A particle of mass in is made to move with uniform speed `v_0` along the perimeter of a regular hexagon, inscribed in a circle of radius `R`. The magnitude of impulse applied at each corner of the hexagon is

Options

A. `2mv_(0)"sin" pi/6`
B. `mv_(0)"sin"pi/6`
C. `mv_(0)"sin"pi/3`
D. `2mv_(0)"sin"pi/3`

Explanation

To solve the problem of finding the magnitude of impulse applied at each corner of a regular hexagon when a particle of mass \( m \) moves with uniform speed \( v_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle moves along the perimeter of a regular hexagon inscribed in a circle of radius \( R \). The corners of the hexagon are the points where the direction of the particle's velocity changes.

2. **Identifying the Change in Momentum**:
Impulse is defined as the change in momentum. When the particle reaches a corner, its velocity changes direction, but its speed remains constant. The change in momentum can be expressed as:
\[ \Delta \vec{p} = m \vec{v}_{final} - m \vec{v}_{initial} \]

3. **Calculating the Initial and Final Velocities**:
At each corner of the hexagon, the particle's velocity changes direction by \( 60^\circ \) (since the internal angle of a regular hexagon is \( 120^\circ \), the external angle is \( 60^\circ \)). If we denote the initial velocity vector as \( \vec{v}_{initial} \) and the final velocity vector as \( \vec{v}_{final} \), we can express them in terms of their magnitudes and angles.

4. **Using the Law of Cosines**:
The magnitude of the change in momentum can be calculated using the law of cosines:
\[ |\Delta \vec{p}| = m \sqrt{|\vec{v}_{initial}|^2 + |\vec{v}_{final}|^2 - 2 |\vec{v}_{initial}| |\vec{v}_{final}| \cos(60^\circ)} \]
Since \( |\vec{v}_{initial}| = |\vec{v}_{final}| = v_0 \):
\[ |\Delta \vec{p}| = m \sqrt{v_0^2 + v_0^2 - 2 v_0^2 \cdot \frac{1}{2}} = m \sqrt{2v_0^2 - v_0^2} = m \sqrt{v_0^2} = mv_0 \]

5. **Final Expression for Impulse**:
The magnitude of impulse \( I \) applied at each corner is equal to the change in momentum:
\[ I = |\Delta \vec{p}| = mv_0 \]

### Conclusion:
The magnitude of impulse applied at each corner of the hexagon is \( mv_0 \).