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(B) Let $m_m = 100\,kg$ be the mass of the man and $m_p = 200\,kg$ be the mass of the platform.Since the system is on a smooth ice surface,there is no

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$10$

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(B) Let $m_m = 100\,kg$ be the mass of the man and $m_p = 200\,kg$ be the mass of the platform.Since the system is on a smooth ice surface,there is no external horizontal force acting on the system.Therefore,the velocity of the center of mass remains constant. Initially,the system is at rest,so $v_{com} = 0$.Let $v_p$ be the velocity of the platform relative to the ice (ground) in the backward direction.Let $v_m$ be the velocity of the man relative to the ice (ground) in the forward direction.The velocity of the man relative to the platform is given as $v_{m/p} = 30\,m/s$.By relative velocity definition,$v_{m/p} = v_m - (-v_p) = v_m + v_p$.So,$v_m = 30 - v_p$.Using the conservation of momentum,the total momentum of the system must be zero:$m_m v_m - m_p v_p = 0$$100(30 - v_p) - 200 v_p = 0$$3000 - 100 v_p - 200 v_p = 0$$3000 = 300 v_p$$v_p = 10\,m/s$.Thus,the platform recoils with a velocity of $10\,m/s$ relative to the ice.

$A$ man of mass $100\,kg$ is standing on a platform of mass $200\,kg$ which is kept on a smooth ice surface. If the man starts moving on the platform with a speed $30\,m/s$ relative to the platform,then calculate with what velocity relative to the ice the platform will recoil.

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