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 respect to the boat. The final velocity of the boat is ( in ${ms}^{- 1})$

Question

Dr. Abubacker · Accepted Answer

(a) Let $v$ be the velocity of the boat with respect to the water, then from conservation of linear momentum.
$(200 + 50)v + 50 	imes 2 = 50 	imes 0 + 200 	imes 0$
$250v = - 100$
$v = - \frac{100}{250} = - \frac{2}{5}\ {ms}^{- 1}$

Q. 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 respect to the boat. The final velocity of the boat is ( in ${ms}^{- 1})$

Explanation

Reviewed by Dr. Abubacker

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