Question

A man of mass $40\text{kg}$ is standing on a platform of $12\text{kg}$. If the man starts running at $6\text{m/s}$ relative to the platform, the velocity of the platform relative to the ground is

• A. $3\text{m/s}$
• B. $4.6\text{m/s}$
• C. $2\text{m/s}$
• D. $6\text{m/s}$

Answer 1

The correct option is B $4.6\text{m/s}$


Mass of the platform $m_1=12\text{kg}$
Mass of the man $m_2=40\text{kg}$
Taking rightward direction as +ve,
Let velocity of platform relative to the ground be $v$ towards left.
$v_1=-v$
Then, velocity of man relative to the ground is $(6-v)$
$v_2=6-v$
Using momentum conservation,
$0=m_1{v}_1+m_2{v}_2$ (initially, both at rest)
$40(6-v)=12v$
$240=52v$
$v=\frac{240}{52}$
$\Rightarrow v=4.6\text{m/s}$ (towards left)