Here, $\mathrm{m}_{1}=\mathrm{m}, \mathrm{m}_{2}=2 \mathrm{m}$

$\mathrm{u}_{1}=2 \mathrm{m} / \mathrm{s}, \mathrm{u}_{2}=0$

coefficient of restitution, $\mathrm{e}=0.5$

$ Let $$v_{1}$ and $v_{2}$ be their respective velocities after collision,

Applying the law of conservation of linear momentum, we get

${\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}}$

${\mathrm{m} \times 2+2 \mathrm{m} \times 0=\mathrm{m} \times \mathrm{v}_{1}+2 \mathrm{m} \times \mathrm{v}_{2}}$

or $2 \mathrm{m}=\mathrm{mv}_{1}+2 \mathrm{mv}_{2}$

or $2=\left(\mathrm{v}_{1}+2 \mathrm{v}_{2}\right)$         $…(i)$

By definition of coefficient of restitution,

$\mathrm{e}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}$

or $\mathrm{e}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)=\mathrm{v}_{2}-\mathrm{v}_{1}$

$0.5(2-0) =v_{2}-v_{1}$             $…(ii)$

$1 =v_{2}-v_{1}$

Solving equations $(i)$ and $(ii),$ we get

$\mathrm{v}_{1}=0 \mathrm{m} / \mathrm{s}, \mathrm{v}_{2}=1 \mathrm{m} / \mathrm{s}$