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from protons produced results like Rutherford had obtained for the nucleus nearly 60 years earlier. The SLAC scattering experiments showed unambiguously that there were three point-like (meaning they had sizes considerably smaller than the probe’s wavelength) charges inside the proton as seen in Figure 23.12. This evidence made all but the most skeptical admit that there was validity to the quark substructure of hadrons. Access for free at openstax.org. 23.2 • Quarks 781 Figure 23.12 Scattering of high-energy electrons from protons at facilities like SLAC produces evidence of three point-like charges consistent with proposed quark properties. This experiment is analogous to Rutherford’s discovery of the small size of the nucleus by scattering α particles. High-energy electrons are used so that the probe wavelength is small enough to see details smaller than the proton. The inclusion of the strange quark with Zweig and Gell-Mann’s model concerned physicists. While the up and down quarks demonstrated fairly clear symmetry and were present in common fundamental particles like protons and neutrons, the strange quark did not have a counterpart of its own. This thought, coupled with the four known leptons at the time, caused scientists to predict that a fourth quark, yet to be found, also existed. In 1974, two groups of physicists independently discovered a particle with this new quark, labeled charmed. This completed the second exoticquark pair, strange (s) and charmed (c). A final pair of quarks was proposed when a third pair of leptons was discovered in 1975. The existence of the bottom (b) quark and the top (t) quark was verified through experimentation in 1976 and 1995, respectively. While it may seem odd that so much time would elapse between the original quark discovery in 1967 and the verification of the top quark in 1995, keep in mind that each quark discovered had a progressively larger mass. As a result, each new quark has required more energy to discover. TIPS FOR SUCCESS Note that a very important tenet of science occurred throughout the period of quark discovery. The charmed, bottom, and top quarks were all speculated on, and then were discovered some time later. Each of their discoveries helped to verify and strengthen the quark model. This process of speculation and verification continues to take place today and is part of what drives physicists to search for evidence of the
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graviton and Grand Unified Theory. One of the most confounding traits of quarks is their electric charge. Long assumed to be discrete, and specifically a multiple of the elementary charge of the electron, the electric charge of an individual quark is fractional and thus seems to violate a presumed tenet of particle physics. The fractional charge of quarks, which are, are the only structures and found in nature with a nonintegral number of charge. However, note that despite this odd construction, the fractional value of the quark does not violate the quantum nature of the charge. After all, free quarks cannot be found in nature, and all quarks are bound into arrangements in which an integer number of charge is constructed. Table 23.3 shows the six known quarks, in addition to their antiquark components, as will be discussed later in this section. Flavor Symbol Antiparticle Charge[1][2] Up Down Strange Charmed [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks 782 Chapter 23 • Particle Physics Flavor Symbol Antiparticle Charge[1][2] Bottom Top [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks While the term flavoris used to differentiate between types of quarks, the concept of color is more analogous to the electric charge in that it is primarily responsible for the force interactions between quarks. Note—Take a moment to think about the electrostatic force. It is the electric charge that causes attraction and repulsion. It is the same case here but with a colorcharge. The three colors available to a quark are red, green, and blue, with antiquarks having colors of anti-red (or cyan), anti-green (or magenta), and anti-blue (or yellow). Why use colors when discussing quarks? After all, the quarks are not actually colored with visible light. The reason colors are used is because the properties of a quark are analogous to the three primary and secondary colors mentioned above. Just as different colors of light can be combined to create white, different colorsof quark may be combined to construct a particle like a proton or neutron. In
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fact, for each hadron, the quarks must combine such that their color sums to white! Recall that two up quarks and one down quark construct a proton, as seen in Figure 23.12. The sum of the three quarks’ colors—red, green, and blue—yields the color white. This theory of color interaction within particles is called quantum chromodynamics, or QCD. As part of QCD, the strong nuclear force can be explained using color. In fact, some scientists refer to the color force, not the strong force, as one of the four fundamental forces. Figure 23.13 is a Feynman diagram showing the interaction between two quarks by using the transmission of a colored gluon. Note that the gluon is also considered the charge carrier for the strong nuclear force. Figure 23.13 The exchange of gluons between quarks carries the strong force and may change the color of the interacting quarks. While the colors of the individual quarks change, their flavors do not. Note that quark flavor may have any color. For instance, in Figure 23.13, the down quark has a red color and a green color. In other words, colors are not specific to a particle quark flavor. Hadrons and Leptons Particles can be revealingly grouped according to what forces they feel between them. All particles (even those that are massless) are affected by gravity since gravity affects the space and time in which particles exist. All charged particles are affected by the electromagnetic force, as are neutral particles that have an internal distribution of charge (such as the neutron with its magnetic moment). Special names are given to particles that feel the strong and weak nuclear forces. Hadrons are particles that feel the strong nuclear force, whereas leptons are particles that do not. All particles feel the weak nuclear force. This means that hadrons are distinguished by being able to feel both the strong and weak nuclear forces. Leptons and hadrons are distinguished in other ways as well. Leptons are fundamental particles that have no measurable size, while hadrons are composed of quarks and have a diameter on the order of 10 to 15 m. Six particles, including the electron and neutrino, make up the list of known leptons. There are hundreds of complex particles in the hadron class, a few of which (including the proton and neutron) are listed in Table 23.4. Access for free at openst
