Question

Find the momentum p, in the laboratory frame of reference, of a proton moving with a speed of 0.805

Answer 1

The relativistic formula for the momentum of a particle of mass m and moving at speed v is
$p= \frac{mv}{ \sqrt{1- \frac{v^2}{c^2} } }$
where c is the speed of light.

In our problem, the speed in terms of c is $v=0.805 c$, and the mass (still written in terms of c) is $m=938 Mev/c^2$. Substituting these values in the previous formula, we can find the value of the momentum of the proton:
$p= \frac{(938 MeV/c^2)(0.805 c)}{ \sqrt{1-( \frac{0.899c}{c} )^2} } =1273 MeV/c$