Asked by srikala

a man of 50kg is standing on one end on a boat of length 25m and mass 200 kg .if he starts running and when he reaches the other end ,he has a velocity 2m/s with respect to the boat .the final velocity of the boat is?
Answered by drwls
Total momentum (relative to land) must remain zero. Let V be the final boat velocity (positive backwards). The man's final velocity with respect to land will be (V -2)m/s.

200*V + 25*(V-2) = 0
175 V = 50
V = 0.286 m/s
Answered by Anonymous
0.23
Answered by Anaswara
according to law of conservation of linear momentum
m1u1+m2u2=m1v1+m2v2

velocity of man with respect to man =Vm-Vb
2=Vm-Vb
therefore ,iniitally at rest so before =o
0=200Vb+100+50Vb
250Vb= -100
Vb= -2/5
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