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. Example 1.18 analyzes the upward part of the motion of an object thrown upward, whereas Example 1.19 analyzes the same object’s downward motion. Example 1.18 A clown throws a ball upward at 10.00 m/s. Find (a) the maximum height the ball reaches above its launch height (b) the time it takes to do so Given Consider up to be positive. 10.00 m/s [up] 10.00 m/s vi a 9.81 m/s2 [down] 9.81 m/s2 Required (a) maximum height above launch height (d) (b) time taken to reach maximum height (t) Analysis and Solution (a) When you throw an object up, as its height increases, its speed decreases because the object is accelerating downward due to gravity. The ball, travelling upward away from its initial launch height, reaches its maximum height when its vertical velocity is zero. In other words, the object stops for an instant at the top of its path up, so vf neglecting air friction, use the equation vf substitute scalar quantities. 0.00 m/s. To find the object’s maximum height, 2 2ad and 2 vi Practice Problem 1. The Slingshot drops riders 27 m from rest before slowing them down to a stop. How fast are they moving before they start slowing down? Answer 1. 23 m/s e WEB Can you shoot an object fast enough so that it does not return to Earth? Research escape velocity. Is it the same regardless of the size of an object? How do you calculate it? Write a brief summary of your findings. To learn more about escape velocity, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 59 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 60 d 2 vi 2 vf 2a m m 2 2 10.00 0.00 s s m 29.81 s2 5.10 m v, where a is the (b) To find the time taken, use the equation a t acceleration due to gravity. Substitute scalar quantities because you are dividing vectors. t vi vf a m m 10.00 0.00 s s m 9.81 s2 1.02 s Paraphrase (a) The ball’s maximum height is 5.10
m above its launch height. (b) It takes the ball 1.02 s to reach maximum height. The next example is a continuation of the previous example: It analyzes the same ball’s motion as it falls back down from its maximum height. Example 1.19 A clown throws a ball upward at 10.00 m/s. Find (a) the time it takes the ball to return to the clown’s hand from maximum height (b) the ball’s final velocity Given Consider up to be positive. v 10.00 m/s [up] 10.00 m/s a 9.81 m/s2 [down] 9.81 m/s2 i Practice Problems 1. A pebble falls from a ledge 20.0 m high. (a) Find the velocity with which it hits the ground. (b) Find the time it takes to hit the ground. Answers 1. (a) 19.8 m/s [down] (b) 2.02 s 60 Unit I Kinematics Required (a) time taken to land (t) (b) final velocity (v f) Analysis and Solution (a) For an object starting from rest at maximum height and accelerating downward due to gravity, its motion is described by the equation d v i t a(t)2, where 1 2 i 0 (at maximum height). For downward motion, the v ball’s displacement and acceleration are in the same direction, so use the scalar form of the equation. For d, substitute 5.10 m (from Example 1.18(a)). Rearrange this equation and substitute the values. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 61 t 2d a 2(5.10 m) m 9.81 s2 1.02 s Compare this time to the time taken to reach maximum height (Example 1.18(b)). (b) The ball’s final velocity (starting from maximum height) when it info BIT At terminal velocity, parachuters no longer accelerate but fall at a constant speed. Humans have a terminal velocity of about 321 km/h [down] when curled up and about 201 km/h [down] with arms and legs fully extended to catch the wind. v lands on the ground is v a i t at v f f i v 0.00 m/s (9.81 m/s2)(1.02
s) 10.0 m/s The negative sign means that the direction is downward. Paraphrase (a) It takes the ball 1.02 s to return to the clown’s hand. (b) The final velocity at the height of landing is 10.0 m/s [down]. Concept Check (a) Why does it make sense that the time taken to travel up to the maximum height is equal to the time to fall back down to the starting height? (b) What variables determine how long a projectile is in the air? Does the answer surprise you? Why or why not? Position vs. Time ) ].00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 0.50 1.00 2.00 2.50 1.50 Time (s) You can use the data calculated in Examples 1.18 and 1.19 to plot a position-time graph of the ball’s motion. Because the ball experiences uniformly accelerated motion, the graph is a parabola (Figure 1.64). Figure 1.64 The positiontime graph of a ball thrown vertically upward is a parabola. A Graphical Representation of a Vertical Projectile You can now represent the motion of the juggler’s ball on a position-time graph. Remember that the ball’s motion can be divided into three different stages: Its speed decreases, becomes zero, and then increases. However, the velocity is uniformly decreasing. The graphs that correspond to these three stages of motion are shown in Figure 1.65. Position vs. Time Position vs. Time Position vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.65(a) Consider up to be positive. The ball rises until it stops. Figure 1.65(b) momentarily at maximum height. The ball stops Figure 1.65(c) back down to its launch height. The ball falls Chapter 1 Graphs and equations describe motion in one dimension. 61 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 62 Position vs. Time ) ] Time (s) Figure 1.66 A ball thrown straight up in the air illustrates uniformly accelerated motion. Now put these three graphs together to generate the complete positiontime graph of the ball’s motion. Remember that the ball is actually moving straight up and down, and not in a parabolic path (Figure 1.66
). Why is the graph of its motion a parabola rather than a straight vertical line? To generate a corresponding velocity-time graph from the positiontime graph in Figure 1.66, draw a series of tangents at specific time instances. Choosing strategic points will make your task easier. The best points to choose are those that begin and end a stage of motion because they define that stage (Figure 1.67(a)). Position vs. Time (a) 6.00 5.00 4.00 3.00 2.00 1.00 ) ].00 0.00 0.50 1.00 1.50 2.00 2.50 Time (s) 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 (b 10.00 15.00 Figure 1.67 To generate the velocity-time graph in (b) corresponding to the positiontime graph in (a), draw tangents at strategic points. info BIT In 1883, the Krakatoa volcano in Indonesia hurled rocks 55 km into the air. This volcanic eruption was 10 000 times more powerful than the Hiroshima bomb! Can you find the time it took for the rocks to reach maximum height? In Figure 1.67(a), notice that the initial slope of the tangent on the positiontime graph is positive, corresponding to an initial positive (upward) velocity on the velocity-time graph below (Figure 1.67(b)). The last tangent has a negative slope, corresponding to a final negative velocity on the velocity-time graph. The middle tangent is a horizontal line (slope equals zero), which means that the ball stopped momentarily. Remember that the slope of a velocity-time graph represents acceleration. Concept Check What should be the value of the slope of the velocity-time graph for vertical projectile motion? 62 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 63 1.6 Check and Reflect 1.6 Check and Reflect Knowledge 1. Define a projectile. 2. What determines how long it will take an object to reach the ground when released with an initial velocity of zero? Applications 11. A penny is dropped from a cliff of height 190 m. Determine the time it takes for the penny to hit the bottom of the cliff. 12. A coin tossed straight up into the air takes 2.
75 s to go up and down from its initial release point 1.30 m above the ground. What is its maximum height? 3. A student drops a bran muffin from the roof of the school. From what height is the muffin dropped if it hits the ground 3.838 s later? 13. If a diver starts from rest, determine the amount of time he takes to reach the water’s surface from the 10-m platform. 4. During a babysitting assignment, a babysitter is constantly picking up toys dropped from the infant’s highchair. If the toys drop from rest and hit the floor 0.56 s later, from what height are they being dropped? 5. A rock takes 1.575 s to drop 2.00 m down toward the surface of the Moon. Determine the acceleration due to gravity on the Moon. 6. At the beginning of a game, a referee throws a basketball vertically upward with an initial speed of 5.0 m/s. Determine the maximum height above the floor reached by the basketball if it starts from a height of 1.50 m. 7. A student rides West Edmonton Mall’s Drop of Doom. If the student starts from rest and falls due to gravity for 2.6 s, what will be his final velocity and how far will he have fallen? 8. If the acceleration due to gravity on Jupiter is 24.8 m/s2 [down], determine the time it takes for a tennis ball to fall 1.75 m from rest. 9. If a baseball popped straight up into the air has a hang time (length of time in the air) of 6.25 s, determine the distance from the point of contact to the baseball’s maximum height. 10. Jumping straight up, how long will a red kangaroo remain in the air if it jumps through a height of 3.0 m? 14. A person in an apartment building is 5.0 m above a person walking below. She plans to drop some keys to him. He is currently walking directly toward a point below her at 2.75 m/s. How far away is he if he catches the keys 1.25 m above the ground? Extensions 15. A rocket launched vertically upward accelerates uniformly for 50 s until it reaches a velocity of 200 m/s [up]. At that instant, its fuel runs out. (a) Calculate the rocket’s acceleration. (b) Calculate the height
of the rocket when its fuel runs out. (c) Explain why the rocket continues to gain height for 20 s after its fuel runs out. (d) Calculate the maximum height of the rocket. 16. A ball is dropped from a height of 60.0 m. A second ball is thrown down 0.850 s later. If both balls reach the ground at the same time, what was the initial velocity of the second ball? e TEST To check your understanding of projectiles and acceleration due to gravity, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 63 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 64 CHAPTER 1 SUMMARY Key Terms and Concepts distance displacement velocity uniform motion at rest acceleration non-uniform motion instantaneous velocity tangent uniformly accelerated motion projectile motion projectile acceleration due to gravity kinematics origin position scalar quantity vector quantity Key Equations v f v i)t d vv 1 t a(t)2 2 i d v 1 t a(t)2 2 f 2 vi vf 2 2ad Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. kinematics uniform motion at rest constant velocity accelerated motion uniform displacement : zero dt graph— vt graph— at graph— equations: none displacement : equal changes over equal time intervals dt graph— vt graph— horizontal line (/) at graph— displacement : changes dt graph— curve vt graph— at graph— horizontal line (/) equations: equations: 1 d (vf vi )t 2 vf 2 vi 2 2ad 64 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 65 CHAPTER 1 REVIEW Knowledge 1. (1.1) State two ways in which a vector quantity differs from a scalar quantity. Give an example of each. 2. (1.2) Complete a position-time data table for the motion described by the ticker tape given below. 3. (1.3) Determine the velocity of each object whose motion is represented by the graphs below. (a) (b) (c 12 10 8 6 4 2 0 Position vs. Time 0 2 4 6 8 10 12 Time (s) Position vs. Time
Time (min) 2 4 6 8 10 12 Position vs. Time Time (s) 5 10 15 20 ) ] 10 5 0 5 10 15 20 25 ) ] 10 5 0 5 10 15 20 25 30 4. (1.5) What is a vehicle’s displacement if it travels at a velocity of 30.0 m/s [W] for 15.0 min? 5. (1.5) How long will it take a cross-country skier, travelling 5.0 km/h, to cover a distance of 3.50 km? 6. (1.2) Determine the average speed, average velocity, and net displacement from the position-time graph below. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 Time (s) 7. (1.2) Explain how a person standing still could have the same average velocity but different average speed than a person running around a circular track. 8. (1.6) If an object thrown directly upwards remains in the air for 5.6 s before it returns to its original position, how long did it take to reach its maximum height? 9. (1.6) If an object thrown directly upwards reaches its maximum height in 3.5 s, how long will the object be in the air before it returns to its original position? Assume there is no air resistance. 10. (1.6) What is the initial vertical velocity for an object that is dropped from a height, h? Applications 11. A scuba diver swims at a constant speed of 0.77 m/s. How long will it take the diver to travel 150 m at this speed? 12. In 1980, during the Marathon of Hope, Terry Fox ran 42 km [W] a day. Assuming he ran for 8.0 h a day, what was his average velocity in m/s? 13. Explain how the point of intersection of two functions on a position-time graph differs from the point of intersection of two functions on a velocity-time graph. 14. A thief snatches a handbag and runs north at 5.0 m/s. A police officer, 20 m to the south
, sees the event and gives chase. If the officer is a good sprinter, going 7.5 m/s, how far will she have to run to catch the thief? Chapter 1 Graphs and equations describe motion in one dimension. 65 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 66 Maintaining that acceleration, how long will it take the police car to catch up with the speeding motorist? At what speed would the police car be moving? Explain whether or not this scenario is likely to happen. 25. Two cars pass one another while travelling on a city street. Using the velocity-time graph below, draw the corresponding position-time graph and determine when and where the two cars pass one another. Velocity vs. Time ) ] 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 26. Calculate displacement and acceleration from the graph below. Velocity vs. Time Time (h) 1.0 2.0 3.0 4.0 5..0 0.0 1.0 2.0 3.0 4.0 5.0 27. Built in Ontario, the world’s fastest fire truck, the Hawaiian Eagle, can accelerate at 9.85 m/s2 [forward]. Starting from rest, how long will it take the Hawaiian Eagle to travel a displacement of 402 m [forward]? 28. A vehicle is travelling at 25.0 m/s. Its brakes provide an acceleration of 3.75 m/s2 [forward]. What is the driver’s maximum reaction time if she is to avoid hitting an obstacle 95.0 m away? 29. Off-ramps are designed for motorists to decrease their vehicles’ velocities to move seamlessly into city traffic. If the off-ramp is 1.10 km long, calculate the magnitude of a vehicle’s acceleration if it reduces its speed from 110.0 km/h to 60.0 km/h. 15. Calculate the magnitude of a bullet’s acceleration if it travels at a speed of 1200 m/s and stops within a bulletproof vest that is 1.0 cm thick. 16. From the velocity-time graph below, determine how far an elk will travel in 30 min. Velocity vs
. Time for an Elk ) ] 80 70 60 50 40 30 20 10 0 0.0 0.5 1.0 Time (h) 1.5 2.0 17. The world record for a speedboat is 829 km/h. Heading south, how far will the boat travel in 2.50 min? 18. How much faster is an airliner than a stagecoach if the stagecoach takes 24 h to travel 300 km and the airliner takes 20 min? 19. A car’s odometer reads 22 647 km at the start of a trip and 23 209 km at the end. If the trip took 5.0 h, what was the car’s average speed in km/h and m/s? 20. A motorcycle coasts downhill from rest with a constant acceleration. If the motorcycle moves 90.0 m in 8.00 s, find its acceleration and velocity after 8.00 s. 21. A cyclist crosses a 30.0-m bridge in 4.0 s. If her initial velocity was 5.0 m/s [N], find her acceleration and velocity at the other end of the bridge. 22. An object with an initial velocity of 10.0 m/s [S] moves 720 m in 45.0 s along a straight line with constant acceleration. For the 45.0-s interval, find its average velocity, final velocity, and acceleration. 23. During qualifying heats for the Molson Indy, a car must complete a 2.88-km lap in 65 s. If the car goes 60 m/s for the first half of the lap, what must be its minimum speed for the second half to still qualify? 24. A car travelling 19.4 m/s passes a police car at rest. As it passes, the police car starts up, accelerating with a magnitude of 3.2 m/s2. 66 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 67 30. Calgary’s CTrain leaves the 10 St. S.W. station at 4:45 p.m. and arrives at 3 St. S.E. at 4:53 p.m. If the distance between the two stops is 3.2 km, determine the CTrain’s average velocity for the trip. 37. A contractor drops a bolt from the top of a roof located 8.52 m above the ground. How long does it
take the bolt to reach the ground, assuming there is no air resistance? McKnight – Westwinds (2010) Whitehorn Rundle Marlborough 38. An improperly installed weathervane falls from the roof of a barn and lands on the ground 1.76 s later. From what height did the weathervane fall and how fast was it travelling just before impact? Dalhousie Brentwood N CTrain Map University Banff Trail Lions Park Sunnyside BOWRIVER SAIT/ ACAD/ Jubllee 7 St. S W PRINCE’S ISLAND PARK 4 St. S W 1 St. S W Bridgeland/ M e m orial Zoo Franklin Barlo w/ M ax Bell Olym pic Plaza 3 St. SE City Hall 10 St. S W 8 St. S W 6 St. S W 3 St. S W Centre St Erlton/Stampede Victoria Park/Stampede B O W RIVER 201 Somerset/Bridlewood/Dalhousie 202 10 St./Whitehorn 7 Avenue Free Fare Zone 201 & 202 Transfer Points Chinook Southland Canyon Meadows NEW Shawnessy 39 Avenue Heritage Anderson Fish Creek – Lacombe Somerset/Bridlewood NEW 31. Describe the motion of the truck from the velocity-time graph below. When is the truck at rest? travelling with a uniform velocity? When is its acceleration the greatest? Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 4.0 6.0 Time (s) 8.0 10.0 32. A racecar accelerates uniformly from 17.5 m/s [W] to 45.2 m/s [W] in 2.47 s. Determine the acceleration of the racecar. 33. How long will it take a vehicle travelling 80 km/h [W] to stop if the average stopping distance for that velocity is 76.0 m? 34. The Slingshot, an amusement park ride, propels its riders upward from rest with an acceleration of 39.24 m/s2. How long does it take to reach a height of 27.0 m? Assume uniform acceleration. 35. Starting from rest, a platform diver hits the water with a speed of 55 km/h. From what height did she start her descent into the pool? 36. A circus performer can land safely on the ground at speeds
up to 13.5 m/s. What is the greatest height from which the performer can fall? 39. Attempting to beat the record for tallest Lego structure, a student drops a piece from a height of 24.91 m. How fast will the piece be travelling when it is 5.0 m above the ground and how long will it take to get there? Extension 40. Weave zones are areas on roads where vehicles are changing their velocities to merge onto and off of busy expressways. Suggest criteria a design engineer must consider in developing a weave zone. Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed on page 64 and the Learning Outcomes on page 4. 1. Create a flowchart to describe the changes in position, velocity, and acceleration for both uniform and accelerated motion. 2. Write a paragraph explaining the two main functions of graphing in kinematics. Share your report with another classmate. Think About It Review your answers to the Think About It questions on page 5. How would you answer each question now? e TEST To check your understanding of motion in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 67 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 68 Vector components describe motion in two dimensions. Figure 2.1 The motion of Canada’s Snowbird precision flight squad can be described using vectors. Imagine being a pilot for the Canadian Snowbirds (Figure 2.1). This precision flight team, composed of highly trained military personnel, performs at air shows across the country. Unlike the flight crew in the cockpit of a commercial airliner, these pilots execute aerobatic manoeuvres that require motion in both horizontal and vertical directions, while being acutely aware of their positions relative to the ground and to each other. In this chapter, you will study motion in one and two dimensions by building on the concepts you learned in Chapter 1. You will use vectors to define position, velocity, and acceleration, and their interrelationships. The vector methods you will learn will allow you to study more complex motions. C H A P T E R 2 Key Concepts In this chapter, you will learn about: two-
dimensional motion vector methods Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain two-dimensional motion in a horizontal or vertical plane interpret the motion of one object relative to another Science, Technology, and Society explain that scientific knowledge is subject to change as new evidence comes to light and as laws and theories are tested, restricted, revised, or reinforced 68 Unit I 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 69 2-1 QuickLab 2-1 QuickLab Taking a One-dimensional Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m tape measure field marker (tape or flag) Procedure 1 Starting at the centre of a football field (or gymnasium), work out a path sequence using six forward and backward displacements to move from your starting position to the goal line. At least two of your six displacements must be oriented in the direction opposite to the direction of the goal line. 2 On the field, mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacements of the path sequence chosen in step 1 while holding the end of the measuring tape. Mark your partner’s endpoint after each displacement (Figure 2.2). 4 Continue the journey, using the measuring tape, until you have walked all the displacements. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. 7 Repeat steps 2–6 using two different sequences of the six displacements you originally chose. 5 m [forward] Figure 2.2 Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? Think About It 1. How does the order of a series of displacements affect the final position of an object? 2. In order to cross a river in the shortest possible time, is it better to aim yourself upstream so that you end up swimming straight across or to aim straight across and swim at an angle downstream? 3. Why does it take longer to fly across Canada from east to west rather than west to east in the same airplane? 4. How does the angle of a throw affect the time a ball spends in the air? 5. Two objects start from the same height at the same time. One is dropped while the other is given an initial horizontal velocity. Which one hits the ground first? Discuss your answers in
a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 2 Vector components describe motion in two dimensions. 69 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 70 2.1 Vector Methods in One Dimension One of the fastest-growing sports in the world today is snowshoeing (Figure 2.3). The equipment required is minimal and the sport is easy to learn — you need to move forward in a straight line. Despite its simplicity, snowshoeing has great cardiovascular benefits: You can burn up to 1000 calories per hour, which makes it the ultimate cross-training program for athletes. It also allows athletes to explore different terrains and gain a greater appreciation of the outdoors, as well as to test their limits, especially by participating in endurance races! The motions in a showshoe race can be broken up into one-dimensional vector segments. In this section, you will study motion in one dimension using vectors. Figure 2.3 Snowshoeing is an excellent way of enjoying the great outdoors in winter while improving your health. Vector Diagrams In Chapter 1, you used variables and graphs to represent vector quantities. You can also represent vector quantities using vector diagrams. In a diagram, a line segment with an arrowhead represents a vector quantity. Its point of origin is called the tail, and its terminal point (arrowhead) is the tip (Figure 2.4). If the magnitude of a vector is given, you can draw the vector to scale. The length of the line segment depends on the vector’s magnitude. The arrowhead indicates direction. Drawing vector diagrams to represent motion helps you to visualize the motion of an object. Properly drawn, vector diagrams enable you to accurately add vectors and to determine an object’s position. tail tip Figure 2.4 A vector has a tail and a tip. 70 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 71 Choosing Reference Coordinates When describing the motion of an object, there are many ways to describe its direction. You could use adjectives such as forward or backward, up or down, into or out of, and left or right. You can also use compass directions, such as north [N], south [S], east [E], and west [W]. When drawing a vector diagram, it
