Problem 7.74 A meteor whose mass was about 2.0×108kgstruck the Earth ( m E=6.0×1

Question

Problem 7.74

A meteor whose mass was about 2.0×108kgstruck the Earth (mE=6.0×1024kg) with a speed of about 25 km/s and came to rest in the Earth.

Part A

What was the Earth's recoil speed (relative to Earth at rest before the collision)?

Express your answer to two significant figures and include the appropriate units.

Part B

What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth?

Express your answer using two significant figures.

Part C

By how much did the Earth's kinetic energy change as a result of this collision?

Express your answer to two significant figures and include the appropriate units.

Problem 7.74

A meteor whose mass was about 2.0×108kgstruck the Earth (mE=6.0×1024kg) with a speed of about 25 km/s and came to rest in the Earth.

Part A

What was the Earth's recoil speed (relative to Earth at rest before the collision)?

Express your answer to two significant figures and include the appropriate units.

vE =

Part B

What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth?

Express your answer using two significant figures.

KEE/KEm =

Part C

By how much did the Earth's kinetic energy change as a result of this collision?

Express your answer to two significant figures and include the appropriate units.

KEE =

Explanation / Answer

a) Use conservation of momentum for a completely inelastic collision

so (M +m)*V = m*v where M represents the Earth and m the meteor

therefore V= m*v/(M + m) = 2*10^8*25000/(6.0x10^24 + 2*10^8) = 8.33x10^-13m/s

b) So the meteor's K = 1/2*m*v^2 = 1/2*2*10^8*25000^2 = 6.25x10^16J

K of Earth = 1/2*6.0x10^24*(8.33x10^-13)^2 = 2.081J

so the fraction of K is 2.081/9.8x10^15*100 = 2.12x10^18%

c) K of Earth = 1/2*6.0x10^24*(8.33x10^-13)^2 = 2.081J

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