Physics Primer

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 1/42 Physics Primer Due: 11:59pm on Friday, September 22, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy A Welcome to the Physics Primer! Students who successfully complete this unit will be able to: Describe the purpose of the Physics Primer. What is in the Physics Primer? The Physics Primer is not a comprehensive online mathematics textbook. It is a focused set of mathematics topics that can give new physics students trouble. Remember, mathematics is only part of the physics problem-solving process, but it is a part that often trips up students. Launch the short video called "Thinking Like a Physicist" to learn more about how the math skills covered in this primer relate to success in a physics course. The Physics Primers include: Learning goals and the necessary prerequisite math skills (if applicable). Succinct explanations of the topic that are pertinent to a physics course. The proper amount of math needed to apply to relevant physics topics. Hints and feedback that are based on common mistakes that students make. Short video presentations that supplement some of the topics. How do you use the Physics Primer? Your physics instructor will assign some, or even all, topics based on the needs of your specific course. These may be assigned at the beginning of your course or throughout your course as appropriate. When beginning one of the topics, carefully read through the introductory material before attempting the problems that follow. If you encounter trouble along the way, don't hesitate to go back and re-read the material as well as take advantage of the available hints, which are there to help guide you in the right direction. In addition, depending on the topic, there may be an accompanying video that goes into further detail with a worked example, so be sure to utilize them as well! Click here to watch a video on the relationship between physics and mathematics. Part A - Applying what you just learned What is the Physics Primer? Hint 1. Locating information

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 2/42 Read the introductory information to help you answer this question. ANSWER: Correct Once you master all of the mathematics in the Physics Primer, you will be better prepared to solve physics problems and apply physics concepts to real world situations. But remember, there is more to physics and physics problem solving than doing mathematics. Scientific Notation Students who successfully complete this primer will be able to: Convert both very large and very small numbers into scientific notation. Convert numbers from scientific notation into regular notation. Multiply and divide numbers written in scientific notation. Before working on this primer, you may need to review: Writing values as powers of 10 For additional practice, you may want to review Conversion to Scientific Notation . Why write numbers in scientific notation Understanding physics will help you describe in great detail how the world works. However, the characteristics of the world span an enormous range of numbers. The mass of an electron is 0.000000000000000000000000000000911 kg (incredibly tiny!). The mass of the Sun is 1,989,000,000,000,000,000,000,000,000,000 kg (incredibly massive!). If you had to write, interpret, or use these numbers as written, you could very easily end up making a minor math error. Scientific notation was invented to easily express numbers that are too large or too small to be conveniently written in decimal form. The advantages of scientific notation are it 1) is compact, 2) helps articulate significant figures, and 3) works with numbers of any size. The traditional format for numbers written in scientific notation is m × 10 n is where m is a number between 1 and 10 and n is an integer (either positive or negative). For the examples above, the mass of an electron can be written as 9.11 × 10 -31 kg and the mass of the Sun can be written as 1.989 × 10 30 kg . Converting decimals to scientific notation A day has 86,400 second (s) . Notice that the decimal point is to the far right of the number. The first part of scientific notation is a number between 1 and 10. Move the decimal point to the left to obtain 8.64, which equals m in our traditional notation. The value for n is given by the number of places the decimal point was moved from its location in 86,400 to 8.64. Thus, 86,400 s = 8.64 × An detailed explanation of all of the mathematics you will use in your introcuctory physics course. A tutorial of the most important physics concepts covered in introductory physics courses. A list of the most important formulas used in introductory physics courses. A set of mathematics topics that are relevant to introductory physics courses.

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 4/42 For a value greater than 1, the exponent is positive. ANSWER: Correct You had to move the decimal point 11 places to the left to obtain 1.5. Part C - Write this small number in scientific notation. Determine the values of m and n when the following charge of a proton is written in scientific notation: 0.000000000000000000160 C . Enter m and n , separated by commas. Hint 1. Moving the decimal point Move the decimal point to the right so you end up with a number between 1 and 10. That's the value for m . Hint 2. Finding n Count the number of place values you moved the decimal point. Hint 3. Sign of the exponent For a value between 0 and 1, the exponent is negative. ANSWER: Correct You had to move the decimal point 19 places to the right to obtain 1.60. Part D - Write this small number in scientific notation. Determine the values of m and n when the following average magnetic field strength of the Earth is written in scientific notation: 0.0000451 T . Enter m and n , separated by commas. Hint 1. Moving the decimal point Move the decimal point to the right so you end up with a number between 1 and 10. That's the value for m . Hint 2. Finding n Count the number of place values you moved the decimal point. Hint 3. Sign of the exponent m , n = 1.5,11 m , n = 1.60,-19

