Saved Bookmarks
| 1. |
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» 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) |
|