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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
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The correct answer is:
0,1
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
$$
\begin{array}{r}
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\
\Rightarrow \quad \mathrm{v}_1=\left[\frac{\mathrm{m}_1-\mathrm{em}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{(1+\mathrm{e}) \mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2
\end{array}
$$
Substituting $\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}$ and
$$
\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5
$$
we get
$$
\begin{array}{lc}
\text { we get } & \mathrm{v}_1=\left[\frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2 \mathrm{~m}}\right] \times 2 \\
\Rightarrow & \mathrm{v}_1=0
\end{array}
$$
Similarly,
$$
\begin{aligned}
\mathrm{v}_2 & =\left[\frac{(1+\mathrm{e}) \mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{\mathrm{m}_2-\mathrm{em}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2 \\
& =\left[\frac{1.5 \times \mathrm{m}}{3 \mathrm{~m}}\right] \times 2 \\
& =1 \mathrm{~ms}^{-1}
\end{aligned}
$$
$$
\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}
$$
From the law of conservation of linear momentum
$$
\begin{array}{r}
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\
\Rightarrow \quad \mathrm{v}_1=\left[\frac{\mathrm{m}_1-\mathrm{em}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{(1+\mathrm{e}) \mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2
\end{array}
$$
Substituting $\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}$ and
$$
\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5
$$
we get
$$
\begin{array}{lc}
\text { we get } & \mathrm{v}_1=\left[\frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2 \mathrm{~m}}\right] \times 2 \\
\Rightarrow & \mathrm{v}_1=0
\end{array}
$$
Similarly,
$$
\begin{aligned}
\mathrm{v}_2 & =\left[\frac{(1+\mathrm{e}) \mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{\mathrm{m}_2-\mathrm{em}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2 \\
& =\left[\frac{1.5 \times \mathrm{m}}{3 \mathrm{~m}}\right] \times 2 \\
& =1 \mathrm{~ms}^{-1}
\end{aligned}
$$
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