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ax.org. Category Leptons Hadrons – Mesons[2] Hadrons – Baryons[3] Particle Name Electron Neutrino (e) Muon Neutrino (μ) Tau Neutrino (τ) Pion Kaon Eta Proton Neutron Lambda Omega 23.2 • Quarks 783 Symbol Antiparticle Rest Mass Mean Lifetime (s) [1] [1] Stable Stable Stable [1] Stable Stable 882 0.511 105.7 1,777 139.6 135.0 493.7 497.6 547.9 938.3 939.6 1,115.7 1,672.5 Self Self p n [1]Neutrino masses may be zero. Experimental upper limits are given in parentheses. [2]Many other mesons known [3]Many other baryons known Table 23.4 List of Leptons and Hadrons. There are many more leptons, mesons, and baryons yet to be discovered and measured. The purpose of trying to uncover the smallest indivisible things in existence is to explain the world around us through forces and the interactions between particles, galaxies and objects. This is why a handful of scientists devote their life’s work to smashing together small particles. What internal structure makes a proton so different from an electron? The proton, like all hadrons, is made up of quarks. A few examples of hadron quark composition can be seen in Figure 23.14. As shown, each hadron is constructed of multiple quarks. As mentioned previously, the fractional quark charge in all four hadrons sums to the particle’s integral value. Also, notice that the color composition for each of the four particles adds to white. Each of the particles shown is constructed of up, down, and their antiquarks. This is not surprising, as the quarks strange, charmed, top, and bottom are found in only our most exotic particles. 784 Chapter 23 • Particle Physics Figure 23.14 All baryons, such as the proton and neutron shown here, are composed of three quarks. All mesons, such as the pions shown here, are composed of a quark–antiquark pair. Arrows represent the spins of the quarks. The colors are such that they need to add to white for any possible combination
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of quarks. You may have noticed that while the proton and neutron in Figure 23.14 are composed of three quarks, both pions are comprised of only two quarks. This refers to a final delineation in particle structure. Particles with three quarks are called baryons. These are heavy particles that can decay into another baryon. Particles with only two quarks—a-quark–anti-quark pair—are called mesons. These are particles of moderate mass that cannot decay into the more massive baryons. Before continuing, take a moment to view Figure 23.15. In this figure, you can see the strong force reimagined as a color force. The particles interacting in this figure are the proton and neutron, just as they were in Figure 23.6. This reenvisioning of the strong force as an interaction between colored quarks is the critical concept behind quantum chromodynamics. Figure 23.15 This Feynman diagram shows the interaction between a proton and a neutron, corresponding to the interaction shown in Figure 23.6. This diagram, however, shows the quark and gluon details of the strong nuclear force interaction. Matter and Antimatter Antimatter was first discovered in the form of the positron, the positively charged electron. In 1932, American physicist Carl Anderson discovered the positron in cosmic ray studies. Through a cloud chamber modified to curve the trajectories of cosmic Access for free at openstax.org. rays, Anderson noticed that the curves of some particles followed that of a negative charge, while others curved like a positive charge. However, the positive curve showed not the mass of a proton but the mass of an electron. This outcome is shown in Figure 23.16 and suggests the existence of a positively charged version of the electron, created by the destruction of solar photons. 23.2 • Quarks 785 Figure 23.16 The image above is from the Fermilab 15 foot bubble chamberand shows the production of an electron and positron (or antielectron) from an incident photon. This event is titled pair production and provides evidence of antimatter, as the two repel each other. Antimatter is considered the opposite of matter. For most antiparticles, this means that they share the same properties as their original particles with the exception of their charge. This is why the positron can be considered a positive electron while the antiproton is considered a negative proton. The
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idea of an opposite charge for neutral particles (like the neutron) can be confusing, but it makes sense when considered from the quark perspective. Just as the neutron is composed of one up quark and two down quarks (of charge, respectively), the antineutron is composed of one anti–up quark and two anti–down and quarks (of charge and, respectively). While the overall charge of the neutron remains the same, its constituent particles do not! A word about antiparticles: Like regular particles, antiparticles could function just fine on their own. In fact, a universe made up of antimatter may operate just as our own matter-based universe does. However, we do not know fully whether this is the case. The reason for this is annihilation. Annihilation is the process of destruction that occurs when a particle and its antiparticle interact. As soon as two particles (like a positron and an electron) coincide, they convert their masses to energy through the equation makes it very difficult for scientists to study antimatter. That said, scientists have had success creating antimatter through highenergy particle collisions. Both antineutrons and antiprotons were created through accelerator experiments in 1956, and an anti–hydrogen atom was even created at CERN in 1995! As referenced in, the annihilation of antiparticles is currently used in medical studies to determine the location of radioisotopes.. This mass-to-energy conversion, which typically results in photon release, happens instantaneously and Completing the Standard Model of the Atom The Standard Model of the atom refers to the current scientific view of the fundamental components and interacting forces of matter. The Standard Model (Figure 23.17) shows the six quarks that bind to form all hadrons, the six lepton particles already considered fundamental, the four carrier particles (or gauge bosons) that transmit forces between the leptons and quarks, and the recently added Higgs boson (which will be discussed shortly). This totals 17 fundamental particles, combinations of which are responsible for all known matter in our entire universe! When adding the antiquarks and antileptons, 31 components make up the Standard Model. 786 Chapter 23 • Particle Physics Figure 23.17 The Standard Model of elementary particles shows an organized view of all fundamental particles, as currently known: six quarks, six leptons, and four gauge bosons (or carrier particles). The Higgs boson, first observed in 2012, is a new addition to
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the Standard Model. Figure 23.17 shows all particles within the Standard Model of the atom. Not only does this chart divide all known particles by color-coded group, but it also provides information on particle stability. Note that the color-coding system in this chart is separate from the red, green, and blue color labeling system of quarks. The first three columns represent the three familiesof matter. The first column, considered Family 1, represents particles that make up normal matter, constructing the protons, neutrons, and electrons that make up the common world. Family 2, represented from the charm quark to the muon neutrino, is comprised of particles that are more massive. The leptons in this group are less stable and more likely to decay. Family 3, represented by the third column, are more massive still and decay more quickly. The order of these families also conveniently represents the order in which these particles were discovered. TIPS FOR SUCCESS Look for trends that exist within the Standard Model. Compare the charge of each particle. Compare the spin. How does mass relate to the model structure? Recognizing each of these trends and asking questions will yield more insight into the organization of particles and the forces that dictate