Question

Two equal masses are situated at a separation `r_(0)` One of them is imparted a velocity `v_(0)=sqrt((GM)/r_(0))` perpendicular to the line joining them both are free to move Treating motion only under mutual gravitational force find the ratio of maximum and minimum separation between them. [Hint Solve in CM frame]

Answer 1

Correct Answer - `3`
w.r.t COM K.E. `=1/2 (red mass) v_(rel)^(2)`
w.r.t. COM Angular momentum `=(mr)/2 v_(rel) `
:.Equating energy
`1/2 m/2 v_(0)^(2)-(Gm^(2))/r^(0)=1/2 m/2 v_(rel)^(2)-(Gm^(2))/r`
(Here `v_(rel)` is relative velocity `bot` to line as `v_(rel)` along the line joining is zero when separation is either min. or max.)
Angular momentum conservation
`(mr_(0))/2v_(0)=(mr)/2 v_(rel)`
:.`v_(rel) = (r_(0)v_(0))/r`
solving `3r^(2)-4rr_(0)+r_(0)^(2)=0`
:.`r_(max)=r_(0) r_(min)=r_(0)/3`
ratio=3 (ans)