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

Deepak Kumar
31 Points
```							Mass of man = M = 80Kg.Mass of boat = m = 40Kg.Relative velocity of man with respect to boat = 6m/s.Let velocity of man = Vm.      Velocity of boat = -Vb (negative sign indicate that direction is opposite to man).Since there is no external force in Jumping direction so we can apply Momentum consercation.Initial momentum = final momentumMm*0 + Mb*0 = Mm*Vm + Mb*(-Vb). So, 2Vm = Vb.Now Vm/b = Vm – Vb         6 = Vm - (-2Vm) = 3Vm.        Vm = 2m/s.Hence, Velocity of man towards shore = 2m/s.
```
2 years ago
downey zer
73 Points
```							let velocity of man =+|v|then velocity of boat =-|v1| (since if man jumps forward boat will go backwards)as given  |v|-(-|v1|)=6 and 40|v1|=80|v|solving we get |v|=2thats our answer
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions