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Question

# A man of mass m1 is standing on a platform of masses m2 kept on a smooth on a horizontal surface . If the man starts moving on the platform with a velocity v relative to the platform, then recoil velocity of platform is

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Solution

## There are small changes in variables taken.just follow this,you will get your answer. Here, m1=m & m2=M Consider the situation. Suppose the man moves at a speed w towards right and the platform recoils at a speed V towards left, both relative to the ice(smooth surface). Hence, the speed of the man relative to the platform is V + w 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 MV=m(v−V). MV=m(v−V) v=Mv/(M+m)

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