Question

Consider a two particle system with particles 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 the second particle be moved so as to keep the centre of mass at the same position?

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

Answer 1

Answer: (A)

To keep the centre of mass at the same position,velocity of centre of mass is zero, so
$\frac{m_1v_1 + m_2v_2}{m_1 + m_2} = 0$
(where, $v_1$ and $v_2$ are the velocities of particles $1$ and $2$ respectively)
$\Rightarrow m_1 \frac{dr_1}{dt} + m_2 \frac{dr_2}{dt} = 0$
$[\because v_1 = \frac{dr_1}{dt} $ and $v_2 = \frac{dr_2}{dt} $ and $ m_1 m_2 \ne 0]$
$\Rightarrow m_1dr_1 + m_2 dr_2 = 0$
($dr_1$ and $dr_2$ represent the small changes in displacement, so that $dr_2 \to 0$ and $dr_2 \to 0$ of particles)
Let $2nd$ particle has been displaced by distance $X$, then
$m_1(d) + m_2(X) = 0$
$X = -\frac{m_1d}{m_2}$
Negative sign shows that both the particle have to move in opposite directions
So, $\frac{m_1}{m_2} d$ is the distance moved by $2nd$ particle to keep position of centre of mass unchanged.