A proton in an accelerator attains a speed of 0.88c. What

is the magnitude of the momentum of the proton?

The given data can be listed below,

Special relativity's concept of momentum in special relativity is conserved for several observers in various reference frames by employing the so-called four-velocity (or appropriate velocity).

The relativistic momentum of the proton is given by,

$\overrightarrow{\mathrm{p}}=\mathrm{\gamma m}\overrightarrow{\mathrm{v}}$ …(i)

Here, m is the mass of the proton whose value is ${\mathrm{m}}_{\mathrm{p}}=1.7\times {10}^{-27}\mathrm{kg},\overrightarrow{\mathrm{v}}$, is the velocity of the proton and $\gamma$is the relativistic factor.

The value of relativistic factor is given by,

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

Here, v is the velocity of the proton and c is the speed of light.

Substitute values in the above,

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.88c}{c}}\right)}^2}} \\ =2.11 \end{gathered}$$

The magnitude of momentum is calculated by substituting all the values in equation (i).

$$\begin{gathered} \overrightarrow{p}=\left(2.11\right)\left(1.7\times {10}^{-27}kg\right)\left(0.88c\right) \\ =3.58\times {10}^{-27}\mathrm{kg}\left(0.88\times 3\times {10}^8\mathrm{m}/\mathrm{s}\right) \\ =9.47\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the magnitude of momentum of the proton is$9.47\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}$ .