Question

A man of mass $m_1$ is standing on a platform of mass $m_2$ kept on a smooth horizontal surface. The man starts moving on the platform with a velocity $v_r$, relative to the platform. Find the recoil velocity of platform.

• A. $\frac{m_2{v}_r}{m_1+m_2}$
• B. $\frac{m_1{v}_r}{m_1+m_2}$
• C. $\frac{2m_1{v}_r}{m_1+m_2}$
• D. $\frac{2m_2{v}_r}{m_1+m_2}$

Answer 1

The correct option is B $\frac{m_1{v}_r}{m_1+m_2}$

Absolute velocity of man $=v_r-v$, where $v=$ recoil velocity of platform.


Taking the platform $+$ man as a system, net external force on the system in horizontal direction is zero.

The linear momentum of the system remains constant.

Initially both the man and the platform were at rest.

Hence,

$0=m_1(v_r-v)-m_{2}v$

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



Alternate method:


Using the equation,

$\sum m_R{v}_R=\sum m_L{v}_L$

$\Rightarrow m_1(v_r-v)=m_{2}v$

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

Hence, $\left(B\right)$ is the correct option.