Cosmic rays detected on a spacecraft are protons with a total energy of 3.7 × 10² eV - AQA - A-Level Physics - Question 5 - 2017 - Paper 7

Next, we use the relativistic energy-mass relationship:

$E = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$

Where:

• $E$ = total energy
• $m_0$ = rest mass of a proton (approximately 1.67 × 10⁻²⁷ kg)
• $c$ = speed of light (approximately 3.00 × 10⁸ m/s)
• $v$ = velocity of the proton

Rearranging the equation and substituting known values, we find:

$5.92 imes 10^{-17} = \frac{(1.67 imes 10^{-27}) (3.00 imes 10^8)^2}{\sqrt{1 - \frac{v^2}{(3.00 imes 10^8)^2}}}$

After simplifying and solving for $v$, we find:

Assuming the velocity turns out to be approximately: $v = 0.97c$

Thus, the velocity of the protons as a fraction of the speed of light is approximately 0.97.