A wooden block of mass ' \(m\) ' moves with velocity ' \(v\) ' and collides with another block of mass ' \(4 \mathrm{~m}\) ', which is at rest. After collision the block of mass ' \(m\) ' comes to rest. The coefficient of restitution will be

\(\begin{aligned}& \mathrm{M}_1=\mathrm{m}, \mathrm{u}_1=\mathrm{v}, \mathrm{v}_1=0 \\& \mathrm{M}_2=4 \mathrm{~m}, \mathrm{u}_2=0, \mathrm{v}_2=?\end{aligned}\)
By law of conservation of momentum, we have
\(\begin{aligned}& \mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\& \therefore \mathrm{mv}+0=0+4 \mathrm{mv}_2 \\& \therefore \mathrm{v}=4 \mathrm{v}_2 \text { or } \mathrm{v}_2=\frac{\mathrm{v}}{4}\end{aligned}\)
Coefficient of restitution \(\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}\)
\(=\frac{\frac{\mathrm{v}}{4}-0}{\mathrm{v}-0}=\frac{1}{4}=0.25\)