Discussion
A man of mass m in an initially stationary boat of mass M gets off the boat by jumping to the left in an exact horizontal direction. Immediately after the jump, the boat is observed to be moving to the right at speed v. How much total work did the man do ? (Neglect friction)
Answer: D
The man does work \( \frac{1}{2}\left(M + \frac{M^{2}}{m}\right)v^{2} \).
Momentum is conserved horizontally, so the man's speed \( u \) is fixed by \( mu = Mv \). The work he does appears as the kinetic energy of both bodies:
\[ \begin{aligned} u &= \frac{Mv}{m} \\ W &= \frac{1}{2}Mv^{2} + \frac{1}{2}m\left(\frac{Mv}{m}\right)^{2} \\ &= \frac{1}{2}\left(M + \frac{M^{2}}{m}\right)v^{2} \end{aligned} \]
\( \frac{1}{2}Mv^{2} \) counts only the boat's energy. \( \frac{1}{2}mv^{2} \) gives the man the boat's speed, and \( \frac{1}{2}(M + m)v^{2} \) gives both bodies the speed \( v \), which momentum conservation forbids unless \( m = M \).
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