Question

The centre of mass of a system of particles (arranged in a straight line) is at the origin. From this we conclude that

• A. The number of particles on positive $x-$ axis is equal to the number of particles on negative $x-$ axis
• B. The total mass of the particles on positive $x-$ axis is same as the total mass on negative $x-$ axis
• C. The number of particles on $x-$ axis should be equal to the number of particles on $y-$ axis.
• D. If there is a particle on the positive $x-$ axis, there must be at least one particle on the negative $x-$ axis.

Answer 1

The correct option is D If there is a particle on the positive $x-$ axis, there must be at least one particle on the negative $x-$ axis.

$\text{C.O.M}$ lies at origin $\text{i.e}{x}_{\text{CM}}=0$.


$\Rightarrow 0=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$m_1{x}_1+m_2{x}_2=0$
$m_1{d}_1+m_2{d}_2=0$
$\therefore d_1=-\frac{m_2}{m_1}{d}_2$
$\Rightarrow$if mass $m_2$ lies on $+vex$ axis, then $d_2$ is $+ve$, hence $d_1$ will be $-ve$ which shows that $m_1$ will lie along $-vex-$axis.

$\therefore$ If there is a particle on the positive $x-$ axis, there must be at least one particle on the negative $x-$ axis.
Such that centre of mass will lie at origin.

Number of particles and mass of particles on either side of axis will be equal, only if each particle is identical and all the particles are distributed symetrically about the origin.

Hence, only option $\left(d\right)$ is the correct answer.