Question

In a system two particles of masses $m_1 = 3$ kg and $m_2 = 2$ kg are placed at 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 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 _______ cm.

Answer 1

3

Answer 2

For the center of mass to remain stationary, the condition $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$ must hold. Given $m_1 = 3$ kg, $m_2 = 2$ kg, and $\Delta x_1 = -2$ cm (since it moves towards the center of mass). Substituting these values: $(3 \text{ kg})(-2 \text{ cm}) + (2 \text{ kg})(\Delta x_2) = 0$ $-6 \text{ kg} \cdot \text{cm} + (2 \text{ kg})(\Delta x_2) = 0$ $(2 \text{ kg})(\Delta x_2) = 6 \text{ kg} \cdot \text{cm}$ $\Delta x_2 = \frac{6 \text{ kg} \cdot \text{cm}}{2 \text{ kg}} = 3 \text{ cm}$.