is important to choose which directions are positive and to include these directions on every vector diagram. As you learned in section 1.1 (Figure 1.6), in this unit, forward, up, right, north, and east are usually designated as positive, whereas their opposites are usually considered negative. You may choose your own designation of positive and negative when solving problems, but make sure your reference direction is consistent within each problem and clearly communicated at the beginning of your solution. Practise drawing vectors in the next Skills Practice exercise Representing a Vector Using an appropriate scale and direction convention, draw each of the following vectors. (a) 5 m [forward] (b) 20 m [down] (c) 30 km [north] (d) 150 km [left] Adding Vectors in One Dimension Motion in one dimension involves vectors that are collinear. Collinear vectors lie along the same straight line. They may point in the same or in opposite directions (Figure 2.5). collinear: along the same straight line, either in the same or in opposite directions (a) (b) Figure 2.5 (a) Collinear vectors in the same direction (b) Collinear vectors in opposite directions When more than one vector describes motion, you need to add the vectors. You can add and subtract vectors graphically as well as algebraically, provided they represent the same quantity or measurement. As in mathematics, in which only like terms can be added, you can only add vectors representing the same types of quantities. For example, you can add two or more position vectors, but not a position vector, 5 m [E], to a velocity vector, 5 m/s [E]. In addition, the unit of measurement must be the same. For example, before adding the position vectors 5 m [E] and 10 km [E], you must convert the units of one of the vectors so that both vectors have the same units. In the next example, determine the sum of all the vector displacements graphically by adding them tip to tail. The sum of a series of vectors is called the resultant vector. resultant vector: a vector drawn from the tail of the first vector to the tip of the last vector Chapter 2 Vector components describe motion in two dimensions. 71 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 72 Example 2.1 Contestants in a snowshoe race must move forward 10.0 m, untie a
series of knots, move forward 5.0 m, solve a puzzle, and finally move forward 25.0 m to the finish line (Figure 2.6). Determine the resultant vector by adding the vectors graphically. Figure 2.6 Analysis and Solution 1. Choose an appropriate scale and reference direction. 1.0 cm : 5.0 m, forward is positive. 2. Draw the first vector and label its magnitude and direction (Figure 2.7). 10.0 m [forward] Figure 2.7 3. Place the tail of the second vector at the tip of the first vector. Continue to place all the remaining vectors in order, tip to tail (Figure 2.8). 10.0 m [forward] 5.0 m [forward] 25.0 m [forward] Figure 2.8 4. Connect the tail of the first vector to the tip of the last vector. This new vector, which points toward the tip of the last vector, is the resultant vector, R (the purple arrow in Figure 2.9). 72 Unit I Kinematics 02-PearsonPhys20-Chap02 7/25/08 7:57 AM Page 73 Tail (origin) of first vector Tip (terminal point) of last vector Figure 2.9 Find the magnitude of the resultant vector by measuring with a ruler, then convert the measured value using the scale. Remember to include the direction. d 8.0 cm [forward] 5.0 m 1.0 cm 40 m [forward] Practice Problems 1. The coach of the high-school rugby team made the team members run a series of sprints: 5.0 m [forward], 10 m [backward], 10 m [forward], 10 m [backward], 20 m [forward], 10 m [backward], 40 m [forward], and 10 m [backward]. (a) What is their total distance? (b) What is their displacement? Answers 1. (a) 115 m (b) 35 m [forward] In summary, you can see that adding vectors involves connecting them tip to tail. The plus sign in a vector equation tells you to connect the vectors tip to tail in the vector diagram. 2 d d 1: Add the negative of d To subtract collinear vectors graphically (Figure 2.10(a)), you may use one of two methods. For the first method, find d using the equation d 2, tip to tail, as you did in Example 2.1. The
negative of a vector creates a new vector that points in the opposite direction of the original vector (Figure 2.10(b)). For the second method, connect the vectors tail to tail. This time, d starts at the tip of d 1 and ends at the tip of d 2 (Figure 2.10(c)). 1 to d (a) (b) (c) d2 d1 d d2 d1 d2 d d1 Figure 2.10 d (a) To subtract two collinear vectors, d 2 or 1 to d 1, graphically, (b) add the negative of d 2 (c) connect the vectors tail to tail and draw the resultant connecting the tip of d 1 to the tip of d 2. Chapter 2 Vector components describe motion in two dimensions. 73 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 74 Recall that the definition of displacement is final position minus initial position, or d i. The next example reviews the algebraic subtraction of vectors, which you learned in Chapter 1, Example 1.1, and also shows you how to subtract vectors graphically. d d f Example 2.2 A sailboat that is initially 15 m to the right of a buoy sails 35 m to the left of the buoy (Figure 2.11). Determine the sailboat’s displacement (a) algebraically and (b) graphically. Figure 2.11 35 m [left] origin 15 m [right] Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. Using vector diagrams, determine the sprinter’s displacement from his initial position. 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. Using vector diagrams, determine the player’s displacement from the starting position. 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, use vector diagrams to calculate the distance and displacement her hand travels during that time. Check your answers against those in Example 1.1 Practice Problems 1-3. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m 74 Unit
I Kinematics Given Consider right to be positive. 15 m [right] 15 m d 35 m [left] 35 m d i f Required displacement (d ) Analysis and Solution (a) To find displacement algebraically, use the equation dd d d f i 35 m (15 m) 35 m 15 m 50 m The sign is negative, so the direction is to the left. (b) To find displacement graphically, subtract the two position vectors. Draw the vectors tail to tail and draw the resultant from the tip of the initial position vector to the tip of the final position vector (Figure 2.12). scale: 1.0 cm : 10 m df di d Figure 2.12 d 5.0 cm [left] 10 m 1.0 cm 50 m [left] Paraphrase The sailboat’s displacement is 50 m [left]. 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 75 For collinear vectors, find displacement by subtracting initial position from final position. Subtract vectors graphically by connecting them tail to tail or by reversing the direction of the initial position vector. Recall from Chapter 1 that direction for displacement is given with respect to initial position. 2.1 Check and Reflect 2.1 Check and Reflect Knowledge 1. Describe the similarities and differences between the two vectors drawn below. 5 m [E] 5 m [W] 2. Using the same scale and reference coordinates, compare the vectors 5 m [N] and 10 m [S]. 3. If the scale vector diagram of 5.0 m [S] is 6.0 cm long, what is the length of the scale vector diagram of 20 m [S]? 4. What scale is being used if 5.0 cm represents 100 km? Applications 5. The scale on a National Geographic world map is 1.0 cm : 520 km. On the map, 4.0 cm separates Alberta’s north and south provincial boundaries. What is the separation in kilometres? 6. During a tough drill on a field of length 100 yards, players run to each 10-yard line and back to the starting position until they reach the other end of the field. (a) Write a vector equation that includes all the legs of the run. (b) What is the players’ final displacement? (c) How far did they run? 7. A car drives north 500 km. It then drives three sequential displacements south, each of
which is 50 km longer than the previous displacement. If the final position of the car is 50 km [N], find the three displacements algebraically. 8. Are vectors A and B equal? Why or why not? y A B x 9. A bouncy ball dropped from a height of 10.0 m bounces back 8.0 m, then drops and rebounds 4.0 m and finally 2.0 m. Find the distance the ball travels and its displacement from the drop point. e TEST To check your understanding of vectors in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 75 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 76 2.2 Motion in Two Dimensions From the boot, the ball flies across the grass into the net and the crowd roars. The enormously successful FIFA Under-19 Women’s World Championship, held in 2002, raised the profile of women’s soccer in Canada and drew crowds totalling almost 200 000 to venues in Edmonton, Vancouver, and Victoria (Figure 2.13). Stars like Charmaine Hooper, Brittany Timko, and Christine Sinclair continue to amaze. From World Championship team members to the Under-6s on the local soccer pitch, performance depends on understanding and coordinating the movement of players and ball across the surface of the field. Figure 2.13 Motion in sports such as soccer can be described by vectors in two dimensions. Playbooks are available for fast-paced games such as hockey and soccer to allow coaches and players to plan the strategies that they hope will lead to success. Sometimes a team can charge straight up the rink or field, but, more often, a series of angled movements is needed to advance the puck or ball (Figure 2.14). For everyone to understand the play, a system is needed to understand motion in two dimensions. Components of Vectors Imagine that you are at one corner of a soccer field and you have to get to the far opposite corner. The shortest path from one corner to the other is a straight diagonal line. Figure 2.15 shows this path to be 150 m. Another way to describe this motion is to imagine an x-axis and a y-axis placed onto the soccer field, with you standing at the point (0, 0). You could move along the length of the field 120 m and then across the field 90 m and end
up at the same spot (Figure 2.15). 4 3 2 1 Figure 2.14 This page is taken from a soccer playbook. How many players are involved in this wall pass-in-succession manoeuvre? 76 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 77 Figure 2.15 The diagonal distance from one corner to the opposite corner of a soccer field is 150 m. 150 m 90 m 120 m In this example, the sideline of the soccer field could be considered the x-axis, and the goal line could be the y-axis. The diagonal motion vector can then be separated, or resolved, into two perpendicular parts, or components: the x component and the y component. The diagonal vector is the resultant vector. If you walked along the sideline or x-axis, you would move through a distance of 120 m. This distance is the x component of the diagonal vector. The second part of the walk along the goal line, parallel to the y-axis, is the y component of the diagonal vector. This motion has a distance of 90 m. Figure 2.16 shows the x and y components of the diagonal motion across the soccer field. y components: perpendicular parts into which a vector can be separated Figure 2.16 The resultant vector representing the diagonal walk across the soccer field can be resolved into x and y components. 150 m 90 m y component (width) x (0, 0) 120 m x component (length) Vector Directions Recall that a vector must have a magnitude and a direction. You have just studied how to resolve a vector into its components. Before going further, you need to know how to indicate the direction of vectors in two dimensions. There are two methods commonly used to show direction for vector quantities in two dimensions: the polar coordinates method and the navigator method. Both methods are considered valid ways to describe the direction of a vector. Chapter 2 Vector components describe motion in two dimensions. 77 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 78 info BIT A third method for measuring direction is the bearing method, in which angles are measured clockwise from north, 0° to 360°, so east is 90°, south is 180°, and west is 270°. 90° y II 180° 240° 5 m [240°] III 270° x 0° (360°) I IV N S W 60° 5 m [60° S
of W] E info BIT Sailors can now create their sailing plans with a click of a mouse. Digitized maps and global positioning satellites have been combined to allow sailors to create a plan by using the mouse to place vectors on the desired path on screen. The computer calculates the total distance, identifies directions, and estimates the time required for the trip. Figure 2.17 The polar coordinates method for stating vector direction Figure 2.18 The navigator method for stating vector direction Polar Coordinates Method With the polar coordinates method, the positive x-axis is at 0° and angles are measured by moving counterclockwise about the origin, or pole. One complete rotation is 360° — a complete circle. In Figure 2.17, the displacement vector, 5 m [240°], is located in quadrant III in the Cartesian plane. This vector is rotated 240° counterclockwise starting from the positive x-axis. Navigator Method Another method for indicating vector direction is the navigator method. This method uses the compass bearings north [N], south [S], east [E], and west [W] to identify vector directions. In Figure 2.18, the displacement vector 5 m [60° S of W] is between the west and south compass bearings. To draw this vector, start with the second compass bearing you are given in square brackets, west, then move 60° in the direction of the first compass bearing you are given, south. The type of problem will determine the method you use for stating vector directions. Often, it will be clear from the context of the problem which method is preferred. For example, if the question is about a boat sailing [30° N of W], then use the navigator method. If a plane has a heading of 135°, then use the polar coordinates method. In the problems below, you can practise identifying and drawing vectors using the two methods Directions 2. For each vector in question 1, state the direction using the alternative method. Then draw each vector using an appropriate scale and reference coordinates. 1. For each of the following vectors, identify the method used for indicating direction. Then draw each vector in your notebook, using an appropriate scale and reference coordinates. (a) 3 m [0°] (b) 17 m/s [245°] (c) 7 m [65°] (d) 8 m/s [35° W of N] (e) 2 m [98°] (f) 12 m/s [30° S of
E] 78 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 79 Concept Check Write the direction [60° S of W] another way using a different starting axis but keeping the angle less than 90°. 2-2 QuickLab 2-2 QuickLab Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m measuring tape large chalkboard protractor field marker (tape or flag) Procedure 1 Using a tree diagram, determine the number of pathways you could take to walk the series of displacement vectors below. Assume that you will start on the centre line of a football field and mark it 0°. (a) 5 m [0°] (b) 12 m [270°] (c) 15 m [90°] (d) 3 m [180°] 7 Use the protractor to estimate the angle of displacement. 8 Repeat steps 3–7 for all the pathways you determined in step 1. 5 m [0°] Figure 2.19 NOTE: Use the same method for determining direction throughout the lab. 2 On a football or large school field, mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacement given in (a) while you hold the end of the measuring tape. Mark your partner’s endpoint (Figure 2.19). Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? 4 Continue the journey, using the protractor and measuring tape, until you have walked all the vectors. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. Chapter 2 Vector components describe motion in two dimensions. 79 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 80 Adding Two-dimensional Vectors Graphically To sink the eight ball in the front side pocket of a billiard table, you must cause the ball to travel down and across the table (Figure 2.20). The ball’s motion occurs in a plane, or two dimensions, even though its path is linear (Figure 2.21). Figure 2.20 Playing billiards involves two-dimensional motion. Figure 2.21 The path of the billiard ball is linear, but it occurs in two dimensions. Recall from section 2.1 that the plus sign ()
in a vector equation indicates that you need to connect the vectors tip to tail. Up to this point, you have added collinear vectors only. In this section, you will learn how to add non-collinear vectors. The plus sign still indicates you need to connect the vectors tip to tail while keeping track of their directions. Adding Non-collinear Vectors In section 2.1, you learned that vectors that lie along the same straight line are collinear. Vectors that are not along the same straight line are non-collinear (Figure 2.22). To determine the magnitude and direction of the sum of two or more non-collinear vectors graphically, use an accurately drawn scale vector diagram. Imagine you are walking north a distance of 40 m. Your initial 1. You stop, head west a distance position from your starting point is d 2. To find your disof 30 m, and stop again. Your final position is d placement, you cannot simply subtract your initial position from your final position because the vectors are not collinear. To find your displacement in two dimensions, you need to add the two position vectors: d d d 1 2. From Figure 2.23, you can see that the answer is not 70 m. You would obtain the answer 70 m if you walked in the same direction for both parts of your walk. Because you did not, you cannot directly substitute values into the displacement equation d 2. Instead, you must draw the vectors to scale, connect them tip to tail (because of the plus sign), and measure the magnitude of the resultant. Since d is a vector quantity, you must also indicate its direction. You can find the direction of the resultant using a protractor (Figure 2.24). d d 1 non-collinear: not along a straight line 45° Figure 2.22 Non-collinear vectors lie along different lines. 80 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 81 30 m d 2 W N S E d 30 m 2 W N S E d 50 m 40 m d 1 d 50 m d 40 m 1 scale 10 m 37º scale 10 m d 50 m [37º W of N] Figure 2.23 What is the sum of 2? 1 and d d Figure 2.24 When adding non-collinear vectors graphically, use a protractor to find the direction of the resultant.
Eight Steps for Adding Non-collinear Vectors Graphically To find the resultant vector in a non-collinear vector addition statement using the graphical method, follow these eight steps (see Figure 2.25): 1. Create an appropriate scale. 2. Choose a set of reference coordinates. 3. Draw vector 1 to scale. Measure its direction from the tail. 4. Draw vector 2 to scale. Draw its tail at the tip (arrowhead) of vector 1. 5. Draw the resultant vector by connecting the tail of vector 1 to the tail tip of vector 2. W N S E tip resultant vector 2 vector 1 tail tip 6. Measure the magnitude (length) of the resultant. Measure the direction Figure 2.25 Adding vectors (angle) of the resultant from its tail. 7. Use your scale to convert the magnitude of the resultant to its original units. 8. State the resultant vector. Remember to include both magnitude and direction. This method also works for more than two vectors. You can add the vectors in any order. The next example shows you how to add more than two non-collinear vectors graphically. Example 2.3 A camper left her tent to go to the lake. She walked 0.80 km [S], then 1.20 km [E] and 0.30 km [N]. Find her resultant displacement. 1 Given d d d 2 3 0.80 km [S] 1.20 km [E] 0.30 km [N] Required resultant displacement (d R) Chapter 2 Vector components describe motion in two dimensions. 81 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 82 Practice Problems 1. For Example 2.3, add the vectors in two different orders and obtain the resultant for each case. 2. A student runs through a field 100 m [E], then 200 m [S], and finally 50 m [45° S of E]. Find her final position relative to her starting point. Answers 1. 1.30 km [67° E of S] 2. 272 m [60° S of E] Analysis and Solution The three vectors are non-collinear, so add them tip to tail to find the resultant (Figure 2.26). tent 67° d1 0.80 km dR d1 d2 d3 Figure 2.26 scale 0.20 km W N S E dR 1.30 km d3 0.30 km d2 1.
20 km Paraphrase The camper’s resultant displacement is 1.30 km [67° E of S]. Distance, Displacement, and Position Figure 2.27 shows the distances a bicycle courier travelled in going along the path from A to D, passing through B and C on the way. Use the information in the diagram, a ruler calibrated in mm, and a protractor to complete the distance, displacement, and position information required in Table 2.1. Assume the bicycle courier’s reference point is A. Complete Table 2.1, then draw and label the displacement vectors AB, BC, and CD, and the position vectors AB, AC, and AD. ▼ Table 2.1 Distance, Displacement, and Position scale 80 m W N S E d 600 m D C Displacement Δd (m) [direction] Distance Δd (m) d Final position (m) [direction] reference point AB BC CD AC AD 82 Unit I Kinematics d 630 m B d 560 m Figure 2.27 d 280 m A 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 83 Determining Components drawn in a Cartesian plane and its two Figure 2.28 shows a vector R components, Rx and Ry, in the x and y directions. The Greek letter theta,, denotes the angle between R and its x and y components form a right triangle. and the x-axis. Vector R y 6.0 4.0 2.0 R 10 km/h 6.0 km/h Ry θ 37° 2.0 x 6.0 8.0 4.0 8.0 km/h Rx Figure 2.28 Drawing components Because the triangle is a right triangle, you can determine components algebraically by using the trigonometric functions sine, cosine, and tangent. You can define each of the trigonometric functions in terms of the sides of a right triangle, like the one in Figure 2.29. Knowing these definitions, you can use the trigonometric functions to help you solve for the components of a vector. To calculate Rx, use the cosine function: Rx R adjacent hypotenuse R cos or Rx cos In Figure 2.28, the x component is: Rx (10 km/h)(cos 37°) 8.0 km/h To calculate Ry, use the sine function: sin opposite hypotenuse Ry R or Ry R
sin hypotenuse opposite In Figure 2.28, the y component is: Ry (10 km/h)(sin 37°) 6.0 km/h θ adjacent Figure 2.29 Labelled sides of a right triangle Example 2.4 shows the steps for finding the velocity components of a car travelling in a northeasterly direction using trigonometry. This example uses the navigator method to indicate the direction of the velocity vector. Note that the east direction [E] is the same as the positive x direction in the Cartesian plane, and north [N] is the same as the positive y direction. So, for any vector R, the x component is the same as the east component, and the y component is the same as the north component. PHYSICS INSIGHT a c θ b For a right triangle with sides a and b forming the right angle, c is the hypotenuse. The Pythagorean theorem states that a2 b2 c2. You can find the angle,, in one of three ways: sin a c opposite hypotenuse cos b c adjacent hypotenuse tan a b opposite adjacent Chapter 2 Vector components describe motion in two dimensions. 83 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 84 Example 2.4 Determine the north and east velocity components of a car travelling at 100 km/h [25° N of E]. y 1 0 0 k m /h 25° vx vy x Figure 2.30 Given v 100 km/h [25° N of E] Required velocity component north (y component, vy) velocity component east (x component, vx) Practice Problems 1. A hiker’s displacement is 15 km [40° E of N]. What is the north component of his displacement? 2. A cyclist’s velocity is 10 m/s [245°]. Determine the x and y components of her velocity. 3. A snowmobile travels 65 km [37° E of S]. How far east does it travel? Answers 1. 11 km [N] 4.2 m/s, vy 2. vx 3. 39 km [E] 9.1 m/s Analysis and Solution The vector lies between the north and east directions, so the x and y components are both positive. Since the north direction is parallel to the y-axis, use the sine function, R sin, to find the north component.
