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Chapter 1-7 MASTER NOTES MASTER CLASS NOTES FOR Chapter 1,2,3,5,6 & 7.

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Presentation on theme: "Chapter 1-7 MASTER NOTES MASTER CLASS NOTES FOR Chapter 1,2,3,5,6 & 7."— Presentation transcript:

1 Chapter 1-7 MASTER NOTES MASTER CLASS NOTES FOR Chapter 1,2,3,5,6 & 7

2 Chapter 1 Table of Contents Chapter 1 The Science of Physics
Chapter 2 Motion in One Dimension Chapter 3 Two- Dimensional Motion and Vectors Chapter 4 Forces and the Laws of Motion Chapter 7 Rotational Motion and the Law of Gravity

3 Chapter 1 The Topics of Physics
Section 1 What is Physics Chapter 1 The Topics of Physics Physics is simply the study of the physical world.

4 Chapter 1 The areas of Physics
Section 1 What is Physics Chapter 1 The areas of Physics 1. Mechanics - The study of motion and its causes. Falling objects, friction, weight, spinning objects. 2. Thermodynamics – The study of heat and temperature. Melting and Freezing processes, engines, refrigerators. 3. Vibration and Wave Phenomena – The study of specific types of repetitive motion. Springs, pendulums, sound

5 The areas of Physics (cont)
Section 1 What is Physics Chapter 1 The areas of Physics (cont) 4. Optics – The study of light. Mirrors, lenses, color, astronomy 5. Electromagnetism – The study of electricity, magnetism, and light. Electrical charge, circuitry, permanent magnets, electromagnets. 6. Relativity – The study of particles moving at any speed, including very high speed. Particle collisions, particle accelerators, nuclear energy.

6 The areas of Physics (cont.)
Section 1 What is Physics Chapter 1 The areas of Physics (cont.) 7. Quantum Mechanics – The study of submicroscopic particles. The atom and its parts

7 Chapter 1 Types of observations
Qualitative- descriptive, but not true measurements Hot Large Quantitative- describe with numbers and units 100C 15 meters

8 Chapter 1 The Scientific Method Section 1 What is Physics
The scientific method is a way to ask and answer scientific questions by making observations and doing experiments. Steps of the scientific : Observation (Ask a Question) Collect Data (Do Background Research) Construct a Hypothesis (Educated guess) Test Your Hypothesis by Doing Experiments Analyze Your Data and Draw a Conclusion The conclusion is only valid if it can be verified by other people. Communicate Your Results

9 Section 1 What is Physics?
Chapter 1 The Scientific Method

10 The Scientific Method (cont)
Section 1 What is Physics Chapter 1 The Scientific Method (cont) System – A set of items or interactions considered a distinct physical entity for the purpose of study. Decide what to study and eliminate everything else that has minimal or no effect on the problem. Draw a diagram of what remains (Model) Models – A replica or description designed to show the structure or workings of an object, system, or concept. Models help guide experimental design

11 Section 1 What is Physics?
Chapter 1 The System

12 Section 1 What is Physics?
Chapter 1 The Scientific Model

13 The Scientific Method (cont)
Section 1 What is Physics Chapter 1 The Scientific Method (cont) Hypothesis – A reasonable explanation for observations, one that can be tested with additional experiments. The hypothesis must be tested in a controlled experiment. Controlled Experiment- Only one variable at a time is changed to determine what influences the phenomenon you are observing.

14 Numbers As Measurements
Section 2 Measurements in Experiments Chapter 1 Numbers As Measurements Numerical measurements in science contain the value (number) and Dimension. Dimension is the physical quantity being measured (length, mass, time, temperature, electric current) Each dimension is measured using units and prefixes from the SI system. The dimension must match the unit. (ex. If you are measuring length, use the meter(m), not the kilogram(kg)

15 Chapter 1 SI is the standard measurement system for science.
Section 2 Measurements in Experiments Chapter 1 SI is the standard measurement system for science. Used so that scientists can communicate with the same language. There are seven base units. They are: Meter(m) – length kilogram(kg) – Mass Second(s) – Time Kelvin(K) – Temperature Ampere(A) – current Mole(mol) – amount of substance Candela(cd) – luminous intensity

16 Chapter 1 Common Metric Prefixes:
Section 2 Measurements in Experiments Chapter 1 Common Metric Prefixes: -See handout or visit reference section of website -Be able to convert between any prefix and another.

17 How good are the measurements?
Scientists use two word to describe how good the measurements are: Accuracy- how close the measurement is to the actual value. Precision- how well can the measurement be repeated.

18 Differences Accuracy can be true of an individual measurement or the average of several. Problems with accuracy are due to error Precision requires several measurements before anything can be said about it. Precision describes the limitation of the measuring instrument.

19 Percent Error Percent error =
(Experimental Value – Accepted value) x 100 Accepted Value Percent error can be negative.

20 Significant Figures Scientific Notation Accuracy and Precision

21 Atlantic Pacific Present Absent If the decimal point is absent, start at the Atlantic (right), find the first non zero, and count all the rest of the digits 230000 1750

22 Atlantic Pacific Present Absent If the decimal point is PRESENT, start at the Pacific (left), find the first non zero, and count all the rest of the digits 0.045 1.2300

23 Sig Figs for Addition 27.93 6.4 + 27.93 6.4 + 27.93 6.4 34.33
First line up the decimal places 27.93 6.4 + 27.93 6.4 Then do the adding.. Find the estimated numbers in the problem. 34.33 This answer must be rounded to the tenths place.

24 Multiplication and Division
Rule is simpler Same number of sig figs in the answer as the least in the question 3.6 x 653 2350.8 3.6 has 2 s.f. 653 has 3 s.f. answer can only have 2 s.f. 2400

25 Multiplication and Division
Same rules for division. practice 4.5 / 6.245 4.5 x 6.245 x .043 3.876 / 1980 16547 / 710

26 The Metric System

27 The Metric System Easier to use because it is a decimal system.
Every conversion is by some power of 10. A metric unit has two parts. A prefix and a base unit. prefix tells you how many times to divide or multiply by 10.

28 Base Units Length - meter - m Mass - grams - g Time - second - s
Temperature - Kelvin K Energy - Joules- J Volume - Liter - L Amount of substance - mole - mol

29 Prefixes kilo k 1000 times deci d 1/10 centi c 1/100 milli m 1/1000
micro μ 1/ nano n 1/

30 Volume calculated by multiplying L x W x H
Liter the volume of a cube 1 dm (10 cm) on a side 1L = 1 dm3 so 1 L = 10 cm x 10 cm x 10 cm 1 L = 1000 cm3 1/1000 L = 1 cm3 1 mL = 1 cm3

31 Mass 1 gram is defined as the mass of 1 cm3 of water at 4 ºC.
1000 g = 1000 cm3 of water 1 kg = 1 L of water

32 Converting k h D d c m how far you have to move on this chart, tells you how far, and which direction to move the decimal place. The box is the base unit, meters, Liters, grams, etc.

33 k h D d c m 5 6 starts at the base unit and move three to the right.
Conversions k h D d c m Change 5.6 m to millimeters starts at the base unit and move three to the right. move the decimal point three to the right 5 6

34 k h D d c m Conversions convert 25 mg to grams convert 0.45 km to mm
convert 35 mL to liters It works because the math works, we are dividing or multiplying by 10 the correct number of times.

35 What about micro- and nano-?
k h D d c m μ n 3 3 The jump in between is 3 places Convert μm to m Convert cm to nm

36 Chapter 2 Table of Contents Section 1 Displacement and Velocity
Motion in One Dimension Table of Contents Section 1 Displacement and Velocity Section 2 Acceleration Section 3 Falling Objects

37 One Dimensional Motion
Section 1 Displacement and Velocity Chapter 2 One Dimensional Motion To simplify the concept of motion, we will first consider motion that takes place in one direction. One example is the motion of a commuter train on a straight track. To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.

38 displacement = final position – initial position
Section 1 Displacement and Velocity Chapter 2 Displacement Displacement is a change in position. Displacement is not always equal to the distance traveled. The SI unit of displacement is the meter, m. Dx = xf – xi displacement = final position – initial position

39 Positive and Negative Displacements
Section 1 Displacement and Velocity Chapter 2 Positive and Negative Displacements

40 Chapter 2 Average Velocity
Section 1 Displacement and Velocity Chapter 2 Average Velocity Average velocity is the total displacement divided by the time interval during which the displacement occurred. In SI, the unit of velocity is meters per second, abbreviated as m/s.

41 Chapter 2 Velocity and Speed
Section 1 Displacement and Velocity Chapter 2 Velocity and Speed Velocity describes motion with both a direction and a numerical value (a magnitude). Speed has no direction, only magnitude. Average speed is equal to the total distance traveled divided by the time interval.

