100 Instructive Trig-based Physics Examples Volume 1: The Laws of Motion

214 Pages·2017·3.605 MB·English

by  Chris McMullen

100 Instructive Trig-based Physics Examples
Volume 1: The Laws of Motion
Fundamental Physics Problems
Chris McMullen, Ph.D.
Physics Instructor
Northwestern State University of Louisiana
Copyright © 2016 Chris McMullen, Ph.D.
Updated edition: February, 2017
www.monkeyphysicsblog.wordpress.com
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www.chrismcmullen.wordpress.com
Zishka Publishing
All rights reserved.

ISBN: 978-1-941691-16-8

Textbooks > Science > Physics
Study Guides > Workbooks> Science
CONTENTS
Introduction 4
Chapter 1 – Algebra Essentials 5
Chapter 2 – One-dimensional Uniform Acceleration 13
Chapter 3 – Geometry Essentials 19
Chapter 4 – Motion Graphs 23
Chapter 5 – Two Objects in Motion 27
Chapter 6 – Net and Average Values 33
Chapter 7 – Trigonometry Essentials 37
Chapter 8 – Vector Addition 53
Chapter 9 – Projectile Motion 59
Chapter 10 – Newton’s Laws of Motion 65
Chapter 11 – Applications of Newton’s Second Law 71
Chapter 12 – Hooke’s Law 83
Chapter 13 – Uniform Circular Motion 87
Chapter 14 – Uniform Circular Motion with Newton’s Second Law 91
Chapter 15 – Newton’s Law of Gravity 99
Chapter 16 – Satellite Motion 105
Chapter 17 – Work and Power 109
Chapter 18 – Conservation of Energy 117
Chapter 19 – One-dimensional Collisions 137
Chapter 20 – Two-dimensional Collisions 147
Chapter 21 – Center of Mass 153
Chapter 22 – Uniform Angular Acceleration 159
Chapter 23 – Torque 163
Chapter 24 – Static Equilibrium 169
Chapter 25 – Moment of Inertia 179
Chapter 26 – A Pulley Rotating without Slipping 189
Chapter 27 – Rolling without Slipping 195
Chapter 28 – Conservation of Angular Momentum 201
INTRODUCTION
This book includes fully-solved examples with detailed explanations for over 100 standard
physics problems. There are also a few examples from relevant math subjects, including
algebra trigonometry, which are essential toward mastering physics.
Each example breaks the solution down into terms that make it easy to understand.
The written explanations between the math help describe exactly what is happening, one
step at a time. These examples are intended to serve as a helpful guide for solving similar
standard physics problems from a textbook or course.
The best way to use this book is to write down the steps of the mathematical solution
on a separate sheet of paper while reading through the example. Since writing is a valuable
memory aid, this is an important step. In addition to writing down the solution, try to think
your way through the solution. It may help to read through the solution at least two times:
The first time, write it down and work it out on a separate sheet of paper as you solve it.
The next time, think your way through each step as you read it.
Math and science books aren’t meant to be read like novels. The best way to learn
math and science is to think it through one step at a time. Read an idea, think about it, and
then move on. Also write down the solutions and work them out on your own paper as you
read. Students who do this tend to learn math and science better.
•
Note that these examples serve two purposes:
They are primarily designed to help students understand how to solve standard
physics problems. This can aid students who are struggling to figure out homework
•
problems, or it can help students prepare for exams.
These examples are also the solutions to the problems of the author’s other book,
Essential Trig-based Physics Study Guide Workbook , ISBN 978-1-941691-14-4.
That study guide workbook includes space on which to solve each problem.
100 Instructive Trig-based Physics Examples
1 REVIEW OF ESSENTIAL ALGEBRA SKILLS

