please try this one, its nice, you are going to have fun with it!

consider instead that the balls on collision just-i.e. instead of rebounding,they just 'go through each other'.then since all the balls are moving at the same speed,after a certain time all the balls,clearly will be at exactly where they started after one rotation. in the present situation, since the collisions are perfectly elastic(which we need to assume) instead of exactly the same balls we have a permutation of the balls but in exactly the same spots as the inital starting point,and also moving in the same direction as the balls initially went. it then follows that after some time,the initial situation will repeat.[/list]

to clarify:i was thinking in terms of balls colliding instead of cars moving round..... (i think i have solved this problem before.....!)

Ok, I know, every period, the cars form a permutation, How do you guarantee that the initial permutation is going to appear again?, i know it is just a detail and my solution is the same as yours but obviously with this detail solved....

I've also seen this problem before (and its solution; I didn't solve it though, it was shown to me ), and I didn't want to spoil the pleasure of those who hadn't . When the labels of the cars reach the initial positions, it's just as if the whole thing started all over again, so by the next time they reach the initial position, the exact same permutation $\pi$ will have been performed on them. This means that we have to wait for $k$ turns until the cars are all in their initial positions, where $\pi^k=e$ ($e$ is the identity permutation; for any $\pi$ there is such an $k$).

We solve this problem through a few observations. Label the cars in integers from $0$ to $n-1$ counterclockwise starting from an arbitrary car. Observation 1. The collision condition given in the problem statement is equivalent to two cars passing through each other and switching numbers, continuing in the same speed and direction. Since every car has the same angular speed, they will all return to their original position regardless of direction after a full revolution. Thus, it is shown that the cars will eventually end in the same position at the same time, possibly having different numbers (but not necessarily). $\square$ Observation 2. A car labeled $k$ for $0 \leq k \leq n - 1, k \in \mathbb{Z}$ must be located between the cars labeled $k + 1 \pmod n$ and $k - 1 \pmod n$ at all times in the same order. (In other words, a car cannot somehow pass through the cars directly adjacent to it while keeping the same number, using the the artificial collision condition used in Observation 1.) This implies that the permutation of the numbers of the cars after a revolution will be a rotation of the original permutation of the numbers. $\square$ In this manner, after the cars travel one revolution we can generate that new permutation/rotation by adding some integer $a$ to every car's number, taking $\pmod n$ if a car's number exceeds $n - 1$. After $n$ revolutions we will have added $na$ to every car's number. Since $na + k \equiv k \pmod n$, after $n$ revolutions the cars will be back in their original permutation and location, which completes the proof. $\blacksquare$