Question

Two point masses of mass 2 kg and 1 kg have
position vectors (î - 2) - k) and (-4 +5j - 2k)
respectively. The position vector of centre of mass
[COLPYH

Answer 1

The position vector of centre of mass is$(-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{4}{3}\hat{k})$.

Explanation:

1. Centre of mass of discrete system

  $\vec{R}=\frac{m_{1}\times \vec{r_{1}}+m_{2}\times \vec{r_{2}}}{m_{1}+m_{2}}$       ...1)

 Where

  $\vec{R}$ = position vector of centre of mass of discrete system

  $m_{1}$ = mass of first object = 2 kg

  $m_{2}$ = mass of second object = 1 kg

  $\vec{r_{1}}$ = position vector of first mass= $\hat{i}-2\hat{j}-\hat{k}$

   $\vec{r_{2}}$ = position vector of second mass= $-4\hat{i}+5\hat{j}-2\hat{k}$

2.  From equation 1)

  $\vec{R}= \frac{2(\hat{i}-2\hat{j}-\hat{k})+1(-4\hat{i}+5\hat{j}-2\hat{k})}{2+1}$

On Solving

$\vec{R}= \frac{-2\hat{i}+\hat{j}-4\hat{k}}{3}$

3.

 So position vector of centre of mass is

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

 

Answer 2

Answer:

The position vector of centre of mass is.

Explanation:

1. Centre of mass of discrete system

        ...1)

Where

  = position vector of centre of mass of discrete system

  = mass of first object = 2 kg

  = mass of second object = 1 kg

  = position vector of first mass=  

   = position vector of second mass=  

2.  From equation 1)

 

On Solving

3.

So position vector of centre of mass is

 =

Explanation: