Solution
Consider the ball and the piston as a single combined system. The ball bounces elastically, transferring momentum to the piston. The piston moves downwards against a constant frictional force $F$.
Eventually, the system will come to a state of rest (or minimal energy) where the potential energy lost by the masses is fully dissipated as work done against friction. Since the collisions are perfectly elastic, no energy is lost in the impact itself; energy is only lost due to the sliding friction of the piston.
We apply the Work-Energy Theorem. The total loss in gravitational potential energy of the system is equal to the work done against the constant frictional force.
Let $y$ be the total distance the piston moves down.
- Loss in PE of Piston: The piston mass $M$ moves down by distance $y$.
$\Delta PE_{\text{piston}} = Mgy$. - Loss in PE of Ball: The ball mass $m$ starts at height $h$ above the piston and ends up at the piston’s final level. Total vertical displacement is $h + y$.
$\Delta PE_{\text{ball}} = mg(h + y)$. - Work done against Friction: The frictional force $F$ acts over the sliding distance $y$.
$W_{\text{friction}} = F \cdot y$.
Equating the energy loss to the work done: $$ \Delta PE_{\text{total}} = W_{\text{friction}} $$ $$ mg(h + y) + Mgy = Fy $$
Expand and group the terms containing $y$: $$ mgh + mgy + Mgy = Fy $$ $$ mgh + (m + M)gy = Fy $$ $$ mgh = Fy – (m + M)gy $$ $$ mgh = y [ F – (m + M)g ] $$
Solving for $y$: $$ y = \frac{mgh}{F – (m + M)g} $$
The piston moves down by a distance $$ \frac{mgh}{F – (m + M)g} $$.