particle relationships. Our understanding of the Standard Model is still young, and the questions you may have in analyzing the Standard Model may be some of the same questions that particle physicists are searching for answers to today! The Standard Model also summarizes the fundamental forces that exist as particles interact. A closer look at the Standard Model, as shown in Figure 23.18, reveals that the arrangement of carrier particles describes these interactions. Access for free at openstax.org. 23.2 • Quarks 787 Figure 23.18 The revised Standard Model shows the interaction between gauge bosons and other fundamental particles. These interactions are responsible for the fundamental forces, three of which are described through the chart’s shaded areas. Each of the shaded areas represents a fundamental force and its constituent particles. The red shaded area shows all particles involved in the strong nuclear force, which we now know is due to quantum chromodynamics. The blue shaded area corresponds to the electromagnetic force, while the green shaded area corresponds to the weak nuclear force, which affects all quarks and leptons. The electromagnetic force and weak nuclear force are considered united by the electroweak force within the Standard Model. Also, because definitive evidence of the graviton is yet to be found, it is not included in the
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Standard Model. The Higgs Boson One interesting feature of the Standard Model shown in Figure 23.18 is that, while the gluon and photon have no mass, the Z and W bosons are very massive. What supplies these quickly moving particles with mass and not the gluons and photons? Furthermore, what causes some quarks to have more mass than others? In the 1960s, British physicist Peter Higgs and others speculated that the W and Z bosons were actually just as massless as the gluon and photon. However, as the W and Z bosons traveled from one particle to another, they were slowed down by the presence of a Higgs field, much like a fish swimming through water. The thinking was that the existence of the Higgs field would slow down the bosons, causing them to decrease in energy and thereby transfer this energy to mass. Under this theory, all particles pass through the Higgs field, which exists throughout the universe. The gluon and photon travel through this field as well but are able to do so unaffected. The presence of a force from the Higgs field suggests the existence of its own carrier particle, the Higgs boson. This theorized boson interacts with all particles but gluons and photons, transferring force from the Higgs field. Particles with large mass (like the top quark) are more likely to receive force from the Higgs boson. While it is difficult to examine a field, it is somewhat simpler to find evidence of its carrier. On July 4, 2012, two groups of scientists at the LHC independently confirmed the existence of a Higgs-like particle. By examining trillions of proton–proton collisions at energies of 7 to 8 TeV, LHC scientists were able to determine the constituent particles that created the protons. In this data, scientists found a particle with similar mass, spin, parity, and interactions with other particles that matched the Higgs boson predicted decades prior. On March 13, 2013, the existence of the Higgs boson was tentatively confirmed by CERN. Peter Higgs and Francois Englert received the Nobel Prize in 2013 for the “theoretical discovery of a mechanism that contributes to our understanding of the origin and mass of subatomic particles.” WORK IN PHYSICS Particle Physicist If you have an innate desire to unravel life’s great mysteries and further understand the nature of the physical world, a career in particle physics may be for
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you! Particle physicists have played a critical role in much of society’s technological progress. From lasers to computers, televisions to space missions, splitting the atom to understanding the DNA molecule to MRIs and PET scans, much of our modern society is based on the work done by particle physicists. While many particle physicists focus on specialized tasks in the fields of astronomy and medicine, the main goal of particle physics is to further scientists’ understanding of the Standard Model. This may mean work in government, industry, or 788 Chapter 23 • Particle Physics academics. Within the government, jobs in particle physics can be found within the National Institute for Standards and Technology, Department of Energy, NASA, and Department of Defense. Both the electronics and computer industries rely on the expertise of particle physicists. College teaching and research positions can also be potential career opportunities for particle physicists, though they often require some postgraduate work as a prerequisite. In addition, many particle physicists are employed to work on high-energy colliders. Domestic collider labs include the Brookhaven National Laboratory in New York, the Fermi National Accelerator Laboratory near Chicago, and the SLAC National Accelerator Laboratory operated by Stanford University. For those who like to travel, work at international collider labs can be found at the CERN facility in Switzerland in addition to institutes like the Budker Institute of Nuclear Physics in Russia, DESY in Germany, and KEK in Japan. Shirley Jackson became the first African American woman to earn a Ph.D. from MIT back in 1973, and she went on to lead a highly successful career in the field of particle physics. Like Dr. Jackson, successful students of particle physics grow up with a strong curiosity in the world around them and a drive to continually learn more. If you are interested in exploring a career in particle physics, work to achieve good grades and SAT scores, and find time to read popular books on physics topics that interest you. While some math may be challenging, recognize that this is only a tool of physics and should not be considered prohibitive to the field. High-level work in particle physics often requires a Ph.D.; however, it is possible to find work with a master’s degree. Additionally, jobs in industry and teaching can be achieved with solely an undergraduate degree. GRASP CHECK What is the primary goal of all work in particle physics? a. The primary goal is to further our understanding of the Standard Model. b. The primary goal is to further our understanding of Rutherford
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’s model. c. The primary goal is to further our understanding of Bohr’s model. d. The primary goal is to further our understanding of Thomson’s model. Check Your Understanding 8. In what particle were quarks originally discovered? a. b. c. d. the electron the neutron the proton the photon 9. Why was the existence of the charm quark speculated, even though no direct evidence of it existed? a. The existence of the charm quark was symmetrical with up and down quarks. Additionally, there were two known leptons at the time and only two quarks. b. The strange particle lacked the symmetry that existed with the up and down quarks. Additionally, there were four known leptons at the time and only three quarks. c. The bottom particle lacked the symmetry that existed with the up and down quarks. Additionally, there were two known leptons at the time and only two quarks. d. The existence of charm quarks was symmetrical with up and down quarks. Additionally, there were four known leptons at the time and only three quarks. 10. What type of particle is the electron? a. The electron is a lepton. b. The electron is a hadron. c. The electron is a baryon. d. The electron is an antibaryon. 11. How do the number of fundamental particles differ between hadrons and leptons? a. Hadrons are constructed of at least three fundamental quark particles, while leptons are fundamental particles. b. Hadrons are constructed of at least three fundamental quark particles, while leptons are constructed of two fundamental particles. c. Hadrons are constructed of at least two fundamental quark particles, while leptons are constructed of three Access for free at openstax.org. 23.2 • Quarks 789 fundamental particles. d. Hadrons are constructed of at least two fundamental quark particles, while leptons are fundamental particles. 12. Does antimatter exist? a. no b. yes 13. How does the deconstruction of a photon into an electron and a positron uphold the principles of mass and charge conservation? a. The sum of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the charges on an electron and a positron is equal to the zero charge of the photon. b. The sum
