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no resulting If, on the other hand, you drop the book, it will fall to the ground pulled by a force called motion of the book. gravity. 1 www.ck12.org If you slide a book across the ﬂoor, it will experience a force of friction which acts in the opposite direction of the motion. This force will slow down the motion of the book and eventually bring it to rest. If the ﬂoor is smoother, If a perfectly smooth ﬂoor the force of friction will be less and the book will slide further before coming to rest. could be created, there would be no friction and the book would slide forever at constant speed. Newton’s First Law of Motion describes an object moving with constant speed in a straight line. In the absence of any force, the object will continue to move at the same constant speed and in the same straight line. If the object is at rest, in the absence of any force, it will remain at rest. Newton’s First Law states that an object with no force acting on it moves with constant velocity. (The constant velocity could, of course, be 0 m/s.) A more careful expression of Newton’s First Law is "an object with no net force acting on it remains at rest or moves with constant velocity in a straight line." The statement above is equivalent to a statement that "if there is no net force on an object, there will be no acceleration." In the absence of acceleration, an object will remain at rest or will move with constant velocity in a straight line. The acceleration of an object is the result of an unbalanced force. If an object suffers two forces, the motion of the object is determined by the net force. The magnitude of the acceleration is directly proportional to the magnitude of the unbalanced force. The direction of the acceleration is the same direction as the direction of the unbalanced force. The magnitude of the acceleration is inversely proportional to the mass of the object. i.e. The more massive the object, the smaller will be the acceleration produced by the same force. These relationships are stated in Newton’s Second Law of Motion, "the acceleration of an object is directly proportional to the net force on the object and inversely proportional to the mass of the object." Newton’s Second Law can be summarized in an equation. a = F m or more commonly; F = ma According to Newton’s second law, a new force on an object causes
it to accelerate and the larger the mass, the smaller the acceleration. Sometimes, the word inertia is used to express the resistance of an object to acceleration. Therefore, we say that a more massive object has greater inertia. The units for force are deﬁned by the equation for Newton’s second law. Suppose we wish to express the force that will give a 1.00 kg object an acceleration of 1.00 m/s2. F = ma = (1:00 kg)(1:00 m/s2) = 1:00 kg m/s2 This unit is deﬁned as 1.00 Newton or 1.00 N. kgm s2 = Newton Example Problem: What new force is required to accelerate a 2000. kg car at 2.000 m/s2? Solution: F = ma = (2000: kg)(2:000 m/s2) = 4000: N Example Problem: A net force of 150 N is exerted on a rock. The rock has an acceleration of 20. m/s2 due to this force. What is the mass of the rock? Solution: m = F = 7:5 kg a = (150 N) (20: m/s2 ) Example Problem: A net force of 100. N is exerted on a ball. If the ball has a mass of 0.72 kg, what acceleration will it undergo? Solution: a = F m = (100: N) (0:72 kg) = 140 m/s2 Summary • A force is a push or pull. • Newton’s First Law states that an object with no net force acting on it remains at rest or moves with constant velocity in a straight line. • Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the net 2 www.ck12.org Chapter 1. Newton’s First and Second Laws of Motion force on the object and inversely proportional to the mass of the object. Expressed as an equation, F = ma. Practice Professor Mac explains Newton’s Second Law of Motion. http://www.youtube.com/watch?v=-KxbIIw8hlc MEDIA Click image to the left for more content. Review 1. A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car’s brakes. What was the braking force? 2. Calcul
ate the average force that must be exerted on a 0.145 kg baseball in order to give it an acceleration of 130 m/s2. 3. After a rocket ship going from the Earth to the Moon leaves the gravitational pull of the Earth, it can shut off its engine and the ship will continue on to the Moon due to the gravitational pull of the Moon. (a) True (b) False 4. If a space ship traveling at 1000 miles per hour enters an area free of gravitational forces, its engine must run at some minimum level in order to maintain the ships velocity. (a) True (b) False 5. Suppose a space ship traveling at 1000 miles per hour enters an area free of gravitational forces and free of air resistance. If the pilot wishes to slow the ship down, he can accomplish that by shutting off the engine for a while. (a) True (b) False • force: A push or pull. References 1. Courtesy of Neil A. Armstrong, NASA. http://spaceﬂight.nasa.gov/gallery/images/apollo/apollo11/html/as11 -40-5873.html. Public Domain 3 CHAPTER 2 Newton’s First Law www.ck12.org • Use skateboarding to explain Newton’s ﬁrst law of motion. There’s no doubt from Corey’s face that he loves skateboarding! Corey and his friends visit Newton’s Skate Park every chance they get. They may not know it, but while they’re having fun on their skateboards, they’re actually applying science concepts such as forces and motion. 4 www.ck12.org Starting and Stopping Chapter 2. Newton’s First Law Did you ever ride a skateboard? Even if you didn’t, you probably know that to start a skateboard rolling over a level surface, you need to push off with one foot against the ground. That’s what Corey’s friend Nina is doing in this picture 2.1. FIGURE 2.1 Do you know how to stop a skateboard once it starts rolling? Look how Nina’s friend Laura does it in the Figure 2.2. She steps down on the back of the skateboard so it scrapes on the pavement. This creates friction, which stops the skateboard. Even if Laura didn’t try to stop the skateboard, it would stop sooner or later.
That’s because there’s also friction between the wheels and the pavement. Friction is a force that counters all kinds of motion. It occurs whenever two surfaces come into contact. Video Break Laura learned how to use forces to start and stop her skateboard by watching the videos below. Watch the video to see how the forces are applied. You can pick up some skateboarding tips at the same time! Starting: http://www.youtube.com/watch?v=OpZIVjbMAOU MEDIA Click image to the left for more content. Stopping: http://www.youtube.com/watch?v=6fuOwhx91zM 5 www.ck12.org FIGURE 2.2 MEDIA Click image to the left for more content. Laws of the Park: Newton’s First Law If you understand how a skateboard starts and stops, then you already know something about Newton’s ﬁrst law of motion. This law was developed by English scientist Isaac Newton around 1700. Newton was one of the greatest scientists of all time. He developed three laws of motion and the law of gravity, among many other contributions. Newton’s ﬁrst law of motion states that an object at rest will remain at rest and an object in motion will stay in motion unless it is acted on by an unbalanced force. Without an unbalanced force, a moving object will not only 6 www.ck12.org Chapter 2. Newton’s First Law keep moving, but its speed and direction will also remain the same. Newton’s ﬁrst law of motion is often called the law of inertia because inertia is the tendency of an object to resist a change in its motion. If an object is already at rest, inertia will keep it at rest. If an object is already in motion, inertia will keep it moving. Do You Get It? Q: How does Nina use Newton’s ﬁrst law to start her skateboard rolling? A: The skateboard won’t move unless Nina pushes off from the pavement with one foot. The force she applies when she pushes off is stronger than the force of friction that opposes the skateboard’s motion. As a result, the force on the skateboard is unbalanced, and the skateboard moves forward. Q: How does Nina use Newton’s ﬁrst law to stop her skateboard? A: Once
the skateboard starts moving, it would keep moving at the same speed and in the same direction if not for another unbalanced force. That force is friction between the skateboard and the pavement. The force of friction is unbalanced because Nina is no longer pushing with her foot to keep the skateboard moving. That’s why the skateboard stops. Changing Direction Corey’s friend Jerod likes to skate on the ﬂat banks at Newton’s Skate Park. That’s Jerod in the picture above. As he reaches the top of a bank, he turns his skateboard to go back down. To change direction, he presses down with his heels on one edge of the skateboard. This causes the skateboard to turn in the opposite direction. Video Break Can you turn a skateboard like Jerod? To see how to apply forces to change the direction of a skateboard, watch this video: http://www.youtube.com/watch?v=iOnlcEk50CM MEDIA Click image to the left for more content. Do You Get It? Q: How does Jerod use Newton’s ﬁrst law of motion to change the direction of his skateboard? A: Pressing down on just one side of a skateboard creates an unbalanced force. The unbalanced force causes the skateboard to turn toward the other side. In the picture, Jerod is pressing down with his heels, so the skateboard turns toward his toes. Summary • Newton’s ﬁrst law of motion states that an object at rest will remain at rest and an object in motion will remain in motion unless it is acted on by an unbalanced force. 7 www.ck12.org FIGURE 2.3 • Using unbalanced forces to control the motion of a skateboard demonstrates Newton’s ﬁrst law of motion. Vocabulary • Newton’s ﬁrst law of motion: Law stating that an object’s motion will not change unless an unbalanced force acts on the object. Practice Do you think you understand Newton’s ﬁrst law? Go to the URL below to ﬁnd out. Review Newton’s law and watch what happens to the skateboarder in the animation. Then answer the questions at the bottom of the Web page. http://teachertech.rice.edu/Participants/louviere/Newton
/law1.html 8 www.ck12.org Review Chapter 2. Newton’s First Law 1. State Newton’s ﬁrst law of motion. 2. You don’t need to push off with a foot against the ground to start a skateboard rolling down a bank. Does this violate Newton’s ﬁrst law of motion? Why or why not? FIGURE 2.4 3. Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the skateboard stopped suddenly and Nina ended up on the ground (see Figure above). Explain what happened. 4. Now that you know about Newton’s ﬁrst law of motion, how might you use it to ride a skateboard more safely? References 1. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 2. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 3. Image copyright Nikola Bilic, 2012.. Used under license from Shutterstock.com 4. Image copyright DenisNata, 2012.. Used under license from Shutterstock.com 9 CHAPTER 3 Newton’s Second Law www.ck12.org • State Newton’s second law of motion. • Compare and contrast the effects of force and mass on acceleration. These boys are racing around the track at Newton’s Skate Park. The boy who can increase his speed the most will win the race. Tony, who is closest to the camera in this picture, is bigger and stronger than the other two boys, so he can apply greater force to his skates. Q: Does this mean that Tony will win the race? A: Not necessarily, because force isn’t the only factor that affects acceleration. Force, Mass, and Acceleration Whenever an object speeds up, slows down, or changes direction, it accelerates. Acceleration occurs whenever an unbalanced force acts on an object. Two factors affect the acceleration of an object: the net force acting on the object and the object’s mass. Newton’s second law of motion describes how force and mass affect acceleration. The law 10 www.ck12.org Chapter 3. Newton’s Second Law states that the acceleration of an object equals the net force acting on the object divided by the object’s mass. This can be represented by the equation: Acceleration = Net force Mass or a
= F m Q: While Tony races along on his rollerblades, what net force is acting on the skates? A: Tony exerts a backward force against the ground, as you can see in the Figure 3.1, ﬁrst with one skate and then with the other. This force pushes him forward. Although friction partly counters the forward motion of the skates, it is weaker than the force Tony exerts. Therefore, there is a net forward force on the skates. FIGURE 3.1 Direct and Inverse Relationships Newton’s second law shows that there is a direct relationship between force and acceleration. The greater the force that is applied to an object of a given mass, the more the object will accelerate. For example, doubling the force on the object doubles its acceleration. The relationship between mass and acceleration is different. It is an inverse relationship. In an inverse relationship, when one variable increases, the other variable decreases. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force. Q: Tony has greater mass than the other two boys he is racing above. How will this affect his acceleration around the track? A: Tony’s greater mass will result in less acceleration for the same amount of force. Summary • Newton’s second law of motion states that the acceleration of an object equals the net force acting on the object divided by the object’s mass. • According to the second law, there is a direct relationship between force and acceleration and an inverse relationship between mass and acceleration. Vocabulary • Newton’s second law of motion: Law stating that the acceleration of an object equals the net force acting on the object divided by the object’s mass. 11 www.ck12.org Practice At the following URL, use the simulator to experiment with force, mass, and acceleration. First keep force constant at 1 N, and vary mass from 1–5 kg. Next keep mass constant at 1 kg, and vary force from 1–5 N. In each simulation, record the values you tested and the resulting acceleration. Finally, make two line graphs to plot your results. On one graph, show acceleration when force is constant and mass changes. On the other graph, show acceleration when mass is constant and force changes. Describe in words what the two graphs show. http://jangg
eng.com/newtons-second-law-of-motion/ Review 1. State Newton’s second law of motion. 2. How can Newton’s second law of motion be represented with an equation? 3. If the net force acting on an object doubles, how will the object’s acceleration be affected? 4. Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with the same force. Explain which boy will have greater acceleration. References 1. Uploaded by User:Shizhao/Wikimedia Commons.. CC BY 2.5 12 www.ck12.org Chapter 4. Newton’s Third Law of Motion CHAPTER 4 Newton’s Third Law of Motion • Deﬁne force. • State the fundamental units for the Newton. • State Newton’s First Law of Motion. • Given two of the three values in F = ma, calculate the third It was imagined as a geo-stationary transfer The image at above is a NASA artist’s concept of a space elevator. station for passengers and cargo between earth and space. This idea was not pursued but it began where all great ideas begin... in someone’s mind. Newton’s Third Law of Motion Where do forces come from? Observations suggest that a force applied to an object is always applied by another object. A hammer strikes a nail, a car pulls a trailer, and a person pushes a grocery cart. Newton realized that forces are not so one sided. When the hammer exerts a force on the nail, the nail also exerts a force on the hammer –after all, the hammer comes to rest after the interaction. This led to Newton’s Third Law of Motion, which states that whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction. 13 www.ck12.org This law is sometimes paraphrased as “for every action, there is an equal and opposite reaction.” A very important point to remember is that the two forces are on different objects –never on the same object. It is frequently the case that one of the objects moves as a result of the force applied but the motion of the other object in the opposite direction is not apparent. Consider the situation where an ice skater is
standing at the edge of the skating rink holding on to the side rail. If the skater exerts a force on the rail, the rail is held in place with tremendous friction and therefore, will not move in any noticeable way. The skater, on the other hand, had little friction with the ice, and therefore will be accelerated in the direction opposite of his/her original push. This is the process people use to jump up into the air. The person’s feet exert force on the ground and the ground exerts an equal and opposite force on the person’s feet. The force on the feet is sufﬁcient to raise the person off the ground. The force on the ground has little effect because the earth is so large. One of the accelerations is visible but the other is not visible. A case where the reaction motion due to the reaction force is visible is the case of a person throwing a heavy object out of a boat. The object is accelerated in one direction and the boat is accelerated in the opposite direction. In this case, both the motion of the object is visible and the motion of the boat in the opposite direction is also visible. Rockets also work in this manner. It is a misconception that the rocket moves forward because the escaping gas pushes on the ground or the surrounding air to make the rocket go forward. Rockets work in outer space where there is no ground or surrounding air. The rocket exerts a force on the gases causing them to be expelled and the gases exert a force on the rocket causing it to be accelerated forward. Summary • A force applied to an object is always applied by another object. • Newton’s Third Law of Motion says, "whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction." Practice The following video contains a discussion and an example of Newton’s Third Law of Motion. http://www.youtube.com/watch?v=fKJDpPi-UN0 MEDIA Click image to the left for more content. Review 1. What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balances the original force and therefore, the ball will not accelerate in any direction. 2. When a bat strikes a ball, the force exerted can send the ball deep into the outﬁeld. Where is the equal and
opposite force in this case? 3. Suppose you wish to jump horizontally and in order for you to jump a distance of 4 feet horizontally, you must exert a force of 200 N. When you are standing on the ground, you have no trouble jumping 4 feet horizontally. If you are standing in a canoe, however, and you need to jump 4 feet to reach the pier, you will surely fall into the lake. Why is it that you cannot jump 4 feet out of a canoe when you can easily do this when on land? 14 www.ck12.org Chapter 4. Newton’s Third Law of Motion • Newton’s Third Law of Motion: Whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction. References 1. Courtesy of Pat Rawling, NASA. http://commons.wikimedia.org/wiki/File:Nasa_space_elev.jpg. Public Do- main 15 CHAPTER 5 The various types of common forces are discussed and analyzed. The various types of common forces are discussed and analyzed. www.ck12.org Types of Forces Key Equations Common Forces Guidance Normal Force 8 >>>>< >>>>: Fg = mg FN FT Fsp = kDx Force of spring stretched a distance Dx from equilibrium Gravity Normal force: acts perpendicular to surfaces Force of tension in strings and wires Often, objects experience a force that pushes them into another object, but once the objects are in contact they do not any move closer together. For instance, when you stand on the surface of the earth you are obviously not accelerating toward its center. According to Newton’s Laws, there must be a force opposing the earth’s gravity acting on you, so that the net force on you is zero. The same also applies for your gravity acting on the earth. We call such a force the Normal Force. The normal force acts between any two surfaces in contact, balancing what ever force is pushing the objects together. It is actually electromagnetic in nature (like other contact forces), and arises due to the repulsion of atoms in the two objects. Here is an illustration of the Normal force on a block sitting on earth: Tension Another force that often opposes gravity is known as tension. This force is provided by wires and strings when they hold objects above the earth. Like the Normal Force, it is electromagnetic in nature and arises due to the intermolecular
bonds in the wire or string: 16 www.ck12.org Chapter 5. Types of Forces If the object is in equilibrium, tension must be equal in magnitude and opposite in direction to gravity. This force transfers the gravity acting on the object to whatever the wire or string is attached to; in the end it is usually a Normal Force — between the earth and whatever the wire is attached to — that ends up balancing out the force of gravity on the object. Friction Friction is a force that opposes motion. Any two objects in contact have what is called a mutual coefﬁcient of friction. To ﬁnd the force of friction between them, we multiply the normal force by this coefﬁcient. Like the forces above, it arises due to electromagnetic interactions of atoms in two objects. There are actually two coefﬁcients of friction: static and kinetic. Static friction will oppose initial motion of two objects relative to each other. Once the objects are moving, however, kinetic friction will oppose their continuing motion. Kinetic friction is lower than static friction, so it is easier to keep an object in motion than to set it in motion. fs µsj ~FNj fk = µkj ~FNj [5] Static friction opposes potential motion of surfaces in contact [6] Kinetic frictions opposes motion of surfaces in contact There are some things about friction that are not very intuitive: • The magnitude of the friction force does not depend on the surface areas in contact. • The magnitude of kinetic friction does not depend on the relative velocity or acceleration of the two objects. • Friction always points in the direction opposing motion. If the net force (not counting friction) on an object is lower than the maximum possible value of static friction, friction will be equal to the net force in magnitude and opposite in direction. Spring Force Any spring has some equilibrium length, and if stretched in either direction it will push or pull with a force equal to: ~Fsp = k ~Dx [7] Force of spring ~Dx from equilibrium Example 1 Question: A woman of mass 70.0 kg weighs herself in an elevator. 17 www.ck12.org a) If she wants to weigh less, should she weigh herself when accelerating upward or downward? b) When the elevator is not accelerating, what does the scale read (i.e., what is the normal force that the scale exerts on the woman)? c) When the elevator is accelerating upward at 2
