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Grade: 11
        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

Answers : (2)

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 momentum
Mm*0 + Mb*0 = Mm*Vm + Mb*(-Vb).
 So, 2Vm = Vb.
Now Vm/b = Vm – V
        6 = V- (-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|=2
thats our answer
2 years ago
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