A particle of mass 2m moving with velocity v strikes a stationary particle of mass 3m and sticks to it . The speed of the system will be

Options

A. `0.8 v`
B. `0.2 v`
C. `0.6v`
D. `0.4v`

Answer

To solve the problem, we will use the principle of conservation of momentum. Here are the steps:

### Step-by-Step Solution:

1. **Identify the masses and their velocities**:
- Mass of the first particle, \( m_1 = 2m \) (moving with velocity \( v \)).
- Mass of the second particle, \( m_2 = 3m \) (stationary, so velocity \( u = 0 \)).

2. **Calculate the initial momentum**:
- The initial momentum of the system can be calculated using the formula:
\[ \text{Initial Momentum} = m_1 \cdot v + m_2 \cdot u \]
- Substituting the values:
\[ \text{Initial Momentum} = (2m) \cdot v + (3m) \cdot 0 = 2mv \]

3. **Determine the final mass after collision**:
- After the collision, the two masses stick together, so the total mass \( m_f \) is:
\[ m_f = m_1 + m_2 = 2m + 3m = 5m \]

4. **Let the final velocity of the combined mass be \( V_1 \)**:
- The final momentum of the system is given by:
\[ \text{Final Momentum} = m_f \cdot V_1 = (5m) \cdot V_1 \]

5. **Apply the conservation of momentum**:
- According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
- Thus, we have:
\[ 2mv = 5m \cdot V_1 \]

6. **Solve for \( V_1 \)**:
- Dividing both sides by \( 5m \):
\[ V_1 = \frac{2mv}{5m} = \frac{2v}{5} \]

7. **Final result**:
- Therefore, the speed of the system after the collision is:
\[ V_1 = 0.4v \]

### Conclusion:
The speed of the system after the collision is \( 0.4v \).

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