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Question

A man is standing on a cart of 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. Find the work done by man during the process of jumping.

A
mv23
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B
mv2
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C
mv26
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D
none of the above
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Solution

The correct option is A mv23
Let the velocity of man after jumping be u towards right. Then speed of cart is vu towards left.
From conversation of momentum
mu=2m(vu)
u=2v3
Hence work done by man = change in K.E of system

=12mu2+122m(vu)2
=12m(2v3)2+122m(v3)2=mv23

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