Question

A body of mass $4\text{kg}$ moving with a speed of $6\text{m/s}$ collides with another body of mass $8\text{kg}$ at rest. The lighter body comes to rest immediately after the collision. Find the coefficient of restitution $\left(e\right)$ for the collision.

• A. $0.25$
• B. $0.6$
• C. $0.3$
• D. $0.5$

Answer 1

The correct option is D $0.5$

Let $m_1=4\text{kg}$, $m_2=8\text{kg}$
$u_1=+6\text{m/s}{u}_2=0\text{m/s}$


After collision, $4\text{kg}$ mass stops, hence $v_1=0\text{m/s}$
Let $v_2$ be the velocity of body of mass $8\text{kg}$ after collision.


Applying the conservation of linear momentum for system of colliding bodies,
$P_i=P_f$
$\Rightarrow m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2...\left(i\right)$
$\Rightarrow (4\times 6)+(8\times 0)=(4\times 0)+(8\times v_2)$
$\therefore v_2=\frac{24}{8}=3\text{m/s}$

Coefficient of resititution is given by:
$e=\frac{\text{Speed of separation}}{\text{Speed of approach}}...\left(ii\right)$
Clearly, from the figures shown,
$\text{Speed of separation}=v_2,\text{Speed of approach}=u_1$

From Eq $\left(ii\right)$, $e=\frac{v_2}{u_1}$
$\therefore e=\frac{3}{6}=0.5$