Question

A $4\text{kg}$ ball moving to the right at $6\text{m/s}$ collides with a $5\text{kg}$ ball ahead of it and moving to the right at speed $2\text{m/s}$ . Find the magnitude of final velocity of the $4\text{kg}$ and $5\text{kg}$ balls respectively, if coefficient of restitution is $0.5$.

• A. $4.4\text{m/s},3.4\text{m/s}$
• B. $2.66\text{m/s},4.66\text{m/s}$
• C. $3.33\text{m/s},5.33\text{m/s}$
• D. $5\text{m/s},7\text{m/s}$

Answer 1

The correct option is B $2.66\text{m/s},4.66\text{m/s}$

Let $m_1=4\text{kg},m_2=5\text{kg}$


Applying conservation of momention before and after collision:


Applying PCLA on the system of two balls,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
or, $(4\times 6)+(5\times 2)=4v_1+5v_2$
$\therefore 4v_1+5v_2=34...\left(i\right)$

For coefficient of restitution $\left(e\right)$,
$e=\frac{\text{Speed of separation}}{\text{Speed of approach}}$
or $0.5=\frac{v_2-v_1}{6-2}$
$\Rightarrow v_2-v_1=0.5\times 4=2...\left(ii\right)$
Multiplying by $4$ in Eq. $\left(ii\right)$ and adding to Eq $\left(i\right)$,
$\Rightarrow 9v_2=34+8=42$
$\therefore v_2=\frac{42}{9}=4.66\text{m/s}$
Putting in Eq $\left(ii\right)$,
$v_1=v_2-2$
$\therefore v_1=2.66\text{m/s}$