A man weighing 80 , kg is standing in a troll | vedclass

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(C) Let $m = 80 \, kg$ be the mass of the man and $M = 320 \, kg$ be the mass of the trolley.Since the system is on frictionless rails,there is no ext

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$3.2$

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(C) Let $m = 80 \, kg$ be the mass of the man and $M = 320 \, kg$ be the mass of the trolley.Since the system is on frictionless rails,there is no external horizontal force acting on the system. Therefore,the momentum of the system remains conserved.Initially,the system is at rest,so the total initial momentum is $0$.Let $v_m$ be the velocity of the man relative to the ground and $v_t$ be the velocity of the trolley relative to the ground.Given that the velocity of the man relative to the trolley is $v_{rel} = 1 \, m/s$,we have $v_m - v_t = 1$,which implies $v_m = 1 + v_t$.By the law of conservation of momentum: $m v_m + M v_t = 0$.Substituting the values: $80(1 + v_t) + 320 v_t = 0$.$80 + 80 v_t + 320 v_t = 0 \implies 400 v_t = -80 \implies v_t = -0.2 \, m/s$.The negative sign indicates the trolley moves in the opposite direction of the man.The velocity of the man relative to the ground is $v_m = 1 + (-0.2) = 0.8 \, m/s$.The displacement of the man relative to the ground in $t = 4 \, s$ is $d = v_m 	imes t = 0.8 	imes 4 = 3.2 \, m$.

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