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of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the same charges on an electron and a positron is equal to the charge on a photon. c. During the particle production the total energy of the photon is converted to the mass of an electron and a positron. The sum of the opposite charges on the electron and positron is equal to the zero charge of the photon. d. During particle production, the total energy of the photon is converted to the mass of an electron and a positron. The sum of the same charges on an electron and a positron is equal to the charge on a photon. 14. How many fundamental particles exist in the Standard Model, including the Higgs boson and the graviton (not yet observed)? 12 a. 15 b. 13 c. 19 d. 15. Why do gluons interact only with particles in the first two rows of the Standard Model? a. The leptons in the third and fourth rows do not have mass, but the gluons can interact between the quarks through gravity only. b. The leptons in the third and fourth rows do not have color, but the gluons can interact between quarks through color interactions only. c. The leptons in the third and fourth rows do not have spin, but the gluons can interact between quarks through spin interactions only. d. The leptons in the third and fourth rows do not have charge, but the gluons can interact between quarks through charge interactions only. 16. What fundamental property is provided by particle interaction with the Higgs boson? charge a. b. mass spin c. color d. 17. Considering the Higgs field, what differentiates more massive particles from less massive particles? a. More massive particles interact more with the Higgs field than the less massive particles. b. More massive particles interact less with the Higgs field than the less massive particles. 18. What particles were launched into the proton during the original discovery of the quark? a. bosons b. electrons c. neutrons d. photons 790 Chapter 23 • Particle Physics 23.3 The Unification of Forces Section Learning Objectives By the end of the section, you will be able to do the following: • Define a grand unified theory and its importance • Explain the evolution of the four fundamental forces from the Big Bang onward • Explain how grand unification theories can be
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tested Section Key Terms Big Bang Inflationary Epoch Electroweak Epoch electroweak theory Grand Unification Epoch Grand Unified Theory Planck Epoch Quark Era superforce Theory of Everything Understanding the Grand Unified Theory Present quests to show that the four basic forces are different manifestations of a single unified force that follow a long tradition. In the nineteenth century, the distinct electric and magnetic forces were shown to be intimately connected and are now collectively called the electromagnetic force. More recently, the weak nuclear force was united with the electromagnetic force. As shown in Figure 23.19, carrier particles transmit three of the four fundamental forces in very similar ways. With these considerations in mind, it is natural to suggest that a theory may be constructed in which the strong nuclear, weak nuclear, and electromagnetic forces are all unified. The search for a correct theory linking the forces, called the Grand Unified Theory (GUT), is explored in this section. In the 1960s, the electroweak theory was developed by Steven Weinberg, Sheldon Glashow, and Abdus Salam. This theory proposed that the electromagnetic and weak nuclear forces are identical at sufficiently high energies. At lower energies, like those in our present-day universe, the two forces remain united but manifest themselves in different ways. One of the main consequences of the electroweak theory was the prediction of three short-range carrier particles, now known as the and and that of the predicted characteristics, including masses having those predicted values as given in. boson was predicted to be 90 GeV/c2. In 1983, these carrier particles were observed at CERN with the bosons. Not only were three particles predicted, but the mass of each boson was predicted to be 81 GeV/c2, and How can forces be unified? They are definitely distinct under most circumstances. For example, they are carried by different particles and have greatly different strengths. But experiments show that at extremely short distances and at extremely high energies, the strengths of the forces begin to become more similar, as seen in Figure 23.20. Figure 23.19 The exchange of a virtual particle (boson) carries the weak nuclear force between an electron and a neutrino in this Feynman diagram. This diagram is similar to the diagrams in Figure 23.6 and for the electromagnetic and strong nuclear forces. As discussed earlier, the short ranges and large masses of the weak carrier bosons require correspondingly high energies to create them. Thus, the energy scale on the horizontal axis of Figure 23.20 also corresponds to
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shorter and shorter distances Access for free at openstax.org. 23.3 • The Unification of Forces 791 (going from left to right), with 100 GeV corresponding to approximately 10−18 m, for example. At that distance, the strengths of the electromagnetic and weak nuclear forces are the same. To test this, energies of about 100 GeV are put into the system. When this occurs, the, and carrier particles become less and less relevant, and the further energy is added, the similar to photons and gluons. carrier particles are created and released. At those and higher energies, the masses of the boson in particular resembles the massless, chargeless photon. As particles are further transformed into massless carrier particles even more, and,, Figure 23.20 The relative strengths of the four basic forces vary with distance, and, hence, energy is needed to probe small distances. At ordinary energies (a few eV or less), the forces differ greatly. However, at energies available in accelerators, the weak nuclear and electromagnetic (EM) forces become unified. Unfortunately, the energies at which the strong nuclear and electroweak forces become the same are unreachable in any conceivable accelerator. The universe may provide a laboratory, and nature may show effects at ordinary energies that give us clues about the validity of this graph. The extremely short distances and high energies at which the electroweak force becomes identical with the strong nuclear force are not reachable with any conceivable human-built accelerator. At energies of about 1014 GeV (16,000 J per particle), distances of about 10 to 30 m can be probed. Such energies are needed to test the theory directly, but these are about 1010 times higher than the maximum energy associated with the LHC, and the distances are about 10 to 12 smaller than any structure we have direct knowledge of. This would be the realm of various GUTs, of which there are many, since there is no constraining evidence at these energies and distances. Past experience has shown that anytime you probe so many orders of magnitude further, you find the unexpected. While direct evidence of a GUT is not presently possible, that does not rule out the ability to assess a GUT through an indirect process. Current GUTs require various other events as a consequence of their theory. Some GUTs require the existence of magnetic monopoles, very massive individual north- and south-pole particles, which have not yet been proven to exist, while others require the use
