having masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ then according to law of conservation of momentum,
$\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$
or $\mathrm{m}_{1}\left(\mathrm{u}_{1}-\mathrm{v}_{2}\right)=\mathrm{m}_{2}\left(\mathrm{v}_{2}-\mathrm{u}_{2}\right)$ $...(i)$
According to law of conservation of $\mathrm{KE}$
$\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}$
Or $m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)$ $...(ii)$
Dividing eqn. $(ii)$ by eqn. $(i),$ we have,
$\mathrm{u}_{1}+\mathrm{v}_{1}=\mathrm{v}_{2}+\mathrm{u}_{2} \quad \text { or } \quad\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)=\left(\mathrm{v}_{2}-\mathrm{v}_{1}\right)$
or $\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}=1$
i.e, $\mathrm{e}=1$ (condition for collision to be perfectly elastic), i.e., there is no differnce between elastic and perfectly elastic collision by definition.
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