Consider a two particle system with particles | vedclass

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Question

Initially,

$0 = \frac{{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}}}{{{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}\,\,\,\,\,...\left( 1 \rig

Vedclass · Accepted Answer

$\;\frac{{m_1}}{{m_2}}$$d$

Vedclass · Answer

Initially,

$0 = \frac{{{m_1}\left( { - {x_1}} ight) + {m_2}{x_2}}}{{{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}\,\,\,\,\,...\left( 1 ight)$

Finally, 

The centre of mass is at the origin

$	herefore 0 = \frac{{{m_1}\left( {d - {x_1}} ight) + {m_2}\left( {{x_2} - d'} ight)}}{{{m_1} + {m_2}}}$

$\begin{array}{l}
 \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'\
 \Rightarrow d' = \frac{{{m_1}}}{{{m_2}}}d\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[Form\,\left( 1 ight).]
\end{array}$

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

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