1.

A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg. If he starts running and when he reaches the other end, he has a velocity 2 ms^(-1) with respectto the boat. The final velocity of the boat is (in ms^(-1))

Answer»

`2/5`
`2/3`
`8/5`
`8/3`

Solution :From the Law of CONSERVATION of Linear momentum
Let L be the velocity of the boat. Then,
`(200 + 50) K+50 XX 2 = 0 `
`250 V+ 100 = 0`
`V=-100/250= -2/5 ms^(-1)`.


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