Since the east Ry direction lies along the x-axis, use the cosine function, Rx R cos, to find the east component. Ry vy Rx vx R sin (100 km/h)(sin 25°) 42.3 km/h R cos (100 km/h)(cos 25°) 90.6 km/h Paraphrase The north component of the car’s velocity is 42.3 km/h and the east component is 90.6 km/h. Concept Check For a vector R in quadrant I (Cartesian method), are Rx and Ry always positive? Determine whether Rx and Ry are positive or negative for vectors in quadrants II, III, and IV. Display your answers in a chart. 84 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 85 Adding Vectors Using Components You can write the magnitude of any two-dimensional vector as the sum of its x and y components. Note that x and y components are perpendicular. Because motion along the x direction is perpendicular to motion along the y direction, a change in one component does not affect the other component. Whether it is movement across a soccer field or any other type of two-dimensional motion, you can describe the motion in terms of x and y components. In Example 2.4, you learned how to determine the x and y com- ponents, Rx and Ry, for a general vector R. In some situations, you already know Rx and Ry, and you must find the magnitude and direc- tion of the resultant vector R. For example, a toy moves 9.0 m right and then 12.0 m across a classroom floor (Figure 2.31). What is the toy’s displacement? Solving this problem algebraically requires two steps: Step 1: Find the magnitude of R. To find the magnitude of the resultant vector, use the Pythagorean theorem. You can use this theorem because the two components, Rx and Ry, form a right triangle with the resultant vector. You are given that Rx R2 Rx R (9.0 m)2 (12.0m)2 9.0 m and Ry 2 Ry 12.0 m. y 2 15 m Step 2: Find the angle of R To find the angle of R, use the tangent function:. R tan opposite adjacent 12.0 m 9.0 m 1.33 tan1(1.33) 53
.1° 12.0 9.0 6.0 3.0 12.0 m Ry θ 3.0 Rx 6.0 9.0 m x 9.0 Figure 2.31 Vector components of the movement of a toy across a classroom floor Using the polar coordinates method, the resultant vector direction is [53.1°]. Using the navigator method, the direction is [53.1° N of E]. Using Components 1. Find Rx and Ry for the following vectors: (a) A boat travelling at 15 km/h [45° N of W] (b) A plane flying at 200 km/h [25° E of S] (c) A mountain bike travelling at 10 km/h [N] 2. Find R (a) Rx (b) Rx (c) Rx 7 m and for the following Rx and Ry values: 12 m, Ry 40 km/h, Ry 30 cm, Ry 55 km/h 10 cm Chapter 2 Vector components describe motion in two dimensions. 85 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 86 In general, most vector motion involves adding non-collinear vectors. Consider the following scenario. During a lacrosse game, players pass the ball from one person to another (Figure 2.32). The ball can then be redirected for a shot on goal. Each of the displacements could involve different angles. In order to find the net displacement, you would use the following sequence of calculations. Four Steps for Adding Non-collinear Vectors Algebraically 1. Determine the x and y components of each vector. 2. Add all components in the x direction. Add all components in the y direction. The sums of the x and y components are the two (perpendicular) components of the resultant vector. 3. To find the magnitude of the resultant vector, use the Pythagorean theorem. 4. To find the angle of the resultant vector, use trigonometric ratios. (See Physics Insight on page 83.) The following example illustrates how to apply these steps. In a lacrosse game (Figure 2.33(a)), player A passes the ball 12.0 m to player B at an angle of 30°. Player B relays the ball to player C, 9.0 m away, at an angle of 155°. Find the ball’s resultant displacement. y B d2 9.0 m 155° 12.0 m
d1 30° x C A Figure 2.33(a) on the lacrosse field The path of the ball Figure 2.33(b) as vectors The path of the ball Figure 2.33(b) shows the path of the lacrosse ball as vectors. This problem is different from previous examples because the two vectors are not at right angles to each other. Even with this difference, you can follow the same general steps to solve the problem. Step 1: Determine the x and y components of each vector. Since you are solving for displacement, resolve each displacement vector into its components (Figure 2.34). Table 2.2 shows how to calculate the x and y components. In this case, designate up and right as positive directions. Figure 2.32 The movement of the players and the ball in a lacrosse game could be tracked using vectors. PHYSICS INSIGHT To simplify calculations for finding components, use acute ( 90°) angles. To determine the acute angle when given an obtuse ( 90°) angle, subtract the obtuse angle from 180°. 25° 155° 180° 155° 25° For an angle greater than 180°, subtract 180° from the angle. For example, 240° 180° 60° 86 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 87 ▼ Table 2.2 Resolution of Components in Figure 2.34 x direction y direction d1x d2x (12.0 m)(cos 30°) 10.39 m (9.0 m)(cos 25°) 8.16 m d1y d2y (12.0 m)(sin 30°) 6.00 m (9.0 m)(sin 25°) 3.80 m (Note that d2x is negative because it points to the left, and up and right were designated as positive.) y d2y 30° d2 d2x d1 d1x 25° 155° d1y x Figure 2.34 The path of the lacrosse ball Step 2: Add the x components and the y components separately. Add all the x components together, then add all the y components (see Table 2.3 and Figure 2.35). ▼ Table 2.3 Adding x and y Components in Figure 2.35 x direction y direction dx d2x d1x 10.39 m (8.16 m) 10.39 m 8.16
m 2.23 m dy d2y d1y 6.00 m 3.80 m 9.80 m dx d2x d1x d2y d1y dy Figure 2.35 Add the x and y components separately first to obtain two perpendicular vectors. Step 3: Find the magnitude of the resultant, d To find the magnitude of the resultant, use the Pythagorean theorem (Figure 2.36).. d2 (dx)2 (dy)2 d (dx)2 (dy)2 (2.23m)2 (9.80 m)2 10 m y d dy θ dx x Figure 2.36 The component method allows you to convert non-perpendicular vectors into perpendicular vectors that you can then combine using the Pythagorean theorem. Chapter 2 Vector components describe motion in two dimensions. 87 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 88 y d 77° d2 d1 Figure 2.37 d resultant displacement of the ball. is the Step 4: Find the angle of d. Use the tangent function to find the angle (Figure 2.36). tan opposite adjacent 9.80 m 2.23 m 4.39 x tan1(4.39) 77° The ball’s displacement is, therefore, 10 m [77°], as shown in Figure 2.37. The following example illustrates another situation where the dis- placement vectors are not at right angles. Example 2.5 Practice Problems 1. Find the displacement of a farmer who walked 80.0 m [0°] and then 60.0 m [335°]. 2. Find the displacement of a soccer player who runs 15 m [15° N of E] and then 13 m [5° W of N]. 3. While tracking a polar bear, a wildlife biologist travels 300 m [S] and then 550 m [75° N of E]. What is her displacement? Answers 1. 137 m [349°] 2. 21 m [52° N of E] 3. 272 m [58° N of E] Use components to determine the displacement of a crosscountry skier who travelled 15.0 m [220°] and then 25.0 m [335°] (Figure 2.38). y x 15.0 m [220°] d1 25.0 m [335°] d2 Figure 2.38 Given d d 1 2 15.0 m [220°]
25.0 m [335°] Required displacement (d ) Analysis and Solution Step 1: Use Rx its x and y components. Designate up and to the right as positive. Work with acute angles (Figure 2.39). R sin to resolve each vector into R cos and Ry y y 220° 40° x 335° 25° x d2y d2 d2x d1y d1 d1x Figure 2.39 88 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 89 x direction: d1x (15.0 m)(cos 40°) 11.49 m d2x (25.0 m)(cos 25°) 22.66 m y direction: d1y (15.0 m)(sin 40°) 9.642 m d2y (25.0 m)(sin 25°) 10.57 m Step 2: Add the x and y components. dx dy d1x + d2x 11.49 m 22.66 m 11.17 m d2y d1y 9.642 m (10.57 m) 20.21 m 11.17 m θ 20.21 m d y θ x d d1 d2 Figure 2.40 Figure 2.41 Step 3: To find the magnitude of the resultant, calculate d using the Pythagorean theorem (Figure 2.40). d2 (dx)2 (dy)2 (11.17 m)2 (20.21 m)2 d (11.17 m)2 (20.21 m)2 23.09 m Figure 2.41 shows that the resultant lies in quadrant IV. Step 4: To find the angle, use the tangent function (see Figure 2.40). tan opposite adjacent 20.21 m 11.17 m 1.810 tan1(1.810) 61° From Figure 2.41, note that the angle,, lies below the positive x-axis. Using the polar coordinates method, the angle is 299°. Paraphrase The cross-country skier’s displacement is 23.1 m [299°]. e SIM Practise the numerical addition of two or more vectors. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 89 02-PearsonPhys20-Chap02 7/24/08 10:17
AM Page 90 In summary, in order to solve a two-dimensional motion problem, you need to split the motion into two one-dimensional problems by using the vectors’ x and y components. Then add the x and y components separately. To find the magnitude of the resultant, use the Pythagorean theorem. To find the angle of the resultant, use the tangent function. 2.2 Check and Reflect 2.2 Check and Reflect Knowledge 1. What trigonometric functions can be used to determine the x or horizontal component of a vector? Draw diagrams to illustrate your answers. 2. Are the following statements true or false? Justify your answer. (a) The order in which vectors are added is important. (b) Displacement and distance are always equal. 3. Describe when you would use the navigator method to indicate the direction of a vector. Applications 4. A student has created a short computer program that calculates components of vectors drawn with a computer mouse. To demonstrate his program, he drags the mouse to create a vector at 55 cm [30° W of S]. What are the components of the vector? 5. Determine the distance travelled and the displacement for each of the following. (a) Blading through Fish Creek Park in Calgary takes you 5.0 km [W], 3.0 km [N], 2.0 km [E], and 1.5 km [S]. (b) A swimmer travels in a northerly direction across a 500-m-wide lake. Once across, the swimmer notices that she is 150 m east of her original starting position. (c) After leaving her cabin, a camper snowshoes 750 m [90°] and then 2.20 km [270°]. 90 Unit I Kinematics 6. A boat sails 5.0 km [45° W of N]. It then changes direction and sails 7.0 km [45° S of E]. Where does the boat end up with reference to its starting point? 7. A pellet gun fires a pellet with a velocity of 355 m/s [30°]. What is the magnitude of the vertical component of the velocity at the moment the pellet is fired? 8. Tourists on a jet ski move 1.20 km [55° N of E] and then 3.15 km [70° S of E]. Determine the jet ski’s displacement. 9. A jogger runs with a velocity of 6.
ion is motion measured with respect to an observer. Concept Check An observer is on a train moving at a velocity of 25 m/s [forward]. A ball rolls at 25 m/s [forward] with respect to the floor of the moving train. What is the velocity of the ball relative to the observer on the train? What is the velocity of the ball relative to an observer standing on the ground? What happens if the ball moves 25 m/s [backward]? How does a moving medium affect the motion of a table tennis ball? relative motion: motion measured with respect to an observer Chapter 2 Vector components describe motion in two dimensions. 91 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 92 2-3 QuickLab 2-3 QuickLab Table Tennis in the Wind Problem How does air movement affect the motion of a table tennis ball? Materials large upright fan table tennis table paddles table tennis ball Procedure 1 With a partner, practise hitting the table tennis ball to each other (Figure 2.43). Figure 2.43 Questions 1. When did the ball move the fastest? the slowest? 2. When the air movement was perpendicular to the ball’s path, did it change the ball’s speed? Did it change the ball’s velocity? Explain. 2 Set up the fan on one side of the table tennis table. 3. Given your results, speculate as to why golfers 3 Hit the table tennis ball straight across the length of the table (a) against the wind (b) with the wind release a tuft of grass into the wind before driving the ball. 4. Describe how wind direction might influence a beach volleyball player’s serve. (c) perpendicular to the wind’s direction e LAB 4 Record how the moving air influences the motion of the ball in each case. For a probeware activity, go to www.pearsoned.ca/school/physicssource. Relative Motion in the Air ground velocity: velocity relative to an observer on the ground air velocity: an object’s velocity relative to still air wind velocity: velocity of the wind relative to the ground A flight from Edmonton to Toronto takes about 3.5 h. The return flight on the same aircraft takes 4.0 h. If the plane’s air speed is the same in both directions, why does the trip east take less time? The reason is that, when travelling eastward from Edmonton, a
tailwind (a wind that blows from the rear of the plane, in the same direction as the plane’s motion) increases the airplane’s ground velocity (velocity relative to an observer on the ground), hence reducing the time of travel and, therefore, fuel consumption and cost. scale 100 km/h A Canadian regional jet travels with an air velocity (the plane’s velocity in still air) of 789 km/h [E]. The jet encounters a wind velocity (the wind’s velocity with respect to the ground) of 56.3 km/h [E] (Figure 2.44). (This wind is a west wind, blowing eastward from the west.) What is the velocity of the airplane relative to an observer on the ground? The resultant velocity of the airplane, or ground velocity, is the vector sum of the plane’s air velocity and the wind velocity (Figure 2.45). Let the positive direction be east. vair 789 km/h [E] vwind 56.3 km/h [E] Figure 2.44 The air velocity and wind velocity are in the same direction. 92 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 93 v ground air wind v v 789 km/h 56.3 km/h 845 km/h The sign is positive, so the ground velocity is 845 km/h [E]. scale 100 km/h vair 789 km/h [W] vwind 56.3 km/h [E] scale 100 km/h vwind Figure 2.45 vair vground scale 100 km/h vair vground vwind Figure 2.47 Figure 2.46 The air velocity and wind velocity are in opposite directions. If the jet heads west, from Toronto to Edmonton (Figure 2.46), its v ground resultant velocity becomes v 789 km/h 56.3 km/h 733 km/h v wind air (See Figure 2.47.) The sign is negative, so the ground velocity is 733 km/h [W]. The plane’s speed decreases due to the headwind (wind that approaches from the front). Non-collinear Relative Motion Suppose the jet travelling west from Toronto encounters a crosswind of 56.3 km [N] (Figure 2.48). W N S E vwind= 56.3 km/h vair= 789
km/h vwind= 56.3 km/h vground= 791 km/h θ vair= 789 km/h W N S E Figure 2.48 A plane flies in a crosswind. Figure 2.49 A plane that flies in a crosswind needs to adjust its direction of motion. air ground v v In this case, the velocity of the plane is not aligned with the wind’s velocity. The defining equation for this case is still the same as for the collinear case: v wind. From section 2.2, recall that the plus sign in a two-dimensional vector equation tells you to connect the vectors tip to tail. The resultant vector is the ground velocity, v ground. The ground velocity indicates the actual path of the plane (Figure 2.49). To solve for the ground velocity, notice that the triangle formed is a right triangle, meaning that you can use the Pythagorean theorem to solve for the magnitude of the ground velocity. Chapter 2 Vector components describe motion in two dimensions. 93 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 94 (vground)2 (vair)2 (vwind)2 vground (vair)2 (vw ind)2 (789 km/h)2 (56.3 km/h)2 791 km/h Using the tangent function, the direction of the ground velocity is tan opposite adjacent 56.3 km/h 789 km/h 0.07136 tan1(0.07136) 4.1° From Figure 2.49, the wind blows the airplane off its westerly course in the northerly direction. Hence, the airplane’s ground velocity is 791 km/h [4.1° N of W]. The pilot must take into consideration the effect of the wind blowing the airplane off course to ensure that the plane reaches its destination. What path would the pilot have to take to arrive at a point due west of the point of departure? Remember that, if the vectors are not perpendicular, resolve them into components first before adding them algebraically. Example 2.6 A plane flies west from Toronto to Edmonton with an air speed of 789 km/h. (a) Find the direction the plane would have to fly to compensate for a wind velocity of 56.3 km/h [N]. (b) Find the plane’s speed relative to the ground. W N S
E vground vwind Figure 2.50 vair PHYSICS INSIGHT To determine the angle, substitute the magnitudes of the relative velocities into the tangent function. To determine the direction, refer to the vector diagram for the problem. 94 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 95 wind 56.3 km/h [N] Given v v direction of ground velocity is west 789 km/h air Required (a) the plane’s direction (direction of air velocity) (b) ground speed (vground) Analysis and Solution (a) First construct a diagram based on the defining equation, v v v ground wind air vground θ vair vwind Figure 2.51 air and v wind tip to tail. The rules of vector addition tell you to connect the vectors v To find the direction required in order to compensate for the wind velocity, find the angle,. Because the connection of the vectors forms a right triangle, and you know the magnitude of the opposite side (v wind) and the hypotenuse (v air), you can use the sine function to find the angle (Figure 2.51). te i s o p sin t se u n e o hy p p o Practice Problems 1. A swimmer can swim at a speed of 1.8 m/s. The river is 200 m wide and has a current of 1.2 m/s [W]. If the swimmer points herself north, directly across the river, find (a) her velocity relative to the ground. (b) the time it takes her to cross. 2. For a river flowing west with a current of 1.2 m/s, a swimmer decides she wants to swim directly across. If she can swim with a speed of 1.8 m/s, find (a) the angle at which she must direct herself. (b) the time it takes her to cross if the river is 200 m wide. Answers 1. (a) 2.2 m/s [34° W of N] (b) 1.1 102 s 2. (a) [42° E of N] (b) 1.5 102 s o p p p hy sin1 sin1 4.1 se te From Figure 2.51, the angle is [4.1° S of W]. (b) To find the magnitude of the ground velocity, use the Pyth
agorean theorem. From Figure 2.51, note that the hypotenuse in this case is the air velocity, v air. (vair)2 (vwind)2 (vground)2 (vground)2 (vair)2 (vwind)2 (789 km/h)2 (56.3 km/h)2 6.1935 105 (km/h)2 787 km/h vground Notice that there is a small change in the magnitude of the ground velocity from the previous example of the plane heading west. As the magnitude of the wind velocity increases, the magnitude of the ground velocity and the compensating angle will significantly change. Paraphrase (a) The plane’s heading must be [4.1° S of W]. (b) The plane’s ground speed is 787 km/h. Chapter 2 Vector components describe motion in two dimensions. 95 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 96 PHYSICS INSIGHT For simplicity, Edmonton was assumed to be directly west of Toronto, which, of course, it is not! However, the calculation is still valid because this problem involves straightline motion. To calculate the time it takes to fly from Toronto to Edmonton, use the d equation v. The distance between Edmonton and Toronto is about t 2335 km, but you must decide which value for velocity to use: air velocity or ground velocity. The displacement and velocity vectors in this equation must be aligned. Since you are assuming that displacement is in the west direction, the appropriate velocity to use is the one that is in the westerly direction. In this example, it is the ground velocity, v ground (Figure 2.52). d vground θ vair vwind Figure 2.52 For calculating time, choose the velocity vector that matches the direction of the displacement vector. Consider west to be positive. Since both vectors are in the same direction (west), use the scalar form of the equation to solve for time. t d v 2335 km m 787 k h 2.97 h It takes 2.97 h to fly from Toronto to Edmonton. In the following example, the three velocity vectors do not form a right triangle. In order to solve the problem, you will need to use components. PHYSICS INSIGHT When substituting values into a vector equation, make sure that the values have the same direction (are collinear). If the vectors
are not collinear, you need to use graphical or algebraic methods to find the answer. Example 2.7 As a pilot of a small plane, you need to transport three people to an airstrip 350.0 km due west in 2.25 h. If the wind is blowing at 40.0 km/h [65° N of W], what should be the plane’s air velocity in order to reach the airstrip on time? d = 350.0 km [W] vground vwind = 40.0 km/h [65º N of W] Figure 2.53 96 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 97 40.0 km/h [65 N of W] wind Given v d t 2.25 h 350.0 km [W] Required plane’s air velocity (v air) Analysis and Solution First draw a vector diagram of the problem (Figure 2.54). W N S E vwind Figure 2.54 vground vair Designate north and west as the positive directions. Then calculate the ground velocity from the given displacement and time. If the plane must fly 350.0 km [W] in 2.25 h, its ground velocity is d t ground v Practice Problems 1. An airplane can fly with a maximum air velocity of 750 km/h [N]. If the wind velocity is 60 km/h [15 E of N], what must be the plane’s ground velocity if it is to remain on a course going straight north? 2. What is the air velocity of a jetliner if its ground velocity is 856 km/h [25.0 W of S] and the wind velocity is 65.0 km/h [S]? 3. How long will it take a plane to travel 100 km [N] if its ground velocity is 795 km/h [25 W of N]? Answers 1. 8.1 102 km/h [1 W of N] 2. 798 km/h [27.0 W of S] 3. 0.139 h 350.0 km [W] 2.25 h 155.6 km/h [W] Now find the components of the wind velocity (Figure 2.55). x direction: vwindx (40.0 km/h)(cos 65) 16.9 km/h y direction: vwindy (40.0 km/
h)(sin 65) 36.25 km/h The ground velocity is directed west, so its x component is 155.6 km/h and its y component is zero. v Since v v v v wind, rearrange this equation to solve for v v air. ground air air ground wind vwindy vwind 65° vwindx Figure 2.55 Use this form of the equation to solve for the components of the air velocity. Add the x (west) components: vairx vgroundx vwindx 155.6 km/h 16.9 km/h 138.7 km/h Add the y (north) components: vairy vgroundy vwindy 0 36.25 km/h 36.25 km/h Use the Pythagorean theorem to find the magnitude of the air velocity. vair vairx 2 vairy2 (138.7 km/h)2 (36.25 km/h)2 143 km/h Chapter 2 Vector components describe motion in two dimensions. 97 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 98 To find the direction of air velocity, use the tangent function (Figure 2.56). tan opposite adjacent 36.25 km/h 138.7 km/h 0.261 tan1(0.261) 15 138.7 km/h θ 36.25 km/h vair Figure 2.56 The x component is positive, so its direction is west. Since the y component is negative, its direction is to the south. Thus, the direction of the air velocity is [15 S of W]. Paraphrase The airplane’s air velocity is 143 km/h [15 S of W]. Relative Motion in the Water Whereas wind velocity affects the speed and direction of flying objects, watercraft and swimmers experience currents. As with flying objects, an object in the water can move with the current (ground velocity increases), against the current (ground velocity decreases), or at an angle (ground velocity increases or decreases). When the object moves at an angle to the current that is not 90, both the object’s speed and direction change. The following example illustrates how to use components to find velocity. Example 2.8 The Edmonton Queen paddleboat travels north on the Saskatchewan River at a speed of 5.00 knots or 9.26 km/h. If the Queen’s ground velocity is 10.1 km/h [23° E