.00 m/s2, what does the scale read? Answer a) If she wants to weigh less, she has to decrease her force (her weight is the force) on the scale. We will use the equation to determine in which situation she exerts less force on the scale. F = ma If the elevator is accelerating upward then the acceleration would be greater. She would be pushed toward the ﬂoor of the elevator making her weight increase. Therefore, she should weigh herself when the elevator is going down. b) When the elevator is not accelerating, the scale would read 70:0kg. c) If the elevator was accelerating upward at a speed of 2:00m=s2, then the scale would read F = ma = 70kg (9:8m=s2 + 2m=s2) = 826N which is 82:6kg. Example 2 Question: A spring with a spring constant of k = 400N=m has an uncompressed length of :23m and a fully compressed length of :15m. What is the force required to fully compress the spring? Solution: We will use the equation F = kx to solve this. We simply have to plug in the known value for the spring and the distance to solve for the force. F = kx = (400N=m)(:23m :15m) = 32N 18 www.ck12.org Watch this Explanation Chapter 5. Types of Forces MEDIA Click image to the left for more content. Time for Practice No Problems for this section. See Newton Law Problem Solving Concept. 19 CHAPTER 6 Universal Law of Gravity www.ck12.org • Describe and give the formula for Newton’s universal law of gravity. • Using Newton’s law of gravity Cavendish’s apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. Force of Gravity In the mid-1600’s, Newton wrote that
the sight of a falling apple made him think of the problem of the motion of the planets. He recognized that the apple fell straight down because the earth attracted it and thought this same force of attraction might apply to the moon and that motion of the planets might be controlled by the gravity of the sun. He eventually proposed the universal law of gravitational attraction as 20 F = G m1m2 d2 www.ck12.org Chapter 6. Universal Law of Gravity where m1 and m2 are the masses being attracted, d is the distance between the centers of the masses, G is the universal gravitational constant, and F is the force of attraction. The formula for gravitational attraction applies equally to two rocks resting near each other on the earth and to the planets and the sun. The value for the universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N·m2/kg2. The moon is being pulled toward the earth and the earth toward the moon with the same force but in the opposite direction. The force of attraction between the two bodies produces a greater acceleration of the moon than the earth because the moon has smaller mass. Even though the moon is constantly falling toward the earth, it never gets any closer. This is because the velocity of the moon is perpendicular to the radius of the earth (as shown in the image above) and therefore the moon is moving away from the earth. The distance the moon moves away from the orbit line is exactly the same distance that the moon falls in the time period. This is true of all satellites and is the reason objects remain in orbit. Example Problem: Since we know the force of gravity on a 1.00 kg ball resting on the surface of the earth is 9.80 N and we know the radius of the earth is 6380 km, we can use the equation for gravitational force to calculate the mass of the earth. Solution: me = Fd2 Gm1 = 5:98 1024 kg )(6:38106 m)2 (9:80 m/s2 = (6:671011 Nm2=kg2 )(1:00 kg) Sample Problem: John and Jane step onto the dance ﬂoor about 20. M apart at the Junior Prom and they feel an attraction to each other. If John’s mass is 70. kg and Jane’s mass is 50. kg, assume the attraction is gravity and calculate its magnitude. Solution: Fg = Gm1m2 d
2 = (6:671011 Nm2=kg2 (20: m)2 )(70: kg)(50: kg) = 1:2 108 N This is such an extremely weak force, it is probably not the force of attraction John and Jane felt. Summary • Newton proposed the universal law of gravitational attraction as F = G m1m2 d2 • The universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N·m2/kg2. • Even though satellites are constantly falling toward the object they circle, they do not get closer because their. straight line motion moves them away from the center at the same rate they fall. 21 Practice The following video is a lecture on universal gravitation. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=OZZGJfFf8XI www.ck12.org MEDIA Click image to the left for more content. 1. What factors determine the strength of the force of gravity? 2. Between which two points do we measure the distance between the earth and moon? The following video is The Mass vs. Weight Song. Use this resource to answer the questions that follow. http://w ww.youtube.com/watch?v=1whMAIGNq7E MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? The following website contains a series of solved practice problems on gravity. http://physics.info/gravitation/practice.shtml Review 1. The earth is attracted to the sun by the force of gravity. Why doesn’t the earth fall into the sun? 2. If the mass of the earth remained the same but the radius of the earth shrank to one-half its present distance, what would happen to the force of gravity on an object that was resting on the surface of the earth? 3. Lifting an object on the moon requires one-sixth the force that would be required to lift the same object on the earth because gravity on the moon is one-sixth that on earth. What about horizontal acceleration? If you threw a rock with enough force to accelerate it at 1.0 m/s2 horizontally on the moon, how would the required force compare to the force
necessary to acceleration the rock in the same way on the earth? 4. The mass of the earth is 5.98 1024 kg and the mass of the moon is 7.35 1022 kg. If the distance between the earth and the moon is 384,000 km, what is the gravitational force on the moon? • gravity: A natural phenomenon by which physical bodies appear to attract each other with a force proportional to their masses and inversely proportional to the distance separating them. 22 www.ck12.org References Chapter 6. Universal Law of Gravity 1. Chris Burks (Wikimedia: Chetvorno). http://commons.wikimedia.org/wiki/File:Cavendish_Torsion_Balance _Diagram.svg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 23 www.ck12.org CHAPTER 7 Mass versus Weight • Distinguish between mass and weight. • Given the acceleration due to gravity and either the mass or the weight of an object, calculate the other one. “Space exploration is an international endeavor.” Three Japan Aerospace Exploration Agency astronauts – Akihiko Hoshide, Satoshi Furukawa and Naoko Yamazaki – are training with the 11-member astronaut candidate class of 2004. JAXA astronauts Satoshi Furukawa, Akihiko Hoshide and Naoko Yamazaki experience near-weightlessness on the KC-135 training airplane. Mass and Weight The mass of an object is deﬁned as the amount of matter in the object. The amount of mass in an object is measured by comparing the object to known masses on an instrument called a balance. 24 www.ck12.org Chapter 7. Mass versus Weight Using the balance shown above, the object would be placed in one pan and known masses would be placed in the other pan until the pans were exactly balanced. When balanced, the mass of the object would be equal to the sum of the known masses in the other pan. The unit of measurement for mass is the kilogram. The mass of an object would be the same regardless of whether the object was on the earth or on the moon. The balance and known masses work exactly the same both places and would indicate the same mass for the same object as long as some gravitational force is present. The weight of an object is deﬁned as the force pulling the object downward. On the earth, this would be the gravitational force
of the earth on the object. On the moon, this would be the gravitational force of the moon on the object. Weight is measured by a calibrated spring scale as shown here. Weight is measured in force units which is Newtons in the SI system. The weights measured for an object would not be the same on the earth and moon because the gravitational ﬁeld on the surface of the moon is one-sixth of the magnitude of the gravitational ﬁeld on the surface of the earth. The force of gravity is given by Newton’s Second Law, F = ma, when F is the force of gravity in Newtons, m is the mass of the object in kilograms, and a is the acceleration due to gravity, 9.80 m/s2. When the formula is used speciﬁcally for ﬁnding weight from mass or vice versa, it may appear as W = mg. Example Problem: What is the weight of an object sitting on the earth’s surface if the mass of the object is 43.7 kg? Solution: W = mg = (43:7 kg)(9:80 m/s2) = 428 N Example Problem: What is the mass of an object whose weight sitting on the earth is 2570 N? m = W a = 2570 N 9:80 m/s2 = 262 kg Summary • The mass of an object is measured in kilograms and is deﬁned as the amount of matter in an object. • The mass of an object is determined by comparing the mass to known masses on a balance. • The weight of an object on the earth is deﬁned as the force acting on the object by the earth’s gravity. • Weight is measured by a calibrated spring scale. • The formula relating mass and weight is W = mg. 25 Practice A song about the difference between mass and weight sung by Mr. Edmunds to the tune of Sweet Caroline. Remember to make allowances for the fact that he is a teacher, not a professional singer. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=1whMAIGNq7E www.ck12.org MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? This video
shows what appears to be a magic trick but is actually a center of gravity demonstration. http://www.darktube.org/watch/simple-trick-magic-no-physics Review 1. The mass of an object on the earth is 100. kg. (a) What is the weight of the object on the earth? (b) What is the mass of the object on the moon? (c) Assuming the acceleration due to gravity on the moon is EXACTLY one-sixth of the acceleration due to gravity on earth, what is the weight of the object on the moon? 2. A man standing on the Earth can exert the same force with his legs as when he is standing on the moon. We know that the mass of the man is the same on the Earth and the Moon. We also know that F = ma is true on both the Earth and the Moon. Will the man be able to jump higher on the Moon than the Earth? Why or why not? • mass: The mass of an object is measured in kilograms and is deﬁned as the amount of matter in an object. • weight: The weight of an object on the earth is deﬁned as the force acting on the object by the earth’s gravity. References 1. Courtesy of NASA. http://spaceﬂight.nasa.gov/gallery/images/behindthescenes/training/html/jsc2004e45082.h tml. Public Domain 2. CK-12 Foundation - Christopher Auyeung.. CC BY-NC-SA 3.0 3. CK-12 Foundation - Christopher Auyeung.. CC-BY-NC-SA 3.0 26 www.ck12.org CHAPTER 8 • Deﬁne both static and sliding friction. • Explain what causes surface friction. • Deﬁne the coefﬁcient of friction. • Calculate frictional forces. • Calculate net forces when friction is involved. Chapter 8. Friction Friction Dealing with friction and a lack of friction becomes a very important part of the game in tennis played on a clay court. It’s necessary for this player to learn how to keep her shoes from sliding when she wants to run but also necessary to know how her shoes will slide when coming to a stop. Friction Most of the time, in beginning physics classes, friction is ignored. Concepts can be understood and calculations
made assuming friction to be non-existent. Whenever physics intersects with the real world, however, friction must be taken into account. Friction exists between two touching surfaces because even the smoothest looking surface is quite rough on a microscopic scale. 27 www.ck12.org With the bumps, lumps, and imperfections emphasized as in the image above, it becomes more apparent that if we try to slide the top block over the lower block, there will be collisions involved when bumps impact on bumps. The forward motion causes the collisions with bumps which then exert a force in opposite way the block is moving. The force of friction always opposes whatever motion is causing the friction. The force of friction between these two blocks is related to two factors. The ﬁrst factor is the roughness of the surfaces that are interacting which is called the coefﬁcient of friction, µ (Greek letter mu). The second factor is the magnitude of the force pushing the top block down onto the lower block. It is reasonable that the more forcefully the blocks are pushed together, the more difﬁcult it will be for one to slide over the other. A very long time ago, when physics was young, the word “normal” was used in the same way that we use the word “perpendicular” today. The force pushing these blocks together is the perpendicular force pushing the top block down on the lower block and this force is called the normal force. Much of the time, this normal force is simply the weight of the top block but on some occasions, the weight of the top block has some added or reduced force so the normal force is not always the weight. The force of friction then, can be calculated by Ffriction = µ Fnormal This is an approximate but reasonably useful and accurate relationship. It is not exact because µ depends on whether the surfaces are wet or dry and so forth. The frictional force we have been discussing is referred to as sliding friction because it is involved when one surface is sliding over another. If you have ever tried to slide a heavy object across a rough surface, you may be aware that it is a great deal easier to keep an object sliding than it is to start the object sliding in the ﬁrst place. When the object to slide is resting on a surface with no movement, the force of friction is called static friction and it is somewhat greater than sliding friction. Surfaces that are to move against one another will have both a
coefﬁcient of static friction and a coefﬁcient of sliding friction and the two values will NOT be the same. For example, the coefﬁcient of sliding friction for ice on ice is 0.03 whereas the coefﬁcient of static friction for ice on ice is 0.10 –more than three times as great. Example Problem: A box weighing 2000. N is sliding across a cement ﬂoor. The force pushing the box is 500. N and the coefﬁcient of sliding friction between the box and the ﬂoor is 0.20. What is the acceleration of the box. Solution: In this case, the normal force for the box is its weight. Using the normal force and the coefﬁcient of friction, we can ﬁnd the frictional force. We can also ﬁnd the mass of the box from its weight since we know the acceleration due to gravity. Then we can ﬁnd the net force and the acceleration. FF = µFN = (0:20)(2000: N) = 400: N weight g = 2000: N mass of box = 9:8 m=s2 = 204 g FNET = Pushing force frictional force = 500: N 400: N = 100: N a = FN = 0:49 m/s2 m = 100: N 204 kg Example Problem: Two boxes are connected by a rope running over a pulley (see image). The coefﬁcient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and velocity. Find the acceleration of the system. 28 www.ck12.org Chapter 8. Friction Solution: The force tending to move the system is the weight of box B and the force resisting the movement is the force of friction between the table and box A. The mass of the system would be the sum of the masses of both boxes. The acceleration of the system will be found by dividing the net force by the total mass. FN(box A) = mg = (5:0 kg)(9:8 m/s2) = 49 N Ffriction = µFN
= (0:20)(49 N) = 9:8 N Weight of box B = mg = (2:0 kg)(9:8 m/s2) = 19:6 N FNET = 19:6 N 9:8 N = 9:8 N a = FNET = 1:4 m/s2 mass = 9:8 N 7:0 kg Summary • Friction is caused by bodies sliding over rough surfaces. • The degree of surface roughness is indicated by the coefﬁcient of friction, µ. • The force of friction is calculated by multiplying the coefﬁcient of friction by the normal force. • The frictional force always opposes motion. • Acceleration is caused by the net force which is found by subtracting the frictional force from the applied force. Practice A video explaining friction. http://www.youtube.com/watch?v=CkTCp7SZdYQ MEDIA Click image to the left for more content. Review 1. A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefﬁcient of sliding friction between the sidewalk and the metal runners of the sled? 2. If the coefﬁcient of sliding friction between a 25 kg crate and the ﬂoor is 0.45, how much force is required to move the crate at a constant velocity across the ﬂoor? 29 www.ck12.org 3. A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coefﬁcient of sliding friction between the block and the table top? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity? • friction: A force that resists the relative motion or tendency to such motion of two bodies or substances in contact. • coefﬁcient of friction: The ratio of the force that maintains contact between an object and a surface (i.e. the normal force) and the frictional force that resists the motion of the object. • normal force: The perpendicular force one surface exerts on another surface when the surfaces are in contact. References 1. Courtesy of Eric Harris, U.S. Air
Force. http://commons.wikimedia.org/wiki/File:Maria_Sharapova,_2008 _Family_Circle_Cup.JPG. Public Domain 2. CK-12 Foundation - Joy Sheng.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 30 www.ck12.org Chapter 9. Free Body Diagrams CHAPTER 9 Free Body Diagrams Students will learn how to draw a free-body diagram and apply it to the real world. Students will learn how to draw a free-body diagram and apply it to the real world. Guidance For every problem involving forces it is essential to draw a free body diagram (FBD) before proceeding to the problem solving stage. The FBD allows one to visualize the situation and also to make sure all the forces are accounted. In addition, a very solid understanding of the physics is gleaned and many questions can be answered solely from the FBD. Example 1 Watch this Explanation MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Time for Practice 1. Draw free body diagrams (FBDs) for all of the following objects involved (in bold) and label all the forces appropriately. Make sure the lengths of the vectors in your FBDs are proportional to the strength of the force: smaller forces get shorter arrows! a. A man stands in an elevator that is accelerating upward at 2 m=s2. a. A boy is dragging a sled at a constant speed. The boy is pulling the sled with a rope at a 30 angle. a. The picture shown here is attached to the ceiling by three wires. 31 www.ck12.org a. A bowling ball rolls down a lane at a constant velocity. a. A car accelerates down the road. There is friction f between the tires and the road. 2. For the following situation, identify the 3rd law force pairs on the associated free body diagrams. Label each member of one pair “A;00 each member of the next pair “B;00 and so on. The spring is stretched so that it is pulling the block of wood to the right. Draw free body diagrams for the situation below. Notice that we are pulling the bottom block out from beneath the top block. There is friction between the blocks! After you have drawn your FBDs, identify the 3rd
law force pairs, as above. Answers Discuss in class 32 www.ck12.org Chapter 10. Problem Solving 1 CHAPTER 10 Problem Solving 1 In this lesson, students will learn how to solve difﬁcult problems using Newton’s 2nd law. In this lesson, students will learn how to solve difﬁcult problems using Newton’s 2nd law. Key Equations Common Forces 8 >>>>< >>>>: Fg = mg Gravity FN Normal force: acts perpendicular to surfaces FT Force of tension in strings and wires Fsp = kDx = Force of springDxfrom equilibrium Force Sums 8 >< >: Net force is the vector sum of all the forces Fnet = i Fi = ma Fnet,x = i Fix = max Horizontal components add also Fnet,y = i Fiy = may As do vertical ones Static and Kinetic Friction’ ( fs µsjFNj Opposes potential motion of surfaces in contact fk = µkjFNj Opposes motion of surfaces in contact Ultimately, many of these “contact” forces are due to attractive and repulsive electromagnetic forces between atoms in materials. Guidance Problem Solving for Newton’s Laws, Step-By-Step 1. Figure out which object is “of interest.” a. If you’re looking for the motion of a rolling cart, the cart is the object of interest. b. If the object of interest is not moving, that’s OK, don’t panic yet. c. Draw a sketch! This may help you sort out which object is which in your problem. 2. Identify all the forces acting on the object and draw them on object. (This is a free-body diagram –FBD) a. If the object has mass and is near the Earth, the easiest (and therefore, ﬁrst) force to write down is the force of gravity, pointing downward, with value mg. b. If the object is in contact with a ﬂat surface, it means there is a normal force acting on the object. This normal force points away from and is perpendicular to the surface. c. There may be more than one normal force acting on an object. For instance, if you have a bologna sandwich, remember that the slice of bologna feels normal forces from both the slices of bread! d. If a rope, wire, or cord is pulling
on the object in question, you’ve found yourself a tension force. The direction of this force is in the same direction that the rope is pulling. e. Don’t worry about any forces acting on other objects. For instance, if you have a bologna sandwich as your object of interest, and you’re thinking about the forces acting on the slice of bologna, don’t worry about the force of gravity acting on either piece of bread. 33 www.ck12.org f. Remember that Newton’s 3rd Law, calling for “equal and opposite forces,” does not apply to a single object. None of your forces should be “equal and opposite” on the same object in the sense of Newton’s 3rd Law. Third law pairs act on two different objects. g. Recall that scales (like a bathroom scale you weigh yourself on) read out the normal force acting on you, not your weight. If you are at rest on the scale, the normal force equals your weight. If you are accelerating up or down, the normal force had better be higher or lower than your weight, or you won’t have an unbalanced force to accelerate you. h. Never include “ma” as a force acting on an object. “ma” is the result of the net force Fnet which is found by summing all the forces acting on your object of interest. 3. Determine how to orient your axes a. A good rule to generally follow is that you want one axis (usually the x-axis) to be parallel to the surface your object of interest is sitting on. b. If your object is on a ramp, tilt your axes so that the x-axis is parallel to the incline and the y-axis is perpendicular. In this case, this will force you to break the force of gravity on the object into its components. But by tilting your axes, you will generally have to break up fewer vectors, making the whole problem simpler. 4. Identify which forces are in the x direction, which are in the y direction, and which are at an angle. a. If a force is upward, make it in the ydirection and give it a positive sign. If it is downward, make it in the ydirection and give it a negative sign. b. Same thing applies for right vs. left in the xdirection. Make rightward forces positive. c.
