In Praise of Simple Physics: The Science and Mathematics behind Everyday Questions

In Praise of Simple Physics: The Science and Mathematics behind Everyday Questions

by Paul J. Nahin


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Fun puzzles that use physics to explore the wonders of everyday life

Physics can explain many of the things that we commonly encounter. It can tell us why the night is dark, what causes the tides, and even how best to catch a baseball. With In Praise of Simple Physics, popular math and science writer Paul Nahin presents a plethora of situations that explore the science and math behind the wonders of everyday life. Roaming through a diverse range of puzzles, he illustrates how physics shows us ways to wring more energy from renewable sources, to measure the gravity in our car garages, to figure out which of three light switches in the basement controls the light bulb in the attic, and much, much more.

How fast can you travel from London to Paris? How do scientists calculate the energy of an atomic bomb explosion? How do you kick a football so it stays in the air and goes a long way downfield? Nahin begins with simpler problems and progresses to more challenging questions, and his entertaining, accessible, and scientifically and mathematically informed explanations are all punctuated by his trademark humor. Readers are presumed to have some background in beginning differential and integral calculus. Whether you simply have a personal interest in physics' influence in the world or you're an engineering and science student who wants to gain more physics know-how, this book has an intriguing scenario for you.

In Praise of Simple Physics proves that if we look carefully at the world around us, physics has answers for the most astonishing day-to-day occurrences.

Product Details

ISBN-13: 9780691166933
Publisher: Princeton University Press
Publication date: 05/24/2016
Series: Princeton Puzzlers
Pages: 272
Product dimensions: 6.20(w) x 9.30(h) x 1.10(d)

About the Author

Paul J. Nahin is professor emeritus of electrical engineering at the University of New Hampshire. He is the author of many best-selling popular math books, including Digital Dice, Chases and Escapes, Dr. Euler's Fabulous Formula, When Least Is Best, Duelling Idiots and Other Probability Puzzlers, and An Imaginary Tale (all Princeton).

Read an Excerpt

In Praise of Simple Physics

The Science and Mathematics behind Everyday Questions



Copyright © 2016 Princeton University Press
All rights reserved.
ISBN: 978-1-4008-8051-5


How's Your Math?

What would life be without arithmetic, but a scene of horrors?

Sydney Smith (in a letter dated July 22, 1835)

In this opening chapter I'll discuss several examples of the kind of mathematics we'll encounter in "simple physics" questions that may (or anyway could) occur in "ordinary life." These are questions whose intent, I think, anybody can understand but that require at least some analytical thinking to answer. The math examples are very different from one another, with their only "unifying" (if I may use that word) feature being a progressively increasing sophistication. The central question to ask yourself as you read each example is, do I follow the arguments? If you can say yes, even if you can't initially work through the detailed analysis yourself, then your understanding is sufficient for the book.

Example 1

Our first example of analytical thinking requires no formal math but, rather, logic and a bit of everyday knowledge (lit lightbulbs get hot). Think about it as you work through the rest of the examples, and, as with the wind-and-airplane problem at the end of the preface, I'll give you the answer at the end of the chapter.

Imagine that you are in a multistory house with three electrical switches in the basement and a 100-watt lightbulb in the attic. All three switches have two positions, labeled ON and OFF, but only one of the switches controls the lightbulb. You don't know which one. All three switches are initially OFF. One way to determine the controlling switch is with the following obvious procedure: Flip any one of the switches to ON, and then go up to the attic to see if the bulb is lit. If it is, you are done. If it isn't, go back to the basement, turn one of the other OFF switches to ON, and then go back up to the attic to see if the bulb is lit. If it is, then the switch you just turned ON controls the bulb. If the bulb is not lit, then the switch that has never been ON controls the bulb. So, you can obviously figure out which switch controls the lightbulb with at most two trips to the attic.

There is, however, another procedure that guarantees your being able to make that determination with just one trip up to the attic. What is it?

