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Question

A man of 80kg attempts to jump from a small boat of mass 40kg on to the shore. He can generate a relative velocity of 6m/s between himself and boat. His velocity towards the shore is:

A
4m/s
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B
8m/s
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C
2m/s
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D
3m/s
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Solution

The correct option is B 2m/s
Mm=80kg
MB=40kg
VmB=6m/s
let velocity of man is Vm
velocity of boat = VB
Since there is no external force in jumping direction so we can apply momentum conservation.
Initial momentum = final momentum
Mm×(0)+MB×(0)=Mm×(Vm)+MB×(VB)
so,
2Vm=VB
Now
VmB=VmVB
Vm=2m/s
Hence velocity of man towards shore =2m/s
hence correct option is (C)


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