Question

The position vector of four particles of masses $m_1=2\text{kg},m_2=4\text{kg},m_3=6\text{kg},m_4=8\text{kg}$ are $\overrightarrow{r_1}=2\hat{i}+3\hat{j}+0\hat{k},\overrightarrow{r_2}=0\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{r_3}=2\hat{i}+0\hat{j}+3\hat{k};\overrightarrow{r_4}=2\hat{i}+3\hat{j}+3\hat{k}$. The position vector of their centre of mass is given by

• A. $\frac{16\hat{i}+19\hat{j}+27\hat{k}}{11}$
• B. $1.6\hat{i}+1.9\hat{j}+2.7\hat{k}$
• C. $8\hat{i}+9\hat{j}+13\hat{k}$
• D. $\frac{32\hat{i}+38\hat{j}+54\hat{k}}{11}$

Answer 1

The correct option is B $1.6\hat{i}+1.9\hat{j}+2.7\hat{k}$

Given $m_1=2\text{kg},m_2=4\text{kg},m_3=6\text{kg},m_4=8\text{kg}$
$\overrightarrow{r_1}=2\hat{i}+3\hat{j}+0\hat{k};\overrightarrow{r_2}=0\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{r_3}=2\hat{i}+0\hat{j}+3\hat{k};\overrightarrow{r_4}=2\hat{i}+3\hat{j}+3\hat{k}$


Let $r_{cm}$ be the COM of the system of particles
Then, $\overrightarrow{r_{cm}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}+m_4\overrightarrow{r_4}}{m_1+m_2+m_3+m_4}$
$=\frac{2(2\hat{i}+3\hat{j}+0\hat{k})+4(0\hat{i}+2\hat{j}+3\hat{k})+6(2\hat{i}+0\hat{j}+3\hat{k})+8(2\hat{i}+3\hat{j}+3\hat{k})}{20}$
$=\frac{(4\hat{i}+6\hat{j})+(8\hat{j}+12\hat{k})+(12\hat{i}+18\hat{k})+(16\hat{i}+24\hat{j}+24\hat{k})}{20}$
$=\frac{32\hat{i}+38\hat{j}+54\hat{k}}{20}=1.6\hat{i}+1.9\hat{j}+2.7\hat{k}$