COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 10, Problem 46QAP
To determine

(a)

The speed of the asteroid by which it strikes the moon's surface when it is falling with speed of 2.00×102m/sat the altitude of 1.00×103m.

Expert Solution
Check Mark

Answer to Problem 46QAP

Speed of the asteroid by which it strikes the surface of moon is 1436m/s.

Explanation of Solution

Given:

Mass of the moon = M=7.35×1022kg

Radius of the moon = R=1.737×106m

Altitude of the asteroid = h=1.00×103km=1.00×106m

Mass of the asteroid = m=1.00×102kg

Formula used:

Gravitation potential energy is defined as

  U=GMmrr= distance between center to center of the two objects.

Kinetic energy of an object is written as,

  K.E=12mv2Here,vis the speed of the object.

Calculation:

Consider the speed at which asteroid strike to the surface of moon is vsurface

By conservation of mechanical energy,

  (K.E+P.E)at the surface of moon=(K.E+P.E)at given altitude12mvsurface2GMmR=12mv2GMm(R+h)mv2surface2=GMmRGMm(R+h)+12mv2v2surface2=GM(1R1(R+h))+12v2vsurface=2GM(1R1(R+h))+v2vsurface=2GM(R+hRR(R+h))+v2vsurface=2GMh(R+h)R+v2

Now, let's plug all these values,

  vsurface=2×6.67×1011Nm2/kg2×7.35×1022kg×1.00×106m(1.737×106m+1.00×106m)(1.737×106m)+(2.00×102m/s)2vsurface=1436m/s

Conclusion:

Thus, the speed of the asteroid when it strikes to the surface of the moon is 1436m/s.

To determine

(b)

The work done on asteroid by the moon when it strikes to the surface of the moon and stops there

Expert Solution
Check Mark

Answer to Problem 46QAP

The work done on the asteroid during collision 1.03×108J.

Explanation of Solution

Given:

Mass of the asteroid = m=1.00×102kg

Speed of the asteroid just before colliding with the surface of moon = vsurface=1436m/s

Speed of the asteroid after stopping = vstopping=0

Calculation:

Since, asteroid is coming to stop and its displacement inside the surface of moon is very small comparison to the radius of the moon. So, in this case, change in potential energy would be zero (ΔU=0).

Thus, total work done on the asteroid by the moon consumes in the change (loss) in kinetic energy.

  W=ΔK.EW=12mv2stopping12mv2surfaceW=12m(0)12×1.00×102kg×(1436m/s)2W=1.03×108J

Conclusion:

Thus, the work done on the asteroid by the moon is 1.03×108J.

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