of N], what is the velocity of the Saskatchewan River? vcurrent W N S E vboat vground 23° vcurrent W N S E vboat= 9.26 km/h vground = 10.1 km/h 23º Figure 2.57(a) Figure 2.57(b) 98 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 99 boat ground 9.26 km/h [N] 10.1 km/h [23 E of N] Given v v (Note that the angle is given with respect to the vertical (y) axis (Figure 2.57(a).) vgroundx vgroundy vground Required velocity of current (v current) 23° Analysis and Solution Let north and east be positive. Calculate the current’s velocity using components. First find the components of the ground velocity (Figure 2.58(a)). Figure 2.58(a) x direction: vgroundx vgroundsin (10.1 km/h)(sin 23) 3.946 km/h y direction: vgroundy vgroundcos (10.1 km/h)(cos 23) 9.297 km/h Practice Problems 1. Determine a Sea Doo’s ground velocity if it travels with a constant velocity of 4.50 m/s [W] and encounters a current of 2.0 m/s [20 W of N]. 2. A jogger runs with a velocity of 3.75 m/s [20 N of E] on an Alaskan cruise ship heading north at 13 m/s. What is the jogger’s ground velocity? 3. A ship travelling 55 [W of N] is 65.0 km farther north after 3.0 h. What is the ship’s velocity? Answers 1. 5.5 m/s [20 N of W] 2. 15 m/s [76 N of E] 3. 38 km/h [55 W of N] Since the boat’s velocity is directed north, its y component is 9.26 km/h and its x component is zero. v You are asked to find v v v v v v current ground boat ground boat current current, so rearrange the vector equation accordingly: Use this form of the equation to solve for the components of the current’s velocity. vcurrentx vboatx vcurrenty v
groundx 3.946 km/h 0 3.946 km/h vboaty vgroundy 9.297 km/h 9.26 km/h 0.037 km/h To find the magnitude of the current’s velocity, use the Pythagorean theorem. vcurrent nty)2 ntx)2 (vcurre (vcurre (3.946 km/h)2 (0.037 km/h)2 3.946 km/h To find the direction of the current’s velocity, use the tangent function (Figure 2.58(b)). opposite adjacent tan 0.03700 km/h vcurrent θ tan1 0.037 km/h 3.946 km/h 0.5 3.946 km/h Figure 2.58(b) Since both the x and y components are positive, the directions are east and north, respectively. Therefore, the current’s direction is [0.5 N of E]. Paraphrase The current’s velocity is 3.95 km/h [0.5 N of E]. Chapter 2 Vector components describe motion in two dimensions. 99 02-PearsonPhys20-Chap02 7/25/08 8:05 AM Page 100 In order to find the time required to cross the river, you need to use the velocity value that corresponds to the direction of the object’s displacement, as you will see in the next example. Example 2.9 From Example 2.8, if the river is 200 m wide and the banks run from east to west, how much time, in seconds, does it take for the Edmonton Queen to travel from the south bank to the north bank? Practice Problems 1. A river flows east to west at 3.0 m/s and is 80 m wide. A boat, capable of moving at 4.0 m/s, crosses in two different ways. (a) Find the time to cross if the boat is pointed directly north and moves at an angle downstream. (b) Find the time to cross if the boat is pointed at an angle upstream and moves directly north. Answers 1. (a) 20 s (b) 30 s Given v width of river 200 m 0.200 km 10.1 km/h [23° E of N] ground Required time of travel (t) Analysis and Solution Determine the distance, d, the boat travels in the direction
of the boat’s ground velocity, 23° E of N. From Figure 2.59, d 0.200 km cos 23° 0.2173 km N 0.200 km d 23° Figure 2.59 The boat’s ground velocity is 10.1 km/h [23° E of N]. vground t 10.1 km/h d vground 0.2173 km 10.1 km/h 0.02151 h 60 min 1 h 60 s 1 min 77.4 s Paraphrase It takes the Edmonton Queen 77.4 s to cross the river. Relative motion problems describe the motion of an object travelling in a medium that is also moving. Both wind and current can affect the magnitude and direction of velocity. To solve relative motion problems in two dimensions, resolve the vectors into components and then add them using trigonometry. 100 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 101 2.3 Check and Reflect 2.3 Check and Reflect Knowledge 1. Describe a situation where a wind or current will increase an object’s ground speed. 2. Describe a situation where a wind or current will change an object’s direction relative to the ground, but not its speed in the original direction (i.e., the velocity component in the original, intended direction of motion). 3. Describe a situation when a wind or current will cause zero displacement as seen from the ground. 4. Provide an example other than in the text that illustrates that perpendicular components of motion are independent of one another. Applications 5. A swimmer needs to cross a river as quickly as possible. The swimmer’s speed in still water is 1.35 m/s. (a) If the river’s current speed is 0.60 m/s and the river is 106.68 m wide, how long will it take the swimmer to cross the river if he swims so that his body is angled slightly upstream while crossing, and he ends up on the far bank directly across from where he started? (b) If he points his body directly across the river and is therefore carried downstream, how long will it take to get across the river and how far downstream from his starting point will he end up? 6. A small plane can travel with a speed of 265 km/h with respect to the air. If the plane heads north, determine its resultant velocity if it encounters
(a) a 32.0-km/h headwind (b) a 32.0-km/h tailwind (c) a 32.0-km/h [W] crosswind 7. The current in a river has a speed of 1.0 m/s. A woman swims 300 m downstream and then back to her starting point without stopping. If she can swim 1.5 m/s in still water, find the time of her round trip. 8. What is the ground velocity of an airplane if its air velocity is 800 km/h [E] and the wind velocity is 60 km/h [42 E of N]? 9. A radio-controlled plane has a measured air velocity of 3.0 m/s [E]. If the plane drifts off course due to a light wind with velocity 1.75 m/s [25° W of S], find the velocity of the plane relative to the ground. If the distance travelled by the plane was 3.2 km, find the time it took the plane to travel that distance. 10. An airplane is observed to be flying at a speed of 600 km/h. The plane’s nose points west. The wind’s velocity is 40 km/h [45 W of S]. Find the plane’s velocity relative to the ground. 11. A canoe can move at a speed of 4.0 m/s [N] in still water. If the velocity of the current is 2.5 m/s [W] and the river is 0.80 km wide, find (a) the velocity of the canoe relative to the ground (b) the time it takes to cross the river e TEST To check your understanding of relative motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 101 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 102 2.4 Projectile Motion Sports are really science experiments in action. Consider golf balls, footballs, and tennis balls. All of these objects are projectiles (Figure 2.60). You know from personal experience that there is a relationship between the distance you can throw a ball and the angle of loft. In this section, you will learn the theory behind projectile motion and how to calculate the values you need to throw the fastball or hit the target dead on. Try the next QuickLab
and discover what factors affect the trajectory of a projectile. Figure 2.60 Sports is projectile motion in action. 102 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 103 2-4 QuickLab 2-4 QuickLab Projectiles Problem What factors affect the trajectory of a marble? Materials wooden board (1 m 1 m) hammer paint marble newspaper white paper to cover the board two nails elastic band spoon brick masking tape gloves Procedure 1 Spread enough newspaper on the floor so that it covers a larger workspace than the wooden board. 2 Hammer two nails, 7.0 cm apart, at the bottom left corner of the board. Stretch the elastic between them. 3 Cover the board with white paper and affix the 6 Pull the elastic band back at an angle and rest the marble in it. 7 Release the elastic band and marble. Label the marble’s trajectory on the paper track 1. 8 Repeat steps 5–7 for different launch angles and extensions of the elastic band. in Figure 2.61 Questions 1. What is the shape of the marble’s trajectory, paper to the board using masking tape. regardless of speed and angle? 4 Prop the board up on the brick (Figure 2.61). 2. How did a change in the elastic band’s extension 5 Wearing gloves, roll the marble in a spoonful of paint. affect the marble’s path? 3. How did a change in launch angle affect the marble’s path? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Galileo studied projectiles and found that they moved in two directions at the same time. He determined that the motion of a projectile, neglecting air resistance, follows the curved path of a parabola. The parabolic path of a projectile is called its trajectory (Figure 2.62). The shape of a projectile’s trajectory depends on its initial velocity — both its initial speed and direction — and on the acceleration due to gravity. To understand and analyze projectile motion, you need to consider the horizontal (x direction) and vertical (y direction) components of the object’s motion separately. viy vix dy trajectory: the parabolic motion of a projectile vi Figure 2.62 A projectile has a parabolic trajectory. dx Chapter 2 Vector components describe motion in two dimensions. 103 02-PearsonPhys20-Chap02 7/
24/08 10:18 AM Page 104 2-5 QuickLab 2-5 QuickLab Which Lands First? Problem What is the relationship between horizontal and vertical motion of objects on a ramp? Materials Galileo apparatus (Figure 2.63) steel balls Procedure 1 Set up the Galileo apparatus at the edge of a lab bench. 2 Place a steel ball at the top of each ramp. 3 Release the balls at the same time. 4 Listen for when each ball hits the ground. 5 Using a different ramp, repeat steps 1–4. Questions 1. Which ball landed first? 2. Did the balls’ initial velocity affect the result? If so, how? 3. What inference can you make about the relationship between horizontal and vertical motion? Figure 2.63 From section 1.6, you know that gravity influences the vertical motion of a projectile by accelerating it downward. From Figure 2.64, note that gravity has no effect on an object’s horizontal motion. So, the two components of a projectile’s motion can be considered independently. As a result, a projectile experiences both uniform motion and uniformly accelerated motion at the same time! The horizontal motion of a projectile is an example of uniform motion; the projectile’s horizontal velocity component is constant. The vertical motion of a projectile is an example of uniformly accelerated motion. The object’s acceleration is the constant acceleration due to gravity or 9.81 m/s2 [down] (neglecting friction). Concept Check In a table, classify the horizontal and vertical components of position, velocity, and acceleration of a horizontally launched projectile as uniform or non-uniform motion. Objects Launched Horizontally Suppose you made a new game based on a combination of shuffleboard and darts. The goal is to flick a penny off a flat, horizontal surface, such as a tabletop, and make it land on a target similar to a dartboard beyond the table. The closer your penny lands to the bull’s eye, the more points you score (Figure 2.65). Figure 2.64 Gravity does not affect the horizontal motion of a projectile because perpendicular components of motion are independent. PHYSICS INSIGHT When a projectile is launched, for a fraction of a second, it accelerates from rest to a velocity that has x and y components. 104 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 105 v table horizontal (x) component dy trajectory padded dartboard dx Figure
2.65 An object launched horizontally experiences uniform horizontal motion and uniformly accelerated vertical motion. In the game, once the penny leaves the tabletop, it becomes a projectile and travels in a parabolic path toward the ground. In section 1.6, you studied motion that was caused by acceleration due to gravity. The velocity of an object falling straight down has no horizontal velocity component. In this game, the penny moves both horizontally and vertically, like the ball on the right in Figure 2.64. In this type of projectile motion, the object’s initial vertical velocity is zero. Because the projectile has a horizontal velocity component, it travels a horizontal distance along the ground from its initial launch point. This distance is called the projectile’s range (Figure 2.66). The velocity component in the y direction increases because of the acceleration due to gravity while the x component remains the same. The combined horizontal and vertical motions produce the parabolic path of the projectile. Concept Check (a) What factors affecting projectile motion in the horizontal direction are being neglected? (b) What causes the projectile to finally stop? (c) If the projectile’s initial velocity had a vertical component, would the projectile’s path still be parabolic? Give reasons for your answer. PHYSICS INSIGHT For a projectile to have a non-zero velocity component in the vertical direction, the object must be thrown up, down, or at an angle relative to the horizontal, rather than sideways. range: the distance a projectile travels horizontally over level ground y maximum height range x Solving Projectile Motion Problems In this chapter, you have been working with components, so you know how to solve motion problems by breaking the motion down into its horizontal (x) and vertical (y) components. Figure 2.66 The range of a projectile is its horizontal distance travelled. Chapter 2 Vector components describe motion in two dimensions. 105 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 106 Before you solve a projectile motion problem, review what you already know (Figure 2.67). vi vx vi vy 0 Figure 2.67(a) The projectile is given an initial horizontal velocity. viy vi sin θ vi vix vi cos θ vi cos θ vix vi sin θ viy vi Figure 2.67(b) The projectile is given an initial horizontal velocity and an upward vertical velocity. Figure 2.67(c) The projectile is given an initial horizontal velocity and
a downward vertical velocity. x direction – There is no acceleration in this direction, so ax 0. In this text, ax will always be zero. The projectile undergoes uniform motion in the x direction. – The general equation for the initial x component of the velocity can be determined using trigonometry, e.g., vix vi cos. – The range is dx. – Because the projectile is moving in both the horizontal and vertical directions at the same time, t is a common variable. y direction – If up is positive, the acceleration due to gravity is down or negative, so ay 9.81 m/s2. – The y component of the initial velocity can be determined using trigonometry, e.g., viy vi sin. – The displacement in the y direction is dy. – Time (t) is the same in both the x and y directions. ▼ Table 2.4 Projectile Problem Setup x direction y direction ax vix 0 vi cos dx vx t 9.81 m/s2 ay vi sin viy can be positive or negative viy depending on the direction of v i. dy t 1 ay(t)2 2 viy If you check the variables, you can see that they are vi, t, d, and a, 1 a(t)2. In the t all of which are present in the equation d 2 v i horizontal direction, the acceleration is zero, so this equation simplifies t. The next example shows you how to apply these equations. to d v i e MATH To explore and graph the relationship between the velocity and position of an object thrown vertically into the air, visit www.pearsoned.ca/school/ physicssource. 106 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 107 Example 2.10 Head-Smashed-In Buffalo Jump, near Fort Macleod, Alberta, is a UNESCO heritage site (Figure 2.68). Over 6000 years ago, the Blackfoot people of the Plains hunted the North American bison by gathering herds and directing them over cliffs 20.0 m tall. Assuming the plain was flat so that the bison ran horizontally off the cliff, and the bison were moving at their maximum speed of 18.0 m/s at the time of the fall, determine how far from the base of the cliff the bison landed. Practice Problems 1. A coin rolls off a table with an initial horizontal
speed of 30 cm/s. How far will the coin land from the base of the table if the table’s height is 1.25 m? 2. An arrow is fired horizontally with a speed of 25.0 m/s from the top of a 150.0-m-tall cliff. Assuming no air resistance, determine the distance the arrow will drop in 2.50 s. 3. What is the horizontal speed of an object if it lands 40.0 m away from the base of a 100-m-tall cliff? Answers 1. 15 cm 2. 30.7 m 3. 8.86 m/s Figure 2.68 y d y x d x Figure 2.69 Given For convenience, choose forward and down to be positive because the motion is forward and down (Figure 2.69). x direction vix 18.0 m/s y direction ay dy 9.81 m/s2 [down] 9.81 m/s2 20.0 m Required distance from the base of the cliff (dx) Analysis and Solution Since there is no vertical component to the initial velocity of the 0 m/s. Therefore, the bison experience uniformly bison, viy accelerated motion due to gravity in the vertical direction but uniform motion in the horizontal direction resulting from the run. From the given values, note that, in the y direction, you have all the variables except for time. So, you can solve for time in the y direction, which is the time taken to fall. PHYSICS INSIGHT For projectile motion in two dimensions, the time taken to travel horizontally equals the time taken to travel vertically. e SIM Analyze balls undergoing projectile motion. Follow the eSim links at www.pearsoned.ca/school/ physicssource. y direction: dy t 1 viy ay(t)2 2 0 1 ay 2 t2 Chapter 2 Vector components describe motion in two dimensions. 107 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 108 t 2dy ay 2(20.0 m) 9.81 m/s2 2.019 s x direction: The time taken for the bison to fall vertically equals the time they travel horizontally. Substitute the value for time you found in the y direction to find the range. Since the bison had a uniform horizontal speed of 18.0 m/s, use the equation dx dx vix t. (18.0 m/s)(2
.019 s) 36.3 m Paraphrase The bison would land 36.3 m from the base of the cliff. Figure 2.70 Baseball is all about projectile motion. Objects Launched at an Angle Baseball is a projectile game (Figure 2.70). The pitcher throws a ball at the batter, who hits it to an open area in the field. The outfielder catches the ball and throws it to second base. The runner is out. All aspects of this sequence involve projectile motion. Each sequence requires a different angle on the throw and a different speed. If the player miscalculates one of these variables, the action fails: Pitchers throw wild pitches, batters strike out, and outfielders overthrow the bases. Winning the game depends on accurately predicting the components of the initial velocity! vy v sin θ v θ vx v cos θ For objects launched at an angle, such as a baseball, the velocity of the object has both a horizontal and a vertical component. Any vector quantity can be resolved into x and y components using the trigonometric ratios R sin, when is measured relative to Rx the x-axis. To determine the horizontal and vertical compo v cos and nents of velocity, this relationship becomes vx vy R cos and Ry v sin, as shown in Figure 2.71. Solving problems involving objects launched at an angle is similar to solving problems involving objects launched horizontally. The object experiences uniform motion in the horizontal direction, so use the equation dx t. In the vertical direction, the object experiences uniformly accelerated motion. The 1 ay(t)2 still applies, but in this case, viy t general equation dy 2 is not zero. The next example shows you how to apply these equations to objects launched at an angle. vix viy Figure 2.71 The horizontal and vertical components of velocity 108 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 109 Example 2.11 Baseball players often practise their swing in a batting cage, in which a pitching machine delivers the ball (Figure 2.72). If the baseball is launched with an initial velocity of 22.0 m/s [30.0°] and the player hits it at the same height from which it was launched, for how long is the baseball in the air on its way to the batter? Figure 2.72 Given v i 22.0 m/s [30.0°] Practice Problems 1. A
ball thrown horizontally at 10.0 m/s travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. 2. A ball is thrown with a velocity of 20.0 m/s [30] and travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. (c) the maximum height of the ball. Answers 1. (a) 30 m (b) 44 m 2. (a) 52 m (b) 14 m (c) 19 m Required time (t) Analysis and Solution Choose forward and up to be positive (Figure 2.73). First find the components of the baseball’s initial velocity. y viy vi 30.0° vix x Figure 2.73 x direction vix vi cos (22.0 m/s)(cos 30.0) 19.05 m/s y direction viy vi sin (22.0 m/s)(sin 30.0) 11.00 m/s Since the ball returns to the same height from which it was launched, dy 0. With this extra known quantity, you now have enough information in the y direction to find the time the ball spent in the air. PHYSICS INSIGHT Be careful to follow the sign convention you chose. If you chose up as positive, ay becomes 9.81 m/s2. info BIT The world’s fastest bird is the peregrine falcon, with a top vertical speed of 321 km/h and a top horizontal speed of 96 km/h. e WEB The fastest speed for a projectile in any ball game is approximately 302 km/h in jai-alai. To learn more about jai-alai, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 109 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 110 PHYSICS INSIGHT Since the vertical velocity of the ball at maximum height is zero, you can also calculate the time taken to go up and multiply the answer by two. If down is positive, t vi vf a 0 m/s (11.00 m/s) 9.81 m/s2 11.00 m s m 9.81 s2 1
.121 s The total time the baseball is in the air is 2 1.121 s 2.24 s. info BIT The longest speedboat jump was 36.5 m in the 1973 James Bond movie Live and Let Die. The boat practically flew over a road. 110 Unit I Kinematics dy viy 1 t ay(t)2 2 0 (11.00 m/s)t (9.81 m/s2)(t)2 1 2 Isolate t and solve. (4.905 m/s2)(t)2 (11.00 m/s)(t) t m 11.00 s m 4.905 s2 2.24 s Paraphrase The baseball is in the air for 2.24 s. How far would the baseball in Example 2.11 travel horizontally if the batter missed and the baseball landed at the same height from which it was launched? Since horizontal velocity is constant, dx t vix (19.05 m/s)(2.24 s) 42.7 m The baseball would travel a horizontal distance of 42.7 m. In the next example, you are given the time and are asked to solve for one of the other variables. However, the style of solving the problem remains the same. In any problem that you will be asked to solve in this course, you will always be able to solve for one quantity in either the x or y direction, and then you can substitute your answer to solve for the remaining variable(s). Example 2.12 A paintball directed at a target is shot at an angle of 25.0. If paint splats on its intended target at the same height from which it was launched, 3.00 s later, find the distance from the shooter to the target. Given Choose down and right to be positive. a ay 25.0 t 3.00 s 9.81 m/s2 [down] 9.81 m/s2 y vi 25.0° Figure 2.74 x 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 111 Required range (dx) Analysis and Solution Use the equation dy viy t ay(t)2. Since the height 1 2 of landing is the same as the launch height, dy y direction: 0. dy viy 1 t ay(t)2 2 0 viy 1 t ay(t)2 2 viy 1 t ay(t)2 2 viy t 1 ay 2 1 9
.81 2 m s2 (3.00 s) 14.7 m/s Since down is positive, the negative sign means that the direction of the vertical component of initial velocity is up. x direction: Find the initial horizontal speed using the tangent function. Because there is no acceleration in the x direction, the ball’s horizontal speed remains the same during its flight: ax 0. PHYSICS INSIGHT Alternatively, the time taken to reach maximum height is the same time taken to fall back down to the same height. So, the paintball is at its maximum height at 1.50 s. The speed at maximum height is zero. If up is positive at 0 m/s at (9.81 m/s2)(t) (9.81 m/s2)(1.50 s) 14.7 m/s v i f The sign is positive, so the direction is up. tan opposite adjacent adjacent opposite tan 14.7 m/s tan 25.0° 31.56 m/s vi vix 25.0° Figure 2.75 14.7 m/s From Figure 2.75, the adjacent side is vix and it points to the right, so vix Now find the horizontal distance travelled. dx 31.56 m/s. t vix (31.56 m/s)(3.00 s) 94.7 m Practice Problems 1. Determine the height reached by a baseball if it is released with a velocity of 17.0 m/s [20]. 2. A German U2 rocket from the Second World War had a range of 300 km, reaching a maximum height of 100 km. Determine the rocket’s maximum initial velocity. Paraphrase The distance that separates the target from the shooter is 94.7 m. Answers 1. 1.72 m 2. 1.75 103 m/s [53.1] Chapter 2 Vector components describe motion in two dimensions. 111 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 112 The points below summarize what you have learned in this section. – To solve problems involving projectiles, first resolve the motion into its components using the trigonometric functions, then apply the kinematics equations. – Perpendicular components of motion are independent of one another. – Horizontal motion is considered uniform and is described by the equation d vt, whereas vertical motion is a special case of uniformly accelerated motion, where the acceleration is the