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of extra dimensions. However, not all theories result in the same consequences. For example, disproving the existence of magnetic monopoles will not disprove all GUTs. Much of the science we accept in our everyday lives is based on different models, each with their own strengths and limitations. Although a particular model may have drawbacks, that does not necessarily mean that it should be discounted completely. One consequence of GUTs that can theoretically be assessed is proton decay. Multiple current GUTs hypothesize that the stable proton should actually decay at a lifetime of 1031 years. While this time is incredibly large (keep in mind that the age of the universe is less than 14 billion years), scientists at the Super-Kamiokande in Japan have used a 50,000-ton tank of water to search for its existence. The decay of a single proton in the Super-Kamiokande tank would be observed by a detector, thereby providing support for the predicting GUT model. However, as of 2014, 17 years into the experiment, decay is yet to be found. This time span equates to a minimum limit on proton life of grand unifying theories, an acceptable model may still exist. years. While this result certainly does not support many TIPS FOR SUCCESS The Super-Kamiokande experiment is a clever use of proportional reasoning. Because it is not feasible to test for 1031 years in order for a single proton to decay, scientists chose instead to manipulate the proton–time ratio. If one proton decays in 1031 792 Chapter 23 • Particle Physics years, then in one year 10−31 protons will decay. With this in mind, if scientists wanted to test the proton decay theory in one year, they would need 1031 protons. While this is also unfeasible, the use of a 50,000-ton tank of water helps to bring both the wait time and proton number to within reason. The Standard Model and the Big Bang Nature is full of examples where the macroscopic and microscopic worlds intertwine. Newton realized that the nature of gravity on Earth that pulls an apple to the ground could explain the motion of the moon and planets so much farther away. Decays of tiny nuclei explain the hot interior of the Earth. Fusion of nuclei likewise explains the energy of stars. Today, the patterns in particle physics seem to be explaining the evolution and character of the universe. And the nature of the universe has implications for
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unexplored regions of particle physics. In 1929, Edwin Hubble observed that all but the closest galaxies surrounding our own had a red shift in their hydrogen spectra that was proportional to their distance from us. Applying the Doppler Effect, Hubble recognized that this meant that all galaxies were receding from our own, with those farther away receding even faster. Knowing that our place in the universe was no more unique than any other, the implication was clear: The space within the universe itself was expanding. Just like pen marks on an expanding balloon, everything in the universe was accelerating away from everything else. Figure 23.21 shows how the recession of galaxies looks like the remnants of a gigantic explosion, the famous Big Bang. Extrapolating backward in time, the Big Bang would have occurred between 13 and 15 billion years ago, when all matter would have been at a single point. From this, questions instantly arise. What caused the explosion? What happened before the Big Bang? Was there a before, or did time start then? For our purposes, the biggest question relating to the Big Bang is this: How does the Big Bang relate to the unification of the fundamental forces? Figure 23.21 Galaxies are flying apart from one another, with the more distant ones moving faster, as if a primordial explosion expelled the matter from which they formed. The most distant known galaxies move nearly at the speed of light relative to us. To fully understand the conditions of the very early universe, recognize that as the universe contracts to the size of the Big Bang, changes will occur. The density and temperature of the universe will increase dramatically. As particles become closer together, they will become too close to exist as we know them. The high energies will create other, more unusual particles to exist in greater abundance. Knowing this, let’s move forward from the start of the universe, beginning with the Big Bang, as illustrated in Figure 23.22. Access for free at openstax.org. 23.3 • The Unification of Forces 793 Figure 23.22 The evolution of the universe from the Big Bang onward (from left to right) is intimately tied to the laws of physics, especially those of particle physics at the earliest stages. Theories of the unification of forces at high energies may be verified by their shaping of the universe and its evolution. The Planck Epoch —Though scientists are unable to model the conditions of the Planck Epoch in the laboratory, GeV necessary to unify gravity speculation is that at
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this time compressed energy was great enough to reach the immense with all other forces. As a result, modern cosmology suggests that all four forces would have existed as one force, a hypothetical superforce as suggested by the Theory of Everything. The Grand Unification Epoch —As the universe expands, the temperatures necessary to maintain the superforce decrease. As a result, gravity separates, leaving the electroweak and strong nuclear forces together. At this time, the electromagnetic, weak, and strong forces are identical, matching the conditions requested in the Grand Unification Theory. The Inflationary Epoch —The separation of the strong nuclear force from the electroweak force during this time is thought to have been responsible for the massive inflation of the universe. Corresponding to the steep diagonal line on the left side of Figure 23.22, the universe may have expanded by a factor of great during this time that it actually occurred faster than the speed of light! Unfortunately, there is little hope that we may be GeV, vastly greater than the limits of modern able to test the inflationary scenario directly since it occurs at energies near accelerators. or more in size. In fact, the expansion was so The Electroweak Epoch —Now separated from both gravity and the strong nuclear force, the electroweak force exists as a singular force during this time period. As stated earlier, scientists are able to create the energies at this stage in the universe’s expansion, needing only 100 GeV, as shown in Figure 23.20. W and Z bosons, as well as the Higgs boson, are released during this time. The Quark Era —During the Quark Era, the universe has expanded and temperatures have decreased to the 794 Chapter 23 • Particle Physics point at which all four fundamental forces have separated. Additionally, quarks began to take form as energies decreased. As the universe expanded, further eras took place, allowing for the existence of hadrons, leptons, and photons, the fundamental particles of the standard model. Eventually, in nucleosynthesis, nuclei would be able to form, and the basic building blocks of atomic matter could take place. Using particle accelerators, we are very much working backwards in an attempt to understand the universe. It is encouraging to see that the macroscopic conditions of the Big Bang align nicely with our submicroscopic particle theory. Check Your Understanding 19. Is there one grand unified theory or multiple grand unifying theories? a. one grand unifying theory b. multiple grand
