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

In this case of `.x_(cm).` is the position of centre 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))`
`Delta x_(cm)=(m_(1)Delta x_(1)+m_(2)Delta x_(2))/(m_(1)+m_(2))`
`0=(10xx2+30 Delta x_(2))/(40) " " therefore Delta x_(2)=-2//3`
Therefore the second block should be moved left through a distance of 2/3 cm to keep the position of centre of mass unchanged.