Question

Consider a two - particle system with the particle having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should be the second particle be moved to keep the centre of mass at the same position ?

• A. $\frac{m_1}{m_2}d$
• B. $\frac{m_2}{m_1}d$
• C. $\frac{m_1+m_2}{m_2}d$
• D. $\frac{m_1-m_2}{m_1+m_2}d$

Answer 1

The correct option is A $\frac{m_1}{m_2}d$

Consider the following diagram where we are having the two masses $m_1$ and $m_2$ and the centre of mass is at $\text{C}$.


Let $x_1$ be the distance of $m_1$ from point $\text{C}$ and $x_2$ be the distance of $m_2$ from $\text{C}$.

Now let $m_1$ be pushed towards point $\text{C}$ through a distance $d$ and let $m_2$ be pushed through a distance $d^{\prime }$ to keep the centre of mass at $\text{C}$.


In the first case we have centre mass equal because of equal masses and it is at the point $C$.

$m_1{x}_1=m_2{x}_2......\left(1\right)$

In the second case we have, we moved both the masses through distances $d$ and $d^{\prime }$,

$m_1(x_1-d)=m_2(x_2-d^{\prime })$

$m_1{x}_1-m_{1}d=m_2{x}_2-m_2{d}^{\prime }......\left(2\right)$

Subtracting eq.$\left(2\right)$ from eq.$\left(1\right)$ we get,

$m_1{x}_1-m_1{x}_1+m_{1}d=m_2{x}_2-m_2{x}_2+m_2{d}^{\prime }$

$\Rightarrow m_{1}d=m_2{d}^{\prime }$

$\Rightarrow d^{\prime }=\frac{m_1}{m_2}d$

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