question_answer

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 :             [JEE Main 8-4-2019 Morning]

A) \[\frac{a}{5}\left( \hat{i}-\hat{j} \right)\]      

B) \[\frac{a}{5}\left( \hat{i}+\hat{j} \right)\]

C) Zero                

D) \[a\left( \hat{i}+\hat{j} \right)\]

Correct Answer: A

Solution :

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