The magnitude of the momentum if the meteorite is 4.411*10^11 kgm/s
Solution
Relativistic energy E is given by:
E = mc^2
c = 3*10^8 m/s
E = [tex]\sqrt{p^2c^2 + m_{0}^2c^4}[/tex]
therefore,
p = [tex]m_{0}v/\sqrt{1-v^2/c^2}[/tex]
since v = 0.700c
p = [tex]1500.7(3*10^8)/\sqrt{1-(0.7c)^2/c^2}[/tex]
p = 4.411*10^11 kgm/s
Momentum is the result of a particle's mass and velocity. A vector quantity with both magnitude and direction is momentum. Isaac Newton's second equation of motion states that the force applied to a particle is equivalent to the time rate of change of momentum.
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