Question

A two particle system having masses, $M_{1}$ and $M_{2}$ are placed in a gravity-free space. If the first particle is shifted towards the centre of mass of the system by a distance $d$ , by how much distance should the second particle be shifted so that the centre of mass remains at its initial position?

• A. $\frac{M_{1}}{M_{2}}d$
• B. $\frac{M_{2}}{M_{1}}d$
• C. $\frac{M_{1}}{M_{1} + M_{2}}d$
• D. $d$

Answer 1

Answer: (A)

Let us suppose that first and second particle are pushed by the distances $\Delta x_{1} \& \Delta x_{2}$ respectively.
Then, the change in the position of the centre of mass, $\Delta x_{cm}=\frac{M_{1} \Delta x_{1} + M_{2} \Delta x_{2 }}{M_{1} + M_{2}}$
As given, $\Delta x_{cm}=0,$ we have $\frac{M_{1} \Delta x_{1} + M_{2} \Delta x_{2}}{M_{1} + M_{2}}=0$
or $M_{1}\Delta x_{1}+M_{2}\Delta x_{2}=0$
Here, $\Delta x_{1}=d$
$\therefore M_{1}\times d+M_{2}\Delta x_{2}=0$
$\Delta x_{2}=-\frac{M_{1}}{M_{2}}d$
The negative sign shows that mass, $\textit{M}_{2 }$ will move in opposite direction.