Example 2

This question also requires no real math but, again, logical reasoning (although it does require an elementary understanding of kinetic and potential energy). Suppose you fire a gun, sending a bullet straight up into the air. Taking air resistance into account, is the time interval during which the bullet is traveling upward greater than or less than the time interval during which the bullet falls back to Earth? You might think you need to know the details of the air-resistance drag-force law, but that is not so. All you need to know is that air resistance exists. You may assume that the Earth's gravitational field is constant over the entire up-down path of the bullet (it remains the same, independent of the bullet's altitude). As in Example 1, think about this question as you work through the rest of the examples, and I'll give you the answer at the end of the chapter. Hint: Potential energy is the energy of position (taking the Earth's surface as the zero reference level, a mass m at height h above the surface has potential energy mgh, where g is the acceleration of gravity, about 32 feet/seconds-squared), and kinetic energy is the energy of motion (a mass m moving at speed v has kinetic energy 1/2 mυ2).

Example 3

This question does require some math, but it's really only arithmetic involving a lot of multiplying and dividing of really big numbers. In the 1956 science fiction story "Expedition" by Fredric Brown (1906–1972), the following situation is the premise. There are 30 seats available on the first rocket ship trip to colonize Mars, with the seats to be filled by selecting at random 30 people from a pool of 500 men and 100 women. What is the probability that the result (as in the story) is one man and 29 women?

We start by imagining the 30 seats lined up, side by side, from left to right, and then we compute the total number of distinguishable (we assume each person is uniquely identifiable) ways to fill the seats without regard to gender from the pool of 600 people. That number, N,1 is


Next, if N2 is the total number of distinguishable ways to fill the 30 seats with exactly one man and 29 women, then the probability we seek is P = N2/N1. We calculate N2 as follows:

there are 30 different ways to pick the seat for the one man


there are 500 different ways to pick the one man for that seat.


N2 = (30)(500)(100)(99)(98) ... (72) = 15,000 100!/71!.

The formal answer to our question is then


I use the word formal because we still don't have a single number for P.

The factorials in this expression are all huge numbers, numbers that are far too large for direct calculation on a hand calculator (my calculator first fails at 70!). So, to make things more manageable, I'll use Stirling's asymptotic approximation for n!: n! ~ [square root of 2πne]-nnn. Then,


Each of the factors in the curly brackets is easily computed on a hand calculator, and the result is

P = 1.55 × 10-23.

The premise in Brown's story is therefore highly unlikely. No matter, though, because while it is so unlikely as to be verging on the "just can't happen," it is not impossible, and besides, it's a very funny story and well worth a willing suspension of disbelief.

Example 4

Quadratic equations are routinely encountered in mathematical physics (you'll see an example of this in Chapter 9), and here's an example of a quadratic in the form of a type of problem that many readers will recall from a high school algebra class. Readers may take some comfort in learning that it was incorrectly solved by Marilyn vos Savant in her Parade Magazine column of June 22, 2014 (but, to her credit, she quickly admitted her slip in the July 13 column after some attentive readers set her straight).

Brad and Angelina, working together, take 6 hours to complete a project. Working alone, Brad would take 4 hours longer to do the project than would Angelina if she did it by herself. How long would it take each to do the project by themselves?

If we denote Angelina's time by x, then Brad's time is x + 4. Thus, Angelina's rate of clearing the project is 1/x per hour, and Brad's rate is 1/[x+4]. So, in six hours Angelina finishes the fraction 6/x of the project, and Brad finishes the fraction 6/[x+4] of the project. These two fractions must total the finished project (that is, must add to 1), and so 6/x + 6/[x+4]. Cross-multiplying, we get 6(x + 4) + 6x = x (x + 4) = x2 + 4x or, 12x + 24 = x2 + 4x or,

x2 – 8x – 24 = 0.