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 5/42 For a value between 0 and 1, the exponent is negative. ANSWER: Correct You had to move the decimal point 5 places to the right to obtain 4.51. Rule for multiplying numbers in scientific notation When multiplying two numbers of the form m × 10 n , the product is: ( m 1 × 10 n 1 )( m 2 × 10 n 2 ) = m 1 m 2 × 10 n 1 + n 2 . Part E - Practice multiplying numbers in scientific notation The mass of a high speed train is 4.5 × 10 5 kg , and it is traveling forward at a velocity of 8.3 × 10 1 m/s . Given that momentum equals mass times velocity, determine the values of m and n when the momentum of the train (in kg m/s ) is written in scientific notation. Enter m and n , separated by commas. Hint 1. Rule for multiplying orders of magnitude Multiply the orders of magnitude by adding the exponents. Hint 2. Rule for finding m Don’t forget that m must be between 1 and 10. ANSWER: Correct Rule for dividing numbers in scientific notation When dividing two numbers of the form m × 10 n , the quotient is: ( m 1 × 10 n 1 ) ÷ ( m 2 × 10 n 2 ) = m 1 m 2 × 10 n 1 − n 2 . Part F - Practice dividing numbers in scientific notation Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of light to get from the Sun to the Earth (in s ) is written in scientific notation. Note: the speed of light is approximately 3.0 × 10 8 m/s . Enter m and n , separated by commas. Hint 1. Sun to Earth distance m , n = 4.51,-5 m , n = 3.7,7

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 7/42 Change sign key (note that on some calculators the change sign key looks very similar to the subtraction key; these two keys do NOT do the same thing so you must be careful to distinguish between them) Square root ( √ ) , square ( x 2 ) , inverse (1/ x or x − 1 ) , and exponent ( x y or ) operations. You may need to use a shift (or 2nd ) key on the calculator for some of these operations. Changing the angle units between degrees (deg) and radians (rad) (some calculators display the angle units currently set, others don’t, although you should be able to change these units in your calculator’s settings) Trigonometric functions: sine (sin) , cosine (cos) , and tangent (tan) , as well as inverse trigonometric functions: arcsine (arcsin , or sin − 1 ) , arccosine (arccos , or cos − 1 ) , and arctangent (arctan , or tan − 1 ) . You may need to use a shift (or 2nd) key for some of these operations. (Note: make sure you must use the correct setting of angle units for both trigonometric functions and inverse trigonometric functions) Scientific notation: you need to know the key that you use to enter numbers in scientific notation (e.g. E , EE , or EXP ), and the way in which your calculator displays numbers that are expressed in scientific notation. Other keys that may also be useful: parentheses (which can be used to group terms and reduce precedence of operation errors), constants (such as π ), memory to store intermediate results, logarithm (log) and natural logarithm (ln) , statistics ( STAT ), and more. Learn your calculator (example 1) Let's evaluate the quantity ˉ v = 15 meter + 5 meter 2 second + 3 second . You can also do this evaluation in your head. That’s an important strategy for verifying that you are using your calculator correctly! Let’s apply the one-step-at-a-time strategy. First, perform the addition in the numerator using the calculator: ˉ v = 20 meter 2 second + 3 second . Then, use the calculator to perform the addition in the denominator: ˉ v = 20 meter 5 second . Finally, use the calculator to perform the division to obtain the correct final numerical answer: ˉ v = 4 meter/second . Note that if you enter the values and operations into your calculator in exactly the same order that you read them in the expression, 15 + 5 ÷ 2 + 3 , your calculator will display the wrong answer for the expression. This would be a precedence of operation error! An alternative approach in a case like this is to place parentheses around the terms in the numerator and denominator separately in your calculator: (15 + 5) ÷ (2 + 3) . When entered this way, your calculator, which follows the order of operations, gives the correct answer. Now try a problem on your own. Part A - Average acceleration Use your calculator to evaluate ˉ a = − 3.7 meter / second − 13.9 meter / second 21.4 second − 7.2 second . Hint 1. Change sign vs subtraction key Make sure you identify which calculator key is the change sign key and which key performs the subtraction operation. Verify that you are using the change sign key correctly by performing operations with it for which you know the answer. Hint 2. One-step-at-a-time Use the one-step-at-a-time strategy (or properly utilize parentheses)