acceleration due to gravity or 9.81 m/s2 [down]. – A projectile’s path is a parabola. – In the vertical direction, a projectile’s velocity is greatest at the instant of launch and just before impact, whereas at maximum height, vertical velocity is zero. 2.4 Check and Reflect 2.4 Check and Reflect Knowledge 1. Platform divers receive lower marks if they enter the water a distance away from the platform, whereas speed swimmers dive as far out into the pool as they can. Compare and contrast the horizontal and vertical components of each type of athlete’s motion. 2. For a fixed speed, how does the range depend on the angle,? 3. (a) For a projectile, is there a location on its trajectory where the acceleration and velocity vectors are perpendicular? Explain. (b) For a projectile, is there a location on its trajectory where the acceleration and velocity vectors are parallel? Explain. 4. Water safety instructors tell novice swimmers to put their toes over the edge and jump out into the pool. Explain why, using concepts from kinematics and projectile motion. Applications 5. Participants in a road race take water from a refreshment station and throw their empty cups away farther down the course. If a runner has a forward speed of 6.20 m/s, how far in advance of a garbage pail should he release his water cup if the vertical distance between the lid of the garbage can and the runner’s point of release is 0.50 m? 6. A baseball is thrown with a velocity of 27.0 m/s [35]. What are the components of the ball’s initial velocity? How high and how far will it travel? 112 Unit I Kinematics 7. A football is thrown to a moving receiver. The football leaves the quarterback’s hands 1.75 m above the ground with a velocity of 17.0 m/s [25]. If the receiver starts 12.0 m away from the quarterback along the line of flight of the ball when it is thrown, what constant velocity must she have to get to the ball at the instant it is 1.75 m above the ground? 8. At the 2004 Olympic Games in Athens, Dwight Phillips won the gold medal in men’s long jump with a jump of 8.59 m. If the angle of his jump was 23, what was his takeoff speed? 9. A projectile is fired with an initial speed of 120 m/
s at an angle of 55.0 above the horizontal from the top of a cliff 50.0 m high. Find (a) the time taken to reach maximum height (b) the maximum height with respect to the ground next to the cliff (c) the total time in the air (d) the range (e) the components of the final velocity just before the projectile hits the ground Extension 10. Design a spreadsheet to determine the maximum height and range of a projectile with a launch angle that increases from 0 to 90 and whose initial speed is 20.0 m/s. e TEST To check your understanding of projectile motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 02-PearsonPhys20-Chap02 7/25/08 8:15 AM Page 113 CHAPTER 2 SUMMARY Key Terms and Concepts collinear resultant vector components polar coordinates method navigator method non-collinear relative motion ground velocity air velocity wind velocity trajectory range Key Equations v a t d v 1 t a(t)2 2 i d vt Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. projectile motion Examples of Two-dimensional Motion relative motion Split given values into horizontal (x) and vertical components (y) Set up problem in x and y : x ax 0 vix vi ___ θ y ay 9.81 m/s2 vi ____ θ viy Time is common to both x and y directions. Use the equation d vi Solve in the x direction, then in the y direction, OR solve in the y direction, then in the x direction. t (t)2 a 1 2 State given values: vground _____ [ ] vair _____ [ ] vwind _____ [ ] Split given values into components. Use the equation v_____ vwind vair Use the algebraic vector method to solve for the unknown value. Two-dimensional motion Addition of vectors Graphical method Algebraic method Draw vectors to scale, connect them _________. Split given values into x and y components. Draw a new vector from starting point of diagram to end point. Use Rx R cosθ and _________. Measure length of vector and convert using scale. Solve separate equations in x and y. Measure angle at start of vector. Combine component answers using the Pythagorean theorem: Determine the angle using
__________. R x2 y2 √ The answer is called the ___________. It has both magnitude and _________. Chapter 2 Vector components describe motion in two dimensions. 113 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 114 CHAPTER 2 REVIEW Knowledge 1. (2.2) During the Terry Fox Run, a participant 8. (2.4) For an object thrown vertically upward, what is the object’s initial horizontal velocity? travelled from A to D, passing through B and C. Copy and complete the table using the information in the diagram, a ruler calibrated in millimetres, and a protractor. In your notebook, draw and label the displacement vectors AB, BC, and CD and the position vectors AB, AC, and AD. Assume the participant’s reference point is A. d = 5.0 km [E] C 2.0 km scale D d = 3.0 km [E] d = 2.0 km [N] A B Distance Δd (km) Displacement Δd (km) [direction] Final position d (km) [direction] reference point AB BC CD AC AD 2. (2.2) Determine the x and y components of the displacement vector 55 m [222]. 3. (2.4) What is the vertical component for velocity at the maximum height of a projectile’s trajectory? 4. (2.4) During a field goal kick, as the football rises, what is the effect on the vertical component of its velocity? 5. (2.1) Fort McMurray is approximately 500 km [N] of Edmonton. Using a scale of 1.0 cm : 50.0 km, draw a displacement vector representing this distance. 6. (2.1) Give one reason why vector diagrams must be drawn to scale. 7. (2.2) Using an appropriate scale and reference coordinates, graphically solve each of the following: (a) 5.0 m [S] and 10.0 m [N] (b) 65.0 cm [E] and 75.0 cm [E] (c) 1.0 km [forward] and 3.5 km [backward] (d) 35.0 km [right] 45.0 km [left] 114 Unit I Kinematics Applications 9. The air medivac, King Air 200, flying at 250 knots (
1 knot 1.853 km/h), makes the trip between Edmonton and Grande Prairie in 50 min. What distance does the plane travel during this time? 10. A golf ball is hit with an initial velocity of 30.0 m/s [55]. What are the ball’s range and maximum height? 11. Off the tee box, a professional golfer can drive a ball with a velocity of 80.0 m/s [10]. How far will the ball travel horizontally before it hits the ground and for how long is the ball airborne? 12. A canoeist capable of paddling north at a speed of 4.0 m/s in still water wishes to cross a river 120 m wide. The river is flowing at 5.0 m/s [E]. Find (a) her velocity relative to the ground (b) the time it takes her to cross 13. An object is thrown horizontally off a cliff with an initial speed of 7.50 m/s. The object strikes the ground 3.0 s later. Find (a) the object’s vertical velocity component when it reaches the ground (b) the distance between the base of the cliff and the object when it strikes the ground (c) the horizontal velocity of the object 1.50 s after its release 14. If a high jumper reaches her maximum height as she travels across the bar, determine the initial velocity she must have to clear a bar set at 2.0 m if her range during the jump is 2.0 m. What assumptions did you make to complete the calculations? 15. An alligator wishes to swim north, directly across a channel 500 m wide. There is a current of 2.0 m/s flowing east. The alligator is capable of swimming at 4.0 m/s. Find (a) the angle at which the alligator must point its body in order to swim directly across the channel (b) its velocity relative to the ground (c) the time it takes to cross the channel 16. A baseball player throws a ball horizontally at 45.0 m/s. How far will the ball drop before reaching first base 27.4 m away? 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 115 17. How much time can you save travelling diagonally instead of walking 750 m [N] and then 350 m [E] if your walking speed is 7.0 m/s? 18. How long will an arrow
be in flight if it is shot at an angle of 25 and hits a target 50.0 m away, at the same elevation? 19. A pilot of a small plane wishes to fly west. The plane has an airspeed of 100 km/h. If there is a 30-km/h wind blowing north, find 27. An airplane is approaching a runway for landing. The plane’s air velocity is 645 km/h [forward], moving through a headwind of 32.2 km/h. The altimeter indicates that the plane is dropping at a constant velocity of 3.0 m/s [down]. If the plane is at a height of 914.4 m and the range from the plane to the start of the runway is 45.0 km, does the pilot need to make any adjustments to her descent in order to land the plane at the start of the runway? Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed on page 113 and the Learning Outcomes on page 68. 1. Create a flowchart to describe the different components required to analyze motion in a horizontal plane and in a vertical plane. 2. Write a paragraph describing the similarities and differences between motion in a horizontal plane and motion in a vertical plane. Share your thoughts with another classmate. Think About It Review your answers to the Think About It questions on page 69. How would you answer each question now? e TEST To check your understanding of two-dimensional motion, follow the eTest links at www.pearsoned.ca/school/physicssource. (a) the plane’s heading (b) the plane’s ground speed 20. At what angle was an object thrown if its initial launch speed is 15.7 m/s, it remains airborne for 2.15 s, and travels 25.0 m horizontally? 21. A coin rolls off a 25.0° incline on top of a 2.5-m-high bookcase with a speed of 30 m/s. How far from the base of the bookcase will the coin land? 22. Starting from the left end of the hockey rink, the goal line is 3.96 m to the right of the boards, the blue line is 18.29 m to the right of the goal line,
the next blue line is 16.46 m to the right of the first blue line, the goal line is 18.29 m right, and the right board is 3.96 m right of the goal line. How long is a standard NHL hockey rink? 23. A plane with a ground speed of 151 km/h is moving 11 south of east. There is a wind blowing at 40 km/h, 45 south of east. Find (a) the plane’s airspeed (b) the plane’s heading, to the nearest degree 24. How long will a soccer ball remain in flight if it is kicked with an initial velocity of 25.0 m/s [35.0]? How far down the field will the ball travel before it hits the ground and what will be its maximum height? 25. At what angle is an object launched if its initial vertical speed is 3.75 m/s and its initial horizontal speed is 4.50 m/s? Extensions 26. During the Apollo 14 mission, Alan Shepard was the first person to hit a golf ball on the Moon. If a golf ball was launched from the Moon’s surface with a velocity of 50 m/s [35] and the acceleration due to gravity on the Moon is 1.61 m/s2, (a) how long was the golf ball in the air? (b) what was the golf ball’s range? Chapter 2 Vector components describe motion in two dimensions. 115 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 116 UNIT I PROJECT Are Amber Traffic Lights Timed Correctly? Scenario The Traffic Safety Act allows law enforcement agencies in Alberta to issue fines for violations using evidence provided by red light cameras at intersections. The cameras photograph vehicles that enter an intersection after the traffic lights have turned red. They record the time, date, location, violation number, and time elapsed since the light turned red. The use of red light cameras and other technology reduces the amount of speeding, running of red lights, and collisions at some intersections. The length of time a traffic light must remain amber depends on three factors: perception time, reaction time, and braking time. The sum of perception time and reaction time is the time elapsed between the driver seeing the amber light and applying the brakes. The Ministry of Infrastructure and Transportation’s (MIT) Basic Licence Driver’s Handbook allows for a perception time of 0.75 s and
a reaction time of 0.75 s. The braking time is the time it takes the vehicle to come to a full stop once the brakes are applied. Braking time depends on the vehicle’s initial speed and negative acceleration. The MIT’s predicted braking times are based on the assumption that vehicles travel at the posted speed limit and have a uniform acceleration of 3.0 m/s2. Other factors that affect acceleration are road conditions, vehicle and tire performance, weather conditions, and whether the vehicle was travelling up or down hill. If drivers decide to go through an intersection safely (go distance) after a light has turned amber, they must be able to travel not only to the intersection but across it before the light turns red. The go distance depends on the speed of the vehicle, the length of the intersection, and the amount of time the light remains amber. If the driver decides to stop (stop distance), the vehicle can safely do so only if the distance from the intersection is farther than the distance travelled during perception time, reaction time, and braking time. As part of a committee reporting to the Ministry of Infrastructure and Transportation, you must respond to concerns that drivers are being improperly fined for red light violations because of improper amber light timing. You are to decide how well the amber light time matches the posted speed limit and intersection length. Assume throughout your analysis that drivers travel at the posted speed limits. Planning Research or derive equations to determine Assessing Results After completing the project, assess its success based on a rubric* designed in class that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team’s presentation (a) a car’s displacement during reaction time (b) stop distance (c) go distance (d) amber light time (e) displacement after brakes are applied (f) amount of time elapsed after the brakes are applied Materials • measuring tape, stopwatch Procedure 1 Design a survey to measure the amber light times at 10 different intersections near your school. For each intersection, record its length. Use caution around intersections due to traffic! You may wish to estimate the length of the intersection by counting the number of steps it takes you to cross and measuring the length of your stride. 2 Apply suitable equations to determine appropriate amber light times for the 10 different intersections. 3 Calculate stop distances and go distances for a range (10 km/h) of posted speed limits for each intersection and plot graphs of stop distance and go distance against posted speed. Thinking Further 1.
Research the effectiveness of red light cameras in reducing accidents, speeding, and red light violations. Using your research, recommend a course of action to increase vehicle-rail safety at light-controlled railway crossings. 2. Based on your surveys and investigation, recommend whether existing amber light times should be increased, decreased, or left alone. Consider posted speeds against actual speeds and wet against dry surface conditions. 3. Prepare a presentation to the other members of your committee. Include graphs and diagrams. *Note: Your instructor will assess the project using a similar assessment rubric. 116 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 117 UNIT I SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 1 Summary Resources and Skill Building Graphs and equations describe motion in one dimension. Scalar and vector quantities 1.1 The Language of Motion A scalar quantity consists of a number and a unit. Distance, position, and displacement A vector quantity consists of a number, a unit, and a direction. Distance is the length of the path taken to travel from one position to another. Position is the straight-line distance from the origin to the object’s location. Displacement is the change in position. Slope of a position-time graph Slope of a velocity-time graph Position-time, velocity-time, acceleration–time graphs representing accelerated motion 1.2 Position-time Graphs and Uniform Motion A position-time graph for an object at rest is a straight line with zero slope. A position-time graph for an object moving at a constant velocity is a straight line with non-zero slope. The greater the slope of a position-time graph, the faster the object is moving. 1.3 Velocity-time Graphs: Uniform and Non-uniform Motion A velocity-time graph for an object experiencing uniform motion is a horizontal line. The slope of a velocity-time graph represents acceleration. The position-time graph for an object undergoing uniformly accelerated motion is a curve. The corresponding velocity-time graph is a straight line with non-zero slope. The corresponding acceleration-time graph is a horizontal line. Section 1.1 Section 1.1 Figure 1.5 Figures 1.3, 1.4, 1.5, Example 1.1 Figures 1.3, 1.4, 1.5, Example 1.1 Figure 1.14 Figures 1.12, 1.15(b), 1-3 Inquiry Lab
Examples 1.2, 1.3, 1-3 Inquiry Lab Figure 1.24 Figures 1.24, 1.30, Example 1.5 Figures 1.28–1.31, Example 1.5 Instantaneous velocity The slope of the tangent on a position-time curve gives instantaneous velocity. Figure 1.29, Example 1.5 Area under and slope of a velocity-time graph Average velocity Velocity-time graphs 1.4 Analyzing Velocity-time Graphs The area under a velocity-time graph represents displacement; slope represents acceleration. Average velocity represents total displacement divided by time elapsed. You can draw acceleration-time and position-time graphs by calculating and plotting slope and area, respectively, of a velocity-time graph. Figure 1.41, Examples 1.6, 1.8, 1.9 Figure 1.45, Examples 1.7, 1.9 Examples 1.10, 1.11 Kinematics equations 1.5 The Kinematics Equations When solving problems in kinematics, choose the equation that contains all the given variables in the problem as well as the unknown variable. Figures 1.53–1.55 Examples 1.12–1.16 Projectile motion straight up and down Maximum height 1.6 Acceleration Due to Gravity Gravity causes objects to accelerate downward. At maximum height, a projectile’s vertical velocity is zero. The time taken to reach maximum height equals the time taken to fall back down to the original height. 1-6 QuickLab, 1-7 Inquiry Lab, 1-8 QuickLab, Examples 1.17–1.19 Figures 1.63–1.65 Examples 1.18, 1.19 Chapter 2 Vector components describe motion in two dimensions. Adding and subtracting vectors 2.1 Vector Methods in One Dimension Add vectors by connecting them tip to tail. Subtract vectors by connecting them tail to tail. Examples 2.1, 2.2 Components Relative motion Projectile motion in two dimensions 2.2 Motion in Two Dimensions To add vectors in two dimensions, draw a scale diagram, or resolve them into their components and use trigonometry to find the resultant. Examples 2.3–2.5, Figures 2.31, 2.33–2.37 2.3 Relative Motion To solve relative motion problems, use trigonometry, with ground velocity as the resultant. If the vectors are not perpendicular, resolve them into their components first. Examples 2.6–2.9 2.4 Projectile
Motion The shape of a projectile’s trajectory is a parabola. Horizontal and vertical components of projectile motion are independent. To solve projectile problems in two dimensions, resolve them into their horizontal and vertical components. Then use the kinematics equations. The time taken to travel horizontally equals the time taken to travel vertically. 2-4 QuickLab, Figure 2.62 2-5 QuickLab, Figure 2.64 Examples 2.10–2.12, Figures 2.67, 2.71 Unit I Kinematics 117 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 118 UNIT I REVIEW Vocabulary 1. Using your own words, define these terms: acceleration acceleration due to gravity air velocity at rest collinear components displacement distance ground velocity instantaneous velocity kinematics navigator method non-collinear non-uniform motion origin polar coordinates method position projectile projectile motion range relative motion resultant vector scalar quantity tangent trajectory uniform motion uniformly accelerated motion vector quantity velocity wind velocity Knowledge CHAPTER 1 2. Describe how scalar quantities differ from vector quantities. CHAPTER 2 5. Using a scale of 1.0 cm : 3.5 km, determine the magnitude and direction of the vector below. W N S Applications E R 6. A wildlife biologist records a moose’s position as it swims away from her. Using the graph below, determine the moose’s velocity. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 5.0 0 0 2.0 4.0 8.0 6.0 Time (s) 10.0 12.0 7. Sketch a position-time graph for each statement below. Assume that right is positive. (a) object accelerating to the right (b) object accelerating to the left (c) object travelling at a constant velocity left (d) object at rest (e) object travelling with constant velocity right 8. Hockey pucks can be shot at speeds of 107 km/h. If a puck is shot at an angle of 30, determine how long the puck is in the air, how far it will travel, and how high it will be at the peak of its trajectory. 3. Resolve the following vectors into their 9. Sketch two different position-time graphs for objects with a negative velocity. 10. Sketch two different velocity-time graphs for objects with a negative acceleration. components: (a) 5
.0 m [90°] (b) 16.0 m/s [20 S of W] 4. Using an appropriate scale and reference coordinates, draw the following vectors: (a) 5.0 m/s [0] (b) 25.0 m/s2 [60 N of E] (c) 1.50 km [120] 118 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 119 11. From the position-time graph below, determine 16. (a) What is the change in velocity in 10.0 s, which object has the greatest velocity. Time (s) 0.0 2.0 4.0 6.0 8.0 10.0 Time (s 12. Solve each of the following equations for initial velocity, v i, algebraically. v v (a) a i t f (b) d v 1 t at2 2 i (c) d 1 2 (v i v f)t 13. The longest kickoff in CFL history is 83.2 m. If the ball remains in the air for 5.0 s, determine its initial speed. 14. Determine the speed of a raven that travels 48 km in 90 min. 15. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1..0 4.0 8.0 6.0 Time (s) 10.0 12.0 as illustrated in the acceleration-time graph below? (b) If the object had an initial velocity of 10 m/s [90], what is its final velocity after 10.0 s? Acceleration vs. Time ) ] ˚.0 3.0 2.0 1.0 0.0 17. How far will a crow fly at 13.4 m/s for 15.0 min? 18. How long will it take a car to travel from Valleyview to Grande Prairie if its speed is 100 km/h? The map’s scale is 1 cm : 118 km. Wood Buffalo National Park 58 High Level Fort Chipewyan Fort Vermilion 35 ALBERTA Fort McMurray 88 Peace River 63 Grande Prairie Slave Lake 2 40 Valleyview 43 Edmonton 19. A baseball player hits a baseball with a velocity of 30 m/s [25]. If an outfielder is
85.0 m from the ball when it is hit, how fast will she have to run to catch the ball before it hits the ground? 20. Determine the magnitude of the acceleration of a Jeep Grand Cherokee if its stopping distance is 51.51 m when travelling at 113 km/h. 21. What is the velocity of an aircraft with respect to the ground if its air velocity is 785 km/h [S] and the wind is blowing 55 km/h [22 S of W]? 22. An object undergoing uniformly accelerated motion has an initial speed of 11.0 m/s and travels 350 m in 3.00 s. Determine the magnitude of its acceleration. 23. Improperly installed air conditioners can occasionally fall from apartment windows down onto the road below. How long does a pedestrian have to get out of the way of an air conditioner falling eight stories (24 m)? Unit I Kinematics 119 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 120 24. An object is launched from the top of a building with an initial velocity of 15 m/s [32]. If the building is 65.0 m high, how far from the base of the building will the object land? 25. Two friends walk at the same speed of 4.0 km/h. One friend steps onto a travelator moving at 3.0 km/h. If he maintains the same initial walking speed, (a) how long will it take him to reach the end of the 100-m-long travelator? (b) what must be the magnitude of the acceleration of the other friend to arrive at the end of the travelator at the same time? 26. How far will a vehicle travel if it accelerates uniformly at 2.00 m/s2 [forward] from 2.50 m/s to 7.75 m/s? 27. An object is thrown into the air with a speed of 25.0 m/s at an angle of 42°. Determine how far it will travel horizontally before hitting the ground. 28. Determine the average velocity of a truck that travels west from Lloydminster to Edmonton at 110 km/h for 1.0 h and 20 min and then 90 km/h for 100 min. 29. What distance will a vehicle travel if it accelerates uniformly from 15.0 m/s [S] to 35.0 m/s [S] in 6.0 s?