The well-known formula for the quadratic equation gives


Since x must be positive, we use the + sign (the minus sign gives x< 0), and so x = 4+2[square root of 10] = 10.32. Thus, Angelina can do the project by herself in 10.32 hours, and Brad can do the project by himself in 14.32 hours.

The underlying assumption in this analysis is that when working together, Brad and Angelina work independently and without interference. This is not necessarily the case, depending on the nature of the project. For example, suppose "the project" is making a truck delivery. If Brad can drive a truck from A to B by himself in one hour, and if Angelina can drive the same truck from A to B by herself in one hour, how long does it take for the two of them together to drive that same truck from A to B? It's still one hour! An even more outrageous abuse of logic is the belief that if one soldier can dig a foxhole in 30 minutes, then 1,800 soldiers can dig a foxhole in one second!

Example 5

A real battery (with internal resistance r > 0 ohms), with a potential difference between its terminals of V volts (when no current is flowing in the battery), is connected to a resistor of R ohms as shown in Figure 1.1. What should R be so that maximum power is delivered to R? This problem is usually solved in textbooks with differential calculus, but that's mathematical overkill, because simple algebra is all that is required.

The current I that flows is (by Ohm's law — see note 1 in Chapter 8 if this isn't clear)

I = V/[r + R].

The power P dissipated (as heat) in R is (where E is the voltage drop across R)

P = EI = (IR) I = I2R,

and so


Obviously, P = 0 when R = 0, and P = 0 when R = ∞. Thus, there is some R between zero and infinity for which P reaches its greatest value. This value can easily be found with calculus (differentiate P with respect to R and set the result to zero), but all that is needed is algebra. Here's how:


We clearly maximize P by minimizing the denominator of the rightmost side of this equation, which just as clearly occurs for R = r (because that makes the first term in the denominator — which is never negative — as small as possible, that is, equal to zero). Thus, R = r, and the maximum power in R is V2/4R.

Example 6

In this example you'll see how simple geometry, combined with physics, allows measuring the distance from the Earth to the Moon with fantastic precision. To establish the physics first, all we'll need is the idea that a ray of light incident on a mirror reflects from that mirror at an angle equal to the angle of incidence, as shown in Figure 1.2. This phenomenon was first noted by Euclid, in the third century BC; however, it was not explained until a few hundred years later, in the first century AD, when Heron of Alexandria (in his book on mirrors, Catoptrica) observed that the reflection law is a consequence of assuming the ray path ARB is the minimum reflected length path. That is, if the point R on the mirror was such that θi ≠ θr, then the resulting total path length would be increased. Heron's observation was the first occurrence of a minimum principle in mathematical physics; such principles play central roles in modern theoretical physics.

Here's a simple geometric proof of Heron's explanation of the mirror reflection law. If B, the destination point, is distance d above the mirror, then B's reflected point (B') is distance d "below" the mirror. RB and RB' are therefore the equal-length hypotenuses of two congruent right triangles, which means that θ'r = θr (referring again to Figure 1.2). Now, the total light path is AR + RB = AR + RB', and this last sum is the path length from A to B'. The shortest path from A to B' (and so the shortest length for the reflected path, too) is along a straight line, and so θ'r = θi, which says that θi = θr. That's it!

The law of reflection has the following application in an optical device called a corner reflector (see Figure 1.3). This gadget allowed the Apollo 11 astronauts to participate in the 1969 measurement of the distance from the Earth to the Moon to within 2.5 meters! The path of an incoming ray of light to mirror 1 has the vector description (rx, ry), and the path of the reflected ray has the vector description (rx,ry). That is, one component of the path vector is reversed, while the other is not; mirror 1, lying along the x-axis, reverses the y-component. The reflected ray continues on to mirror 2, lying along the y-axis, and there the x-component of the path vector is reversed, giving a path vector description of the reflected ray off mirror 2 of (–rx,ry) = – (rx, ry), which is the total reversal of the original incoming ray's path vector. Notice that this means the reflected ray from mirror 2 is perfectly parallel to the incident ray on mirror 1, is laterally offset and reversed in direction, and these conditions are independent of the value of the angle α.

