Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:

Let the $$1.0\text{ kg}$$ mass be at the origin $$(0,0)$$.

Given particle positions from the geometry:

$$(x_1, y_1) = (0, 0) \text{ for } m_1 = 1.0\text{ kg}$$

$$(x_2, y_2) = (3, 0) \text{ for } m_2 = 1.5\text{ kg}$$

$$(x_3, y_3) = (0, 4) \text{ for } m_3 = 2.5\text{ kg}$$

Total mass of the system: $$M = m_1 + m_2 + m_3 = 1.0 + 1.5 + 2.5 = 5.0\text{ kg}$$

X-coordinate of the center of mass: $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{1.0(0) + 1.5(3) + 2.5(0)}{5.0} = \frac{4.5}{5.0} = 0.9\text{ cm}$$

Y-coordinate of the center of mass: $$y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{1.0(0) + 1.5(0) + 2.5(4)}{5.0} = \frac{10}{5.0} = 2.0\text{ cm}$$