Question

A ball of mass $m$ 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 $\frac{3}{4}^{t h}$ of the original. What is the coefficient of restitution?

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

Answer: (B)

Since we know that in the case, when $m_{1}=m_{2} \, and \, v_{1}=0$
then, $v_{1}^{′}=\left(\frac{1 + e}{2}\right)v_{2} \, and \, v_{2}^{′}=\left(\frac{1 - e}{2}\right)v_{2}$
Given that, $K_{F}=\frac{3}{4}K_{i}$
Using energy conservation
$\Rightarrow \, \, \frac{1}{2}mv^{′}_{1}^{2}+\frac{1}{2}mv^{′}_{2}^{2}=\frac{3}{4}\left(\frac{1}{2} m v^{2}\right)$
Substituting the values, we get
$\left(\frac{1 + e}{2}\right)^{2}+\left(\frac{1 - e}{2}\right)^{2}=\frac{3}{4}$
$\Rightarrow \, \, \left(1 + e\right)^{2}+\left(1 - e\right)^{2}=3$
$\Rightarrow \, \, \, 2+2e^{2}=3\Rightarrow e^{2}=\frac{1}{2}\Rightarrow e=\frac{1}{\sqrt{2}}$