Which of the following statement is true?

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Question

If we consider the elastic collision of two bodies

having masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ then according to law of conservation of mo

Vedclass · Accepted Answer

By definition there is no difference between elastic and perfectly elastic collisions.

Vedclass · Answer

If we consider the elastic collision of two bodies

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 $\mathrm{m}_{1}\left(\mathrm{u}_{1}-\mathrm{v}_{2}ight)=\mathrm{m}_{2}\left(\mathrm{v}_{2}-\mathrm{u}_{2}ight)$     $...(i)$

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 $m_{1}\left(u_{1}^{2}-v_{1}^{2}ight)=m_{2}\left(v_{2}^{2}-u_{2}^{2}ight)$         $...(ii)$

Dividing eqn. $(ii)$ by eqn. $(i),$ we have,

$\mathrm{u}_{1}+\mathrm{v}_{1}=\mathrm{v}_{2}+\mathrm{u}_{2} \quad 	ext { or } \quad\left(\mathrm{u}_{1}-\mathrm{u}_{2}ight)=\left(\mathrm{v}_{2}-\mathrm{v}_{1}ight)$

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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