42 Interpreting Velocity Graphically
Section 1 Displacement and Velocity Chapter 2 Interpreting Velocity Graphically For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph. If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity. Object 1: positive slope = positive velocity Object 2: zero slope= zero velocity Object 3: negative slope = negative velocity

43 Interpreting Velocity Graphically, continued
Section 1 Displacement and Velocity Chapter 2 Interpreting Velocity Graphically, continued The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path. The instantaneous velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

44 Chapter 2 Changes in Velocity
Section 2 Acceleration Chapter 2 Changes in Velocity Acceleration is the rate at which velocity changes over time. An object accelerates if its speed, direction, or both change. Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

45 Changes in Velocity, continued
Section 2 Acceleration Chapter 2 Changes in Velocity, continued Consider a train moving to the right, so that the displacement and the velocity are positive. The slope of the velocity-time graph is the average acceleration. When the velocity in the positive direction is increasing, the acceleration is positive, as at A. When the velocity is constant, there is no acceleration, as at B. When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.

46 Velocity and Acceleration
Section 2 Acceleration Chapter 2 Velocity and Acceleration

47 Motion with Constant Acceleration
Section 2 Acceleration Chapter 2 Motion with Constant Acceleration When velocity changes by the same amount during each time interval, acceleration is constant. The relationships between displacement, time, velocity, and constant acceleration are expressed by the equations shown on the next slide. These equations apply to any object moving with constant acceleration. These equations use the following symbols: Dx = displacement vi = initial velocity vf = final velocity Dt = time interval

48 Equations for Constantly Accelerated Straight-Line Motion
Section 2 Acceleration Chapter 2 Equations for Constantly Accelerated Straight-Line Motion

49 Chapter 2 Sample Problem Final Velocity After Any Displacement
Section 2 Acceleration Chapter 2 Sample Problem Final Velocity After Any Displacement A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

50 Sample Problem, continued
Section 2 Acceleration Chapter 2 Sample Problem, continued 1. Define Given: vi = 0 m/s a = m/s2 Dx = 4.75 m Unknown: vf = ? Diagram: Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller, as shown above. The positive direction is to the right.

51 Sample Problem, continued
Section 2 Acceleration Chapter 2 Sample Problem, continued 2. Plan Choose an equation or situation: Because the initial velocity, acceleration, and displacement are known, the final velocity can be found using the following equation: Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf .

52 Sample Problem, continued
Section 2 Acceleration Chapter 2 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: Tip: Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity. So, the final velocity must be positive. 4. Evaluate The stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

53 Section 3 Falling Objects
Chapter 2 Free Fall Free fall is the motion of a body when only the force due to gravity is acting on the body. The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration. Free-fall acceleration is denoted with the symbols ag (generally) or g (on Earth’s surface).

54 Free-Fall Acceleration
Section 3 Falling Objects Chapter 2 Free-Fall Acceleration Free-fall acceleration is the same for all objects, regardless of mass. This book will use the value g = 9.81 m/s2. Free-fall acceleration on Earth’s surface is –9.81 m/s2 at all points in the object’s motion. Consider a ball thrown up into the air. Moving upward: velocity is decreasing, acceleration is –9.81 m/s2 Top of path: velocity is zero, acceleration is –9.81 m/s2 Moving downward: velocity is increasing, acceleration is –9.81 m/s2

55 Chapter 2 Sample Problem Falling Object
Section 3 Falling Objects Chapter 2 Sample Problem Falling Object Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?

56 Sample Problem, continued
Section 3 Falling Objects Chapter 2 Sample Problem, continued 1. Define Given: Unknown: vi = +6.0 m/s Dt = ? a = –g = –9.81 m/s2 Dy = –2.0 m Diagram: Place the origin at the Starting point of the ball (yi = 0 at ti = 0).

57 Sample Problem, continued
Section 3 Falling Objects Chapter 2 Sample Problem, continued 2. Plan Choose an equation or situation: Both ∆t and vf are unknown. Therefore, first solve for vf using the equation that does not require time. Then, the equation for vf that does involve time can be used to solve for ∆t. Rearrange the equation to isolate the unknown: Take the square root of the first equation to isolate vf. The second equation must be rearranged to solve for ∆t.

58 Sample Problem, continued
Section 3 Falling Objects Chapter 2 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: First find the velocity of the ball at the moment that it hits the floor. Tip: When you take the square root to find vf , select the negative answer because the ball will be moving toward the floor, in the negative direction.

59 Sample Problem, continued
Section 3 Falling Objects Chapter 2 Sample Problem, continued Next, use this value of vf in the second equation to solve for ∆t. 4. Evaluate The solution, 1.50 s, is a reasonable amount of time for the ball to be in the air.

60 Chapter 2 Multiple Choice Use the graphs to answer questions 1–3.
Standardized Test Prep Multiple Choice Use the graphs to answer questions 1–3. 1. Which graph represents an object moving with a constant positive velocity? A. I C. III B. II D. IV

61 Chapter 2 Multiple Choice Use the graphs to answer questions 1–3.
Standardized Test Prep Multiple Choice Use the graphs to answer questions 1–3. 1. Which graph represents an object moving with a constant positive velocity? A. I C. III B. II D. IV

62 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 2. Which graph represents an object at rest? F. I H. III G. II J. IV

63 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 2. Which graph represents an object at rest? F. I H. III G. II J. IV

64 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 3. Which graph represents an object moving with a constant positive acceleration? A. I C. III B. II D. IV

65 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 3. Which graph represents an object moving with a constant positive acceleration? A. I C. III B. II D. IV

66 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of 73 km/h to the south.What is the bus’s displacement? F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/h to the south

67 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of 73 km/h to the south.What is the bus’s displacement? F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/h to the south

68 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 5. What is the squirrel’s displacement at time t = 3.0 s? A. –6.0 m B. –2.0 m C m D m

69 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 5. What is the squirrel’s displacement at time t = 3.0 s? A. –6.0 m B. –2.0 m C m D m

70 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 6. What is the squirrel’s average velocity during the time interval between 0.0 s and 3.0 s? F. –2.0 m/s G. –0.67 m/s H. 0.0 m/s J m/s

71 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 6. What is the squirrel’s average velocity during the time interval between 0.0 s and 3.0 s? F. –2.0 m/s G. –0.67 m/s H. 0.0 m/s J m/s

72 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 7. Which of the following statements is true of acceleration? A. Acceleration always has the same sign as displacement. B. Acceleration always has the same sign as velocity. C. The sign of acceleration depends on both the direction of motion and how the velocity is changing. D. Acceleration always has a positive sign.

73 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 7. Which of the following statements is true of acceleration? A. Acceleration always has the same sign as displacement. B. Acceleration always has the same sign as velocity. C. The sign of acceleration depends on both the direction of motion and how the velocity is changing. D. Acceleration always has a positive sign.

74 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 8. A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move during this time? F. 12 m G. 93 m H. 120 m J. 190 m

75 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 8. A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move during this time? F. 12 m G. 93 m H. 120 m J. 190 m

76 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 9. Which of the following statements is true for a ball thrown vertically upward? A. The ball has a negative acceleration on the way up and a positive acceleration on the way down. B. The ball has a positive acceleration on the way up and a negative acceleration on the way down. C. The ball has zero acceleration on the way up and a positive acceleration on the way down. D. The ball has a constant acceleration throughout its flight.

77 Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued 9. Which of the following statements is true for a ball thrown vertically upward? A. The ball has a negative acceleration on the way up and a positive acceleration on the way down. B. The ball has a positive acceleration on the way up and a negative acceleration on the way down. C. The ball has zero acceleration on the way up and a positive acceleration on the way down. D. The ball has a constant acceleration throughout its flight.

78 Chapter 2 Short Response
Standardized Test Prep Short Response 10. In one or two sentences, explain the difference between displacement and distance traveled.

79 Chapter 2 Short Response
Standardized Test Prep Short Response 10. In one or two sentences, explain the difference between displacement and distance traveled. Answer: Displacement measures only the net change in position from starting point to end point. The distance traveled is the total length of the path followed from starting point to end point and may be greater than or equal to the displacement.

80 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity: a. for the time interval from t = 0.0 min to t = 10.0 min b. for the time interval from t = 10.0 min to t = 20.0 min c. for the time interval from t = 20.0 min to t = 30.0 min d. for the entire run

81 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity. Answers will vary but should be approximately as follows: a. for t = 0.0 min to t = 10.0 min Answer: m, +4.0 m/s b. for t = 10.0 min to t = 20.0 min Answer: m, +2.5 m/s c. for t = 20.0 min to t = 30.0 min Answer: +900 m, +2 m/s d. for the entire run Answer: m, +2.7 m/s

82 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 12. For an object moving with constant negative acceleration, draw the following: a. a graph of position vs. time b. a graph of velocity vs. time For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin.

83 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 12. For an object moving with constant negative acceleration, draw the following: a. a graph of position vs. time b. a graph of velocity vs. time For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin. Answers:

84 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s. a. If the snowmobile accelerates at a rate of m/s2 for 7.0 s, what is its final velocity? b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop?

85 Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s. a. If the snowmobile accelerates at a rate of m/s2 for 7.0 s, what is its final velocity? b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop? Answers: a m/s b. 5.0 s

86 Chapter 2 Extended Response
Standardized Test Prep Extended Response 14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s. a. What is the car’s average acceleration? b. What is the car’s average velocity? c. How far did the car move while accelerating? Show all of your work for these calculations.

