A man of mass $40\text{kg}$ is standing on a platform of mass $60\text{kg}$ kept on a smooth horizontal surface. The man starts moving on the platform with a velocity $20\text{m/s}$ relative to the platform. The recoil velocity of the platform is

Given, mass of man, $\left(m_1\right)=40\text{kg}$

mass of platform, $\left(m_2\right)=60\text{kg}$


From the figure, the speed of the man relative to platform is
$u+v=v_r$
$\Rightarrow u=v_r-v...\left(1\right)$

Here, the man and platform together act as s system. There is no external horizontal force on the system.

Initially, both man and platform were at rest, so on applying conservation of linear momentum (taking rightward +ve),

$0=-m_{2}v+m_{1}u$

$\Rightarrow m_{2}v=m_{1}u$

$\Rightarrow m_{2}v=m_1(v_r-v)$[from (1)]

$\Rightarrow v=\frac{m_1{v}_r}{m_1+m_2}$

$\Rightarrow v=\frac{40\times 20}{40+60}=8\text{m/s}$

So, recoil velocity of the platform is $8\text{m/s}$.