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A man is standing on a cart having mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ‘v’ horizontally towards right with respect to cart. The work done by man during the process of jumping is

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

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

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

$m v^{2}$

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 mu = 2m(v-u)

$∴ u = \frac{2 v}{3}$

Hence work done by man = change in KE of system

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

$w o r k d o n e b y m a n = \frac{1}{2} m {(\frac{2 v}{3})}^{2} + \frac{1}{2} 2 m {(\frac{v}{3})}^{2} = \frac{m v^{2}}{3}$

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