Pi Day Special

It's 31415

Thanks for the precisions! The link appears not to be working, a better one is this one , found on the same website.

This is not true. In the notation of my solution to your posted problem, there will be $n$ ball-ball collisions if $(2n-1)\alpha \; < \; \pi \; \le \; (2n+1)\pi$ where $\tan\alpha = \sqrt{\tfrac{m}{M}}$ . In the subcase that $2n\alpha < \pi \le (2n+1)\pi$ , the $m$ particle will hit the wall, but not the $M$ particle after the $n$ th ball-ball collision, so that there will be $2n$ collisions altogether.

As a concrete example, suppose that $m = 4M$ , so that $\tan\alpha = 2$ . Then $\tfrac13\pi < \alpha < \tfrac12\pi$ , so $2\alpha < \pi < 3\alpha$ , and my formula says that there should be $2$ collisions. Before the first collision, the $M$ particle has velocity $v$ , and the $m$ particle has velocity $0$ , towards the wall. After the first ball-ball collision, the $M$ particle has velocity $-\tfrac35v$ , and the $m$ particle has velocity $\tfrac25v$ , towards the wall.. After the ball-wall collision, the particles have velocities $-\tfrac35v$ , $-\tfrac25v$ towards the wall, and so the $m$ particle will not catch up with the $M$ particle as it heads away from the wall. There have been exactly $2$ collisions.

After the final ball-ball collision, the $M$ particle will always be headed away from the wall (since $(2n+1)\alpha \ge \pi$ , $2n\alpha > \tfrac12\pi$ , and so $\cos2n\alpha < 0$ ). The $m$ particle might have velocity away from the wall (and therefore not hit the wall again), but it might have velocity towards the wall, and hit the wall, but not bounce back fast enough to catch the $M$ particle.