Question

At what speed, as a fraction of $c$, is a particle’s momentum twice its Newtonian value?

Short answer

$v=0.866c$

Step-by-step solution

Given Information

We have to find the speed speed as a fraction of $c$, is a particle’s momentum twice its Newtonian value.

Simplify

We use formula

$\sqrt{\left(1-\frac{v^2}{c^2}\right)}=0.5$

From the above formula, we will solve for speed $v$,

$$\begin{gathered} \sqrt{\left(1-\frac{v^2}{c^2}\right)}=0.5 \\ 1-\frac{v^2}{c^2}=0.25 \\ \frac{v^2}{c^2}=0.75 \\ v=c\sqrt{0.75} \\ v=0.866c \end{gathered}$$