AP Physics C – Mechanics Summer Assignmenthurrellphysics.com/APphysicsCMechFiles/APPhysicsCSummerAssignment.pdffor students majoring in the physical sciences or engineering. ...

AP Physics C – Mechanics

Summer Assignment 2018 – 2019 School Year

Welcome to AP Physics C, an exciting and intensive introductory college physics course for students

majoring in the physical sciences or engineering. Students must have prior exposure to physics topics

(i.e. Academic Physics), be proficient in both algebra and trigonometry, and have experience or be

concurrently enrolled in the AP Calculus AB course.

AP Physics C is comprised of two courses: Mechanics, which is equivalent to the first semester of a

university course; and Electricity & Magnetism, which is equivalent to the second semester of a

university course. During the 2018-2019 students will cover the Mechanics material, focusing on

Newtonian Mechanics to include:

• Kinematics

• Newton’s Laws of Motion

• Work-Energy-Power

• Linear Momentum

• Circular Motion

• Oscillations

• Gravity

While succeeding in AP Physics C is not entirely dependent upon being an expert mathematician, being

proficient with the practices and math skills addressed in this assignment will get you rolling in the right

direction.

This assignment is due the first day of school, is worth 100 pts, and your grade, teacher, parents, etc.…,

will love you for completing it.

Have a wonderful summer and I look forward to seeing you in late August. If you have any questions,

please feel free to contact me at [email protected]. I’ll do my best to get back to you as soon as

possible.

Mr. Hurrell

mailto:[email protected]

AP Physics C Summer Assignment 2018-2019 School Year

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Table of Contents Topic Page (s)

Conversions

Topic Introduction 3 – 6

Problems 7 – 9

Scientific Notation

Topic Introduction 10

Problems 11 – 12

The Metric/SI System


Problems 16

Manipulating Algebraic Equations


Problems 21 – 22

Significant Figures


Problems 25

Trigonometry Basis

Topic Introduction 26

Problems 27

Vector Basis


Problems 31


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Conversions

We perform conversions if we need to change the units of a quantity. For example, 1 foot equals 12

inches, the amount is the same, but how we express the amount is different.

To do a conversion you need a “conversion factor”.

Conversion factor nuts and bolts

• Conversion factors come from commonly used equalities

• Any equality can make two conversion factors

o Example 1

▪ Equality: 12 in = 1 ft

▪ Two conversion factors: (12 𝑖𝑛

1 𝑓𝑡) 𝑜𝑟 (

1 𝑓𝑡

12 𝑖𝑛)

o Example 2

▪ Equality: 1 hr = 60 min

▪ Two conversion factors: (1 ℎ𝑟

60 𝑚𝑖𝑛) 𝑜𝑟 (

60 𝑚𝑖𝑛

1 ℎ𝑟)

• Since anything divided by itself is 1, a conversion factor also equals 1.

Step-by-step Conversion Guide

Step # Description

1 Write what you are given as a fraction with one unit on the top (aka numerator) and one unit on the bottom (aka denominator)

2 In parenthesis and without numbers, write the units you want to discard diagonal from itself. In the other part of the fraction write what you are converting to.

3 Put numbers into the parenthesis so that the top equals the bottom. This will be a common equality that you know.

4 Cancel out the units but not the numbers

5 Do the math. Multiply the numbers if they are both on top. Divide if the second one is on the bottom.


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Single conversion examples

Example 1: Convert 55 mi/hr to mi/min

Step # Description Details


55 𝑚𝑖

1 ℎ𝑟


55 𝑚𝑖

1 𝒉𝒓(

𝒉𝒓

𝑚𝑖𝑛)

3 Put numbers into the parenthesis so that the top equals the bottom Common equality used: 1 hr = 60 min

55 𝑚𝑖

1 𝒉𝒓(

1 𝒉𝒓

60 𝑚𝑖𝑛)


