Question

A meteor whose mass was about 1.5 * 108 kg struck the Earth AmE = 6.0 * 1024 kgB with a speed of about 25 km????s and came to rest in the Earth. (a) What was the Earth’s recoil speed (relative to Earth at rest before the collision)? (b) What fraction of the meteor’s kinetic energy was trans- formed to kinetic energy of the Earth? (c) By how much did the Earth’s kinetic energy change as a result of this collision?

Answer 1

Answer:

a.$6.25\times10^{-13}\ m/s$

b.$2.5\times10^{-17}\%$

c.1.1719J

Explanation:

a.Given the mass of the meteor $m=1.5\times 10^{28}kg$ and of earth $mE=6.04\times10^{24}kg$ and the speed of the meteor as $25000m/s$, we can use the law of Conservation of Momentum to determine Earth's relative recoil speed:

# Note Earth's initial speed, $v_1=0m/s$

$m_1v_1+m_2v_1=m_1v_2+m_2v_2\\\\1.5\times10^{8}\times 25000+6\times10^{24}\times 0=1.5\times10^{8}\times0+6\times10^{24}v_2\\\\v_2=6.25\times10^{-13}m/s$

Hence, Earth's recoils speed is $6.25\times10^{-13}\ m/s$

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

#Due to Earth's huge size compared to the meteor, the change in mass of the Earth is negligible:

$=\frac{0.5(mv^2)_E}{(0.5mv^2)_m}\\\\=\frac{0.5\times6\times10^{24}\times6.25\times10^{-13}\ m/s}{0.5\times1.5\times10^{8}\times 25000}\\\\=2.5\times10^{-17}\%$

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

Kinetic Energy change in earth is calculated as;

$\bigtriangleup K_E=0.5mv_2^2\\\\=0.5\times6.0\times10^{24}\times6.25\times10^{-13}\\\\=1.1719J$

#Hence Earth's Kinetic energy change is 1.1719J