Question

A "cosmic-ray" proton hits the upper atmosphere with a speed 0.9999c, where c is the speed of light. What is the magnitude of the momentum of this proton? Note that $\left|\overrightarrow{v}\right|\mathbf{/}\mathit{c}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9999}$; you don't actually need to calculate the speed$\left|v\right|$.

Short answer

The magnitude of the momentum of the proton is $3.3937\times {10}^{-17}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Definition of momentum

Momentum is a vector quantity that is, it has both magnitude and direction.

The formula for relativistic momentum is,

$p=\gamma mv$

Here, mis mass of the proton, vis velocity, and $\gamma$is relativistic factor which is given by,

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

Here, vis relative velocity and c is speed of light in vacuum.

Finding the momentum

From the given equation, v = 0.9999c.

Substitute v = 0.9999c into the formula of relativistic factor $\gamma$.

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.9999c}{c}}\right)}^2}} \\ =\frac{1}{\sqrt{1-{(0.9999)}^2}} \\ =70.71 \end{gathered}$$

The proton's speed is so close to that of light.

Substitute $\gamma =70.71,m=1.6\times {10}^{-27}\mathrm{kg}$and v = 0.9999c into the formula of momentum.

$$\begin{gathered} p=(70.71)(1.6\times {10}^{-27})(0.9999c) \\ =(1.13136\times {10}^{-25})(2.9997\times {10}^8) \\ =3.3937\times {10}^{-17}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the momentum of the proton is $3.3937\times {10}^{-17}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.