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 final speed of $3\text{kg}$ and $4\text{kg}$ balls respectively in $\text{m/s}$.

• A. $6.67\text{m/s}$ and $8.27\text{m/s}$
• B. $1.17\text{m/s}$ and $2.27\text{m/s}$
• C. $4\text{m/s}$ and $5\text{m/s}$
• D. $2.17\text{m/s}$ and $3.37\text{m/s}$

Answer 1

The correct option is D $2.17\text{m/s}$ and $3.37\text{m/s}$


Before collision, the balls are moving with the 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 $\left(ii\right)$
$v_1=3.37-1.2=2.17\text{m/s}$
Hence $2.17\text{m/s}$ and $3.37\text{m/s}$ are the final speeds of $3\text{kg}$ and $4\text{kg}$ balls respectively.