# Four particles A, B, C and D with masses $m_{A}=m, m_{B}=2m, m_{C}=3m\; and\; m_{D}=4m$ are 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 : Option 1) $\frac{a}{5}(\widehat{i}-\widehat{j})$   Option 2) $a(\widehat{i}+\widehat{j})$ Option 3) $Zero$   Option 4) $\frac{a}{5}(\widehat{i}+\widehat{j})$

$\vec{a}_{cm}=\frac{ \sum m\vec{a}}{ \sum m}=\frac{m(-a\hat{i}\: )+(2m)(a\hat{j}\: )+(3m)(a\hat{i}\: )+4m(-a\hat{j}\: )}{m+2m+3m+4m}$

$\vec{a}_{cm}=\frac{2(\; \hat{i}-\hat{j}\: )a}{10}=\frac{1}{5}(\: \hat{i}-\hat{j}\: )a$

$\vec{a}_{cm}=\frac{a}{5}(\: \hat{i}-\hat{j}\; )$

Option 1)

$\frac{a}{5}(\widehat{i}-\widehat{j})$

Option 2)

$a(\widehat{i}+\widehat{j})$

Option 3)

$Zero$

Option 4)

$\frac{a}{5}(\widehat{i}+\widehat{j})$

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