Question

The position vector of three particles of masses $m_1=1\text{kg},m_2=2\text{kg}$ and $m_3=3\text{kg}$ are ${\overrightarrow{r}}_1=(\hat{i}+4\hat{j}+\hat{k})\text{m},{\overrightarrow{r}}_2=(\hat{i}+\hat{j}+\hat{k})\text{m}$ and ${\overrightarrow{r}}_3=(2\hat{i}-\hat{j}-2\hat{k})\text{m}$ respectively. The position vector of their centre of mass is

• A. $\frac{1}{2}(3\hat{i}+\hat{j}-\hat{k})\text{m}$
• B. $\frac{1}{3}(2\hat{i}+\hat{j}+\hat{k})\text{m}$
• C. $(3\hat{i}+\hat{j}-\hat{k})\text{m}$
• D. $\frac{1}{3}(3\hat{i}+\hat{j}-\hat{k})\text{m}$

Answer 1

The correct option is A $\frac{1}{2}(3\hat{i}+\hat{j}-\hat{k})\text{m}$

The position vector of COM of the three particles is given by

${\overrightarrow{r}}_{COM}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}}{m_1+m_2+m_3}$

Substituting the values, we get

${\overrightarrow{r}}_{COM}=\frac{\left(1\right)(\hat{i}+4\hat{j}+\hat{k})+\left(2\right)(\hat{i}+\hat{j}+\hat{k})+\left(3\right)(2\hat{i}-\hat{j}-2\hat{k})}{1+2+3}$

${\overrightarrow{r}}_{COM}=\frac{9\hat{i}+3\hat{j}-3\hat{k}}{6}$

${\overrightarrow{r}}_{COM}=\frac{1}{2}(3\hat{i}+\hat{j}-\hat{k})\text{m}$

Therefore, option $\left(A\right)$ is correct.

Why this question ? This question is aimed at making the students comfortable with the vector form of position of COM.