Question

Two men of the same mass $m=50\text{kg}$ jump from a stationary buggy of mass $M=50\text{kg}$ simultaneously, with a velocity of $u=20\text{m/s}$ towards left. Find the velocity of the buggy after they jump.

• A. $20\text{m/s}$ towards left
• B. $\frac{40}{3}\text{m/s}$ towards right
• C. $10\text{m/s}$ towards right
• D. $\frac{20}{3}\text{m/s}$ towards left

Answer 1

The correct option is B $\frac{40}{3}\text{m/s}$ towards right

Given,
Masses of men $=m=50\text{kg}$ each
Mass of the buggy $=M=50\text{kg}$

Velocity of each man w.r.t buggy $u=20\text{m/s}$ (towards left)
Let $v=$ velocity of buggy after men jump off (towards right)


There is no external force. Hence momentum will be conserved. Taking leftward direction +ve,
$P_i=P_f$
$\Rightarrow 0=m(u-v)+m(u-v)-Mv$
$\Rightarrow 2m(u-v)=Mv$
$\Rightarrow 2mu=(2m+M)v$
$\Rightarrow v=\left(\frac{2\times 50\times 20}{2\times 50+50}\right)$
$\Rightarrow v=\frac{40}{3}\text{m/s}$ towards right