A uniform rod is resting freely over a smooth horizontal plane. A particle moving horizontally strikes at one end of the rod normally and gets stuck. Then

Options

A. the momentum of the particle is shared between the particle and the rod and remains conserved.
B. the angular momentum about the mid-point of the rod before and after the collision is equal.
C. the angular momentum about the centre of mass of the combination before and after the collision is equal.
D. . the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle.

Explanation

To solve the problem step by step, we will analyze the situation involving the uniform rod and the particle that strikes it.

### Step 1: Understand the System
We have a uniform rod resting on a smooth horizontal plane and a particle moving horizontally strikes one end of the rod normally. After the collision, the particle gets stuck to the rod.

### Step 2: Analyze Momentum Conservation
Since there are no external forces acting on the system (the rod is on a smooth horizontal plane, implying no friction), the total linear momentum of the system (rod + particle) before the collision is equal to the total linear momentum after the collision.

**Initial Momentum**:
- The momentum of the particle before the collision is \( mv \) (where \( m \) is the mass of the particle and \( v \) is its velocity).
- The rod is initially at rest, so its momentum is \( 0 \).

Thus, the total initial momentum \( P_i = mv + 0 = mv \).

**Final Momentum**:
- After the collision, let the combined mass of the rod and particle be \( M + m \) (where \( M \) is the mass of the rod) and let their common velocity be \( V' \).
- Therefore, the total final momentum \( P_f = (M + m)V' \).

By conservation of momentum:
\[ P_i = P_f \implies mv = (M + m)V' \]

### Step 3: Analyze Angular Momentum Conservation
Next, we will consider angular momentum about different points.

1. **About the midpoint of the rod**:
- Before the collision, the angular momentum of the particle about the midpoint of the rod can be calculated as \( L_{initial} = mv \cdot \frac{L}{2} \) (where \( L \) is the length of the rod).
- After the collision, the angular momentum of the system about the midpoint must remain the same since there are no external torques acting on it.

2. **About the center of mass of the system**:
- The center of mass will shift after the collision, but the angular momentum about the center of mass will also remain conserved due to the absence of external torques.

### Step 4: Analyze the Motion of the Center of Mass
The center of mass of the rod-particle system will start to move translationally with the momentum equal to the original momentum of the particle.

### Conclusion
Based on the analysis:
- **Statement A**: True, momentum is conserved.
- **Statement B**: True, angular momentum about the midpoint is conserved.
- **Statement C**: True, angular momentum about the center of mass is conserved.
- **Statement D**: True, the center of mass moves with the original momentum of the particle.

Thus, all statements A, B, C, and D are true.