87 Chapter 2 Extended Response
Standardized Test Prep Extended Response 14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s. a. What is the car’s average acceleration? b. What is the car’s average velocity? c. How far did the car move while accelerating? Answers: a. 1.6 m/s2 eastward b. 24 m/s c. 240 m

88 Extended Response, continued
Chapter 2 Standardized Test Prep Extended Response, continued 15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m. a. How long does it take the ball to reach its highest point? b. How long is the ball in the air? Show all of your work for these calculations.

89 Extended Response, continued
Chapter 2 Standardized Test Prep Extended Response, continued 15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m. a. How long does it take the ball to reach its highest point? b. How long is the ball in the air? Show all of your work for these calculations. Answers: a s b s

90 Chapter 3 Two-Dimensional Motion and Vectors

91 Chapter 3 Table of Contents Section 1 Introduction to Vectors
Two-Dimensional Motion and Vectors Table of Contents Section 1 Introduction to Vectors Section 2 Vector Operations Section 3 Projectile Motion Section 4 Relative Motion

92 Chapter 3 Scalars and Vectors
Section 1 Introduction to Vectors Chapter 3 Scalars and Vectors A scalar is a physical quantity that has magnitude but no direction. Examples: speed, volume, the number of pages in your textbook A vector is a physical quantity that has both magnitude and direction. Examples: displacement, velocity, acceleration In this book, scalar quantities are in italics. Vectors are represented by boldface symbols.

93 Graphical Addition of Vectors
Section 1 Introduction to Vectors Chapter 3 Graphical Addition of Vectors A resultant vector represents the sum of two or more vectors. Vectors can be added graphically. A student walks from his house to his friend’s house (a), then from his friend’s house to the school (b). The student’s resultant displacement (c) can be found by using a ruler and a protractor.

94 Triangle Method of Addition
Section 1 Introduction to Vectors Chapter 3 Triangle Method of Addition Vectors can be moved parallel to themselves in a diagram. Thus, you can draw one vector with its tail starting at the tip of the other as long as the size and direction of each vector do not change. The resultant vector can then be drawn from the tail of the first vector to the tip of the last vector.

95 Chapter 3 Properties of Vectors Vectors can be added in any order.
Section 1 Introduction to Vectors Chapter 3 Properties of Vectors Vectors can be added in any order. To subtract a vector, add its opposite. Multiplying or dividing vectors by scalars results in vectors.

96 Coordinate Systems in Two Dimensions
Section 2 Vector Operations Chapter 3 Coordinate Systems in Two Dimensions One method for diagraming the motion of an object employs vectors and the use of the x- and y-axes. Axes are often designated using fixed directions. In the figure shown here, the positive y-axis points north and the positive x-axis points east.

97 Determining Resultant Magnitude and Direction
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction In Section 1, the magnitude and direction of a resultant were found graphically. With this approach, the accuracy of the answer depends on how carefully the diagram is drawn and measured. A simpler method uses the Pythagorean theorem and the tangent function.

98 Determining Resultant Magnitude and Direction, continued
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction, continued The Pythagorean Theorem Use the Pythagorean theorem to find the magnitude of the resultant vector. The Pythagorean theorem states that for any right triangle, the square of the hypotenuse—the side opposite the right angle—equals the sum of the squares of the other two sides, or legs.

99 Determining Resultant Magnitude and Direction, continued
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction, continued The Tangent Function Use the tangent function to find the direction of the resultant vector. For any right triangle, the tangent of an angle is defined as the ratio of the opposite and adjacent legs with respect to a specified acute angle of a right triangle.

100 Chapter 3 Sample Problem Finding Resultant Magnitude and Direction
Section 2 Vector Operations Chapter 3 Sample Problem Finding Resultant Magnitude and Direction An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30  102 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?

101 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 1. Define Given: Dy = 136 m Dx = 1/2(width) = 115 m Unknown: d = ? q = ? Diagram: Choose the archaeologist’s starting position as the origin of the coordinate system, as shown above.

102 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 2. Plan Choose an equation or situation: The Pythagorean theorem can be used to find the magnitude of the archaeologist’s displacement. The direction of the displacement can be found by using the inverse tangent function. Rearrange the equations to isolate the unknowns:

103 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 3. Calculate Evaluate Because d is the hypotenuse, the archaeologist’s displacement should be less than the sum of the height and half of the width. The angle is expected to be more than 45 because the height is greater than half of the width.

104 Resolving Vectors into Components
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components You can often describe an object’s motion more conveniently by breaking a single vector into two components, or resolving the vector. The components of a vector are the projections of the vector along the axes of a coordinate system. Resolving a vector allows you to analyze the motion in each direction.

105 Resolving Vectors into Components, continued
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components, continued Consider an airplane flying at 95 km/h. The hypotenuse (vplane) is the resultant vector that describes the airplane’s total velocity. The adjacent leg represents the x component (vx), which describes the airplane’s horizontal speed. The opposite leg represents the y component (vy), which describes the airplane’s vertical speed.

106 Resolving Vectors into Components, continued
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components, continued The sine and cosine functions can be used to find the components of a vector. The sine and cosine functions are defined in terms of the lengths of the sides of right triangles.

107 Adding Vectors That Are Not Perpendicular
Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular Suppose that a plane travels first 5 km at an angle of 35°, then climbs at 10° for 22 km, as shown below. How can you find the total displacement? Because the original displacement vectors do not form a right triangle, you can not directly apply the tangent function or the Pythagorean theorem. d2 d1

108 Adding Vectors That Are Not Perpendicular, continued
Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular, continued You can find the magnitude and the direction of the resultant by resolving each of the plane’s displacement vectors into its x and y components. Then the components along each axis can be added together. As shown in the figure, these sums will be the two perpendicular components of the resultant, d. The resultant’s magnitude can then be found by using the Pythagorean theorem, and its direction can be found by using the inverse tangent function.

109 Chapter 3 Sample Problem Adding Vectors Algebraically
Section 2 Vector Operations Chapter 3 Sample Problem Adding Vectors Algebraically A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement

110 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 1 . Select a coordinate system. Then sketch and label each vector. Given: d1 = 27.0 km q1 = –35° d2 = 41.0 km q2 = 65° Tip: q1 is negative, because clockwise movement from the positive x-axis is negative by convention. Unknown: d = ? q = ?

111 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 2 . Find the x and y components of all vectors. Make a separate sketch of the displacements for each day. Use the cosine and sine functions to find the components.

112 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 3 . Find the x and y components of the total displacement. 4 . Use the Pythagorean theorem to find the magnitude of the resultant vector. This slide is part of SAMPLE PROBLEM custom show.

113 Sample Problem, continued
Section 2 Vector Operations Chapter 3 Sample Problem, continued 5 . Use a suitable trigonometric function to find the angle.

114 Section 3 Projectile Motion
Chapter 3 Projectiles Objects that are thrown or launched into the air and are subject to gravity are called projectiles. Projectile motion is the curved path that an object follows when thrown, launched,or otherwise projected near the surface of Earth. If air resistance is disregarded, projectiles follow parabolic trajectories.

115 Projectiles, continued
Section 3 Projectile Motion Chapter 3 Projectiles, continued Projectile motion is free fall with an initial horizontal velocity. The yellow ball is given an initial horizontal velocity and the red ball is dropped. Both balls fall at the same rate. In this book, the horizontal velocity of a projectile will be considered constant. This would not be the case if we accounted for air resistance.

116 Kinematic Equations for Projectiles
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles How can you know the displacement, velocity, and acceleration of a projectile at any point in time during its flight? One method is to resolve vectors into components, then apply the simpler one-dimensional forms of the equations for each component. Finally, you can recombine the components to determine the resultant.

117 Kinematic Equations for Projectiles, continued
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles, continued To solve projectile problems, apply the kinematic equations in the horizontal and vertical directions. In the vertical direction, the acceleration ay will equal –g (–9.81 m/s2) because the only vertical component of acceleration is free-fall acceleration. In the horizontal direction, the acceleration is zero, so the velocity is constant.

118 Kinematic Equations for Projectiles, continued
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles, continued Projectiles Launched Horizontally The initial vertical velocity is 0. The initial horizontal velocity is the initial velocity. Projectiles Launched At An Angle Resolve the initial velocity into x and y components. The initial vertical velocity is the y component. The initial horizontal velocity is the x component.

119 Chapter 3 Sample Problem Projectiles Launched At An Angle
Section 3 Projectile Motion Chapter 3 Sample Problem Projectiles Launched At An Angle A zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole,which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s,will the dart hit the monkey, the banana, or neither one?

120 Sample Problem, continued
Section 3 Projectile Motion Chapter 3 Sample Problem, continued 1 . Select a coordinate system. The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m.

121 Sample Problem, continued
Section 3 Projectile Motion Chapter 3 Sample Problem, continued 2 . Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis.

122 Sample Problem, continued
Section 3 Projectile Motion Chapter 3 Sample Problem, continued 3 . Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate the unknown Dt, which is the time the dart takes to travel the horizontal distance.

