A man of mass m is standing on a platform of | vedclass

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(A) Let the velocity of the platform relative to the ice be $V$ in the backward direction and the velocity of the man relative to the ice be $w$ in th

Vedclass · Accepted Answer

$\frac{mv}{M + m}$

Vedclass · Answer

(A) Let the velocity of the platform relative to the ice be $V$ in the backward direction and the velocity of the man relative to the ice be $w$ in the forward direction.Given that the velocity of the man relative to the platform is $v$,we have $w + V = v$,which implies $w = v - V$.Since there is no external horizontal force acting on the system (man + platform),the linear momentum of the system is conserved.Initially,both are at rest,so the initial momentum is $0$.Applying the law of conservation of linear momentum: $MV - mw = 0$.Substituting $w = v - V$ into the equation: $MV - m(v - V) = 0$.$MV - mv + mV = 0$.$V(M + m) = mv$.Therefore,the velocity of the platform relative to the ice is $V = \frac{mv}{M + m}$.

$A$ man of mass $m$ is standing on a platform of mass $M$ kept on a smooth ice surface. If the man starts moving on the platform with a speed $v$ relative to the platform,with what velocity does the platform recoil relative to the ice?

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