Q103 / 200 · NEET 2004 · Physics

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?

  • A
    $\frac{m_1}{m_1+m_2} d$
  • Correct answer
    B
    $\frac{m_1}{m_2} d$
  • C
    $d$
  • 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. $=\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}$ $$ \begin{aligned} 0 & =\frac{m_1 d-m_2 d_2}{m_1+m_2} \\ \Rightarrow d_2 & =\frac{m_1}{m_2} d \end{aligned} $$

Hence, the correct answer is option: 'B'.

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