gravitation + centre of mass..gud 1

EQUATE th energies....

initial = PE at sep R0 + KE takin the given vel.

Now put v=0 and find the PE to get max sep.

The ratio is 1 . Ithe mass is rotate in a circle

oye tapanmast ! as a matter of fact, velocity can never be zero for both... !!!!

and the initial separation is NOT the minimum !
think of the centre of mass !!

is it ∞??

cuz ... V_com = v₀/2 (upwards)

hence the particles will dfinitely collide at sum time... hence min. sep = 0 ...

it is an easy problem ans=3

heres the method ..(in short)

at extremum distance total sum of both the velocites along the line joining them=0

so let the velocitis along the line be v and v in the same direction

let the velocities be u1 and u2 in the perpendicular direction

conserving angular momentum

u1+u2=v₀r₀/r
conserving momentum along x direction (no force on comand initial velocity only along y)
and conserving momentum along y direction=mv₀

we get

(u1²+u2²)2=v_{0[/ss²cos²Î¸+v0²r0²/r²}

and 2v=v₀sinθ

then we conserve energy and theta gets eliminated we get a quadratic in r

the roots are r₀ and r₀/3 (Dont forget the initial velocity value in terms of Gm etc.)

hence the answer

sry that is

v₀²cos²Î¸ +v₀²r₀²/r²