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Originally posted by Identity Crisis:
Oh, I see. You are confused.

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BTW, this is _exactly_ the same as if the particle had hit and bounced off a barrier in the y plane. (i.e. a 180deg collision). The only reason I made it two 90deg collisions was because people were complaining about the exhaust particles going back into the impulse drive.

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I find that incomprehensible.

Point One:



In the above, the top section represents a standard rocket firing. Scene 1 (so as not to be confused with your steps) shows the rocket just sitting there. In Scene 2, the red particle is ejected from the rear of the rocket, and the rocket flies forward (blue).

Below the green line, we see a new version of events. The rocket has a flat plate doohickey attached. Scenes 1 and 2 are identical.

In Scene 3, the particle has collided with the plate. This is my version of what happens . . . note how the two objects are no longer mobile. Their momentums have cancelled.

Scene 4 shows your version of what happens, as judged by your statements. The particle has rebounded off of the plate and the rocket is now moving backwards (at least until they collide again, as is about to occur . . . but, if the plate rebound occurred and produced reverse ship motion, then the particle should rebound off the rocket just as well, and we'd end up with a bouncing-back-and-forth action).

And now that I have the visual and a proper analogy in mind, I see that your version is actually correct, I think. The analogy would be to one of those swinging ball thingies. If you pick up one ball from each side and drop them, they both "bounce" as opposed to stopping entirely.

Funny . . . I knew that. I actually used that very concept on one of my pages several months ago, too. But, in reference to the ship example, I missed it. Must've been the headaches from all of this.

This leads us straight into what was originally going to be:

Point Two:



Here we have two particles . . . one red, one yellow . . . striking a stationary object (or more precisely, stationary to our frame of reference) at different times. Due to the angles of the strikes (with the momentum transfer direction identified by similar colors to the striking particle), the net effect on the object would be the same.

And here we come to the reason why I was disagreeing with you. And in case this is too hard to read, pretend the rest of this is in italics:

It was all based on my ship-plate concept that was wonky. Because I was thinking that the momentum would cancel and the two would become stationary, it struck me as absurd to conclude that two 90 degree collisions could produce rearward motion where one 180 had failed.

As near as I can figure, I must've had an inelastic collision on the brain in reference to the 180, instead of an elastic one.

In any case, as you can see in the pic above, the yellow (180) particle smacks the stationary doohickey and gets reflected back. The total momentum on the doohickey would be 2mv, since it has to both stop the particle and send it back on its way as any good elastic collision should produce. When I was thinking the wrong thing about the ship for whatever reason, I was thinking ("in effect" only, not quite consciously) that this should only be 1mv against the doohickey.

Hence my disagreement with you over the 1mv leftward and 1mv downward in relation to the ship.

On the other hand, I'm still confused by something. The bit about "two 90 degree collisions could produce rearward motion where one 180 had failed" was only part of it.

Earlier in the thread, I used this pic:



Now first, we have the red particle smacking into the gray angled wall. We know the actual transfer should occur along the blue (and green) diagonal line perpendicular to the wall.

If that is so, then we still cannot have -1mv acting on the wall on the x plane, since the only way to get the particle to deflect upward with 1mv would be for the particle and wall to transfer .7071mv between one another along that perpendicular line. Then the vectors add up, as shown by the green lines. That would still leave the wall (or ship) with .5mv and .5mv along the horizontal and vertical axes (the blue lines). For a second collision that sent the particle right-ward, we'd still only get a total of 1mv against the ship.

(Incidentally, a collision simulator I was using seems to agree with this view, insofar as it shows a .5-.5 momentum transfer from the light particle against a stationary wall.)

To my mind, we're left with two possibilities:

1. The momentum transfers involved in getting the particle deflected upward count toward the total, meaning that the two-hit thrust reverser wouldn't work because a whole 1mv gets effectively wasted by the act of getting the particle up and out of the way so it won't hit the ship on its way out.

2. If two 90 degree collisions are to be the same as a single 180 (i.e. no wasted effort as suggested above), then by damn the ship will get 2mv leftward somehow-or-other. The only way I can imagine this working within the confines of my diagram would be for us to simply add the two .7071mv diagonal lines into a single 1.414mv diagonal, which would then require 1mv left and 1mv down for the ship. However, that doesn't strike me as a proper maneuver off-hand . . . though it sounds similar to geometric vector addition.

Given my "Before After" diagram, I must assume that such a maneuver is the way to go. But now I'm just curious about how to think about it.

Suggestions?

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G2k's ST v. SW Tech Assessment