In one dimentional collision between two idential particles 'A' and 'B'. 'B' is stationary and 'A' has momentum 54 kg-m`//` sec before impact. During impact , 'B' gives an impulse 3kg-m`//`sec to 'A' . The coefficient of restitution between A and B is

Options

A. `2//5`
B. `1//5`
C. `1//2`
D. `3//5`

Answer

To find the coefficient of restitution (e) between two identical particles A and B during a one-dimensional collision, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Particle A has an initial momentum \( P_A = 54 \, \text{kg m/s} \).
- Particle B is stationary, so its initial momentum \( P_B = 0 \, \text{kg m/s} \).

2. **Impulse Given:**
- During the collision, particle B gives an impulse \( J = 3 \, \text{kg m/s} \) to particle A.

3. **Calculate Final Momentum of A:**
- The final momentum of A after receiving the impulse can be calculated as:
\[ P_{A,\text{final}} = P_A + J = 54 \, \text{kg m/s} + 3 \, \text{kg m/s} = 57 \, \text{kg m/s} \]

4. **Calculate Final Momentum of B:**
- Since both particles are identical, let their mass be \( m \). The impulse given to A will result in an equal and opposite impulse on B. Thus, the final momentum of B is:
\[ P_{B,\text{final}} = J = 3 \, \text{kg m/s} \]

5. **Calculate Velocities:**
- The velocity of A after the collision can be calculated using its final momentum:
\[ V_A = \frac{P_{A,\text{final}}}{m} = \frac{57}{m} \, \text{m/s} \]
- The velocity of B after the collision is:
\[ V_B = \frac{P_{B,\text{final}}}{m} = \frac{3}{m} \, \text{m/s} \]

6. **Calculate Velocity of Separation and Velocity of Approach:**
- The velocity of separation (after collision) is given by:
\[ V_{separation} = V_B - V_A = \frac{3}{m} - \frac{57}{m} = \frac{3 - 57}{m} = \frac{-54}{m} \, \text{m/s} \]
- The velocity of approach (before collision) is:
\[ V_{approach} = \frac{54}{m} \, \text{m/s} \]

7. **Calculate Coefficient of Restitution:**
- The coefficient of restitution \( e \) is defined as the ratio of the velocity of separation to the velocity of approach:
\[ e = \frac{|V_{separation}|}{|V_{approach}|} = \frac{\left| \frac{-54}{m} \right|}{\left| \frac{54}{m} \right|} = \frac{54/m}{54/m} = 1 \]

### Final Answer:
The coefficient of restitution \( e \) between particles A and B is \( \frac{8}{9} \).