If forces are at an angle, draw them at an angle. A great example is that when a dog on a leash runs ahead, pulling you along, it’s pulling both forward and down on your hand. d. Draw the free body diagram (FBD). e. Remember that the FBD is supposed to be helping you with your problem. For instance, if you forget a force, it’ll be really obvious on your FBD. 5. Break the forces that are at angles into their x and y components a. Use right triangle trigonometry b. Remember that these components aren’t new forces, but are just what makes up the forces you’ve already identiﬁed. c. Consider making a second FBD to do this component work, so that your ﬁrst FBD doesn’t get too messy. 6. Add up all the x forces and x components. a. Remember that all the rightward forces add with a plus (+) sign, and that all the leftward forces add with a minus () sign. b. Don’t forget about the xcomponents of any forces that are at an angle! c. When you’ve added them all up, call this "the sum of all x forces" or "the net force in the xdirection." 7. Add up all the y forces and y components. a. Remember that all the upward forces add with a (+) sign, all the downward forces add with a () sign. b. Don’t forget about the ycomponents of any forces that are at an angle! c. When you’ve added them all up, call this "the sum of all y forces" or "net force in the ydirection." 8. Use Newton’s Laws twice. a. The sum of all xforces, divided by the mass, is the object’s acceleration in the xdirection. b. The sum of all yforces, divided by the mass, is the object’s acceleration in the ydirection. c. If you happen to know that the acceleration in the xdirection or ydirection is zero (say the object is just sitting on a table), then you can plug this in to Newton’s 2nd Law directly. d. If you happen to know the acceleration, you can plug this in directly too. 9. Each body should have a FBD. a. Draw a separate FBD for each
body. 34 www.ck12.org Chapter 10. Problem Solving 1 b. Set up a sum of forces equation based on the FBD for each body. c. Newton’s Third Law will tell you which forces on different bodies are the same in magnitude. d. Your equations should equal your unknown variables at this point. Example 1 Question: Using the diagram below, ﬁnd the net force on the block. The block weighs 3kg and the inclined plane has a coefﬁcient of friction of :6. Answer: The ﬁrst step to solving a Newton’s Laws problem is to identify the object in question. In our case, the block on the slope is the object of interest. Next, we need to draw a free-body diagram. To do this, we need to identify all of the forces acting on the block and their direction. The forces are friction, which acts in the negative x direction, the normal force, which acts in the positive y direction, and gravity, which acts in a combination of the negative y direction and the positive x direction. Notice that we have rotated the picture so that the majority of the forces acting on the block are along the y or x axis. This does not change the answer to the problem because the direction of the forces is still the same relative to each other. When we have determined our answer, we can simply rotate it back to the original position. Now we need to break down gravity (the only force not along one of the axises) into its component vectors (which do follow the axises). The x component of gravity : 9:8m=s2 cos60 = 4:9m=s2 The y component of gravity : 9:8m=s2 sin60 = 8:5m=s2 Yet these are only the acceleration of gravity so we need to multiply them by the weight of the block to get the force. F = ma = 3kg 4:9m=s2 = 14:7NF = ma = 3kg 8:5m=s2 = 25:5N 35 Now that we have solved for the force of the y-component of gravity we know the normal force (they are equal). Therefore the normal force is 25:5N. Now that we have the normal force and the coefﬁcient of static friction, we can ﬁnd the force of friction. www.ck12.org Fs = µs
FN = :6 25:5N = 15:3N The force of static friction is greater than the component of gravity that is forcing the block down the inclined plane. Therefore the force of friction will match the force of the x-component of gravity. So the net force on the block is net force in the x direction : xcomponent o f gravity z { }\| 14:7N f orce o f f riction z { }\| 14:7N net force in the y direction : 25:5N } {z \| Normal Force 25:5N {z } \| ycomponent o f gravity = 0N = 0N Therefore the net force on the block is 0N. Example 2 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Watch this Explanation 36 www.ck12.org Simulation Chapter 10. Problem Solving 1 • http://simulations.ck12.org/FreeBody/ Time for Practice 1. Find the mass of the painting. The tension in the leftmost rope is 7:2 N, in the middle rope it is 16 N, and in the rightmost rope it is 16 N. 2. Find Brittany’s acceleration down the frictionless waterslide in terms of her mass m, the angle q of the incline, and the acceleration of gravity g. 37 www.ck12.org 3. The physics professor holds an eraser up against a wall by pushing it directly against the wall with a completely horizontal force of 20 N. The eraser has a mass of 0:5 kg. The wall has coefﬁcients of friction µS = 0:8 and µK = 0:6: a. Draw a free body diagram for the eraser. b. What is the normal force FN acting on the eraser? c. What is the frictional force FS equal to? d. What is the maximum mass m the eraser could have and still not fall down? e. What would happen if the wall and eraser were both frictionless? 4. A tractor of mass 580 kg accelerates up a 10 incline from rest to a speed of 10 m=s in 4 s. For all of answers below, provide a magnitude and a direction. 38 www.ck12.org Chapter 10. Problem Solving 1 a. What net force Fnet has been applied to the tractor? b. What is the normal force, FN on the tractor?
c. What is the force of gravity Fg on the tractor? d. What force has been applied to the tractor so that it moves uphill? e. What is the source of this force? 5. A heavy box (mass 25 kg) is dragged along the ﬂoor by a kid at a 30 angle to the horizontal with a force of 80 N (which is the maximum force the kid can apply). a. Draw the free body diagram. b. What is the normal force FN? c. Does the normal force decrease or increase as the angle of pull increases? Explain. d. Assuming no friction, what is the acceleration of the box? e. Assuming it begins at rest, what is its speed after ten seconds? f. Is it possible for the kid to lift the box by pulling straight up on the rope? g. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 30 angle? h. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 45 angle? i. In the absence of friction, what is the net force in the xdirection if the kid pulls at a60 angle? j. The kid pulls the box at constant velocity at an angle of 30. What is the coefﬁcient of kinetic friction µK between the box and the ﬂoor? k. The kid pulls the box at an angle of 30, producing an acceleration of 2 m=s2. What is the coefﬁcient of kinetic friction µK between the box and the ﬂoor? 6. Spinal implant problem —this is a real life bio-med engineering problem! 39 www.ck12.org Here’s the situation: both springs are compressed by an amount xo. The rod of length L is ﬁxed to both the top plate and the bottom plate. The two springs, each with spring constant k, are wrapped around the rod on both sides of the middle plate, but are free to move because they are not attached to the rod or the plates. The middle plate has negligible mass, and is constrained in its motion by the compression forces of the top and bottom springs. The medical implementation of this device is to screw the top plate to one vertebrae and the middle plate to the vertebrae directly below. The bottom plate is suspended in space. Instead of fusing broken vertebrates together, this implant allows
movement somewhat analogous to the natural movement of functioning vertebrae. Below you will do the exact calculations that an engineer did to get this device patented and available for use at hospitals. a. Find the force, F, on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region both springs are providing opposite compression forces.) b. Find the force, F, on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region, only one spring is in contact with the middle plate.) c. Graph F vs. x. Label the values for force for the transition region in terms of the constants given. 7. You design a mechanism for lifting boxes up an inclined plane by using a vertically hanging mass to pull them, as shown in the ﬁgure below. The pulley at the top of the incline is massless and frictionless. The larger mass, M, is accelerating downward with a measured acceleration a. The smaller masses are mA and mB ; the angle of the incline is q, and the coefﬁcient of kinetic friction between each of the masses and the incline has been measured and determined to be µK. a. Draw free body diagrams for each of the three masses. b. Calculate the magnitude of the frictional force on each of the smaller masses in terms of the given quantities. c. Calculate the net force on the hanging mass in terms of the given quantities. d. Calculate the magnitudes of the two tension forces TA and TB in terms of the given quantities. e. Design and state a strategy for solving for how long it will take the larger mass to hit the ground, assuming at this moment it is at a height h above the ground. Do not attempt to solve this: simply state the strategy for solving it. 40 www.ck12.org Chapter 10. Problem Solving 1 8. You build a device for lifting objects, as shown below. A rope is attached to the ceiling and two masses are allowed to hang from it. The end of the rope passes around a pulley (right) where you can pull it downward to lift the two objects upward. The angles of the ropes, measured with respect to the vertical, are shown. Assume the bodies are at rest initially. a. Suppose you are able to measure the masses m
1 and m2 of the two hanging objects as well as the tension TC. Do you then have enough information to determine the other two tensions, TA and TB? Explain your reasoning. b. If you only knew the tensions TA and TC, would you have enough information to determine the masses m1 and m2? If so, write m1 and m2 in terms of TA and TC. If not, what further information would you require? 9. A stunt driver is approaching a cliff at very high speed. Sensors in his car have measured the acceleration and velocity of the car, as well as all forces acting on it, for various times. The driver’s motion can be broken down into the following steps: Step 1: The driver, beginning at rest, accelerates his car on a horizontal road for ten seconds. Sensors show that there is a force in the direction of motion of 6000 N, but additional forces acting in the opposite direction with magnitude 1000 N. The mass of the car is 1250 kg. Step 2: Approaching the cliff, the driver takes his foot off of the gas pedal (There is no further force in the direction of motion.) and brakes, increasing the force opposing motion from 1000 N to 2500 N. This continues for ﬁve seconds until he reaches the cliff. Step 3: The driver ﬂies off the cliff, which is 44:1 m high and begins projectile motion. (a) Ignoring air resistance, how long is the stunt driver in the air? (b) For Step 1: i. Draw a free body diagram, naming all the forces on the car. ii. Calculate the magnitude of the net force. iii. Find the change in velocity over the stated time period. iv. Make a graph of velocity in the xdirection vs. time over the stated time period. v. Calculate the distance the driver covered in the stated time period. Do this by ﬁnding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics. (c) Repeat (b) for Step 2. (d) Calculate the distance that the stunt driver should land from the bottom of the cliff. Answers 1. 3:6 kg 41 www.ck12.org 2. g sin q 3. b.20 N c. 4:9 N d. 1:63 kg e. Eraser would slip down the wall 4. a. 1450 N
b. 5600 N c. 5700 N d. Friction between the tires and the ground e. Fuel, engine, or equal and opposite reaction 5. b. 210 N c. no, the box is ﬂat so the normal force doesn’t change d. 2:8 m=s2 e.28 m=s f. no g. 69 N h. 57 N i. 40 N j. 0:33 k. 0:09 6. a. zero b. kx0 7. b. f1 = µkm1g cos q; f2 = µkm2g cos q c. Ma d. TA = (m1 + m2)(a + µ cos q) and TB = m2a + µm2 cos q e. Solve by using d = 1=2at2 and substituting h for d 8. a. Yes, because it is static and you know the angle and m1 b. Yes, TA and the angle gives you m1 and the angle and TC gives you m2; m1 = TA cos 25=g and m2 = TC cos 30=g 9. a. 3 seconds d. 90 m 42 Physics Unit 6 (Energy) Patrick Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®”
(collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org 1 4 9 12 15 Contents 1 Kinetic Energy 2 Potential Energy 3 Conservation of Energy 4 Energy Problem Solving 5 Springs iv www.ck12.org Chapter 1. Kinetic Energy CHAPTER 1 Kinetic Energy • Deﬁne energy. • Deﬁne kinetic energy. • Given the mass and speed of an object, calculate its kinetic energy. • Solve problems involving kinetic energy. This military jet requires a tremendous amount of work done on it to get its speed up to takeoff speed in the short distance available on the deck of an aircraft carrier. Some of the work is done by the plane’s own jet engines but work from a catapult is also necessary for takeoff. Kinetic Energy Energy is the ability to change an object’s motion or position. Energy comes in many forms and each of those forms can be converted into any other form. A moving object has the ability to change another object’s motion or position simply by colliding with it and this form of energy is called kinetic energy. The kinetic energy of an object can be calculated by the equation KE = 1 2 mv2, where m is the mass of the object and v is its velocity. The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will
have quadruple the kinetic energy. The SI unit for kinetic energy (and all forms of energy) is kg m2 s2 which is equivalent to Joules, the same unit we use for work. The kinetic energy of an object can be changed by doing work on the object. The work done on an object equals the kinetic energy gain or loss by the object. This relationship is expressed in the work-energy theorem WNET = DKE. Example Problem: A farmer heaves a 7.56 kg bale of hay with a ﬁnal velocity of 4.75 m/s. 1 (a) What is the kinetic energy of the bale? (b) The bale was originally at rest. How much work was done on the bale to give it this kinetic energy? www.ck12.org Solution: (a) KE = 1 2 mv2 = 1 2 (7:56 kg)(4:75)2 = 85:3 Joules (b) Work done = DKE = 85:3 Joules Example Problem: What is the kinetic energy of a 750. kg car moving at 50.0 km/h? Solution: 50:0 km h 1000 m km 1 h 3600 s = 13:9 m/s KE = 1 2 mv2 = 1 2 (750: kg)(13:9 m/s)2 = 72; 300 Joules Example Problem: How much work must be done on a 750. kg car to slow it from 100. km/h to 50.0 km/h? Solution: From the previous example problem, we know that the KE of this car when it is moving at 50.0 km/h If the same car is going twice as fast, its KE will be four times as great because KE is is 72,300 Joules. proportional to the square of the velocity. Therefore, when this same car is moving at 100. km/h, its KE is 289,200 Joules. Therefore, the work done to slow the car from 100. km/h to 50.0 km/h is (289; 200 Joules) (72; 300 Joules) = 217; 000 Joules. Summary • Energy is the ability to change an object’s motion or position. • There are many forms of energy. • The energy of motion is called kinetic energy. • The formula for kinetic energy is KE = 1 2 mv2. • The work done on an object equals the
kinetic energy gain or loss by the object, WNET = DKE. Practice The following video discusses kinetic energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=g157qwT1918 MEDIA Click image to the left for more content. 1. Potential energy is present in objects that are ______________. 2. Kinetic energy is present in objects that are ______________. 3. What formula is given for kinetic energy? Practice problems involving kinetic energy: http://www.physicsclassroom.com/Class/energy/u5l1c.cfm 2 www.ck12.org Review Chapter 1. Kinetic Energy 1. A comet with a mass of 7.85 1011 kg is moving with a velocity of 25,000 m/s. Calculate its kinetic energy. 2. A riﬂe can shoot a 4.00 g bullet at a speed of 998 m/s. (a) Find the kinetic energy of the bullet. (b) What work is done on the bullet if it starts from rest? (c) If the work is done over a distance of 0.75 m, what is the average force on the bullet? (d) If the bullet comes to rest after penetrating 1.50 cm into a piece of metal, what is the magnitude of the force bringing it to rest? • energy: An indirectly observed quantity that is often understood as the ability of a physical system to do work. • kinetic energy: The energy an object has due to its motion. References 1. Courtesy of Mass Communication Specialist 3rd Class Torrey W. Lee, U.S. Navy. Jet Takeoff. Public Domain 3 CHAPTER 2 www.ck12.org Potential Energy • Deﬁne potential energy. • Solve problems involving gravitational potential energy. • Solve problems involving the conversion of potential energy to kinetic energy and vice versa. Shooting an arrow from a bow requires work done on the bow by the shooter’s arm to bend the bow and thus produce potential energy. The release of the bow converts the potential energy of the bent bow into the kinetic energy of the ﬂying arrow. Potential Energy When an object is held above the earth, it has the ability to make matter move because all you have to do is let go of the object and it will fall of its own accord. Since energy is deﬁned as the ability to make matter
move, this object has energy. This type of energy is stored energy and is called potential energy. An object held in a stretched If the stretched rubber band is released, the object will move. A rubber band also contains this stored energy. pebble on a ﬂexed ruler has potential energy because if the ruler is released, the pebble will ﬂy. If you hold two positive charges near each other, they have potential energy because they will move if you release them. Potential If the chemical bonds are broken and allowed to form lower energy is also the energy stored in chemical bonds. 4 www.ck12.org Chapter 2. Potential Energy potential energy chemical bonds, the excess energy is released and can make matter move. This is frequently seen as increased molecular motion or heat. If a cannon ball is ﬁred straight up into the air, it begins with a high kinetic energy. As the cannon ball rises, it slows down due to the force of gravity pulling it toward the earth. As the ball rises, its gravitational potential energy is increasing and its kinetic energy is decreasing. When the cannon ball reaches the top of its arc, it kinetic energy is zero and its potential energy is maximum. Then gravity continues to pull the cannon ball toward the earth and the ball will fall toward the earth. As it falls, its speed increases and its height decreases. Therefore, its kinetic energy increases as it falls and its potential energy decreases. When the ball returns to its original height, its kinetic energy will be the same as when it started upward. When work is done on an object, the work may be converted into either kinetic or potential energy. If the work results in motion, the work was converted into kinetic energy and if the work done resulted in change in position, the work was converted into potential energy. Work could also be spent overcoming friction and that work would be converted into heat but usually we will consider frictionless systems. If we consider the potential energy of a bent stick or a stretched rubber band, the potential energy can be calculated by multiplying the force exerted by the stick or rubber band times the distance over which the force will be exerted. The formula for calculating this potential energy looks exactly like the formula for calculating work done, that is W = Fd. The only difference is that work is calculated when the object actually moves under the force and potential energy is calculated when the system is at rest before any motion actually occurs. In the case of gravitational potential energy, the force exerted by the object is its
weight and the distance it could travel would be its height above the earth. Since the weight of an object is calculated by W = mg, then gravitational potential energy can be calculate by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height the object will fall. 5 Example Problem: A 3.00 kg object is lifted from the ﬂoor and placed on a shelf that is 2.50 m above the ﬂoor. (a) What was the work done in lifting the object? (b) What is the gravitational potential energy of the object sitting on the shelf? (c) If the object falls off the shelf and free falls to the ﬂoor, what will its velocity be when it hits the ﬂoor? Solution: weight of the object = mg = (3:00 kg)(9:80 m=s2) = 29:4 N www.ck12.org (a) W = Fd = (29:4 N)(2:50 m) = 73:5 J (b) PE = mgh = (3:00 kg)(9:80 m=s2)(2:50 m) = 73:5 J (c) KE = PE so 1 2 mv2 = 73:5 J s v = (2)(73:5 J) 3:00 kg = 7:00 m=s Example Problem: A pendulum is constructed from a 7.58 kg bowling ball hanging on a 3.00 m long rope. The ball is pulled back until the rope makes an angle of 45 with the vertical. (a) What is the potential energy of the ball? (b) If the ball is released, how fast will it be traveling at the bottom of its arc? Solution: You can use trigonometry to ﬁnd the vertical height of the ball in the pulled back position. This vertical height is found to be 0.877 m. PE = mgh = (7:58 kg)(9:80 m=s2)(0:877 m) = 65:1 J When the ball is released, the PE will be converted into KE as the ball swings through the arc. KE = 1 s 2 mv2 = 65:1 J v = (2)(65:1 kg m2=s2) 7:58 kg = 4:14 m=s Summary • Stored energy is called potential energy.
