Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed `v` towards 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 occured?

Options

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

Answer

To determine the type of collision that occurred between the two billiard balls, we can analyze the problem step by step using the principles of conservation of momentum and the definition of elastic and inelastic collisions.

### Step-by-Step Solution:

1. **Identify the Masses and Initial Velocities:**
- Let the mass of ball 1 be \( m_1 = 2m \) (twice the mass of ball 2).
- Let the mass of ball 2 be \( m_2 = m \).
- The initial velocity of ball 1 is \( v \) and ball 2 is at rest, so its initial velocity \( u_2 = 0 \).

2. **Final Velocities After Collision:**
- After the collision, ball 1 moves with a velocity \( v_1' = \frac{v}{3} \).
- Let the final velocity of ball 2 be \( v_2' \).

3. **Apply Conservation of Momentum:**
- The total initial momentum before the collision is:
\[ p_{\text{initial}} = m_1 v + m_2 u_2 = (2m)v + (m)(0) = 2mv \]
- The total final momentum after the collision is:
\[ p_{\text{final}} = m_1 v_1' + m_2 v_2' = (2m) \left(\frac{v}{3}\right) + m v_2' = \frac{2mv}{3} + mv_2' \]
- Setting initial momentum equal to final momentum:
\[ 2mv = \frac{2mv}{3} + mv_2' \]

4. **Solve for \( v_2' \):**
- Rearranging the equation:
\[ 2mv - \frac{2mv}{3} = mv_2' \]
- Finding a common denominator:
\[ \frac{6mv}{3} - \frac{2mv}{3} = mv_2' \]
- Simplifying:
\[ \frac{4mv}{3} = mv_2' \]
- Dividing by \( m \):
\[ v_2' = \frac{4v}{3} \]

5. **Determine the Type of Collision:**
- The velocities after the collision are \( v_1' = \frac{v}{3} \) and \( v_2' = \frac{4v}{3} \).
- Calculate the relative velocity of separation and approach:
- Velocity of separation: \( v_2' - v_1' = \frac{4v}{3} - \frac{v}{3} = \frac{3v}{3} = v \)
- Velocity of approach: \( v - 0 = v \)
- The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} = \frac{v}{v} = 1 \]

6. **Conclusion:**
- Since the coefficient of restitution \( e = 1 \), this indicates that the collision is perfectly elastic.

### Final Answer:
The type of collision that occurred is **elastic collision**.