A wooden block of mass ' $m$ ' moves with velocity ' $v$ ' and collides with another block of mass ' $4…
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
$0.70$
$0.25$
$0.40$
$0.50$
Solution
$\begin{aligned}
& \mathrm{M}_1=\mathrm{m}, \mathrm{u}_1=\mathrm{v}, \mathrm{v}_1=0 \\
& \mathrm{M}_2=4 \mathrm{~m}, \mathrm{u}_2=0, \mathrm{v}_2=?
\end{aligned}$
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 $\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}$
$=\frac{\frac{\mathrm{v}}{4}-0}{\mathrm{v}-0}=\frac{1}{4}=0.25$