Question 68 of 130

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:

A$\dfrac{a}{5}\left(\hat{i}+\hat{j}\right)$B$\dfrac{a}{5}\left(\hat{i}-\hat{j}\right)$C$a\left(\hat{i}+\hat{j}\right)$DZero

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Answer: (B) $\dfrac{a}{5}\left(\hat{i}-\hat{j}\right)$

${\vec{a}}_{A}=-a\hat{i}$
${\vec{a}}_{B}=a\hat{j}$
${\vec{a}}_{C}=a\hat{i}$
${\vec{a}}_{D}=-a\hat{j}$
${\vec{a}}_{cm}=\dfrac{{m}_{a}{\vec{a}}_{a}+{m}_{b}{\vec{a}}_{b}+{m}_{c}{\vec{a}}_{c}+{m}_{d}{\vec{a}}_{d}}{{m}_{a}+{m}_{b}+{m}_{c}+{m}_{d}}$
${\vec{a}}_{cm}=\dfrac{-ma\hat{i}+2ma\hat{j}+3ma\hat{i}-4ma\hat{j}}{10m}$
$=\dfrac{2ma\hat{i}-2ma\hat{j}}{10,}=\dfrac{a}{5}\hat{i}-\dfrac{a}{5}\hat{j}=\dfrac{a}{5}\left(\hat{i}-\hat{j}\right)$