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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

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

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 0.20 . 2 then the distance travelled by the ball before the second collision at the centre is nRS\dfrac{nR}{S}. Find the value of nn.A smooth tunnel is dug along the radius of the Earth that ends at the centre. A ball is released from the surface of

Correct answer

7

Step-by-step solution

a=−GMR3xa = - \dfrac{GM}{{R}^{3}}x

a=−ω2xa = - {\omega}^{2}x

vc=\omegaA=GMR3×R{v}_{c} = \omegaA = \sqrt{\dfrac{GM}{{R}^{3}}} \times R

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

After collision velocity of the ball towards A,

vC=evC=0.2GMR=15GMR{v}_{C} = e{v}_{C} = 0.2\sqrt{\dfrac{GM}{R}} = \dfrac{1}{5}\sqrt{\dfrac{GM}{R}}

Let now amplitude be A, then

A′=vCω=15GMRGMR3=R5{A}^{′} = \dfrac{{v}_{C}}{\omega} = \dfrac{\dfrac{1}{5}\sqrt{\dfrac{GM}{R}}}{\sqrt{\dfrac{GM}{{R}^{3}}}} = \dfrac{R}{5}

Net distance =R+(R5)+(R5)=7R5= R + \left(\dfrac{R}{5}\right) + \left(\dfrac{R}{5}\right) = \dfrac{7R}{5}
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