30. From the graph below, determine the instantaneous (b) velocity of the object at 5.0 s, 10.0 s, and 15.0 s. Position vs. Time 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0. 33. Determine the displacement of the blue jay from the velocity-time graph below. Velocity vs. Time for a Blue Jay ) ] 32.0 28.0 24.0 20.0 16.0 12.0 8.0 4.0 0.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) 34. Sketch a position-time graph for an object that travels at a constant velocity of 5.0 m/s for 10 s, stops for 10 s, then travels with a velocity of 2.0 m/s for 20 s. 35. Determine the height reached by a projectile if it is released with a velocity of 18.0 m/s [20]. 36. The bishop is a chess piece that moves diagonally along one colour of square. Assuming the first move is toward the left of the board, determine (a) the minimum number of squares the bishop covers in getting to the top right square. the bishop's displacement from the start if the side length of each square is taken as 1 unit and each move is from the centre of a square to the centre of another square. End 0.0 5.0 10.0 Time (s) 15.0 20.0 31. A speedboat’s engine can move the boat at a velocity of 215 km/h [N]. What is the velocity of the current if the boat’s displacement is 877 km [25 E of N] 3.5 h later? 32. An object starts from rest and travels 50.0 m along a frictionless, level surface in 2.75 s. What is the magnitude of its acceleration? 120 Unit I Kinematics Start 37. A wildlife biologist notes that she is 350 m [N] from the park ranger station at 8:15 a.m. when she spots a polar bear. At 8:30 a.m., she is 1.75 km [N] of the ranger station. Determine the biologist’s average velocity. 38. A bus travels 500 m [N], 200 m [E], and then 750 m [S]. Determine its
displacement from its initial position. 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 121 39. Match the motion with the correct position-time graph given below. Identify the motion as at rest, uniform motion, or uniformly accelerated motion. (a) an airplane taking off (b) an airplane landing (c) passing a car on the highway 43. A motorcycle stunt rider wants to jump a 20.0-mwide row of cars. The launch ramp is angled at 30 and is 9.0 m high. The landing ramp is also angled at 30 and is 6.0 m high. Find the minimum launch velocity required for the stunt rider to reach the landing ramp. (d) waiting at the red line at Canada Customs (e) standing watching a parade Skills Practice 44. Draw a Venn diagram to compare and contrast (f) travelling along the highway on cruise control vector and scalar quantities. Position vs. Time 45. Draw a Venn diagram to illustrate the concepts 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10. ii iii iv 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 40. Determine the magnitude of the acceleration of a Jeep Grand Cherokee that can reach 26.9 m/s from rest in 4.50 s. Extensions 41. A penny is released from the top of a wishing well and hits the water’s surface 1.47 s later. Calculate (a) the velocity of the penny just before it hits the water’s surface (b) the distance from the top of the well to the water’s surface 42. A balloonist drops a sandbag from a balloon that is rising at a constant velocity of 3.25 m/s [up]. It takes 8.75 s for the sandbag to reach the ground. Determine of graphical analysis. 46. A swimmer wants to cross from the east to the west bank of the Athabasca River in Fort McMurray. The swimmer’s speed in still water is 3.0 m/s and the current’s velocity is 4.05 m/s [N]. He heads west and ends up downstream on the west bank. Draw a vector diagram for this problem. 47. For an experiment to measure the velocity of an object, you have a radar
gun, probeware, and motion sensors. Explain to a classmate how you would decide which instrument to use. 48. Design an experiment to determine the acceleration of an object rolling down an inclined plane. 49. Construct a concept map for solving a twodimensional motion problem involving a projectile thrown at an angle. 50. Explain how you can use velocity-time graphs to describe the motion of an object. Self-assessment 51. Describe to a classmate which kinematics concepts and laws you found most interesting when studying this unit. Give reasons for your choices. 52. Identify one issue pertaining to motion studied in this unit that you would like to investigate in greater detail. 53. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? 54. As a future voter, what legislation would you support to improve vehicular and road safety? (a) the height of the balloon when the sandbag 55. Assess how well you are able to graph the is dropped (b) the height of the balloon when the sandbag reaches the ground (c) the velocity with which the sandbag hits the ground motion of an object. Explain how you determine a reference point. e TEST To check your understanding of kinematics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit I Kinematics 121 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 122 U N I T II Dynamics Dynamics The design of equipment used in many activities, such as ice climbing, involves understanding the cause of motion. How does gravity affect the climber and the icy cliff? How can understanding the cause of motion help you predict motion? e WEB Explore the physics principles that apply to ice and mountain climbing. Write a summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 122 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 123 Unit at a Glance C H A P T E R 3 Forces can change velocity. 3.1 The Nature of Force 3.2 Newton’s First Law 3.3 Newton’s Second Law 3.4 Newton’s Third Law 3.5 Friction Affects Motion C H A P T E R 4 Gravity extends throughout the universe. 4.1 Gravitational Forces due to Earth 4
.2 Newton’s Law of Universal Gravitation 4.3 Relating Gravitational Field Strength to Gravitational Force Unit Themes and Emphases • Change and Systems • Social and Environmental Contexts • Problem-Solving Skills Focussing Questions In this study of dynamics and gravitation, you will investigate different types of forces and how they change the motion of objects and affect the design of various technological systems. As you study this unit, consider these questions: • How does an understanding of forces help humans interact with their environment? • How do the principles of dynamics affect mechanical and other systems? • What role does gravity play in the universe? Unit Project Tire Design, Stopping Distance, and Vehicle Mass • By the time you complete this unit, you will have the skills to evaluate how tire treads, road surfaces, and vehicle mass affect stopping distances. You will need to consider human reaction times and the amount of moisture on road surfaces to investigate this problem. Unit II Dynamics 123 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 124 C H A P T E R Forces can change velocity. 3 Key Concepts In this chapter, you will learn about: vector addition Newton’s laws of motion static and kinetic friction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain that a non-zero net force causes a change in velocity calculate the net force apply Newton’s three laws to solve motion problems explain static and kinetic friction Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that science and technology develop to meet societal needs explain that science develops through experimentation Screeching tires on the road and the sound of metal and fibreglass being crushed are familiar sounds of a vehicle collision. Depending on the presence of airbags and the correct use of seat belts and headrests, a motorist may suffer serious injury. In order to design these safety devices, engineers must understand what forces are and how forces affect the motion of an object. When a driver suddenly applies the brakes, the seat belts of all occupants lock. If the vehicle collides head-on with another vehicle, airbags may become deployed. Both seat belts and airbags are designed to stop the forward motion of motorists during a head-on collision (Figure 3.1). Motorists in a stationary vehicle that is rear-ended also experience forces. The car seats move forward quickly, taking the lower part of each person
’s body with it. But each person’s head stays in the same place until yanked forward by the neck. It is this sudden yank that causes whiplash. Adjustable headrests are designed to prevent whiplash by supporting the head of each motorist. In this chapter, you will investigate how forces affect motion and how to explain and predict the motion of an object using Newton’s three laws. Figure 3.1 To design cars with better safety features, accident researchers use dummies to investigate the results of high-speed collisions. 124 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 125 3-1 QuickLab 3-1 QuickLab Accelerating a Cart Problem If you pull an object with a force, how do force and mass affect the acceleration of the object? loaded cart spring scale Materials dynamics cart with hook two 200-g standard masses one 1-kg standard mass spring scale (0–5 N) smooth, flat surface (about 1.5 m long) Procedure 1 Place the 200-g standard masses on the cart and attach the spring scale to the hook on the cart. Figure 3.2 Questions 1. Why do you think it was difficult to apply a constant force when pulling the cart each time? 2. Describe how the acceleration of the cart changed from what it was in step 2 when (a) you used the 1-kg standard mass instead of the 2 Pull the spring scale so that the cart starts to 200-g masses, accelerate forward (Figure 3.2). Make sure that the force reading on the spring scale is 2 N and that the force remains as constant as possible while pulling the cart. Observe the acceleration of the cart. 3 Replace the 200-g masses on the cart with the 1-kg standard mass. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 4 Remove all the objects from the cart. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 5 Repeat step 4 but this time pull with a force of 1 N. (b) you removed all the objects from the cart, (c) you decreased the pulling force to 1 N, and (d) you increased the pulling force to 3 N. 3. What force was required to start the cart moving in step 7? 4. Suppose, instead of hooking a spring
scale to the cart in steps 2 to 4, you gave the cart a push of the same magnitude each time. (a) Which cart would you expect to travel the farthest distance? (b) Which cart would you expect to slow down sooner? 6 Repeat step 4 but this time pull with a force of 3 N. (c) What force do you think makes the cart 7 Repeat step 4 but now only pull with just enough force to start the cart moving. Measure the force reading on the spring scale. eventually come to a stop? Think About It 1. Describe the motion of a large rocket during liftoff using the concept of force. Include diagrams in your explanation. 2. Is a plane during takeoff accelerating or moving with constant velocity? Explain in words and with diagrams. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 3 Forces can change velocity. 125 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 126 info BIT Statics is a branch of dynamics that deals with the forces acting on stationary objects. Architecture is primarily a practical application of statics. dynamics: branch of mechanics dealing with the cause of motion 3.1 The Nature of Force The Petronas Twin Towers in Kuala Lumpur, Malaysia, are currently the world’s tallest twin towers. Including the spire on top, each tower measures 452 m above street level. To allow for easier movement of people within the building, architects designed a bridge to link each tower at the 41st floor. What is interesting is that this bridge is not stationary. In order for the bridge to not collapse, it must move with the towers as they sway in the wind (Figure 3.3). In Unit I, you learned that kinematics describes the motion of an object without considering the cause. When designing a structure, the kinematics quantities that an architect considers are displacement, velocity, and acceleration. But to predict how and explain why a structure moves, an architect must understand dynamics. Dynamics deals with the effects of forces on objects. Structures such as bridges and buildings are required to either remain stationary or move in appropriate ways, depending on the design, so that they are safe to use. Architects must determine all the forces that act at critical points of the structure. If the forces along a particular direction do not balance, acceleration will occur. Before you
can predict or explain the motion of an object, it is important to first understand what a force is and how to measure and calculate the sum of all forces acting on an object. Figure 3.3 The design of tall buildings involves understanding forces. Towering buildings are susceptible to movement from the wind. 126 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 127 Force Is a Vector Quantity You experience a force when you push or pull an object. A push or a pull can have different magnitudes and can be in different directions. For this reason, force is a vector quantity. In general, any force acting on an object can change the shape and/or velocity of the object (Figure 3.4). If you want to deform an object yet keep it stationary, at least two forces must be present. (a) force exerted by hand on ball (b) force exerted by hand on ball force exerted by hand on ball (c) force: a quantity measuring a push or a pull on an object force exerted by table on ball distorting starting stopping (d) force exerted by hand on ball (e) force exerted by hand on ball (f) force exerted by hand on ball info BIT One newton is roughly equal to the magnitude of the weight of a medium-sized apple or two golf balls. speeding up slowing down changing the direction of motion Figure 3.4 Different forces acting on a ball change either the shape or the motion of the ball. (a) The deformation of the ball is caused by both the hand and the table applying opposing forces on the ball. (b)–(f) The motion of the ball is changed, depending on the magnitude and the direction of the force applied by the hand. The symbol of force is F and the SI unit for force is the newton (N), named in honour of physicist Isaac Newton (1642–1727). One newton is equal to one kilogram-metre per second squared (1 kgm/s2), which is the force required to move a 1-kg object with an acceleration of 1 m/s2. The direction of a force is described using reference coordinates that you choose for a particular situation. You may use [forward] or [backward], compass directions, or polar coordinates. When stating directions using polar coordinates, measure angles counterclockwise from the positive x-axis (Figure 3.5). (a) y 1 0 N 30.0
° x (b) y II III 330° 10 N I x IV Figure 3.5 Two vectors of the same magnitude but with different directions. (a) 10 N [30] (b) 10 N [330] Chapter 3 Forces can change velocity. 127 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 128 Measuring Force One way you could measure forces involves using a calibrated spring scale. To measure the force of gravity acting on an object, attach the object to the end of a vertical spring and observe the stretch of the spring. The weight of an object is the force of gravity acting on the object. The symbol g. of weight is F When the spring stops stretching, the gravitational and elastic forces acting on the object balance each other (Figure 3.6). At this point, the elastic force is equal in magnitude to the weight of the object. So you can determine the magnitude of the weight of an object by reading the pointer position on a calibrated spring scale once the spring stops stretching. pointer elastic force vector Find out the relationship between the stretch of a spring and the weight of an object by doing 3-2 QuickLab. object 10 6 N gravitational force vector 6 N Figure 3.6 A spring scale is one type of instrument that can be used to measure forces. 3-2 QuickLab 3-2 QuickLab Measuring Force Using a Spring Scale Problem How is the amount of stretch of a calibrated spring related to the magnitude of the force acting on an object? 3 Hang additional objects from the spring, up to a total mass of 1000 g. Each time, record the mass and the magnitudes of the corresponding gravitational and elastic forces. Materials set of standard masses with hooks spring scale (0–10 N) Procedure 1 Hold the spring scale vertically and make sure the pointer reads zero when nothing is attached. 2 Gently suspend a 100-g standard mass from the spring. Use a table to record the mass and the magnitudes of the gravitational and elastic forces acting on the object. Questions 1. What was the reading on the spring scale when the 100-g mass was attached? 2. What happened to the stretch of the spring when the mass of the object attached to the spring scale (a) doubled? (b) tripled? (c) changed by a factor of 10? 3. Why is a spring scale ideal for measuring force? 128 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08
10:37 AM Page 129 Representing Forces Using Free-Body Diagrams A free-body diagram is a powerful tool that can be used to analyze situations involving forces. This diagram is a sketch that shows the object by itself, isolated from all others with which it may be interacting. Only the force vectors exerted on the object are included and, in this physics course, the vectors are drawn with their tails meeting at the centre of the object (Figure 3.7). However, it does not necessarily mean that the centre of the object is where the forces act. When drawing a free-body diagram, it is important to show the reference coordinates that apply to the situation in a given problem. Remember to always include which directions you will choose as positive. Figure 3.8 shows the steps for drawing free-body diagrams. free-body diagram: vector diagram of an object in isolation showing all the forces acting on it Fs elastic force exerted by spring on mass + up down Fg gravitational force exerted by Earth on mass Identify the object for the free-body diagram Sketch the object in isolation with a dot at its centre Identify all the forces acting on the object Include in free-body diagram Choose appropriate reference coordinates and include which directions are positive Forces on other objects? Ignore Figure 3.7 The free-body diagram for the object in Figure 3.6 that is suspended from the spring scale. The spring scale is not included here because it is not the object being studied. N, S, E, W up, down, forward, backward x, y If possible, choose an appropriate scale Draw forces proportionally extending out from the dot at the centre of the object Figure 3.8 Flowchart summarizing the steps for drawing a free-body diagram Chapter 3 Forces can change velocity. 129 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 130 normal force: force on an object that is perpendicular to a common contact surface Some Types of Forces There are different types of forces and scientists distinguish among them by giving these forces special names. When an object is in contact with another, the objects will have a common surface of contact, and the two objects will exert a normal force on each other. The normal force, F N, is a force that is perpendicular to this common surface. Depending on the situation, another force called friction, F f, may be present, and this force acts parallel to the common surface. The adjective “normal” simply means perpendicular. Figure
3.9 (a) shows a book at rest on a level table. The normal force exerted by the table on the book is represented by the vector directed upward. If the table top were slanted and smooth as in Figure 3.9 (b), the normal force acting on the book would not be directed vertically upward. Instead, it would be slanted, but always perpendicular to the contact surface. (a) (b) FN Fg FN Ff Fg θ Figure 3.9 Forces acting on (a) a stationary book on a level table and on (b) a book accelerating down a smooth, slanted table. app, if, say, a A stationary object may experience an applied force, F person pushes against the object (Figure 3.10). In this case, the force of friction acting on the object will oppose the direction of impending motion. FN Ff Fapp Fg Figure 3.10 The forces acting on a stationary box Example 3.1 demonstrates how to draw a free-body diagram for a car experiencing different types of forces. In this situation, the normal force acting on the car is equal in magnitude to the weight of the car. 130 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 131 Example 3.1 g, of 10 000 N [down] is coasting on a level road. A car with a weight, F N, of 10 000 N [up], a force of air The car experiences a normal force, F f, exerted air, of 2500 N [backward], and a force of friction, F resistance, F by the road on the tires of 500 N [backward]. Draw a free-body diagram for this situation. Analysis and Solution While the car is coasting, there is no forward force acting on the car. The free-body diagram shows four forces (Figure 3.11). 2000 N up down backward forward FN Ff Fair Fg Figure 3.11 Practice Problems 1. The driver in Example 3.1 sees a pedestrian and steps on the brakes. The force of air resistance is 2500 N [backward]. With the brakes engaged, the force of friction exerted on the car is 5000 N [backward]. Draw a free-body diagram for this situation. 2. A car moving at constant velocity starts to speed up. The weight of the car is 12 000 N [down]. The force of air resistance is 3600 N [backward].
