Question

a. What are the momentum and total energy of a proton with speed $0.99\mathrm{c}$?

b. What is the proton’s momentum in a different reference frame in which $\mathrm{E}\text{'}=5.0\times {10}^{-10}\mathrm{J}$?

Short answer

a. Energy of the proton is $10.6\times {10}^{-10}\mathrm{J}$and momentum is $3.5\times {10}^{-18}\mathrm{kg}\mathrm{m}/\mathrm{s}.$

b. Proton's momentum in difference reference frame is$1.58\times {10}^{-18}\mathrm{kg}\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

Speed of the proton = $0.99\mathrm{c}$

We have to find the momentum and energy of the proton.

Simplify

Since we know the mass of proton is,

${\mathrm{m}}_{\mathrm{p}}=1.67\times {10}^{-27}\mathrm{kg}$

Then, energy is given by,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{mc}}^2}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{E}=\frac{(1.67\times {10}^{-27}\mathrm{kg}){(3\times {10}^8\mathrm{m}/\mathrm{s})}^2}{\sqrt{1-{\displaystyle \frac{{(0.99\mathrm{c})}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{E}=\frac{15.03\times {10}^{-11}}{\sqrt{0.0199}}\mathrm{J} \\ \mathrm{E}=10.6\times {10}^{-10}\mathrm{J} \end{gathered}$$

Then,

the momentums will be,

$$\begin{gathered} \mathrm{E}=\mathrm{pc} \\ \mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}} \\ \mathrm{p}=\frac{10.6\times {10}^{-10}\mathrm{J}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ \mathrm{p}=3.5\times {10}^{-18}\mathrm{kg}\mathrm{m}/\mathrm{s} \end{gathered}$$

Part 9b) Step 1: Given information

We have given,

$\mathrm{E}\text{'}=5.0\times {10}^{-10}\mathrm{J}$

we have to find the momentum of the proton.

Simplify

We know the relativistic energy is write as,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{mc}}^2}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}} \\ \sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}=\frac{{\mathrm{mc}}^2}{\mathrm{E}} \\ \mathrm{v}=\mathrm{c}\sqrt{(1-\frac{{\mathrm{m}}^2{\mathrm{c}}^4}{{\mathrm{E}}^2})} \\ \mathrm{v}=(3\times {10}^8\mathrm{m}/\mathrm{s})(\sqrt{1-\frac{{(1.67\times {10}^{-27}\mathrm{kg})}^2(3\times {10}^8{)}^4}{{(5\times {10}^{-10})}^2}} \\ \mathrm{v}=2.86\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Then momentum will be,

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{mv}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{p}=\frac{(1.67\times {10}^{-27}\mathrm{kg})(2.86\times {10}^8\mathrm{m}/\mathrm{s})}{\sqrt{1-{\displaystyle \frac{{(2.86)}^2}{9}}}} \\ \mathrm{p}=1.58\times {10}^{-18}\mathrm{kg}\mathrm{m}/\mathrm{s} \end{gathered}$$