A boy is standing at the center of a boat which is free to move on water. The distance of the boy from the shore is $100 m$ . If mass of the boy and boat are $40 K g$ and $60 K g$ respectively and boy moves with a velocity $1 m / s e c$ towards the shore, what is the distance of the boy from the shore at the end of 5th second?

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Solution

Velocity of center mass will be zero.
Velocity $v$ of boat is $60 v - 40 \times 1 = 0$
$\Rightarrow v = \frac{2}{3} m / s$
After $5$ sec boy move distance $= 5 \times 1$
$= 5 m$
Boat move distance $= \frac{2}{3} \times 5$
$= \frac{10}{3}$
Distance of boy from share $= 100 - 5 + \frac{10}{3}$
$= 95 + \frac{10}{3}$
$= 98.33 m .$
Hence, solved.

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Velocity of center mass will be zero. Velocity v of boat is 60v−40×1=0⇒v=23m/sAfter 5 sec boy move distance =5×1 =5mBoat move distance =23×5 =103Distance of boy from share =100−5+103 =95+103 =98.33m.Hence, solved.