With the engine engaged, the force of friction exerted by the road on the tires is 7200 N [forward]. Draw a free-body diagram for this situation. Answers 1. and 2. See page 898. Using Free-Body Diagrams to Find Net Force Free-body diagrams are very useful when you need to calculate the net force, F net, on an object. The net force is a vector sum of all the forces acting simultaneously on an object. The force vectors can be added using either a scale vector diagram or using components. net force: vector sum of all the forces acting simultaneously on an object Concept Check Can the net force on an object ever equal zero? Explain using an example and a free-body diagram. Chapter 3 Forces can change velocity. 131 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 132 e SIM Learn how to use free-body diagrams to find the net force on an object. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Adding Collinear Forces Vectors that are parallel are collinear, even if they have opposite directions. Example 3.2 demonstrates how to find the net force on an object given two collinear forces. In this example, a canoe is dragged using two ropes. The magnitude of the force F T exerted by a rope on an object at the point where the rope is attached to the object is called the tension in the rope. In this physics course, there are a few assumptions that you need to make about ropes or cables to simplify calculations. These assumptions and the corresponding inferences are listed in Table 3.1. Note that a “light” object means that it has negligible mass. Table 3.1 Assumptions about Ropes or Cables Assumption Inference The mass of the rope is negligible. The rope has a negligible thickness. The tension is uniform throughout the length of the rope. F T acts parallel to the rope and is directed away from the object to which the rope is attached. The rope is taut and does not stretch. Any objects attached to the rope will have the same magnitude of acceleration as the rope. Example 3.2 Two people, A and B, are dragging a canoe out of a lake onto a beach using light ropes (Figure 3.12). Each person applies a force of 60.0 N [forward] on the rope. The force of friction exerted by the beach on the
canoe is 85.0 N [backward]. Starting with a free-body diagram, calculate the net force on the canoe. forward backward Practice Problems 1. Two dogs, A and B, are pulling a sled across a horizontal, snowy surface. Dog A exerts a force of 200 N [forward] and dog B a force of 150 N [forward]. The force of friction exerted by the snow on the sled is 60 N [backward]. The driver attempts to slow down the sled by pulling on it with a force of 100 N [backward]. Starting with a free-body diagram, calculate the net force on the sled. 132 Unit II Dynamics Given F F T1 T2 F f 60.0 N [forward] 60.0 N [forward] 85.0 N [backward] Figure 3.12 Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.13). backward forward FT1 FT2 Ff Figure 3.13 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 133 backward forward FT1 Fnet Figure 3.14 FT2 Ff Add the force vectors shown in the vector addition diagram (Figure 3.14). F f F Ff F T2 FT2 F T1 FT1 60.0 N 60.0 N (85.0 N) 60.0 N 60.0 N 85.0 N 35.0 N 35.0 N [forward] net Fnet F net Paraphrase The net force on the canoe is 35.0 N [forward]. Practice Problems 2. In a tractor pull, four tractors are connected by strong chains to a heavy load. The load is initially at rest. Tractors A and B pull with forces of 5000 N [E] and 4000 N [E] respectively. Tractors C and D pull with forces of 4500 N [W] and 3500 N [W] respectively. The magnitude of the force of friction exerted by the ground on the load is 1000 N. (a) Starting with a free-body diagram, calculate the net force on the load. (b) If the load is initially at rest, will it start moving? Explain. Answers 1. 190 N [forward] 2. (a) 0 N, (b) no Adding Non-Collinear Forces Example 3.3 demonstrates how to find the net force on an object if the forces acting on
it are neither parallel nor perpendicular. By observing the relationship between the components of the force vectors, you can greatly simplify the calculations. Example 3.3 Refer to Example 3.2 on page 132. Person A thinks that if A and B each pull a rope forming an angle of 20.0° with the bow, the net force on the canoe will be greater than in Example 3.2 (Figure 3.15). The canoe is being dragged along the beach using ropes that are parallel to the surface of the beach. Starting with a free-body diagram, calculate the net force on the canoe. Is person A’s thinking correct? 20.0° θ 1 20.0° θ 2 Figure 3.15 Given F 60.0 N [along rope] 85.0 N [backward] T1 F f F T2 1 60.0 N [along rope] 2 20.0 y Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.16). x Ff Figure 3.16 FT1 20.0° 20.0° FT2 Chapter 3 Forces can change velocity. 133 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 134 Practice Problems 1. Refer to Example 3.3. Suppose person A pulls a rope forming an angle of 40.0 with the bow and person B pulls a rope forming an angle of 20.0 with the bow. Each person applies a force of 60.0 N on the rope. The canoe and ropes are parallel to the surface of the beach. If the canoe is being dragged across a horizontal, frictionless surface, calculate the net force on the canoe. 2. Two people, A and B, are dragging a sled on a horizontal, icy surface with two light ropes. Person A applies a force of 65.0 N [30.0] on one rope. Person B applies a force of 70.0 N [300] on the other rope. The force of friction on the sled is negligible and the ropes are parallel to the icy surface. Calculate the net force on the sled. Answers 1. 1.04 102 N [10.0] 2. 95.5 N [343] Separate all forces into x and y components. T1 Vector F F F T2 f x component y component (60.0 N)(cos 20.0) (60.0 N)(cos 20.0
) 85.0 N (60.0 N)(sin 20.0) (60.0 N)(sin 20.0) 0 From the chart, FT1y F nety Fnety F T1y FT1y So y FT2y. F T2y FT2y 0 N FT1 20.0° Fnet 20.0° FT2 Ff x Figure 3.17 Figure 3.17 Add the x components of all force vectors in the vector addition diagram (Figure 3.17). netx Fnetx F f Ff F T2x FT2x x direction F F T1x FT1x (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) (85.0 N) (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) 85.0 N 27.8 N 27.8 N [0] F net Paraphrase The net force is 27.8 N [0°]. Since the net force in Example 3.3 is less than that in Example 3.2, person A’s thinking is incorrect. Applying Free-Body Diagrams to Objects in Equilibrium At the beginning of this section, you learned that architects consider the net force acting at critical points of a building or bridge in order to prevent structure failure. Example 3.4 demonstrates how free-body diagrams and the concept of net force apply to a stationary object. Stationary objects are examples of objects at equilibrium because the net force acting on them equals zero. Example 3.4 A store sign that experiences a downward gravitational force of 245 N is suspended as shown in Figure 3.18. T1 and F Calculate the forces F exerted at the point at which the sign is suspended. T2 Figure 3.18 θ 1 55.0° FT1 θ 2 FT2 134 Unit II Dynamics 03-Phys20-Chap03.qxd 7/25/08 8:17 AM Page 135 Given F g 245 N [down] Required T2) T1 and F forces (F 1 55.0° 2 90.0° Analysis and Solution Draw a free-body diagram for the sign (Figure 3.19). Resolve all forces into x and y components. T1 Vector F F F T2 g x component FT1 cos 55.0 FT2 0 y component FT1 sin 55.0 0 Fg Since the sign is
at equilibrium, the net force in both the x and y directions is zero. Fnetx Fnety 0 N Add the x and y components of all force vectors separately. x direction F netx Fnetx F T1x FT1x F T2x FT2x 0 FT1 cos 55.0 F T2 FT2 FT1 cos 55.0 y direction F F T1y FT1y nety Fnety F g Fg 0 FT1 sin 55.0 ( Fg) 0 FT1 Fg sin 55.0 FT1 Fg sin 55.0 245 N sin 55.0 kgm s2 299.1 299 N Substitute FT1 into the equation for FT2. FT2 F T2 FT1 cos 55.0 (299.1 N)(cos 55.0) 171.6 N 172 N [0] From Figure 3.19, the direction of F T1 measured counterclockwise from the positive x-axis is 180 55.0 125. F T1 299 N [125] Paraphrase T2 is 172 N [0]. T1 is 299 N [125] and F F 55.0° FT1 55.0° y x FT2 Fg Figure 3.19 Practice Problems 1. If the sign in Example 3.4 had half the weight, how would the forces T2 compare? T1 and F F 2. Suppose the sign in Example 3.4 is suspended as shown in Figure 3.20. Calculate the forces FF T1 T2. and FF θ 1 40.0° FT1 θ 2 FT2 Figure 3.20 3. Refer to the solutions to Example 3.4 and Practice Problem 2 above. (a) As 1 decreases, what happens T2? T1 and F to F (b) Explain why 1 can never equal zero. Answers 1. directions of F T1 and F T2 would remain the same as before, but the respective magnitudes would be half 3.81 102 N [140°] 2.92 102 N [0°] T1 2. F F 3. (a) F T2 T1 and F (b) magnitude of F T2 increase in value T1y must always equal Fg Chapter 3 Forces can change velocity. 135 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 136 3.1 Check and Reflect 3.1 Check and Reflect Knowledge 1. (
a) Explain what a force is, and state the SI unit of force. (b) Why is force a dynamics quantity and not a kinematics quantity? Applications 2. Sketch a free-body diagram for (a) a bicycle moving west on a level road with decreasing speed (b) a ball experiencing forces of 45 N [12.0], 60 N [100], and 80 N [280] simultaneously 3. The total weight of a biker and her motorbike is 1800 N [down]. With the engine engaged, the force of friction exerted by the road on the tires is 500 N [forward]. The air resistance acting on the biker and bike is 200 N [backward]. The normal force exerted by the road on the biker and bike is 1800 N [up]. (a) Consider the biker and bike as a single object. Draw a free-body diagram for this object. (b) Calculate the net force. 4. If two forces act on an object, state the angle between these forces that will result in the net force given below. Explain using sketches. (a) maximum net force (b) minimum net force 5. Two people, A and B, are pulling on a tree with ropes while person C is cutting the tree down. Person A applies a force of 80.0 N [45.0] on one rope. Person B applies a force of 90.0 N [345] on the other rope. Calculate the net force on the tree. 6. Three forces act simultaneously on an 2 is 80 N [115], 1 is 65 N [30.0], F object: F 3 is 105 N [235]. Calculate the net and F force acting on the object. Extensions 7. A sign that experiences a downward gravitational force of 245 N is suspended, as shown below. Calculate the forces F T2. and F T1 θ 1 55.0° θ 2 55.0° FT1 FT2 8. The blanket toss is a centuries-old hunting technique that the Inuit used to find herds of caribou. During the toss, several people would hold a hide taut while the hunter would jump up and down, much like on a trampoline, increasing the jump height each time. At the top of the jump, the hunter would rotate 360 looking for the herd. Draw a free-body diagram for a hunter of weight 700 N [down] while (a) standing at rest on the taut hide
just before a jump (b) at the maximum jump height 9. Construct a flowchart to summarize how to add two or more non-collinear forces using components. Refer to Figure 3.8 on page 129 or Student References 4: Using Graphic Organizers on page 869. e TEST To check your understanding of forces, follow the eTest links at www.pearsoned.ca/school/ physicssource. 136 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 137 info BIT The skeleton is an Olympic sledding sport believed to have originated in St. Moritz, Switzerland. 3.2 Newton’s First Law At the 2006 Winter Olympics in Turin, Italy, Duff Gibson charged head-first down a 1.4-km icy track with 19 challenging curves. He reached speeds well over 125 km/h and ended up clinching the gold medal in the men’s skeleton event (Figure 3.21). Gibson’s success had a lot to do with understanding the physics of motion. Imagine an ideal situation in which no friction acts on the sled and no air resistance acts on the athlete. Scientist Galileo Galilei (1564–1642) thought that an object moving on a level surface would continue moving forever at constant speed and in the same direction if no external force acts on the object. If the object is initially stationary, then it will remain stationary, provided no external force acts on the object. In the real world, friction and air resistance are external forces that act on all moving objects. So an object that is in motion will eventually slow down to a stop, unless another force acts to compensate for both friction and air resistance. Galileo recognized the existence of friction, so he used thought experiments, as well as experiments with controlled variables, to understand motion. Thought experiments are theoretical, idealized situations that can be imagined but cannot be created in the real world. Figure 3.21 In the 2006 Winter Olympics, Calgary residents Duff Gibson (shown in photo) and Jeff Pain competed in the men’s skeleton. Gibson edged Pain by 0.26 s to win the gold medal. Pain won the silver medal. Chapter 3 Forces can change velocity. 137 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 138 inertia: property of an object that resists acceleration The Concept of Inertia Since ancient times, many thinkers attempted to understand how and why objects moved.
But it took thousands of years before satisfactory explanations were developed that accounted for actual observations. A major stumbling block was not identifying friction as a force that exists in the real world. In his study of motion, Galileo realized that every object has inertia, a property that resists acceleration. A stationary curling stone on ice requires a force to start the stone moving. Once it is moving, the curling stone requires a force to stop it. To better understand the concept of inertia, try the challenges in 3-3 QuickLab. Concept Check Compare the inertia of an astronaut on Earth’s surface, in orbit around Earth, and in outer space. Can an object ever have an inertia of zero? Explain. 3-3 QuickLab 3-3 QuickLab Challenges with Inertia Problem What role does inertia play in each of these challenges? 3 Set up a stack of loonies and the ruler as shown in Figure 3.22 (c). Use the ruler to remove the loonie at the very bottom without toppling the stack. Materials glass tumbler small, stiff piece of cardboard several loonies empty soft-drink bottle plastic hoop (about 2 cm wide, cut from a large plastic bottle) small crayon with flat ends ruler (thinner than thickness of one loonie) Procedure 1 Set up the tumbler, piece of cardboard, and loonie as shown in Figure 3.22 (a). Remove the cardboard so the loonie falls into the tumbler. 2 Set up the soft-drink bottle, plastic hoop, and a crayon as shown in Figure 3.22 (b). Remove the hoop so the crayon falls into the bottle. Caution: Keep your eyes well above the coin stack. Questions 1. (a) What method did you use to successfully perform each step in the procedure? (b) Apply the concept of inertia to explain why each of your procedures was effective. (a) (b) (c) Figure 3.22 138 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 139 Newton’s First Law and Its Applications Newton modified and extended Galileo’s ideas about inertia in a law, called Newton’s first law of motion (Figure 3.23). (a) Fnet 0 An object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force. v 0
If F net 0, then v 0. So if you want to change the motion of an object, a non-zero net (b) Fnet 0 force must act on the object. Concept Check The Voyager 1 and 2 space probes are in interstellar space. If the speed of Voyager 1 is 17 km/s and no external force acts on the probe, describe the motion of the probe (Figure 3.24). v constant Figure 3.23 If the net force on an object is zero, (a) a stationary object will remain at rest, and (b) an object in motion will continue moving at constant speed in the same direction. Figure 3.24 The Voyager planetary mission is NASA’s most successful in terms of the number of scientific discoveries. Newton’s First Law and Sliding on Ice Many winter sports involve a person sliding on ice. In the case of the skeleton event in the Winter Olympics, an athlete uses a sled to slide along a bobsled track. In hockey, a player uses skates to glide across the icy surface of a rink. Suppose a person on a sled is sliding along a horizontal, icy surface. If no external force acts on the person-sled system, then according to Newton’s first law, the person would maintain the same speed. In fact, the person would not stop at all (Figure 3.25). In real life, the external forces of friction and air resistance act on all moving objects. So the system would eventually come to a stop. FN Fg Figure 3.25 Free-body diagram of a person-sled system sliding on a horizontal, frictionless surface Chapter 3 Forces can change velocity. 139 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 140 Newton’s First Law and Vehicle Safety Devices When you are in a moving car, you can feel the effects of your own inertia. If the car accelerates forward, you feel as if your body is being pushed back against the seat, because your body resists the increase in speed. If the car turns a corner, you feel as if your body is being pushed against the door, because your body resists the change in the direction of motion. If the car stops suddenly, you feel as if your body is being pushed forward, because your body resists the decrease in speed (Figure 3.26). (a) Inertia of motorist makes her feel like she is being pushed backward. direction of motion (c
) Inertia of motorist makes her feel like she is being thrown forward. direction of acceleration of vehicle (b) Inertia of motorist makes her feel like she wants to continue moving in a straight line. direction of acceleration of vehicle Figure 3.26 The inertia of a motorist resists changes in the motion of a vehicle. (a) The vehicle is speeding up, (b) the vehicle is changing direction, and (c) the vehicle is stopping suddenly. When a car is rear-ended, a motorist’s body moves forward suddenly as the car seat moves forward. However, the motorist’s head resists moving forward. A properly adjusted headrest can minimize or prevent whiplash, an injury resulting from the rapid forward accelerations in a rear-end collision (Figure 3.27). Research shows that properly adjusted headrests can reduce the risk of whiplash-related injuries by as much as 40%. A poorly adjusted headrest, however, can actually worsen the effects of a rear-end collision on the neck and spine. When a car is involved in a head-on collision, the motorist continues to move forward. When a seat belt is worn properly, the forward motion of a motorist is safely restricted. If a head-on collision is violent enough, sodium azide undergoes a rapid chemical reaction to produce non-toxic nitrogen gas, which inflates an airbag. The inflated airbag provides a protective cushion to slow down the head and body of a motorist (Figure 3.28). 5–10 cm Figure 3.27 The ideal position for a headrest Figure 3.28 Airbag systems in vehicles are designed to deploy during vehicle collisions. 140 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 141 3-4 Decision-Making Analysis 3-4 Decision-Making Analysis The Airbag Debate The Issue Front airbags were introduced in the 1990s to help prevent injury to motorists, especially during head-on collisions. Side airbags can also help. Yet front airbags have also been the cause of serious injury, even death. Furthermore, airbags add to the cost of a vehicle. Airbag advocates want both front and side airbags installed, better airbags, and greater control over their operation. Opponents want airbags removed from cars altogether. Background Information Airbags are connected to sensors that detect sudden changes in acceleration. The process of triggering and inflating an air
bag occurs in about 40 ms. It is in that instant that arms and legs have been broken and children have been killed by the impact of a rapidly inflating airbag. Tragically, some of these deaths occurred during minor car accidents. Manufacturers have placed on/off switches for airbags on some vehicles, and some engineers are now developing “smart” airbags, which can detect the size of a motorist and the distance that person is sitting from an airbag. This information can then be used to adjust the speed at which the airbag inflates. Required Skills Defining the issue Developing assessment criteria Researching the issue Analyzing data and information Proposing a course of action Justifying the course of action Evaluating the decision Analysis 1. Identify the different stakeholders involved in the airbag controversy. 2. Research the development and safety history of airbags in cars. Research both front and side airbags. Consider head, torso, and knee airbags. Analyze your results, and identify any trends in your data. 3. Propose a solution to this issue, based on the trends you identified. 4. Propose possible changes to current airbag design that could address the issues of safety and cost. 5. Plan a class debate to argue the pros and cons of airbag use. Identify five stakeholders to represent each side in the debate. Support your position with research. Participants will be assessed on their research, organizational skills, debating skills, and attitudes toward learning. Concept Check Use Newton’s first law to explain why (a) steel barriers usually separate the cab of a truck from the load (Figure 3.29), (b) trucks carrying tall loads navigate corners slowly, and (c) customers who order take-out drinks are provided with lids. e TECH Explore the motion of an object that experiences a net force of zero. Follow the eTECH links at www.pearsoned.ca/school/ physicssource. Figure 3.29 Chapter 3 Forces can change velocity. 141 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 142 3.2 Check and Reflect 3.2 Check and Reflect Knowledge 6. Imagine you are the hockey coach for a 1. In your own words, state Newton’s first law. 2. Give two examples, other than those in the text, that illustrate the property of inertia for both a stationary and a moving object. 3. Use Newton’s first law to
describe the motion of (a) a car that attempts to go around an icy curve too quickly, and (b) a lacrosse ball after leaving the lacrosse stick. 4. Apply Newton’s first law and the concept of inertia to each of these situations. team of 10-year-olds. At a hockey practice, you ask the players to skate across the ice along the blue line (the line closest to the net), and shoot the puck into the empty net. Most of the shots miss the net. The faster the children skate, the more they miss. Newton’s first law would help the players understand the problem, but a technical explanation might confuse them. (a) Create an explanation that would make sense to the 10-year-olds. (b) With the aid of a diagram, design a drill for the team that would help the players score in this type of situation. (a) How could you remove the newspaper without toppling the plastic beaker? Extensions 7. Research why parents use booster seats for young children using information from Safe Kids Canada. Summarize the “seat belt test” that determines whether a child is big enough to wear a seat belt without a booster seat. Begin your search at www.pearsoned.ca/school/physicssource. 8. During a sudden stop or if a motorist tries to adjust a seat belt suddenly, the seat belt locks into position. Research why seat belts lock. Write a brief report, including a diagram, of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 9. Make a web diagram to summarize concepts and ideas associated with Newton’s first law. Label the oval in the middle as “Newton’s first law.” Cluster your words or ideas in other ovals around it. Connect these ovals to the central one and one another, where appropriate, with lines. See Student References 4: Using Graphic Organizers on page 869 for an example. e TEST To check your understanding of inertia and Newton’s first law, follow the eTest links at www.pearsoned.ca/school/physicssource. (b) While moving at constant speed on a level, snowy surface, a snowmobiler throws a ball vertically upward. If the snowmobile continues moving at constant velocity, the ball returns to the driver. Why does the ball land ahead of the driver if the snow
mobile stops? Assume that the air resistance acting on the ball is negligible. x y vs vby vbx Applications 5. Design an experiment using an air puck on an air table or spark air table to verify Newton’s first law. Report your findings. Caution: A shock from a spark air table can be dangerous. 142 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 143 3.3 Newton’s Second Law If a speed skater wants to win a championship race, the cardiovascular system and leg muscles of the athlete must be in peak condition. The athlete must also know how to effectively apply forces to propel the body forward. World-class speed skaters such as Cindy Klassen know that maximizing the forward acceleration requires understanding the relationship among force, acceleration, and mass (Figure 3.30). Newton spent many years of his life trying to understand the motion of objects. After many experiments and carefully analyzing the ideas of Galileo and others, Newton eventually found a simple mathematical relationship that models the motion of an object. This relationship, known as Newton’s second law, relates the net force acting on an object, the acceleration of the object, and its mass. Begin by doing 3-5 Inquiry Lab to find the relationship between the acceleration of an object and the net force acting on it. info BIT Cindy Klassen won a total of five medals during the 2006 Winter Olympics, a Canadian record, and is currently Canada’s most decorated Olympian with six medals. Figure 3.30 Cindy Klassen, originally from Winnipeg but now a Calgary resident, won the gold medal in the 1500-m speed skating event in the 2006 Winter Olympics in Turin, Italy. Chapter 3 Forces can change velocity. 143 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 144 3-5 Inquiry Lab 3-5 Inquiry Lab Relating Acceleration and Net Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 1 2 it a (t)2 relates The kinematics equation d v displacement d, initial velocity v and acceleration aa. If v dd 1 a (t)2. If you solve for acceleration, you get 2 2d a. Remember to use the scalar form of this equation ( )2 t i 0, the equation simplifies to i, time interval t, when solving for acceleration. Question
How is the acceleration of an object related to the net force acting on the object? Hypothesis State a hypothesis relating acceleration and net force. Remember to write an “if/then” statement. Variables The variables involved in this lab are the mass of the system, the applied force acting on the system, friction acting on the system, time interval, the distance the system travels, and the acceleration of the system. Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment C-clamp dynamics cart three 100-g standard masses pulley smooth, flat surface (about 1.5 m) string (about 2 m) recording tape recording timer with power supply metre-stick masking tape graph paper e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 144 Unit II Dynamics Procedure 1 Copy Tables 3.2 and 3.3 from page 145 into your notebook. 2 Measure the mass of the cart. Record the value in Table 3.2. 3 Set up the recording timer, pulley, and cart loaded with three 100-g standard masses on a lab bench (Figure 3.31). Make a loop at each end of the string. Hook one loop to the end of the cart and let the other loop hang down over the pulley. recording tape recording timer 100 g 100 g string pulley 100 g 1 N Figure 3.31 Caution: Position a catcher person near the edge of the lab bench. Do not let the cart or hanging objects fall to the ground. 4 Attach a length of recording tape to the cart and thread it through the timer. 5 While holding the cart still, transfer one 100-g mass from the cart to the loop of string over the pulley and start the timer. When you release the cart, the hanging 100-g mass will exert a force of about 1 N on the system. Release the cart but stop it before it hits the pulley. Label the tape “trial 1; magnitude of F app 1 N.” 6 Repeat steps 4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the first hanging object. Label the tape “trial 2; magnitude of F app = 2 N.” By transferring objects from the cart to the end of the string hanging over the pulley, the mass of the system remains constant but the net force acting on the system varies. 7 Repeat steps
4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the two hanging masses. Label the tape “trial 3; magnitude of F app = 3 N.” 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 145 Analysis 1. Calculate the mass of the system, mT. Record the value in Table 3.2. 2. Using the tape labelled “trial 3,” label the dot at the start t 0 and mark off a convenient time interval. If the 1 s, a time interval of 30 dot period of the timer is 0 6 1 s 0.5 s). Record spaces represents 0.5 s (30 0 6 the time interval in Table 3.2. 3. Measure the distance the system travelled during this time interval. Record this value in Table 3.2. 2 d to calculate the magnitude 4. Use the equation a ( )2 t of the acceleration of the system. Record the value in Tables 3.2 and 3.3. 5. Using the same time interval, repeat questions 3 and 4 for the tapes labelled “trial 1” and “trial 2.” 6. Why is it a good idea to choose the time interval using the tape labelled “trial 3”? 7. Plot a graph of the magnitude of the acceleration vs. the magnitude of the applied force (Table 3.3). 8. (a) Describe the graph you drew in question 7. (b) Where does the graph intersect the x-axis? Why? What conditions would have to be present for it to pass through the origin? (c) For each trial, subtract the x-intercept from the applied force to find the net force. Record the values in Table 3.3. Then plot a graph of the magnitude of the acceleration vs. the magnitude of the net force. 9. When the magnitude of the net force acting on the system is doubled, what happens to the magnitude of the acceleration of the system? 10. What is the relationship between the magnitude of the acceleration and the magnitude of the net force? Write this relationship as a proportionality statement. Does this relationship agree with your hypothesis? Table 3.2 Mass, Time, Distance, and Acceleration Trial Mass of Cart mc (kg) Mass of Load on Cart ml (kg) Mass of Load Hanging over Pulley mh (kg) mT 1 2 3 0.
