A particle of mass `m_(1)` moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after collision, velocity of the system

Options

A. increases
B. decreases
C. remains constant
D. becomes zero

Answer

To solve the problem of a particle of mass \( m_1 \) moving with velocity \( v \) colliding with a mass \( m_2 \) at rest, and then getting embedded, we will use the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Identify the Initial Momentum:**
- The initial momentum of the system before the collision can be calculated using the formula:
\[ P_{\text{initial}} = m_1 \cdot v + m_2 \cdot 0 \]
- Since mass \( m_2 \) is at rest, its contribution to the initial momentum is zero. Therefore:
\[ P_{\text{initial}} = m_1 \cdot v \]

2. **Identify the Final Momentum:**
- After the collision, the two masses \( m_1 \) and \( m_2 \) get embedded and move together with a common velocity \( v' \). The final momentum of the system is:
\[ P_{\text{final}} = (m_1 + m_2) \cdot v' \]

3. **Apply Conservation of Momentum:**
- According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
\[ P_{\text{initial}} = P_{\text{final}} \]
- Substituting the expressions for initial and final momentum:
\[ m_1 \cdot v = (m_1 + m_2) \cdot v' \]

4. **Solve for the Final Velocity \( v' \):**
- Rearranging the equation to solve for \( v' \):
\[ v' = \frac{m_1 \cdot v}{m_1 + m_2} \]

5. **Interpret the Result:**
- The final velocity \( v' \) is a fraction of the initial velocity \( v \). Since \( m_1 + m_2 \) is greater than \( m_1 \), it follows that:
\[ v' < v \]
- This indicates that the velocity of the system after the collision is less than the initial velocity of \( m_1 \).

### Final Answer:
The velocity of the system just after the collision is:
\[ v' = \frac{m_1 \cdot v}{m_1 + m_2} \]