Given a thin homogenous disc of radius a and mass `m_(1)`. A particle of mass `m_(2)` is placed at a distance `l` from the disc on it's axis of symmetry. Initially both are motionless in free space but they ultimately collide because of gravitational attraction. find the relative velocity at the time of collision. assume `altlt1`.

Options

A. `[2G(m_(1)+m_(2))(2/a-1/l)]`
B. `[2G(m_(1)+m_(2))(2/a-1/l)]^(1/2)`
C. `[2G(m_(1)+m_(2))(2/a-1/l)^(2)]`
D. `[2G(m_(1)+m_(2))^(2)(1/a-1/l)]`

Answer

B

Explanation

Potential at centre of disc is
`V=int_(0)^(a)-(G(dm))/r=int_(0)^(a)-(G((m_(1))/(pia^(2)))(2pir.dr))/r`
`=-(2Gm_(1))/a`
Now applying conservation of mechanical energy, decreases in gravitational `P.E.`=increases in `K.E.` or `1/2muv_(r)^(2)=U_(i)-U_(f)`, here `mu`=reduced mass `=(m_(1)m_(2))/(m_(1)+m_(2))` and `v_(r)`=relative speed.
`:. 1/2((m_(1)m_(2))/(m_(1)+m_(2)))v_(r)^(2)=[-(Gm_(1)m_(2))/l-((-2Gm_(1)m_(2))/a)]`
`:. v_(r)=sqrt(2G(m_(1)+m_(2))(2/a-1/l))`