Solve the following :

Question:

A ball of $m$ moving at a speed $v$ makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourths of the original. Find the coefficient of restitution.

Solution:

Use C.O.L.M

v=initial velocity

$v_{1}, v_{2} \rightarrow_{\text {final velocities }}$

$m v=m\left(v_{1}+v_{2}\right) \Longrightarrow v=v_{1}+v_{2}-(1)$

Given

$\frac{1}{2} m\left(v_{1}^{2}+v_{2}^{2}\right)=\frac{3}{4} \cdot \frac{1}{2} m v^{2}$

$v_{1}^{2}+v_{2}^{2}=\frac{3}{4} v^{2}$

Solve (1) and (2)

$v^{2}=\left(v_{1}+v_{2}\right)^{2}$

$v^{2}=v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2}$

$v_{1} v_{2}=\frac{v^{2}-\left(v_{1}^{2}+v_{2}^{2}\right)}{2}$

$=\frac{v^{2}}{8}$

 $\left(v_{1}-v_{2}\right)^{2}=v_{1}^{2}+v_{2}^{2}-2 v_{1} v_{2}=\frac{3}{4} v^{2}-\frac{v^{2}}{4}=\frac{2 v^{2}}{4}$

$v_{1}-v_{2}=\frac{v}{\sqrt{2}}$

$e=\frac{v_{1}-v_{2}}{v}=\frac{1}{\sqrt{2}}$

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