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 8/42 Hint 3. Verify proper use of the calculator Verify that you are using the calculator correctly by using it with the example above and making sure you get the correct answer. ANSWER: Correct When you are unsure about the operation of your calculator, verify it using operations with simple numbers for which you know the answer. Learn your calculator (example 2) Imagine we wish to evaluate the quantity 1 2 (1.67 × 10 − 27 kilogram)(3.00 × 10 5 meter/second) 2 . Start by verifying the operations of your calculator to work with scientific notation with an expression for which you know the answer: 2 × (5 × 10 1 ) , and then verify the squaring operation of your calculator with 3 2 . Finally, once you've completed this, use your calculator to evaluate the original expression, which should come out to 7.52 × 10 − 17 kilogram meter 2 /second 2 . Learn your calculator (example 3) Imagine we wish to evaluate the quantity 2 × 4.71 meter second 2 × 12.9 meter . Start by verifying the square root operation with √ 100 . On some calculators, you enter the square root operator first and then the number (sometimes followed by a closed parenthesis), while on others you enter the number first and then the square root operator. The same may be true for other operations as well (e.g. trigonometric and inverse trigonometric functions, inverse functions, and so on). Make sure you learn how your calculator works with these important functions and operators! Remember, also, units are central to physics –propagate them into your answer . Finally, once you've completed this, use your calculator to evaluate the original expression, which should come out to 11.0 meter/second . Try another problem now on your own. Part B - Astronomy calculation In analyzing distances by applying the physics of gravitational forces, an astronomer has obtained the expression R = 1 1 149 × 10 9 meter 2 − 1 298 × 10 9 meter 2 Use your calculator to evaluate R . Hint 1. Verify proper use of the calculator −0.718 meter/second 2 −1.24 meter/second 2 11.5 meter/second 2 0.718 meter/second 2 √ √ ( ) ( )

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 10/42 Make sure you verify how your calculator performs the necessary operations using it to calculate expressions for which you know the answer. Hint 3. One-step-at-a-time Use the one-step-at-a-time strategy, where one step will be to multiply θ by 2 before you evaluate the sine function. ANSWER: Correct When you are unsure about the operation of your calculator, verify it using operations with simple numbers for which you know the answer. Also, check the angle units of your calculator before using trigonometric and inverse trigonometric functions. Part D - Capstone problem A group of physics students hypothesize that for an experiment they are performing, the speed of an object sliding down an inclined plane will be given by the expression v = 2 gd (sin( θ ) − μ k cos( θ )) . For their experiment, d = 0.725 meter , θ = 45.0 , μ k = 0.120 , and g = 9.80 meter/second 2 . Use your calculator to obtain the value that their hypothesis predicts for v . Hint 1. Inserting values into the equation Write down the expression with the given values and units replacing the variables. Hint 2. Verify proper use of the calculator Make sure you verify how your calculator performs the necessary operations using it calculate expressions for which you know the answer. Hint 3. One-step-at-a-time Use the one-step-at-a-time strategy. ANSWER: 69.5 meter 31.5 meter 66.7 meter 10.5 meter √

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 11/42 Correct When you are unsure about the operation of your calculator, verify it using operations with simple numbers for which you know the answer. Unit Conversions Students who successfully complete this primer will be able to: Identify factors associated with physically important metric prefixes. Convert units between quantities possessing simple/compound units, multiple conversion factors, and metric prefixes. Before working on this primer you may need to review: Scientific notation For additional practice, you may want to review Conversion Factors . Overview of unit conversion steps 1. Identify the equality relationship between the quantities that are being converted (e.g. 1 mile = 5280 foot ). Metric prefixes (see the table below) are just a shorthand for an equality relationship between units, and are based on powers of 10 (e.g. 1 millimeter = 10 − 3 meter ) 2. Use the equality relationship to obtain a conversion factor that is a ratio of the units. Since the numerator and denominator of a conversion factor are equal, it will always have a value of exactly 1. Therefore, when you multiply a quantity by a conversion factor, it will not change its inherent value, rather just the associated units. The conversion factor can have either of the units in the numerator, with the other in the denominator. The placement of the units depends on the quantity being converted . (e.g. 5280 foot 1 mile or 1 mile 5280 foot ) 3. Multiply the quantity to be converted by the appropriate factor obtained in the step above. If the same unit appears in the numerator and denominator of a fraction, it cancels. Therefore, write your conversion factor such that the unit being converted cancels out. Be sure to write these conversion expressions with both the numerical quantities and the units so that they can be correctly evaluated (e.g. 14, 411 foot × 1 mile 5280 foot = 2.7294 mile ) 4. To convert quantities with compound units, apply multiple factors using the steps above. To check that the units being converted have all cancelled out and only the desired units remain, you may find it helpful to cross off the cancelled units along the way (e.g. 1.20 mile minute × 5280 foot 1 mile × 1 minute 60 second = 106 foot second or, for units raised to a power 7.29 × 10 6 millimeter 2 × 10 − 3 meter 1millimeter × 10 − 3 meter 1millimeter = 7.29 meter 2 ) 6.25 meter/second 2.97 meter/second 3.15 meter/second 2.43 meter/second

