A particle of mass $m$ bounces back and forth between two walls of mass $M$ moving towards each other in one dimension. The walls are separated by a distance $\ell(t)$. The wall on the left has velocity $+V(t)$ and the wall on the right has velocity $-V(t)$. The particle has speed $v(t)$. Friction is negligible and the particle-wall collisions are elastic.

Consider a collision between the particle and the wall on the right. Show that the centre-of-mass velocity of the particle-wall system is $v_{\mathrm{cm}}=(m v-M V) /(m+M)$. Calculate the particle's speed following the collision.

Assume that the particle is much lighter than the walls, i.e., $m \ll M$. Show that the particle's speed increases by approximately $2 V$ every time it collides with a wall.

Assume also that $v \gg V$ (so that particle-wall collisions are frequent) and that the velocities of the two walls remain nearly equal and opposite. Show that in a time interval $\Delta t$, over which the change in $V$ is negligible, the wall separation changes by $\Delta \ell \approx-2 V \Delta t$. Show that the number of particle-wall collisions during $\Delta t$ is approximately $v \Delta t / \ell$ and that the particle's speed increases by $\Delta v \approx-(\Delta \ell / \ell) v$ during this time interval.

Hence show that under the given conditions the particle speed $v$ is approximately proportional to $\ell^{-1}$.