Three particles of masses $1 kg$ , $\frac{3}{2} kg$ , 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

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$(\frac{5 a}{9}, \frac{2 a}{\sqrt{3}})$

$(\frac{5 a}{9}, \frac{2 a}{3 \sqrt{3}})$

$(\frac{2 a}{3 \sqrt{3}}, \frac{5 a}{9})$

$(\frac{2 a}{\sqrt{3}}, \frac{5 a}{9})$

answer is A.

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Detailed Solution

Let the masses $1 kg$ , $\frac{3}{2} kg$ and $2 kg$ are located at the vertices $A, B$ and $C$ as shown in figure. The coordinates of points $A, B$ and $C$ are $(0, 0), (a, 0)$ and $\frac{a}{2}$ , $\frac{\sqrt{3} a}{2}$ respectively. The coordinates of centre of mass are $X_{CM} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

$= \frac{[1 \times 0 + \frac{3}{2} \times a + 2 \times \frac{a}{2}]}{[1 + \frac{3}{2} + 2]} = \frac{5 a}{9} Y_{CM} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$

$= \frac{[1 \times 0 + \frac{3}{2} \times 0 + 2 \times \frac{\sqrt{3} a}{2}]}{[1 + \frac{3}{2} + 2]} = \frac{2 \sqrt{3} a}{9} = \frac{2 a}{3 \sqrt{3}}$