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 3/4 of the original kinetic energy. Calculate the coefficient of restitution.

Answer 1

Mass of first ball $=m$

Speed of first ball $=v$

Mass of second ball $=m$

Let final velocities of the first and second ball are $v_1$ and $v_2$ respectively.

Refer above image.

final K.E $=\frac{3}{4}$ initial K.E

$\Rightarrow \frac{3}{4}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$

$=\frac{3}{4}{v}^2=v_1^2+v_2^2$...........(i)

Applying conservation of momentum

$p_i=p_f$

$mv=m{v}_2^{\prime }+m{v}_1^{\prime }$

$v=v_2^{\prime }+v_1^{\prime }$............(ii)

as coefficient of restitution $e=\frac{\text{velocity of separation}}{\text{velocity of approach}}$

$e=\frac{v_2^{\prime }-v_1^{\prime }}{v}$

$\Rightarrow ev=v_2^{\prime }-v_1^{\prime }$..........(iii)

(i) + (ii) $\Rightarrow (1+e)v=2v_2^1$

$v_2^{\prime }=\left(\frac{1+e}{2}\right)v$

Hence $v_1^{\prime }=\left(\frac{1-e}{2}\right)v$

putting the value of $v_1^1$ and $v_2^1$ in (1)

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

$\frac{3}{4}=\frac{(1-e{)}^2}{4}+\frac{(1+e{)}^2}{4}$

$2+2e^2=3$

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

Hence coefficient of restitution $=\frac{1}{\sqrt{2}}$