Question

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

• A. $\frac{m_1}{m_1+m_2}d$
• B. $\frac{m_1}{m_2}d$
• C. $d$
• D. $\frac{m_2}{m_1}d$

Answer 1

Answer: (B)

$\frac{m_1}{m_2}d$

the 2 masses are m1and m2

let x and y be the distance of m1and m2 from the centre of mass respectively...

now,$m_{1}x=m_{2}y$

the mass m1 is moved by a distance d,

let the mass m2 be moved by a distance D

therefore, $m_1(x-d)=m_2(y-D)$

$m_{1}x-m_{1}d$=$m_{2}y-m_{2}D$

by eqn (i) $m_{1}x=m_{2}y$

=>$-m_{1}d=-m_{2}D$

= $D=\frac{m_{1}d}{m2}$