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 ______ \( \, \text{cm} \).
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 ______ \( \, \text{cm} \).
Updated On: Jan 13, 2026
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Correct Answer: 3
Approach Solution - 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: \(3 \, \text{cm}\)
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Approach Solution -2
Step 1: Understand the problem
We have two particles of masses \( m_1 = 3 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \) placed at some distance apart. The system’s center of mass (C.M.) is initially at a certain position. The first mass \( m_1 \) moves toward the C.M. by 2 cm, and we must find how much \( m_2 \) should move (also toward the C.M.) so that the C.M. of the system remains unchanged.
Step 2: Use the condition for no change in center of mass
For the center of mass to remain fixed, the weighted displacement of each particle must cancel each other:
\[ m_1 x_1 = m_2 x_2 \] where \( x_1 \) and \( x_2 \) are the magnitudes of displacements of \( m_1 \) and \( m_2 \) toward the center of mass respectively.
Step 3: Substitute the known values
\[ 3 \times 2 = 2 \times x_2 \] \[ 6 = 2x_2 \] \[ x_2 = 3 \, \text{cm} \]
Step 4: Interpret the result
This means that to maintain the center of mass at its original position, the lighter particle must move by a greater distance toward the center of mass, balancing the motion of the heavier particle.
Final Answer:
We have two particles of masses \( m_1 = 3 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \) placed at some distance apart. The system’s center of mass (C.M.) is initially at a certain position. The first mass \( m_1 \) moves toward the C.M. by 2 cm, and we must find how much \( m_2 \) should move (also toward the C.M.) so that the C.M. of the system remains unchanged.
Step 2: Use the condition for no change in center of mass
For the center of mass to remain fixed, the weighted displacement of each particle must cancel each other:
\[ m_1 x_1 = m_2 x_2 \] where \( x_1 \) and \( x_2 \) are the magnitudes of displacements of \( m_1 \) and \( m_2 \) toward the center of mass respectively.
Step 3: Substitute the known values
\[ 3 \times 2 = 2 \times x_2 \] \[ 6 = 2x_2 \] \[ x_2 = 3 \, \text{cm} \]
Step 4: Interpret the result
This means that to maintain the center of mass at its original position, the lighter particle must move by a greater distance toward the center of mass, balancing the motion of the heavier particle.
Final Answer:
3
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