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unifying theories 20. In what manner is considered a precursor to the Grand Unified Theory? a. The grand unified theory seeks relate the electroweak and strong nuclear forces to one another just as related energy and mass. b. The grand unified theory seeks to relate the electroweak force and mass to one another just as related energy and mass. c. The grand unified theory seeks to relate the mass and strong nuclear forces to one another just as related energy and mass. d. The grand unified theory seeks to relate gravity and strong nuclear force to one another, just as related energy and mass. 21. List the following eras in order of occurrence from the Big Bang: Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Planck Epoch, Quark Era. a. Quark Era, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Planck Epoch b. Planck Epoch, Inflationary Epoch, Grand Unification Epoch, Electroweak Epoch, Quark Era c. Planck Epoch, Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Quark Era d. Planck Epoch, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Quark Era 22. How did the temperature of the universe change as it expanded? a. The temperature of the universe increased. b. The temperature of the universe decreased. c. The temperature of the universe first decreased and then increased. d. The temperature of the universe first increased and then decreased. 23. Under current conditions, is it possible for scientists to use particle accelerators to verify the Grand Unified Theory? a. No, there is not enough energy. b. Yes, there is enough energy. 24. Why are particles and antiparticles made to collide as shown in this image? a. Particles and antiparticles have the same mass. b. Particles and antiparticles have different mass. c. Particles and antiparticles have the same charge. d. Particles and antiparticles have opposite charges. 25. The existence of what particles were predicted as a consequence of the electroweak theory? a. fermions b. Higgs bosons Access for free at openstax.org. leptons c. d. W+, W-, and Z0 bosons 23.3 • The Unification of Forces 795 796 Chapter 23 • Key Terms KEY TERMS boson positive carrier particle
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och the time period from 10−43 to 10−34 seconds after the Big Bang, when Grand Unification Theory, in which all forces except gravity are identical, governed the universe Grand Unified Theory theory that shows unification of the strong and electroweak forces graviton hypothesized particle exchanged between two particles of mass, transmitting the gravitational force between them hadron particles composed of quarks that feel the strong and weak nuclear force Quark Era the time period from 10–11 to 10–6 seconds at which all four fundamental forces are separated and quarks begin to exit Standard Model an organization of fundamental particles and forces that is a result of quantum chromodynamics and electroweak theory strange quark the third lightest of all quarks superforce the unification of all four fundamental forces into one force synchrotron a version of a cyclotron in which the frequency of the alternating voltage and the magnetic field strength are increased as the beam particles are accelerated Theory of Everything the theory that shows unification of all four fundamental forces top quark a quark flavor up quark the lightest of all quarks weak nuclear force fundamental force responsible for Higgs boson a massive particle that provides mass to the weak bosons and provides validity to the theory that particle decay Access for free at openstax.org. SECTION SUMMARY 23.1 The Four Fundamental Forces • The four fundamental forces are gravity, the electromagnetic force, the weak nuclear force, and the strong nuclear force. • A variety of particle accelerators have been used to explore the nature of subatomic particles and to test predictions of particle theories. 23.2 Quarks • There are three types of fundamental particles—leptons, quarks, and carrier particles. • Quarks come in six flavors and three colors and occur only in combinations that produce white. • Hadrons are thought to be composed of quarks, with baryons having three quarks and mesons having a quark and an antiquark. • Known particles can be divided into three major groups—leptons, hadrons, and carrier particles (gauge bosons). • All particles of matter have an antimatter counterpart that has the opposite charge and certain other quantum CHAPTER REVIEW Concept Items 23.1 The Four Fundamental Forces 1. What forces does the inverse square law describe? the electromagnetic and weak nuclear force the electromagnetic force and strong nuclear force the electromagnetic force and gravity the strong nuclear force and gravity a. b. c. d. 2. Do the carrier particles explain the loss of mass in nuclear decay
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? a. no b. yes 3. What happens to the rate of voltage oscillation within a synchrotron each time the particle completes a loop? a. The rate of voltage oscillation increases as the particle travels faster and faster on each loop. b. The rate of voltage oscillation decreases as the particle travels faster and faster on each loop. c. The rate of voltage oscillation remains the same each time the particle completes a loop. d. The rate of voltage oscillation first increases and then remains constant each time the particle completes a loop. 4. Which of the four forces is responsible for ionic bonding? a. electromagnetic force Chapter 23 • Section Summary 797 numbers. These matter–antimatter pairs are otherwise very similar but will annihilate when brought together. • The strong force is carried by eight proposed particles called gluons, which are intimately connected to a quantum number called color—their governing theory is thus called quantum chromodynamics (QCD). Taken together, QCD and the electroweak theory are widely accepted as the Standard Model of particle physics. 23.3 The Unification of Forces • Attempts to show unification of the four forces are called Grand Unified Theories (GUTs) and have been partially successful, with connections proven between EM and weak forces in electroweak theory. • Unification of the strong force is expected at such high energies that it cannot be directly tested, but it may have observable consequences in the as-yet-unobserved decay of the proton. Although unification of forces is generally anticipated, much remains to be done to prove its validity. b. gravity c. strong force d. weak nuclear force 5. What type of particle accelerator uses oscillating electric fields to accelerate particles around a fixed radius track? a. LINAC b. c. SLAC d. Van de Graaff accelerator synchrotron 23.2 Quarks 6. How does the charge of an individual quark determine hadron structure? a. Since the hadron must have an integral value, the individual quarks must be combined such that the average of their charges results in the value of a quark. b. Since the hadron must have an integral value, the individual atoms must be combined such that the sum of their charges is less than zero. c. The individual quarks must be combined such that the product of their charges is equal to the total charge of the hadron structure. d. Since the hadron must have an integral value of charge, the individual qu