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 13/42 Part B - Converting compound units You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram meter 3 and the density of silicon in other units: 2.33 gram centimeter 3 . You decide to convert the density of silicon into units of kilogram meter 3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram centimeter 3 to perform the unit conversion? Hint 1. Units raised to a power? When converting quantites that have some units raised to a power, the conversion factor for those units must be applied once for each power. Each power of the unit must be converted. ANSWER: Correct Part C - Capstone unit conversion You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the base need to know the acceleration of a freely falling object at the surface of the planet in order to adequately design the structures. The Omicronians have told you that the value is g OP7 = 7.29 flurg grom 2 , but your architects use the units meter second 2 , and from your previous experience you know that both the Omicronians and your architects are terrible at unit conversion. Thus, it's up to you to do the unit conversion. Fortunately, you know the unit equality relationships: 5.24 flurg = 1 meter and 1 grom = 0.493 second . What is the value of g OP7 in the units your architects will use, in meter second 2 ? ANSWER: Correct 1 kilogram 10 3 gram × 1 centimeter 10 − 2 meter × 1 centimeter 10 − 2 meter × 1 centimeter 10 − 2 meter 1 kilogram 10 3 gram × 1 centimeter 10 − 2 meter 10 3 gram 1 kilogram × 10 − 2 meter 1 centimeter × 10 − 2 meter 1 centimeter × 10 − 2 meter 1 centimeter 10 3 gram 1 kilogram × 1 centimeter 10 − 2 meter g OP7 = 5.72 meter second 2

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 14/42 Working with Fractions Students who successfully complete this primer will be able to: Add, subtract, multiply and divide physical quantities that are expressed as fractions. Evaluate the numerical value of physical quantities expressed as fractions. For additional practice, you may want to review Adding Fractions . Problem solving with variables First, it is worth pointing out the concept of solving problems using variables. Successful physics students develop the habit of performing algebraic operations when solving a problem using variables (i.e. symbols) as much as possible, rather than substituting in numerical values given in a problem and solving . There are a several very important reasons for this. Performing the algebraic operations to solve a problem using variables is easier, faster, and less prone to errors. Students who haven’t developed the habit of using variables sometimes don’t believe this because they are less familiar with the approach, but with just a little effort and practice using variables rather than numerical values, they will find it true. can provide some very easy to use and important checks that a problem has been solved correctly. gives results that are much more informative than results found by early substitution of the numerical values. Overview of working with fractions Below are some important properties to keep in mind related to fractions: Fraction notation: Numerator Denominator Units in fractions: units in fractions are treated similarly to variables or numerical values with respect to multiplication, division and cancellation, and it is very important to correctly carry out the operations with the units as well as the variables or numerical values. Multiplying two fractions: The numerator of the resulting fraction is the product of the numerators of the two fractions, while the denominator of the resulting fraction is the product of the denominators of the two fractions. (e.g. a g m 1 m 2 = a m 1 g m 2 ) Note that a quantity that is not written in fraction form can be easily written in fraction form by writing the quantity as the numerator of a fraction whose denominator is 1. Example 1 : m v 2 r = m 1 v 2 r = mv 2 r Example 2 : 2 kilogram 2 meter 7 second = 4 kilogram meter 7 second Dividing one fraction by another: Invert the fraction in the denominator and then multiply it by the fraction in the numerator. Example 1 : 11 meter 15 second 2 second 5 = 11 meter 15 second 5 2 second = 55 meter 30 second 2 = 11 meter 6 second 2 Example 2 : x t 1 t 2 = x t 1 t 2 1 = x t 1 1 t 2 = x t 1 t 2 Adding or subtracting two fractions: Multiply each fraction by a factor that equals 1 so that they have the same denominator (i.e., find a common denominator). A simple and general way to do this is to multiply the first fraction by a fraction that has the denominator of the second fraction in both its numerator and denominator, and multiply