55 𝑚𝑖

1 𝒉𝒓(

1 𝒉𝒓

60 𝑚𝑖𝑛)


55 𝑚𝑖

1 𝒉𝒓(

1 𝒉𝒓

60 𝑚𝑖𝑛) =

55 𝑚𝑖

60 𝑚𝑖𝑛= 𝟎. 𝟗𝟐

𝒎𝒊

𝒎𝒊𝒏

Example 2: Convert 25 m/s to ft/s

Step # Description Details


25 𝑚

1 𝑠


25 𝒎

1 𝑠(

𝑓𝑡

𝒎)

3 Put numbers into the parenthesis so that the top equals the bottom Common equality used: 3.3 ft = 1 m

25 𝒎

1 𝑠(

3.3 𝑓𝑡

1 𝒎)


25 𝒎

1 𝑠(

3.3 𝑓𝑡

1 𝒎)


25 𝒎

1 𝑠(

3.3 𝑓𝑡

1 𝒎) =

82.5 𝑓𝑡

1 𝑠= 𝟖𝟐. 𝟓

𝒇𝒕

𝒔


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Multiple conversion examples

Notes:

• When performing more than one conversion, you can do:

o Method 1: Perform each conversion independently

OR

o Method 2: Perform all of the conversions in one long “chain”

• Both methods will provide the same answer, but once you master the “chain” approach, I think

you will find it to be the most efficient

Convert 700 hours to weeks performing one conversion at a time

Conversion 1: (hours to days)

Note: Since our starting point is a single unit (hours), just write it over 1.

700 ℎ𝑟

1 (

1 𝑑𝑎𝑦𝑠

24 ℎ𝑟) = 29.17 𝑑𝑎𝑦𝑠

Conversion 2: (days to weeks)

29.17 𝑑𝑎𝑦𝑠

1 (

1 𝑤𝑒𝑒𝑘

7 𝑑𝑎𝑦𝑠) = 4.17 𝑤𝑒𝑒𝑘𝑠

Convert 700 hours to weeks using the “chain” method

700 ℎ𝑟

1 (

1 𝑑𝑎𝑦𝑠

24 ℎ𝑟) (

1 𝑤𝑒𝑒𝑘

7 𝑑𝑎𝑦𝑠) =

700 𝑤𝑒𝑒𝑘𝑠

24 ∗ 7= 4.17 𝑤𝑒𝑒𝑘𝑠


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Railroad Tracks Method (My Favorite)

• This method of converting units is sometimes called the “Factor Label” method, but engineers

prefer trains so they changed the name to the "Railroad Tracks" method (woo woo).

• Very similar to the “chains” method, just a different look.

• Draw up a set of "Railroad Tracks" as shown below

(Train is not required )

• With the original value in the top "row" of the first "column" and the original units in the top or

bottom row of this column just the way you would normally put them above or below the line.

• Then write each conversion factor into the following columns in such a way that the units cancel

out, leaving the units you want.

Note: Each time you add a conversion factor you are actually multiplying by 1.0 because the top

and bottom are equal, just in different units.

• Then multiply or divide all the conversion factors depending on whether they are above or

below the line.

Railroad Tracks Examples:

Convert 700 hours to weeks

700 hr 1 days 1 week = 700 weeks = 4.17 weeks 1 24 hr 7 days 24 * 7

Convert 25 m/s to mi/hr

25 m 1 mi 60 s 60 min = 25* 60 * 60 mi = 55.94 mi

s 1609 m 1 min 1 hr 1609 hr hr


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Perform the following conversions given the following equalities:

• 5280 ft = 1 mi

• 12 in = 1 ft

• 1 in = 2.54 cm (0.0254 m)

• 1 m = 3.3 ft

Note:

ft = feet

mi = miles

in = inches

cm = centimeters

m = meters

SHOW ALL WORK

(1) Convert 7.5 mi to ft

(2) Convert 7.5 mi to in

(3) Convert 5 ft to m

(4) Convert 7 weeks to days


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(5) Convert 3500 seconds to minutes

