= 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> $u$:
$$
3mgl(1-\cos\theta)=\frac12(3m)u^2,
\qquad u=\sqrt{2gl(1-\cos\theta)}.
$$
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> give
$$
3mu=3mU+mw,\qquad w-U=u.
$$
Thus $U=u/2$, and the light ball's outgoing <speed> is
$$
\boxed{w=\frac32\sqrt{2gl(1-\cos\theta)}
=3\sqrt{gl}\,\left|\sin\frac\theta2\right|.}
$$
The 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.
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