Answer:
a.[tex]6.25\times10^{-13}\ m/s[/tex]
b.[tex]2.5\times10^{-17}\%[/tex]
c.1.1719J
Explanation:
a.Given the mass of the meteor [tex]m=1.5\times 10^{28}kg[/tex] and of earth [tex]mE=6.04\times10^{24}kg[/tex] and the speed of the meteor as [tex]25000m/s[/tex], we can use the law of Conservation of Momentum to determine Earth's relative recoil speed:
# Note Earth's initial speed, [tex]v_1=0m/s[/tex]
[tex]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[/tex]
Hence, Earth's recoils speed is [tex]6.25\times10^{-13}\ m/s[/tex]
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:
[tex]=\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}\%[/tex]
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;
[tex]\bigtriangleup K_E=0.5mv_2^2\\\\=0.5\times6.0\times10^{24}\times6.25\times10^{-13}\\\\=1.1719J[/tex]
#Hence Earth's Kinetic energy change is 1.1719J
