Question

Consider a system of two particle having masses m1 and m2. If the partile of mass m1 is pushed towards the centre of mass of particle through a distance d, by what distance would be particle of mass m2 move so as to keep the centre of mass 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)

$\frac{m_{1}}{m_{2}}d$

Answer 2

Initial position of centre of mass

$r_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}$ ...(i)

If the particle of mass m1 is pushed towards the centre of mass of the system through distance d and to keep the centre of mass at the original position let second particle displaced through distance d' away from the centre of mass.

Now $r_{cm} = \frac{m_{1}(x_{1} + d) + m_{2}(x_{2} + d')}{m_{1} + m_{2}}$ ...(ii)

Equating (i) and (ii)

$\frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}} = \frac{m_{1}(x_{1} + d) + m_{2}(x_{2} + d')}{m_{1} + m_{2}}$

By solving$d' = - \frac{m_{1}}{m_{2}}d$

Negative sign shows that particle m2 should be displaced towards the centre of mass of the system.