 Solution

Measure $\theta$ from the downward vertical. The falling mass loses gravitational potential energy $3mgl(1-\cos \theta )$. The fixed pivot does no work, so conservation of energy gives its incident speed $u$:At the bottom, the rod is vertical and the impact is horizontal. Its impulsive reaction therefore has no horizontal component. Let $U$ and $w$ be the subsequent horizontal velocities of the heavy and light balls. Horizontal momentum conservation and unit coefficient of restitution giveThus $U=u/2$, and the light ball's outgoing speed isThe calculation neglects the sizes of the balls, treats the rod as massless, and uses an elastic collision. The expression vanishes for release at the bottom; a ball placed exactly at the unstable upright equilibrium requires a perturbation before it can fall.