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A ball moving with velocity $2\, m/s$ collides head-on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities (in $m/s$) after collision will be
$0, 1$
$1, 1$
$1, 0.5$
$0, 2$
Solution
Here, $m_{1}=m, m_{2}=2 m$
$\mathrm{u}_{1}=2 \mathrm{m} / \mathrm{s}, \mathrm{u}_{2}=0$
Coefficient of restitution, $\mathrm{e}=0.5$
Let $v_{1}$ and $v_{2}$ be their respective velocities afte collision.
Applying the law of conservation of linea 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 $2=\left(v_{1}+2 v_{2}\right)$ $…(i)$
By definition of coefficient of restitution,
${\mathrm{e}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}}$
Or ${\mathrm{e}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)=\left(\mathrm{v}_{2}-\mathrm{v}_{1}\right)}$
$0.5(2-0)=\left(v_{2}-v_{1}\right)$ $…(ii)$
$1=v_{2}-v_{1}$
Solving equations $(i)$ and $(ii),$ we get,
$\mathrm{v}_{1}=0 \mathrm{m} / \mathrm{s}, \mathrm{v}_{2}=1 \mathrm{m} / \mathrm{s}$