Example 1. Use the quadratic formula to solve for 𝑥𝑥 in the following equation.
2
2𝑥𝑥 −2𝑥𝑥−40 = 0
Solution. This is a quadratic equation because it has one term with the variable squared
2 2
(2𝑥𝑥 contains 𝑥𝑥 ), one linear term (−2𝑥𝑥 is linear because it includes 𝑥𝑥), and one constant
term (−40 is constant because it doesn’t include a variable). The standard form of the
quadratic equation is:
2
𝑎𝑎𝑥𝑥 +𝑏𝑏𝑥𝑥+𝑐𝑐 = 0
2
Note that the given equation is already in standard form, since the squared term (2𝑥𝑥 ) is
first, the linear term (−2𝑥𝑥) is second, and the constant term (−40) is last. Identify the
2 2
constants 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 by comparing 2𝑥𝑥 −2𝑥𝑥−40 = 0 with 𝑎𝑎𝑥𝑥 +𝑏𝑏𝑥𝑥 +𝑐𝑐 = 0.
𝑎𝑎 = 2 , 𝑏𝑏 = −2 , 𝑐𝑐 = −40
Note that 𝑏𝑏 and 𝑐𝑐 are both negative. Plug these values into the quadratic formula.
2 2
−𝑏𝑏±√𝑏𝑏 −4𝑎𝑎𝑐𝑐 −(−2)±�(−2) −4(2)(−40)
𝑥𝑥 = =
2𝑎𝑎 2(2)
Note that −(−2) = +2 since the two minus signs make a plus sign. Also note that the
2 2
minus sign doesn’t matter in (−2) , since the minus sign gets squared: (−2) = +4. It’s a
2 2
common mistake for students to incorrectly type −2 or −(2) on their calculator when the
2 2
correct thing to type is (−2) , which is the same thing as 2 . Similarly, the two minus signs
inside the squareroot make a plus sign: −4(2)(−40) = +320.
2±√4+320 2±√324 2±18
𝑥𝑥 = = =
4 4 4
There are two solutions for 𝑥𝑥. We must work these out separately.
2+18 2−18
𝑥𝑥 = or 𝑥𝑥 =
4 4
20 −16
𝑥𝑥 = or 𝑥𝑥 =
4 4
𝑥𝑥 = 5 or 𝑥𝑥 = −4
The two answers are 𝑥𝑥 = 5 and 𝑥𝑥 = −4.

Check. We can check our answers by plugging them into the original equation.
2 2 
2𝑥𝑥 −2𝑥𝑥 −40 = 2(5) −2(5)−40 = 2(25)−10−40 = 50−50 = 0
2 2
2𝑥𝑥 −2𝑥𝑥 −40 = 2(−4) −2(−4)−40 = 2(16)+8−40 = 32+8−40 = 0

Example 2. Use the quadratic formula to solve for 𝑦𝑦.
2
3𝑦𝑦−27+2𝑦𝑦 = 0
2
Solution. Reorder the terms in standard form. Put the squared term (2𝑦𝑦 ) first, the linear
term (3𝑦𝑦) next, and the constant term (−27) last.
2
2𝑦𝑦 +3𝑦𝑦−27 = 0
5
Chapter 1 – Review of Essential Algebra Skills

2 2
Identify the constants 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 by comparing 2𝑦𝑦 +3𝑦𝑦−27 = 0 with 𝑎𝑎𝑦𝑦 +𝑏𝑏𝑦𝑦+𝑐𝑐 = 0.
𝑎𝑎 = 2 , 𝑏𝑏 = 3 , 𝑐𝑐 = −27
Plug these values into the quadratic formula.
2 2
−𝑏𝑏±√𝑏𝑏 −4𝑎𝑎𝑐𝑐 −3±�3 −4(2)(−27)
𝑦𝑦 = =
2𝑎𝑎 2(2)
Note that the two minus signs make a plus sign: −4(2)(−27) = +216.
−3±√9+216 −3±√225 −3±15
𝑦𝑦 = = =
4 4 4
−3+15 −3−15
𝑦𝑦 = or 𝑦𝑦 =
4 4
12 −18
𝑦𝑦 = or 𝑦𝑦 =
4 4
18 9
Note that − 4 reduces to −2 if you divide both the numerator and denominator by 2. That
18 18÷2 9
is, − 4 = − 4÷2 = −2. Simplifying the previous equations, we get:
9
𝑦𝑦 = 3 or 𝑦𝑦 = −
2
9
The two answers are 𝑦𝑦 = 3 and 𝑦𝑦 = −2.