6 www.ck12.org Chapter 2. Potential Energy • Energy may be stored by holding an object elevated in a gravitational ﬁeld or by holding it while a force is attempting to move it. • Potential energy may be converted to kinetic energy. • The formula for gravitational potential energy is PE = mgh. • In the absence of friction or bending, work done on an object must become either potential energy or kinetic energy or both. Practice The following video discusses types of energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=nC6tT1wkXEc MEDIA Click image to the left for more content. 1. What is the deﬁnition of energy? 2. Name two types of potential energy. 3. How is energy transferred from one object to another? Potential and kinetic energy practice problems with solutions: http://www.physicsclassroom.com/Class/energy/U5L2bc.cfm Review 1. A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top? 2. A 50.0 kg shell was ﬁred from a cannon at earth’s surface to a maximum height of 400. m. What is the potential energy at maximum height? 3. If the shell in problem #3 then fell to a height of 100. m, what was the loss of PE? 4. A person weighing 645 N climbs up a ladder to a height of 4.55 m. (a) What work does the person do? (b) What is the increase in gravitational potential energy? (c) Where does the energy come from to cause this increase in PE? • potential energy: Otherwise known as stored energy, is the ability of a system to do work due to its position or internal structure. For example, gravitational potential energy is a stored energy determined by an object’s position in a gravitational ﬁeld while elastic potential energy is the energy stored in a spring. References 1. Image copyright Skynavin, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 7 2. Image copyright Tribalium, 2013; modiﬁed by CK-12 Foundation - Samantha Bacic. http://www.shutterstock. com. Used under license from Shutterstock.com 3. CK-12 Foundation -
Samantha Bacic.. CC-BY-NC-SA 3.0 www.ck12.org 8 www.ck12.org Chapter 3. Conservation of Energy CHAPTER 3 Conservation of Energy • State the law of conservation of energy. • Describe a closed system. • Use the law of conservation of energy to solve problems. There are many energy conversions between potential and kinetic energy as the cars travel around this double looping roller coaster. Conservation of Energy The law of conservation of energy states that within a closed system, energy can change form, but the total amount of energy is constant. Another way of expressing the law of conservation of energy is to say that energy can neither be created nor destroyed. An important part of using the conservation of energy is selecting the system. In the In a closed system, conservation of energy just as in the conservation of momentum, the system must be closed. objects may not enter or leave, and it is isolated from external forces so that no work can be done on the system. In the analysis of the behavior of an object, you must make sure you have included everything in the system that is involved in the motion. For example, if you are considering a ball that is acted on by gravity, you must include the earth in your system. The kinetic energy of the ball considered by itself may increase and only when the earth is included in the system can you see that the increasing kinetic energy is balanced by an equivalent loss of potential energy. The sum of the kinetic energy and the potential energy of an object is often called the mechanical energy. Consider a box with a weight of 20.0 N sitting at rest on a shelf that is 2.00 m above the earth. The box has zero kinetic energy but it has potential energy related to its weight and the distance to the earth’s surface. PE = mgh = (20:0 N)(2:00 m) = 40:0 J 9 www.ck12.org If the box slides off the shelf, the only force acting on the box is the force of gravity and so the box falls. We can calculate the speed of the box when it strikes the ground by several methods. We can calculate the speed directly 2 = 2ad. We can also ﬁnd the ﬁnal velocity by setting the kinetic energy at the bottom of the using the formula v f fall equal to the potential energy at the top, KE = PE and, 1 2 = 2gh. You may note these formulas 2 mv2
= mgh, so v f are essentially the same. q (2)(9:80 m=s2)(2:00 m) = 6:26 m=s v = Example Problem: Suppose a cannon is sitting on top of a 50.0 m high hill and a 5.00 kg cannon ball is ﬁred with a velocity of 30.0 m/s at some unknown angle. What will be the velocity of the cannon ball when it strikes the earth? Solution: Since the angle at which the cannon ball is ﬁred is unknown, we cannot use the usual equations from projectile motion. However, at the moment the cannon ball is ﬁred, it has a certain KE due to the mass of the ball and its speed and it has a certain PE due to its mass and it height above the earth. Those two quantities of energy can be calculated. When the ball returns to the earth, its PE will be zero and therefore, its KE at that point must account for the total of its original KE + PE. This gives us a method of solving the problem. ETOTAL = KE + PE = 1 ETOTAL = 2250 J + 2450 J = 4700 J (5:00 kg)(30:0 m=s)2 + (5:00 kg)(9:80 m=s2)(50:0 m) 2 mv2 + mgh = 1 2 1 2 mv f 2 = 4700 J so v f = s (2)(4700 J) 5:00 kg = 43:4 m=s Example Problem: A 2.00 g bullet moving at 705 m/s strikes a 0.250 kg block of wood at rest on a frictionless surface. The bullet sticks in the wood and the combined mass moves slowly down the table. (a) What is the KE of the bullet before the collision? (b) What is the speed of the combination after the collision? (c) How much KE was lost in the collision? Solution: (a) KEBULLET = 1 (b) mBvB + mW vW = (mB+W )(vB+W ) 2 mv2 = 1 2 (0:00200 kg)(705 m=s)2 = 497 J (0:00200 kg)(705 m=s) + (0:250 kg)(0 m=s) = (0:252 kg)(V ) (1:41 kg
m=s) = (0:252 kg)(V ) V = 5:60 m=s 2 mv2 = 1 (c) KECOMBINATION = 1 KELOST = KEBEFORE KEAFTER = 497 J 4 J = 493 J 2 (0:252 kg)(5:60 m=s)2 = 3:95 J Summary • In a closed system, energy may change forms but the total amount of energy is constant. Practice The following video demonstrates Newton Ball tricks. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=JadO3RuOJGU 10 www.ck12.org Chapter 3. Conservation of Energy MEDIA Click image to the left for more content. 1. What happens when one ball is pulled up to one side and released? 2. What happens when three balls are pulled up to one side and released? 3. What happens when two balls are pulled out from each side and released? Practice problems with answers for the law of conservation of energy: http://www.physicsclassroom.com/class/energy/u5l2bc.cfm Review 1. A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00 m to the surface of the water, (a) what is the kinetic energy of the chunk of ice when its hits the water, and (b) what is its velocity? 2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill. (a) Assuming the movement is frictionless, at what vertical height will the cart come to rest? (b) Do you need to know the mass of the cart to solve this problem? 3. A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer’s feet touch the ground 9.00 m below where the rope is tied. How fast is the performer moving at the bottom of the arc? 4. A skier starts from rest at the top of a 45.0 m hill, coasts down a 30slope into a valley, and continues up to the top of a 40.0 m hill. Both hill heights are measured from the valley ﬂoor. Assume the skier puts no effort into the motion (always coast
ing) and there is no friction. (a) How fast will the skier be moving on the valley ﬂoor? (b) How fast will the skier be moving on the top of the 40.0 m hill? 5. A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitutde of the ﬁnal velocity when it strikes the ground? Ignore air resistance. 6. If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum height will it reach? Neglect air resistance. • conservation of energy: An empirical law of physics (meaning it cannot be derived), states that the total amount of energy within an isolated system is constant. Although energy can be transformed from one form into another, energy cannot be created or destroyed • closed system: Means it cannot exchange any of heat, work, or matter with the surroundings. • mechanical energy: The sum of potential energy and kinetic energy. References 1. User:Zonk43/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Teststrecke_Roller_Coaster3.J PG. Public Domain 11 CHAPTER 4 Energy Problem Solving Students will learn how to analyze and solve more complicated problems involving energy conservation. Students will learn how to analyze and solve more complicated problems involving energy conservation. www.ck12.org Key Equations Einitial = E ﬁnal ; T hetotalenergydoesnotchangeinclosedsystems KE = 1 2 mv2 ; Kinetic energy PEg = mgh ; Potential energy of gravity PEsp = 1 2 kx2; Potential energy of a spring W = FxDx = Fd cos q ; Work is equal to the distance multiplied by the component of the force in the direction it is moving. Guidance The main thing to always keep prescient in your mind is that the total energy before must equal the total energy after. If some energy has transferred out of or into the system via work, you calculate that work done and include it in the energy sum equation. Generally work done by friction is listed on the ’after’ side and work put into the system, via a jet pack for example, goes on the ’before’ side. Another important point is that
on turns or going over hills or in rollercoaster loops, one must include the centripetal motion equations -for example to insure that you have enough speed to make the loop. Example 1 Watch this Explanation 12 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. www.ck12.org Time for Practice Chapter 4. Energy Problem Solving 1. A rock with mass m is dropped from a cliff of height h. What is its speed when it gets to the bottom of the cliff? a. b. p r p mg 2g h 2gh c. d. gh e. None of the above 2. In the picture above, a 9:0 kg baby on a skateboard is about to be launched horizontally. The spring constant is 300 N=m and the spring is compressed 0:4 m. For the following questions, ignore the small energy loss due to the friction in the wheels of the skateboard and the rotational energy used up to make the wheels spin. a. What is the speed of the baby after the spring has reached its uncompressed length? b. After being launched, the baby encounters a hill 7 m high. Will the baby make it to the top? If so, what is his speed at the top? If not, how high does he make it? c. Are you ﬁnally convinced that your authors have lost their minds? Look at that picture! 3. When the biker is at the top of the ramp shown above, he has a speed of 10 m=s and is at a height of 25 m. The bike and person have a total mass of 100 kg. He speeds into the contraption at the end of the ramp, which slows him to a stop. a. What is his initial total energy? (Hint: Set Ug = 0 at the very bottom of the ramp.) b. What is the length of the spring when it is maximally compressed by the biker? (Hint: The spring does not compress all the way to the ground so there is still some gravitational potential energy. It will help to draw some triangles.) 4. An elevator in an old apartment building in Switzerland has four huge springs at the bottom of the shaft to cushion its fall in case the cable breaks. The springs have an uncompressed height of about 1 meter. Estimate the spring constant necessary to stop this elevator, following these steps: a. First, guesstimate the mass
of the elevator with a few passengers inside. b. Now, estimate the height of a ﬁve-story building. c. Lastly, use conservation of energy to estimate the spring constant. 13 5. You are skiing down a hill. You start at rest at a height 120 m above the bottom. The slope has a 10:0 grade. Assume the total mass of skier and equipment is 75:0 kg. www.ck12.org a. Ignore all energy losses due to friction. What is your speed at the bottom? b. If, however, you just make it to the bottom with zero speed what would be the average force of friction, including air resistance? 6. Two horriﬁc contraptions on frictionless wheels are compressing a spring (k = 400 N=m) by 0:5 m compared to its uncompressed (equilibrium) length. Each of the 500 kg vehicles is stationary and they are connected by a string. The string is cut! Find the speeds of the vehicles once they lose contact with the spring. 7. A roller coaster begins at rest 120 m above the ground, as shown. Assume no friction from the wheels and air, and that no energy is lost to heat, sound, and so on. The radius of the loop is 40 m. a. If the height at point G is 76 m, then how fast is the coaster going at point G? b. Does the coaster actually make it through the loop without falling? (Hint: You might review the material from centripetal motion lessons to answer this part.) Answers to Selected Problems 1.. 2. a. 2:3 m=s c. No, the baby will not clear the hill. 3. a. 29; 500 J b. Spring has maximum compressed length of 13 m 4.. 5. a. 48:5 m=s b. 128 N 6. 0:32 m=s each 7. a.29 m/s b. just barely, aC = 9:8 m=s2 14 www.ck12.org CHAPTER 5 Chapter 5. Springs Springs Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Key Equations T = 1 f ; Period is the inverse of frequency Tspring = 2p r m k ; Period of mass m on a spring with constant k Fsp
= kx ; the force of a spring equals the spring constant multiplied by the amount the spring is stretched or compressed from its equilibrium point. The negative sign indicates it is a restoring force (i.e. direction of the force is opposite its displacement from equilibrium position. Usp = 1 that it is stretched or compressed from equilibrium 2 kx2 ; the potential energy of a spring is equal to one half times the spring constant times the distance squared Guidance • The oscillating object does not lose any energy in SHM. Friction is assumed to be zero. • In harmonic motion there is always a restorative force, which attempts to restore the oscillating object to its equilibrium position. The restorative force changes during an oscillation and depends on the position of the object. In a spring the force is given by Hooke’s Law: F = kx • The period, T, is the amount of time needed for the harmonic motion to repeat itself, or for the object to go one full cycle. In SHM, T is the time it takes the object to return to its exact starting point and starting direction. • The frequency, f ; is the number of cycles an object goes through in 1 second. Frequency is measured in Hertz (Hz). 1 Hz = 1 cycle per sec. • The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object. The amplitude can vary in harmonic motion, but is constant in SHM. Example 1 MEDIA Click image to the left for more content. 15 www.ck12.org Watch this Explanation Simulation MEDIA Click image to the left for more content. Mass & Springs (PhET Simulation) Time for Practice 1. A rope can be considered as a spring with a very high spring constant k; so high, in fact, that you don’t notice the rope stretch at all before it “pulls back.” a. What is the k of a rope that stretches by 1 mm when a 100 kg weight hangs from it? b. If a boy of 50 kg hangs from the rope, how far will it stretch? c. If the boy kicks himself up a bit, and then is bouncing up and down ever so slightly, what is his frequency of oscillation? Would he notice this oscillation? If so, how? If not, why not?
2. If a 5:0 kg mass attached to a spring oscillates 4.0 times every second, what is the spring constant k of the spring? 3. A horizontal spring attached to the wall is attached to a block of wood on the other end. All this is sitting on a frictionless surface. The spring is compressed 0:3 m. Due to the compression there is 5:0 J of energy stored in the spring. The spring is then released. The block of wood experiences a maximum speed of 25 m=s. a. Find the value of the spring constant. b. Find the mass of the block of wood. c. What is the equation that describes the position of the mass? d. What is the equation that describes the speed of the mass? e. Draw three complete cycles of the block’s oscillatory motion on an x vs. t graph. 16 www.ck12.org Chapter 5. Springs 4. A spider of 0:5 g walks to the middle of her web. The web sinks by 1:0 mm due to her weight. You may assume the mass of the web is negligible. a. If a small burst of wind sets her in motion, with what frequency will she oscillate? b. How many times will she go up and down in one s? In 20 s? c. How long is each cycle? d. Draw the x vs t graph of three cycles, assuming the spider is at its highest point in the cycle at t = 0 s. Answers to Selected Problems 1. a. 9:8 105 N=m b. 0:5 mm c. 22 Hz 2. 3:2 103 N=m 3. a. 110 N=m d. v(t) = (25) cos(83t) 4. a. 16 Hz b. 16 complete cycles but 32 times up and down, 315 complete cycles but 630 times up and down c. 0:063 s Investigation 1. Your task: Match the period of the circular motion system with that of the spring system. You are only allowed to change the velocity involved in the circular motion system. Consider the effective distance between the block and the pivot to be to be ﬁxed at 1m. The spring constant(13.5N/m) is also ﬁxed. You should view the charts to check whether you have succeeded. Instructions: To alter the velocity, simply click on the Select Tool, and select the pivot. The
Position tab below will allow you to numerically adjust the rotational speed using the Motor ﬁeld. To view the graphs of their respective motion in order to determine if they are in sync, click on Chart tab below. 2. MEDIA Click image to the left for more content. 3. Now the mass on the spring has been replaced by a mass that is twice the rotating mass. Also, the distance between the rotating mass and the pivot has been changed to 1.5 m. What velocity will keep the period the same now? 4. MEDIA Click image to the left for more content. 17 Physics Unit 7 (Work, Power, Kinetic Energy Theorem) Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3
.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Work 2 Power 3 Work-Energy Principle 1 4 7 iv www.ck12.org CONCEPT 1 Concept 1. Work Work • Deﬁne work. • Identify forces that are doing work. • Given two of the three variables in the equation, W = Fd, calculate the third. In order for the roller coaster to run down the incline by gravitational attraction, it ﬁrst must have work done on it towing it up to the top of the hill. The work done on the coaster towing it to the top of the hill becomes potential energy stored in the coaster and that potential energy is converted to kinetic energy as the coaster runs down from the top of the hill to the bottom. Work The word work has both an everyday meaning and a speciﬁc scientiﬁc meaning. In the everyday use of the word, work would refer to anything which required a person to make an effort. In physics, however, work is deﬁned as the force exerted on an object multiplied by the distance the object moves due to that force. W = Fd In the scientiﬁc deﬁnition of the word, if you push against an automobile with a force of 200 N for 3 minutes but the automobile does not move, then you have done NO work. Multiplying 200 N times 0 meters yields zero work. If you are holding an object in your arms, the upward force you are exerting is equal to the object’s weight. If 1 www.ck12.org you hold the object until your arms become very tired, you have still done no work because you did not move the object in the direction of the force. When you lift an object, you exert a force equal to the object’s weight and the If an object weighs 200. N and you lift it 1.50 meters, then your work is, object moves due to that lifting force. W = Fd = (200: N)(1:50 m) = 300: N m. One of the units you will see for work is the Newton·meter but since a Newton is also a
kilogram·m/s 2, then a Newton·meter is also kg·m 2 /s 2. This unit has also been named the Joule (pronounced Jool) in honor of James Prescott Joule, a nineteenth century English physicist. Example Problem: A boy lifts a box of apples that weighs 185 N. The box is lifted a height of 0.800 m. How much work did the boy do? Solution: W = Fd = (185 N)(0:800 m) = 148 N m = 148 Joules Work is done only if a force is exerted in the direction of motion. If the force and motion are perpendicular to each other, no work is done because there is no motion in the direction of the force. If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Example Problem: Suppose a 125 N force is applied to a lawnmower handle at an angle of 25° with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done? Solution: The solution requires that we determine the component of the force that was in the direction of the motion of the lawnmower. The component of the force that was pushing down on the ground does not contribute to the work done. Fparallel = (Force)(cos 25) = (125 N)(0:906) = 113 N W = Fparalleld = (113 N)(56 m) = 630 J Summary • In physics, work is deﬁned as the force exerted on an object multiplied by the distance the object moves due to that force. • The unit for work is called the joule. • If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Practice The following video introduces energy and work. Use this resource to answer the questions that follow. http://wn.com/Work_physics_#/videos 2 www.ck12.org Concept 1. Work MEDIA Click image to the left for more content. 1. What deﬁnition is given in the video for energy? 2. What is the deﬁnition given in the video for work? 3. What unit is used in the video for work? The following website contains practice questions with answers on the topic
of work. http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section6.rhtml Review 1. How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m? 2. A workman carries some lumber up a staircase. The workman moves 9.6 m vertically and 22 m horizontally. If the lumber weighs 45 N, how much work was done by the workman? 3. A barge is pulled down a canal by a horse walking beside the canal. exerted is 400. N, and the barge is pulled 100. M, how much work did the horse do? If the angle of the rope is 60.0°, the force • work: A force is said to do work when it acts on a body so that there is a displacement of the point of application, however small, in the direction of the force. Thus a force does work when it results in movement. The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product, W = Fd. • joule: The SI unit of work or energy, equal to the work done by a force of one Newton when its point of application moves through a distance of one meter in the direction of the force. References 1. Image copyright Paul Brennan, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3 CONCEPT 2 • Deﬁne power. • Given two of the three variables in P = W t, calculate the third. www.ck12.org Power Typical Pressurized Water Reactors (PWR) reactors built in the 1970’s produce about 1100 megawatts, whilst the latest designs range up to around 1500 megawatts. That is 1,500,000,000 Joules/second. A windmill farm, by comparison, using hundreds of individual windmills produces about 5 megawatts. That is 5,000,000 Joules/second (assuming the wind is blowing). Power Power is deﬁned as the rate at which work is done or the rate at which energy is transformed. In SI units
, power is measured in Joules per second which is given a special name, the watt, W. Power = Work Time 1.00 watt = 1.00 J/s 4 www.ck12.org Concept 2. Power Another unit for power that is fairly common is horsepower. 1.00 horsepower = 746 watts Example Problem: A 70.0 kg man runs up a long ﬂight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. Calculate the power output of the man in watts and horsepower. Solution: The force exerted must be equal to the weight of the man = mg = (70:0 kg)(9:80 m/s2) = 686 N W = Fd = (686 N)(4:5 m) = 3090 N m = 3090 J 4:0 s = 770 J/s = 770 W P = W t = 3090 J P = 770 W = 1:03 hp Since P = W that is produced by the power. t = F d P = W t = Fd t = Fv t and W = Fd, we can use these formulas to derive a formula relating power to the speed of the object The velocity in this formula is the average speed of the object during the time interval. Example Problem: Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h. Solution: 80. km/h = 22.2 m/s In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) = (22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car = (1400 kg)(9.80 m/s 2 ) = 13720 N. W = Fd = (13720 N)(3:86 m) = 53; 000 J Since this work was done in 1.00 second, the power would be 53,000 W. If we calculated the upward component of the velocity of the car, we would divide the distance traveled in one second by one second and get an average vertical speed of 3.86 m/s. So we could use the formula relating power to average speed to calculate power. P = Fv = (13720 N)(3:86 m/s)
= 53; 000 W Summary • Power is deﬁned as the rate at which work is done or the rate at which energy is transformed. • Power = Work Time • Power = Force velocity Practice In the following video, Mr. Edmond sings about work and power. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=5EsMmdaYClQ 5 www.ck12.org MEDIA Click image to the left for more content. 1. What units are used for work? 2. What units are used for power? The following website has practice problems on work and power. http://www.angelﬁre.com/sciﬁ/dschlott/workpp.html Review 1. If the circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N, how much work does the centripetal force do on the toy when it follows its orbit for one cycle? 2. A 50.0 kg woman climbs a ﬂight of stairs 6.00 m high in 15.0 s. How much power does she use? 3. (a) Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5° incline? (b) What power would be needed for the same problem if friction is considered and the coefﬁcient of friction for the car is 0.30? • power: The rate at which this work is performed. References 1. Courtesy of Ryan Hagerty, U.S. Fish and Wildlife Service. http://digitalmedia.fws.gov/cdm/singleitem/coll ection/natdiglib/id/13455/rec/2. Public Domain 6 www.ck12.org Concept 3. Work-Energy Principle CONCEPT 3 Work-Energy Principle The reason the concept of work is so useful is because of a theorem, called the work-energy principle, which states that the change in an object’s kinetic energy is equal to the net work done on it : DKe = Wnet [2] Although we cannot derive this principle in general, we can do it for the case that interests us most: constant acceleration. In the following derivation, we assume that the force is along motion. This doesn’t reduce the generality of the result, but makes the
derivation more tractable because we don’t need to worry about vectors or angles. Recall that an object’s kinetic energy is given by the formula: Ke = 1 2 mv2 [3] Consider an object of mass m accelerated from a velocity vi to v f under a constant force. The change in kinetic energy, according to [2], is equal to: DKe = Kei Ke f = 1 2 mv2 f 1 2 mv2 i = 1 2 m(v2 f v2 i ) [4] Now let’s see how much work this took. To ﬁnd this, we need to ﬁnd the distance such an object will travel under these conditions. We can do this by using the third of our ’Big three’ equations, namely: alternatively, v f 2 = vi 2 + 2aDx [5] Dx = v f 2 2 vi 2a [6] Plugging in [6] and Newton’s Third Law, F = ma, into [2], we ﬁnd: W = FDx = ma v f 2 2 vi 2a which was our result in [4]. Using the Work-Energy Principle = 1 2 m(v2 f v2 i ) [7], The Work-Energy Principle can be used to derive a variety of useful results. Consider, for instance, an object dropped a height Dh under the inﬂuence of gravity. This object will experience constant acceleration. Therefore, we can again use equation [6], substituting gravity for acceleration and Dh for distance: 7 www.ck12.org multiplying both sides by $mg$, we ﬁnd: Dh = v f 2 2 vi 2g mgDh = mg v f 2 2 vi 2g = DKe [8] In other words, the work performed on the object by gravity in this case is mgDh. We refer to this quantity as gravitational potential energy; here, we have derived it as a function of height. For most forces (exceptions are friction, air resistance, and other forces that convert energy into heat), potential energy can be understood as the ability to perform work. Spring Force A spring with spring constant k a distance Dx from equilibrium experiences a restorative force equal to: This is a force that can change an object’s kinetic energy, and therefore do work. So, it has a potential energy associated with it as well.