123 Sample Problem, continued
Section 3 Projectile Motion Chapter 3 Sample Problem, continued 4 . Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases. For the banana, vi = 0. Thus: Dyb = ½ay(Dt)2 = ½(–9.81 m/s2)(0.215 s)2 = –0.227 m The dart has an initial vertical component of velocity equal to vi sin q, so: Dyd = (vi sin q)(Dt) + ½ay(Dt)2 Dyd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81 m/s2)(0.215 s)2 Dyd = 3.99 m – m = 3.76 m

124 Sample Problem, continued
Section 3 Projectile Motion Chapter 3 Sample Problem, continued 5 . Analyze the results. Find the final height of both the banana and the dart. ybanana, f = yb,i+ Dyb = 5.00 m + (–0.227 m) ybanana, f = 4.77 m above the ground ydart, f = yd,i+ Dyd = 1.00 m m ydart, f = 4.76 m above the ground The dart hits the banana. The slight difference is due to rounding.

125 Section 4 Relative Motion
Chapter 3 Objectives Describe situations in terms of frame of reference. Solve problems involving relative velocity.

126 Chapter 3 Frames of Reference
Section 4 Relative Motion Chapter 3 Frames of Reference If you are moving at 80 km/h north and a car passes you going 90 km/h, to you the faster car seems to be moving north at 10 km/h. Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north. This simple example demonstrates that velocity measurements depend on the frame of reference of the observer.

127 Frames of Reference, continued
Section 4 Relative Motion Chapter 3 Frames of Reference, continued Consider a stunt dummy dropped from a plane. (a) When viewed from the plane, the stunt dummy falls straight down. (b) When viewed from a stationary position on the ground, the stunt dummy follows a parabolic projectile path.

128 Chapter 3 Relative Velocity Section 4 Relative Motion
When solving relative velocity problems, write down the information in the form of velocities with subscripts. Using our earlier example, we have: vse = +80 km/h north (se = slower car with respect to Earth) vfe = +90 km/h north (fe = fast car with respect to Earth) unknown = vfs (fs = fast car with respect to slower car) Write an equation for vfs in terms of the other velocities. The subscripts start with f and end with s. The other subscripts start with the letter that ended the preceding velocity: vfs = vfe + ves

129 Relative Velocity, continued
Section 4 Relative Motion Chapter 3 Relative Velocity, continued An observer in the slow car perceives Earth as moving south at a velocity of 80 km/h while a stationary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/h. In equation form: ves = –vse Thus, this problem can be solved as follows: vfs = vfe + ves = vfe – vse vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n A general form of the relative velocity equation is: vac = vab + vbc

130 Chapter 3 Multiple Choice
Standardized Test Prep Multiple Choice 1. Vector A has a magnitude of 30 units. Vector B is perpendicular to vector A and has a magnitude of 40 units. What would the magnitude of the resultant vector A + B be? A. 10 units B. 50 units C. 70 units D. zero

131 Chapter 3 Multiple Choice
Standardized Test Prep Multiple Choice 1. Vector A has a magnitude of 30 units. Vector B is perpendicular to vector A and has a magnitude of 40 units. What would the magnitude of the resultant vector A + B be? A. 10 units B. 50 units C. 70 units D. zero

132 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 2. What term represents the magnitude of a velocity vector? F. acceleration G. momentum H. speed J. velocity

133 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 2. What term represents the magnitude of a velocity vector? F. acceleration G. momentum H. speed J. velocity

134 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 3. What is the direction of the resultant vector A + B? A. 15º above the x-axis B. 75º above the x-axis C. 15º below the x-axis D. 75º below the x-axis

135 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 3. What is the direction of the resultant vector A + B? A. 15º above the x-axis B. 75º above the x-axis C. 15º below the x-axis D. 75º below the x-axis

136 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 4. What is the direction of the resultant vector A – B? F. 15º above the x-axis G. 75º above the x-axis H. 15º below the x-axis J. 75º below the x-axis

137 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 4. What is the direction of the resultant vector A – B? F. 15º above the x-axis G. 75º above the x-axis H. 15º below the x-axis J. 75º below the x-axis

138 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 5. What is the resultant velocity relative to an observer on the shore ? A. 3.2 m/s to the southeast B. 5.0 m/s to the southeast C. 7.1 m/s to the southeast D m/s to the southeast

139 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 5. What is the resultant velocity relative to an observer on the shore ? A. 3.2 m/s to the southeast B. 5.0 m/s to the southeast C. 7.1 m/s to the southeast D m/s to the southeast

140 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 6. If the river is 125 m wide, how long does the boat take to cross the river? F. 39 s G. 25 s H. 17 s J. 12 s

141 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 6. If the river is 125 m wide, how long does the boat take to cross the river? F. 39 s G. 25 s H. 17 s J. 12 s

142 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 7. The pilot of a plane measures an air velocity of 165 km/h south relative to the plane. An observer on the ground sees the plane pass overhead at a velocity of 145 km/h toward the north.What is the velocity of the wind that is affecting the plane relative to the observer? A. 20 km/h to the north B. 20 km/h to the south C. 165 km/h to the north D. 310 km/h to the south

143 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 7. The pilot of a plane measures an air velocity of 165 km/h south relative to the plane. An observer on the ground sees the plane pass overhead at a velocity of 145 km/h toward the north.What is the velocity of the wind that is affecting the plane relative to the observer? A. 20 km/h to the north B. 20 km/h to the south C. 165 km/h to the north D. 310 km/h to the south

144 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 8. A golfer takes two putts to sink his ball in the hole once he is on the green. The first putt displaces the ball 6.00 m east, and the second putt displaces the ball 5.40 m south. What displacement would put the ball in the hole in one putt? F m southeast G m at 48.0º south of east H m at 42.0º south of east J m at 42.0º south of east

145 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued 8. A golfer takes two putts to sink his ball in the hole once he is on the green. The first putt displaces the ball 6.00 m east, and the second putt displaces the ball 5.40 m south. What displacement would put the ball in the hole in one putt? F m southeast G m at 48.0º south of east H m at 42.0º south of east J m at 42.0º south of east

146 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 9. What is the initial speed of the girl’s ball relative to the boy? A. 1.0 m/s C. 2.0 m/s B. 1.5 m/s D. 3.0 m/s

147 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 9. What is the initial speed of the girl’s ball relative to the boy? A. 1.0 m/s C. 2.0 m/s B. 1.5 m/s D. 3.0 m/s

148 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 10. If air resistance is disregarded, which ball will hit the ground first? F. the boy’s ball H. neither G. the girl’s ball J. cannot be determined

149 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 10. If air resistance is disregarded, which ball will hit the ground first? F. the boy’s ball H. neither G. the girl’s ball J. cannot be determined

150 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground? A. the boy’s ball C. neither B. the girl’s ball D. cannot be determined

151 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground? A. the boy’s ball C. neither B. the girl’s ball D. cannot be determined

152 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 12. What is the speed of the girl’s ball when it hits the ground? F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s

153 Multiple Choice, continued
Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 12. What is the speed of the girl’s ball when it hits the ground? F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s

154 Chapter 3 Short Response
Standardized Test Prep Short Response 13. If one of the components of one vector along the direction of another vector is zero, what can you conclude about these two vectors?

155 Chapter 3 Short Response
Standardized Test Prep Short Response 13. If one of the components of one vector along the direction of another vector is zero, what can you conclude about these two vectors? Answer: They are perpendicular.

156 Short Response, continued
Chapter 3 Standardized Test Prep Short Response, continued 14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move horizontally and vertically?

157 Short Response, continued
Chapter 3 Standardized Test Prep Short Response, continued 14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move horizontally and vertically? Answer: 31.5 m horizontally, 26.4 m vertically

158 Short Response, continued
Chapter 3 Standardized Test Prep Short Response, continued 15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically?

159 Short Response, continued
Chapter 3 Standardized Test Prep Short Response, continued 15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically? Answer: 29.4 m/s

160 Chapter 3 Extended Response
Standardized Test Prep Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work.

161 Chapter 3 Extended Response
Standardized Test Prep Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work. Answer: 10.8 m

162 Extended Response, continued
Chapter 3 Standardized Test Prep Extended Response, continued Read the following passage to answer question 17. Three airline executives are discussing ideas for developing flights that are more energy efficient. Executive A: Because the Earth rotates from west to east, we could operate “static flights”—a helicopter or airship could begin by rising straight up from New York City and then descend straight down four hours later when San Francisco arrives below. Executive B: This approach could work for one-way flights, but the return trip would take 20 hours. continued on the next slide

163 Extended Response, continued
Chapter 3 Standardized Test Prep Extended Response, continued Executive C: That approach will never work. Think about it.When you throw a ball straight up in the air, it comes straight back down to the same point. Executive A: The ball returns to the same point because Earth’s motion is not significant during such a short time. 17. State which of the executives is correct, and explain why.

