The distance between two particles of masses `m_(1)andm_(2)` is r. If the distance of these particles from the centre of mass of the system are `r_(1)andr_(2)` respectively, then show that `r_(1)=r((m_(2))/(m_(1)+m_(2)))andr_(2)=r((m_(1))/(m_(1)+m_(2)))`

Explanation

Position vector of c.m.
`vec(r)_(cm)=(m_(1)vec(r_(1))+m_(2)vec(r_(2)))/(m_(1)+m_(2))" "....(1)`
Taking c.m. at the origin `vec(r_(cm))=vec0`
`therefore 0=m_(1)vec(r_(1))+m_(2)vec(r_(2))[because" From eq. " (1)]`
Now c.m. is on the line joining `m_(1)andm_(2)` and the directions of `vec(r_(1))andvec(r_(2))` are mutually opposite.
`therefore 0=-m_(1)r_(1)+m_(2)r_(2)`
`therefore m_(1)r_(1)=m_(2)r_(2)" "......(2)`
`therefore m_(1)(r_(1))=m_(2)(r-r_(1)) [because r=r_(1)+r_(2)]`
`therefore m_(1)r_(1)=m_(2)r-m_(2)r_(1)`
`therefore (m_(1)+m_(2))r_(1)=m_(2)r`
`therefore r_(1)=r[(m_(2))/(m_(1)+m_(2))]`
Again taking `r_(1)=r-r_(2)" in "m_(1)r_(1)=m_(2)r_(2)`
`therefore m_(1)(r-r_(2))=m_(2)r_(2a)`
`therefore m_(1)r-m_(1)r_(2)=m_(2)r_(2)`
`therefore m_(1)r=(m_(1)+m_(2))r_(2)`
`therefore r_(2)=r[(m_(1))/(m_(1)+m_(2))]`