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

• A. $m v^{2}$
• B. $\frac{m v^{2}}{2}$
• C. $\frac{m v^{2}}{4}$
• D. $\frac{m v^{2}}{3}$

Answer 1

Answer: (D)

Let the velocity of man after jumping be ' $u$ ' towards right.
Then speed of cart is $v-u$ towards left.
From conservation of momentum $m u=2 m(v-u)$
$\therefore u=\frac{2 v}{3}$

Hence, work done by man = change in $KE$ of system
$=\frac{1}{2} m u^{2}+\frac{1}{2} 2 m(v-u)^{2}$
$=\frac{1}{2} m\left(\frac{2 v}{3}\right)^{2}+\frac{1}{2} 2 m\left(\frac{v}{3}\right)^{2}=\frac{m v^{2}}{3}$