A particle is moving on a circular path with constant speed v then the change in its velocity after it has desceibed an angle of `60^(@)` will be

Options

A. `vsqrt2`
B. 0
C. v
D. 2v

Explanation

To solve the problem of finding the change in velocity of a particle moving in a circular path after it has described an angle of 60 degrees, we can follow these steps:

### Step 1: Understand the Initial and Final Velocities
- The particle is moving with a constant speed \( v \) in a circular path.
- Initially, let’s denote the initial velocity vector as \( \vec{u} \). At the starting point, this velocity is directed tangentially to the circle.

### Step 2: Determine the Initial Velocity Vector
- Assume the particle starts at point A on the circle. The initial velocity vector \( \vec{u} \) can be represented as:
\[ \vec{u} = v \hat{i} \]
where \( \hat{i} \) is the unit vector in the horizontal direction.

### Step 3: Find the Final Velocity Vector after 60 Degrees
- After the particle has moved 60 degrees along the circular path, it will reach point B. The final velocity vector \( \vec{v} \) will also have a magnitude \( v \) but will be directed tangentially at point B.
- The angle between the initial and final positions is 60 degrees. Thus, the final velocity vector can be broken down into components:
\[ \vec{v} = v \cos(30^\circ) \hat{i} + v \sin(30^\circ) \hat{j} \]

### Step 4: Calculate the Components of the Final Velocity
- Using trigonometric values:
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)
- Therefore, the final velocity vector becomes:
\[ \vec{v} = v \left(\frac{\sqrt{3}}{2}\right) \hat{i} + v \left(\frac{1}{2}\right) \hat{j} = \frac{\sqrt{3}v}{2} \hat{i} + \frac{v}{2} \hat{j} \]

### Step 5: Calculate the Change in Velocity
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v} - \vec{u} \]
- Substituting the vectors:
\[ \Delta \vec{v} = \left(\frac{\sqrt{3}v}{2} \hat{i} + \frac{v}{2} \hat{j}\right) - (v \hat{i}) \]
\[ = \left(\frac{\sqrt{3}v}{2} - v\right) \hat{i} + \frac{v}{2} \hat{j} \]
\[ = \left(\frac{\sqrt{3}v}{2} - \frac{2v}{2}\right) \hat{i} + \frac{v}{2} \hat{j} \]
\[ = \left(\frac{\sqrt{3}v - 2v}{2}\right) \hat{i} + \frac{v}{2} \hat{j} \]
\[ = \left(\frac{(\sqrt{3} - 2)v}{2}\right) \hat{i} + \frac{v}{2} \hat{j} \]

### Step 6: Calculate the Magnitude of the Change in Velocity
- The magnitude of the change in velocity \( |\Delta \vec{v}| \) is given by:
\[ |\Delta \vec{v}| = \sqrt{\left(\frac{(\sqrt{3} - 2)v}{2}\right)^2 + \left(\frac{v}{2}\right)^2} \]
\[ = \sqrt{\frac{(\sqrt{3} - 2)^2 v^2}{4} + \frac{v^2}{4}} \]
\[ = \sqrt{\frac{((\sqrt{3} - 2)^2 + 1)v^2}{4}} \]
\[ = \frac{v}{2} \sqrt{(\sqrt{3} - 2)^2 + 1} \]

### Step 7: Simplify the Expression
- Calculate \( (\sqrt{3} - 2)^2 + 1 \):
\[ = 3 - 4\sqrt{3} + 4 + 1 = 8 - 4\sqrt{3} \]
- Thus, the magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = \frac{v}{2} \sqrt{8 - 4\sqrt{3}} \]

### Final Answer
The change in velocity after the particle has described an angle of 60 degrees is:
\[ |\Delta \vec{v}| = v \]