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The centre of mass of three particles of...

The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (2,2, 2). The position of the fourth mass of 4 kg to be placed in the system as that the new centre of mass is at (0, 0, 0) is

A

`(-3, -3, -3)`

B

`(-3, 3, -3)`

C

`(2, 3, -3)`

D

`(2, -2, 3)`

Text Solution

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The correct Answer is:
To find the position of the fourth mass (4 kg) such that the new center of mass of the system is at (0, 0, 0), we can follow these steps: ### Step 1: Understand the Given Data We have three masses: - \( m_1 = 1 \, \text{kg} \) - \( m_2 = 2 \, \text{kg} \) - \( m_3 = 3 \, \text{kg} \) The center of mass of these three particles is given as \( (2, 2, 2) \). ### Step 2: Calculate the Total Mass The total mass of the three particles is: \[ M = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \, \text{kg} \] ### Step 3: Use the Center of Mass Formula The formula for the center of mass \( (x_{cm}, y_{cm}, z_{cm}) \) of a system of particles is given by: \[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \] Similarly for \( y_{cm} \) and \( z_{cm} \). Given that the center of mass is at \( (2, 2, 2) \), we can write: \[ m_1 x_1 + m_2 x_2 + m_3 x_3 = 12 \quad \text{(for x-coordinate)} \] \[ m_1 y_1 + m_2 y_2 + m_3 y_3 = 12 \quad \text{(for y-coordinate)} \] \[ m_1 z_1 + m_2 z_2 + m_3 z_3 = 12 \quad \text{(for z-coordinate)} \] ### Step 4: Introduce the Fourth Mass Let the position of the fourth mass \( m_4 = 4 \, \text{kg} \) be \( (x_4, y_4, z_4) \). The new center of mass after adding the fourth mass should be at \( (0, 0, 0) \). Thus, we can write: \[ x_{cm}' = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4}{m_1 + m_2 + m_3 + m_4} = 0 \] \[ y_{cm}' = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{m_1 + m_2 + m_3 + m_4} = 0 \] \[ z_{cm}' = \frac{m_1 z_1 + m_2 z_2 + m_3 z_3 + m_4 z_4}{m_1 + m_2 + m_3 + m_4} = 0 \] ### Step 5: Set Up the Equations Substituting the known values into the equations: \[ \frac{12 + 4x_4}{10} = 0 \quad \Rightarrow \quad 12 + 4x_4 = 0 \quad \Rightarrow \quad 4x_4 = -12 \quad \Rightarrow \quad x_4 = -3 \] \[ \frac{12 + 4y_4}{10} = 0 \quad \Rightarrow \quad 12 + 4y_4 = 0 \quad \Rightarrow \quad 4y_4 = -12 \quad \Rightarrow \quad y_4 = -3 \] \[ \frac{12 + 4z_4}{10} = 0 \quad \Rightarrow \quad 12 + 4z_4 = 0 \quad \Rightarrow \quad 4z_4 = -12 \quad \Rightarrow \quad z_4 = -3 \] ### Step 6: Conclusion Thus, the position of the fourth mass should be: \[ (x_4, y_4, z_4) = (-3, -3, -3) \]
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