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A 50kg man is standing at one end of a 25m long boat. He starts running towards the other end. On reaching the other end his velocity is $2 m s^{- 1}$ . If the mass of the boat is 200 kg, final velocity of the boat is ( in $m s^{- 1}$ )

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$\frac{8}{5}$

$\frac{8}{3}$

$\frac{2}{5}$

$\frac{2}{3}$

answer is A.

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Detailed Solution

$A \underline{C C}$ to L.C.L.M

$m (V_{m} - V_{b}) = M V_{b}$

$m V_{m} = (M + m) V_{b}$

$V_{b} = \frac{m V_{m}}{M + m} = \frac{50 \times 2}{250} = \frac{2}{5} m s^{- 1}$

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