A ball of mass `m` moving at a speed `v` collides with another ball of mass `3m` at rest. The lighter block comes to rest after collisoin. The coefficient of restitution is

Options

A. (a) `1/2`
B. (b) `2/3`
C. (c) `1/4`
D. (d) None of these

Explanation

To solve the problem, we will use the principles of conservation of momentum and the definition of the coefficient of restitution.

### Step-by-Step Solution:

1. **Identify the masses and initial velocities:**
- Mass of the first ball (m1) = m (moving with speed v)
- Mass of the second ball (m2) = 3m (at rest, so initial speed = 0)

2. **Write the equation for conservation of momentum:**
The total momentum before the collision must equal the total momentum after the collision.
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ mv + 3m \cdot 0 = m \cdot 0 + 3m \cdot v' \]
Simplifying this gives:
\[ mv = 3mv' \]

3. **Solve for the final velocity of the second ball (v'):**
\[ v' = \frac{mv}{3m} = \frac{v}{3} \]

4. **Identify the velocities after the collision:**
- The first ball comes to rest, so its final velocity = 0.
- The second ball moves with a final velocity \( v' = \frac{v}{3} \).

5. **Calculate the coefficient of restitution (e):**
The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
- Velocity of separation = final velocity of second ball - final velocity of first ball = \( \frac{v}{3} - 0 = \frac{v}{3} \)
- Velocity of approach = initial velocity of first ball - initial velocity of second ball = \( v - 0 = v \)

Therefore,
\[ e = \frac{\frac{v}{3}}{v} = \frac{1}{3} \]

### Final Answer:
The coefficient of restitution is \( \frac{1}{3} \).