Question

Three bodies of mass $1\text{kg},2\text{kg}$ & $4\text{kg}$ have position vectors $3\hat{i},2\hat{i}-\hat{j},2\hat{i}+\hat{j}$ respectively. Then, centre of mass of this system has a position vector :

• A. $2\hat{i}+15\hat{j}$
• B. $15\hat{i}-2\hat{j}$
• C. $\left(\frac{2\hat{i}+15\hat{j}}{7}\right)$
• D. $\left(\frac{15\hat{i}+2\hat{j}}{7}\right)$

Answer 1

The correct option is D $\left(\frac{15\hat{i}+2\hat{j}}{7}\right)$

Masses $m_1,m_2,m_3$ with their respective position vectors can be written as:
$m_1=1\text{kg},\overrightarrow{r_1}=3\hat{i}$
$m_2=2\text{kg},\overrightarrow{r_2}=2\hat{i}-\hat{j}$
$m_3=4\text{kg},\overrightarrow{r_3}=2\hat{i}+\hat{j}$

Applying the formula for position vector of COM of system:
${\overrightarrow{r}}_{\text{COM}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}}{m_1+m_2+m_3}$

$=\frac{(1\times 3\hat{i})+2(2\hat{i}-\hat{j})+4(2\hat{i}+\hat{j})}{1+2+4}$

$\therefore {\overrightarrow{r}}_{\text{CM}}=\frac{15\hat{i}-2\hat{j}+4\hat{j}}{7}=\left(\frac{15\hat{i}+2\hat{j}}{7}\right)$