Question

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?

• 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 ${}^{\prime }{u}^{\prime }$ towards right. Then speed of cart is $v-u$ towards left. From conservation of momentum $mu=2m\left(v-u\right)$
$\therefore u=\frac{2v}{3}$
Hence work done by man $=$ change in KE of system
$=\frac{1}{2}m{u}^2+\frac{1}{2}2m{\left(v-u\right)}^2$
$=\frac{1}{2}m{\left(\frac{2v}{3}\right)}^2+\frac{1}{2}2m{\left(\frac{v}{3}\right)}^2=\frac{m{v}^2}{3}$