Question

A boy of mass 50 kg is standing at one end of a boat 5 m long floating on water. The water offers no friction. The boy walks slowly till the other end of the boat of mass 200 kg and in this process the boat recoils (backward). Displacement of the boy with respect to the fixed ground is

Answer 1

From law of conservation of momentum
Initial momentum = final momentum
As initially boy and boat were at rest , initial momentum is zero.
So, $0=m_{boy}{V}_{boy}+\left(m_{boy}+m_{boat}\right)V_{boat}$
0 = $50\times \frac{5}{t}+\left(50+200\right)\frac{x}{t}$
or, $$\begin{gathered} \frac{250\mathrm{x}}{\mathrm{t}}=\frac{-250}{\mathrm{t}} \\ \mathrm{x}=\frac{-25}{25}\mathrm{m} \\ \mathrm{x}=1\mathrm{m}\mathrm{Backward}\mathrm{displacement}\mathrm{of}\mathrm{the}\mathrm{boat} \\ \mathrm{Boy}\mathrm{will}\mathrm{also}\mathrm{displaced}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{dirstion}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{ground} \end{gathered}$$