Question

Two particles of equal mass have co-ordinates $(2,4,6)\text{m}$ and $(6,2,8)\text{m}$. One particle has a velocity ${\overrightarrow{v}}_1=\left(2\hat{i}\right)\text{m/s}$ and another particle has velocity ${\overrightarrow{v}}_2=\left(2\hat{j}\right)\text{m/s}$ at time $t=0$. The coordinates of the centre of mass of the system at time $t=1\text{s}$ will be

• A. $(4,4,7)$
• B. $(5,4,7)$
• C. $(2,4,6)$
• D. $(4,5,4)$

Answer 1

The correct option is B $(5,4,7)$

For particle 1 (mass $m_1=m$)
${\overrightarrow{r}}_1=(2\hat{i}+4\hat{j}+6\hat{k})$ w.r.t origin
${\overrightarrow{v}}_1=\left(2\hat{i}\right)\text{m/s}$
For particle 2 (mass $m_2=m$)
${\overrightarrow{r}}_2=(6\hat{i}+2\hat{j}+8\hat{k})$ w.r.t origin
${\overrightarrow{v}}_2=\left(2\hat{j}\right)\text{m/s}$

Let initial velocity of C.O.M of the system is $v_{\text{com}}$

${\overrightarrow{v}}_{\text{com}}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}=\frac{m\left(2\hat{i}\right)+m\left(2\hat{j}\right)}{m+m}=(\hat{i}+\hat{j})$

As both particles are moving with constant velocity, therefore acceleration of both particles are zero.
$\Rightarrow {\overrightarrow{a}}_{com}=0$
$\therefore \text{The velocity of C.O.M will remain constant.}$

Let initial position of C.O.M of the system is $r_{{\text{com}}_1}$ .

${\overrightarrow{r}}_{{\text{com}}_1}=\frac{m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2}{m_1+m_2}=\frac{m(2\hat{i}+4\hat{j}+6\hat{k})+m(6\hat{i}+2\hat{j}+8\hat{k})}{m+m}$

$=4\hat{i}+3\hat{j}+7\hat{k}$

After time $t=1\text{sec}$, let the position of C.O.M of the system is $r_{{\text{com}}_2}$. As $v_{com}$ is constant.
$\Rightarrow {\overrightarrow{r}}_{{\text{com}}_2}={\overrightarrow{r}}_{{\text{com}}_1}+{\overrightarrow{v}}_{\text{com}}\times t$
$=4\hat{i}+3\hat{j}+7\hat{k}+(\hat{i}+\hat{j})\times 1$
$=(5\hat{i}+4\hat{j}+7\hat{k})$ w.r.t. origin
$\therefore$ Coordinate of centre of mass at time $t$ $=(5,4,7)$