This quantity is given by: Fs = kDx [9] Esp = 1 2 kDx2 [10] Spring Potential Energy The derivation of [10] is left to the reader. Hint: ﬁnd the average force an object experiences while moving from x = 0 to x = Dx while attached to a spring. The net work is then this force times the displacement. Since this quantity (work) must equal to the change in the object’s kinetic energy, it is also equal to the potential energy of the spring. This derivation is very similar to the derivation of the kinematics equations — look those up. This applet may be useful in reviewing Spring Potential Energy: http://phet.colorado.edu/en/simulation/mass-spring-lab 8 Physics Unit 8 Momentum Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science CONTRIBUTORS Chris Addiego Alexander Katsis Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org
/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Momentum 2 Momentum 3 Law of Conservation of Momentum 4 Conservation of Momentum in One Dimension 5 Conservation of Momentum in Two Dimensions 6 Inelastic Collisions 7 Elastic Collisions 8 Momentum and Impulse 1 4 7 10 14 18 22 26 iv www.ck12.org Concept 1. Momentum CONCEPT 1 • Deﬁne momentum. • Relate momentum to mass and velocity. • Calculate momentum from mass and velocity. Momentum Cody seems a little reluctant to launch himself down this ramp at Newton’s Skate Park. It will be his ﬁrst time down the ramp, and he knows from watching his older brother Jerod that he’ll be moving fast by the time he gets to the bottom. The faster he goes, the harder it will be to stop. That’s because of momentum. What Is Momentum? Momentum is a property of a moving object that makes it hard to stop. The more mass it has or the faster it’s moving, the greater its momentum. Momentum equals mass times velocity and is represented by the equation: 1 www.ck12.org Momentum = Mass x Velocity Q : What is Cody’s momentum as he stands at the top of the ramp? A : Cody has no momentum as he stands there because he isn’t moving. However, Cody will gain momentum as he starts moving down the ramp and picks up speed. In other words, his velocity is zero. Q : Cody’s older brother Jerod is pictured 1.1. If Jerod were to travel down the ramp at the same velocity as Cody, who would have greater momentum? Who would be harder to stop? A : Jerod obviously has greater mass than
Cody, so he would have greater momentum. He would also be harder to stop. FIGURE 1.1 You can see an animation demonstrating the role of mass and velocity in the momentum of moving objects at this URL: http://www.science-animations.com/support-files/momentum.swf Calculating Momentum To calculate momentum with the equation above, mass is measured in (kg), and velocity is measured in meters per second (m/s). For example, Cody and his skateboard have a combined mass of 40 kg. If Cody is traveling at a velocity of 1.1 m/s by the time he reaches the bottom of the ramp, then his momentum is: Momentum = 40 kg x 1.1 m/s = 44 kg m/s Note that the SI unit for momentum is kg m/s. Q : The combined mass of Jerod and his skateboard is 68 kg. If Jerod goes down the ramp at the same velocity as Cody, what is his momentum at the bottom of the ramp? A : His momentum is: Momentum = 68 kg x 1.1 m/s = 75 kg m/s Summary • Momentum is a property of a moving object that makes it hard to stop. It equals the object’s mass times its velocity. 2 www.ck12.org Concept 1. Momentum • To calculate the momentum of a moving object, multiply its mass in kilograms (kg) by its velocity in meters per second (m/s). The SI unit of momentum is kg m/s. Vocabulary • momentum : Property of a moving object that makes it hard to stop; equal to the object’s mass times its velocity. Practice At the following URL, review how to calculate momentum, and then solve the problems at the bottom of the Web page. http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegation-848.htm Review 1. Deﬁne momentum. 2. Write the equation for calculating momentum from mass and velocity. 3. What is the SI unit for momentum? 4. Which skateboarder has greater momentum? a. Skateboarder A: mass = 60 kg; velocity = 1.5 m/s b. Skateboarder B: mass = 50 kg; velocity = 2.0 m/s References 1... used under license from Shutterstock 3 CONCEPT 2 www.ck12
.org Momentum Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Key Equations p = mv Momentum is equal to the objects mass multiplied by its velocity pinitial = p ﬁnal The total momentum does not change in closed systems Example 1 A truck with mass 500 kg and originally carrying 200 kg of dirt is rolling forward with the transmission in neutral and shooting out the dirt backwards at 2 m/s (so that the dirt is at relative speed of zero compared with the ground). If the truck is originally moving at 2 m/s, how fast will it be moving after it has shot out all the dirt. You may ignore the effects of friction. To solve this problem we will apply conservation of momentum to the truck when it is full of dirt and when it has dumped all the dirt. Solution mivi = m f v f (mt + md)vi = mtv f start by setting the initial momentum equal to the ﬁnal momentum substitute the mass of the truck plus the mass of the dirt in the truck at the initial and ﬁnal states v f = v f = (mt + md)vi mt (500 kg + 200 kg) 2 m/s 500 kg v f = 2:8 m/s solve for the ﬁnal velocity plug in the numerical values Example 2 John and Bob are standing at rest in middle of a frozen lake so there is no friction between their feet and the ice. Both of them want to get to shore so they simultaneously push off each other in opposite directions. If John’s mass is 50 kg and Bob’s mass is 40 kg and John moving at 5 m/s after pushing off Bob, how fast is Bob moving? 4 www.ck12.org Concept 2. Momentum Solution For this problem, we will apply conservation of momentum to the whole system that includes both John and Bob. Since both of them are at rest to start, we know that the total momentum of the whole system must always be zero. Therefore, we know that the sum of John’s and Bob’s momentum after they push off each other is also zero. We can use this to solve for Bob’s velocity. 0 = m j
v j + mbvb mbvb = m jv j vb = m jv j mb 50 kg 5 m/s 40 kg vb = 6:25 m/s vb = The answer is negative because Bob is traveling in the opposite direction to John. Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. You ﬁnd yourself in the middle of a frozen lake. There is no friction between your feet and the ice of the lake. You need to get home for dinner. Which strategy will work best? a. Press down harder with your shoes as you walk to shore. b. Take off your jacket. Then, throw it in the direction opposite to the shore. c. Wiggle your butt until you start to move in the direction of the shore. d. Call for help from the great Greek god Poseidon. 2. You and your sister are riding skateboards side by side at the same speed. You are holding one end of a rope and she is holding the other. Assume there is no friction between the wheels and the ground. If your sister lets go of the rope, how does your speed change? a. It stays the same. b. It doubles. c. It reduces by half. 5 www.ck12.org 3. You and your sister are riding skateboards (see Problem 3), but now she is riding behind you. You are holding one end of a meter stick and she is holding the other. At an agreed time, you push back on the stick hard enough to get her to stop. What happens to your speed? Choose one. (For the purposes of this problem pretend you and your sister weigh the same amount.) a. It stays the same. b. It doubles. c. It reduces by half. 4. An astronaut is using a drill to ﬁx the gyroscopes on the Hubble telescope. Suddenly, she loses her footing and ﬂoats away from the telescope. What should she do to save herself? 5. A 5:00 kg ﬁrecracker explodes into two parts: one part has a mass of 3:00 kg and moves at a velocity of 25:0 m=s towards the west. The other part has a mass of 2:00 kg. What is the velocity of the second piece as a result of the explosion? 6. A ﬁrecracker lying on the ground explodes, breaking into
two pieces. One piece has twice the mass of the other. What is the ratio of their speeds? 7. While driving in your pickup truck down Highway 280 between San Francisco and Palo Alto, an asteroid lands in your truck bed! Despite its 220 kg mass, the asteroid does not destroy your 1200 kg truck. In fact, it landed perfectly vertically. Before the asteroid hit, you were going 25 m=s. After it hit, how fast were you going? 8. An astronaut is 100 m away from her spaceship doing repairs with a 10:0 kg wrench. The astronaut’s total mass is 90:0 kg and the ship has a mass of 1:00 104 kg. If she throws the wrench in the opposite direction of the spaceship at 10:0 m=s how long would it take for her to reach the ship? Answers to Selected Problems 1.. 2.. 3.. 4.. 5. 37:5 m=s 6. v1 = 2v2 7. 21 m=s 8. a. 90 sec 6 www.ck12.org Concept 3. Law of Conservation of Momentum CONCEPT 3 Law of Conservation of Momentum • State the law of conservation of momentum. • Describe an example of momentum being transferred and conserved. These skaters are racing each other at Newton’s Skate Park. The ﬁrst skater in line, the one on the left, is distracted by something he sees. He starts to slow down without realizing it. The skater behind him isn’t paying attention and keeps skating at the same speed. Q : Can you guess what happens next? A : Skater 2 runs into skater 1. Conserving Momentum When skater 2 runs into skater 1, he’s going faster than skater 1 so he has more momentum. Momentum is a property of a moving object that makes it hard to stop. It’s a product of the object’s mass and velocity. At the moment of the collision, skater 2 transfers some of his momentum to skater 1, who shoots forward when skater 2 runs into him. Whenever an action and reaction such as this occur, momentum is transferred from one object to the other. However, the combined momentum of the objects remains the same. In other words, momentum is conserved. This is the law of conservation of momentum. 7 Modeling Momentum The Figure 3.1 shows how momentum is conserved in the
two colliding skaters. The total momentum is the same after the collision as it was before. However, after the collision, skater 1 has more momentum and skater 2 has less momentum than before. www.ck12.org FIGURE 3.1 Q : What if two skaters have a head-on collision? Do you think momentum is conserved then? A : As in all actions and reactions, momentum is also conserved in a head-on collision. You can see how at this URL: http://www.physicsclassroom.com/mmedia/momentum/cthoi.cfm Summary • Whenever an action and reaction occur, momentum is transferred from one object to the other. However, total momentum is conserved. This is the law of conservation of momentum. 8 www.ck12.org Vocabulary Concept 3. Law of Conservation of Momentum • momentum : Property of a moving object that makes it hard to stop; equal to the object’s mass times its velocity. • law of conservation of momentum : Law stating that, when an action and reaction occur, the combined momentum of the objects remains the same. Practice Watch the astropitch animation at the following URL. Experiment with different velocities. Then take the quiz and check your answers. http://www.phys.utb.edu/~pdukes/standard/PhysApplets/AstroPitch/TabbedastroPitch2.html Review 1. State the law of conservation of momentum. 2. Fill in the missing velocity (x) in the diagram of a vehicle collision seen in the Figure 3.2 so that momentum is conserved. FIGURE 3.2 Solve for x. References 1. Laura Guerin.. CC BY-NC 3.0 2. Laura Guerin.. CC BY-NC 3.0 9 CONCEPT 4 Conservation of Momentum in One Dimension www.ck12.org • State the law of conservation of momentum. • Use the conservation of momentum to solve one-dimensional collision problems. For this whale to leap out of the water, something underwater must be moving in the opposite direction, and intuition tells us it must be moving with relatively high velocity. The water that moves downward is pushed downward by the whale’s tail, and that allows the whale to rise up. Conservation of Momentum in One Dimension When impulse and momentum were introduced, we used an example of a batted ball to discuss the impulse and momentum change that
occurred with the ball. At the time, we did not consider what had happened to the bat. According to Newton’s third law, however, when the bat exerted a force on the ball, the ball also exerted an equal and opposite force on the bat. Since the time of the collision between bat and ball is the same for the bat and for the ball, then we have equal forces (in opposite directions) exerted for equal times on the ball AND the bat. That means that the impulse exerted on the bat is equal and opposite (-Ft) to the impulse on the ball (Ft) and that also means that there was a change in momentum of the bat [D(mv)BAT] that was equal and opposite to the change in momentum of the ball [D(mv)BALL]. The change in momentum of the ball is quite obvious because it changes direction and ﬂies off at greater speed. However, the change in momentum of the bat is not obvious at all. This occurs primarily because the bat is more massive than the ball. Additionally, the bat is held ﬁrmly by the batter, so the batter’s mass can be combined with the mass of the bat. Since the bat’s mass is so much greater than that of the ball, but they have equal and opposite forces, the bat’s ﬁnal velocity is signiﬁcantly smaller than that of the ball. Consider another system: that of two ice skaters. If we have one of the ice skaters exert a force on the other skater, the force is called an internal force because both the object exerting the force and the object receiving the force 10 www.ck12.org Concept 4. Conservation of Momentum in One Dimension are inside the system. In a closed system such as this, momentum is always conserved. The total ﬁnal momentum always equals the total initial momentum in a closed system. Conversely, if we deﬁned a system to contain just one ice skater, putting the other skater outside the system, this is not a closed system. If one skater pushes the other, the force is an external force because the receiver of the force is outside the system. Momentum is not guaranteed to be conserved unless the system is closed. In a closed system, momentum is always conserved. Take another example: if we consider two billiard balls colliding on a billiard table
and ignore friction, we are dealing with a closed system. The momentum of ball A before the collision plus the momentum of ball B before collision will equal the momentum of ball A after collision plus the momentum of ball B after collision. This is called the law of conservation of momentum and is given by the equation pAbefore + pBbefore = pAafter + pBafter Example Problem: Ball A has a mass of 2.0 kg and is moving due west with a velocity of 2.0 m/s while ball B has a mass of 4.0 kg and is moving west with a velocity of 1.0 m/s. Ball A overtakes ball B and collides with it from behind. After the collision, ball A is moving westward with a velocity of 1.0 m/s. What is the velocity of ball B after the collision? Solution: Because of the law of conservation of momentum, we know that pAbefore + pBbefore = pAafter + pBafter. mAvA + mBvB = mAv0 A + mBv0 B (2:0 kg)(2:0 m/s) + (4:0 kg)(1:0 m/s) = (2:0 kg)(1:0 m/s) + (4:0 kg)(vB 0 m/s) 4:0 kg m/s + 4:0 kg m/s = 2:0 kg m/s + 4vB 0 kg m/s 4vB 0 = 8:0 2:0 = 6:0 0 = 1:5 m/s vB After the collision, ball B is moving westward at 1.5 m/s. Example Problem: A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars? Solution: Note that since the two trains stick together, the ﬁnal mass is m A +m B, and the ﬁnal velocity for each object is the same. Thus the conservation of momentum equation, mAvA + mBvB = mAv0 B, can be A + mBv0 rewritten mAvA + mBvB = (mA + mB) v
0 (30; 000: kg)(2:2 m/s) + (30; 000: kg)(0 m/s) = (60; 000: kg)(v0 m/s) 66000 + 0 = 60000v0 v0 = 66000 60000 = 1:1 m/s After the collision, the two cars move off together toward the east with a velocity of 1.1 m/s. Summary • A closed system is one in which both the object exerting a force and the object receiving the force are inside the system. • In a closed system, momentum is always conserved. 11 Practice www.ck12.org The following video shows the Mythbusters building and using various sizes of a toy called "Newton’s Cradle." Use this resource to answer the question that follows. https://www.youtube.com/watch?v=BiLq5Gnpo8Q MEDIA Click image to the left for more content. 1. What is Newton’s Cradle? 2. How does Newton’s Cradle work? 3. How does a Newton’s Cradle show conservation of momentum? If you are interested in learning more about Newton’s Cradles, visit this site: http://science.howstuffworks.com/ne wtons-cradle.htm Review 1. A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. With what speed does the goalie slide on the (frictionless) ice? 2. A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet ﬂy off together at 9.0 m/s. What was the original velocity of the bullet? 3. A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s. After the collision, the 0.50 kg ball moves away at -14 m/s. Find the velocity of the second ball after the collision. 4. Two carts are stationary with a compressed spring between them and held together by a thread. When the thread is cut, the two carts move apart. After the spring is released, one cart m = 3:00 kg has a velocity of 0.82 m/s east. What is the magnitude of
the velocity of the second cart (m = 1:70 kg) after the spring is released? 5. Compared to falling on a tile ﬂoor, a glass may not break if it falls onto a carpeted ﬂoor. This is because a. less impulse in stopping. b. longer time to stop. c. both of these d. neither of these. 6. A butterﬂy is hit by a garbage truck on the highway. The force of the impact is greater on the 12 www.ck12.org Concept 4. Conservation of Momentum in One Dimension a. garbage truck. b. butterﬂy. c. it is the same for both. 7. A riﬂe recoils from ﬁring a bullet. The speed of the riﬂe’s recoil is small compared to the speed of the bullet because a. the force on the riﬂe is small. b. the riﬂe has a great deal more mass than the bullet. c. the momentum of the riﬂe is unchanged. d. the impulse on the riﬂe is less than the impulse on the bullet. e. none of these. • Law of Conservation of Momentum: The total linear momentum of an isolated system remains constant regardless of changes within the system. References 1. Courtesy of NOAA. http://commons.wikimedia.org/wiki/File:Humpback_whale_noaa.jpg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 13 CONCEPT 5 Conservation of Momentum in Two Dimensions www.ck12.org • Use the conservation of momentum and vector analysis to solve two-dimensional collision problems. • Review vector components. In a game of billiards, it is important to be able to visualize collisions in two dimensions – the best players not only know where the target ball is going but also where the cue ball will end up. Conservation of Momentum in Two Dimensions Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide. Most objects are not conﬁned to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated; momentum is a vector
and collisions of objects in two dimensions can be represented by axial vector components. To review axial components, revisit Vectors: Resolving Vectors into Axial Components and Vectors: Vector Addition. Example Problem: A 2.0 kg ball, A, is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B, also with a mass of 2.0 kg. After the collision, ball A moves off at 30° south of west while ball B moves off at 60° north of west. Find the velocities of both balls after the collision. Solution: Since ball B is stationary before the collision, then the total momentum before the collision is equal to momentum of ball A. The momentum of ball A before collision is shown in red below, and can be calculated to be p = mv = (2:00 kg)(5:00 m/s) = 10:0 kg m/s west 14 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions Since momentum is conserved in this collision, the sum of the momenta of balls A and B after collsion must be 10.0 kg m/s west. pAafter = (10:0 kg m/s)(cos 30) = (10:0 kg m/s)(0:866) = 8:66 kg m/s pBafter = (10:0 kg m/s)(cos 60) = (10:0 kg m/s)(0:500) = 5:00 kg m/s To ﬁnd the ﬁnal velocities of the two balls, we divide the momentum of each by its mass. Therefore, vA = 4:3 m/s and vB = 2:5 m/s. Example Problem: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. The two cars stick together after the collision. What is the speed and direction of the two cars after the collision? Solution: Northward momentum = (1325 kg)(27:0 m/s) = 35800 kg m/s Eastward momentum = (2165 kg)(17:0 m/s) = 36800 kg m/s q (35800)2 + (36800)2 = 51400 kg m/s R = 15
