A particle is moving on a circular path of radius 10m with uniform speed of `4ms^(-1)`.The magnitude of change in velocity of particle when it completes a semi circular path is

Options

A. `4ms^(-1)`
B. `8ms^(-1)`
C. `10ms^(-1)`
D. zero

Answer

To solve the problem of finding the magnitude of change in velocity of a particle moving on a circular path when it completes a semi-circular path, we can follow these steps:

### Step 1: Understand the Initial and Final Velocities
- The particle is moving in a circular path of radius \( r = 10 \, \text{m} \) with a uniform speed of \( v = 4 \, \text{m/s} \).
- When the particle starts at point A (the top of the semicircle), its velocity is directed tangentially to the path. We can assume it is moving in the positive y-direction. Thus, the initial velocity \( \vec{v}_A = 4 \hat{j} \, \text{m/s} \).

### Step 2: Determine the Final Velocity
- After completing a semi-circular path, the particle reaches point B (the bottom of the semicircle). At this point, the velocity is still \( 4 \, \text{m/s} \) but directed tangentially in the negative y-direction. Thus, the final velocity \( \vec{v}_B = -4 \hat{j} \, \text{m/s} \).

### Step 3: Calculate the Change in Velocity
- The change in velocity \( \Delta \vec{v} \) can be calculated as:
\[ \Delta \vec{v} = \vec{v}_B - \vec{v}_A \]
Substituting the values:
\[ \Delta \vec{v} = (-4 \hat{j}) - (4 \hat{j}) = -4 \hat{j} - 4 \hat{j} = -8 \hat{j} \, \text{m/s} \]

### Step 4: Find the Magnitude of Change in Velocity
- The magnitude of the change in velocity is given by:
\[ |\Delta \vec{v}| = |\Delta v| = \sqrt{(-8)^2} = \sqrt{64} = 8 \, \text{m/s} \]

### Conclusion
- The magnitude of the change in velocity of the particle when it completes a semi-circular path is \( 8 \, \text{m/s} \).