Consider a system of two particle having masses $m_1$ and $m_2$. If the particle of mas $m_1$ is pushed…
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?
$\frac{m_1}{m_1+m_2} d$
$\frac{m_1}{m_2} d$
$d$
$\frac{m_2}{m_1} d$
Solution
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. $=\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}$
$\begin{aligned}
0 & =\frac{m_1 d-m_2 d_2}{m_1+m_2} \\
\Rightarrow d_2 & =\frac{m_1}{m_2} d
\end{aligned}$