Question

a man of mass 72 kg jumps from a small boat on to the lake shore with a forward velocity of 9.0m/s.if the mass of the boat is 216 kg,calculate the initial backward velocityof the boat

Answer 1

Answer:

Using Law of conservation of momentum,

 m1u1+m2u2=m1v1+m2v2

72*9+216*0=72*0+216*v2

  648=216v2

therefore, velocity of the boat v2=648/216=3 m/s

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