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

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: $3 \, \text{cm}$