Question

A ball of mass $m$ moving at a speed $\nu$ makes a head-on collision with an identical ball at rest. the kinetic energy of the balls after collision is ${\frac{3}{4}}^{th}$ of the original. Find the coefficient of restitution.

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

Answer 1

Answer: (B)

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

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

let,

$m_1=m_2=m$ identical mass

$v=$ speed of first ball

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

Using law of conservation of momentum ,

$m(v_1+v_2)=mv$

$v_1+v_2=v$. . . . . . .(1)

and $v_1-v_2=v$

Given,

Final kinetic energy$=3/4$ Initial kinetic energy

$\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{3}{4}\times \frac{1}{2}m{v}^2$

$v_1^2+v_2^2=\frac{3}{4}{v}^2$

$\left[\right(v_1+v_2{)}^2+(v_1-v_2{)}^2]/2=3v^2/4$

$(1+e^2)v^2/2=3v^2/4$

$1+e^2=3/2$

$e^2=1/2$

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