Download 3.1.2 In Class or Homework Exercise 1. The brakes of a car apply a

3.1.2 In Class or Homework Exercise
1. The brakes of a car apply a force of 640 N. The car weighs 15,680 N and is
moving at 20.0 m/s. The car finally stops. How long does the braking force
act on the car to bring it to a halt?
We will use the car’s original direction as the positive direction:
F f  640 N
Fg  15680 N
Fg  mg
vi  20.0m / s
15680  m(9.80)
vf  0
m  1600kg
t  ?
Since change in momentum is equal to the impulse applied, we will start
with that:
p  J
mv  F t
1600(0  20.0)  640t
t  50. s
2. A snowmobile has a mass of 250 kg. A constant force is exerted on it for
60.0 s. The snowmobile’s initial velocity is 6.0 m/s and its final velocity is
28.0 m/s. What is the magnitude of the force exerted on it?
Using the snowmobile’s original direction as positive
m  250kg
p  J
t  60.0 s
mv  F t
vi  6.0m / s
250(28.0  6.0)  F (60.0)
v f  28.0m / s
F  92 N
F ?
3. Small rockets are used to make small adjustments in the speed of satellites.
One such rocket has a thrust of 35 N. If it is fired to change the velocity of a
72000 kg spacecraft by 63 cm/s, how long should it be fired?
F  35 N
m  72000kg
v  63cm / s  0.63m / s
t  ?
UNIT 3 Momentum and Energy
p  J
mv  F t
(72000)(0.63)  (35) t
t  1300 s
RRHS PHYSICS 11
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4. A car moving at 50.0 km/h crashes into a barrier and stops in 0.25 m.
a. Find the time required to stop the car.
Using the car’s original direction as positive,
vi  50.0km / h  13.9m / s
d  0.25m
vf  0
t  ?
This can be solved by going back and using one of our kinematics
equations:
d 
vi  v f
t
2
13.9  0
0.25 
t
2
t  0.036 s
b. If a 10.0 kg child were to be stopped in the same time as the car,
what average force must be exerted?
m  10.0kg
p  J
t  0.036 s
mv  F t
vi  13.9m / s
(10.0)(0  13.9)  F (0.036)
vf  0
F  3900 N
F ?
c. Approximately what is the mass of an object whose weight equals
the force in part b? Could you lift such a mass with your arms?
Fg  mg
3900  m(9.80)
m  400kg
No, the average person would no be able to lift this mass!
d. What does your answer to part c say about holding an infant on your
lap instead of using a separate infant restraint?
Since the force required is much more than the average person can
exert, it is not safe to hold an infant in your lap – you would not be
able to exert the force required to keep the infant in the car if there
were an accident.
5. If you jump off a table, as your feet hit the floor you let your legs bend at
the knees. Explain why.
UNIT 3 Momentum and Energy
RRHS PHYSICS 11
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You bend your knees in order to lengthen the time over which you come to
a stop. Doing this reduces the average force applied to your body ( J  F t )
while applying the required impulse to stop you. If you were to land without
bending your knees, the time would be short. This would require a large
force and would be quite painful.
6. An archer shoots arrows at a target. Some arrows stick in the target while
others bounce off. Assuming the mass and velocity are the same, which
arrows give a bigger impulse to the target?
Remember that impulse is equal to change in momentum. The arrows that
bounce off undergo a larger change in velocity (they go from a positive
velocity to a negative velocity rather than the same positive velocity to zero)
and therefore have a larger change in momentum ( p  mv ). Since the
change in momentum is larger, the impulse on (and given by) these arrows
must be larger (Remember that the force exerted by the target on the
arrows is the same as the force exerted by the arrows on the target
according to Newton’s Third Law)
UNIT 3 Momentum and Energy
RRHS PHYSICS 11
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