Question
n the theory of relativity, the mass of a particle with speed v is m=f(v)=m01−v2/c2√, where m0 is the rest mass of the particle and c is the speed of light in a vacuum. Letting m0=1, find f−1(10).
Answers
I believe the formula is
f(v) = m<sub>0</sub> / √(1-v^2/c^2)
so, plugging your numbers, if I read it right, you want f<sup> -1</sup>(10), meaning you want to know how fast to go to make the mass increase by a factor of 10.
10 = 1/√(1 - v^2/c^2)
100 = 1/(1 - v^2/c^2)
1 - v^2/c^2 = 1/100
c^2 - v^2 = c^2/100
v^2 = 99c^2/100
v = √99/10 c = .995c
f(v) = m<sub>0</sub> / √(1-v^2/c^2)
so, plugging your numbers, if I read it right, you want f<sup> -1</sup>(10), meaning you want to know how fast to go to make the mass increase by a factor of 10.
10 = 1/√(1 - v^2/c^2)
100 = 1/(1 - v^2/c^2)
1 - v^2/c^2 = 1/100
c^2 - v^2 = c^2/100
v^2 = 99c^2/100
v = √99/10 c = .995c
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