Question

Consider a two-particle system with the particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass by 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. ($\frac{m_1{m}_2}{m_1+m_2}$)d
• B. $\frac{m_2}{m_1}$d
• C. $\frac{m_1}{m_2}$d
• D. ($\frac{m_1-m_2}{m_1+m_2}$)d

Answer 1

The correct option is C

$\frac{m_1}{m_2}$d

Again, as we aleady know ,
$△$$X_{CM}$ = $\frac{m_1△X_1+m_2△X_2}{m_1+m_2}$
$△$$X_{CM}$ = 0
Assuming leftards as negative and rightwards as potive
$0=\frac{m_{1}d+m_2{d}^{\prime }}{m_1+m_2}$
d' = $\frac{-m_{1}d}{m_2}$
The negative sign indicates that $m_2$ should be shifted by
($\frac{m_1}{m_2}$) d leftwards in order to keep the COM stationary.