Question

particle’s kinetic energy is ntimes its rest energy, what is its speed?

Short answer

The speed is$\mathrm{u}=\frac{\sqrt{\mathrm{n}(\mathrm{n}+2)}}{\mathrm{n}+1}\mathrm{c}$.

Step-by-step solution

Expression for the particle’s kinetic energy:

Let the speed of a particle be u, and the rest mass be m.

It is given that the particle’s kinetic energy is n times its rest energy, write the equation based on the given condition$$\begin{gathered} \mathrm{kinetic}\mathrm{energy}=\mathrm{n}\times \mathrm{Rest}\mathrm{energy} \\ {\mathrm{E}}_{\mathrm{k}\mathrm{in}}=\mathrm{n}\times {\mathrm{E}}_{\mathrm{rest}} \end{gathered}$$ …… (1)

Write the energy for the kinetic energy.

${\mathbf{E}}_{\mathbf{kin}}\mathbf{=}\frac{{\mathbf{mc}}^{\mathbf{2}}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}\mathbf{-}{\mathbf{mc}}^{\mathbf{2}}$

Write the expression for the rest energy.

${\mathbf{E}}_{\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{t}}\mathbf{=}\mathit{m}{\mathit{c}}^{\mathbf{2}}$

Determine the speed of a particle:

Substitute $$\begin{gathered} \frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}-m{c}^{2}for{E}_{kin}andm{c}^{2}for{E}_{rest}inequation\left(1\right). \\ \frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}-m{c}^2=n\times m{c}^2 \\ m{c}^2\left(\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}-1\right)=n\times m{c}^2 \\ \frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}-1=n \\ \frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}=n+1 \end{gathered}$$

On further solving,

$$\begin{gathered} \sqrt{1-\frac{u^2}{c^2}}=\frac{1}{n+1} \\ 1-\frac{u^2}{c^2}={\left(\frac{1}{n+1}\right)}^2 \\ 1-\frac{u^2}{c^2}=\frac{1}{{(n+1)}^2} \\ -\frac{u^2}{c^2}=\frac{1}{{(n+1)}^2}-1 \end{gathered}$$

Solve the R.H.S:

$$\begin{gathered} -\frac{u^2}{c^2}=\frac{1-{(n+1)}^2}{{(n+1)}^2} \\ -\frac{u^2}{c^2}=\frac{1-(n^2+1+2n)}{{(n=1)}^2} \\ u=\frac{\sqrt{n(n+2)}}{n+1}c \\ Therefore,thespeedisu=\frac{\sqrt{n(n+2}}{n+1}c. \end{gathered}$$