(6) Convert 22 m/sec to m/min

(7) Convert 60 mph (miles/hr) to m/hr

(8) Convert 180 m/min to m/hr

(9) Convert 15 in/min to ft/min


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(10) Convert 15 in/min to ft/sec

(11) Convert 650 cm/min to cm/sec

(12) Convert 650 cm/min to in/sec

(13) Convert 12 mph to m/s


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Scientific Notation Scientific Notation is a neat and compact way of writing large and small numbers while using only

significant digits (addressed in the next section) and powers of ten. The conventional form for scientific

notation is:

𝑵 × 10𝑷

Where:

N = a number in the form of x.xx (digit in the ones, tenths, and hundredths place)

P = the powers of ten telling you which direction to move the decimal place and how many times

• Examples

Expanded Form Scientific Notation Comments

0.00347 3.47 x 10-3 The negative three power of ten indicates that the decimal point should be moved three places to the left

324,000 3.24 x 105 The positive power of five indicates that the decimal point should be moved five places to the right. In this case, zeros are needed as placeholders

• When dividing, divide the decimals while keeping the correct number of significant figures.

When dividing powers of ten, subtract the bottom power of ten’s exponent from the top power

of ten’s exponent. If multiplying, add powers of ten.

9.6 × 107

3.2 × 104= 3.0 × 103

(3.2 × 104) ∙ (2.0 × 102) = 6.4 × 106

• Before adding or subtracting numbers in scientific notation, all numbers need to have the same

power of ten.

(4.5 × 10−2) + (8.2 × 10−3) = 4.5 × 10−2 + 0.82 × 10−2 = 5.3 × 10−2


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Write the following numbers in scientific notation:

(1) 5800 m

(2) 0.0345 mm

(3) 0.00890 kg

(4) 560 g

(5) 4,320,000 cm2

(6) 0.00065 m/s

(7) 101.35 m


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(8) (5.8 x 10-6) · (2 x 104)

(9) (0.8 x 104) · (1.28 x 106)

(10) (3.8 x 10-6) · (2.37 x 10-3)

(11) 5.3×103

7.65×105

(12) 4.6×102

5.01×10−3

(13) (4.0 × 103) − (2.0 × 102)

(14) (3.2 × 104) + (7.0 × 103)


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SI/Metric System

The International System of Units, universally abbreviated SI (from the French Le Système International

d’Unités), is the modern metric system of measurement. There are two quantities required for every

measurement:

• Dimension: The physical quantity being measured, such as length, mass, or force. The

dimension of a measurement is indicated by the unit following the number.

• Unit: This indicates the size of the measurement that is made, for example a mile is larger than

a foot which is larger than an inch.

The significant difference between the Metric system and British system of measurement is the way in

which they indicate the size of a measurement. In the British system, different units are used to

measure larger amounts. When an inch becomes too small a foot is used instead. In the metric system

of measurement, instead of changing the units, prefixes are applied to base units to indicate larger or

smaller amounts.

The SI system is a component of the metric system; the only difference is that the SI system uses certain

standard units for measurements. This is meant to better standardize the measurements throughout

different sciences. These are referred to as SI Units and are used to make all other units of

measurement in different combinations. The SI UNITS are shown below with the corresponding English

(aka British) for reference.


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All other quantities in the metric system are different combinations of SI units and are called derived

quantities. Below is a table listing some other quantities that are defined in terms of the seven base

units.


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Below is a list of some derived quantities that have special names

The common SI prefixes that will be used in AP Physics to form decimal multiples and submultiples of SI

units are provided below.


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Below is a complete list of SI prefixes.

