Some bumper cars are moving around a circular track at the same constant speed. However, they are not all going in the same direction. Collisions are perfectly elastic, so that two colliding cars instantaneously change directions (and continue at the same speed).

Show that at some point in the future, all the cars will be back to their starting positions and directions. Assume that each car has no length.

Consider any particular car A at time 0. It is either going clockwise or counterclockwise at constant speed S, and it starts at location X. Thus we can predict the time T that, if uninterrupted, it would complete one circuit of the track (arrive at location X, still with speed S). Well, if car A were the only car on the track, it would, and the problem would be trivial, so let's look at what happens when it bumps into car A'. The moment after they bump into each other, car A' is moving in the same direction car A was moving a moment before, and at the same speed. So if uninterrupted, car A' will now arrive at X at time T, travelling at speed S, since all momentum of car A was transferred to A'. Likewise, any car that car A' bumps into will now be the one scheduled to arrive at X at time T. So we can guarantee that regardless of how many bumps there are, there will be SOME car that arrives at X at time T, and that car will be travelling at speed S. If this car is A, then we've just shown that each car arrives back at its starting position at time T. (Note: I think this is the case, but I'm not sure how to prove it.) If not, then each car is now at the starting position and direction of another car, essentially a permutation of the original positions. Hence, since there are only a finite number of cars and therefore a finite number of permutations, at some multiple of T time, the cars will all be back to their starting positions and directions.

Posted by Avin on 2004-12-30 18:40:25