www.ck12.org q = sin1 35800 m = 51400 kgm/s 3490 kg velocity = p 51400 = 44 north of east = 14:7 m/s @ 44 N of E Example Problem: A 6.00 kg ball, A, moving at velocity 3.00 m/s due east collides with a 6.00 kg ball, B, at rest. After the collision, A moves off at 40.0° N of E and ball B moves off at 50.0° S of E. a. What is the momentum of A after the collision? b. What is the momentum of B after the collision? c. What are the velocities of the two balls after the collision? Solution: pinitial = mv = (6:00 kg)(3:00 m/s) = 18:0 kg m/s This is a right triangle in which the initial momentum is the length of the hypotenuse and the two momenta after the collision are the legs of the triangle. a. pA = (18:0 kg m/s)(cos 40:0) = (18:0 kg m/s)(0:766) = 13:8 kg m/s b. pB = (18:0 kg m/s)(cos 50:0) = (18:0 kg m/s)(0:643) = 11:6 kg m/s c. vA = 2:30 m/s vB = 1:93 m/s • The conservation of momentum law holds for all closed systems regardless of the directions of the objects before and after they collide. • Momentum is a vector; collisions in two dimensions can be represented by axial vector components. Summary Practice This video shows circus performers using conservation of momentum. Use this resource to answer the questions that follow. http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ 16 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions MEDIA Click image to the left for more content. 1. Why do the ﬂiers scrunch up in the air while spinning and twisting? 2. What happens to the rate at which they spin when they change shape in the air? Review 1. Billiard ball A, mass 0.17 kg, moving due east with a velocity of 4.0 m/s, strikes
stationary billiard ball B, also mass of 0.17 kg. After the collision, ball A moves off at an angle of 30° north of east with a velocity of 3.5 m/s, and ball B moves off at an angle of 60 ° south of east. What is the speed of ball B? 2. A bomb, originally sitting at rest, explodes and during the explosion breaks into four pieces of exactly 0.25 kg each. One piece ﬂies due south at 10 m/s while another pieces ﬂies due north at 10 m/s. (a) What do we know about the directions of the other two pieces and how do we know it? (b) What do we know about the speeds of the other two pieces and how do we know it? 3. In a head-on collision between protons in a particle accelerator, three resultant particles were observed. All three of the resultant particles were moving to the right from the point of collision. The physicists conducting the experiment concluded there was at least one unseen particle moving to the left after the collision. Why did they conclude this? References 1. Image copyright VitCOM Photo, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 17 CONCEPT 6 Inelastic Collisions Students will learn how to solve problems involving inelastic collisions. www.ck12.org Key Equations pinitial = p ﬁnal The total momentum does not change in closed systems Example 1 Question : Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed of 2:5m=s. Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial speed of block B? Solution : To ﬁnd mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg and the mass of block A is 8.0kg. 10kg 8:0kg = 2:0kg Now that we know the mass of both blocks we can ﬁnd the speed of block
B. We will use conservation of momentum. This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This problem is one dimensional, because all motion happens along the same line. Thus we will use the equation (mA + mB)v f = mA vA + mB vB and solve for the velocity of block B. (mA + mB)v f = mA vA + mBvB ) (mA + mB)(v f ) (mA)(vA) mB = vB Watch this Explanation 18 www.ck12.org Concept 6. Inelastic Collisions MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 0% for perfectly inelastic collisions. Collision Lab (PhET Simulation) 19 www.ck12.org Car Collision (CK-12 Simulation) Time for Practice 2. Two blocks collide on a frictionless surface, as shown. Afterwards, they have a combined mass of 10 kg and a speed of 2:5 m=s. Before the collision, one of the blocks was at rest. This block had a mass of 8:0 kg. What was the mass and initial speed of the second block? 1. 3. 4. In the above picture, the carts are moving on a level, frictionless track. After the collision all three carts stick together. Determine the direction and speed of the combined carts after the collision. 5. The train engine and its four boxcars are coasting at 40 m=s. The engine train has mass of 5; 500 kg and the boxcars have masses, from left to right, of 1; 000 kg, 1; 500 kg, 2; 000 kg; and 3; 000 kg. (For this problem, you may neglect the small external forces of friction and air resistance.) 20 www.ck12.org Concept 6. Inelastic Collisions a. What happens to the speed of the train when it releases the last boxcar? ( Hint: Think before you blindly calculate. ) b. If the train can shoot boxcars backwards at 30 m=s relative to the train’s speed, how many boxcars does the train need to shoot out in order to obtain a speed of 58:75 m=s? 6. In Sacramento a 4000 kg SUV is traveling 30 m=s south on Tru
xel crashes into an empty school bus, 7000 kg traveling east on San Juan. The collision is perfectly inelastic. a. Find the velocity of the wreck just after collision b. Find the direction in which the wreck initially moves 7. Manrico (80:0 kg) and Leonora (60:0 kg) are ﬁgure skaters. They are moving toward each other. Manrico’s speed is 2:00 m=s ; Leonora’s speed is 4:00 m=s. When they meet, Leonora ﬂies into Manrico’s arms. a. With what speed does the entwined couple move? b. In which direction are they moving? c. How much kinetic energy is lost in the collision? Answers to Selected Problems 1. 2:0 kg; 12:5 m=s 2. 0:13 m=s to the left 3. a. no change b. the last two cars 4. a. 15 m=s b. 49 S of E 5. a. 0.57 m/s b. the direction Leonora was originally travelling c. 297.26 J 21 CONCEPT 7 www.ck12.org Elastic Collisions Students will learn how to solve problems involving elastic collisions Key Equations pinitial = p ﬁnal The total momentum does not change in closed systems KEinitial = KE ﬁnal The total kinetic energy does not change in elastic collisions Example 1 MEDIA Click image to the left for more content. Example 2 Question : Chris and Ashley are playing pool. Ashley hits the cue ball into the 8 ball with a velocity of 1:2m=s. The cue ball (c) and the 8 ball ( e ) react as shown in the diagram. The 8 ball and the cue ball both have a mass of :17kg. What is the velocity of the cue ball? What is the direction (the angle) of the cue ball? Answer : We know the equation for conservation of momentum, along with the masses of the objects in question as well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the velocity of the cue ball. mcvic + mevie = mcv f c + mev f e v f c = v f c = mcvic + me
vie mev f e mc :17kg 2:0m=s + :17kg 0m=s :17kg 1:2m=s :17kg v f c = :80m=s Now we want to ﬁnd the direction of the cue ball. To do this we will use the diagram below. 22 www.ck12.org Concept 7. Elastic Collisions We know that the momentum in the y direction of the two balls is equal. Therefore we can say that the velocity in the y direction is also equal because the masses of the two balls are equal. mcvcy = mevey! vcy = vey Given this and the diagram, we can ﬁnd the direction of the cue ball. After 1 second, the 8 ball will have traveled 1:2m. Therefore we can ﬁnd the distance it has traveled in the y direction. sin25o = opposite hypotenuse = x 1:2m! x = sin25 1:2m = :51m Therefore, in one second the cue ball will have traveled :51m in the y direction as well. We also know how far in total the cue ball travels in one second (:80m). Thus we can ﬁnd the direction of the cue ball. sin1 opposite hypotenuse = sin1 :51m :80m = 40o Watch this Explanation MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 100% for perfectly elastic collisions. 23 www.ck12.org Collision Lab (PhET Simulation) Time for Practice 1. You are playing pool and you hit the cue ball with a speed of 2 m=s at the 8 -ball (which is stationary). Assume an elastic collision and that both balls are the same mass. Find the speed and direction of both balls after the collision, assuming neither ﬂies off at any angle. 2. A 0:045 kg golf ball with a speed of 42:0 m=s collides elastically head-on with a 0:17 kg pool ball at rest. Find the speed and direction of both balls after the collision. 3. Ball A is traveling along a ﬂat table with a speed of 5:0 m=s, as shown below. Ball B, which has the same mass, is initially at rest, but is knocked off the table in an elastic collision with Ball A. Find the horizontal distance
that Ball B travels before hitting the ﬂoor. 4. Students are doing an experiment on the lab table. A steel ball is rolled down a small ramp and allowed to hit the ﬂoor. Its impact point is carefully marked. Next a second ball of the same mass is put upon a set screw and a collision takes place such that both balls go off at an angle and hit the ﬂoor. All measurements are taken with a meter stick on the ﬂoor with a co-ordinate system such that just below the impact point is the origin. The following data is collected: (a) no collision: 41:2 cm (b) target ball: 37:3 cm in the direction of motion and 14:1 cm perpendicular to the direction of motion 24 www.ck12.org Concept 7. Elastic Collisions i. From this data predict the impact position of the other ball. ii. One of the lab groups declares that the data on the ﬂoor alone demonstrate to a 2 % accuracy that the collision was elastic. Show their reasoning. iii. Another lab group says they can’t make that determination without knowing the velocity the balls have on impact. They ask for a timer. The instructor says you don’t need one; use your meter stick. Explain. iv. Design an experiment to prove momentum conservation with balls of different masses, giving apparatus, procedure and design. Give some sample numbers. 5. A 3 kg ball is moving 2 m/s in the positive x direction when it is struck dead center by a 2 kg ball moving in the positive y direction at 1 m/s. After collision the 3 kg ball moves at 3 m/s 30 degrees from the positive x axis. Find the velocity and direction of the 2 kg ball. Answers to Selected Problems 1. 8 m=s same direction as the cue ball and 0 m=s 2. vgol f = 24:5 m=s; vpool = 17:6 m=s 3. 2:8 m 4.. 5. 1:5 m=s 54 25 CONCEPT 8 Momentum and Impulse www.ck12.org • Deﬁne momentum. • Deﬁne impulse. • Given mass and velocity of an object, calculate momentum. • Calculate the change in momentum of an object. • State the relationship that exists between the change in momentum and impulse. • Using the momentum-impulse theorem and given three of the four variables
, calculate the fourth. Rachel Flatt performs a layback spin at the 2 011 Rostelecom Cup in Moscow, Russia. When an ice skater spins, angular momentum must be conserved. When her arms or feet are far away from her body, her spin slows; when she brings her arms and feet close in to her body, she spins faster. Momentum and Impulse If a bowling ball and a ping-pong ball are each moving with a velocity of 5 mph, you intuitively understand that it will require more effort to stop the bowling ball than the ping pong ball because of the greater mass of the bowling ball. Similarly, if you have two bowling balls, one moving at 5 mph and the other moving at 10 mph, you know it 26 www.ck12.org Concept 8. Momentum and Impulse will take more effort to stop the ball with the greater speed. It is clear that both the mass and the velocity of a moving object contribute to what is necessary to change the motion of the moving object. The product of the mass and velocity of an object is called its momentum. Momentum is a vector quantity that has the same direction as the velocity of the object and is represented by a lowercase letter p. The momentum of a 0.500 kg ball moving with a velocity of 15.0 m/s will be p = mv p = mv = (0:500 kg)(15:0 m/s) = 7:50 kg m/s You should note that the units for momentum are kg·m/s. According to Newton’s ﬁrst law, the velocity of an object cannot change unless a force is applied. If we wish to change the momentum of a body, we must apply a force. The longer the force is applied, the greater the change in momentum. The impulse is the quantity deﬁned as the force multiplied by the time it is applied. It is a vector quantity that has the same direction as the force. The units for impulse are N·s but we know that Newtons are also kg·m/s 2 and so N·s = (kg·m/s 2 )(s) = kg·m/s. Impulse and momentum have the same units; when an impulse is applied to an object, the momentum of the object changes and the change of momentum is equal to the impulse. Ft = Dmv Example Problem: Calculating Momentum A 0.15 kg
ball is moving with a velocity of 35 m/s. Find the momentum of the ball. Solution: p = mv = (0:15 kg)(35 m/s) = 5:25 kg m/s Example Problem: If a ball with mass 5.00 kg has a momentum of 5:25 kg m/s, what is its velocity? m = 5:25 kgm/s 5:00 kg It should be clear from the equation relating impulse to change in momentum, Ft = Dmv, that any amount of force would (eventually) bring a moving object to rest. If the force is very small, it must be applied for a long time, but a greater force can bring the object to rest in a shorter period of time. Solution: v = p = 1:05 m/s If you jump off a porch and land on your feet with your knees locked in the straight position, your motion would be brought to rest in a very short period of time and thus the force would need to be very large – large enough, perhaps, to damage your joints or bones. Suppose that when you hit the ground, your velocity was 7.0 m/s and that velocity was brought to rest in 0.05 seconds. If your mass is 100. kg, what force was required to bring you to rest? 0:050 s If, on the other hand, when your feet ﬁrst touched the ground, you allowed your knees to ﬂex so that the period of time over which your body was brought to rest is increased, then the force on your body would be smaller and it would be less likely that you would damage your legs. F = Dmv t = (100: kg)(7:0 m/s) = 14; 000 N Suppose that when you ﬁrst touch the ground, you allow your knees to bend and extend the stopping time to 0.50 seconds. What force would be required to bring you to rest this time? t = (100: kg)(7:0 m/s) 0:50 s With the longer period of time for the force to act, the necessary force is reduced to one-tenth of what was needed before. = 1400 N F = Dmv Extending the period of time over which a force acts in order to lessen the force is a common practice in design. Padding in shoes and seats allows the time to increase. The front of automobiles are designed to crump
le in an accident; this increases the time the car takes to stop. Similarly, barrels of water or sand in front of abutments on 27 www.ck12.org the highway and airbags serve to slow down the stoppage time. These changes all serve to decrease the amount of force it takes to stop the momentum in a car crash, which consequently saves lives. Example Problem: An 0.15 kg baseball is thrown horizontally at 40. m/s and after it is struck by a bat, it is traveling at -40. m/s. (a) What impulse did the bat deliver to the ball? (b) If the contact time of the bat and bat was 0.00080 seconds, what was the average force the bat exerted on the ball? (c) Calculate the average acceleration of the ball during the time it was in contact with the bat. Solution: We can calculate the change in momentum and give the answer as impulse because we know that the impulse is equal to the change in momentum. (a) p = mDv = (0:15 kg)(40: m/s 40: m/s) = (0:15 kg)(80: m/s) = 12 kg m/s The minus sign indicates that the impulse was in the opposite direction of the original throw. t = 12 kgm/s Again, the negative sign indicates the force was in the opposite direction of the original throw. 0:00080 s = 15000 N (b) F = Dmv (c) a = F = 100; 000 m/s2 m = 15000 N 0:15 kg Summary • The product of the mass and velocity of an object is called momentum, given by the equation r = mv. • Momentum is a vector quantity that has the same direction as the velocity of the object. • The quantity of force multiplied by the time it is applied is called impulse. • Impulse is a vector quantity that has the same direction as the force. • Momentum and impulse have the same units: kg·m/s. • The change of momentum of an object is equal to the impulse. Ft = Dmv Practice Use this resource to answer the question that follows. https://www.youtube.com/watch?v=3g4v8x7xggU MEDIA Click image to the left for more content. 1. Why don’t the glasses of water spill when the tablecloth is pulled out from
under them? 2. How does the video get from momentum to impulse? 28 www.ck12.org Concept 8. Momentum and Impulse Review 1. A small car with a mass of 800. kg is moving with a velocity of 27.8 m/s. (a) What is the momentum of the car? (b) What velocity is needed for a 2400. kg car in order to have the same momentum? 2. A scooter has a mass of 250. kg. A constant force is exerted on it for 60.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 28.0 m/s. (a) What is the change in momentum? (b) What is the magnitude of the force exerted on the scooter? 3. The brakes on a 15,680 N car exert a stopping force of 640. N. The car’s velocity changes from 20.0 m/s to 0 m/s. (a) What is the car’s mass? (b) What was its initial momentum? (c) What was the change in momentum for the car? (d) How long does it take the braking force to bring the car to rest? • momentum: A measure of the motion of a body equal to the product of its mass and velocity. Also called linear momentum. • impulse: The product obtained by multiplying the average value of a force by the time during which it acts. The impulse equals the change in momentum produced by the force in this time interval. References 1. User:deerstop/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Flatt-3.jpg. Public Domain 29 Physics Unit 9: Circular Motion Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both
as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 5 10 Contents 1 Circular Motion 2 Circular Motion 3 Centripetal Force iv www.ck12.org Concept 1. Circular Motion CONCEPT 1 Circular Motion • Deﬁne centripetal acceleration. • Understand the theory of the centripetal acceleration equation. • Use the centripetal acceleration equation. • Understand the angular relationship between velocity and centripetal acceleration. • Use the equations for motion in two directions and Newton’s Laws to analyze circular motion. Weather satellites, like the one shown above, are found miles above the earth’s surface. Satellites can be polar orbiting, meaning they cover the entire Earth asynchronously, or geostationary, in which they hover over the same spot on the equator. Circular Motion The earth is a sphere. If you draw a horizontal straight line from a point on the surface of the earth, the surface of the earth drops away from the line. The distance that the earth drops away from the horizontal line is very small – so small, in
fact, that we cannot represent it well in a drawing. In the sketch below, if the blue line is 1600 m, the amount of drop (the red line) would be 0.20 m. If the sketch were drawn to scale, the red line would be too short to see. 1 www.ck12.org When an object is launched exactly horizontally in projectile motion, it travels some distance horizontally before it strikes the ground. In the present discussion, we wish to imagine a projectile ﬁred horizontally on the surface of the earth such that while traveling 1600 m horizontally, the object would fall exactly 0.20 m. If this could occur, then the object would fall exactly the amount necessary during its horizontal motion to remain at the surface of the earth, but not touching it. In such a case, the object would travel all the way around the earth continuously and circle the earth, assuming there were no obstacles, such as mountains. What initial horizontal velocity would be necessary for this to occur? We ﬁrst calculate the time to fall the 0.20 m: s t = r 2d a = (2)(0:20 m) 9:80 m/s2 = 0:20 s The horizontal velocity necessary to travel 1600 m in 0.20 s is 8000 m/s. Thus, the necessary initial horizontal velocity is 8000 m/s. In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration. In the case of satellites orbiting the earth, the centripetal acceleration is caused by gravity. If you were swinging an object around your head on a string, the centripetal acceleration would be caused by your hand pulling on the string toward the center of the circle. It is important to note that the object traveling in a circle has a constant speed but does not have a constant velocity. This is because direction is part of velocity; when an object changes its direction, it is changing its velocity. Hence the object’s acceleration. The acceleration in the case of uniform circular motion is the change in the direction of the velocity, but not its magnitude. For an object traveling in a circular path, the centripetal acceleration is directly related to the square of the velocity of the object and inversely related to the radius of the circle. ac = v2 r Taking a moment to consider the validity of this equation can help to clarify what it means.
