A smooth tunnel is dug along the radius of the Earth that ends at the centre. A ball is released from the…

Solution

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

$a= -{\omega }^{2}x$

$v_c=\omega A= \sqrt{\frac{GM}{R^3}}\times R$



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 $=R+\left(\frac{R}{5}\right)+\left(\frac{R}{5}\right)=\frac{7R}{5}$

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