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

Options

A. Kinetic energy changes 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 of mass \( M \) moving in a horizontal circle of radius \( R \) with uniform speed \( V \). We want to determine the changes that occur when the particle moves from one point to a diametrically opposite point.

### Step-by-Step Solution:

1. **Understanding the Initial and Final Positions**:
- The particle starts at point A and moves to point B, which is diametrically opposite to point A in the circular path.

2. **Velocity at Both Points**:
- At point A, the velocity \( \vec{V}_A \) is directed tangentially along the circle.
- At point B, the velocity \( \vec{V}_B \) is also directed tangentially but in the opposite direction since it is diametrically opposite to point A.

3. **Kinetic Energy Calculation**:
- The kinetic energy \( KE \) of the particle at both points can be calculated using the formula:
\[ KE = \frac{1}{2} M V^2 \]
- Thus, the kinetic energy at point A is:
\[ KE_A = \frac{1}{2} M V^2 \]
- The kinetic energy at point B is:
\[ KE_B = \frac{1}{2} M V^2 \]
- Since both kinetic energies are equal, the change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_B - KE_A = 0 \]

4. **Momentum Calculation**:
- The momentum \( P \) of the particle at point A is given by:
\[ P_A = M \vec{V}_A \]
- Assuming the direction of \( \vec{V}_A \) is positive, we have:
\[ P_A = M V \]
- At point B, the velocity is in the opposite direction, so:
\[ P_B = M \vec{V}_B = M (-V) = -M V \]

5. **Change in Momentum**:
- The change in momentum \( \Delta P \) is calculated as:
\[ \Delta P = P_B - P_A = (-M V) - (M V) = -M V - M V = -2 M V \]
- However, since we are interested in the magnitude of the change in momentum, we take:
\[ |\Delta P| = 2 M V \]

6. **Final Result**:
- The change in momentum when the particle moves from one point to the diametrically opposite point is \( 2 M V \).

### Summary:
- The change in kinetic energy is \( 0 \).
- The change in momentum is \( 2 M V \).