Chapter 21, Problem 86Q

(a)

To determine

The length of a pencil or a pen moving at a speed specified by the following situations:

(a) a bicycle at constant speed.

(b) a car travelling on a highway.

(c) a commercial jet liner at cruising altitude.

Answer to Problem 86Q

Solution:

In all the cases, there is a negligible change in the length of a pencil that is 15cms long.

Explanation of Solution

Given data:

A pencil or a pen travelling at a speed equal to that of a bicycle rider, a car travelling on a highway and a commercial jetliner at cruising altitude.

Formula used:

Write the expression for Lorentz transformation of length.

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

Here, $L$ is length of the object moving along the direction of motion, $L_o$ is length of the object at rest, $v$ is the speed of the moving object and $c$ is the speed of light.

Explanation:

The speed of light is $3\times {10}^8\text{m/s}$.

Assume the length of pen to be $15\text{ cm}$.

Assume the speed of a bicycle to be $4.30\text{ m/s}$, speed of car travelling on highway to be $11.17\text{ m/s}$ and speed of a commercial jet liner at cruising altitude to be $244.4\text{m/s}$

Calculate the length of a pen, moving at a speed equal to the speed of a bicycle.

Refer to the expression for Lorentz transformation of length.

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

Substitute the known values: $3\times {10}^8\text{m/s}$ for $c$, $15\text{ cm}$ for $L_o$ and $4.30\text{ m/s}$ for $v$.

$\begin{array}{c}L=\left(15\text{ cm}\right)\left(\sqrt{1-{\left(\frac{4.30\text{ m/s}}{3\times {10}^8\text{m/s}}\right)}^2}\right)\\ =14.99999\text{ cms}\end{array}$

Now, calculate the length of a pen, moving at a speed equal to the speed of a car travelling on the highway.

Refer to the expression for Lorentz transformation of length.

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

Substitute the known values: $3\times {10}^8\text{m/s}$ for $c$, $15\text{ cm}$ for $L_o$ and $11.17\text{ m/s}$ for $v$.

$\begin{array}{c}L=\left(15\text{ cm}\right)\left(\sqrt{1-{\left(\frac{11.17\text{ m/s}}{3\times {10}^8\text{m/s}}\right)}^2}\right)\\ =14.99999\text{ cms}\end{array}$

Now, calculate the length of a pen moving at a speed equal to the speed of a commercial jet liner at cruising altitude.

Refer to the expression for Lorentz transformation of length.

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

Substitute the known values: $3\times {10}^8\text{m/s}$ for $c$, $15\text{ cm}$ for $L_o$ and $244.4\text{m/s}$ for $v$.

$\begin{array}{c}L=\left(15\text{ cm}\right)\left(\sqrt{1-{\left(\frac{244.4\text{m/s}}{3\times {10}^8\text{m/s}}\right)}^2}\right)\\ =14.99998\text{ cms}\end{array}$

Conclusion:

Hence, the length of the pen is nearly same in all the three situations.

(b)

To determine

The length of a pencil or a pen travelling at a speed equal to the speed of the beam emitted by a spaceship at a speed of $200000\text{ kms/s}$.

Answer to Problem 86Q

Solution:

$11.18\text{ m/s}$

Explanation of Solution

Given data:

A pen travelling at a speed equal to the speed of a beam emitted by a spaceship towards the other at $200000\text{ kms/s}$.

Formula used:

Write the expression for Lorentz transformation of length.

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

Here, $L$ is length of the object moving along the direction of motion, $L_o$ is length of the object at rest, $v$ is the speed of the moving object and $c$ is the speed of light.

Explanation:

The speed of light is $3\times {10}^8\text{m/s}$

Assume the length of the pen to be $15\text{ cm}$.

The speed of the pen is given to be $200000\text{ km/s}$.

Calculate the length of the pen.

Refer to the expression for Lorentz transformation of length

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

Substitute $3\times {10}^8\text{m/s}$ for $c$, $15\text{ cm}$ for $L_o$ and $200000\text{ km/s}$ for $v$.

$\begin{array}{c}L=\left(15\text{ cm}\right)\left(\sqrt{1-{\left(\frac{200000\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)}{3\times {10}^8\text{m/s}}\right)}^2}\right)\\ =11.18\text{ cms}\end{array}$

Conclusion:

Hence, the length of the pen or the pencil is $11.18\text{ cm}$.