Two particles of masses 10 kg and 30 kg are lying on a straight line. The 10 kg mass is shifted towards the 30 kg mass by a distance of 2cm. By what distance should the 30kg mass be shifted so that the position of their centre of mass does not change

Explanation

mass of the first block, `m_(1)=10kg`
mass of the second block, `m_(2)=30kg`
Let `x_(1)` and `x_(1).` are positions of `m_(1)`
`x_(2)` and `x_(1).` are positions of `m_(2)`
initially and later respectively

In this case if `x_(cm)` is the position of centre of mass then `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`
then the new position of CM when blocks are shifted `x_(cm).=(m_(1)x_(1).+m_(2)x_(2))/(m_(1)+m_(2))`
subtracting the above equations
`x_(cm).-x_(cm)=(m_(1)(x_(1).-x_(1))+m_(2)(x_(2).-x_(2)))/(m_(1)+m_(2))`
`Deltax_(cm)=(m_(1)Deltax_(1)+m_(2)Deltax_(2))/(m_(1)+m_(2))`
`0=(10xx2+30Deltax_(2))/(40)` `:.Deltax_(2)=-(2)/(3)`
Therefore the second block should be moved left through a distance `(2)/(3)cm` to keep the position of centre of mass unchanged.