Dynamics | Part IA, 2006

A particle of mass mm bounces back and forth between two walls of mass MM moving towards each other in one dimension. The walls are separated by a distance (t)\ell(t). The wall on the left has velocity +V(t)+V(t) and the wall on the right has velocity V(t)-V(t). The particle has speed v(t)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 vcm=(mvMV)/(m+M)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., mMm \ll M. Show that the particle's speed increases by approximately 2V2 V every time it collides with a wall.

Assume also that vVv \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 Δt\Delta t, over which the change in VV is negligible, the wall separation changes by Δ2VΔt\Delta \ell \approx-2 V \Delta t. Show that the number of particle-wall collisions during Δt\Delta t is approximately vΔt/v \Delta t / \ell and that the particle's speed increases by Δv(Δ/)v\Delta v \approx-(\Delta \ell / \ell) v during this time interval.

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

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