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Question: Answered & Verified by Expert
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 nRS. Find the value of n.
PhysicsCenter of Mass Momentum and CollisionJEE Main
Solution:
1663 Upvotes Verified Answer
The correct answer is: 7
a= -GMR3x

a= -ω2x

vc=ωA= GMR3×R



After collision velocity of the ball towards A,

vC=evC=0.2GMR=15GMR

Let now amplitude be A, then

A=vCω=15GMRGMR3=R5

Net distance =R+R5+R5=7R5

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