Question

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

is the magnitude of the momentum of the proton?

Short answer

The magnitude of momentum of the proton is $9.47\times {10}^{-19}\mathrm{kg}/\mathrm{ms}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The accelerating speed of the proton is,v=0.88c

Concept/Significance of relativistic momentum

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).

Determination of the magnitude of the momentum of the proton

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$9.47\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}$ .