Physics 101- Spring 2007
Momentum solutions
1.
You are helping to investigate a mishap in a mountain railroad. One night, a railroad car was left
parked at the top of a hill above another car of the same mass. In the morning, both cars are found
coupled together in a lake. The supervisor claims that the engineer must have not set the brake on the
upper car, and that during the night it rolled down, hit and coupled with the lower car, carrying both
into the lake. The engineer disputes this, saying that someone must have released the brake and then
pushed the upper car. After you study the map of the terrain (which also shows the positions of the
cars the night before), you submit your findings which say that…?
24m
7m
64m
Lake
48m
This is a problem where we’ll have to break up the motion into different time
intervals. In some of these intervals momentum will be conserved and in some energy
will be conserved. The key for momentum conservation will be whether or not the net
force on the system is zero. For energy we’ll have to check whether or not there are any
non-conservative forces (in other words, mechanical energy converted to other forms).
There are couple ways we can approach this problem (try to answer the question
of the engineer’s guilt). The way I’ll go about it in this model is to assume that the first
car started at rest (nobody pushed it), and then see if the two cars make it to the top of the
second hill. If they do, I’ll assume that they then make it to the lake. If the model shows
that the cars do not make it to the top of the second hill, then the engineer is most likely
not at fault. In other words, this would show that the only way the two cars could make it
to the lake is if they had more energy. This would most likely be due to a push at the top
of the first hill.
I would guess that this is going to be a close call- the cars might just make it over
the hill. Certainly the two cars won’t go as high as where the first one stared, but it
seems possible that they could make it up 7m. (In fact, one might even think that their
final height is just 1/2 the original. The 1/2 guess is due to the doubling of the mass.)
For the time between when the first car starts at the first hill, until just before it collides
with the second car, let’s assume:
•
•
•
•
Friction & air resistance is negligible -> energy is conserved. (This is ignoring the
reality of friction within the axle. This may be a stretch, but it must be done so the
problem is doable.)
There are no other forces acting on the car besides gravity & the normal force.
The initial point will be when the car is at the top of the hill, at rest. The final point
will be when the car is at the bottom of the valley (just short of the second car).
(Momentum is not conserved, due to the net external force on the car.)
1
mgy i = mv 2
2
with yi= 24m
We can solve for the speed of the car, just before the collision
!
v = 2y i g
Now that we know the speeds of the two car before the collision, we can analyze the
collision.
!
Let’s assume:
• The second car is at rest before the collision.
• Momentum is conserved due to a lack of any net external force. This essentially
means we’re taking the time of the collision to be small and/ or the ground to be flat.
(Also, we are still assuming that the friction is negligible.)
• (Mechanical energy is not conserved as this is a perfectly inelastic collision.)
Equating the momentum before and after the collision:
mv = 2mv f
If we substitute in the speed from the first portion, we can then write the speed after the
crash (vf) in terms of the first hill height.
!
vf =
1
2y i g
2
Now, we return to energy conservation. This last portion of the run is just like the first,
just in reverse (converting kinetic into potential energy). We can know answer the
!go up the second hill.
question of how high the cars
1
(2m)v 2f = (2m) gh
2
h will give us the vertical height that the two cars reach.
!
% 2 1 " 2y i g % y i
1 "1
h = $ 2y i g ' = $
'=
& 2g # 4 & 4
2g # 2
Since yi= 24m, h= 6m.
! do not make it up the second hill with our model. The only way for
It looks like the cars
the cars to reach the lake would to add energy to the system, most likely with an initial
push. So, it appears as though the engineer isn’t guilty. (If we would have added
friction, the cars wouldn’t have even reached 6m, so it seems pretty safe to say that the
engineer isn’t guilty.)
2.
This problem refers to the demo that Dr. Jeff did in class last Friday. Explain why the tennis ball
bounces so much higher when it is on top of the basketball than when it is dropped by itself. Make an
estimate for the tennis ball’s maximum height if the pair is dropped from 1.5m above the ground.
The first task, much like the previous problem is to identify different points in time that
might serve as “initial” or “final “ moments.
vT3
vT1
vT2
vB3
vB1
vB2
1
2
3
1. represents when both balls are falling and haven’t yet hit the floor.
2. is just after the basketball has hit the floor and is now moving upward, but the tennis
ball is still falling. This difference in direction is due to the space between the two
balls, which admittedly is tiny.
