Question

A block of mass $2\text{kg}$ moving with speed $5\text{m/s}$ collides with another block of mass $4\text{kg}$ at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution $\left(e\right)$. (Neglect all other external force)

• A. $0.6$
• B. $0.5$
• C. $0.4$
• D. $0.3$

Answer 1

The correct option is B $0.5$


Applying conservation of linear momentum before and after collision.
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$2\times 5+4\times 0=2\times 0+4\times v_2$
$10=4\times v_2$
$[v_2=\frac{5}{2}\text{m/s}]$
Coefficient of restitution $\left(e\right)=\frac{\text{Velocity of separation}}{\text{Velocity of approach}}$
$e=\frac{v_2-v_1}{u_1-u_2}=\frac{\frac{5}{2}-0}{5-0}=\frac{5/2}{5}$
$e=\frac{1}{2}$ or $e=0.5$