Solution to 1996 Problem 66

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.

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