How Many Collisions Occur in a Particle Collision Problem on the Unit Interval?

[alec_tronn]

Homework Statement

Suppose that five particles are traveling back and forth on the unit interval [0,1]. Initially, all five particles move to the right with the same speed. (The initial placement of the particles does not matter as long as they are not at the endpoints.) When a particle reaches 0 or 1, it reverses direction but maintains its speed. When two particles collide, they both reverse direction (and maintain speeds). How many particle-particle collisions occur before the particles once again occupy their original positions and are moving to the right?

The Attempt at a Solution

I've drawn out the process if all particles are equal distances apart (1st one at 1\6, 2nd at 2\6, 3rd one at 3\6, 4th one at 4\6, and 5th one at 5\6), and counted the collisions. I've decided the answer is 20, but I don't know how to prove it. If I just write down the case I tried, then it only proves it for that one case, and that's no good. Does anybody have any suggestions as to where to start on this proof?

 

[AKG]

Two particles colliding then switching directions is equivalent to the two particles passing right through each other. In that situation, you get a train of 5 particles going to the right, then each one changing direction as it reaches 1, then changing direction again upon reaching 0, and then moving right toward their original places. Thinking of it this way, you just need to count the number of times particles pass each other. You'll see that the rightmost passes the other 4 once each when its heading left and once again for each as its heading back right. So the rightmost is involved in 8 "collisions". The second right-most similarly passes through the three to the left of it once on its way left and once again on its way right, making 6. You'll see the answer is 8 + 6 + 4 + 2 = 20.

 

[alec_tronn]

I think I'm following you, but I have a couple questions. Physically, yes bouncing off of each other is the same as passing through, but then the names for the particles have switched, and we're required to have the original particles in their original positions, so how to I know\prove that they're not all mixed up and in the same positions? In other words, if it started like (1st one at 1\6, 2nd at 2\6, 3rd one at 3\6, 4th one at 4\6, and 5th one at 5\6) that it isn't now (2nd one at 1\6, 5thnd at 2\6, 3rd one at 3\6, 4th one at 4\6, and 1st one at 5\6)?

 

[AKG]

Since they're really bouncing off each other and not passing through each other, the rightmost particle will remain rightmost. So the 5th cannot be at 2/6 if the 3rd is at 3/6. The rightmost particle will not remain rightmost throughout the entire journey IF you regard them as passing through each other, but in the situation that counts, the colliding-situation and passing-through-situation turn out to be the same. You'll see this if you think about it for a bit. In other words, the desired ending condition occurs when the particles are thought as colliding iff the desired ending condition occurs while the particles are thought of as passing through each other.

 

[alec_tronn]

Oooooooh. Okay Thanks a lot. I got it now.