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A man is standing on a cart of mass double the mass of the man. Initially, the cart is at rest on the smooth ground. Now man jumps with relative velocity v horizontally towards the right with respect to cart. What will be the work done by the man during the process of jumping?

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$m v^{2}$

$\frac{m v^{2}}{2}$

$\frac{m v^{2}}{4}$

$\frac{m v^{2}}{3}$

answer is D.

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Detailed Solution

Let the velocity of man after jumping be ‘u’ ' towards right. Then speed of cart is v - u towards left. From conservation of momentum

$m u = 2 m (v u)$

$∴ u = \frac{2 v}{3}$

Hence work done by man = change in KE of system

$= \frac{1}{2} m u^{2} + \frac{1}{2} 2 m (v - u)^{2}$

$= \frac{1}{2} m {(\frac{2 v}{3})}^{2} + \frac{1}{2} 2 m {(\frac{v}{3})}^{2} = \frac{m v^{2}}{3}$

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