Question

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 center of mass of the system is at a point:

• A. 0.6 cm right and 2.0 cm above 1kg mass
• B. 2.0 cm right and 0.9 cm above 1 kg mass.
• C. 0.9 cm right and 2.0 cm above 1kg mass
• D. 1.5 cm right and 1.2 cm above 1kg mass

Answer 1

Answer: (C)

0.9 cm right and 2.0 cm above 1kg mass

Answer 2

To find the center of mass of the system, we need to define a coordinate system and assign coordinates to each mass. Let's place the 1.0 kg mass at the origin (0,0).

1. Coordinates:

   • 1. 0 kg mass ($m_1$) is at (0, 0).
   • 2. 5 kg mass ($m_2$) is at (3.0 cm, 0).
   • 3. 5 kg mass ($m_3$) is at (0, 4.0 cm).

2. Total Mass: $M = m_1 + m_2 + m_3 = 1.0 \text{ kg} + 1.5 \text{ kg} + 2.5 \text{ kg} = 5.0 \text{ kg}$.

3. X-coordinate of the center of mass: $X_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M} = \frac{(1.0 \text{ kg})(0 \text{ cm}) + (1.5 \text{ kg})(3.0 \text{ cm}) + (2.5 \text{ kg})(0 \text{ cm})}{5.0 \text{ kg}} = 0.9 \text{ cm}$

4. Y-coordinate of the center of mass: $Y_{CM} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M} = \frac{(1.0 \text{ kg})(0 \text{ cm}) + (1.5 \text{ kg})(0 \text{ cm}) + (2.5 \text{ kg})(4.0 \text{ cm})}{5.0 \text{ kg}} = 2.0 \text{ cm}$

Therefore, the center of mass of the system is at (0.9 cm, 2.0 cm) relative to the 1 kg mass.