Check. We can check our answers by plugging them into the original equation.
2 2
3𝑦𝑦−27+2𝑦𝑦 = 3(3)−27+2(3) = 9−27+2(9) = 9−27+18 = 0
2
2 9 9 27 81
3𝑦𝑦−27+2𝑦𝑦 = 3�− �−27+2� � = − −27+2� �
2 2 2 4
To add or subtract fractions, make a common denominator. We can make a common
27 2 4
denominator of 4 by multiplying − 2 by 2 and multiplying −27 by 4.
27 81 27 2 4 81
− −27+2� � = − � �−27� �+2� �
2 4 2 2 4 4
54 108 162 −54−108+162
= − − + = = 0
4 4 4 4

Example 3. Use the quadratic formula to solve for 𝑡𝑡.
2
6𝑡𝑡 = 8−2𝑡𝑡
Solution. Reorder the terms in standard form. Use algebra to bring all of the terms to the
2
same side of the equation (we will put them on the left side). Put the squared term (−2𝑡𝑡 )
first, the linear term (6𝑡𝑡) next, and the constant term (8) last. Note that the sign of a term
will change if it is brought from the right-hand side to the left-hand side (we’re subtracting
2
8 from both sides and we’re adding 2𝑡𝑡 to both sides of the equation).
2
2𝑡𝑡 +6𝑡𝑡−8 = 0
2 2
Identify the constants 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 by comparing 2𝑡𝑡 +6𝑡𝑡−8 = 0 with 𝑎𝑎𝑡𝑡 +𝑏𝑏𝑡𝑡+𝑐𝑐 = 0.
𝑎𝑎 = 2 , 𝑏𝑏 = 6 , 𝑐𝑐 = −8
Plug these values into the quadratic formula.
6
100 Instructive Trig-based Physics Examples

2 2
−𝑏𝑏±√𝑏𝑏 −4𝑎𝑎𝑐𝑐 −6±�6 −4(2)(−8)
𝑡𝑡 = =
2𝑎𝑎 2(2)
Note that the two minus signs make a plus sign: −4(2)(−8) = +64.
−6±√36+64 −6±√100 −6±10
𝑡𝑡 = = =
4 4 4
−6+10 −6−10
𝑡𝑡 = or 𝑡𝑡 =
4 4
4 −16
𝑡𝑡 = or 𝑡𝑡 =
4 4
𝑡𝑡 = 1 or 𝑡𝑡 = −4
The two answers are 𝑡𝑡 = 1 and 𝑡𝑡 = −4.
Check. We can check our answers by plugging them into the original equation. For each
2
answer, we’ll compare the left-hand side (6𝑡𝑡) with the right-hand side (8−2𝑡𝑡 ). First plug
2 
𝑡𝑡 = 1 into both sides of 6𝑡𝑡 = 8−2𝑡𝑡 .
2 2
6𝑡𝑡 = 6(1) = 6 and 8−2𝑡𝑡 = 8−2(1) = 8−2 = 6
2 
Now plug in 𝑡𝑡 = −4 into both sides of 6𝑡𝑡 = 8−2𝑡𝑡 .
2 2
6𝑡𝑡 = 6(−4) = −24 and 8−2𝑡𝑡 = 8−2(−4) = 8−2(16) = 8−32 = −24

Example 4. Use the quadratic formula to solve for 𝑥𝑥.
2 2
1+25𝑥𝑥 −5𝑥𝑥 = 8𝑥𝑥 −3𝑥𝑥 +9
Solution. Reorder the terms in standard form. Use algebra to bring all of the terms to the
2
same side of the equation (we will put them on the left side). Put the squared terms (−5𝑥𝑥
2
and −3𝑥𝑥 ) first, the linear terms (25𝑥𝑥 and 8𝑥𝑥) next, and the constant terms (1 and 9) last.
Note that the sign of a term will change if it is brought from the right-hand side to the left-
2
hand side (we’re subtracting 8𝑥𝑥 from both sides, adding 3𝑥𝑥 to both sides, and subtracting
9 from both sides of the equation).
2 2
−5𝑥𝑥 +3𝑥𝑥 +25𝑥𝑥 −8𝑥𝑥+1−9 = 0
2
Combine like terms: Combine the two 𝑥𝑥 terms, the two 𝑥𝑥 terms, and the two constants.
2
−2𝑥𝑥 +17𝑥𝑥 −8 = 0
2 2
Identify the constants 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 by comparing −2𝑥𝑥 +17𝑥𝑥 −8 = 0 with 𝑎𝑎𝑥𝑥 +𝑏𝑏𝑥𝑥 +𝑐𝑐 = 0.
𝑎𝑎 = −2 , 𝑏𝑏 = 17 , 𝑐𝑐 = −8
Plug these values into the quadratic formula.
2 2
−𝑏𝑏±√𝑏𝑏 −4𝑎𝑎𝑐𝑐 −17±�17 −4(−2)(−8)
𝑥𝑥 = =
2𝑎𝑎 2(−2)
Note that the three minus signs make a minus sign: −4(−2)(−8) = −64.
−17±√289−64 −17±√225 −17±15
𝑥𝑥 = = =
−4 −4 −4
−17+15 −17−15
𝑥𝑥 = or 𝑥𝑥 =
−4 −4
7
Chapter 1 – Review of Essential Algebra Skills