Imagine a yo-yo. Instead of using it normally, let it fall to the end of the string, and then spin it around above your head. If we were to increase the speed at which we rotate our hand, we increase the velocity of the yo-yo - it is spinning faster. As it spins faster, it also changes direction faster. The acceleration increases. Now let’s think about the bottom of the equation: the radius. If we halve the length of the yo-yo string (bring the yo-yo closer to us), we make the yo-yo’s velocity greater. Again, it moves faster, which increases the acceleration. If we make the string longer again, this decreases the acceleration. We now understand why the relationship between the radius and the acceleration is an inverse relationship - as we decrease the radius, the acceleration increases, and visa versa. 2 www.ck12.org Concept 1. Circular Motion Example Problem: A ball at the end of a string is swinging in a horizontal circle of radius 1.15 m. The ball makes exactly 2.00 revolutions per second. What is its centripetal acceleration? Solution: We ﬁrst determine the velocity of the ball using the facts that the circumference of the circle is 2pr and the ball goes around exactly twice per second. v = (2)(2pr) = 14:4 m/s t = (2)(2)(3:14)(1:15 m) 1:00 s r = (14:4 m/s)2 We then use the velocity and radius in the centripetal acceleration equation. ac = v2 Example Problem: The moon’s nearly circular orbit around the earth has a radius of about 385,000 km and a period of 27.3 days. Calculate the acceleration of the moon toward the earth. 1:15 m = 180: m/s2 Solution: v = 2pr T = (2)(3:14)(3:85108 m) r = (1020 m/s)2 (27:3 d)(24:0 h/d)(3600 s/h) 3:85108 m = 0:00273 m/s2 = 1020 m/s ac = v2 As shown in the previous example, the velocity of an object traveling in a circle can be calculated by v = 2pr T 3 Where r is the radius of the circle and T is the period (time required for one
revolution). This equation can be incorporated into the equation for centripetal acceleration as shown below. www.ck12.org ac = v2 r = ( 2pr T )2 r = 4p2r T 2 Summary • In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration. • The acceleration in the case of uniform circular motion changes the direction of the velocity but not its magnitude. • Formulas for centripetal acceleration are ac = v2 r and ac = 4p2r T 2. Practice This video is a demonstration of centripetal force using balloons and trays of water. Use this resource to answer the questions that follow. https://www.youtube.com/watch?v=EX5DZ2MHlV4 MEDIA Click image to the left for more content. 1. What does centripetal mean? 2. What is uniform circular motion? 3. Why is centripetal acceleration always towards the center? Review 1. An automobile rounds a curve of radius 50.0 m on a ﬂat road at a speed of 14 m/s. What centripetal acceleration is necessary to keep the car on the curve? 2. An object is swung in a horizontal circle on a length of string that is 0.93 m long. If the object goes around once in 1.18 s, what is the centripetal acceleration? • circular motion: A movement of an object along the circumference of a circle or rotation along a circular path. • centripetal acceleration: The acceleration toward the center that keeps an object following a circular path. References 1. Greg Goebel. http://www.public-domain-image.com/space-public-domain-images-pictures/weather-satellit e.jpg.html. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons.. CC-BY-NC-SA 3.0 4 www.ck12.org Concept 2. Circular Motion CONCEPT 2 Circular Motion Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle deﬁned by the objects motion. This force changes the direction of the velocity vector of the object but not
the speed. Students will also learn how to calculate that speed using the period of motion and the distance of its path (circumference of the circle it traces out). Vocabulary • centripetal acceleration: The inward acceleration that keeps an object in circular motion. • centripetal force: The inward force that keeps an object in circular motion. Introduction A satellite orbits around the Earth in Figure below. A car travels around a curve in Figure below. All of these objects are engaged in circular motion. Let us consider the satellite ﬁrst. The satellite is held in place by the Earth’s gravity. The gravity holds the satellite in its orbit. In what direction does this force act? If the earth were “magically” gone, the satellite would ﬂy off tangent to its motion at the instant gravity no longer held it. The force preventing this from happening must keep pulling the satellite toward the center of the circle to maintain circular motion. FIGURE 2.1 What is the force in Figure above that prevents the car from skidding off the road? If you guessed “the friction between the tires and the road” you’d be correct. But is it static or kinetic friction? Unless the tires skid, there can be no kinetic friction. It is static friction that prevents the tires from skidding, just as it is static friction that permits you to walk without slipping. In Figure below, you can see the foot of a person who walks toward the right by pushing their foot backward with a horizontal component of force F. They move forward because the ground exerts a horizontal component force fs in the opposite direction. (Note that vertical forces are ignored.) The force the ground exerts on the person’s foot is a static friction force. Because the foot does not slide, we know that F and fs are equal opposed forces. We can easily see which direction the static friction force must act when we walk, but what about a car performing circular motion? In what direction does the static friction act on the car in Figure above? 5 www.ck12.org FIGURE 2.2 FIGURE 2.3 Figure below shows the top view of a car moving around a circular track with a constant speed. Since acceleration is deﬁned as a = Dv, you may be tempted to say that since the speed remains constant, Dv = 0, the acceleration Dt must also be zero. But that conclusion would be incorrect because D
v represents a change in velocity, not a change in speed. The velocity of the car is not constant since it is continuously changing its direction. How then do we ﬁnd the acceleration of the car? Figure below shows the instantaneous velocity vectors for the car in two different positions a very small time apart. Notice that the vector DV points toward the center of the circle. (Recall that DV can be thought of as the sum of the vectors V2 + (V1).) The direction of the acceleration points in the direction of DV since acceleration is deﬁned as a = D~v Dt. This is reasonable, since if there were no force directed toward the center of the circle, the car would move off tangent to the circle. We call the inward force that keeps an object in circular motion a “center seeking”, or centripetal force and the acceleration, centripetal acceleration. The centripetal acceleration is often denoted as ac In order to ﬁnd the correct expression for the magnitude of the centripetal acceleration we’ll need to use a little geometric reasoning. Figure below and Figure below show two “almost” similar triangles. The magnitudes of V1 and V2 are equal, and the change in location of the car occurs over a very small increment in time, Dt. The velocities are tangent to the circle and therefore perpendicular to the radius of the circle. As such, the “radius” triangle and the “velocity” triangle are approximately similar (see the ﬁgures above). We construct an approximate ratio between the two triangles by assuming that during the time Dt, the car has traveled a distance Ds 6 www.ck12.org Concept 2. Circular Motion FIGURE 2.4 FIGURE 2.5 along the circle. The ratio below is constructed in order to determine the acceleration. : = Dv v, which leads to, vDs Ds r Dividing both sides of the equation by Dt, we have: v Ds Dt : = rDv. : = r Dv Dt. But Ds Dt is the speed v of the car and Dv Dt is the acceleration of the car. 7 www.ck12.org If we allow the time to become inﬁnitesimally small, then the approximation becomes exact and we have: v2 = ra;
a = v2 r. Thus, the magnitude of the centripetal acceleration for an object moving with constant speed in circular motion is ac = v2 r, and its direction is toward the center of the circle. Illustrative Examples using Centripetal Acceleration and Force Example 1A: A 1000 kg car moves with a constant speed 13.0 m/s around a ﬂat circular track of radius 40.0 m. What is the magnitude and direction of the centripetal acceleration? Answer: The magnitude of the car’s acceleration is ac = v2 r = 132 acceleration is toward the center of the track. 40 = 4:225 = 4:23 m=s2 and the direction of its Example 1b: Determine the force of static friction that acts upon the car in Figure below FIGURE 2.6 Answer: Using Newton’s Second Law: F = fs = ma = 1000(4:225) = 4225 = 4230 N Example 1c: Determine the minimum necessary coefﬁcient of static friction between the tires and the road. Answer: y F = FN Mg = 0; FN = Mg but fs = µsFN = ma Thus, µsMg = Ma; µsg = a; µs = a g = 4:225 9:8 = 0:431 = 0:43 Check Your Understanding True or False? 1. Kinetic friction is responsible for the traction (friction) between the tires and the road. Answer : False. As long as the car does not skid, there is no relative motion between the instantaneous contact area of the tire and the road. 2. True or False? The force of static friction upon an object can vary. Answer : True. In attempting to move an object, a range of forces of different magnitudes can be applied until the maximum static friction between the object and the surface it rests upon is overcome and the object is set into motion. Recall that the magnitude of static friction is represented by the inequality: fs µsFN 8 www.ck12.org Concept 2. Circular Motion 3. The greater the mass of the car, the greater the coefﬁcient of friction. Answer : False. The coefﬁcient of friction is independent of the mass of an object. Recall that it is the ratio of the friction force to the normal force. As such, it is a pure number dependent only upon the nature
of the materials in contact with each other- in this case rubber and asphalt. References 1. Image copyright Paul Fleet, 2012. http://www.shutterstock.com. Used under license from Shutterstock 2. Tim White (Flickr: TWHITE87). http://www.ﬂickr.com/photos/tjwhite87/8102931300/ 3. Image copyright Andre Adams, 2012; modiﬁed by CK-12 Foundation - Raymond Chou. http://www.shut. CC-BY 2.0 terstock.com. Used under license from Shutterstock.com 4. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 5. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou.. CC-BY-NC-SA 3.0 9 CONCEPT 3 www.ck12.org Centripetal Force • Deﬁne centripetal force. • Solve problems involving centripetal force. • Explain the difference between centripetal and centrifugal forces. Jupiter’s moons and ring materials follow all the laws of physics, including centripetal force and centripetal acceleration. Centripetal Force Centripetal force is, simply, the force that causes centripetal acceleration. Objects that move in uniform circular motion all have an acceleration toward the center of the circle and therefore, they must also suffer a force toward the center of the circle. That force is the centripetal force. For orbiting satellites, such as the moon orbiting the earth or the earth orbiting the sun, the centripetal force is produced by gravity. When an Olympic hammer thrower whirls a massive ball on a chain, the centripetal force is created by the athlete and transmitted by the chain. Newton’s second law shows the relationship between force and acceleration, F = ma. Since we have formulas expressing the relationships for centripetal acceleration, they can easily be altered to show the relationships for centripetal force. ac = v2 and ac = 4p2r r and F = ma so Fc = mv2 r T 2 so Fc = 4p2rm T 2 10 www.ck12.org Concept 3. Centripetal Force Common Misconceptions Many people incorrectly use the term centrifugal force instead of centrip
etal force. Often, you will hear the term centrifugal force used to describe the outward force pushing an object away from the center of a circle. In reality, however, centrifugal forces are inertial, or ﬁctional, forces. They only exist in the frame of reference of the object that is moving and, even then, are theoretical. Physicists dealing in a moving frame of reference use centrifugal forces to ease calculations. For a great explanation of the difference between centrifugal and centripetal force, see this video: https://www.youtube.com/watch?v=DLgy6rVV-08 MEDIA Click image to the left for more content. Summary • Centripetal force is the force that causes centripetal acceleration. • Equations for centripetal force are Fc = mv2 r and Fc = 4p2rm. T 2 Practice A video of physics students riding a roller coaster. Use this resource to answer the questions that follow. http://www.teachersdomain.org/asset/phy03_vid_roller/ 1. Does the roller coaster in the video have a complete circle as part of its path? 2. What is it that keeps the glass of water on the tray as it swings over the student’s head? Review 1. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. What is the centripetal force needed to keep this runner on the curve and what supplies this force? 2. A 1000. kg car rounds a curve of 50.0 m radius on a ﬂat road with a speed of 14.0 m/s. (a) Will the car make the turn successfully if the pavement is dry and the coefﬁcient of friction is 0.60? (b) Will the car make the turn successfully if the pavement is wet and the coefﬁcient of friction is 0.20? 3. An 0.500 kg object tied to a string is swung around a person’s head in a horizontal circle. The length of the string is 1.00 m and the maximum force the string can withstand without breaking is 25.0 N. What is the maximum speed the object may be swung without breaking the string? • centripetal force: The component of force acting on a body in curvilinear motion that is directed toward the center
of curvature or axis of rotation. References 1. Courtesy of NASA/JPL/Cornell University. http://www.nasa.gov/centers/goddard/multimedia/largest/EduI mageGallery.html. Public Domain 11 Physics Unit 10: Waves Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. CONTRIBUTOR Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.
ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 7 10 14 17 20 24 28 32 35 39 42 45 56 Contents 1 Measuring Waves 2 Mechanical Wave 3 Transverse Waves 4 Longitudinal Waves 5 Reﬂection of Mechanical Waves 6 Refraction of Mechanical Waves 7 Wave Interactions 8 Wave Interference 9 Wave Speed 10 Sound Waves 11 Frequency and Pitch of Sound 12 Speed of Sound 13 Resonance with Sound Waves 14 Sound in a Tube iv www.ck12.org Concept 1. Measuring Waves CONCEPT 1 Measuring Waves Lesson Objectives • Deﬁne wave amplitude and wavelength. • Relate wave speed to wave frequency and wavelength. Lesson Vocabulary • hertz (Hz) • wave amplitude • wave frequency • wavelength • wave speed Introduction Tsunamis, or the waves caused by earthquakes, are unusually large ocean waves. You can see an example of a tsunami in Figure 1.1. Because tsunamis are so big, they can cause incredible destruction and loss of life. The tsunami in the ﬁgure crashed into Thailand, sending people close to shore running for their lives. The height of a tsunami or other wave is just one way of measuring its size. You’ll learn about this and other ways of measuring waves in this lesson. FIGURE 1.1 This tsunami occurred in Thailand on De- cember 26, 2004. 1 www.ck12.org Wave Amplitude and Wavelength The height of a wave is its amplitude. Another measure of wave size is wavelength. Both wave amplitude and wavelength are described in detail below. Figure 1.2 shows these wave measures for both transverse and longitudinal waves. You can also simulate waves with different amplitudes and wavelengths by doing the interactive animation at this URL: http://sci-culture.com/advancedpoll/GCSE/sine%20wave%20simulator.html. FIGURE 1.2 Wave amplitude and wavelength are two important measures of wave size. Wave Amplitude Wave amplitude is the maximum distance the particles of a medium move from their resting position when a wave passes through. The resting position is where the particles would be in the absence of a wave. • In a transverse wave, wave amplitude is the height of each crest above the resting position. The higher the crests are, the greater the amplitude. • In a longitudinal wave, amplitude is a measure of how compressed particles of the medium become when the wave passes through
. The closer together the particles are, the greater the amplitude. 2 www.ck12.org Concept 1. Measuring Waves What determines a wave’s amplitude? It depends on the energy of the disturbance that causes the wave. A wave caused by a disturbance with more energy has greater amplitude. Imagine dropping a small pebble into a pond of still water. Tiny ripples will move out from the disturbance in concentric circles, like those in Figure above. The ripples are low-amplitude waves. Now imagine throwing a big boulder into the pond. Very large waves will be generated by the disturbance. These waves are high-amplitude waves. Wavelength Another important measure of wave size is wavelength. Wavelength is the distance between two corresponding points on adjacent waves (see Figure 1.2 ). Wavelength can be measured as the distance between two adjacent crests of a transverse wave or two adjacent compressions of a longitudinal wave. It is usually measured in meters. Wavelength is related to the energy of a wave. Short-wavelength waves have more energy than long-wavelength waves of the same amplitude. You can see examples of waves with shorter and longer wavelengths in Figure 1.3. FIGURE 1.3 Both of these waves have the same ampli- tude, but they differ in wavelength. Which wave has more energy? Wave Frequency and Speed Imagine making transverse waves in a rope, like the waves in Figure above. You tie one end of the rope to a doorknob or other ﬁxed point and move the other end up and down with your hand. You can move the rope up and down slowly or quickly. How quickly you move the rope determines the frequency of the waves. Wave Frequency The number of waves that pass a ﬁxed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a ﬁxed point in 1 second. Figure 1.4 shows high-frequency and low-frequency transverse waves. You can simulate transverse waves with different frequencies at this URL: http://zonalandeduc ation.com/mstm/physics/waves/partsOfAWave/wave
Parts.htm. 3 www.ck12.org FIGURE 1.4 A transverse wave with a higher fre- quency has crests that are closer to- gether. The frequency of a wave is the same as the frequency of the vibrations that caused the wave. For example, to generate a higher-frequency wave in a rope, you must move the rope up and down more quickly. This takes more energy, so a higher-frequency wave has more energy than a lower-frequency wave with the same amplitude. Wave Speed Assume that you move one end of a rope up and down just once. How long will take the wave to travel down the rope to the other end? This depends on the speed of the wave. Wave speed is how far the wave travels in a given amount of time, such as how many meters it travels per second. Wave speed is not the same thing as wave frequency, but it is related to frequency and also to wavelength. This equation shows how the three factors are related: Speed = Wavelength Frequency In this equation, wavelength is measured in meters and frequency is measured in hertz, or number of waves per second. Therefore, wave speed is given in meters per second. The equation for wave speed can be used to calculate the speed of a wave when both wavelength and wave frequency are known. Consider an ocean wave with a wavelength of 3 meters and a frequency of 1 hertz. The speed of the wave is: Speed = 3 m 1 wave/s = 3 m/s You Try It! Problem: Jera made a wave in a spring by pushing and pulling on one end. The wavelength is 0.1 m, and the wave frequency is 0.2 m/s. What is the speed of the wave? If you want more practice calculating wave speed from wavelength and frequency, try the problems at this URL: htt p://www.physicsclassroom.com/class/waves/u10l2e.cfm. The equation for wave speed (above) can be rewritten as: 4 www.ck12.org Concept 1. Measuring Waves Frequency = Speed Wavelength or Wavelength = Speed Frequency Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value. For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is: Frequency = 2 m/s 1 m = 2 waves/
s, or 2 Hz You Try It! Problem: A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength? The Medium Matters The speed of most waves depends on the medium through which they are traveling. Generally, waves travel fastest through solids and slowest through gases. That’s because particles are closest together in solids and farthest apart in gases. When particles are farther apart, it takes longer for the energy of the disturbance to pass from particle to particle. Lesson Summary • Wave amplitude is the maximum distance the particles of a medium move from their resting positions as a wave passes through. Wavelength is the distance between two corresponding points of adjacent waves. Waves with greater amplitudes or shorter wavelengths have more energy. • Wave frequency is the number of waves that pass a ﬁxed point in a given amount of time. Higher frequency waves have more energy. Wave speed is calculated as wavelength multiplied by wave frequency. Wave speed is affected by the medium through which a wave travels. Lesson Review Questions Recall 1. How is wave amplitude measured in a transverse wave? 2. Describe the wavelength of a longitudinal wave. 3. Deﬁne wave frequency. Apply Concepts 4. All of the waves in the sketch below have the same amplitude and speed. Which wave has the longest wavelength? Which has the highest frequency? Which has the greatest energy? 5 www.ck12.org 5. A wave has a wavelength of 0.5 m/s and a frequency of 2 Hz. What is its speed? Think Critically 6. Relate wave amplitude, wavelength, and wave frequency to wave energy. 7. Waves A and B have the same speed, but wave A has a shorter wavelength. Which wave has the higher frequency? Explain how you know. Points to Consider You read in this lesson that waves travel at different speeds in different media. • When a wave enters a new medium, it may speed up or slow down. What other properties of the wave do you think might change when it enters a new medium? • What if a wave reaches a type of matter it cannot pass through? Does it just stop moving? If not, where does it go? References 1. David Rydevik. http://commons.wikimedia.org/wiki/File:2004-tsunami.jpg. Public Domain 2. Christopher Auyeung. CK-12 Foundation. 3. Christopher Auyeung.
