A uniform rod of length `L` lies on a smooth horizontal table. The rod has a mass `M`. A particle of mass `m` moving with speed `v` strikes the rod perpendicularly at one of the ends of the rod sticks to it after collision.
Find the velocity of the particle with respect to `C` before the collision

Options

A. `(Mv)/(M+m)`
B. `(mv)/(M+m)`
C. `(2mv)/(M+m)`
D. `(2Mv)/(M+m)`

Explanation

To solve the problem of finding the velocity of the particle with respect to the center of mass (C) before the collision, we can follow these steps:

### Step-by-step Solution:

1. **Identify the System**:
- We have a uniform rod of mass \( M \) and length \( L \) lying on a smooth horizontal table.
- A particle of mass \( m \) is moving with speed \( v \) and strikes the rod perpendicularly at one of its ends.

2. **Determine the Center of Mass (C)**:
- The center of mass of the rod is located at its midpoint, which is at a distance \( \frac{L}{2} \) from either end.

3. **Calculate the Initial Angular Momentum**:
- The initial angular momentum \( L_{\text{initial}} \) of the system about the center of mass (C) is due to the particle only, as the rod is initially at rest.
- The distance from the center of mass to the point of collision (the end of the rod) is \( \frac{L}{2} \).
- Therefore, the initial angular momentum due to the particle is given by:
\[ L_{\text{initial}} = m \cdot v \cdot \frac{L}{2} \]

4. **Calculate the Moment of Inertia**:
- The moment of inertia \( I \) of the rod about its center of mass is:
\[ I_{\text{rod}} = \frac{1}{12} M L^2 \]
- The moment of inertia of the particle about the center of mass (using the parallel axis theorem) is:
\[ I_{\text{particle}} = m \left(\frac{L}{2}\right)^2 = \frac{mL^2}{4} \]
- Thus, the total moment of inertia \( I_{\text{total}} \) of the system after the collision is:
\[ I_{\text{total}} = \frac{1}{12} M L^2 + \frac{mL^2}{4} \]

5. **Apply Conservation of Angular Momentum**:
- After the collision, the angular momentum must be conserved. Therefore:
\[ L_{\text{initial}} = L_{\text{final}} \]
- The final angular momentum can be expressed as:
\[ L_{\text{final}} = I_{\text{total}} \cdot \omega \]
- Setting the two expressions equal gives:
\[ m \cdot v \cdot \frac{L}{2} = \left(\frac{1}{12} M L^2 + \frac{mL^2}{4}\right) \cdot \omega \]

6. **Relate Angular Velocity to Linear Velocity**:
- The relationship between linear velocity \( v' \) of the center of mass and angular velocity \( \omega \) is given by:
\[ v' = \frac{L}{2} \cdot \omega \]
- Substitute \( \omega \) in terms of \( v' \):
\[ \omega = \frac{2v'}{L} \]

7. **Solve for the Velocity of the Particle**:
- Substitute \( \omega \) back into the angular momentum equation and solve for \( v \):
\[ m \cdot v \cdot \frac{L}{2} = \left(\frac{1}{12} M L^2 + \frac{mL^2}{4}\right) \cdot \frac{2v'}{L} \]
- Rearranging gives:
\[ v = \frac{L}{2} \cdot \frac{\left(\frac{1}{12} M + \frac{m}{4}\right) \cdot 2v'}{m} \]
- Simplifying leads to the final expression for the velocity of the particle before the collision.

### Final Result:
The velocity of the particle with respect to the center of mass before the collision can be expressed as:
\[ v = \frac{m + 3m}{3m} \cdot v' = \frac{4}{3} v' \]