Convert the following metric measurements:

1000 mg = _____ g

198g = _____ kg 8 mm = _____ cm

160 cm = _____ mm

75 mL = _____ L 6.3 cm = _____ mm

109 g = _____ kg

50 cm = _____ m 5.6 m = _____ cm

250 m = _____ km

5 L = _____mL 26,000 cm = _____ m

14 km = _____ m

16 cm = _____mm 56,500 mm = _____ km

1 L = _____ mL

65 g = _____ mg 27.5 mg = _____ g

480 cm = _____ m

2500 m = _____ km 923 cm = _____ m

27 g = _____ kg

355 mL = _____ L 0.025 km = _____ cm


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Manipulating Algebraic Equations

The types of algebraic equations you will be manipulating in AP Physics are ones expressed in terms of

variable symbols (such as d, v, and a) and constants (such as r, g, and π). Often in science and

mathematics you are given an equation and asked to solve it for a particular variable symbol or letter

called the unknown. The symbols which are not the particular variable we are interested in solving for

are called literals, and may represent variables or constants. Literal equations are solved by isolating the

unknown variable on one side of the equation, and all of the remaining literal variables on the other side

of the equation. Sometimes the unknown variable is part of another term. A term is a combination of

symbols such as the products ma or πr2. In this case the unknown (such as r in πr2) must be factored out

of the term before we can isolate it.

The following rules, examples, and exercises will help you review and practice solving literal equations.

PROCEDURE

Note: In general, we solve a literal equation for a particular variable by following the basic procedure

below.

1. Recall the conventional order of operations, that is, the order in which we perform the

operations of multiplication, division, addition, subtraction, etc.:

a. Parenthesis

b. Exponents

c. Multiplication and Division

d. Addition and Subtraction

This means that you should do what is possible within parentheses first, then exponents, then

multiplication and division from left to right, then addition and subtraction from left to right. If

some parentheses are enclosed within other parentheses, work from the inside out.

2. If the unknown is a part of a grouped expression (such as a sum inside parentheses), use the

distributive property to expand the expression.

3. By adding, subtracting, multiplying, or dividing appropriately,

a. move all terms containing the unknown variable to one side of the equation, and

b. move all other variables and constants to the other side of the equation. Combine like

terms when possible.

4. Factor the unknown variable out of its term by appropriately multiplying or dividing both sides

of the equation by the other literals in the term.

5. If the unknown variable is raised to an exponent (such as 2, 3, or ½), perform the appropriate

operation to raise the unknown variable to the first power, that is, so that it has an exponent of

one.


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EXAMPLES

1. F = ma [Solve for a]

F = ma

Step 1: Divide both sides by m

𝐹

𝑚= 𝒂

Step 2: Since the unknown variable (in this case a) is usually placed on the left side of the

equation, we can switch the two sides:

𝒂 =𝐹

𝑚

2. 𝑣 =𝑑

𝒕 [Solve for t]

𝑣 =𝑑

𝒕

Step 1: Multiply each side by t

𝒕𝑣 = 𝑑

Step 2: Divide both sides by v

𝒕 =𝑑

𝑣

3. 𝑈 =1

2𝑘𝒙2 [Solve for x]

𝑈 =1

2𝑘𝒙2

Step 1: Multiply both sides by 2

2𝑈 = 𝑘𝒙2

Step 2: Divide both sides by k

2𝑈

𝑘= 𝒙2


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Step 3: Take the square root of both sides (For many physics problems the positive root is the

only practical solution, therefore we will ignore the negative root in this example)

√2𝑈

𝑘= 𝒙

𝒙 = √2𝑈

𝑘

4. 𝑇 = 2𝜋√𝑳

𝑔 [Solve for L]

𝑇 = 2𝜋√𝑳

𝑔

Step 1: Divide both sides by 2π

𝑇

2𝜋= √

𝑳

𝑔

Step 2: Square both sides

𝑇2

4𝜋2 =𝑳

𝑔

Step 3: Multiply both sides by g

𝑔𝑇2

4𝜋2 = 𝑳

𝑳 =𝑔𝑇2

4𝜋2


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5. 𝐹 =𝐺𝑚1𝑚2

𝒓2 [Solve for r]

