Question:
Four particles A, B, C and D with masses \(m_A = m\), \(m_B = 2m\), \(m_C = 3m\) and \(m_D = 4m\) are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :
Four particles A, B, C and D with masses \(m_A = m\), \(m_B = 2m\), \(m_C = 3m\) and \(m_D = 4m\) are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :
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Use \(\vec{a}_{cm} = \frac{\sum m_i\vec{a}_i}{\sum m_i}\) with the directions from the figure: A along \(-x\), B along \(+y\), C along \(+x\), D along \(-y\).
Updated On: Nov 5, 2026
- \(\frac{a}{5}(\hat{i} + \hat{j})\)
- \(\frac{a}{5}(\hat{i} - \hat{j})\)
- \(a(\hat{i} + \hat{j})\)
- Zero
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The Correct Option is B
Solution and Explanation
From the figure: A (mass \(m\)) accelerates along \(-\hat{i}\); B (\(2m\)) along \(+\hat{j}\); C (\(3m\)) along \(+\hat{i}\); D (\(4m\)) along \(-\hat{j}\).
\(\sum m_i\vec{a}_i = m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j}) = 2ma\hat{i} - 2ma\hat{j}\).
\(\sum m_i = 10m\).
\(\vec{a}_{cm} = \frac{2ma(\hat{i} - \hat{j})}{10m} = \frac{a}{5}(\hat{i} - \hat{j})\).
Answer: (2) \(\frac{a}{5}(\hat{i} - \hat{j})\)
\(\sum m_i\vec{a}_i = m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j}) = 2ma\hat{i} - 2ma\hat{j}\).
\(\sum m_i = 10m\).
\(\vec{a}_{cm} = \frac{2ma(\hat{i} - \hat{j})}{10m} = \frac{a}{5}(\hat{i} - \hat{j})\).
Answer: (2) \(\frac{a}{5}(\hat{i} - \hat{j})\)
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