Question

Two billiard balls undergo a head-on collision. Ball $1$ is twice as heavy as ball $2$ . Initially, the ball $1$ moves with a speed $v$ towards the ball $2$ which is at rest. Immediately after the collision, ball $1$ travels at a speed of $v/3$ in the same direction. What type of collision has occurred?

• A. Inelastic
• B. Elastic
• C. Completely inelastic
• D. Cannot be determined from the information given

Answer 1

Answer: (B)

$\left(\text{v}\right)_{1} = \frac{\left(\left(\text{m}\right)_{1} - \left(\text{m}\right)_{2}\right) \left(\text{u}\right)_{1}}{\left(\text{m}\right)_{1} + \left(\text{m}\right)_{2}} + \frac{2 \left(\text{m}\right)_{2} \left(\text{u}\right)_{2}}{\left(\text{m}\right)_{1} + \left(\text{m}\right)_{2}}$
$\left(\text{v}\right)_{2} = \frac{\left(\left(\text{m}\right)_{2} - \left(\text{m}\right)_{1}\right) \left(\text{u}\right)_{2}}{\left(\text{m}\right)_{1} + \left(\text{m}\right)_{2}} + \frac{2 \left(\text{m}\right)_{1} \left(\text{u}\right)_{1}}{\left(\text{m}\right)_{1} + \left(\text{m}\right)_{2}}$

$\text{u}_{1}=\text{v} \\ \text{m}_{1}=2\text{m} \\ \text{v}_{1}=\frac{\text{v}}{3} \left|\right. \text{u}_{2}=0 \\ \text{m}_{2}=\text{m} \\ \text{v}_{2}=\text{?}$
From conservation of linear momentum
$2 \text{m v} = 2 \text{m} \times \frac{\text{v}}{3} + \text{m} \times \text{v}_{2}$
$\text{v}_{2} = 2 \text{v} - \frac{2 \text{v}}{3} \, \, ⇒ \, \, \text{v}_{2} = \frac{4 \text{v}}{3}$
$\text{e} = \frac{\text{velocity of seperation after collission}}{\text{velocity of apporach before collission}} ⇒ \text{e} = \frac{\frac{4 \text{v}}{3} - \frac{\text{v}}{3}}{\text{v}}$
e = 1