𝐹 =𝐺𝑚1𝑚2

𝒓2

Step 1: Multiply both sides by r2

𝐹𝒓2 = 𝐺𝑚1𝑚2

Step 2: Divide both sides by F

𝒓2 =𝐺𝑚1𝑚2

𝐹

Step 3: Take the square root of both sides

𝒓 = √𝐺𝑚1𝑚2

𝐹

6. 𝐹 = 𝑞𝑣𝐵 sin 𝜽 [Solve for Ɵ]

𝐹 = 𝑞𝑣𝐵 sin 𝜽

Step 1: Divide both sides by qvB

𝐹

𝑞𝑣𝐵= sin 𝜽

Step 2: Take the inverse sine of both sides

𝜽 = sin−1 (𝐹

𝑞𝑣𝐵)


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PROBLEMS

For each of the following equations, solve for the variable in bold print. Be sure to show each step

you take to solve the equation for the bold variable.

1. v = at [Solve for a]

2. 𝑃 =𝐹

𝑨 [Solve for A]

3. 𝜆 =𝒉

𝑝 [Solve for h]

4. 𝐹(𝚫𝒕) = 𝑚∆𝑣 [Solve for ∆t]

5. 𝑈 =𝐺𝒎𝟏𝑚2

𝑟 [Solve for m1]

6. 𝐶 =5

9(𝑭 − 32) [Solve for F]


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7. 𝑥 = 𝑥𝑜 + 𝑣𝑜𝑡 +1

2𝒂𝑡2 [Solve for a]

8. 𝑛1 sin 𝜃1 = 𝑛2 sin 𝜽𝟐 [Solve for Ɵ2]

9. 𝑚𝑔 sin 𝜽 = 𝜇𝑚𝑔 cos 𝜽 (𝑀+𝑚

𝑚) [Solve for Ɵ]

10. 𝑣𝑟𝑚𝑠 = √3𝑅𝑇

𝑴 [Solve for M]

11. 𝐹 =1

4𝜋𝜀𝑜∙

𝐾𝑞1𝑞2

𝒓2 [Solve for r]

12. 1

𝑠𝑖+

1

𝑠𝑜=

1

𝒇 [Solve for f]


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Significant Figures

• Significant figures: More simply known as “number of digits”, represents to what certainty a

quantity is known

• Rules of the road

o Leading zero

▪ When writing a number using decimal notation where all the significant figures

are to the right of the decimal point always include one zero to the left of the

decimal point

• CORRECT 0.00123

• INCORRECT .00123

▪ How to determine how many significant digits there are in a number

• Rules

o The leftmost digit which is not a zero is the most significant digit

o If the number does not have a decimal point, the rightmost digit

which is not a zero is the least significant digit

o If the number does have a decimal point, the rightmost

significant digit is the least significant digit, even if it's a zero

o Every digit between the least and most significant digits should

be counted as a significant digit

▪ Examples: The following numbers all have three significant digits

• 123

• 12.3

• 1.23 × 106

• 1.00

• 0.000123

o When Adding or Subtracting numbers

▪ Steps

• 1: Perform the indicated mathematical operations

• 2: Find the number which is known to the fewest decimal places

• 3: Round the result to that decimal place

▪ Examples

• 1: 21.398 + 405 - 2.9 = 423

• 2: 2355.2342 + 23.24 = 2378.47

• 3: 15600.00 + 172.49 = 15772.49

• 4: 153 + 1.8 + 9.16 = 164


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o When Multiplying or Subtracting numbers

▪ Steps

• 1: Perform the indicated mathematical operations

• 2: Find the number with the fewest significant figures

• 3: Result will have the same number of significant figures as is the

number with the fewest significant figures.

