Two persons 60 kg and 80 kg are sitting at the ends of a boat of mass 60 kg and length 4 m. The boat is stationary. If the men now exchange their positions, then:

Let ${\mathrm{m}}_1,{\mathrm{m}}_2$ are masses of two persons. m₃ is the mass of the boat.
${\mathrm{m}}_1{\overrightarrow{\mathrm{s}}}_1+{\mathrm{m}}_2{\overrightarrow{\mathrm{s}}}_2+{\mathrm{m}}_3{\overrightarrow{\mathrm{s}}}_3=0$
${\overrightarrow{\mathrm{s}}}_3$ is the displacement of boat
Where ${\overrightarrow{\mathrm{s}}}_1\mathrm{\&}{\overrightarrow{\mathrm{s}}}_2$ are displacements of persons
Give that 
$\begin{array}{l}{\overrightarrow{\mathrm{s}}}_{13}=4\widehat{\mathrm{i}}={\overrightarrow{\mathrm{s}}}_1-{\overrightarrow{\mathrm{s}}}_3\\ {\overrightarrow{\mathrm{s}}}_{23}=-4\widehat{\mathrm{i}}={\overrightarrow{\mathrm{s}}}_2-{\overrightarrow{\mathrm{s}}}_3\end{array}$
From the above equations we get 
$\begin{array}{l}60\left(4\widehat{\mathrm{i}}+{\overrightarrow{\mathrm{s}}}_3\right)+80\left({\overrightarrow{\mathrm{s}}}_3-4\widehat{\mathrm{i}}\right)+60{\overrightarrow{\mathrm{s}}}_3=0\\ 200{\overrightarrow{\mathrm{s}}}_3=80\widehat{\mathrm{i}}\\ {\overrightarrow{\mathrm{s}}}_3=0.4\widehat{\mathrm{i}}\end{array}$