164 Extended Response, continued
Chapter 3 Standardized Test Prep Extended Response, continued 17. State which of the executives is correct, and explain why. Answer: Executive C is correct. Explanations should include the concept of relative velocity—when a helicopter lifts off straight up from the ground, it is already moving horizontally with Earth’s horizontal velocity. (We assume that Earth’s motion is constant for the purposes of this scenario and does not depend on time.)

165 Section 3 Projectile Motion
Chapter 3 Projectiles

166 Chapter 5 Table of Contents Section 1 Work Section 2 Energy
Work and Energy Table of Contents Section 1 Work Section 2 Energy Section 3 Conservation of Energy Section 4 Power

167 Chapter 5 Definition of Work
Section 1 Work Chapter 5 Definition of Work Work is done on an object when a force causes a displacement of the object. Work is done only when components of a force are parallel to a displacement.

168 Section 1 Work Chapter 5 Definition of Work

169 Chapter 5 Kinetic Energy Kinetic Energy
Section 2 Energy Chapter 5 Kinetic Energy Kinetic Energy The energy of an object that is due to the object’s motion is called kinetic energy. Kinetic energy depends on speed and mass.

170 Kinetic Energy, continued
Section 2 Energy Chapter 5 Kinetic Energy, continued Work-Kinetic Energy Theorem The net work done by all the forces acting on an object is equal to the change in the object’s kinetic energy. The net work done on a body equals its change in kinetic energy. Wnet = ∆KE net work = change in kinetic energy

171 gravitational PE = mass  free-fall acceleration  height
Section 2 Energy Chapter 5 Potential Energy Potential Energy is the energy associated with an object because of the position, shape, or condition of the object. Gravitational potential energy is the potential energy stored in the gravitational fields of interacting bodies. Gravitational potential energy depends on height from a zero level. PEg = mgh gravitational PE = mass  free-fall acceleration  height

172 Potential Energy, continued
Section 2 Energy Chapter 5 Potential Energy, continued Elastic potential energy is the energy available for use when a deformed elastic object returns to its original configuration. The symbol k is called the spring constant, a parameter that measures the spring’s resistance to being compressed or stretched.

173 Elastic Potential Energy
Section 2 Energy Chapter 5 Elastic Potential Energy

174 Chapter 5 Conserved Quantities
Section 3 Conservation of Energy Chapter 5 Conserved Quantities When we say that something is conserved, we mean that it remains constant.

175 Chapter 5 Mechanical Energy
Section 3 Conservation of Energy Chapter 5 Mechanical Energy Mechanical energy is the sum of kinetic energy and all forms of potential energy associated with an object or group of objects. ME = KE + ∑PE Mechanical energy is often conserved. MEi = MEf initial mechanical energy = final mechanical energy (in the absence of friction)

176 Chapter 5 Sample Problem Conservation of Mechanical Energy
Section 3 Conservation of Energy Chapter 5 Sample Problem Conservation of Mechanical Energy Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0 kg.

177 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 1. Define Given: h = hi = 3.00 m m = 25.0 kg vi = 0.0 m/s hf = 0 m Unknown: vf = ?

178 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 2. Plan Choose an equation or situation: The slide is frictionless, so mechanical energy is conserved. Kinetic energy and gravitational potential energy are the only forms of energy present.

179 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 2. Plan, continued The zero level chosen for gravitational potential energy is the bottom of the slide. Because the child ends at the zero level, the final gravitational potential energy is zero. PEg,f = 0

180 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 2. Plan, continued The initial gravitational potential energy at the top of the slide is PEg,i = mghi = mgh Because the child starts at rest, the initial kinetic energy at the top is zero. KEi = 0 Therefore, the final kinetic energy is as follows:

181 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 3. Calculate Substitute values into the equations: PEg,i = (25.0 kg)(9.81 m/s2)(3.00 m) = 736 J KEf = (1/2)(25.0 kg)vf2 Now use the calculated quantities to evaluate the final velocity. MEi = MEf PEi + KEi = PEf + KEf 736 J + 0 J = 0 J + (0.500)(25.0 kg)vf2 vf = 7.67 m/s

182 Sample Problem, continued
Section 3 Conservation of Energy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 4. Evaluate The expression for the square of the final speed can be written as follows: Notice that the masses cancel, so the final speed does not depend on the mass of the child. This result makes sense because the acceleration of an object due to gravity does not depend on the mass of the object.

183 Mechanical Energy, continued
Section 3 Conservation of Energy Chapter 5 Mechanical Energy, continued Mechanical Energy is not conserved in the presence of friction. As a sanding block slides on a piece of wood, energy (in the form of heat) is dissipated into the block and surface.

184 Rate of Energy Transfer
Section 4 Power Chapter 5 Rate of Energy Transfer Power is a quantity that measures the rate at which work is done or energy is transformed. P = W/∆t power = work ÷ time interval An alternate equation for power in terms of force and speed is P = Fv power = force  speed

185 Chapter 5 Multiple Choice
Standardized Test Prep Multiple Choice 1. In which of the following situations is work not being done? A. A chair is lifted vertically with respect to the floor. B. A bookcase is slid across carpeting. C. A table is dropped onto the ground. D. A stack of books is carried at waist level across a room.

186 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 1. In which of the following situations is work not being done? A. A chair is lifted vertically with respect to the floor. B. A bookcase is slid across carpeting. C. A table is dropped onto the ground. D. A stack of books is carried at waist level across a room.

187 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 2. Which of the following equations correctly describes the relation between power,work, and time?

188 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 2. Which of the following equations correctly describes the relation between power,work, and time?

189 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the graph below to answer questions 3–5. The graph shows the energy of a 75 g yo-yo at different times as the yo-yo moves up and down on its string.

190 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 3. By what amount does the mechanical energy of the yo-yo change after 6.0 s? A. 500 mJ B. 0 mJ C. –100 mJ D. –600 mJ

191 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 3. By what amount does the mechanical energy of the yo-yo change after 6.0 s? A. 500 mJ B. 0 mJ C. –100 mJ D. –600 mJ

192 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 4. What is the speed of the yo-yo after 4.5 s? F. 3.1 m/s G. 2.3 m/s H. 3.6 m/s J. 1.6 m/s

193 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 4. What is the speed of the yo-yo after 4.5 s? F. 3.1 m/s G. 2.3 m/s H. 3.6 m/s J. 1.6 m/s

194 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 5. What is the maximum height of the yo-yo? A m B m C m D m

195 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 5. What is the maximum height of the yo-yo? A m B m C m D m

196 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 6. A car with mass m requires 5.0 kJ of work to move from rest to a final speed v. If this same amount of work is performed during the same amount of time on a car with a mass of 2m, what is the final speed of the second car?

197 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 6. A car with mass m requires 5.0 kJ of work to move from rest to a final speed v. If this same amount of work is performed during the same amount of time on a car with a mass of 2m, what is the final speed of the second car?

198 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 7–8. A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his clothes and Earth is His slide lowers his speed to zero just as he reaches the base. 7. How much mechanical energy is lost because of friction acting on the runner? A J B. 560 J C. 140 J D. 0 J

199 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 7–8. A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his clothes and Earth is His slide lowers his speed to zero just as he reaches the base. 7. How much mechanical energy is lost because of friction acting on the runner? A J B. 560 J C. 140 J D. 0 J

200 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 7–8. A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his clothes and Earth is His slide lowers his speed to zero just as he reaches the base. 8. How far does the runner slide? F m G m H m J. 1.2 m

201 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 7–8. A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his clothes and Earth is His slide lowers his speed to zero just as he reaches the base. 8. How far does the runner slide? F m G m H m J. 1.2 m

202 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 9–10. A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of kg. During one weighing, the spring is stretched a distance of 12 cm from equilibrium. During a second weighing, the spring is stretched a distance of 18 cm. How far does the runner slide?

203 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 9. How much greater is the elastic potential energy of the stretched spring during the second weighing than during the first weighing?

204 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 9. How much greater is the elastic potential energy of the stretched spring during the second weighing than during the first weighing?

205 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 10. If the spring is suddenly released after each weighing, the weighing pan moves back and forth through the equilibrium position. What is the ratio of the pan’s maximum speed after the second weighing to the pan’s maximum speed after the first weighing? Consider the force of gravity on the pan to be negligible.

206 Multiple Choice, continued
Chapter 5 Standardized Test Prep Multiple Choice, continued 10. If the spring is suddenly released after each weighing, the weighing pan moves back and forth through the equilibrium position. What is the ratio of the pan’s maximum speed after the second weighing to the pan’s maximum speed after the first weighing? Consider the force of gravity on the pan to be negligible.

207 Chapter 5 Short Response
Standardized Test Prep Short Response 11. A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is 14.0 m, how much power will the student deliver by climbing the stairs?

208 Short Response, continued
Chapter 5 Standardized Test Prep Short Response, continued 11. A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is 14.0 m, how much power will the student deliver by climbing the stairs? Answer: 206 W

209 Short Response, continued
Chapter 5 Standardized Test Prep Short Response, continued Base your answers to questions 12–13 on the information below. A 75.0 kg man jumps from a window that is 1.00 m above a sidewalk. 12. Write the equation for the man’s speed when he strikes the ground.

