Question

Two carts of equal mass are standing still. Each cart has two men of equal mass standing on them. From the first cart, a man jumps in north direction with some velocity with respect to the cart and then the second man jumps with the same velocity with respect to cart towards the south. The two men on other cart jump simultaneously with the same velocity with respect to cart. The final velocity of both the carts is same.

• A. True
• B. False

Answer 1

The correct option is B

False

Let the mass of carts be $M$ and the mass of the men be $m$. Also, let the velocity of the men after jumping with respect to cart be $u$, velocity of the cart after first jump be $v$ and the velocity of the cart after second jump be $v^{\prime }$.
Conserving momentum of the first cart after first jump:
$(M+2m)\times 0=m\times (u-v)+(M+m)(-v)\Rightarrow 0=mu-mv-Mv-mv\Rightarrow \frac{mu}{M+2m}=v$ ------ (i)
Conserving momentum of the first cart after second jump:
$(M+m)(-v)=m\times (u-v^{\prime })-M{v}^{\prime }\Rightarrow \frac{-(M+m)mu}{M+2m}=mu-(m+M)v^{\prime }\left(from\right(i\left)\right)\Rightarrow v^{\prime }=\frac{mu(2M+3m)}{(M+2m)(M+m)}$
In case of second cart, both men jumps simultaneously so the net effect will be zero. Hence, the final velocity of the second cart will be zero.
Thus the above statement is false.