Meteor - Ans.fyi 參考答案

Meteor

用戶6016

2008-06-26 9:42 pm

A meteor whose mass was about 1.0*10^8kg struck the Earth with a speed of about 15km/s and came to rest in the Earth. (a) What was the Earth's recoil speed? (b) What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth? (c) By how much did the Earth's kinetic energy change as a result of this collision?

回答 (1)

Audrey Hepburn

2008-06-27 1:08 am

✔ 最佳答案

a. Mass of the Earth, M = 5.98 X 1024 kg

Initial speed of meteor, u = 15 km/s = 1.5 X 104 m/s

Let v be the final common speed of the two objects / recoil speed of the Earth

By the law of conservation of momentum,

mu = (M + m)v

(1.0 X 108)(1.5 X 104) = (5.98 X 1024 + 1.0 X 108)v

Recoil speed of the Earth, v = 2.51 X 10-13 m/s

b. Initial K.E. of the meteor, E1 = 1/2 X (1.0 X 108)(1.5 X 104)2 = 1.125 X 1016 J

K.E. of the Earth, E2 = 1/2 X (5.98 X 1024)(2.51 X 10-13)2 = 0.188 J

So, the fraction of K.E. transferred = 0.188 / (1.125 X 1016) ~ 0

c. The kinetic energy change = 0.188 J

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