Three particles of masses `1 kg, 3/2 kg, and 2 kg` are located at the vertices of an equilateral triangle of side `a`. The `x, y` coordinates of the centre of mass are.

Options

A. `((5a)/(9), (2a)/(3sqrt(3)))`
B. `((2a)/(33), (5a)/(9))`
C. `((5a)/(9), (2a)/(sqrt3))`
D. `((2a)/(sqrt(3)), (5a)/(9))`

Explanation

To find the coordinates of the center of mass of the three particles located at the vertices of an equilateral triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Their Positions:**
- Let the masses be:
- \( m_1 = 1 \, \text{kg} \) at vertex \( A \) (0, 0)
- \( m_2 = \frac{3}{2} \, \text{kg} \) at vertex \( B \) (a, 0)
- \( m_3 = 2 \, \text{kg} \) at vertex \( C \) (a/2, \( \frac{\sqrt{3}}{2}a \))

2. **Calculate the Total Mass:**
\[ M = m_1 + m_2 + m_3 = 1 + \frac{3}{2} + 2 = \frac{7}{2} \, \text{kg} \]

3. **Calculate the x-coordinate of the Center of Mass:**
The formula for the x-coordinate of the center of mass is:
\[ x_{cm} = \frac{1}{M} \sum m_i x_i \]
Substituting the values:
\[ x_{cm} = \frac{1}{\frac{7}{2}} \left( 1 \cdot 0 + \frac{3}{2} \cdot a + 2 \cdot \frac{a}{2} \right) \]
Simplifying:
\[ x_{cm} = \frac{2}{7} \left( 0 + \frac{3a}{2} + a \right) = \frac{2}{7} \left( \frac{3a}{2} + \frac{2a}{2} \right) = \frac{2}{7} \left( \frac{5a}{2} \right) = \frac{5a}{7} \]

4. **Calculate the y-coordinate of the Center of Mass:**
The formula for the y-coordinate of the center of mass is:
\[ y_{cm} = \frac{1}{M} \sum m_i y_i \]
Substituting the values:
\[ y_{cm} = \frac{1}{\frac{7}{2}} \left( 1 \cdot 0 + \frac{3}{2} \cdot 0 + 2 \cdot \frac{\sqrt{3}}{2}a \right) \]
Simplifying:
\[ y_{cm} = \frac{2}{7} \left( 0 + 0 + \sqrt{3}a \right) = \frac{2\sqrt{3}a}{7} \]

5. **Final Result:**
The coordinates of the center of mass are:
\[ \left( \frac{5a}{7}, \frac{2\sqrt{3}a}{7} \right) \]