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 16/42 Correct A key step in adding or subtracting fractions is to obtain a common denominator. Part B - Finding acceleration (II) Next, based on the correct answer from Part A, which of the following is the correct symbolic expression for a ? Hint 1. Multiplying fractions To multiply two fractions, the result is a fraction whose numerator is the product of the two numerators, and whose denominator is the product of the two denominators. Hint 2. Combining fractions (denominator) If two fractions are being added and they have a common denominator, the denominator of the result will be the common denominator. Hint 3. Combining fractions (numerator) If two fractions are being added and they have common denominator, the numerator of the result will be the sum of the numerators. ANSWER: Correct To multiply two fractions, the result is a fraction whose numerator is the product of the two numerators and whose denominator is the product of the two denominators. If two fractions are being added or subtracted and they have a common denominator, the denominator of the result will be the common denominator, and the numerator of the result will be the sum or difference of the numerators. Part C - Finding acceleration (III) Finally, what is the numerical vaue of a ? Hint 1. Units in fractions Units in fractions are treated similarly to variables or numerical values with respect to multiplication, division and cancellation. a = m ( a 1 + F ) 2 m a 1 + F m ma 1 + F m ma 1 + F 2 m

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 17/42 ANSWER: Correct Units in fractions are treated similarly to variables or numerical values with respect to multiplication, division and cancellation. Correctly evaluating units in expressions can be as important as evaluating numerical values. Part D - Capstone problem A student solving a physics problem for the velocity of an object has applied appropriate physics principles and obtained the expression v = d t − v 0 , where d = 35.0 meter , t = 9.00 second , and v 0 = 3.00 meter/second . What is the velocity v ? ANSWER: Correct Simplifying Algebraic Expressions Students who successfully complete this primer will be able to: Understand that expert physics problem-solvers associate quantities with variables and solve for unknowns in terms of these variables. Simplify algebraic relationships for expressions that are commonly encountered in physics. For additional practice, you may want to review Rearrangement of Algebraic Expressions . It is widely believed that it is easier to work physics problems by substituting numerical values in at the very beginning of the problem. However, expert physics problem-solvers will typically solve physics problems using variables, and substitute known a = 4.71 meter/second 2 4.71 kilogram meter/second 2 4.71 kilogram 22.7 kilogram meter/second 2 22.7 meter/second 2 32.0 meter 9.00 8.00 meter 9.00 second 32.0 meter 9.00 second 312 second 9.00

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 19/42 Pull out common factors. ANSWER: Correct Technique 3 : Reduce compound fractions (note: compound fractions are fractions that have a fraction in the numerator and/or the denominator). Example : Assume that the following expression is derived by analyzing the physics of a particular system: t = v F m . By reducing the compound fraction, the expression for t can be simplified as: t = mv F . Technique 4 : Consolidate and/or divide out factors (sometimes referred to as "cancelling") if they occur multiple times in a single expression. Example : Assume that the following equation is derived by analyzing the physics of a particular system: d = √ 2 ax 2 x a . By consolidating and canceling variables, the expression for d can be simplified as: d = 2 x . Part C Imagine you derive the following expression by analyzing the physics of a particular system: M = ( mv 2 r ) ( mG r 2 ) . Simplify the expression for M using the techniques mentioned above. Hint 1. Dividing by a fraction Dividing by a fraction is performed by inverting it and then multiplying by it. ANSWER: a = g (sin θ − μ k cos θ ) a = (9.80 meter/second 2 )sin θ − μ k (9.80 meter/second 2 )cos θ a = g sin θ − μ k gcos θ (no simplification should be performed on the expression in this situation) √

12/7/23, 12:11 AM Physics Primer https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10768245 20/42 Correct For some physical systems there can be important cancellations – in this case m – that will simplify not only expressions, but what you need to know to obtain numerical answers to a problem. Part D A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μ k mg cos θ = mg sin θ − ma , where g = 9.80 meter/second 2 , a = 3.60 meter/second 2 , θ = 27.0 , and m is not given. Which of the following represents a simplified expression for μ k ? ANSWER: Correct Solving problems using variables, simplifying expressions, and substituting in known values only at the final step is faster, easier, less mistake-prone, and provides better insight into how the variables relate to each other. In this last problem, the variable m ends up canceling out, and so ultimately it isn't needed to numerically evaluate the expression. Physically, this tells us that the answer does not depend on the variable m . This important realization might have been overlooked if one simply approached the problem with a plug and chug mindset. Solving Two Equations and Two Unknowns Students who successfully complete this primer will be able to: Identify situations that involve solving two equations and two unknowns. Use substitution to obtain one equation and one unknown. Simplify the above result and solve for both of the unknowns. M = ( mv 2 r ) ( mG r 2 ) (no simplification should be performed on the expression in this situation) M = v 2 r G M = m 2 v 2 G r 3 g sin θ − a g cos θ To avoid making mistakes, the expression should not be simplified until the numerical values are substituted. tan θ − a g The single equation has two unknowns and cannot be solved with the information given.