Question

Two men, each of mass $m,$ stand on the edge of a stationary buggy of mass $M.$ Assuming the friction to be negligible, find the velocity of the buggy after both men jump off with the same horizontal velocity $u$ relative to the buggy:
(i) simultaneously;
(ii one after the other.
In what case will the velocity of the buggy be greater and how many times?

Answer 1

(i) Let $\overrightarrow{v_1}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men $+$ buggy), from the conservation of linear momentum,
$M\overrightarrow{v_1}+2m(\overrightarrow{u}+\overrightarrow{v_1})=0$
or, $\overrightarrow{v_1}=\frac{-2m\overrightarrow{u}}{M+2m}$ (1)
(ii) Let ${\overrightarrow{v}}^{\prime }$ be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man $+$ other man) from the conservation of linear momentum,
$0=(M+m){\overrightarrow{v}}^{\prime }+m(\overrightarrow{u}+{\overrightarrow{v}}^{\prime })$ (2)
Let $\overrightarrow{v_2}$ be the sought velocity of the buggy when the second man jump off the buggy; then from conservation of linear momentum of the system (buggy $+$ one man),
$(M+m){\overrightarrow{v}}^{\prime }=M\overrightarrow{v_2}+m(\overrightarrow{u}+\overrightarrow{v_2})$ (3)
Solving equations (2) and (3) we get
$\overrightarrow{v_2}=\frac{m(2M+3m)\overrightarrow{u}}{(M+m)(M+2m)}$ (4)
From (1) and (4)
$\frac{v_2}{v_1}=1+\frac{m}{2(M+m)}>1$
Hence $v_2>v_1$