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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
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The correct answer is:
$0.25$
$$
\begin{aligned}
& \mathrm{M}_1=\mathrm{m}, \mathrm{u}_1=\mathrm{v}, \mathrm{v}_1=0 \\
& \mathrm{M}_2=4 \mathrm{~m}, \mathrm{u}_2=0, \mathrm{v}_2=?
\end{aligned}
$$
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 $\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}$
$$
=\frac{\frac{\mathrm{v}}{4}-0}{\mathrm{v}-0}=\frac{1}{4}=0.25
$$
\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
$$
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