−2 −32
𝑥𝑥 = or 𝑥𝑥 =
−4 −4
Note that a negative number divided by a negative number results in a positive answer.
The minus signs from the numerator and denominator cancel out.
1
𝑥𝑥 = or 𝑥𝑥 = 8
2
1
The two answers are 𝑥𝑥 = 2 and 𝑥𝑥 = 8.
Check. We can check our answers by plugging them into the original equation. For each
2
answer, we’ll compare the left-hand side (1+25𝑥𝑥 −5𝑥𝑥 ) with the right-hand side (8𝑥𝑥 −
2 1 2
3𝑥𝑥 +9). First plug 𝑥𝑥 = 2 into the left-hand side (1+25𝑥𝑥 −5𝑥𝑥 ).
2
2 1 1 25 5
1+25𝑥𝑥 −5𝑥𝑥 = 1+25� �−5� � = 1+ −
2 2 2 4
4 25 2
In order to add and subtract the fractions, multiply 1 by 4 and multiply 2 by 2 to make a
common denominator.
25 5 4 252 5 4 50 5 4+50−5 49
1+ − = 1� �+ − = + − = =
2 4 4 2 2 4 4 4 4 4 4
1 2
Next plug 𝑥𝑥 = 2 into the right-hand side (8𝑥𝑥 −3𝑥𝑥 +9).
2
2 1 1 1 3
8𝑥𝑥 −3𝑥𝑥 +9 = 8� �−3� � +9 = 4−3� �+9 = 13−
2 2 4 4
4
In order to subtract the fraction, multiply 13 by 4 to make a common denominator.
3 4 3 52 3 52−3 49
13− = 13� �− = − = =
4 4 4 4 4 4 4
2 2
Now plug in 𝑥𝑥 = 8 into both sides of 1+25𝑥𝑥 −5𝑥𝑥 = 8𝑥𝑥 −3𝑥𝑥 +9.
2 2 
1+25𝑥𝑥 −5𝑥𝑥 = 1+25(8)−5(8) = 1+200−5(64) = 201−320 = −119
2 2
8𝑥𝑥 −3𝑥𝑥 +9 = 8(8)−3(8) +9 = 64−3(64)+9 = 64−192+9 = −119

Example 5. Use the method of substitution to solve the following system of equations for
each unknown.
3𝑥𝑥+2𝑦𝑦 = 18
8𝑥𝑥−5𝑦𝑦 = 17
Solution. First isolate 𝑦𝑦 in the top equation. Subtract 3𝑥𝑥 from both sides.
2𝑦𝑦 = 18−3𝑥𝑥
Divide both sides of the equation by 2.
18−3𝑥𝑥
𝑦𝑦 =
2
Substitute this expression in parentheses in place of 𝑦𝑦 in the bottom equation.
8𝑥𝑥−5𝑦𝑦 = 17
18−3𝑥𝑥
8𝑥𝑥 −5� � = 17
2
8
100 Instructive Trig-based Physics Examples

