If one proton is travelling through space at 0.3c, what is it's kinetic energy in MeV?

We know the formula for kinetic energy is

KE = (1/2)mv².

If we know that c, the speed of light, is 3x10⁸ms^-1 (you will be given this value in your Data & Formula books in the exam) and the proton is traveling at 0.3c, it has a velocity of

0.3x3x10⁸ = 9x10⁷ms^-1.

Therefore v² will be (9x10⁷)² = 8.1x10¹⁵.

We know that the mass of a proton is 1.67x10^-27kg (again, you will be given this value in the exam).

Therefore kinetic energy in Joules (the SI unit, because we have currently only used SI units) is (1/2) x 1.67x10^-27 x 8.1x10¹⁵ = 6.76x10^-12J

It is known that 1eV (electron volt) is equal to 1.6x10^-19J. Therefore dividing our energy in Joules by this value will give us our energy in eV.

(6.76x10^-12) / (1.6x10^-19) = 4.23x10⁷eV.

As 1 MeV = 1x10⁶eV, we can find our value in MeV by dividing it by 1x10⁶;

(4.23x10⁷) / (1x10⁶) = 42.3MeV.

This is a standard unit used in particle energies, and is quite a common question in any A Level physics exam.