Q.
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?
NTA AbhyasNTA Abhyas 2020
Solution:
Let the velocity of man after jumping be $'u'$ towards right. Then speed of cart is $v-u$ towards left. From conservation of momentum $mu=2m\left(v - u\right)$
$\therefore u=\frac{2 v}{3}$
Hence work done by man $=$ change in KE of system
$=\frac{1}{2}mu^{2}+\frac{1}{2}2m\left(v - u\right)^{2}$
$=\frac{1}{2}m\left(\frac{2 v}{3}\right)^{2}+\frac{1}{2}2m\left(\frac{v}{3}\right)^{2}=\frac{m v^{2}}{3}$
$\therefore u=\frac{2 v}{3}$
Hence work done by man $=$ change in KE of system
$=\frac{1}{2}mu^{2}+\frac{1}{2}2m\left(v - u\right)^{2}$
$=\frac{1}{2}m\left(\frac{2 v}{3}\right)^{2}+\frac{1}{2}2m\left(\frac{v}{3}\right)^{2}=\frac{m v^{2}}{3}$