Chapter 39, Problem 40P

(a)

To determine

The fine for travel at the speed of $1090\text{km}/\text{h}$.

Answer to Problem 40P

The fine for travel at the speed of $1090\text{km}/\text{h}$ is $\$800$.

Explanation of Solution

Given info: The speed limit is $90.0\text{km}/\text{h}$ and the fine for driving at $190E$ is $\$80.0$.

Write the expression for the total fine in the excess momentum.

$F=p_u-p_{90\text{km}/\text{h}}$ (1)

Here,

$F$ is the total fine.

$p_{\text{u}}$ is the momentum of the vehicle.

$p_{90\text{km}/\text{h}}$ is the speed limit momentum.

Write the expression for fine to travel.

$\$80.0=p_{190\text{km}/\text{h}}-p_{90\text{km}/\text{h}}$ (2)

Divide equation (1) and equation (2).

$\frac{F}{\$80.0}=\frac{p_u-p_{90\text{km}/\text{h}}}{p_{190\text{km}/\text{h}}-p_{90\text{km}/\text{h}}}$ (3)

Write the expression for the relativistic momentum of the vehicle.

$p_{\text{u}}=\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}$

Here,

$m$ is the mass of vehicle.

$u$ is the speed of vehicle.

$c$ is the speed of the light.

Write the expression for the relativistic momentum of the vehicle at speed of $90.0\text{km}/\text{h}$.

$p_{90\text{km}/\text{h}}=\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}$

Here,

$u$ is the speed of vehicle travel at speed of $90.0\text{km}/\text{h}$.

Write the expression for the relativistic momentum of the vehicle at speed of $190.0\text{km}/\text{h}$.

$p_{190\text{km}/\text{h}}=\frac{m\left(u_{190\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{190\text{km}/\text{h}}/c\right)}^2}}$

Here,

$u$ is the speed of vehicle travel at speed of $190.0\text{km}/\text{h}$.

Substitute $\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}$ for $p_{\text{u}}$, $\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}$ for $p_{90\text{km}/\text{h}}$ and $\frac{m\left(u_{190\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{190\text{km}/\text{h}}/c\right)}^2}}$ for $p_{190\text{km}/\text{h}}$ in equation (3).

$\frac{F}{\$80.0}=\frac{\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}-\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}}{\frac{m\left(u_{190\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{190\text{km}/\text{h}}/c\right)}^2}}-\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}}$ (4)

Substitute $1090.0\text{km}/\text{h}$ for $u$, $90.0\text{km}/\text{h}$ for $u_{90.0\text{km}/\text{h}}$ and $190.0\text{km}/\text{h}$ for $u_{190.0\text{km}/\text{h}}$ in equation (4).

$\begin{array}{c}\frac{F}{\$80.0}=\frac{\frac{m\left(1090\text{km}/\text{h}\right)}{\sqrt{1-{\left(1090\text{km}/\text{h}/c\right)}^2}}-\frac{m\left(90\text{km}/\text{h}\right)}{\sqrt{1-{\left(u/c\right)}^2}}}{\frac{m\left(190\text{km}/\text{h}\right)}{\sqrt{1-{\left(\frac{190\text{km}/\text{h}}{3\times {10}^8\text{m}/\text{s}\times \frac{3.6\text{km}/\text{h}}{1\text{m}/\text{s}}}\right)}^2}}-\frac{m\left(90\text{km}/\text{h}\right)}{\sqrt{1-{\left(u/c\right)}^2}}}\\ \frac{F}{\$80.0}=\frac{\frac{1090\text{km}/\text{h}}{\sqrt{1-{\left(0.999\right)}^2}}-90\text{km}/\text{h}}{100\text{km}/\text{h}}\\ \frac{F}{\$80.0}\approx \frac{1090\text{km}/\text{h}-90\text{km}/\text{h}}{100\text{km}/\text{h}}\\ F\approx \$800\end{array}$

Conclusion:

Therefore, the fine for travel at the speed of $1090\text{km}/\text{h}$ is $\$800$.

(b)

To determine

The fine for travel at the speed of $1000000090\text{km}/\text{h}$.

Answer to Problem 40P

The fine for travel at the speed of $1000000090\text{km}/\text{h}$ is $\$2.12\times {10}^9$.

Explanation of Solution

Given info: The speed limit is $90.0\text{km}/\text{h}$ and the fine for driving at $190E$ is $\$80.0$.

From equation (4), the equation is given as,

$\frac{F}{\$80.0}=\frac{\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}-\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}}{\frac{m\left(u_{190\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{190\text{km}/\text{h}}/c\right)}^2}}-\frac{m\left(u_{90\text{km}/\text{h}}\right)}{\sqrt{1-{\left(u_{90\text{km}/\text{h}}/c\right)}^2}}}$

Substitute $1000000090.0\text{km}/\text{h}$ for $u$, $90.0\text{km}/\text{h}$ for $u_{90.0\text{km}/\text{h}}$ and $190.0\text{km}/\text{h}$ for $u_{190.0\text{km}/\text{h}}$ in equation (4).

$\begin{array}{c}\frac{F}{\$80.0}=\frac{\frac{m\left(1000000090.0\text{km}/\text{h}\right)}{\sqrt{1-{\left(1000000090.0\text{km}/\text{h}/c\right)}^2}}-\frac{m\left(90\text{km}/\text{h}\right)}{\sqrt{1-{\left(u/c\right)}^2}}}{\frac{m\left(190\text{km}/\text{h}\right)}{\sqrt{1-{\left(\frac{190\text{km}/\text{h}}{3\times {10}^8\text{m}/\text{s}\times \frac{3.6\text{km}/\text{h}}{1\text{m}/\text{s}}}\right)}^2}}-\frac{m\left(90\text{km}/\text{h}\right)}{\sqrt{1-{\left(u/c\right)}^2}}}\\ \frac{F}{\$80.0}=\frac{\frac{1000000090.0\text{km}/\text{h}}{\sqrt{1-{\left(0.999\right)}^2}}-90\text{km}/\text{h}}{100\text{km}/\text{h}}\\ \frac{F}{\$80.0}\approx \frac{1000000090\text{km}/\text{h}-90\text{km}/\text{h}}{100\text{km}/\text{h}}\\ F\approx \$2.12\times {10}^9\end{array}$

Conclusion:

Therefore, the fine for travel at the speed of $1000000090\text{km}/\text{h}$ is $\$2.12\times {10}^9$.