200 0.100 0 0.100 0.200 0.300 Table 3.3 Force and Acceleration Total Mass mc ml (kg) mh Time Interval t (s) Distance d (m) Magnitude of a (m/s2) Trial 1 2 3 Magnitude of F app Acting Magnitude of F net Acting on System (N) on System (N) 1 2 3 Magnitude of a of System (m/s2) Chapter 3 Forces can change velocity. 145 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 146 FN Fair Ff Fg Fapp Fapp Fnet Ff Fair Fnet Fapp Fair Ff Relating Acceleration and Net Force For the system in 3-5 Inquiry Lab, you discovered that there is a linear relationship between acceleration and net force. This relationship can be written as a proportionality statement: Horizontal Forces a Fnet This relationship applies to speed skating. In the short track relay event, a speed skater pushes the next teammate forward onto the track when it is the teammate’s turn to start skating. While the teammate is being pushed, the horizontal forces acting on the skater are the applied push force, friction, and air resistance (Figure 3.32). As long as the applied push force is greater in magnitude than the sum of the force of friction acting on the skates and the air resistance acting on the skater’s body, the net force on the teammate acts forward. Figure 3.32 (left) Free-body diagram showing the forces acting on a speed skater being pushed by a teammate in the short track relay event; (right) vector addition diagram for the horizontal forces. The harder the forward push, the greater will be the forward net force on the teammate (Figure 3.33). So the acceleration of the teammate will be greater. Note that the acceleration is in the same direction as the net force. Find out the relationship between the acceleration of an object and its mass by doing 3-6 Design a Lab. m constant a a Fnet Fnet Concept Check What is the difference between a net force and an applied force? Can a net force ever equal an applied force? Explain using an example and a free-body diagram. Figure 3.33 For the same mass, a greater net force results in a greater acceleration. 146 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM
Page 147 3-6 Design a Lab 3-6 Design a Lab Relating Acceleration and Mass In this lab, you will investigate the relationship between acceleration and mass when the net force acting on the system is constant. The Question How is the acceleration of an object related to the mass of the object? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Design and Conduct Your Investigation • State a hypothesis relating acceleration and mass. • Then use the set-up in Figure 3.31 on page 144 to design an experiment. List the materials you will use as well as a detailed procedure. Use the procedure and questions in 3-5 Inquiry Lab to help you. • Plot a graph of the magnitude of the acceleration vs. the mass of the system. Then plot a graph of the magnitude of the acceleration vs. the reciprocal of the mass of the system. • Analyze your data and form conclusions. How well did your results agree with your hypothesis? e TECH Explore how the net force on an object and its mass affect its acceleration. Follow the eTech links at www.pearsoned.ca/ school/physicssource. Relating Acceleration and Mass In 3-6 Design a Lab, you discovered that the relationship between acceleration and mass is non-linear. But if you plot acceleration as a function of the reciprocal of mass, you get a straight line. This shows that there is a linear relationship between acceleration and the reciprocal of mass. This relationship can be written as a proportionality statement: 1 a m In speed skating, evidence of this relationship is the different accelerations that two athletes of different mass have. Suppose athlete A has a mass of 60 kg and athlete B a mass of 90 kg. If the net force acting on A and B is the same, you would expect A to have a greater acceleration than B (Figure 3.34). This observation makes sense in terms of inertia, because the inertia of B resists the change in motion more so than the inertia of A does. In fact, you observed this relationship in 3-1 QuickLab when you compared the acceleration of an empty cart and a cart loaded with a 1-kg standard mass. m Fnet constant m a a A B Figure 3.34 For the same net force, a more massive person has a smaller acceleration than a less massive one does. Chapter 3 Forces can change velocity. 147 03-Phys20-Chap03.qxd 7/24
/08 10:37 AM Page 148 Newton’s Second Law and Inertial Mass 1 can be combined The proportionality statements a Fnet and a m F F net where k is the proportionality net or a k into one statement, a m m constant. Since 1 N is defined as the net force required to accelerate a 1-kg object at 1 m/s2, k is equal to 1. So F net. the equation can be written as a m This mathematical relationship is Newton’s second law. When an external non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. The equation for Newton’s second law is usually written with F net on the left side: F net ma The Concept of Inertial Mass All objects have mass, so all objects have inertia. From experience, it is more difficult to accelerate a curling stone than to accelerate a hockey puck (Figure 3.35). This means that the inertia of an object is related to its mass. The greater the mass of the object, the greater its inertia. The mass of an object in Newton’s second law is determined by finding the ratio of a known net force acting on an object to the acceleration of the object. In other words, the mass is a measure of the inertia of an object. Because of this relationship, the mass in Newton’s second law is called inertial mass, which indicates how the mass is measured. m Fnet a a constant Fnet m m Fnet Figure 3.35 If the acceleration of the curling stone and the hockey puck is the same, F net on the curling stone would be 95 times greater than F inertial mass of the curling stone is that much greater than the hockey puck. net on the hockey puck because the Concept Check What happens to the acceleration of an object if (a) the mass and net force both decrease by a factor of 4? (b) the mass and net force both increase by a factor of 4? (c) the mass increases by a factor of 4, but the net force decreases by the same factor? (d) the mass decreases by a factor of 4, and the net force is zero? e MATH Use technology to explore the relationship among Fnet, m, and a in Newton’s second law. Follow the
eMath links at www.pearsoned.ca/school/ physicssource to download sample data. inertial mass: mass measurement based on the ratio of a known net force on an object to the acceleration of the object 148 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 149 Applying Newton’s Second Law to Horizontal Motion Example 3.5 demonstrates how to use Newton’s second law to predict the average acceleration of a lacrosse ball. In this situation, air resistance is assumed to be negligible to simplify the problem. Example 3.5 A lacrosse player exerts an average net horizontal force of 2.8 N [forward] on a 0.14-kg lacrosse ball while running with it in the net of his stick (Figure 3.36). Calculate the average horizontal acceleration of the ball while in contact with the lacrosse net. Given F 2.8 N [forward] net m 0.14 kg up down forward backward Figure 3.36 Required average horizontal acceleration of ball (a) Analysis and Solution The ball is not accelerating up or down. So in the vertical direction, F In the horizontal direction, the acceleration of the ball is in the direction of the net force. So use the scalar form of Newton’s second law. 0 N. net Fnet ma F net a m N.8 2 0 4 kg.1 m 2.8 kg s2 0.14 kg 20 m/s2 a 20 m/s2 [forward] Practice Problems 1. The net force acting on a 6.0-kg grocery cart is 12 N [left]. Calculate the acceleration of the cart. 2. A net force of 34 N [forward] acts on a curling stone causing it to accelerate at 1.8 m/s2 [forward] on a frictionless icy surface. Calculate the mass of the curling stone. Answers 1. 2.0 m/s2 [left] 2. 19 kg Paraphrase The average horizontal acceleration of the lacrosse ball is 20 m/s2 [forward]. In Example 3.6, a free-body diagram is used to first help determine the net force acting on a canoe. Then Newton’s second law is applied to predict the average acceleration of the canoe. Chapter 3 Forces can change velocity. 149 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page
150 Example 3.6 Two athletes on a team, A and B, are practising to compete in a canoe race (Figure 3.37). Athlete A has a mass of 70 kg, B a mass of 75 kg, and the canoe a mass of 20 kg. Athlete A can exert an average force of 400 N [forward] and B an average force of 420 N [forward] on the canoe using the paddles. During paddling, the magnitude of the water resistance on the canoe is 380 N. Calculate the initial acceleration of the canoe. up down forward backward Practice Problem 1. In the men’s four-man bobsled event in the Winter Olympics, the maximum mass of a bobsled with two riders, a pilot, and a brakeman is 630 kg (Figure 3.39). Figure 3.37 Given mA F A F f 70 kg 400 N [forward] 380 N [backward] mB F B 75 kg 420 N [forward] mc 20 kg Required initial acceleration of canoe (a) Analysis and Solution The canoe and athletes are a system because they move together as a unit. Find the total mass of the system. mT mc mB mA 70 kg 75 kg 20 kg 165 kg Draw a free-body diagram for the system (Figure 3.38). Figure 3.39 During a practice run, riders A and B exert average forces of 1220 N and 1200 N [forward] respectively to accelerate a bobsled of mass 255 kg, a pilot of mass 98 kg, and a brakeman of mass 97 kg. Then they jump in for the challenging ride down a 1300-m course. During the pushing, the magnitude of the force of friction acting on the bobsled is 430 N. Calculate the average acceleration of the bobsled, pilot, and brakeman. up down forward FN FA backward Ff FB Fg Figure 3.38 FA FB Ff Fneth Answer 1. 4.4 m/s2 [forward] 150 Unit II Dynamics The system is not accelerating up or down. So in the vertical direction, Fnetv Write equations to find the net force on the system in both the horizontal and vertical directions. 0 N. 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 151 neth Fneth horizontal direction F FF f Ff F B FB F A FA 400 N 420 N (380 N) 400 N 420 N 380 N 440 N FF
g vertical direction FF F N 0 netv Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton’s second law to the horizontal direction. Fneth a mTa Fnet h mT 4 N 0 4 5 g k 6 1 440 kg m 2 s 165 kg 2.7 m/s2 a 2.7 m/s2 [forward] Paraphrase The canoe will have an initial acceleration of 2.7 m/s2 [forward]. Applying Newton’s Second Law to Vertical Motion Example 3.7 demonstrates how to apply Newton’s second law to determine the vertical acceleration of a person riding an elevator. To determine the net force on the elevator, use a free-body diagram. Example 3.7 A person and an elevator have a combined mass of 6.00 102 kg (Figure 3.40). The elevator cable exerts a force of 6.50 103 N [up] on the elevator. Find the acceleration of the person. Given 6.00 102 kg mT 6.50 103 N [up] F T g 9.81 m/s2 [down] Required acceleration of person (a ) up down Figure 3.40 Analysis and Solution Draw a free-body diagram for the person-elevator system [Figure 3.41 (a)]. The system is not accelerating left or right. 0 N. So in the horizontal direction, F net Since the person is standing on the elevator floor, both the person and the elevator will have the same vertical acceleration. up down FT Fg Figure 3.41 (a) Chapter 3 Forces can change velocity. 151 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 152 For the vertical direction, write an equation to find the net force on the system [Figure 3.41 (b)]. F T F g F net Practice Problems 1. The person in Example 3.7 rides the same elevator when the elevator cable exerts a force of 5.20 103 N [up] on the elevator. Find the acceleration of the person. 2. An electric chain hoist in a garage exerts a force of 2.85 103 N [up] on an engine to remove it from a car. The acceleration of the engine is 1.50 m/s2 [up]. What is the mass of the engine? Answers 1. 1.14 m/s2 [down] 2. 252 kg Apply Newton’s
second law. ma F F T g ma FT Fg mg ma FT 9.81 m/s2.50 103 kg m 2 s 6.00 102 kg 1.02 m/s2 a 1.02 m/s2 [up] FT Fg Fnet 9.81 m/s2 Figure 3.41 (b) e WEB Air resistance is the frictional force that acts on all objects falling under the influence of gravity. Research how this force affects the maximum speed that an object reaches during its fall. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. F f F g up down Figure 3.42 152 Unit II Dynamics Paraphrase The acceleration of the person is 1.02 m/s2 [up]. In Example 3.8, the force of gravity causes a skydiver to accelerate downward. Since the only motion under consideration is that of the skydiver and the direction of motion is down, it is convenient to choose down to be positive. Example 3.8 A skydiver is jumping out of an airplane. During the first few seconds of one jump, the parachute is unopened, and the magnitude of the air resistance acting on the skydiver is 250 N. The acceleration of the skydiver during this time is 5.96 m/s2 [down]. Calculate the mass of the skydiver. Given F f g 9.81 m/s2 [down] 250 N [up] Required mass of skydiver (m) a 5.96 m/s2 [down] Analysis and Solution Draw a free-body diagram for the skydiver (Figure 3.42). The skydiver is not accelerating left or right. So in the horizontal direction, F net For the vertical direction, write an equation to find the net force on the skydiver (Figure 3.43). F net 0 N. F g F f Fnet Fg Ff Figure 3.43 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 153 Apply Newton’s second law. maa F FF f g ma Fg Ff ma mg (250 N) mg 250 N 250 N mg ma m(g a) 250 N g a m 250 N 9.81 m/s2 5.96 m/s2 250 kgm s2 3.85 m s2 64
.9 kg Paraphrase The mass of the skydiver is 64.9 kg. Practice Problems 1. A 55-kg female bungee jumper fastens one end of the cord (made of elastic material) to her ankle and the other end to a bridge. Then she jumps off the bridge. As the cord is stretching, it exerts an elastic force directed up on her. Calculate her acceleration at the instant the cord exerts an elastic force of 825 N [up] on her. 2. During a bungee jump, the velocity of the 55-kg woman at the lowest point is zero and the cord stretches to its maximum. (a) Compare the direction of her acceleration at the lowest point of the jump to the part of the jump where she is accelerating due to gravity. (b) At this point, what is the direction of her acceleration? Applying Newton’s Second Law to Two-Body Systems When two objects are connected by a light rope as in Example 3.9, applying a force on one of the objects will cause both objects to accelerate at the same rate and in the same direction. In other words, the applied force can be thought to act on a single object whose mass is equivalent to the total mass. Answers 1. 5.2 m/s2 [up] 2. (b) up Example 3.9 Two blocks of identical material are connected by a light rope on a level surface (Figure 3.44). An applied force of 55 N [right] causes the blocks to accelerate. While in motion, the magnitude of the force of friction on the block system is 44.1 N. Calculate the acceleration of the blocks. Given mA mB F app F f 20 kg 10 kg 55 N [right] 44.1 N [left] 20 kg 10 kg Figure 3.44 Required acceleration (a) Analysis and Solution The two blocks move together as a unit with the same acceleration. So consider the blocks to be a single object. Find the total mass of both blocks. mA mB 20 kg 10 kg 30 kg mT Practice Problems 1. Two buckets of nails are hung one above the other and are pulled up to a roof by a rope. Each bucket has a mass of 5.0 kg. The tension in the rope connecting the buckets is 60 N. Calculate the acceleration of the buckets. 2. Refer to Example 3.9. The force of friction on the 10-kg block has a magnitude of 14.7 N
. (a) Calculate the tension in the rope connecting the two blocks. (b) Calculate the tension in the rope between the hand and the 10-kg block. Answers 1. 2.2 m/s2 [up] 2. (a) 37 N (b) 55 N Chapter 3 Forces can change velocity. 153 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 154 Draw a free-body diagram for this single object (Figure 3.45). The single object is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on the single object in both the horizontal and vertical directions. horizontal direction F FF F f app neth Apply Newton’s second law. mTaa F F f mTa Fapp Ff Ff Fap p a m T app 55 N (44.1 N) 30 kg 0.36 m/s2 a 0.36 m/s2 [right] vertical direction F F N 0 F g netv Fnetv FN up right left down Calculations in the vertical direction are not required in this problem. Ff mT Fapp Fg Figure 3.45 Fapp Ff Fneth e SIM Apply Newton’s second law to determine the motion of two blocks connected by a string over a pulley. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Paraphrase The acceleration of the blocks is 0.36 m/s2 [right]. Applying Newton’s Second Law to a Single Pulley System In Example 3.10, two objects are attached by a rope over a pulley. The objects, the rope, and the pulley form a system. You can assume that the rope has a negligible mass and thickness, and the rope does not stretch or break. To simplify calculations in this physics course, you need to also assume that a pulley has negligible mass and has no frictional forces acting on its axle(s). In Example 3.10, the external forces on the system are the gravitational forces acting on the hanging objects. The internal forces on the system are the forces along the string that pull on each object. The magnitude of both the internal and external forces acting on the system are not affected by the pulley. The pulley simply redirects the forces along the string that pulls on each object. Example 3.10 Two
objects, A and B, are connected by a light rope over a light, frictionless pulley (Figure 3.46). A has a mass of 25 kg and B a mass of 35 kg. Determine the motion of each object once the objects are released. Given mA 25 kg g 9.81 m/s2 [down] mB 35 kg Required acceleration of each object (a A and a B) 25 kg 35 kg Figure 3.46 154 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 155 Analysis and Solution The difference in mass between objects A and B will provide the net force that will accelerate both objects. Since mB expect mB to accelerate down while mA accelerates up. The rope has a negligible mass. So the tension in the rope is the same on both sides of the pulley. The rope does not stretch. So the magnitude of aa Find the total mass of both objects. A is equal to the magnitude of a mA, you would B. mT mB mA 25 kg 35 kg 60 kg Choose an equivalent system in terms of mT to analyze the motion [Figure 3.47 (a)]. left right FA mT FB Figure 3.47 (a) F net Fnet F A is equal to the gravitational force acting on mA, and F B is equal to the gravitational force acting on mB. Apply Newton’s second law to find the net force acting on mT [Figure 3.47 (b)]. F F B A FA FB mAg mBg mA)g (mB (35 kg 25 kg)(9.81 m/s2) m 98.1 kg 2 s 98.1 N Figure 3.47 (b) Fnet FA FB Use the scalar form of Newton’s second law to calculate the magnitude of the acceleration. Fnet mTa a F et n m T.1 N 8 9 g k 0 6 98.1 kg m 2 s 60 kg 1.6 m/s2 1.6 m/s2 [up] and a a A 1.6 m/s2 [down] B Paraphrase Object A will have an acceleration of 1.6 m/s2 [up] and object B will have an acceleration of 1.6 m/s2 [down]. Practice Problems 1. Determine the acceleration of the system shown in Example 3.10 for each situation
below. State the direction of motion for each object. Express your answer in terms of g. (a) mA (b) mA (c) mA mB 1 3 2mB mB 2. Use the result of Example 3.10 and a free-body diagram to calculate the tension in the rope. 3. Draw a free-body diagram for each object in Example 3.10. Answers 1 g, mA moves up, mB moves down 1. (a) a 2 1 g, mA moves down, mB moves up (b) a 3 (c) a 0, neither mass moves 2. 2.9 102 N 3. FT a 25 kg FT 35 kg a Fg Fg Chapter 3 Forces can change velocity. 155 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 156 Applying Newton’s Second Law to a Two-Pulley System In Example 3.11, the system is made up of three objects (A, B, and C). As in Example 3.10, the difference in weight between objects B and C will provide the net force that will accelerate the system. Example 3.11 A 20-kg truck tire (object A) is lying on a horizontal, frictionless surface. The tire is attached to two light ropes that pass over light, frictionless pulleys to hanging pails B and C (Figure 3.48). Pail B has a mass of 8.0 kg and C a mass of 6.0 kg. Calculate the magnitude of the acceleration of the system. tire (A) pail B pail C Figure 3.48 Given mA 20 kg g 9.81 m/s2 [down] mB 8.0 kg mC 6.0 kg Required magnitude of the acceleration of the system (a) Analysis and Solution Since mB accelerates up. Since object A will accelerate left, choose left to be positive. mC, you would expect mB to accelerate down while mC The rope has a negligible mass and the rope does not stretch. So the magnitude of a equal to a C. Find the total mass of the system. A is equal to the magnitude of a B, which is also mT mC mB mA 20 kg 8.0 kg 6.0 kg 34 kg Choose an equivalent system in terms of mT to analyze the motion (Figure 3.49). left right Figure 3.49
FB mT FC 156 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 157 F C is equal to B is equal to the gravitational force acting on mB, and F the gravitational force acting on mC. left right FB FC Fnet Figure 3.50 C Fnet Apply Newton’s second law to find the net force acting on mT (Figure 3.50). F F F B net FB FC mBg mCg mC)g (mB (8.0 kg 6.0 kg)(9.81 m/s2) m 19.6 kg 2 s 19.6 N Use the scalar form of Newton’s second law to calculate the magnitude of the acceleration. Fnet a mTa Fnet mT 19.6N 34 kg 19.6 kg m 2 s 34 kg 0.58 m/s2 Paraphrase The system will have an acceleration of magnitude 0.58 m/s2. Practice Problems 1. Calculate the acceleration of the tire in Example 3.11 if the mass of pail B is increased to 12 kg, without changing the mass of pail C. 2. If the tire in Example 3.11 is replaced by a car tire of mass 15 kg, calculate the acceleration of each object. Answers 1. 1.5 m/s2 [left] 2. (A) 0.68 m/s2 [left], (B) 0.68 m/s2 [down], (C) 0.68 m/s2 [up] Chapter 3 Forces can change velocity. 157 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 158 3.3 Check and Reflect 3.3 Check and Reflect Knowledge 1. In your own words, state Newton’s second law. app acting on an object 2. An applied force F of constant mass causes the object to accelerate. Sketch graphs to show the relationship between a and Fapp when friction is (a) present, and (b) absent. Refer to Student References 5.1: Graphing Techniques on pp. 872–873. 3. Sketch a graph to show the relationship between the magnitude of acceleration and mass for constant net force. 4. Explain why vehicles with more powerful engines are able to accelerate faster. Applications 5. A dolphin experiences a force of 320 N [up] when it jumps out of

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