Thre identical masses are kept at the corners of an equilateral triangle ABC. A moves towards B with a velocity V, B moves towards C with velocity V, and C moves towards A with same velocity V. Then the velocity of centre of mass of the system of particles is

Options

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

Answer

To solve the problem, we need to determine the velocity of the center of mass of a system of three identical masses located at the corners of an equilateral triangle, each moving towards the next mass with the same velocity \( V \).

### Step-by-Step Solution:

1. **Identify the Masses and Their Positions**:
- Let the three identical masses be \( m \) located at points \( A \), \( B \), and \( C \) of the equilateral triangle.

2. **Define the Velocities**:
- Mass \( A \) moves towards mass \( B \) with velocity \( V \).
- Mass \( B \) moves towards mass \( C \) with velocity \( V \).
- Mass \( C \) moves towards mass \( A \) with velocity \( V \).

3. **Determine the Center of Mass**:
- The center of mass \( \vec{R}_{cm} \) of a system of particles is given by the formula:
\[ \vec{R}_{cm} = \frac{1}{M} \sum_{i} m_i \vec{r}_i \]
where \( M \) is the total mass and \( \vec{r}_i \) are the position vectors of the masses.

4. **Calculate the Total Mass**:
- The total mass \( M \) of the system is:
\[ M = m + m + m = 3m \]

5. **Express the Velocities as Vectors**:
- Let \( \vec{V}_A \), \( \vec{V}_B \), and \( \vec{V}_C \) be the velocity vectors of masses \( A \), \( B \), and \( C \) respectively. Each of these vectors has a magnitude of \( V \).

6. **Sum of the Velocity Vectors**:
- The velocity vectors can be expressed as:
\[ \vec{V}_A + \vec{V}_B + \vec{V}_C \]
- Since the masses are moving towards each other in a symmetrical manner, the sum of these vectors will yield a null vector:
\[ \vec{V}_A + \vec{V}_B + \vec{V}_C = 0 \]

7. **Calculate the Velocity of the Center of Mass**:
- The velocity of the center of mass \( \vec{V}_{cm} \) is given by:
\[ \vec{V}_{cm} = \frac{1}{M} \left( m \vec{V}_A + m \vec{V}_B + m \vec{V}_C \right) = \frac{1}{3m} \left( m \vec{V}_A + m \vec{V}_B + m \vec{V}_C \right) \]
- Since \( \vec{V}_A + \vec{V}_B + \vec{V}_C = 0 \), we have:
\[ \vec{V}_{cm} = \frac{1}{3m} (0) = 0 \]

### Conclusion:
The velocity of the center of mass of the system of particles is \( 0 \).