Blog: A relativistic proton has a momentum of `1.0 * 10^(-17) kg * m/s` and a rest energy of `0.15` nanojoules. What is the kinetic energy of this...

Hello!

As we know, a mass of a body changes for an observer when an observer moves relative to a body. Denote the speed as $\displaystyle{v}.$

For a body with a rest mass $\displaystyle{m}_{{0}}$ its rest energy is $\displaystyle{E}_{{r}}={m}_{{0}}{{c}}^{{2}}.$ Its full energy is $\displaystyle{E}={m}{{c}}^{{2}},$ where $\displaystyle{m}$ is its mass (not a rest mass), and its kinetic energy is the difference between full and rest energies, $\displaystyle{E}_{{k}}={m}{{c}}^{{2}}-{m}_{{0}}{{c}}^{{2}}.$

Recall the well-known formula for...

Hello!

As we know, a mass of a body changes for an observer when an observer moves relative to a body. Denote the speed as $\displaystyle{v}.$

Recall the well-known formula for a relativistic mass, $\displaystyle{m}=\frac{{m}_{{0}}}{\sqrt{{{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}}}}.$ Therefore a momentum is $\displaystyle{p}={m}{v}=\frac{{{m}_{{0}}{v}}}{\sqrt{{{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}}}}.$

Square this equality and obtain $\displaystyle{{p}}^{{2}}=\frac{{{{m}_{{0}}^{{2}}}{{v}}^{{2}}}}{{{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}}},$ or $\displaystyle{{p}}^{{2}}-{{p}}^{{2}}\frac{{{v}}^{{2}}}{{{c}}^{{2}}}={{m}_{{0}}^{{2}}}{{v}}^{{2}}.$ Solving this for $\displaystyle{{v}}^{{2}}$ we obtain $\displaystyle{{v}}^{{2}}=\frac{{{p}}^{{2}}}{{{{m}_{{0}}^{{2}}}+\frac{{{p}}^{{2}}}{{{c}}^{{2}}}}}=\frac{{{{p}}^{{2}}{{c}}^{{2}}}}{{{{p}}^{{2}}+\frac{{{E}_{{r}}^{{2}}}}{{{c}}^{{2}}}}}$ and therefore $\displaystyle{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}={1}-\frac{{{p}}^{{2}}}{{{{p}}^{{2}}+\frac{{{E}_{{r}}^{{2}}}}{{{c}}^{{2}}}}}=\frac{{{E}_{{r}}^{{2}}}}{{{{c}}^{{2}}{{p}}^{{2}}+{{E}_{{r}}^{{2}}}}}.$

Now we can compute a kinetic energy:

$\displaystyle{E}_{{k}}={m}{{c}}^{{2}}-{m}_{{0}}{{c}}^{{2}}=\frac{{{m}_{{0}}{{c}}^{{2}}}}{\sqrt{{{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}}}}-{E}_{{r}}=\frac{{E}_{{r}}}{{\frac{{E}_{{r}}}{\sqrt{{{{c}}^{{2}}{{p}}^{{2}}+{{E}_{{r}}^{{2}}}}}}}}-{E}_{{r}}=$

$\displaystyle=\sqrt{{{{c}}^{{2}}{{p}}^{{2}}+{{E}_{{r}}^{{2}}}}}-{E}_{{r}}.$

$\displaystyle{c}$ , $\displaystyle{E}_{{r}}$ and $\displaystyle{p}$ are given and we can find the numeric result:

$\displaystyle{E}_{{k}}=\sqrt{{{9}\cdot{{10}}^{{16}}\cdot{{10}}^{{-{34}}}+{2.25}\cdot{{10}}^{{-{20}}}}}-{1.5}\cdot{{10}}^{{-{10}}}=$

$\displaystyle={{10}}^{{-{10}}}\cdot{\left(\sqrt{{{9}\cdot{100}+{2.25}}}-{1.5}\right)}\approx{28.54}\cdot{{10}}^{{-{10}}}{\left({J}\right)},$

or 2.854 nanojoules.

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A relativistic proton has a momentum of $\displaystyle{1.0}\cdot{{10}}^{{-{17}}}{k}{g{\cdot}}\frac{{m}}{{s}}$ and a rest energy of $\displaystyle{0.15}$ nanojoules. What is the kinetic energy of this...

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A relativistic proton has a momentum of and a rest energy of nanojoules. What is the kinetic energy of this...

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A relativistic proton has a momentum of $\displaystyle{1.0}\cdot{{10}}^{{-{17}}}{k}{g{\cdot}}\frac{{m}}{{s}}$ and a rest energy of $\displaystyle{0.15}$ nanojoules. What is the kinetic energy of this...

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