Two points mass `m` and `2m` are kept at a distance `a`. Find the speed of particles and their relative velocity of approach when separation becomes `a//2`.

Explanation

Reduced mass `mu=(m.2m)/(m+2m)=(2m)/(3)`
Changining in potential energy is
`U_(1)-U_(2)=-(Gm.2m)/(a)-(-(Gm.2m)/(a//2))`
`=(2Gm^(2))/(a)`
`(1)/(2)mu v_(1//2)^(2)=(2Gm^(2))/(a)implies(1)/(2)=sqrt((6Gm)/(a))`
`v_(1//2)=v_(1)+v_(2)=sqrt((6Gm)/(a))` ......(i)
`mv_(1)=2mv_(2)`
`v_(1)=2v_(2)`
`v_(1)+v_(2)=3v_(2)=sqrt((6Gm)/(a))implies v_(2)=(1)/(3)sqrt((6Gm)/(a))=sqrt((2Gm)/(3a))`
`v_(1)=2v_(2)=2aqrt((2Gm)/(3a))`