Download Review Questions

Momentum Review Questions Answered
Physics
Review Questions
1. Which has more momentum, a cheetah running at 60 miles/hour or a Greyhound bus driving at 60
miles/hour? Why?
A greyhound bus will have more momentum because it has more mass.
2. Which has more momentum, a mosquito flying at 1 m/s or an elephant standing still? Why?
The mosquito has more momentum because the elephant is not moving and therefore has no
momentum. The moving mosquito has very little momentum because of its low mass, but it is
more than zero.
3. Why is it a good idea to follow through on your golf swing? How does that help you hit the golf ball
further?
Follow through on your golf swing means that the golf club is in contact with the ball for more
time, which means the golf club will impart a greater impulse to the ball. A greater impulse
means that the ball gains more momentum, which is another way of saying that it gets up to
higher speed. Of course, from our study of projectiles we remember that higher launch speed
goes with greater range for an angled projectile.
4. To catch a falling person and absorb their momentum, firefighters often used to use a piece of canvas
stretched on a metal frame. A person can fall from a height of several stories and be unhurt when they
are caught like this. How is this possible? What would happen if the person landed on the sidewalk
instead?
A falling person is gaining momentum all the way down as gravity pulls on them. When they
reach the ground and stop, whatever they hit has to impart just the right amount of impulse to
take away all their momentum. If they had 1000 kg·m/s of momentum, they would need to
receive –1000 N·s of impulse to make them stop.
Since impulse is the product of force and time, there are many ways that they can receive just
the right impulse. The different methods use different sized forces acting for different amounts
of time, but in every case the product of the force and time has to come to exactly the right
amount. This means that the smaller amount of time allowed, the larger the force has to be and
the longer that the impulse lasts, the smaller the force can be. For example, to get the –1000 N·s
impulse, you could have a 1000 N force that lasts 1 seconds or a 10 N force that lasts 100 s.
Clearly, a falling person wants to feel the smallest possible force when stopping, since forces
hurt and break bones. When they land in the canvas stretcher, the material gives and the arms of
the firefighters give to provide more time for the impulse than the ground would (since it doesn’t
give much at all), which means the cloth can apply a smaller force than the ground has to.
5. An 850,000 kg space shuttle, sitting in the launch cradle, fires its
rockets to lift off. The rockets push straight upward on the ship
with 24,000,000 N of force for 8 min.
a) What is the net force on the spaceship? (Don’t forget gravity)
I draw a force diagram – as shown to the right – and add up all the
forces to find the net force.
Fg = mg = (850,000 kg)(9.8 m/s/s) = 8,330,000 kg·m/s
Fnet = FThrust – Fg
= 24,000,000 N – 8,330,000 kg·m/s
= 15,670,000 N
b) How fast is the spaceship moving after firing its rockets for 8 min?
I = Fnet·t = (15,670,000 N)(480 s) = 7,521,600,000 N·s
pi = mvi = (850,000 kg)(0 m/s) = 0 kg·m/s
I = pf – pi
7,521,600,000 N·s = pf – 0 kg·m/s
7,521,600,000 N·s = pf
pf = mvf
7,521,600,000 N·s = 850,000 kg · vf
8849 m/s = vf
c) How high off the ground will the spaceship be at the end of 8 min if it fires its rockets straight up the
whole time?
We need an equation that will solve for the distance the rocket moves. One equation that will
work is:
vi + vf
d =  2 ·t


