# Simple conservation of momentum

Imagine a particle moving right at 10 mph. It enters a cylinder with an open left end and closed right end. The cylinder is moving right at 1 mph. In the frame of reference of the cylinder, the particle is moving right at 9 mph.

Assume a 100% elastic collision with the closed right end of the cylinder. The particle is now moving left at 9 mph with respect to the cylinder frame and left at 8 mph with respect to the original frame. Part of the momentum of the particle must have been passed to the cylinder.

From the perspective of someone inside the cylinder and moving along with it the particle collided with the end of the cylinder and rebounded without moving the cylinder wall. It looked like the cylinder wall was stationary. This assumes that the observer is somehow attached to the cylinder so that the cylinder itself never appears to move.

From the perspective of someone outside the cylinder even though the cylinder may be opaque the outside observer could tell when the collision occurred because the cylinder's velocity to the right increased at that moment. This must be so because the overall momentum of the particle and the cylinder in the original frame must be preserved. Since the particle reversed course, the cylinder must increase its speed to the right to compensate for the change in the particle's momentum.

To the observer inside and attached to the cylinder, she could tell when the collision occurred even without watching because she would briefly feel a force pushing her to the right when the particle strikes the cylinder--similar to the force one feels when an elevator starts moving upward.

Is this right so far?

If so, this also implies that when a particle strikes a fixed wall and rebounds with 100% elasticity, the wall must move to the right. If the reason the wall is fixed is that it is anchored to some larger structure, the larger structure must move to the right. In other words, since the total momentum of the particle and object it strikes must be constant, when the particle rebounds the object being struck must increase its rightward momentum. In other words, there really is no such thing as a fixed wall from which a particle can rebound without transferring some momentum to the wall.

This all sounds simple, but it grew out of some other question, and I'm here trying to explain how I now understand the larger issue. I had phrased my other question as if there can be a fixed wall from which a particle can rebound without affecting the wall. That was a wrong assumption and lead me down some wrong paths.

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This is all correct AFAICT, assuming that the cylinder is perfectly rigid and that its attached observer has some very sensitive equipment to measure acceleration. Going back to the enclosed gas in your previous question, collisions with the right wall accelerate the whole box to the right, but collisions with the left wall accelerate it to the left. We call the net effect of these collisions pressure. If the box is attached to the Earth, it jiggles the entire planet around*, but only by an unimaginably tiny amount. – Nathaniel Jan 22 '12 at 11:57
*actually it doesn't, because the Earth isn't a perfectly rigid body. But if it was then it would do. – Nathaniel Jan 22 '12 at 11:59

One thing that is wrong, is the idea that an observer would experience a jerk when hit by a molecule. The observer may feel heat at the spot where the molecule hit, and then maybe he will feel the heat spreading around slowly.

Another wrong idea is the idea of an elastic collision between a molecule and a macroscopic object. These collisions are always inelastic, how and why these collisions are inelastic becomes clear when we consider that the object does not experience a jerk, but rather some heating up.

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This isn't quite true - the observer attached to the wall will feel some acceleration when the particle hits it. Pressure is the net effect of all these little pushes, so if all a collision of a particle with a macroscopic object did was heat it up, there would be no such thing as pressure. – Nathaniel Jan 22 '12 at 11:49

Basically, you are right, some momentum had to be transferred or else conservation of momentum would be violated, even from the perspective of the original reference frame. This becomes even clearer if you forget bout cylinders or walls and just make both bodies be billiard balls but let the second one have different masses. If the second ball, the one with the observer on it, has practially infinitely larger mass than the first one, its motion will be negligible but it will absorb twice the momentum of the oncoming ball, thus conserving momentum.

It is instructive to think about what would happen if the observer's hit ball had equal mass, and what if it had negligible mass. The fastest sports item in normal American sports is the badminton birdie: right off the racket is is going more than 170 mph, faster than a batted baseball.

It is important to point out that therefore neither reference frame, that of the first billiard ball, nor the second, is « inertial » in the senses of Special Relativity or Galilean Relativity, because both of them undergo an infinite acceleration at the moment of impact.

So now you see there is a slight correction to your acount: if the motion of the infintely massive hit ball with the observer on it really is negligible, the observer will not feel anything. You wouldn't notice it if your racket swatted a fly...

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