▪ Examples

• 1: 13.1 x 2.25 = 29.5

• 2: 13.10 x 2.25 = 29.5

• 3: 13.100 x 2.2500 = 29.475

• 4: 15310 x 2.3 = 35000

• 5: 1.00 x 10.04 = 10.0


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PROBLEMS

How many significant figures in the following numbers?

1. ______ 1,245m 2. _______ 0.030m 3. _______ 10,000m

4._______ 1.340 x 1023m 5. _______ 3.02003 x 1014m 6. _______ 0.0000001m

7. _______ 1,000. 8. _______ 0.10000010

Problems 9 – 13: Perform the following Calculations and record your answers in the proper number of

significant figures.

9. 0.3030 + 0.42 =

10. 2.1m + 0.3889m – 123.22m =

11. 3.1567 x 102 + 9.212 x 104 - 4.677 x 106 =

12. √1.33 𝑥 1053 =

13. 0.303

1.0 𝑥 10−3 =

14. 1.25 𝑥 10−3

(3.2+10+4.9)(1.3 𝑥 10−6)


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Trigonometry Basics

• The focus of this packet will be right-triangle trigonometry

• Examine the right triangle pictured below

o Remember a right triangle has a 90° angle and that the sum of all of the angles in any

triangle is equal to 180°

o The two short sides of this right triangle have been labeled A and B

o The longest side of a right triangle is known as the hypotenuse.

o The right angle is marked and the other two angles are marked with ‘a’ and ‘b’.

o Notice that the right angle is opposite to the hypotenuse and that angle ‘a’ is opposite

to side A and angle ‘b’ is opposite to side B.

o We use the Pythagorean Theorem to calculate the length of any side of a right triangle

when the length of the other two sides is known

Pythagorean’s Theorem

𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒2 = 𝐴2 + 𝐵2

o We use trigonometric relationships to calculate the length of any side of a right triangle

when the length of one side and one angle is known.


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PROBLEMS

Your calculator must be in degree mode! Show all your work.

1. Ɵ = 55° and c = 32, solve for a and b

2. Ɵ = 45° and a = 15, solve for b and c

3. B = 17.8 m and Ɵ = 65°, solve for a and c


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Vector Basics

• General Information

o Some quantities can be described with a single number (with units) giving its size or

magnitude. Such quantities are called scalar quantities. Examples of scalar quantities are

time, temperature, and mass.

o Many quantities not only have a magnitude but also a direction. Such quantities are

called vectors. An example of a vector quantity is displacement. Displacement describes

how far you’ve traveled and in which direction you have traveled. For example, a car has

traveled 2 km due east. Other vector quantities are velocity, acceleration, and force.

o An arrow is used to represent a vector. 2 km due east The length of the arrow

represents the magnitude and the which way the arrow points is the direction of the

quantity.

• Sign Conventions

o Positive and negative signs are typically used to indicate the direction of a vector

mathematically. (In Physics, positive and negative signs do NOT mean positive or

negative numbers, as in a number line.)

▪ Positive sign: to the east or north (right or up)

▪ Negative sign: to the west or south (left or down)

• Adding Vectors

o Head-Tail Method

▪ Graphically adding vectors

▪ Place the tail of the 2nd vector at the head of the 1st vector


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o Adding Mathematically

▪ Vectors that are in the same direction

o Vectors that are perpendicular to each other

Note: Use methods presented in the “Trigonometry Basics” section of this packet


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o Vectors That Are Not Perpendicular to Each Other

▪ All vectors have an x-component and a y-component.

▪ Whenever adding vectors that are not going in the same direction or not

perpendicular to each other, you must determine the x-component and y-

component of each vector and add their components together.


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PROBLEMS

1. Determine the x and y components of each of the force vectors shown below.

2. Are the following quantities vectors or scalars? Explain.

a. The cost of a theater ticket

b. The current in a river

3. Write each combination of vectors as a single vector. See the figure below.

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