A man of $50 \mathrm{~kg}$ is standing at one end on a boat of length $25 \mathrm{~m}$ and mass $200…

A man of $50 \mathrm{~kg}$ is standing at one end on a boat of length $25 \mathrm{~m}$ and mass $200 \mathrm{~kg}$. If he starts running and when he reaches the other end, he has a velocity $2 \mathrm{~ms}^{-1}$ with respect to the boat. The final velocity of the boat is : (in $\mathrm{ms}^{-1}$ )
  1. $\frac{2}{5}$
  2. $\frac{2}{3}$
  3. $\frac{8}{5}$
  4. $\frac{8}{3}$

Solution

Let $v$ be the velocity of the boat with respect to the water, then from conservation of linear momentum. $(200+50) v+50 \times 2=50 \times 0+200 \times 0$ $250 v=-100$ $v=-\frac{100}{250}=-\frac{2}{5} \mathrm{~m} / \mathrm{s}$

Asked in: AP EAMCET 2006

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