Bounding the Masses!

If the masses $M_1,M_2$ have velocities $u_1,u_2$ before a collision and velocities $v_1,v_2$ after a collision, then $M_1v_1 + M_2v_2 \; = \; M_1u_1 + M_2u_2 \hspace{2cm} -v_1 + v_2 \; = \; u_1 - u_2$ and hence $v_1 \; = \; \frac{(M_1-M_2)u_1 + 2M_2u_2}{M_1+M_2} \hspace{2cm} v_2 \; = \; \frac{2M_1u_1 + (M_2-M_1)u_2}{M_1+M_2}$ Thus, after the first collision of the masses, they have speeds $\tfrac{M_1-M_2}{M_1+M_2}u$ and $\tfrac{2M_1}{M_1+M_2}u$ respectively. For there to be a second collision, the $M_1$ mass must now hit the wall, and hence we must have $M_1 < M_2$ . After the collision of the $M_1$ mass with the wall, it now has velocity $\tfrac{M_2-M_1}{M_1+M+2}u$ . Provided that $M_2 - M_1 > 2M_1$ , the two masses will collide again. Thus we get at least three collisions provided that $M_2 > 3M_1$ . After this third collision, the $M_1,M_2$ masses have velocities $\frac{6M_1M_2 - M_1^2 - M_2^2}{(M_1+M_2)^2}u \hspace{2cm} \frac{4M_1(M_2-M_1)}{(M_1+M_2)^2}u$ respectively. There will therefore be exactly three collisions provided that $0 \; \le \; 6M_1M_2-M_1^2-M_2^2 \; = \; 8M_1^2-(M_2-3M_1)^2$ Putting this altogether, there will be exactly three collisions provided that $3M_1 < M_2 \le (3+\sqrt{8})M_1$ . Thus we require $3\pi \; < \; \Gamma \; \le \; (3 + \sqrt{8})\pi$ and hence $\gamma = (6 +\sqrt{8})\pi$ , which makes $\lfloor 100\gamma \rfloor = \boxed{2773}$ .