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arks must be combined such that the sum of their charges results in an 798 Chapter 23 • Chapter Review integral value. 7. Why do leptons not feel the strong nuclear force? a. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are constructed of quarks that do not have gluons. b. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are not constructed of quarks and are not massive. c. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are constructed of the quarks that are not massive. d. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are not constructed of quarks, nor do they have color constituents. 8. What property commonly distinguishes antimatter from its matter analogue? a. mass b. charge c. energy speed d. 9. Can the Standard Model change as new information is gathered? a. yes b. no account for the remaining excess mass in protons compared to electrons. b. The highly energetic photons connecting the quarks account for the remaining excess mass in protons compared to electrons. c. The antiparallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. d. The parallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. 23.3 The Unification of Forces 13. Why is the unification of fundamental forces important? a. The unification of forces will help us understand fundamental structures of the universe. b. The unification of forces will help in the proof of the graviton. c. The unification of forces will help in achieving a speed greater than the speed of light. d. The unification of forces will help in studying antimatter particles. 14. Why are scientists unable to model the conditions of the universe at time periods shortly after the Big Bang? a. The amount of energy necessary to replicate the Planck Epoch is too high. b. The amount of energy necessary to replicate the Planck Epoch is too low. c. The volume of setup necessary to replicate the 10. What is the relationship between the Higgs field and the Planck Epoch is too high. Higgs boson? a. The H
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iggs boson is the carrier that transfers force d. The volume of setup necessary to replicate the Planck Epoch is too low. for the Higgs field. b. The Higgs field is the time duration over which the Higgs particles transfer force to the other particles. c. The Higgs field is the magnitude of momentum transferred by the Higgs particles to the other particles. 15. What role does proton decay have in the search for GUTs? a. Proton decay is a premise of a number of GUTs. b. Proton decay negates the validity of a number of GUTs. d. The Higgs field is the magnitude of torque transfers 16. What is the name for the theory of unification of all four by the Higgs particles on the other particles. 11. What were the original three flavors of quarks discovered? a. up, down, and charm b. up, down, and bottom c. up, down, and strange d. up, down, and top 12. Protons are more massive than electrons. The three quarks in the proton account for only a small amount of this mass difference. What accounts for the remaining excess mass in protons compared to electrons? a. The highly energetic gluons connecting the quarks fundamental forces? a. b. c. d. the theory of everything the theory of energy-to-mass conversion the theory of relativity the theory of the Big Bang 17. Is it easier for scientists to find evidence for the Grand Unified Theory or the Theory of Everything? Explain. a. Theory of Everything, because it requires of energy b. Theory of Everything, because it requires of energy c. Grand Unified Theory, because it requires Access for free at openstax.org. Chapter 23 • Chapter Review 799 of energy d. Grand Unified Theory, because it requires of energy Critical Thinking Items 23.1 The Four Fundamental Forces 18. The gravitational force is considered a very weak force. Yet, it is strong enough to hold Earth in orbit around the Sun. Explain this apparent disparity. a. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because neither the strong nor the weak nuclear force exists at this distance. b. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because both the strong and the weak nuclear force is minimal at this distance. 19. True or False—Given that their carrier particles are massless, some may argue
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that the electromagnetic and gravitational forces should maintain the same value at all distances from their source. However, both forces decrease with distance at a rate of a. b. false true 20. Why is a stationary target considered inefficient in a particle accelerator? a. The stationary target recoils upon particle strike, thereby transferring much of the particle’s energy into its motion. As a result, a greater amount of energy goes into breaking the particle into its constituent components. b. The stationary target contains zero kinetic energy, so it requires more energy to break the particle into its constituent components. c. The stationary target contains zero potential energy, so it requires more energy to break the particle into its constituent components. d. The stationary target recoils upon particle strike, transferring much of the particle’s energy into its motion. As a result, a lesser amount of energy goes into breaking the particle into its constituent components. 21. Compare the total strong nuclear force in a lithium atom. to the total strong nuclear force in a lithium ion a. The total strong nuclear force in a lithium atom is thrice the total strong nuclear force in a lithium ion. b. The total strong nuclear force in a lithium atom is twice the total strong nuclear force in a lithium ion. c. The total strong nuclear force in a lithium atom is the same as the total strong nuclear force in a lithium ion. d. The total strong nuclear force in a lithium atom is half the total strong nuclear force in a lithium ion. 23.2 Quarks 22. Explain why it is not possible to find a particle composed of just two quarks. a. A particle composed of two quarks will have an integral charge and a white color. Hence, it cannot exist. b. A particle composed of two quarks will have an integral charge and a color that is not white. Hence, it cannot exist. c. A particle composed of two quarks will have a fractional charge and a white color. Hence, it cannot exist. d. A particle composed of two quarks will have a fractional charge and a color that is not white. Hence, it cannot exist. 23. Why are mesons considered unstable? a. Mesons are composites of two antiparticles that quickly annihilate each other. b. Mesons are composites of two particles that quickly annihilate each other. c. Mesons are composites of a particle and antiparticle that quickly annihilate each other. d. Mesons are composites of two particles and
