Question

Two identical balls of masses $5\text{kg}$ each are kept on a frictionless surface as shown in the figure below. One of the balls moving at $4\text{m/s}$ makes a head-on collision with the other at rest initially. It is observed that final kinetic energy of the balls is half of the initial kinetic energy. Find the coefficient of restitution.

• A. $e=0$
• B. $e=1$
• C. $e>1$
• D. $0<e<1$

Answer 1

The correct option is A $e=0$

Given, mass of each ball $m_1=5\text{kg};m_2=5\text{kg}$
speed of ball 1 before collision $u_1=4\text{m/s}$
speed of ball 2 before collision $u_2=0\text{m/s}$
Let $v_1$ and $v_2$ be the velocities after collision.
Applying linear momentum conservation principle :
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
or $v_1+v_2=4$ ......(i)
(since $m_1=m_2$)
Also $e=\frac{v_2-v_1}{u_1-u_2}==\frac{v_2-v_1}{4-0}$
or $v_2-v_1=4e$ .......(ii)

Given $K_f=\frac{1}{2}{K}_i$
$\Rightarrow \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}[\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2]$
$\Rightarrow 2(v_1^2+v_2^2)=(u_1^2+u_2^2)$
(since $m_1=m_2$)
$\Rightarrow (v_1+v_2{)}^2+(v_1-v_2{)}^2=4^2+0^2$
$\Rightarrow 16+16e^2=16$ (using (i) and (ii))
$\Rightarrow e=0$