Can the same thing be done in three dimensions? The answer is yes, and that is easily seen once we give the following interpretation to what a reflecting mirror does: the mirror reverses the incident ray's path vector component that is normal to the mirror and leaves the other component(s) unaltered. (Look back at the two-dimensional discussion and you'll see that's what happened there.) So, in the case of a three-dimensional corner reflector (think of the inside corner of a cube made of three mutually perpendicular mirrors, with the corner of the cube defining the origin of an x, y, z-coordinate system), imagine that mirrors 1, 2, and 3 lie along the xy-, the xz-, and the yz-surfaces, respectively. Then, a ray reflecting off mirror 1 has its z-component reversed, a ray reflecting off mirror 2 has its y-component reversed, and a ray reflecting off mirror 3 has its x-component reversed.

After an incident ray has completed three reflections it emerges from the corner cube reflector in an exactly reversed direction. The special cases where the incoming ray hits only one (or two) of the mirrors are simply the cases where the incident ray arrives parallel to one (or two) of the mirrors, and so one (or two) of the path vector components happen to be zero (and, of course, the reversal of zero is zero). The Apollo 11 astronauts placed multiple corner cube reflectors on the Moon's surface, which were then targets for very brief (picosecond) laser pulses from Earth. The corner cube reflector sent reflected pulses back to almost precisely where their transmission had occurred, and the elapsed time for the Earth-to-Moon- to-Earth round trip then gave the separation distance. Such measurements have shown that the Moon is very slowly moving away from Earth (just an inch and a half per year), and in Chapter 10 you'll learn why.

Example 7

Here's a simple example of high school trigonometry at work in an interesting physics setting. In Robert Serber's book on the U.S. atomic bomb project (see note 12 in the preface), mention is made of the occurrence of the equation

x cos(x) = (1 – a) sin(x)

in one of the theoretical problems studied by the Los Alamos scientists, where a is a given constant. For any particular value of a, what are the positive solutions for x (x ≤ 0 solutions were not physically interesting to the bomb designers)?


Excerpted from In Praise of Simple Physics by PAUL J. NAHIN. Copyright © 2016 Princeton University Press. Excerpted by permission of PRINCETON UNIVERSITY PRESS.
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
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Table of Contents

Foreword by T. M. Helliwell ix

Preface with Challenge Problems xi

1 How’s Your Math? 1

2 The Traffic-Light Dilemma 20

3 Energy from Moving Air 25

4 Dragsters and Space Station Physics 32

5 Merry-Go-Round Physics and the Tides 42

6 Energy from Moving Water 51

7 Vectors and Bad Hair Days 63

8 An Illuminating Problem 67

9 How to Measure Depth with a Stopwatch 74

10 Doing the Preface Problems 79

11 The Physics of Stacking Books 92

12 Communication Satellite Physics 103

13 Walking a Ladder Upright 110

14 Why Is the Sky Dark at Night? 115

15 How Some Things Float (or Don’t) 126

16 A Reciprocating Problem 141

17 How to Catch a Baseball (or Not) 146

18 Tossing Balls and Shooting Bullets Uphill 153

19 Rapid Travel in a Great Circle Transit Tube 163

20 Hurtling Your Body through Space 177

21 The Path of a Punt 194

22 Easy Ways to Measure Gravity in Your Garage 200

23 Epilogue Newton’s Gravity Calculation Mistake 218

Postscript 227


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"Many books that explore classical physics problems tend to be relentlessly sequential and completely humorless. In this book, Paul Nahin shows a lighter touch and he gives himself the necessary space to tell the mathematical story. The book encourages readers to engage with the less-than-obvious aspects of physics."—Jens Zorn, professor emeritus of physics, University of Michigan

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