Two particles start simultaneously from a point with equal velocities v each along two straight inclined at `60^(@)`. Find their relative velocity

Options

A. V
B. 2V
C. `(V)/(2)`
D. 3V

Answer

To find the relative velocity of two particles starting from the same point with equal velocities \( v \) and moving along two straight lines inclined at \( 60^\circ \), we can follow these steps:

### Step 1: Define the velocities of the particles
Let the first particle move along the vertical direction (y-axis) and the second particle move at an angle of \( 60^\circ \) with respect to the vertical.

- The velocity of the first particle, \( \vec{v_1} \), can be represented as:
\[ \vec{v_1} = v \hat{j} \]

- The velocity of the second particle, \( \vec{v_2} \), can be resolved into its components:
- The x-component (horizontal) is given by:
\[ v_{2x} = v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2} \]
- The y-component (vertical) is given by:
\[ v_{2y} = v \cos(60^\circ) = v \cdot \frac{1}{2} \]

Thus, the velocity of the second particle can be represented as:
\[ \vec{v_2} = v \cdot \frac{\sqrt{3}}{2} \hat{i} + v \cdot \frac{1}{2} \hat{j} \]

### Step 2: Calculate the relative velocity
The relative velocity of particle 1 with respect to particle 2, denoted as \( \vec{v_{12}} \), is given by:
\[ \vec{v_{12}} = \vec{v_1} - \vec{v_2} \]

Substituting the expressions for \( \vec{v_1} \) and \( \vec{v_2} \):
\[ \vec{v_{12}} = (0 \hat{i} + v \hat{j}) - \left(v \cdot \frac{\sqrt{3}}{2} \hat{i} + v \cdot \frac{1}{2} \hat{j}\right) \]
\[ \vec{v_{12}} = -v \cdot \frac{\sqrt{3}}{2} \hat{i} + \left(v - v \cdot \frac{1}{2}\right) \hat{j} \]
\[ \vec{v_{12}} = -v \cdot \frac{\sqrt{3}}{2} \hat{i} + v \cdot \frac{1}{2} \hat{j} \]

### Step 3: Find the magnitude of the relative velocity
To find the magnitude of the relative velocity \( |\vec{v_{12}}| \), we use the Pythagorean theorem:
\[ |\vec{v_{12}}| = \sqrt{\left(-v \cdot \frac{\sqrt{3}}{2}\right)^2 + \left(v \cdot \frac{1}{2}\right)^2} \]
\[ |\vec{v_{12}}| = \sqrt{v^2 \cdot \frac{3}{4} + v^2 \cdot \frac{1}{4}} \]
\[ |\vec{v_{12}}| = \sqrt{v^2 \cdot \left(\frac{3}{4} + \frac{1}{4}\right)} = \sqrt{v^2} = v \]

### Conclusion
The relative velocity of the two particles is \( v \).