A smooth tunnel is dug along the radius of earth that ends at center. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at center and ball is $0.5$. Calculate the distance traveled by ball just before second collision at center. Given mass of the earth is $M$ and radius of the earth is $R$.

Let mass of the ball be $m$.

$\frac{1}{2}m{v}^2=m\left(V_A-V_B\right)$

          $=m\left[-\frac{GM}{R}-\left(-1.5\frac{GM}{R}\right)\right]$

         $=\frac{GMm}{2R}$

$\therefore v=\sqrt{\frac{GM}{R}}$

Velocity of ball just after collision,

$v^{\text{'}}=ev=\frac{1}{2}\sqrt{\frac{GM}{R}}$

Let $r$ be the distance from the center upto where the ball reaches after collision. Then,

$\frac{1}{2}m{V}^2=m\left[V\right(r)-V\text{ (centre) }$

or  $\frac{1}{8}\frac{GMm}{R}=m\left[\frac{3GM}{2R}-\frac{GM}{R^3}\left(\frac{3R^2}{2}-\frac{r^2}{2}\right)\right]$

or  

  $\frac{1}{8}=\frac{3}{2}-\frac{3}{2}+\frac{r^2}{2R^2}$

$\therefore \frac{r^2}{R^2}=\frac{1}{4}\text{ or }r=\frac{R}{2}$

$\therefore$ The desired distance,

$x=R+\frac{R}{2}+\frac{R}{2}=2R$    Ans.