Question

On a friction less surface, 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 $\frac{3}{4}th$ of the original. Find the coefficient of restitution?

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

Initial kinetic energy is $m{v}^2/2-\left(1\right)$

Let speed of first ball after collision be $v_1$

Let speed of first ball after collision be $v_2$

By conservation of momentum, $mv=m{v}_1+m{v}_2\Rightarrow v_1+v_2=v$

Also, $v_1-v_2=ev$

Solving above equations, $v_1=\left(\frac{1+e}{2}\right)v$ and $v_2=\left(\frac{1-e}{2}\right)v$

Final kinetic energy is $\left(m{v}^2/2\right)\times \left(\right(\frac{1+e}{2}{)}^2+\left(\frac{1-e}{2}{)}^2\right)-\left(2\right)$

As Final kinetic energy is $0.75$ times initial K.E.

Solving the expression $\left(1\right)and\left(2\right)$,

$e=1/\sqrt{2}$