A man of mass 40kg is standing on a platform of 12kg. If the man starts running at 6m/s relative to the platform, the velocity of the platform relative to the ground is
A
3m/s
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B
4.6m/s
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C
2m/s
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D
6m/s
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Solution
The correct option is B4.6m/s
Mass of the platform m1=12kg Mass of the man m2=40kg Taking rightward direction as +ve, Let velocity of platform relative to the ground be v towards left. v1=−v Then, velocity of man relative to the ground is (6−v) v2=6−v Using momentum conservation, 0=m1v1+m2v2 (initially, both at rest) 40(6−v)=12v 240=52v v=24052 ⇒v=4.6m/s (towards left)