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[Phys-L] Re: collision question (revisited)



John Denker wrote:

The cleanest version of the idea is as follows:
Imagine a box-car of mass M, with a trap door
on the back side. For simplicity, assume it
is initially at rest. We shoot it from behind
with a ball of mass m. The ball goes freely
in through the trap door and goes all the way
to the front wall of the box, where it bounces
elastically. It then travels back to the back
of the box, where it cannot get out the trap
door, so it bounces elastically, returns to
the front, et cetera ad infinitum.

This makes a great homework problem: Find the
*average* velocity of the M+m system.

For good students, you should not be explicit
about what you mean by "average". Make them
figure out what should be meant, and make
them tell you. This is good practice for
solving real-world ("Letter to Garcia")
problems. But for the dimmer bulbs, you'll
have to tell them what average you want.

Let me take a crack at this. When the ball is traveling forward, the
velocity of the ball is +v and the velocity of the box is 0. Using
the elastic collision equations, when the ball is traveling backward,
the velocity of the ball is -v' where v' = v*(M-m)/(M+m) and the
velocity of the box is +V where V = v*2m/(M+m). Note that when the
ball is traveling backward, its speed relative to the box is v'+V =
v. Hence the ball takes the same amount of time to travel forward as
backward after each bounce.

CoM velocity for the forward traveling ball is v*m/(M+m) and for the
backward traveling ball is (MV-mv')/(M+m) = v*m/(M+m). Not
surprisingly (by conservation of momentum), these two are found to be
equal. Call it v_CoM.

Now consider the velocity of the box. When the ball is traveling
forward, it is 0, and when the ball is traveling backward, it is
v*2m/(M+m). Since the ball takes the *same* amount of time to go
forward as backward, the *time-averaged* velocity of the box is
v*m/(M+m). This, not surprisingly, is equal to v_CoM.

Note that *on average* no momentum is lost. Before the ball strikes
the box, the system momentum is mv+0. Afterward it is v_CoM*(M+m),
the same value.

Anyway, among other things, this gives a nice
unforgettable mental model of dissipation. If
you draw the world-line of the system in enough
detail, there is no dissipation. But if you
average over a moderately-long timescale, you
see only the average behavior, which is what
we call the dissipative case.

This calls for looking at the system's KE. Initially we have
0.5mv^2+0 = K_i. After the ball strikes the box, we have the same if
we are allowed to look inside the box (since collisions are elastic).
But assuming instead we can only see the box, then we have 0 when the
ball is traveling forward and 0.5MV^2 when the ball is traveling
backward, for an average of 0.25MV^2 = K_i*2mM/(M+m)^2. In the best
case where M=m, this is still only 0.5K_i since the box is stationary
half the time (and the ball is stationary the other half of the
time). More realistically, if M>>m, then the final average KE is very
small. Hence dissipation of KE with no loss of momentum.

The point is that the transferred p cannot be
hidden -- not for long anyway -- not in a closed
system. You can hide some of the momentum some
of the time, but there's a strict upper bound
(tx) on how long you can hide it, and averaging
over any longer timescale will reveal the "hidden"
momentum, forcing it to contribute to the black-
box CoM motion.

For homework, find tx explicitly in terms of m,
M, and the size of the box.

I confess you lost me on this part. In the above example, the box is
stationary half the time *regardless* of the values of m and M. Were
you expecting me to consider a more general case now to get some kind
of uncertainty principle? Please repeat the conditions for this last
problem. Carl
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5040
mailto:mungan@usna.edu http://usna.edu/Users/physics/mungan/
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