3. now the tennis ball has bounced off of the basketball and is moving upward.
To find the maximum height we’ll want the speed of the tennis ball at moment #3. With
this speed, we will be able to either use kinematics or energy to find the maximum height.
What do we know? We should be able to estimate the speeds at moment 1 using
kinematics or energy methods. From the time the balls are dropped until moment 1, the
balls are freefalling.
The trick is figuring out how to connect what we know (moment 1) to what we want
(moment 3). An examination of moment 2 seems in order. The tennis ball’s speed
should be essentially the same as in 1 as it is still freefalling and hasn’t yet interacted
with anything besides gravity. The basketball has bounced upward. But, how does this
speed relate to what the ball had just prior to hitting the floor? The two will be nearly
equal; or, at least, that’s what we’re going to assume. (We can improve this estimate at
the end of the solution.)
It’s starting to look like a doable problem. We should be able to find the ball’s speeds at
moment 2 and use these to find the tennis ball’s speed at moment 3. What is the physics
principle that will guide us? Conservation of momentum. We’ll assume that the
collision occurs over such a short time (distance) that the gravitational force does not
provide a significant impulse.
To find the ball’s velocities at moment 2 we simply use conservation of energy, and
assume that there is no significant air resistance.
mT mass of the tennis ball
mB mass of the basketball
h= 1.5m, initial height of the balls
mT gh = 1 2 mT vT2 2 " vT 2 = # 2gh
mB gh = 1 2 mB v B2 2 " v B 2 = 2gh
!
Notice that the tennis ball’s velocity is negative as it is moving downward.
To find the tennis ball’s speed at moment 3 we need one more piece of information.
Why? Well, we’re about to run into two unknowns (vT3 and vB3) and only one equation
(conservation of momentum). Clearly we need something else. (Notice that even though
aren’t concerned with vB3, it is still an unknown that will appear in the conservation of
momentum equation.
A good approximation to use, since it will make our lives easier and not cost us too much
accuracy, is that the mechanical energy in the tennis ball- basketball collision is
conserved. This will allow us to generate a second equation and then solve for the
desired variable.
Notice that the book has already done this on page 261-262. We simply can rename the
variables and make use of equation 9.20 to find the tennis ball’s speed after the collision.
# m " mB &
# 2mB &
vT 3 = % T
(vT 2 + %
(v B 2
$ mT + mB '
$ mT + mB '
# m " mB
2mB &
= vT 2 % T
"
(
$ mT + mB mT + mB '
# m " mB
2mB &
= " 2gh % T
"
(
$ mT + mB mT + mB '
# 2mB
m " mB &
= 2gh %
" T
(
$ mT + mB mT + mB '
!
We can now make substitutions for h=1.5m, mT≈50g and mB≈500g. We find that vT3≈ 14
m/s.
Employing conservation of energy once more, we find that the maximum height of the
tennis ball is approximately 10m.
Does this make sense? Well, in class the tennis ball easily hit the ceiling, which is about
4m. However, 10m still sounds a bit high as that is roughly 3 stories (height of Seaver).
Did we make a mistake somewhere? No, we simply followed some simplifications that
may be not be 100% accurate. (Thus the name “approximations!”) We assumed that
mechanical energy is conserved in the basketball- floor collision as well as in the
basketball- tennis ball collision. It isn’t in either one. We can see this by simply
dropping either ball and noting that it does not bounce back to the original height.
Mechanical energy is lost.
It’s difficult to say exactly how much is lost, but if we assume that one-fourth of the
mechanical energy is lost in both of the collisions we can alter our mathematics.
For the velocities at moment 2 we would now have
vT 2 = " 2gh
vB 2 #
3
4
2gh
The conservation equations for the tennis ball- basketball collision now look like:
!
!
mT vT 2 + mB v B 2 = mT vT 3 + mB v B 3
3 1
2
2
2
1
1
(
2 mT vT 2 + 2 m B v B 2 ) = 2 mT vT 3 +
4
1
2
mB v B2 3
Algebra (on another piece of paper) shows us that; vT3=9 m/s, so the maximum height is
approximately 4m (13 feet). This essentially the height of the room. Not bad. But, did
we really gain anything? The original solution is correct with within a factor of two, and
was easy to obtain with our book’s help. For a rough problem such as this, that is
sufficient.