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

• A. $\frac{a}{5}(\hat{i}-\hat{j})$
• B. $\frac{a}{5}(\hat{i}+\hat{j})$
• C. Zero
• D. $a(\hat{i}+\hat{j})$

Answer 1

The correct option is A $\frac{a}{5}(\hat{i}-\hat{j})$


${\overrightarrow{a}}_A=-a\hat{i}$
${\overrightarrow{a}}_B=a\hat{j}$
${\overrightarrow{a}}_C=a\hat{i}$
${\overrightarrow{a}}_D=-a\hat{j}$
${\overrightarrow{a}}_{cm}=\frac{m_a{\overrightarrow{a}}_a+m_b{\overrightarrow{a}}_b+m_c{\overrightarrow{a}}_c+m_d{\overrightarrow{a}}_d}{m_a+m_b+m_c+m_d}$
${\overrightarrow{a}}_{cm}=\frac{-ma\hat{i}+2ma\overrightarrow{j}+3ma\hat{i}-4ma\hat{j}}{10m}$
$=\frac{2ma\hat{i}-2ma\hat{j}}{10m}$
$=\frac{a}{5}\hat{i}-\frac{a}{5}\hat{j}$
$=\frac{a}{5}(\hat{i}-\hat{j})$.