Question

A man of mass $m$ is standing on a platform of mass $M$ kept on smooth ice. If the man starts moving on the platform with a speed $v$ relative to the platform, with what velocity relative to the ice does the platform recoil?
A) $\dfrac{{mv}}{{M + m}}$
B) $\dfrac{{Mv}}{{M + m}}$
C) $\dfrac{{mv}}{{M - m}}$
D) $\dfrac{{Mv}}{{M - m}}$

Answer 1

Hint: In order to find the solution of the given question, first of all we need to relate the relative velocity of the man with the platform. Then we need to relate that velocity with the linear momentum of the system. After solving the equation formed we can finally conclude with the correct solution of the given question.

Complete step by step solution:
First of all let us find the velocity of man relative to the platform.
Let us assume that the man moves with a velocity of $u$ on the right side and the platform recoils with a velocity of $V$towards the left side relative to the smooth surface.
So, the velocity of the man relative to the platform can be written as, $u + V$
Now, the equation for the relative velocities can be written as,
$u + V = v$
Or, $u = v - V$………… (i)
We know that linear momentum of a system is constant.
And initially both the man and the platform were at rest.
Now, according to the conservation of momentum, we can write,
$\Rightarrow MV - mu = 0$
$ \Rightarrow MV = mu$
From equation (i), we can write the above equation as,
$ \Rightarrow MV = m(v - V)$
$ \Rightarrow Mv = mv - mV$
$ \Rightarrow Mv + mv = mV$
$ \Rightarrow v(M + m) = mV$
$\therefore v = \dfrac{{mV}}{{M + m}}$
Therefore, the required recoil velocity of the ice is $\dfrac{{mV}}{{M + m}}$.

Hence, option (A), i.e. $\dfrac{{mV}}{{M + m}}$ is the correct choice of the given question.

Note: According to the conservation of momentum, the momentum before collision is equal to the momentum after collision. We define momentum as the product of mass and the velocity of a body. When the bodies are moving in a straight path, the momentum is said to be linear momentum.