Question:
Two particles of masses $m_1$ and $m_2$ ($m_1 > m_2$) are separated by a distance 'd'. When the positions of the two particles are interchanged, the shift in the centre of mass is:
Two particles of masses $m_1$ and $m_2$ ($m_1 > m_2$) are separated by a distance 'd'. When the positions of the two particles are interchanged, the shift in the centre of mass is:
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The CM always shifts towards the heavier mass when they are swapped.
Updated On: Jun 10, 2026
- $\left(\frac{m_1 + m_2}{m_1 - m_2}\right)d$
- $\left(\frac{m_1 - m_2}{m_1 + m_2}\right)d$
- zero
- $\left(\frac{m_1}{m_1 - m_2}\right)d$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Centre of Mass ($CM$) calculation.
Step 2: Analysis
Let $m_1$ be at $0$ and $m_2$ be at $d$. $CM_1 = \frac{m_1(0) + m_2(d)}{m_1+m_2} = \frac{m_2 d}{m_1+m_2}$. After interchange, $m_2$ at $0$ and $m_1$ at $d$. $CM_2 = \frac{m_2(0) + m_1(d)}{m_1+m_2} = \frac{m_1 d}{m_1+m_2}$. Shift = $|CM_2 - CM_1| = \left| [cite_start]\frac{m_1-m_2}{m_1+m_2} \right| d$.
Step 3: Conclusion
The shift is $\left(\frac{m_1 - m_2}{m_1 + m_2}\right)d$.
Final Answer: (B)
Centre of Mass ($CM$) calculation.
Step 2: Analysis
Let $m_1$ be at $0$ and $m_2$ be at $d$. $CM_1 = \frac{m_1(0) + m_2(d)}{m_1+m_2} = \frac{m_2 d}{m_1+m_2}$. After interchange, $m_2$ at $0$ and $m_1$ at $d$. $CM_2 = \frac{m_2(0) + m_1(d)}{m_1+m_2} = \frac{m_1 d}{m_1+m_2}$. Shift = $|CM_2 - CM_1| = \left| [cite_start]\frac{m_1-m_2}{m_1+m_2} \right| d$.
Step 3: Conclusion
The shift is $\left(\frac{m_1 - m_2}{m_1 + m_2}\right)d$.
Final Answer: (B)
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