Question

In a system, two particles of masses \( m_1 = 3 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \) are placed at a certain distance from each other. The particle of mass \( m_1 \) is moved towards the center of mass of the system through a distance \( 2 \, \text{cm} \). In order to keep the center of mass of the system at the original position, the particle of mass \( m_2 \) should move towards the center of mass by the distance ______ \( \, \text{cm} \).

Answer 1

Correct Answer: 3

1. Define Movement of Center of Mass:
To maintain the center of mass position, the total movement of the center of mass (\(\Delta X_\text{C.O.M.}\)) must be zero.
2. Apply Center of Mass Condition:
Let the movement of \(m_2\) be \(x\) cm towards the center of mass. Then:
\[ \Delta X_\text{C.O.M.} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}, \] where \(\Delta x_1 = 2 \, \text{cm}\) (movement of \(m_1\)) and \(\Delta x_2 = -x \, \text{cm}\) (movement of \(m_2\)).
3. Set \(\Delta X_\text{C.O.M.}\) to Zero:
\[ 0 = \frac{3 \times 2 + 2 \times (-x)}{3 + 2}. \] Simplifying,
\[ 6 - 2x = 0. \] \[ x = 3 \, \text{cm}. \]

Answer: \(3 \, \text{cm}\)