A particle of mass m moves along a circle of radius 'R'. The modulus of the average vector of force acting on the particle over the distance equal to a quarter of the circle is:

Options

A. zero if the particle moves with uniform speed v
B. `(sqrt(2)mv^(2))/(piR)` if the particle moves with uniform speed v
C. `(2sqrt(2)mv^(2))/(piR)` - if the particle moves with uniform speed v
D. ma if particle moves with constant tangential acceleration 'a', the initial velocity being equal to zero

Answer

To solve the problem of finding the modulus of the average vector of force acting on a particle moving along a quarter of a circle, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
- We have a particle of mass \( m \) moving along a circular path with radius \( R \).
- We need to find the average force acting on the particle over a distance equal to a quarter of the circle.

2. **Determine the Distance for a Quarter Circle**:
- The circumference of a full circle is given by \( 2\pi R \).
- Therefore, the distance for a quarter of the circle is:
\[ d = \frac{1}{4} \times 2\pi R = \frac{\pi R}{2} \]

3. **Calculate the Velocity**:
- The particle moves with a constant speed \( v \) along the circular path. The time taken to cover the quarter circle can be calculated as:
\[ \Delta t = \frac{\text{distance}}{\text{velocity}} = \frac{\frac{\pi R}{2}}{v} = \frac{\pi R}{2v} \]

4. **Change in Velocity**:
- As the particle moves in a circular path, its velocity vector changes direction. The change in velocity \( \Delta v \) over the quarter circle can be calculated using the geometry of the circle. The initial and final velocities form a right angle.
- The magnitude of the change in velocity can be expressed as:
\[ \Delta v = v \sqrt{2} \]

5. **Change in Momentum**:
- The change in momentum \( \Delta p \) can be expressed as:
\[ \Delta p = m \Delta v = m v \sqrt{2} \]

6. **Average Force**:
- The average force \( F \) can be calculated using the formula:
\[ F = \frac{\Delta p}{\Delta t} \]
- Substituting the values we have:
\[ F = \frac{m v \sqrt{2}}{\frac{\pi R}{2v}} = \frac{2 m v^2 \sqrt{2}}{\pi R} \]

7. **Final Result**:
- The modulus of the average vector of force acting on the particle over the distance equal to a quarter of the circle is:
\[ F = \frac{2 \sqrt{2} m v^2}{\pi R} \]