A smooth tunnel is dug along the radius of the earth (R) that ends at the centre of earth. A ball is released from the surface of earth along the tunnel. If the coefficient of restitution is 0.2 between the end of tunnel and ball then the total distance travelled by the ball before second collision at the centre of earth is $\frac{x R}{5}$ . Then x is____

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

In the tunnel, motion of the particle is simple harmonic with angular frequency $ω = \sqrt{\frac{g}{R}} .$ Just before collision, $V_{0} = A ω = R \sqrt{\frac{g}{R}} = \sqrt{g R}$
Just after collision, $v = e V_{0} = \sqrt{g R} / 5$
Let new amplitude be $A'$ , then
$A' = \frac{v}{ω} = \frac{\sqrt{g R} / 5}{\sqrt{g / R}} = \frac{R}{5}$
Net distance $= R + (R / 5) + (R / 5) = \frac{7 R}{5}$