The velocity of the mass $m$ immediately before the collision is

$\begin{align}v = \sqrt{2 g h}\end{align}$
The velocity of the mass $m$ immediately after the collision is

$\setcounter{equation}{1}\begin{align}v' = \frac{m - 2m}{m + 2m} v = \frac{-1}{3}\sqrt{2 g h}\end{align}$
So, the kinetic energy of mass $m$ after the collision is

$\setcounter{equation}{2}\begin{align}1/2 m v'^2 = m g h/9\end{align}$

So, the ball rises to a final height of $\boxed{h/9}$ . Thus, answer (A) is correct.

Solution to 1996 Problem 7