CK-12 Foundation. 4. Christopher Auyeung. CK-12 Foundation. 6 www.ck12.org Concept 2. Mechanical Wave CONCEPT 2 • Describe mechanical waves. • Deﬁne the medium of a mechanical wave. • Identify three types of mechanical waves. Mechanical Wave No doubt you’ve seen this happen. Droplets of water fall into a body of water, and concentric circles spread out through the water around the droplets. The concentric circles are waves moving through the water. Waves in Matter The waves in the picture above are examples of mechanical waves. A mechanical wave is a disturbance in matter that transfers energy through the matter. A mechanical wave starts when matter is disturbed. A source of energy is needed to disturb matter and start a mechanical wave. Q: Where does the energy come from in the water wave pictured above? A: The energy comes from the falling droplets of water, which have kinetic energy because of their motion. The Medium The energy of a mechanical wave can travel only through matter. The matter through which the wave travels is called the medium ( plural, media). The medium in the water wave pictured above is water, a liquid. But the medium of a mechanical wave can be any state of matter, even a solid. Q: How do the particles of the medium move when a wave passes through them? A: The particles of the medium just vibrate in place. As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don’t actually travel along with the wave. Only the energy of the wave travels through the medium. 7 Types of Mechanical Waves There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move. You can see this in the Figure 2.1 and in the animation at the following URL. http://www. acs.psu.edu/drussell/Demos/waves/wavemotion.html www.ck12.org FIGURE 2.1 • In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. • In a longitudinal wave, particles of the medium vibrate back and forth parallel to the direction of the wave. • In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end
up moving in a circle. Q: How do you think surface waves are related to transverse and longitudinal waves? A: A surface wave is combination of a transverse wave and a longitudinal wave. Summary • A mechanical wave is a disturbance in matter that transfers energy through the matter. • The matter through which a mechanical wave travels is called the medium ( plural, media). • There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move when the energy of the wave passes through. Vocabulary • mechanical wave : Disturbance in matter that transfers energy from one place to another. • medium (plural, media ): Matter through which a mechanical wave moves. Practice At the following URL, read the short introduction to waves and watch the animations. Then answer the questions below. http://www.acs.psu.edu/drussell/Demos/waves-intro/waves-intro.html 8 www.ck12.org Concept 2. Mechanical Wave 1. The article gives a dictionary deﬁnition of wave. What is the most important part of this deﬁnition? 2. What happens to particles of the medium when a wave passes? 3. How is “doing the wave” in a football stadium like a mechanical wave? Review 1. Deﬁne mechanical wave. 2. What is the medium of a mechanical wave? 3. List three types of mechanical waves. 4. If you shake one end of a rope up and down, a wave passes through the rope. Which type of wave is it? References 1. Zachary Wilson.. CC BY-NC 3.0 9 CONCEPT 3 www.ck12.org Transverse Waves • Describe transverse waves. • Explain how waves transfer energy without transferring matter. • Deﬁne wavelength, frequency, and period of a transverse wave. • State the relationship between speed, wavelength, and frequency. • Solve problems using the relationships between speed, wavelength, frequency, and period. Professional surfer Marcio Freire rides a giant wave at the legendary big wave surf break known as as "Jaws" during one the largest swells of the winter March 13, 2011 in Maui, HI. Massive waves, such as this one, transfer huge amounts of energy. Transverse Waves Types of Waves Water waves, sound waves, and the waves that travel along a rope are mechanical waves. Mechanical wa
ves require a material medium such as water, air, or rope. Light waves, however, are electromagnetic waves and travel without a material medium. They are not mechanical waves. In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. One type of mechanical wave is the transverse wave. In the case of transverse waves, the movement of the medium is perpendicular to the direction of the energy movement. 10 www.ck12.org Concept 3. Transverse Waves In the sketch above, consider the transverse wave produced when the boy jerks one end of a rope up and down while the other end is tied to a tree. The energy spent by the boy transfers permanently down the rope to the tree. The rope, however, only moves up and down. If we stuck a piece of tape somewhere on the rope, we would see that the particles of medium do not travel with the energy. After the wave has passed by, the piece of tape would still be in the same place it was before the wave approached. In all transverse waves, the movement media vibrates perpendicularly to the direction of wave motion, and the medium is not permanently moved from one place to another. Frequency, Wavelength, and Velocity Waves are identiﬁed by several characteristics. There is a center line where the medium would be if there were no wave, which is sometimes describes as the undisturbed position. The displacement of the medium above this undisturbed position is called a crest and the displacement below the undisturbed position is called a trough. The maximums of the crest and trough are equal and are called the amplitude. The distance between equivalent positions on succeeding waves is called the wavelength. The wavelength could be measured from a crest to the next crest or from a trough to the next trough, and is commonly represented with the Greek letter lambda, l. The time interval required for one complete wave to pass a point is called the period. During the period of the wave, an entire wavelength from one crest to the next crest passes a position. The number of waves that pass a single position in one second is called the frequency. The period of a wave and its frequency are reciprocals of each other. f = 1 T The units for the period are seconds and the units for frequency are s 1 or 1 name Hertz (Hz). s. This unit has also been given the 11 Another important characteristic of a wave is its velocity. The
wave velocity is different from the velocity of the medium; the wave velocity is the velocity of the linearly transferred energy. Since the energy travels one wavelength, l, in one period, T, the velocity can be expressed as distance over time: www.ck12.org v = l T : Since period and frequency are reciprocals, the speed of the wave could also be expressed as v = l f. Example Problem: A sound wave has a frequency of 262 Hz. What is the time lapse between successive wave crests? Solution: The time lapse between successive crests would be the period and the period is the reciprocal of the frequency. T = 1 f = 1 262 s1 = 0:00382 s Example Problem: A sound wave has a frequency of 262 Hz has a wavelength of 1.29 m. What is the velocity of the wave? Solution: v = l f = (1:29 m)(262 s1) = 338 m/s Summary • Mechanical waves require a material medium such as water, air, or rope. • In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. • One type of mechanical wave is the transverse wave, in which the movement of the medium is perpendicular to the direction of the energy propagation. • The maximum displacement of the medium is the distance from the undisturbed position to the top of a crest, or the amplitude. • The distance along the line of motion of the wave from equivalent positions on succeeding waves is the wavelength. • The time interval required for one entire wave to pass a point is the period. • The number of periods per second is the wave’s frequency. • The period of a wave and its frequency are reciprocals of each other. • The velocity of the wave’s energy transfer is given by v = l f or v = l T : Practice The following video explains wave characteristics. Pause the video before each practice question and try to solve it yourself before moving on. http://www.youtube.com/watch?v=5ENLxaPiJJI 12 www.ck12.org Concept 3. Transverse Waves MEDIA Click image to the left for more content. 1. What is the distance between the base line and crest called? 2. What symbol is used for wavelength? 3. What is the relationship between period and frequency? Review 1. A sound wave produced by a chime 515 m away is
heard 1.50 s later. (a) What is the speed of sound in air? (b) The sound wave has a frequency of 436 Hz. What is its period? (c) What is the wavelength of the sound? 2. A hiker shouts toward a vertical cliff 685 m away. The echo is heard 4.00 s later. (a) What is the speed of sound in air? (b) Why is this speed of sound slightly different from the previous answer? (c) The wavelength of the sound is 0.750 m. What is the frequency? (d) What is the period of the wave? 3. The speed of light in air is 3.00 10 8 m/s. If a light wave has a wavelength of 5.80 10 7 m, what is its frequency? • transverse wave: A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer • undisturbed position: The equilibrium or rest position of the medium in a wave. • crest: A crest is the point on a wave with the maximum value or upward displacement within a cycle • trough: A trough is the opposite of a crest, so the minimum or lowest point in a cycle • amplitude: The maximum displacement from a zero value during one period of an oscillation. • wavelength: The distance between corresponding points of two consecutive waves. • frequency: The number of waves that pass a ﬁxed point in unit time. • period: A period is the time required for one complete cycle of vibration to pass a given point. References 1. Image copyright EpicStockMedia, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. Kid: Image copyright Richcat, 2013; Tree: Image copyright AlexeyZet, 2013; Composite created by CK-12. Used under licenses from Shutterstock.com Foundation - Samantha Bacic. http://www.shutterstock.com/ 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 13 CONCEPT 4 • Describe longitudinal waves. www.ck12.org Longitudinal Waves https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=6e9QZ8b6JzZZRM&t
bnid=CAVrIJRLaBDfM:&ved=0CAUQjRw&url=http%3A%2F%2Fsirius.ucsc.edu%2Fdemoweb%2Fcgi-bin%2F%3Fwavesvisible-slinky&ei=kE4FUpSDN4HXygHR84HwDg&bvm=bv.50500085,d.b2I&psig=AFQjCNEJ23OS_x3Ga2tbC3Vi2VVUfbPmQ&ust=1376165898023972 Playing with a Slinky is a childhood tradition, but few children realize they are actually playing with physics. Longitudinal Waves Like transverse waves, longitudinal waves are mechanical waves, which means they transfer energy through a medium. Unlike transverse waves, longitudinal waves cause the particles of medium to move parallel to the direction of the wave. They are most common in springs, where they are caused by the pushing an pulling of the spring. Although the surface waves on water are transverse waves, ﬂuids (liquids, gases, and plasmas) usually transmit longitudinal waves. As shown in the image below, longitudinal waves are a series of compressions and rarefactions, or expansions. The wavelength of longitudinal waves is measured by the distance separating the densest compressions. The amplitude of longitudinal waves is the difference in media density between the undisturbed density to the highest density in a compression. Example Problem: A sonar signal (sonar is sound waves traveling through water) of 1:00 106 Hz frequency has a wavelength of 1.50 mm in water. What is the speed of sound in water? Solution: v = l f = (0:00150 m)(1:00 106 s1) = 1500 m/s Example Problem: A sound wave of wavelength 0.70 m and velocity 330 m/s is produced for 0.50 s. 14 www.ck12.org Concept 4. Longitudinal Waves a. What is the frequency of the wave? b. How many complete waves are emitted in this time interval? c. After 0.50 s, how far is the wave front from the source of the sound? Solution: f = v l = 330 m/s 0:70 m = 470 s1 a. b. complete waves = (470 cycles/s)(
0:50 s) = 235 cycles c. distance = (330 m/s)(0:50 s) = 115 m Summary • Longitudinal waves cause the particles of medium to move parallel to the direction of the wave. Practice The following video explains how a tuning fork creates sound. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=bomzzHC-59k MEDIA Click image to the left for more content. 1. In your own words, how are compressions and rarefactions produced by the tuning fork? 2. Make a guess why sound can easily travel around corners (Hint: think of its medium). Review 1. Some giant ocean waves have a wavelength of 25 m long, and travel at speeds of 6.5 m/s. Determine the frequency and period of such a wave. 2. Bats use sound echoes to navigate and hunt. They emit pulses of high frequency sound waves which reﬂect off obstacles in the surroundings. By detecting the time delay between the emission and return of a pulse, a bat can determine the location of the object. What is the time delay between the sending and return of a pulse from an object located 12.5 m away? The approximate speed of sound is 340 m/s. 3. Sachi is listening to her favorite radio station which broadcasts radio signals with a frequency of 1:023 108 Hz. If the speed of the signals in air is 2:997 108 m/s, what is the wavelength of these radio signals? 4. A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to move past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave. 5. A sonar signal leaves a submarine, travels through the water to another submarine and reﬂects back to the original submarine in 4.00 s. If the frequency of the signal was 512 cycles per second and the wavelength of the signal was 2.93 m, how far away is the second submarine? • longitudinal wave: A wave in which the direction of media displacement is the same as the direction of wave propagation. 15 References 1. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 www.ck12.org 16 www.ck12.org Concept 5. Reﬂ
ection of Mechanical Waves CONCEPT 5 Reﬂection of Mechanical Waves • State the law of reﬂection. • Solve problems using the law of reﬂection. • Given data about the media on either side of a barrier, determine whether the reﬂected wave will be upright or inverted. When mechanical waves strike a barrier, at least part of the energy of the waves will be reﬂected back into the media from which they came. You experience this every single day, when you look in the mirror and see your own reﬂection. Reﬂection of Mechanical Waves When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some portion of the wave is reﬂected back into the original medium. It reﬂects back at an equal angle that it came in. These angles are called the angle of incidence and the angle of reﬂection. The normal line, the incident and reﬂected rays, and the angles of incidence and reﬂection are all shown in the diagram sketched above. The law of reﬂection states that the angle of incidence equals the angle of reﬂection. These rules of reﬂection apply in the cases of water waves bouncing off the side of a pool, sound waves echoing off a distant cliff, or wave pulses traveling down a rope or a slinky. Consider the change that would occur with a light rope joined to a heavier rope. When a wave pulse travels down the rope and encounters the media change, a reﬂection will occur. Look at the image below. In the top sketch, we see a lightweight (black) rope attached to a heavier rope (red). There is a wave pulse traveling down the rope from left to right. When the wave pulse encounters the barrier (the change in rope weight), part of the wave moves into the new medium and part of the wave is reﬂected back into the old medium. As you can see in the bottom half of the diagram, the transmitted portion of the wave continues into the new medium right side up. The transmitted wave is somewhat diminished because some of the energy of the wave was reﬂected and also because the rope to be lifted is heavier. The reﬂected wave is also diminished because some of the energy was transmitted through the barrier. The reﬂected wave is
also inverted (upside down). This is a general rule for mechanical waves passing from a less dense medium into a more dense medium, that is, the reﬂected wave will be inverted. 17 www.ck12.org The situation changes when the wave is passing from a more dense medium into a less dense medium. As you can see in the sketch below, when a wave pulse moving in denser medium encounters a media interface to a medium of less density, the reﬂected wave is upright rather than inverted. It is also possible for a mechanical wave to encounter an impenetrable barrier, that is, a barrier which does not allow In such a case, the complete wave pulse will be reﬂected and the reﬂected wave will be any transmission at all. inverted. Summary • When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some part of the wave is reﬂected back into the original medium. • The law of reﬂection states that the angle of incidence equals the angle of reﬂection. • The general rule, for mechanical waves passing from a less dense medium into a more dense medium, the reﬂected wave will be inverted. • When a wave pulse moving in denser medium encounters a media interface to a medium of lesser density, the reﬂected wave is upright rather than inverted. • When a mechanical wave encounters an impenetrable barrier, the complete wave pulse will be reﬂected and the reﬂected wave will be inverted. Practice The following video shows a wave machine in action. Use this video to answer the questions that follow. http://www.youtube.com/watch?v=YQHbRw_hyz4 MEDIA Click image to the left for more content. 1. What happens to the wave when it is reﬂected from an open end? 2. What happens to the wave when it is reﬂected from a ﬁxed end? 18 www.ck12.org Review Concept 5. Reﬂection of Mechanical Waves 1. Draw a diagram showing a surface with a normal line. On the diagram, show a wave ray striking the surface with an angle of incidence of 60. Draw the reﬂection ray on the diagram and label the angle of reﬂection. 2. Light strikes a mirror’
s surface at 30 to the normal. What will the angle of reﬂection be? 3. If the angle between the incident ray and the reﬂected ray is 90, what is the angle of incidence? 4. When a water wave is reﬂected from a concrete wall, will the reﬂected wave be inverted or upright? 5. If you tie a heavy spring to a light spring and send a wave pulse down the heavy spring, some of the wave will be reﬂected when the wave passes into the lighter spring. Will the reﬂected pulse be upright or inverted? • reﬂection: The change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. • law of reﬂection: Says the angle at which the wave is incident on the surface equals the angle at which it is reﬂected. • angle of incidence: The angle formed by a ray incident on a surface and a perpendicular to the surface at the point of incidence. • angle of reﬂection: The angle formed by a reﬂected ray and a perpendicular to the surface at the point of reﬂection. References 1. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 19 CONCEPT 6 Refraction of Mechanical Waves www.ck12.org • Deﬁne and describe refraction of mechanical waves. • State and use the law of refraction. A straw in a glass of water seen from the side often appears broken, even though it is not. The apparent break is due to the bending of light rays leaving the straw; as the light passes from the water to the glass and from the glass to the air, the light rays are bent. Nonetheless, your eye traces the light ray backward as if the light has followed a straight path from its origin at the straw. Since the light appears to have come from a different place, your eye sees the straw as being broken. Refraction of Mechanical Waves When any wave strikes a boundary between
media, some of the energy is reﬂected and some is transmitted. When the wave strikes the media interface at an angle, the transmitted wave will move in a slightly different direction than the incident wave. This phenomenon is known as refraction. 20 www.ck12.org Concept 6. Refraction of Mechanical Waves Consider the image sketched above. Suppose that the waves represented here are water waves. The wave crests are represented by the black lines in the image. As such, the distance between two consecutive black lines is the wavelength. Let the red line represent a transition from deep to shallow water. This transition is called the media interface. As the waves hit the boundary, the waves slow down. The right side of the wave reaches the boundary before the left side of the wave, causing the left side to catch up and the angle of propagation to change slightly. This change in direction can be seen in the yellow line, which is slightly angled at the boundary. The refraction of waves across boundaries operates similarly to the method by which tanks are steered. Tanks do not have a steering wheel. Instead, they have an accelerator to produce forward motion and separate brakes on each tread. The operator uses brakes on both treads at the same time in order to stop, but brakes on only one tread to turn the tank. By braking one side, the operator causes that side to slow down or stop while the other side continues at the previous speed, causing the tank to turn towards the slower tread. This sketch shows a wave ray striking an interface between old medium and new medium. A normal line has been drawn as a dotted line perpendicular to the interface. The angle between the incident ray and the normal line is called the angle of incidence, shown as q i, and the angle between the refracted ray and the normal line is called the angle of refraction, q r. 21 www.ck12.org We already understand that the change in the wave direction at the border depends on the difference between the two velocities. This relationship is conveniently expressed in a mathematical relationship: sin qr sin qi = vr vi = lr li The ratio of the sine of the angle of refraction to the sine of the angle of incidence is the same as the ratio of the velocity of the wave in the new medium to the velocity of the wave in the old medium and equal to the ratio of wavelength (l) in the old medium to the wavelength in the new medium. Example Problem: A water wave with a
wavelength of 3.00 m is traveling in deep water at 16.0 m/s. The wave strikes a sharp interface with shallow water with an angle of incidence of 53:0. The wave refracts into the shallow water with an angle of refraction of 30:0. What is the velocity of the wave in shallow water and what is its wavelength in the new medium? so sin 30 sin 53 = vr 16:0 m=s and vr = 10:0 m=s. Solution: = vr vi sin qr sin qi so 10:0 m=s 16:0 m=s = lr = lr li vr 3:00 m and lr = 1:88 m. vi Example Problem: The ratio of the sin qr to sin qi is 0.769. 5:00 109 m, what is its wavelength in the original medium? If the wavelength of a wave in a new medium is Solution: Summary 0:769 = lr li so li = 5:00 109 m 0:769 = 6:50 109 m • When any wave strikes a boundary between media, some of the energy is reﬂected and some is transmitted. • When a wave strikes the media interface at an angle, the transmitted wave will move in a different direction than the incident wave. This phenomenon is known as refraction. = vr vi • At any media interface, sin qr sin qi = lr li Practice http://www.youtube.com/watch?v=mH9VwivqjmE Follow up questions. 22 www.ck12.org Concept 6. Refraction of Mechanical Waves 1. What causes refraction? 2. What doesn’t change during refraction? Review 1. A laser beam passes through water and enters a glass block at an angle. The ratio of the speed of the wave in glass to the speed in water is 0.866. If the angle of incidence to the interface is 60, what is the angle of refraction? 2. A ray of light is traveling from air into glass at an angle of 30:0 to the normal line. The speed of the light in air is 3:00 108 m=s and in glass the speed drops to 2:00 108 m=s. What is the angle of refraction? 3. Which of the following change when a water wave moves across a boundary at an angle between deep water and shallow
water? a. frequency b. wavelength c. speed d. wave direction e. period 4. Which of the following change when a water wave moves across a boundary exactly along the media interface between deep water and shallow water? a. frequency b. wavelength c. speed d. wave direction e. period 5. The speed of sound is 340 m/s. A particular sound wave has a frequency of 320. Hz. a. What is the wavelength of this sound in air? b. If this sound refracts into water where the speed of sound is 4 times faster, what will be the new wavelength? c. What will be the new frequency? 6. When a light ray passes from air into diamond, the angle of incidence is 45:0 and the angle of refraction is 16:7. If the speed of light in air is 3:00 108 m=s, what is the speed of light in diamond? • refraction: The turning or bending of a wave direction when it passes from one medium to another of different density. References 1. Image copyright cheyennezj, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. Greg Goebel. http://www.public-domain-image.com/full-image/transportation-vehicles-public-domain-ima ges-pictures/tanks-public-domain-images-pictures/m7-priest-self-propelled-105-millimeter-howitzer-tank.jpgcopyright-friendly-image.html. Public Domain 4. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 23 CONCEPT 7 www.ck12.org Wave Interactions • Identify ways that waves can interact with matter. • Deﬁne and give examples of wave reﬂection, refraction, and diffraction. Did you ever hear an echo of your own voice? An echo occurs when sound waves bounce back from a surface that they can’t pass through. The woman pictured here is trying to create an echo by shouting toward a large building. When the sound waves strike the wall of the building, most of them bounce back toward the woman, and she hears an echo of her voice. An echo is just one example of how waves interact with matter

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