran the stop sign? Explain your
answer in terms of our momentum studies.
B ran the stop sign. The two cars will go the
farthest in the direction of the fastest car (the one
that ran the sign). See the explanation below. One
can see that the drawing for B running the stop sign
is closest to the accident result.
Sum of the 2
10 mi/hr
If A ran the stop sign.
Sum of the 2
40 mi/hr
b. Sketch a picture of how the cars would have come to rest had the driver’s story that ran the stop
sign been true.
c. Briefly explain why the cars end up where
you drew them.
I figure the car that ran the stop sign will be
traveling much faster before the collision than
the car that stopped. I’ll say the one that ran
the sign is going 40 miles/hour and the one
that stopped 10 miles/hour. After the
collision the two cars will travel in the
direction of the vector sum of the
B
A
momentums. I’m assuming the two have
car A’s path
close to the same mass.
stop
sign
10 mi/hr
stop
sign
car
B’s path
If B ran the stop sign.
40 mi/hr
©Modeling Instruction - AMTA 2013
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U9 Momentum - review v3.1