Answer:
[tex]4.4\cdot 10^{11} kg m/s[/tex]
Explanation:
Since the meteorite is moving at a speed which is a significant fraction of the speed of light, we must use the formula for the relativistic momentum:
[tex]p=\gamma m_0 v[/tex]
where
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] is the relativistic factor
[tex]m_0 = 1500 kg[/tex] is the rest mass of the meteorite
[tex]v=0.700 c[/tex] is the speed of the meteorite
First of all, let's find the relativistic factor
[tex]\gamma = \frac{1}{\sqrt{1-\frac{(0.700 c)^2}{c^2}}}=1.4[/tex]
And since [tex]c=3.0\cdot 10^8 m/s[/tex], the speed of the meteorite is
[tex]v=(0.700)(3\cdot 10^8 m/s)=2.1\cdot 10^8 m/s[/tex]
So now we can find the momentum of the meteorite
[tex]p=\gamma m_0 v=(1.4)(1500 kg)(2.1\cdot 10^8 m/s)=4.4\cdot 10^{11} kg m/s[/tex]
