A ball moving with velocity 2 ~ms^ -1 collides head on with another...

If two bodies collide head on with coefficient of restitution $\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}$ From the law of conservation of linear momentum \begin{array}{r}\mathrm{m}_{1} \mathrm{u}}_{1}+\mathrm{m}_{2} \mathrm{u}}_{2}=\mathrm{m}_{1} \mathrm{v}}_{1}+\mathrm{m}_{2} \mathrm{v}}_{2} \\ \Rightarrow \quad \mathrm{v}_{1}=\left[\frac{\mathrm{m}_1-\mathrm{em}_2}{\mathrm{ m}_1+\mathrm{m}_2}\right] \mathrm{u}}_{1}+\left[\frac{(1+\mathrm{e}) \mathrm{m}_2}{\mathrm{ m}_1+\mathrm{m}_2}\right] \mathrm{u}_{2} \end{array}Substituting \mathrm{u}_1=2 \mathrm{ ms}^{-1}, \mathrm{u}_2=0, \mathrm{ m}_1=\mathrm{m} and \mathrm{m}}_{2}=2\mathrm{ m}, \mathrm{e}=0.5 we get \begin{array}{lc}\text{ we get }& \mathrm{v}_{1}=\left[\frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2 \mathrm{ m}}\right] \times2 \\ \Rightarrow & \mathrm{v}}_{1}=0 \end{array} Similarly, \begin{aligned}\mathrm{v}_{2} & =\left[\frac{(1+\mathrm{e}) \mathrm{m}_1}{\mathrm{ m}_1+\mathrm{m}_2}\right] \mathrm{u}}_{1}+\left[\frac{\mathrm{m}_2-\mathrm{em}_1}{\mathrm{ m}_1+\mathrm{m}_2}\right] \mathrm{u} }_{2} \\ & =\left[\frac{1.5 \times \mathrm{m}}{3 \mathrm{ m}}\right] \times2 \\ & =1 \mathrm{ ms}^{-1} \end{aligned}