Question

Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)

Short answer

The center of mass is given by the weighted average position of the particles. We calculated the X- and Y-coordinates of the center of mass to be \((x_{cm}, y_{cm}) = (2, 2)\). Thus, the correct answer is \(\mathrm{B}(2,2)\).

Step-by-step solution

Calculate the X-coordinate of the center of mass

The X-coordinate of the center of mass, \(x_{cm}\), is given by the weighted average of the X-coordinates of the three particles: \[x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}\] Because the three particles have the same mass, we can write: \[x_{cm} = \frac{1}{3}(x_1 + x_2 + x_3)\] Substitute the X-coordinates of the particles: \[x_{cm} = \frac{1}{3}(1 + 2 + 3) = \frac{1}{3}(6) = 2\]

Calculate the Y-coordinate of the center of mass

Similarly, the Y-coordinate of the center of mass, \(y_{cm}\), is given by the weighted average of the Y-coordinates of the three particles: \[y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}\] Because the three particles have the same mass, we can write: \[y_{cm} = \frac{1}{3}(y_1 + y_2 + y_3)\] Substitute the Y-coordinates of the particles: \[y_{cm} = \frac{1}{3}(1 + 2 + 3) = \frac{1}{3}(6) = 2\]

Identify the correct answer

We found that the \((X, Y)\) coordinates of the center of mass are \((2, 2)\). Therefore, the correct answer is \(\mathrm{B}(2,2)\).