Momentum in the horizontal direction must be conserved, because there are no forces in the horizontal direction. Therefore,
$\begin{align*}M v = m v'\end{align*}$
where $v'$ is the velocity of the man. The total kinetic energy after the leap is
$\begin{align*}\frac{1}{2} M v^2 + \frac{1}{2} m v'^2 = \frac{1}{2} M v^2 + \frac{1}{2} m (M v/m)'^2 = \frac{1}{2} \left(M + M...$
Therefore, answer (D) is correct.

Solution to 1996 Problem 66