Question

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 for the collision between the ball and the surface of the tunnel is $0.2$ then the distance travelled by the ball before the second collision at the centre is $\frac{nR}{S}$ . Find the value of $n$ .

Answer 1

Answer: 7

$a=-\frac{GM}{R^3}x$
$a=-{\omega }^{2}x$
$v_c=\omega A=\sqrt{\frac{GM}{R^3}}\times R$

After collision velocity of the ball towards A,
$v_C=e{v}_C=0.2\sqrt{\frac{GM}{R}}=\frac{1}{5}\sqrt{\frac{GM}{R}}$
Let now amplitude be A, then
$A^{\prime }=\frac{v_C}{\omega }=\frac{\frac{1}{5}\sqrt{\frac{GM}{R}}}{\sqrt{\frac{GM}{R^3}}}=\frac{R}{5}$
Net distance $=R+\left(\frac{R}{5}\right)+\left(\frac{R}{5}\right)=\frac{7R}{5}$