The expression for the relativistic energy [tex]E[/tex] is given by:
[tex]E=mc^{2}[/tex] (1)
being [tex]c=3(10)^{8}m/s[/tex]
This famous equation includes the relativistic Kinetic energy [tex]K[/tex] and the energy at rest [tex]E_{o}[/tex]:
[tex]E=K+E_{o}[/tex] (2)
Where:
[tex]E_{o}=m_{o}c^{2}[/tex] (3)
Being [tex]m_{o}=1500kg[/tex] the mass at rest for the meteorite
[tex]K=E-E_{o}[/tex] (4)
In addition, there is a relation between the relativistic energy and the momentum [tex]p[/tex]:
[tex]E=\sqrt{p^{2}c^{2}+m_{o}^{2}c^{4}}[/tex] (5)
Where:
[tex]p=\frac{m_{o}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex] (6)
Knowing this, let's begin with the answers:
In order to solve this part, equation (6) will be helpful, since we already know the mass of the meteorite and its speed [tex]v=0.700c[/tex]:
[tex]p=\frac{m_{o}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
[tex]p=\frac{(1500kg)(0.7c)}{\sqrt{1-\frac{(0.7c)^{2}}{c^{2}}}}[/tex]
[tex]p=\frac{(1500kg)(0.7(3(10)^{8}m/s))}{\sqrt{1-0.49}}[/tex]
[tex]p=4.411(10)^{11}kg.m/s[/tex] (7) >>>>This is the meteorite's momentum
Remembering equation (5), which relates the total energy with the momentum:
[tex]E=\sqrt{p^{2}c^{2}+m_{o}^{2}c^{4}}[/tex]
We can substitute the value of the momentum found on (7):
[tex]E=\sqrt{(4.411(10)^{11}kg.m/s)^{2}(3(10)^{8}m/s)^{2}+(1500kg)^{2}(3(10)^{8}m/s)^{4}}[/tex]
Then:
[tex]E=1.890(10)^{20}kg.m^{2}/s^{2}[/tex]
Knowing [tex]1kg.m^{2}/s^{2}=1N.m=1J=1 Joule[/tex]:
[tex]E=1.890(10)^{20}J[/tex] (8)>>>This is the total energy of the meteorite
Using equation (3):
[tex]E_{o}=m_{o}c^{2}[/tex]
[tex]E_{o}=1500kg(3(10)^{8}m/s)^{2}[/tex]
[tex]E_{o}=1.35(10)^{20}J[/tex] (9) >>>Meteorite's energy at rest
According to equation (4) the relativistic kinetic energy depends on the total energy and the energy at rest:
[tex]K=E-E_{o}[/tex]
We already know the values of [tex]E[/tex] and [tex]E_{o}[/tex] from (8) and (9). Hence we only have to substitute them on the equation:
[tex]K=1.890(10)^{20}J-1.35(10)^{20}J[/tex]
[tex]K=5.4(10)^{9}J[/tex] >>>Meteorite's Relativistic energy
