A body of mass `m_(1)` is moving with a velocity V. it collides with another stationary body of mass `m_(2)`. They get embedded. At the point of collision, the velocity of the system.

Options

A. increases
B. decreases but does not become zero
C. remains same
D. become zero

Answer

To solve the problem of a body of mass \( m_1 \) colliding with a stationary body of mass \( m_2 \) and getting embedded, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- The mass of the first body is \( m_1 \) and it is moving with a velocity \( V \).
- The mass of the second body is \( m_2 \) and it is stationary, so its initial velocity is \( 0 \).

### Step 2: Write down the initial momentum
The initial momentum \( P_{\text{initial}} \) of the system can be calculated as:
\[ P_{\text{initial}} = m_1 \cdot V + m_2 \cdot 0 = m_1 \cdot V \]

### Step 3: Write down the final momentum
After the collision, the two bodies get embedded and move together with a common velocity \( U \). The final momentum \( P_{\text{final}} \) of the system is given by:
\[ P_{\text{final}} = (m_1 + m_2) \cdot U \]

### Step 4: Apply the conservation of momentum
According to the principle of conservation of momentum, the initial momentum must equal the final momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
Substituting the expressions we found:
\[ m_1 \cdot V = (m_1 + m_2) \cdot U \]

### Step 5: Solve for the common velocity \( U \)
Rearranging the equation to solve for \( U \):
\[ U = \frac{m_1 \cdot V}{m_1 + m_2} \]

### Step 6: Analyze the result
The expression \( U = \frac{m_1 \cdot V}{m_1 + m_2} \) shows that the velocity \( U \) is a fraction of \( V \). Since \( m_1 \) and \( m_2 \) are both positive, \( U \) will always be less than \( V \) (as long as \( m_2 \) is not zero). This indicates that the speed of the combined mass after the collision decreases but does not become zero.

### Conclusion
Thus, the final velocity of the system after the collision is:
\[ U = \frac{m_1}{m_1 + m_2} \cdot V \]
This shows that the velocity decreases but does not become zero.

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