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 is 0.2 between the surface and ball then the total distance travelled by the ball before second collision at the centre.

• A. $\frac{6}{5}R$
• B. $\frac{7}{5}R$
• C. $\frac{9}{5}R$
• D. $\frac{3}{2}R$

Answer 1

The correct option is B $\frac{7}{5}R$

$a=-\frac{GM}{R^3}x$

$a=-{\omega }^{2}x$
so, $\omega =\sqrt{\frac{GM}{R^3}}{V}_c=\omega A=\sqrt{\frac{GM}{R^3}}\times R$
$V_c=\sqrt{\frac{GM}{R}}$
After collision velocity of 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 $=R+\left(\frac{R}{5}\right)+\left(\frac{R}{5}\right)=\left(\frac{7}{5}\right)R$