1. What was the Earth\'s recoil speed (relative to Earth at rest before the coll
ID: 1287068 • Letter: 1
Question
1. What was the Earth's recoil speed (relative to Earth at rest before the collision)?
2. What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth?
3. By how much did the Earth's kinetic energy change as a result of this collision?
A meteor whose mass was about 2.0½108kg ; km/s and came to rest in the Earth. 1. What was the Earth's recoil speed (relative to Earth at rest before the collision)? 2. What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth? 3. By how much did the Earth's kinetic energy change as a result of this collision? (m_{rm{E}} = 6.0 times 10^{24} ;{rm kg}) with a speed of about 25km/s 2.0times10^{8}; kg struck the Earth (mE=6.0½1024kg)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) = 8.3x10^-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*(8.3x10^-13)^2 = 2.06J
so the fraction of K is 2.06/6.25x10^16*100 = 3.30x10^17%
c) K of Earth = 1/2*6.0x10^24*(2.3x10^-13)^2 = 0.16J
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