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

Answer: (A)

$\begin{array}{l}{\vec{a}}_A=-a\hat{j}\\ {\vec{a}}_B=aj\\ {\vec{a}}_C=ai\\ {\vec{a}}_D=-aj\\ {\vec{a}}_{cm}=\frac{m_a{\vec{a}}_a+m_b{\vec{a}}_b+m_c{\vec{a}}_c+m_d{\vec{a}}_d}{m_a+m_b+m_c+m_d}\\ {\vec{a}}_{cm}=\frac{-ma\hat{i}+2ma\vec{j}+3ma\hat{i}-4ma\hat{j}}{10m}\\ =\frac{3ma\hat{ȷ}-2ma\hat{j}}{10m}\\ =\frac{a_{\hat{A}}}{5}-\frac{a_{\hat{\hat{1}}}}{5}\\ =\frac{a}{5}(\hat{i}-\hat{j}).\end{array}$