Question

What is the speed, as a fraction of $c$, of a particle whose momentum is $mc$?

Short answer

$v=\frac{c}{\sqrt{2}}$

Step-by-step solution

Given Information

We have to find the speed as a fraction of $c$, of a particle whose momentum is $mc$.

Simplify

We will use the below formula

$p=\frac{mv}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}}$

From the above formula we will find the speed value,

$$\begin{gathered} p=\frac{mv}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}} \\ mc=\frac{mv}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}} \\ \sqrt{\left(1-\frac{v^2}{c^2}\right)}·mc=\sqrt{\left(1-\frac{v^2}{c^2}\right)}·\frac{mv}{\sqrt{\left(1-\frac{v^2}{c^2}\right)}} \\ \sqrt{\left(1-\frac{v^2}{c^2}\right)}·mc=mv \\ \frac{\sqrt{\left(1-\frac{v^2}{c^2}\right)}·mc}{m}=\frac{mv}{m} \\ \sqrt{\left(1-\frac{v^2}{c^2}\right)}·c=v^2 \\ {c}^2-v^2=v^2 \\ {c}^2=2v^2 \\ {v}^2=\frac{c^2}{2} \\ v=\frac{c}{\sqrt{2}} \end{gathered}$$