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. $\frac{mv}{M+m}$
• B. $\frac{Mv}{M+m}$
• C. $\frac{mv}{M-m}$
• D. $\frac{Mv}{M-m}$

Answer 1

The correct option is A $\frac{mv}{M+m}$

Answer is A.
Consider the situation shown in figure. Suppose the man moves at a speed $w$ towards right and the the platform recoils at a speed $V$ towards left, both relative to the ice. Hence, the speed of the man relative to the platform is $V+w.$.

By the question,

$V+w=v,$ or $w=v-V$ ........ (i)

Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Thus,

$0=MV-mw$ or $MV=m(v-V)$ [Using (i)]

or, $V=\frac{mv}{M+m}$

Hence, the velocity relative to the ice is $V=\frac{mv}{M+m}$.