Question

A $3\text{kg}$ ball moving in rightward direction on a horizontal surface with speed $4\text{m/s}$ collides with a $4\text{kg}$ ball moving ahead of it, in rightward direction with a speed of $2\text{m/s}$. If the coefficient of restitution is $0.6$, find the loss in $KE$ due to collision.

• A. $2.22\text{J}$
• B. $15.12\text{J}$
• C. $4.52\text{J}$
• D. $16.74\text{J}$

Answer 1

The correct option is A $2.22\text{J}$


Before collision, the balls are moving with velocities $u_1,u_2$ along the $+vex-$ direction.


After collision, let the balls be moving with velocities $v_1,v_2$ along the $+vex-$ direction as shown in the figure.

Applying linear momentum conservation along $x-$ direction for the system of balls:
$P_i=P_f$
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$3\times 4+4\times 2=3\times v_1+4\times v_2$
$\Rightarrow 3v_1+4v_2=20...\left(i\right)$
Now, $e=\frac{\text{speed of seperation}}{\text{speed of approach}}$
$\Rightarrow 0.6=\frac{v_2-v_1}{u_1-u_2}$
$\Rightarrow 0.6=\frac{v_2-v_1}{4-2}$
$\Rightarrow v_2-v_1=1.2...\left(ii\right)$
$3\times$ Eq. $\left(ii\right)+$ Eq. $\left(i\right)$ gives:
$7v_2=23.6$
$\therefore v_2=\frac{23.6}{7}=3.37\text{m/s}$
From Eq $\left(ii\right)$
$v_1=3.37-1.2=2.17\text{m/s}$

$\text{Loss in KE}=K{E}_i-K{E}_f$
$\because K{E}_i>K{E}_f$ in case of colliding system.
$\Rightarrow \text{Loss in KE}=(\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2)-(\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2)$
$=(\frac{1}{2}\times 3\times 4^2+\frac{1}{2}\times 4\times 2^2)-(\frac{1}{2}\times 3\times (2.17{)}^2+\frac{1}{2}\times 4\times \left(3.37{)}^2\right)$
$\therefore \text{Loss in KE}=2.22\text{J}$