Question

Two men jump simultaneously from one buggy to another on the same track, towards left direction with a velocity of $u$ . Initially, both the buggies are stationary. Find the final velocities $\left(\text{in m/s}\right)$ of the buggies after they jump. Given: masses of the men $=50\text{kg}$ each, mass of buggy $=200\text{kg}$ and relative velocity of jumping $u=10\text{m/s}$ w.r.t. buggy.

• A. $\frac{10}{3}$, $\frac{20}{9}$
• B. $\frac{50}{3}$, $\frac{25}{9}$
• C. $\frac{25}{13}$, $\frac{25}{18}$
• D. $\frac{50}{3}$, $\frac{20}{9}$

Answer 1

The correct option is A $\frac{10}{3}$, $\frac{20}{9}$

Given:
Mass of men, $m=50\text{kg}$
Mass of buggies, $M=200\text{kg}$
Velocity of man w.r.t. buggy, $u=10\text{m/s}$

There is no external force is acting on the system so momentum will be conserved.
$(p_i=p_f)$

Take +ve direction leftward.
When both men jumped from the buggy. Let $v$ be the velocity of the buggy (towards right)


$0=m(u-v)+m(u-v)-Mv$
$2mu=2mv+Mv$
$v=\frac{2mu}{(2m+M)}=\frac{2\times 50\times 10}{(2\times 50+200)}$ $v=\frac{10}{3}\text{m/s}$

(2) When both men land on the other buggy. Let $v_1$ be the velocity of the other buggy (left)


$\because p_i=p_f$
$2m(u-v)=(2m+M)v_1$
$2\times 50(10-\frac{10}{3})=(2\times 50+200)\times v_1$
$\Rightarrow v_1=\frac{20}{9}\text{m/s}$

Velocity of buggy ($1$) is $v=\frac{10}{3}\text{m/s}$ & buggy ($2$) is $v_1=\frac{20}{9}\text{m/s}$