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Consider a system of two particle having masses $m_1$ and $m_2$. If the particle of mas $m_1$ is pushed towards the mass centre of particle through a distance $d$, by what distance would the particle of mass $m_2$ move so as to keep the mass centre of particles at the original position?
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Verified Answer
The correct answer is:
$\frac{m_1}{m_2} d$
We know that
$$
\mathrm{CM}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}
$$
After changing a position of $m_1$ and to keep the position of C.M. same.
C.M. $=\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}$
$$
\begin{aligned}
0 & =\frac{m_1 d-m_2 d_2}{m_1+m_2} \\
\Rightarrow d_2 & =\frac{m_1}{m_2} d
\end{aligned}
$$
$$
\mathrm{CM}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}
$$
After changing a position of $m_1$ and to keep the position of C.M. same.
C.M. $=\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}$
$$
\begin{aligned}
0 & =\frac{m_1 d-m_2 d_2}{m_1+m_2} \\
\Rightarrow d_2 & =\frac{m_1}{m_2} d
\end{aligned}
$$
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