vi + vf
0 + 8849 m/s
d =  2 ·t = 
 (480 s) = 2,123,760 m
2




6. A 10,000 kg bullet train is accelerated forward from rest by a constant force for 70 seconds and ends
up going 90 m/s.
a) How much momentum does the train gain?
pi = mvi = 10,000 kg · 90 m/s = 900,000 kg·m/s
pf = mvf = 10,000 kg · 0 m/s = 0 kg·m/s
p = pf – pi
p = 900,000 kg·m/s – 0 kg·m/s = 900,000 kg·m/s
b) How big an impulse is applied to the train?
I = pf – pi = 900,000 kg·m/s = 900,000 N·s
c) How large is the force that is accelerating the train? Assume air resistance and friction are
negligible.
I = Fnet · t = p
900,000 N·s = Fnet · 70 s
12857 N = Fnet
7. As Catrina (60 kg) is flying across the continent, a small maintenance oversight leads to a window
popping out and Catrina is shoved out of the window by the pressure difference between the interior of
the cabin and the exterior atmosphere. As she falls, she quickly reaches her terminal velocity of 50 m/s
and then “coasts” down to the ground. As she nears the ground, she sees that she is falling into a
farmer’s field with a haystack, for which she tries to aim.
a) If Catrina hits the haystack, she comes to a stop in 0.08 s. How much force does the hay exert on her
to make her stop?
pi = mvi = 60 kg · 0 m/s = 0 kg·m/s
pf = mvf = 60 kg · 50 m/s = 300 kg·m/s
I = pf – pi
I = 0 kg·m/s – 300 kg·m/s = –300 kg·m/s
I = Fnet · t
– 300 kg·m/s = Fnet · 0.08 s
– 3750 N = Fnet
b) If Catrina lands in the field next to the haystack, she comes to a stop in 0.004 s. How much force
does the ground exert on her to make her stop?
pi = mvi = 60 kg · 0 m/s = 0 kg·m/s
pf = mvf = 60 kg · 50 m/s = 300 kg·m/s
I = pf – pi
I = 0 kg·m/s – 300 kg·m/s = –300 kg·m/s
I = Fnet · t
– 300 kg·m/s = Fnet · 0.004 s
– 75000 N = Fnet
c) Which way does she lose a greater amount of momentum, hitting the hay or hitting the ground?
In each case she loses all of her momentum, since she ends up stopped. Since she starts with the
same amount of momentum wherever she lands, she will lose the same amount of momentum:
pi = mvi = 60 kg · 0 m/s = 0 kg·m/s
pf = mvf = 60 kg · 50 m/s = 300 kg·m/s
p = pf – pi
p = 0 kg·m/s – 300 kg·m/s
p = – 300 kg·m/s
8. Two cars with the same mass and the same velocity are playing chicken and choose not to swerve.
They crash into each other and get tangled together. What are the cars doing after the collision?
Explain based on the law of conservation of momentum.
If the two cars have the same velocity, they will have the same amount of momentum. Because
they are going opposite directions, their momenta will be opposite – one positive and one
negative. When you add up the total momentum, you find that it is 0. That means, that after the
crash the total momentum must still be zero. Because the cars are stuck together, they are
effectively become one object. The only way that that object can have zero momentum is to be
sitting still, so the two cars must come to a complete start.
9. A large (100 kg) hockey player going 4.8 m/s skates into a smaller player (75 kg) who is at rest.
After the elastic collision, the smaller player is going 4 m/s. How fast is the large player moving?
This is a conservation of momentum question, because we don’t care about the details of the
collision, just the final result, so we apply the law of conservation of momentum. All motion is
in the same direction (which is not specified), so we make that direction positive. We will
choose the large hockey player to be object A and the small hockey player to be object B:
piA = mA·viA = (100 kg)(4.8 m/s) = 480 kg·m/s
pfA = mA·vfA = ?
piB = mB·viB = (75 kg)(0 m/s) = 0 kg·m/s
pfB = mB·vfB = (75 kg)(4 m/s) = 300 kg·m/s
pfA + pfB = piA + piB
pfA + 300 kg·m/s = 480 kg·m/s + 0 kg·m/s
pfA = 180 kg·m/s
pfA = mA· vfA
180 kg·m/s = (100 kg)vfA
vfA = 1.80 m/s
10. Nick (85 kg) is standing on his skateboard (2 kg) when Alex (60 kg) runs up from the tail end of the
skateboard at 4 m/s and jumps on. Assuming that they don’t fall off and get scraped up badly, how fast
will they be moving afterwards?
This is a conservation of momentum question, so we apply the law of conservation of
momentum. All motion is in the same direction (even though it is not specified), so we make
that direction positive. We will choose Nick and the skateboard to be object A and Alex to be
object B:
piA = mA·viA = (87 kg)(0 m/s) = 0 kg·m/s
pfA = mA·vfA = ?
piB = mB·viB = (60 kg)(4 m/s) = 240 kg·m/s
pfB = mB·vfB = ?
pfA + pfB = piA + piB
pfA + pfB = 0 kg·m/s + 240 kg·m/s
pfA + pfB = 240 kg·m/s
In an inelastic collision, the objects are connected so they move at the same speed vfA = vfB = vf
pfA + pfB = 240 kg·m/s
mA·vfA + mB·vfB = 240 kg·m/s
mA·vf + mB·vf = 240 kg·m/s
(mA + mB)·vf = 240 kg·m/s
(147 kg)vf = 240 kg·m/s
vf = 1.63 m/s
11. A small train car has a cannon mounted on it and a total mass of 6000 kg. The car is moving
forward along the tracks at 5 m/s, when the cannon accidentally discharges (good thing there were no
other cars in front of it!). A 20 kg cannonball is shot forward at 250 m/s (compared to the ground).
How fast is the train car moving after the cannonball has been fired?
This is a conservation of momentum question, so we apply the law of conservation of
momentum. The cannonball is launched in the same direction that the train is going, so we make
that direction positive. We will choose the car and cannon to be object A and the cannonball to
be object B:
piA = mA·viA = (6000 kg)(5 m/s) = 30,000 kg·m/s
pfA = mA·vfA = ?
piB = mB·viB = (20 kg)(5 m/s) = 100 kg·m/s
pfB = mB·vfB = (20 kg)(250 m/s) = 5000 kg·m/s
pfA + pfB = piA + piB
pfA + 5000 kg·m/s = 30000 kg·m/s + 100 kg·m/s
pfA = 25,100 kg·m/s
pfA = mA· vfA
25,100 kg·m/s = 6,000 kg · vfA
4.18 m/s = vfA
12. Jeff (60 kg) takes up roller dodge ball - playing dodge ball on roller skates. He is standing perfectly
still when a 1 kg ball going 30 m/s hits him in an elastic collision. After being hit, he rolls backwards at
0.6 m/s. What is the ball’s velocity after hitting Heff?
This is a conservation of momentum question, so we apply the law of conservation of
momentum. We choose the direction that the ball is going before it hits Jeff to be the positive
direction. We will choose Jeff to be object A and the ball to be object B:
piA = mA·viA = (60 kg)(0 m/s) = 0 kg·m/s
pfA = mA·vfA = (60 kg)(0.6 m/s) = 36 kg·m/s
piB = mB·viB = (1 kg)(30 m/s) = 30 kg·m/s
pfB = mB·vfB = ?
pfA + pfB = piA + piB
36 kg·m/s + pfB = 0 kg·m/s + 30 kg·m/s
pfB = –6 kg·m/s
pfB = mB· vfB
–6 kg·m/s = 1 kg · vfB
–6 m/s = vfB
vBf = –6m/s