Question

A man of mass $75\text{kg}$ is standing on a platform of mass $15\text{kg}$ kept on a smooth horizontal surface. The man starts moving on the platform with a velocity of $20\text{m/s}$ relative to the platform. Then, magnitude of the recoil velocity of the platform is

• A. $12.33\text{m/s}$
• B. $20\text{m/s}$
• C. $25\text{m/s}$
• D. $16.66\text{m/s}$

Answer 1

The correct option is D $16.66\text{m/s}$

Taking the platform and the man as a system, $F_{\text{ext}}=0$ on the system in horizontal direction.

$\therefore$ The linear momentum of the system remains constant and system was at rest intitially.
$P_i=P_f=0...\left(i\right)$
Mass of man $\left(m_1\right)=75\text{kg}$
Mass of platform $\left(m_2\right)=15\text{kg}$

Given, velocity of man is $20\text{m/s}$ be w.r.t platform (along $+vex$ direction)
Let the recoil velocity of platform be $-v\text{m/s}$ (along $-vex$ axis).
$\therefore$ Velocity of man w.r.t ground is $(20-v)\text{m/s}$
From Eq $\left(i\right)$:
$P_f=0$
$75(20-v)+15(-v)=0$
$90v=1500$
$\therefore v=\frac{1500}{90}=16.66\text{m/s}$