210 Short Response, continued
Chapter 5 Standardized Test Prep Short Response, continued Base your answers to questions 12–13 on the information below. A 75.0 kg man jumps from a window that is 1.00 m above a sidewalk. 12. Write the equation for the man’s speed when he strikes the ground.

211 Short Response, continued
Chapter 5 Standardized Test Prep Short Response, continued Base your answers to questions 12–13 on the information below. A 75.0 kg man jumps from a window that is 1.00 m above a sidewalk. 13. Calculate the man’s speed when he strikes the ground.

212 Short Response, continued
Chapter 5 Standardized Test Prep Short Response, continued Base your answers to questions 12–13 on the information below. A 75.0 kg man jumps from a window that is 1.00 m above a sidewalk. 13. Calculate the man’s speed when he strikes the ground. Answer: 4.4 m/s

213 Chapter 5 Extended Response
Standardized Test Prep Extended Response Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 14. The work done on the projectile by gravity.

214 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 14. The work done on the projectile by gravity. Answer: 1200 J

215 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 15. The change in kinetic energy since the projectile was fired.

216 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 15. The change in kinetic energy since the projectile was fired. Answer: 1200 J

217 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 16. The final kinetic energy of the projectile.

218 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued Base your answers to questions 14–16 on the information below. A projectile with a mass of 5.0 kg is shot horizontally from a height of 25.0 m above a flat desert surface. The projectile’s initial speed is 17 m/s. Calculate the following for the instant before the projectile hits the surface: 16. The final kinetic energy of the projectile. Answer: 1900 J

219 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued 17. A skier starts from rest at the top of a hill that is inclined at 10.5° with the horizontal. The hillside is m long, and the coefficient of friction between the snow and the skis is At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of the snow before coming to rest? Show all of your work.

220 Extended Response, continued
Chapter 5 Standardized Test Prep Extended Response, continued 17. A skier starts from rest at the top of a hill that is inclined at 10.5° with the horizontal. The hillside is m long, and the coefficient of friction between the snow and the skis is At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of the snow before coming to rest? Show all of your work. Answer: 290 m

221 Section 3 Conservation of Energy
Chapter 5 Mechanical Energy

222 Momentum and Collision
Chapter 6 Momentum and Collisions Chapter 6 Momentum and Collision

223 Chapter 6 Momentum Momentum (p) is the quantity of motion in a body. Momentum is a vector. It has a size and a direction. Only moving object have momentum. Calculation: The size of the momentum is equal to the mass of the object(kg) multiplied by the size of the object's velocity (m/s). Equation: p=mv Units: kgm/s The direction of the momentum is the same as the direction of the object's velocity..

224 total initial momentum = total final momentum
The Law of Conservation of Momentum Often physics problems deal with momentum before and after a collision. In such cases the total momentum of the bodies before collision is taken as equal to the total momentum of the bodies after collision. That is to say: momentum is conserved. m1v1,i + m2v2,i = m1v1,f + m2v2,f total initial momentum = total final momentum A collision is an event where momentum or kinetic energy is transferred from one object to another. The two types are elastic and inelastic collisions.

225 Momentum is conserved. Calculated by using:
An inelastic collisions occurs when two objects collide and do not bounce away from each other Momentum is conserved. Calculated by using: m1v1,i + m2v2,i = (m1 + m2)vf Kinetic energy is not conserved. Calculated by using: ∆KE = KEf - KEi

226 A completely inelastic collision:

227 An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. Momentum and kinetic energy are conserved and can be calculated using:

228 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision:

229 Newton’s Laws and Momentum
Momentum refers to inertia in motion. Momentum is a measure of how difficult it is to stop an object; a measure of “how much motion” an object has. Inertia is the tendency of a body at rest to remain at rest or of a body in straight line motion to stay in motion in a straight line unless acted on by an outside force. Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum

230 Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). F=ma Since acceleration is the change in velocity divided by time, (a= ∆V/∆t ) ,you can connect the two concepts with the following relationship: The impulse-momentum theorem states that when a net force is applied to an object over a certain time interval, the force will cause a change in the object’s momentum. F∆t = ∆p = mvf – mvi

231 Momentum and Impulse Momentum is a measure of how difficult it is to stop an object. p=mv change in momentum is given by: Δp = mΔv Impulse occurs as force is applied to an object over a period of time: I = FΔt Newton's Second Law of Motion is the rate of change of momentum: F=ma so F= mΔv/Δt so FΔt=mΔv so I= Δp and F = Δp/t

232 Chapter 6 Sample Problems Calculating Momentum
Calculate the momentum of a 11.35kg wagon rolling down a hill at 12m/s. p = mv p = (11.35kg) (12m/s) p = kgm/s down the hill

233 Chapter 6 Sample Problems Conservation of Momentum
A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right,what is the final velocity of the boat?

234 Sample Problem, continued
Chapter 6 Sample Problem, continued Conservation of Momentum 1. Define Given: m1 = 76 kg m2 = 45 kg v1,i = 0 v2,i = 0 v1,f = 2.5 m/s to the right Unknown: v2,f = ?

235 Sample Problem, continued
Chapter 6 Sample Problem, continued Conservation of Momentum 2. Plan Choose an equation or situation: Because the total momentum of an isolated system remains constant, the total initial momentum of the boater and the boat will be equal to the total final momentum of the boater and the boat. m1v1,i + m2v2,i = m1v1,f + m2v2,f

236 Sample Problem, continued
Chapter 6 Sample Problem, continued Conservation of Momentum 2. Plan, continued Because the boater and the boat are initially at rest, the total initial momentum of the system is equal to zero. Therefore, the final momentum of the system must also be equal to zero. m1v1,f + m2v2,f = 0 Rearrange the equation to solve for the final velocity of the boat.

237 Sample Problem, continued
Chapter 6 Sample Problem, continued Conservation of Momentum 3. Calculate Substitute the values into the equation and solve:

238 Sample Problem, continued
Chapter 6 Sample Problem, continued Conservation of Momentum 4. Evaluate The negative sign for v2,f indicates that the boat is moving to the left, in the direction opposite the motion of the boater. Therefore, v2,f = 4.2 m/s to the left

239 Chapter 6 Sample Problem
Kinetic Energy in Perfectly Inelastic Collisions Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of kg and an initial velocity of 4.00 m/s to the right. The second ball has a mass of kg and an initial velocity of 3.00 m/s to the left. What is the decrease in kinetic energy during the collision?

240 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 1. Define Given: m1= kg m2 = kg v1,i = 4.00 m/s to the right, v1,i = m/s v2,i = 3.00 m/s to the left, v2,i = –3.00 m/s Unknown: ∆KE = ?

241 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 2. Plan Choose an equation or situation: The change in kinetic energy is simply the initial kinetic energy subtracted from the final kinetic energy. ∆KE = KEf - KEi Determine both the initial and final kinetic energy.

242 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 2. Plan, continued Use the equation for a perfectly inelastic collision to calculate the final velocity.

243 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 3. Calculate Substitute the values into the equation and solve: First, calculate the final velocity, which will be used in the final kinetic energy equation.

244 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 3. Calculate, continued Next calculate the initial and final kinetic energy.

245 Sample Problem, continued
Chapter 6 Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 3. Calculate, continued Finally, calculate the change in kinetic energy. 4. Evaluate The negative sign indicates that kinetic energy is lost.

246 Sample Problem, continued
Chapter 6 Sample Problem, continued Elastic Collisions A kg marble moving to the right at m/s makes an elastic head-on collision with a kg shooter marble moving to the left at m/s. After the collision, the smaller marble moves to the left at m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the kg marble after the collision?

247 Sample Problem, continued
Section 3 Elastic and Inelastic Collisions Chapter 6 Sample Problem, continued Elastic Collisions 1. Define Given: m1 = kg m2 = kg v1,i = m/s to the right, v1,i = m/s v2,i = m/s to the left, v2,i = –0.180 m/s v1,f = m/s to the left, v1,i = –0.315 m/s Unknown: v2,f = ?

248 Sample Problem, continued
Chapter 6 Sample Problem, continued Elastic Collisions 2. Plan Choose an equation or situation: Use the equation for the conservation of momentum to find the final velocity of m2, the kg marble. m1v1,i + m2v2,i = m1v1,f + m2v2,f Rearrange the equation to isolate the final velocity of m2.

249 Sample Problem, continued
Chapter 6 Sample Problem, continued Elastic Collisions 3. Calculate Substitute the values into the equation and solve: The rearranged conservation-of-momentum equation will allow you to isolate and solve for the final velocity.

250 Sample Problem, continued
Chapter 6 Sample Problem, continued Elastic Collisions 4. Evaluate Confirm your answer by making sure kinetic energy is also conserved using these values.

251 Standardized Test Prep

252 Chapter 6 Multiple Choice
Standardized Test Prep Multiple Choice 1. If a particle’s kinetic energy is zero, what is its momentum? A. zero B. 1 kg • m/s C. 15 kg • m/s D. negative

253 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 1. If a particle’s kinetic energy is zero, what is its momentum? A. zero B. 1 kg • m/s C. 15 kg • m/s D. negative

254 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 2. The vector below represents the momentum of a car traveling along a road. The car strikes another car, which is at rest, and the result is an inelastic collision. Which of the following vectors represents the momentum of the first car after the collision? F. G. H. J.

