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 ${}^{\prime }{v}^{\prime }$ horizontally towards right with respect to cart. Find the work done by man during the process of jumping.

• A. $\frac{m{v}^2}{3}$
• B. $m{v}^2$
• C. $\frac{m{v}^2}{6}$
• D. none of the above

Answer 1

The correct option is A $\frac{m{v}^2}{3}$

Let the velocity of man after jumping be ${}^{\prime }{u}^{\prime }$ towards right. Then speed of cart is $v-u$ towards left.
From conversation of momentum
$mu=2m(v-u)$
$\therefore u=\frac{2v}{3}$
Hence work done by man = change in K.E of system

$=\frac{1}{2}m{u}^2+\frac{1}{2}2m(v-u{)}^2$
$=\frac{1}{2}m{\left(\frac{2v}{3}\right)}^2+\frac{1}{2}2m{\left(\frac{v}{3}\right)}^2=\frac{m{v}^2}{3}$