Question

Three identical particles each of mass $1\text{kg}$ are placed with their centres on a straight line. Their centres are marked at $A,B$ and$C$ respectively. All the centres $A,B$ and $C$ are placed at equal distance of $2\text{m}$. Find the distance of centre of mass of system from $A$

• A. $1\text{m}$
• B. $0\text{m}$
• C. $2\text{m}$
• D. $1.5\text{m}$

Answer 1

The correct option is C $2\text{m}$

Let the centre of $A$ placed at origin

Also, particles are at $x-$axis only, so $y_{COM}$ will be zero.
Here $m_A=1\text{kg}$
$m_B=1\text{kg}$
$m_C=1\text{kg}$
$x-$ coordinate of COM of system is given by
$x_{com}=\frac{m_A{x}_B+m_B{x}_B+m_C{x}_C}{m_A+m_B+m_C}$
$x_{com}=\frac{\left(1\right)\left(0\right)+1\left(2\right)+1\left(4\right)}{1+1+1}=\frac{2+4}{3}$
$x_{com}=2\text{m}$

Distance between point $A$ & centre of mass is
$x_{com}-x_A=2-0=2\text{m}$


Alternate:
As we can see that the system of particles are symmetric about the position of particle $B$
So, from $A$ it is $2\text{m}$ away
Hence, COM of the system will be $2\text{m}$ away from $A$