Question

Three particles of masses $50 \, g$ , $100 \, g$ and $150 \, g$ are placed at the vertices of an equilateral triangle of side $1 \, m$ (as shown in the figure). The $\left(x , \, y\right)$ coordinates of the centre of mass will be:

• A. $\left(\frac{7}{12} \, m , \, \frac{\sqrt{3}}{4} \, m\right)$
• B. $\left(\frac{7}{12} \, m , \, \frac{\sqrt{3}}{8} \, m\right)$
• C. $\left(\frac{\sqrt{3}}{4} \, m , \, \frac{5}{12} \, m\right)$
• D. $\left(\frac{\sqrt{3}}{8} \, m , \, \frac{7}{12} \, m\right)$

Answer 1

Answer: (A)

Coordinates of $A:\left(0 , \, 0\right)$
$B:\left(\frac{1}{2} \, m , \, \, \frac{\sqrt{3}}{2} \, m\right)$
$C:\left(\right.1 \, m, \, 0\left.\right)$
Coordinates of center of mass
$\left(x_{c m} , y_{c m}\right)=\left(\frac{50 \times 0 + 150 \times \frac{1}{2} + 100 \times 1}{50 + 150 + 100} \, m , \, \frac{50 \times 0 + 150 \times \frac{\sqrt{3}}{2} + 100 \times 0}{50 + 150 + 100} \, m\right)$
$=\left(\frac{7}{12} \, m , \, \frac{\sqrt{3}}{4} \, m\right)$