255 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 2. The vector below represents the momentum of a car traveling along a road. The car strikes another car, which is at rest, and the result is an inelastic collision. Which of the following vectors represents the momentum of the first car after the collision? F. G. H. J.

256 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 3. What is the momentum of a kg baseball thrown with a velocity of 35 m/s toward home plate? A. 5.1 kg • m/s toward home plate B. 5.1 kg • m/s away from home plate C. 5.2 kg • m/s toward home plate D. 5.2 kg • m/s away from home plate

257 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 3. What is the momentum of a kg baseball thrown with a velocity of 35 m/s toward home plate? A. 5.1 kg • m/s toward home plate B. 5.1 kg • m/s away from home plate C. 5.2 kg • m/s toward home plate D. 5.2 kg • m/s away from home plate

258 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 4. What is the final velocity of the second pin if the first pin moves to the right at 0.5 m/s after the collision? F. 2.5 m/s to the left G. 2.5 m/s to the right H. 3.0 m/s to the left J. 3.0 m/s to the right

259 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 4. What is the final velocity of the second pin if the first pin moves to the right at 0.5 m/s after the collision? F. 2.5 m/s to the left G. 2.5 m/s to the right H. 3.0 m/s to the left J. 3.0 m/s to the right

260 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 5. What is the final velocity of the second pin if the first pin stops moving when it hits the second pin? A. 2.5 m/s to the left B. 2.5 m/s to the right C. 3.0 m/s to the left D. 3.0 m/s to the right

261 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 5. What is the final velocity of the second pin if the first pin stops moving when it hits the second pin? A. 2.5 m/s to the left B. 2.5 m/s to the right C. 3.0 m/s to the left D. 3.0 m/s to the right

262 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 6. For a given change in momentum, if the net force that is applied to an object increases, what happens to the time interval over which the force is applied? F. The time interval increases. G. The time interval decreases. H. The time interval stays the same. J. It is impossible to determine the answer from the given information.

263 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 6. For a given change in momentum, if the net force that is applied to an object increases, what happens to the time interval over which the force is applied? F. The time interval increases. G. The time interval decreases. H. The time interval stays the same. J. It is impossible to determine the answer from the given information.

264 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 7. Which equation expresses the law of conservation of momentum? A. p = mv B. m1v1,i + m2v2,i = m1v1,f + m2v2,f C. (1/2)m1v1,i2 + m2v2,i2 = (1/2)(m1 + m2)vf2 D. KE = p

265 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 7. Which equation expresses the law of conservation of momentum? A. p = mv B. m1v1,i + m2v2,i = m1v1,f + m2v2,f C. (1/2)m1v1,i2 + m2v2,i2 = (1/2)(m1 + m2)vf2 D. KE = p

266 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 8. Two shuffleboard disks of equal mass, one of which is orange and one of which is yellow, are involved in an elastic collision. The yellow disk is initially at rest and is struck by the orange disk, which is moving initially to the right at 5.00 m/s. After the collision, the orange disk is at rest. What is the velocity of the yellow disk after the collision? F. zero G m/s to the left H m/s to the right J m/s to the right

267 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued 8. Two shuffleboard disks of equal mass, one of which is orange and one of which is yellow, are involved in an elastic collision. The yellow disk is initially at rest and is struck by the orange disk, which is moving initially to the right at 5.00 m/s. After the collision, the orange disk is at rest. What is the velocity of the yellow disk after the collision? F. zero G m/s to the left H m/s to the right J m/s to the right

268 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 9. What is the large bead’s velocity after the collision? A cm/s to the right B cm/s to the right C cm/s to the right D cm/s to the right

269 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 9. What is the large bead’s velocity after the collision? A cm/s to the right B cm/s to the right C cm/s to the right D cm/s to the right

270 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 10. What is the total kinetic energy of the system after the collision? F  10–4 J G  10–4 J H  10 –4 J J  10 –4 J

271 Multiple Choice, continued
Chapter 6 Standardized Test Prep Multiple Choice, continued Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 10. What is the total kinetic energy of the system after the collision? F  10–4 J G  10–4 J H  10 –4 J J  10 –4 J

272 Chapter 6 Short Response
Standardized Test Prep Short Response 11. Is momentum conserved when two objects with zero initial momentum push away from each other?

273 Short Response, continued
Chapter 6 Standardized Test Prep Short Response, continued 11. Is momentum conserved when two objects with zero initial momentum push away from each other? Answer: yes

274 Short Response, continued
Chapter 6 Standardized Test Prep Short Response, continued 12. In which type of collision is kinetic energy conserved? What is an example of this type of collision?

275 Short Response, continued
Chapter 6 Standardized Test Prep Short Response, continued 12. In which type of collision is kinetic energy conserved? Answer: elastic collision What is an example of this type of collision? Answer: Two billiard balls collide and then move separately after the collision.

276 Chapter 7 Circular Motion and Gravitation

277 Chapter 7 Table of Contents Section 1 Circular Motion
Circular Motion and Gravitation Table of Contents Section 1 Circular Motion Section 2 Newton’s Law of Universal Gravitation Section 3 Motion in Space Section 4 Torque and Simple Machines

278 Section 1 Circular Motion
Chapter 7 Objectives Solve problems involving centripetal acceleration. Solve problems involving centripetal force. Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the centripetal force.

279 Chapter 7 Tangential Speed
Section 1 Circular Motion Chapter 7 Tangential Speed The tangential speed (vt) of an object in circular motion is the object’s speed along an imaginary line drawn tangent to the circular path. Tangential speed depends on the distance from the object to the center of the circular path. When the tangential speed is constant, the motion is described as uniform circular motion.

280 Centripetal Acceleration
Section 1 Circular Motion Chapter 7 Centripetal Acceleration

281 Centripetal Acceleration
Section 1 Circular Motion Chapter 7 Centripetal Acceleration The acceleration of an object moving in a circular path and at constant speed is due to a change in direction. An acceleration of this nature is called a centripetal acceleration.

282 Centripetal Acceleration, continued
Section 1 Circular Motion Chapter 7 Centripetal Acceleration, continued (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes. (b) For short time intervals, ∆v is directed toward the center of the circle. Centripetal acceleration is always directed toward the center of a circle.

283 Centripetal Acceleration, continued
Section 1 Circular Motion Chapter 7 Centripetal Acceleration, continued You have seen that centripetal acceleration results from a change in direction. In circular motion, an acceleration due to a change in speed is called tangential acceleration. To understand the difference between centripetal and tangential acceleration, consider a car traveling in a circular track. Because the car is moving in a circle, the car has a centripetal component of acceleration. If the car’s speed changes, the car also has a tangential component of acceleration.

284 Chapter 7 Centripetal Force Section 1 Circular Motion
Consider a ball of mass m that is being whirled in a horizontal circular path of radius r with constant speed. The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force. This net force, which is is directed toward the center of the circle, is a centripetal force.

285 Centripetal Force, continued
Section 1 Circular Motion Chapter 7 Centripetal Force, continued Newton’s second law can be combined with the equation for centripetal acceleration to derive an equation for centripetal force:

286 Centripetal Force, continued
Section 1 Circular Motion Chapter 7 Centripetal Force, continued Centripetal force is simply the name given to the net force on an object in uniform circular motion. Any type of force or combination of forces can provide this net force. For example, friction between a race car’s tires and a circular track is a centripetal force that keeps the car in a circular path. As another example, gravitational force is a centripetal force that keeps the moon in its orbit.

287 Centripetal Force, continued
Section 1 Circular Motion Chapter 7 Centripetal Force, continued If the centripetal force vanishes, the object stops moving in a circular path. A ball that is on the end of a string is whirled in a vertical circular path. If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. If the string breaks at the top of the ball’s path, as in (b), the ball will move along a parabolic path.

288 Describing a Rotating System
Section 1 Circular Motion Chapter 7 Describing a Rotating System To better understand the motion of a rotating system, consider a car traveling at high speed and approaching an exit ramp that curves to the left. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. What causes the passenger to move toward the door?

289 Describing a Rotating System, continued
Section 1 Circular Motion Chapter 7 Describing a Rotating System, continued As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight path. If a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat. If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath.

290 Section 2 Newton’s Law of Universal Gravitation
Chapter 7 Objectives Explain how Newton’s law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides. Apply Newton’s law of universal gravitation to solve problems.

291 Chapter 7 Gravitational Force Orbiting objects are in free fall.
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force Orbiting objects are in free fall. To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop. Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

292 Gravitational Force, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued The centripetal force that holds the planets in orbit is the same force that pulls an apple toward the ground—gravitational force. Gravitational force is the mutual force of attraction between particles of matter. Gravitational force depends on the masses and on the distance between them.

293 Gravitational Force, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued Newton developed the following equation to describe quantitatively the magnitude of the gravitational force if distance r separates masses m1 and m2: The constant G, called the constant of universal gravitation, equals  10–11 N•m2/kg.

