Question

The momentum pof an electron at speednear the speedof light increases according to the formula $\mathbf{p}\mathbf{=}\frac{\mathbf{mv}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$, whereis a constant (mass of the electron). If an electron is subject to a constant force F, Newton’s second law describing its motion is localid="1659249453669" $\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\frac{\mathbf{m}\mathbf{v}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}\mathbf{=}\mathit{F}\mathbf{.}$

Find $\mathit{v}\left(t\right)$and show that $\mathit{v}\mathbf{\to }\mathit{c}$as $\mathit{t}\mathbf{\to }\mathbf{\infty }$. Find the distance travelled by the electron in timeif it starts from rest.

Short answer

The value of $v\left(t\right)$is $\sqrt{\frac{F^2{t}^2{c}^2}{m^2{c}^2+F^2{t}^2}}$and the distance the electron will travel at any time is $\frac{c}{F}\left[\sqrt{F^2{t}^2+m^2{c}^2-mc}\right]$.

Step-by-step solution

Given Information.

It is given thatan electron is subject to a constant forcei.e.$\frac{\mathrm{dp}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\frac{\mathrm{mv}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}=F.$ .

Meaning of Distance.

The overall movement of an object, regardless of direction, is referred to as distance. We can define distance as the amount of ground covered by an item, regardless of its starting or ending position.

Find v( t)and the distance travelled by the electron in time t if it starts from rest.

In this case when the radius has been reduced 0.1cm, that took 6 months.

So, if the radius reduces 0.25cm, that’s means it will take a time:

6months+6months+3months= 15months.

Find the time when the mothball has a volume equal to half of the original volume.

Because the force is constant, integrate both sides to get the speed of the electron as a function of time

$$\begin{gathered} \int \frac{d}{dx}\frac{mv}{\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}=dt=F\int dt \\ \frac{mv}{\sqrt{1-\raisebox{1ex}{$v{\left(t\right)}^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}=Ft+A \end{gathered}$$

Therefore, $v\left(t\right)=\sqrt{\frac{{\left(Ft+A\right)}^2{c}^2}{m^2{c}^2+{\left(Ft+A\right)}^2}}......\left(1\right)$

Ast$\to \infty$equation (1) become $v\left(t\to \infty \right)\to \sqrt{\frac{Ft{c}^2}{Ft}}=c.$

$$\begin{gathered} Useboundaryconditionv\left(0\right)=0. \\ v\left(t\to 0\right)=0 \\ =\sqrt{\frac{A{c}^2}{m^2{c}^2+A}} \\ A=0 \end{gathered}$$

For finding the distance, integrate eq. (1).

$$\begin{gathered} \int v\left(t\right)dt=\int \sqrt{\frac{F^2{t}^2{c}^2}{m^2{c}^2+F^2{t}^2}}dt \\ x\left(t\right)=Fc\int \sqrt{\frac{t^2}{B^2+t^2}}dtWhere,B=\frac{mc}{F} \end{gathered}$$

From integration table:$\int \frac{u}{\sqrt{u^2+a^2}}du=\sqrt{u^2+a^2}+C$

$x\left(t\right)=c\int \sqrt{t^2+B^2}+D$

Therefore, $=\frac{c}{F}\sqrt{F^2{t}^2+m^2{c}^2}+D$

Assume that the motion starts from the origin, then find the integration constant D.

So, the distance the electron will travel at any time is$$\begin{gathered} x\left(t\right)=\frac{c}{F}\sqrt{F^2{t}^2+m^2{c}^2}-\frac{m{c}^2}{F} \\ =\frac{c}{F}\left[\sqrt{F^2{t}^2+m^2{c}^2}-mc\right] \end{gathered}$$