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3 years ago

What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of 2.7 3 10210 j?

1 answer:

8 0

The equation that is applicable to this problem is expressed as 

p = γmu

where p is the relativistic momentum, m is equal to the proton mass and u is the speed of proton. γ is equal to 1/√(1 - u²/c²) where c is the speed of light.

Another equation to determine p fully is to use this relativistic equation:
E = γmc²

where E is the total kinetic energy including the gamma energy.

We are given with E = 2.7 × 10^(-10) J and from a proton’s data, mc^2 = 938.27201 MeV. Using the conversion from MeV to J which is 1 eV = 1.60218 * 10^(-19) J, we compute for γ

γ=E/mc^2 =2.7 × 10^(-10) J / 938.27201x10^6 eV *(1.60218 * 10^(-19) J/eV) = 1.80

we use this number to determine u from the second equation γ = 1/√(1 - u²/c²)

we square both sides and cross multiply, resulting to 
γ² (1 - u²/c²) = 1

u²/c² = (γ² -1)/γ²

u = c/y sq rt of (γ² -1)

u = 0.83

Substituting eventually to the first equation, we get

p = γmu

p = 1.80 * 1.67262 * 10^(-27) kg * 0.83 * 2.99792 * 10^8 m/s

p = 7.481 * 10^(-19) kg.m/s

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