294 Newton’s Law of Universal Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Newton’s Law of Universal Gravitation

295 Gravitational Force, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued The gravitational forces that two masses exert on each other are always equal in magnitude and opposite in direction. This is an example of Newton’s third law of motion. One example is the Earth-moon system, shown on the next slide. As a result of these forces, the moon and Earth each orbit the center of mass of the Earth-moon system. Because Earth has a much greater mass than the moon, this center of mass lies within Earth.

296 Newton’s Law of Universal Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Newton’s Law of Universal Gravitation

297 Applying the Law of Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation Newton’s law of gravitation accounts for ocean tides. High and low tides are partly due to the gravitational force exerted on Earth by its moon. The tides result from the difference between the gravitational force at Earth’s surface and at Earth’s center.

298 Applying the Law of Gravitation, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued Cavendish applied Newton’s law of universal gravitation to find the value of G and Earth’s mass. When two masses, the distance between them, and the gravitational force are known, Newton’s law of universal gravitation can be used to find G. Once the value of G is known, the law can be used again to find Earth’s mass.

299 Applying the Law of Gravitation, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued Gravity is a field force. Gravitational field strength, g, equals Fg/m. The gravitational field, g, is a vector with magnitude g that points in the direction of Fg. Gravitational field strength equals free-fall acceleration. The gravitational field vectors represent Earth’s gravitational field at each point.

300 Applying the Law of Gravitation, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued weight = mass  gravitational field strength Because it depends on gravitational field strength, weight changes with location: On the surface of any planet, the value of g, as well as your weight, will depend on the planet’s mass and radius.

301 Chapter 7 Objectives Describe Kepler’s laws of planetary motion.
Section 3 Motion in Space Chapter 7 Objectives Describe Kepler’s laws of planetary motion. Relate Newton’s mathematical analysis of gravitational force to the elliptical planetary orbits proposed by Kepler. Solve problems involving orbital speed and period.

302 Section 3 Motion in Space
Chapter 7 Kepler’s Laws Kepler’s laws describe the motion of the planets. First Law: Each planet travels in an elliptical orbit around the sun, and the sun is at one of the focal points. Second Law: An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals. Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun.

303 Kepler’s Laws, continued
Section 3 Motion in Space Chapter 7 Kepler’s Laws, continued Kepler’s laws were developed a generation before Newton’s law of universal gravitation. Newton demonstrated that Kepler’s laws are consistent with the law of universal gravitation. The fact that Kepler’s laws closely matched observations gave additional support for Newton’s theory of gravitation.

304 Kepler’s Laws, continued
Section 3 Motion in Space Chapter 7 Kepler’s Laws, continued According to Kepler’s second law, if the time a planet takes to travel the arc on the left (∆t1) is equal to the time the planet takes to cover the arc on the right (∆t2), then the area A1 is equal to the area A2. Thus, the planet travels faster when it is closer to the sun and slower when it is farther away.

305 Kepler’s Laws, continued
Section 3 Motion in Space Chapter 7 Kepler’s Laws, continued Kepler’s third law states that T2  r3. The constant of proportionality is 4p2/Gm, where m is the mass of the object being orbited. So, Kepler’s third law can also be stated as follows:

306 Kepler’s Laws, continued
Section 3 Motion in Space Chapter 7 Kepler’s Laws, continued Kepler’s third law leads to an equation for the period of an object in a circular orbit. The speed of an object in a circular orbit depends on the same factors: Note that m is the mass of the central object that is being orbited. The mass of the planet or satellite that is in orbit does not affect its speed or period. The mean radius (r) is the distance between the centers of the two bodies.

307 Section 3 Motion in Space
Chapter 7 Planetary Data

308 Chapter 7 Sample Problem Period and Speed of an Orbiting Object
Section 3 Motion in Space Chapter 7 Sample Problem Period and Speed of an Orbiting Object Magellan was the first planetary spacecraft to be launched from a space shuttle. During the spacecraft’s fifth orbit around Venus, Magellan traveled at a mean altitude of 361km. If the orbit had been circular, what would Magellan’s period and speed have been?

309 Sample Problem, continued
Section 3 Motion in Space Chapter 7 Sample Problem, continued 1. Define Given: r1 = 361 km = 3.61  105 m Unknown: T = ? vt = ? 2. Plan Choose an equation or situation: Use the equations for the period and speed of an object in a circular orbit.

310 Sample Problem, continued
Section 3 Motion in Space Chapter 7 Sample Problem, continued Use Table 1 in the textbook to find the values for the radius (r2) and mass (m) of Venus. r2 = 6.05  106 m m = 4.87  1024 kg Find r by adding the distance between the spacecraft and Venus’s surface (r1) to Venus’s radius (r2). r = r1 + r2 r = 3.61  105 m  106 m = 6.41  106 m

311 Sample Problem, continued
Section 3 Motion in Space Chapter 7 Sample Problem, continued 3. Calculate 4. Evaluate Magellan takes (5.66  103 s)(1 min/60 s)  94 min to complete one orbit.

312 Weight and Weightlessness
Section 3 Motion in Space Chapter 7 Weight and Weightlessness To learn about apparent weightlessness, imagine that you are in an elevator: When the elevator is at rest, the magnitude of the normal force acting on you equals your weight. If the elevator were to accelerate downward at 9.81 m/s2, you and the elevator would both be in free fall. You have the same weight, but there is no normal force acting on you. This situation is called apparent weightlessness. Astronauts in orbit experience apparent weightlessness.

313 Weight and Weightlessness
Section 3 Motion in Space Chapter 7 Weight and Weightlessness

314 Chapter 7 Multiple Choice
Standardized Test Prep Multiple Choice 1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object.

315 Chapter 7 Multiple Choice
Standardized Test Prep Multiple Choice 1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object.

316 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? F. 2.4  10-2 m/s2 G m/s2 H. 9.0 m/s2 J. zero

317 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? F. 2.4  10-2 m/s2 G m/s2 H. 9.0 m/s2 J. zero

318 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 3. What is the most direct cause of the car’s centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road

319 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 3. What is the most direct cause of the car’s centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road

320 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 4. Earth (m = 5.97  1024 kg) orbits the sun (m =  1030 kg) at a mean distance of 1.50  1011 m. What is the gravitational force of the sun on Earth? (G =  N•m2/kg2) F  1032 N G  1022 N H  10–2 N J  10–8 N

321 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 4. Earth (m = 5.97  1024 kg) orbits the sun (m =  1030 kg) at a mean distance of 1.50  1011 m. What is the gravitational force of the sun on Earth? (G =  N•m2/kg2) F  1032 N G  1022 N H  10–2 N J  10–8 N

322 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 5. Which of the following is a correct interpretation of the expression ? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretations.

323 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 5. Which of the following is a correct interpretation of the expression ? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretations.

324 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 6. What data do you need to calculate the orbital speed of a satellite? F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbit H. mass of satellite and radius of orbit only J. mass of planet and radius of orbit only

325 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 6. What data do you need to calculate the orbital speed of a satellite? F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbit H. mass of satellite and radius of orbit only J. mass of planet and radius of orbit only

326 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 7. Which of the following choices correctly describes the orbital relationship between Earth and the sun? A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle. C. The sun orbits Earth in an ellipse, with Earth at one focus. D. Earth orbits the sun in an ellipse, with the sun

327 Multiple Choice, continued
Chapter 7 Standardized Test Prep Multiple Choice, continued 7. Which of the following choices correctly describes the orbital relationship between Earth and the sun? A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle. C. The sun orbits Earth in an ellipse, with Earth at one focus. D. Earth orbits the sun in an ellipse, with the sun

328 Chapter 7 Short Response
Standardized Test Prep Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below.

329 Chapter 7 Short Response
Standardized Test Prep Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below. Answer: The water remains in the pail even when the pail is upside down because the water tends to move in a straight path due to inertia.

330 Short Response, continued
Chapter 7 Standardized Test Prep Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth.

331 Short Response, continued
Chapter 7 Standardized Test Prep Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth. Answer: The moon’s tidal forces create two bulges on Earth. As Earth rotates on its axis once per day, any given point on Earth passes through both bulges.

332 Short Response, continued
Chapter 7 Standardized Test Prep Short Response, continued 16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an example.

333 Short Response, continued
Chapter 7 Standardized Test Prep Short Response, continued 16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an example. Answer: You would have to apply the input force over a greater distance. Examples may include any machines that increase output force at the expense of input distance.

334 Chapter 7 Extended Response
Standardized Test Prep Extended Response 17. Mars orbits the sun (m = 1.99  1030 kg) at a mean distance of 2.28  1011 m. Calculate the length of the Martian year in Earth days. Show all of your work. (G =  10–11 N•m2/kg2)

335 Chapter 7 Extended Response
Standardized Test Prep Extended Response 17. Mars orbits the sun (m = 1.99  1030 kg) at a mean distance of 2.28  1011 m. Calculate the length of the Martian year in Earth days. Show all of your work. (G =  10–11 N•m2/kg2) Answer: 687 days


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