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A smooth tunnel is dug along the radius of earth that ends at

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A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is $M$ and radius of the earth is $R$ .

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Book:

Mechanics Volume 2 (DC Pandey)

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Harper Discussed

A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is

M

and radius of the earth is

R

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Text solutionVerified

Let mass of the ball be m.
Solution Image

\frac{1}{2} m v^{2} ∴ v = m (V_{A} - V_{B}) = m [- \frac{GM}{R} - (- 1.5 \frac{GM}{R})] = \frac{GM m}{2 R} = \frac{GM}{R}

Velocity of ball just after collision,

v^{'} = e v = \frac{1}{2} \frac{GM}{R}

Let

r

be the distance from the centre upto where the ball reaches after collision. Then,

\frac{1}{2} m v^{'2} = m [V (r) - V (centre)]

or \frac{1}{8} \frac{GM m}{R} = m [\frac{3 GM}{2 R} - \frac{GM}{R ^{3}} (\frac{3 R ^{2}}{2} - \frac{r ^{2}}{2})]

\frac{1}{8} = \frac{3}{2} - \frac{3}{2} + \frac{r ^{2}}{2 R ^{2}}

∴ \frac{r ^{2}}{R ^{2}} = \frac{1}{4}

r = \frac{R}{2}

∴

The desired distance,

s = R + \frac{R}{2} + \frac{R}{2} = 2 R

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Question Text	A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is $M$ and radius of the earth is $R$ .
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	118

Text solutionVerified

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