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 in the figure. The acceleration of centre of mass of the particles (in ${\text{ms}}^{-2}$) is :

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

Answer 1

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

All $4$ particles are having equal magnitude of acceleration i.e $a$, hence representing the accelerations of the particles in vector form w.r.t the given axes:
${\overrightarrow{a}}_A=-a\hat{j}{\text{ms}}^{-2}$
${\overrightarrow{a}}_B=-a\hat{i}{\text{ms}}^{-2}$
${\overrightarrow{a}}_C=+a\hat{j}{\text{ms}}^{-2}$
${\overrightarrow{a}}_D=+a\hat{i}{\text{ms}}^{-2}$

Now, the acceleration of $\text{COM}$ of the system of particles is given by:
${\overrightarrow{a}}_{\text{CM}}=\frac{m_A{\overrightarrow{a}}_A+m_B{\overrightarrow{a}}_B+m_C{\overrightarrow{a}}_C+m_D{\overrightarrow{a}}_D}{m_A+m_B+m_C+m_D}$
$\Rightarrow {\overrightarrow{a}}_{\text{CM}}=\frac{\left(m\right)\times (-a\hat{j})+\left(2m\right)\times (-a\hat{i})+\left(3m\right)\times (+a\hat{j})+\left(4m\right)\times (+a\hat{i})}{m+2m+3m+4m}$
$\Rightarrow {\overrightarrow{a}}_{\text{CM}}=\frac{2ma(\hat{i}+\hat{j})}{10m}{\text{ms}}^{-2}$
$\therefore {\overrightarrow{a}}_{\text{CM}}=\frac{a}{5}(\hat{i}+\hat{j}){\text{ms}}^{-2}$