    Question

# A man of mass M=50 kg stands at one end of the boat which is floating on still water. The man walks to the other end of the boat. Length of the boat is L=10 m and mass M′=100 kg. Then find the distance moved by the boat relative to the ground

A
103 m
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B
53 m
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C
23 m
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D
0 m
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Solution

## The correct option is A 103 mHere no external force is acting on the system. Hence xcom=0. COM will not change its position. Assume When man reaches the other end boat moved x distance relative to the ground. Hence, Distance travelled by the man w.r.t ground, x1=(L−x)=(10−x) m Distance travelled by the boat w.r.t ground is x m As we know, Δxcom=m1Δx1+m2Δx2m1+m2 Δxcom=0 0=50×(10−x)+(100)(−x)50+100 (−x because we assumed rightward movement as positive) ⇒500−50x−100x=0 ⇒x=500150 ∴x=103 m  Suggest Corrections  0      Similar questions  Explore more