A particle of mass M is moving in a horizontal circle fo radius R with unifomr speed V. When it moves from one point to a diameterically opposite point, its

Options

A. kinetic energy change by `MV^(2)//4`
B. momentum does not change
C. momentum changes by 2 MV
D. kinetic energy changes by `MV^(2)`

Explanation

To solve the problem, we need to analyze the motion of a particle moving in a horizontal circle and determine the changes in kinetic energy and momentum as it moves from one point to the diametrically opposite point.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The particle has a mass \( M \).
- It moves in a circle of radius \( R \) with a uniform speed \( V \).
- The initial position of the particle can be considered at point A.

2. **Calculating Initial Kinetic Energy**:
- The kinetic energy (KE) of the particle at point A is given by the formula:
\[ KE_A = \frac{1}{2} M V^2 \]

3. **Calculating Final Kinetic Energy**:
- When the particle moves to the diametrically opposite point (point B), it continues to move with the same speed \( V \) (since the speed is uniform).
- The kinetic energy at point B is:
\[ KE_B = \frac{1}{2} M V^2 \]

4. **Change in Kinetic Energy**:
- The change in kinetic energy as the particle moves from point A to point B is:
\[ \Delta KE = KE_B - KE_A = \frac{1}{2} M V^2 - \frac{1}{2} M V^2 = 0 \]
- Thus, the change in kinetic energy is **zero**.

5. **Calculating Initial and Final Momentum**:
- The momentum of the particle at point A is given by:
\[ p_A = M V \]
- At point B, the direction of the velocity is opposite, so the momentum is:
\[ p_B = M (-V) = -M V \]

6. **Change in Momentum**:
- The change in momentum as the particle moves from point A to point B is:
\[ \Delta p = p_B - p_A = (-M V) - (M V) = -M V - M V = -2 M V \]
- The magnitude of the change in momentum is:
\[ |\Delta p| = 2 M V \]

### Conclusion:
- The change in kinetic energy is **0**.
- The change in momentum is **2MV**.

### Final Answer:
- The correct option is that the change in momentum is \( 2MV \).