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

$\begin{array}{l}\mathrm{m}=50\mathrm{kg}; \mathrm{v}=2\mathrm{m}/\mathrm{s}.\\ \mathrm{M}=200\mathrm{kg}; {\mathrm{v}}_{\mathrm{b}}=? \mathrm{mv}-{\mathrm{mv}}_{\mathrm{b}}={\mathrm{Mv}}_{\mathrm{b}}\\ \mathrm{velocity} \mathrm{of} \mathrm{the} \mathrm{man} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{boat} \mathrm{v}={\mathrm{v}}_{\mathrm{m}}-{\mathrm{v}}_{\mathrm{b}}\\ \mathrm{velocity} \mathrm{of} \mathrm{the} \mathrm{man} {\mathrm{v}}_{\mathrm{m}}=\mathrm{v}+{\mathrm{v}}_{\mathrm{b}}\\ \mathrm{Applying} \mathrm{conservation} \mathrm{of} \mathrm{momentum}\\ m{v}_m+M{v}_b=0\\ m\left(v+v_b\right)+M{v}_b=0\\ {\mathrm{v}}_{\mathrm{b}}=-\frac{\mathrm{mv}}{\mathrm{M}+\mathrm{m}}=-\frac{50\times 2}{200+50}\\ \Rightarrow {\mathrm{v}}_{\mathrm{b}}=-\frac{2}{5} \mathrm{boat} \mathrm{will} \mathrm{move} \mathrm{in} \mathrm{opposite} \mathrm{direction}\end{array}$