A boy of mass m stands on one end of a wooden plank of length I and mass M. The plank is floating on water. If
the boy walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is given by

. Initially, when the body stands on the plank, the momentum o f boy-plank system is zero because they are at rest. Let the boy walks with a speed v on the plank. As a result, let the speed of the plank in the opposite direction be y. Now the total momentum of the system is $\mathrm{mv}-(\mathrm{M}+\mathrm{m})\mathrm{V}$According to the law of consecrations of momentum $\mathrm{mv}-(\mathrm{M}+\mathrm{m})\mathrm{V}=0$

$$\begin{gathered} \therefore \frac{\mathrm{V}}{\mathrm{V}}=\frac{\mathrm{m}}{(\mathrm{M}+\mathrm{m})} \\ \mathrm{As} \mathrm{the} \mathrm{distance} \mathrm{moved} \mathrm{is} \mathrm{proportional} \mathrm{to} \mathrm{the} \mathrm{speed}, \mathrm{the} \mathrm{displacement} \mathrm{L}\text{'} \mathrm{of} \mathrm{the} \mathrm{plank} \mathrm{is} \mathrm{givea} \mathrm{by} \\ \begin{array}{r}\frac{{\mathrm{L}}^{\mathrm{\prime }}}{\mathrm{L}}=\frac{\mathrm{V}}{\mathrm{V}}=\frac{\mathrm{m}}{(\mathrm{M}+\mathrm{m})}\\ {\mathrm{L}}^{\mathrm{\prime }}=\left(\frac{\mathrm{mL}}{\mathrm{M}+\mathrm{m}}\right)\end{array} \end{gathered}$$