Question

Considering a system having two masses $m_1$ and $m_2$ in which first mass is pushed towards centre of mass by a distance $a$. The distance the second mass must be moved to keep centre of mass at same position is :

• A. $\frac{m_1}{m_2}a$
• B. $\frac{m_1{m}_2}{a}$
• C. $\frac{m_2}{m_1}a$
• D. $\left(\frac{m_2{m}_1}{m_1+m_2}\right)a$

Answer 1

The correct option is A $\frac{m_1}{m_2}a$

If $x_1$ and $x_2$ are the positions of masses $m_1$ and $m_2$, the position of the centre of mass is given by $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

If $x_1$ changes by $\Delta x_1$ and $x_2$ changes by $\Delta x_2$, the change in $x_{cm}$ will be,

$\Delta x_{cm}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$ ............(1)

Given, $\Delta x_{cm}=0$ and $\Delta x_1=a$

Using these values in equation (1), we get $m_{1}a+m_2\Delta x_2=0$

or

$\Delta x_2=\frac{m_1}{m_2}a$

So, the distance moved by $m_2=\frac{m_1}{m_2}a$.