Question

Ball 1 of mass $m$ moving at a speed $v$ makes a head-on collision with an identical ball 2 at rest. The kinetic energy of the balls after the collision is $\frac{3}{4}$th of original. If $e$ is the coefficient of restitution and $v_1,v_2$ are the velocities of the balls after collision, then:

• A. $v_1=\left(\frac{1-\frac{1}{\sqrt{2}}}{2}\right)v$
• B. $v_2=\left(\frac{1+\frac{1}{\sqrt{2}}}{2}\right)v$
• C. $e=\frac{1}{\sqrt{2}}$
• D. $e=\frac{1}{2}$

Answer 1

Answer: (A), (B), (C)

$e=\frac{1}{\sqrt{2}}$

For inelastic collision where ball 2 is at rest initially,
$v_1=\left(\frac{1-e}{2}\right)v$
and $v_2=\left(\frac{1+e}{2}\right)\times v$

Given, $K{E}_f=\frac{3}{4}\times K{E}_i$
$\Rightarrow \frac{3}{4}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}m{\left\{\left(\frac{1-e}{2}\right)v\right\}}^2+\frac{1}{2}m{\left\{\left(\frac{1+e}{2}\right)v\right\}}^2$

$\Rightarrow \frac{3}{4}=\frac{(1-e{)}^2}{4}+\frac{(1+e{)}^2}{4}$
$\therefore e=\frac{1}{\sqrt{2}}$
$\Rightarrow v_1=\left(\frac{1-\frac{1}{\sqrt{2}}}{2}\right)v$
and $v_2=\left(\frac{1+\frac{1}{\sqrt{2}}}{2}\right)v$

Hence, options (a), (b) and (c) are correct.