Question

At what speed, as a fraction of c, is the kinetic energy of a particle twice its Newtonian value?

Short answer

The speed will be$0.78\mathrm{c}.$

Step-by-step solution

Given information,

We have given, the relativistic kinetic energy of a particle is equal to twice its Newtonian value of kinetic energy.

We have to find the speed of the particle.

Simplify 

Since,

$$\begin{gathered} \mathrm{K}.{\mathrm{E}}_{\mathrm{R}}=\mathrm{K}.{\mathrm{E}}_{\mathrm{N}} \\ \frac{{\mathrm{mc}}^2}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}-{\mathrm{mc}}^2=\frac{1}{2}{\mathrm{mv}}^2 \\ 2{\mathrm{c}}^2(1-\sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}})={\mathrm{v}}^2\sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}} \\ \sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}=\frac{{\mathrm{c}}^2}{{\mathrm{v}}^2+{\mathrm{c}}^2} \\ \mathrm{v}=2.35\times {10}^8\mathrm{m}/\mathrm{s} \\ \mathrm{v}=0.78\mathrm{c}. \end{gathered}$$