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. The work done by man during the process of jumping is

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=2v3Hence work done by man = change in KE of system w=12mu2+122m(v−u)2 work done by man =12m(2v3)2+122m(v3)2=mv23