A smooth tunnel is dug along the radius of the earth that ends at the centre. A ball is released from the surface of earth along the tunnel. If the coefficient of restitution is `0.2` between the surface and ball, then the distance travelled by the ball before second collision at the centre is

Options

A. `(6R)/5`
B. `(7R)/5`
C. `(9R)/5`
D. `(3R)/2`

Answer

B

Explanation

Let `V_(0)` be the velocity of the satellite at `A` when it is in a circular orbit. Then
`(GMm)/(r^(2))=(mV_(0)^(2))/r`
`V_(0)^(2)=(Gm)/r`
Due to impulse, the velocity `V_(0)` is increased to `V` such that `V^(2)=KV_(0)^(2)`
According to `LCAM`,
`mVR=mvr`
`V=(vr)/R`
According `LCE`
`-(GMm)/R+1/2mV^(2)=(GMm)/r+1/2mv^(2)`
`((2GM)/r-v^(2))R^(2)-2GMR+v^(2)r^(2)=0`
`:. R=(v^(2)r)/(((2GM)/r-v^(2)))=(V^(2)r)/(2v_(0)^(2)-v^(2))=(rK)/(2-K)`