| 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) (a)\[\frac{a}{5}\left( \hat{i}-\hat{j} \right)\]
B) (b)\[\frac{a}{5}\left( \hat{i}+\hat{j} \right)\]
C) Zero
D) (d)\[a\left( \hat{i}+\hat{j} \right)\]
Correct Answer: A
Solution :
| [a] |
 |
| \[{{\vec{a}}_{A}}=-a\hat{i}\] |
| \[{{\vec{a}}_{B}}=a\hat{j}\] |
| \[{{\vec{a}}_{C}}=a\hat{i}\] |
| \[{{\vec{a}}_{D}}=-a\hat{j}\] |
| \[{{\vec{a}}_{cm}}=\frac{{{m}_{a}}{{{\vec{a}}}_{a}}+{{m}_{b}}{{{\vec{a}}}_{b}}+{{m}_{c}}{{{\vec{a}}}_{c}}+{{m}_{d}}{{{\vec{a}}}_{d}}}{{{m}_{a}}+{{m}_{b}}+{{m}_{c}}+{{m}_{d}}}\] |
| \[{{\vec{a}}_{cm}}=\frac{-ma\hat{i}+2ma\hat{j}+3ma\hat{i}-4ma\hat{j}}{10m}\] |
| \[=\frac{2ma\hat{i}-2ma\hat{j}}{10m}\] |
| \[=\frac{a}{5}\hat{i}-\frac{a}{5}\hat{j}\] |
| \[=\frac{a}{5}\left( \hat{i}-\hat{j} \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
