91Ó°ÊÓ

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,

$\stackrel{→}{\mathrm{p}}=\mathrm{γ³¾}\stackrel{→}{\mathrm{v}}$ …(¾±)

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

The value of relativistic factor is given by,

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

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

Substitute values in the above,

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

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

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

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