Distribute the 5. When you distribute, the two minus signs make a plus.
18 3𝑥𝑥
8𝑥𝑥 −5� �−5�− � = 17
2 2
3𝑥𝑥
8𝑥𝑥 −5(9)+5� � = 17
2
15𝑥𝑥
8𝑥𝑥 −45+ = 17
2
15𝑥𝑥
Combine like terms: 8𝑥𝑥 and 2 are like terms, and −45 and 17 are like terms. Combine the
15𝑥𝑥 2
terms 8𝑥𝑥 + 2 using a common denominator. Multiply 8𝑥𝑥 by 2. In order to combine the
constant terms, add 45 to both sides.
2 15𝑥𝑥
8𝑥𝑥� �+ = 17+45
2 2
16𝑥𝑥 15𝑥𝑥
+ = 62
2 2
31𝑥𝑥
= 62
2
Multiply both sides of the equation by 2.
31𝑥𝑥 = 124
Divide both sides of the equation by 31.
𝑥𝑥 = 4
Now that we have an answer for 𝑥𝑥, we can plug it into one of the previous equations in
order to solve for 𝑦𝑦. It’s convenient to use the equation where 𝑦𝑦 was isolated.
18−3𝑥𝑥 18−3(4) 18−12 6
𝑦𝑦 = = = = = 3
2 2 2 2
The answers are 𝑥𝑥 = 4 and 𝑦𝑦 = 3.

Check. We can check our answers by plugging them into the original equations.

3𝑥𝑥 +2𝑦𝑦 = 3(4)+2(3) = 12+6 = 18
8𝑥𝑥 −5𝑦𝑦 = 8(4)−5(3) = 32−15 = 17

Example 6. Use the method of substitution to solve the following system of equations for
each unknown.
4𝑦𝑦+3𝑧𝑧 = 10
5𝑦𝑦−2𝑧𝑧 = −22
Solution. First isolate 𝑦𝑦 in the top equation. Subtract 3𝑧𝑧 from both sides.
4𝑦𝑦 = 10−3𝑧𝑧
Divide both sides of the equation by 4.
10−3𝑧𝑧
𝑦𝑦 =
4
Substitute this expression in parentheses in place of 𝑦𝑦 in the bottom equation.
5𝑦𝑦−2𝑧𝑧 = −22
9
Chapter 1 – Review of Essential Algebra Skills

10−3𝑧𝑧
5� �−2𝑧𝑧 = −22
4
Distribute the 5.
10 3𝑧𝑧
5� �−5� �−2𝑧𝑧 = −22
4 4
50 15𝑧𝑧
− −2𝑧𝑧 = −22
4 4
15𝑧𝑧 50
Combine like terms: − 4 and −2𝑧𝑧 are like terms, and 4 and −22 are like terms. Combine
4
like terms using a common denominator. Multiply −2𝑧𝑧 and −22 each by 4 to make a
50
common denominator. In order to combine the constant terms, subtract 4 from both sides.
15𝑧𝑧 4 4 50
− −2𝑧𝑧� � = −22� �−
4 4 4 4
15𝑧𝑧 8𝑧𝑧 88 50
− − = − −
4 4 4 4
−15𝑧𝑧−8𝑧𝑧 −88−50
=
4 4
23𝑧𝑧 138
− = −
4 4
Multiply both sides of the equation by 4.
−23𝑧𝑧 = −138
Divide both sides of the equation by −23. The two minus signs will cancel.
𝑧𝑧 = 6
Now that we have an answer for 𝑧𝑧, we can plug it into one of the previous equations in
order to solve for 𝑦𝑦. It’s convenient to use the equation where 𝑦𝑦 was isolated.
10−3𝑧𝑧 10−3(6) 10−18 −8
𝑦𝑦 = = = = = −2
4 4 4 4
The answers are 𝑧𝑧 = 6 and 𝑦𝑦 = −2.

Check. We can check our answers by plugging them into the original equations.

4𝑦𝑦+3𝑧𝑧 = 4(−2)+3(6) = −8+18 = 10
5𝑦𝑦−2𝑧𝑧 = 5(−2)−2(6) = −10−12 = −22

Example 7. Use the method of substitution to solve the following system of equations for
each unknown.
3𝑥𝑥 −4𝑦𝑦+2𝑧𝑧 = 44
5𝑦𝑦+6𝑧𝑧 = 29
2𝑥𝑥 +𝑧𝑧 = 13
Solution. It would be easiest to isolate 𝑧𝑧 in the bottom equation. Subtract 2𝑥𝑥 from both
sides in the bottom equation.
𝑧𝑧 = 13−2𝑥𝑥
Substitute this expression in parentheses in place of 𝑧𝑧 in the top two equations.
10