Q126 / 160 · AP EAMCET 2006 · Physics

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 $\mathrm{ms}^{-1}$ )

  • Correct answer
    A
    $\frac{2}{5}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{8}{5}$
  • D
    $\frac{8}{3}$

Solution

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

Hence, the correct answer is option: 'A'.

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