Question

Two particles of masses $1\text{kg}$ and $2\text{kg}$ have displacement vectors $(7\hat{i}+2\hat{j}+3\hat{k})\text{m}$ and $(2\hat{i}+10\hat{j}+6\hat{k})\text{m}$. Find the displacement vector of the center of mass of the system.

• A. $\frac{1}{3}(22\hat{i}+11\hat{j}+15\hat{k})\text{m}$
• B. $\frac{1}{3}(11\hat{i}+22\hat{j}+15\hat{k})\text{m}$
• C. $\frac{1}{3}(11\hat{i}+15\hat{j}+22\hat{k})\text{m}$
• D. $\frac{1}{3}(15\hat{i}+22\hat{j}+11\hat{k})\text{m}$

Answer 1

The correct option is B $\frac{1}{3}(11\hat{i}+22\hat{j}+15\hat{k})\text{m}$

Given, $m_1=1\text{kg}$, $m_2=2\text{kg}$
The displacement vector of mass $m_1$ is
$\overrightarrow{r_1}=(7\hat{i}+2\hat{j}+3\hat{k})\text{m}$
The displacement vector of mass $m_2$ is
$\overrightarrow{r_2}=(2\hat{i}+10\hat{j}+6\hat{k})\text{m}$

The displacement of $COM$ of the system
$\overrightarrow{r_{COM}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}$
$=\frac{1(7\hat{i}+2\hat{j}+3\hat{k})+2(2\hat{i}+10\hat{j}+6\hat{k})}{1+2}$
$\overrightarrow{r}=\frac{(11\hat{i}+22\hat{j}+15\hat{k})}{3}\text{m}$