How Much Work Does a Man Perform by Jumping Off a Stationary Boat?

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The discussion revolves around calculating the work done by a man jumping off a stationary boat, involving concepts of conservation of momentum and kinetic energy. The man’s horizontal velocity after the jump is derived from the momentum equation, leading to the expression for his kinetic energy. There is confusion regarding the correct formulation of the kinetic energy, particularly in squaring the velocity term. The participants are seeking clarification on the calculations and the correct approach to determine the total work done. The thread emphasizes the importance of accurately applying physics equations to solve the problem.
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Homework Statement


A man of mass m is on a still boat of mass M. the man jumps to the left so he has only horizontal velocity.
Immediately afterwards the boat has velocity v.
What's the work the man made.

Homework Equations


Conservation of momentum: ##m_1v_1=m_2v_2##
Kinetic energy: ##E=\frac{1}{2}mv^2##

The Attempt at a Solution


The man's velocity, vm, in an inertial frame, after the jump:
$$Mv=mv_m\rightarrow v_m=\frac{M}{m}v$$
$$E=\frac{1}{2}\left[ Mv^2+\frac{M}{m}v^2 \right]=\frac{1}{2}\frac{M(m+1)}{m}v^2$$
I don't think it's the right answer
 
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What is the square of ##\frac{M}{m}v##?
 
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$$E=\frac{1}{2}\left[ Mv^2+m\left( \frac{M}{m}v\right)^2 \right]=\frac{1}{2}\left( M+\frac{M^2}{m} \right)v^2$$

Thanks
 
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