A ball moving with velocity 2 ${\mathrm{ms}}^{-1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ${\mathrm{ms}}^{-1}$) after collision will be

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

${\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2 = {\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2$

$\Rightarrow {\mathrm{v}}_1 = \left[\frac{{\mathrm{m}}_1-{\mathrm{em}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right]{\mathrm{u}}_1+\left[\frac{(1+\mathrm{e}){\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right]{\mathrm{u}}_2$

Substituting ${\mathrm{u}}_1 = 2 {\mathrm{ms}}^{-1}, {\mathrm{u}}_2 = 0, {\mathrm{m}}_1 = \mathrm{m} \mathrm{and} {\mathrm{m}}_2 = 2\mathrm{m}, \mathrm{e} = 0.5$

We get ${\mathrm{v}}_1 = \frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2\mathrm{m}}\times 2$

$\begin{array}{c}\Rightarrow {\mathrm{v}}_1=0\\ \text{ Similarly, }{\mathrm{v}}_2=\left[\frac{(1+\mathrm{e}){\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right]{\mathrm{u}}_1+\left[\frac{{\mathrm{m}}_2-{\mathrm{em}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right]{\mathrm{u}}_2\\ =\left[\frac{1.5\times \mathrm{m}}{3\mathrm{m}}\right]\times 2=1{\mathrm{ms}}^{-1}\end{array}$