Question

A man is standing on a cart of mass double the mass of the man. Initially cart is at rest. Now, man jumps horizontally with velocity $u$ relative to cart. Then the work done by the man during the process of jumping will be

• A. $\frac{m{u}^2}{2}$
• B. $\frac{3m{u}^2}{4}$
• C. $m{u}^2$
• D. none of the above

Answer 1

The correct option is D none of the above

$2mv=m(u-v)$

$\therefore v=\frac{u}{3}\therefore u-v=\frac{2u}{3}$

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

$=\frac{1}{3}m{u}^2$

Using Work Energy Theorem

Work Done $=$ Change in kinetic Energy

$W=\frac{1}{3}m{u}^2$