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one antiparticle that quickly annihilate each other. 24. Does antimatter have a negative mass? a. No, antimatter does not have a negative mass. b. Yes, antimatter does have a negative mass. 25. What similarities exist between the Standard Model and the periodic table of elements? a. During their invention, both the Standard Model and the periodic table organized material by mass. b. At the times of their invention, both the Standard Model and the periodic table organized material by charge. c. At the times of their invention, both the Standard Model and the periodic table organized material by interaction with other available particles. d. At the times of their invention, both the Standard Model and the periodic table organized material by size. 26. How were particle collisions used to provide evidence of the Higgs boson? 800 Chapter 23 • Chapter Review a. Because some particles do not contain the Higgs boson, the collisions of such particles will cause their destruction. b. Because only the charged particles contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. c. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will absorb the Higgs boson. d. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. 27. Explain how the combination of a quark and antiquark can result in the creation of a hadron. a. The combination of a quark and antiquark can result in a particle with an integer charge and color of white, therefore satisfying the properties for a hadron. b. The combination of a quark and antiquark must result in a particle with a negative charge and color of white, therefore satisfying the properties for a hadron. c. The combination of a quark and antiquark can result in a particle with an integer charge and color that is not white, therefore satisfying the properties for a hadron. d. The combination of a quark and antiquark can result in particle with a fractional charge and color that is not white, therefore satisfying the properties for a hadron. 23.3 The Unification of Forces 28. Why does the strength of the strong force diminish under high-energy conditions? a. Under high-energy conditions, particles interacting under the strong force will be compressed closer together. As a result, the force between them will decrease
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. b. Under high-energy conditions, particles interacting under the strong force will start oscillating. As a result, the force between them will increase. c. Under high-energy conditions, particles interacting under the strong force will have high velocity. As a result, the force between them will decrease. d. Under high-energy conditions, particles interacting under the strong force will start moving randomly. As a result, the force between them will decrease. 29. If some unknown cause of the red shift, such as light becoming tiredfrom traveling long distances through empty space, is discovered, what effect would there be on cosmology? a. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are moving toward one another. b. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that the galaxies are moving away from one another. c. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are neither moving away from nor moving toward one another. d. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are sometimes moving away from and sometimes moving toward one another. 30. How many molecules of water are necessary if scientists -yr estimate of proton decay wanted to check the within the course of one calendar year? a. b. c. d. 31. As energy of interacting particles increases toward the theory of everything, the gravitational force between them increases. Why does this occur? a. As energy increases, the masses of the interacting particles will increase. b. As energy increases, the masses of the interacting particles will decrease. c. As energy increases, the masses of the interacting particles will remain constant. d. As energy increases, the masses of the interacting particles starts changing (increasing or decreasing). As a result, the gravitational force between the particles will increase. Access for free at openstax.org. Performance Task 23.3 The Unification of Forces 32. Communication is an often overlooked and useful skill for a scientist, especially in a competitive field where financial resources are limited. Scientists are often required to explain their findings or the relevance of their work to agencies within the government in order to maintain funding to continue their research. Let’s say you are an ambitious young particle physicist, heading an expensive project, and you need to justify its existence to the appropriate funding agency
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. Write a brief paper (about one page) explaining why molecularlevel structure is important in the functioning of designed materials in a specific industry. TEST PREP Multiple Choice 23.1 The Four Fundamental Forces 33. Which of the following is not one of the four fundamental forces? a. gravity friction b. strong nuclear c. d. electromagnetic 34. What type of carrier particle has not yet been found? a. gravitons b. bosons c. bosons d. pions 35. What effect does an increase in electric potential have It increases accelerating capacity. It decreases accelerating capacity. on the accelerating capacity of a Van de Graaff generator? a. b. c. The accelerating capacity of a Van de Graaff generator is constant regardless of electric potential. Chapter 23 • Test Prep 801 • First, think of an industry where molecular-level structure is important. • Research what materials are used in that industry as well as what are the desired properties of the materials. • What molecular-level characteristics lead to what properties? One example would be explaining how flexible but durable materials are made up of long-chained molecules and how this is useful for finding more environmentally friendly alternatives to plastics. Another example is explaining why electrically conductive materials are often made of metal and how this is useful for developing better batteries. and the electromagnetic force 23.2 Quarks 37. To what color must quarks combine for a particle to be constructed? a. black b. green red c. d. white 38. What type of hadron is always constructed partially of an antiquark? a. baryon b. lepton c. meson d. photno 39. What particle is typically released when two particles annihilate? a. graviton b. antimatter c. pion d. photno d. Van de Graaff generators do not have the capacity 40. Which of the following categories is not one of the three to accelerate particles. 36. What force or forces exist between a proton and a second proton? a. The weak electrostatic force and strong magnetic force main categories of the Standard Model? a. gauge bosons b. hadrons leptons c. d. quarks b. The weak electrostatic and strong gravitational 41. Analysis of what particles began the search for the Higgs force c. The weak frictional force and strong gravitational force d. The weak nuclear force, the strong nuclear force, boson? a. W and Z bosons b. up and down quarks c
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. mesons and baryons 802 Chapter 23 • Test Prep d. neutrinos and photons 44. After the Big Bang, what was the first force to separate 42. What similarities exist between the discovery of the quark and the discovery of the neutron? a. Both the quark and the neutron were discovered by launching charged particles through an unknown structure and observing the particle recoil. b. Both the quark and the neutron were discovered by launching electrically neutral particles through an unknown structure and observing the particle recoil. c. Both quarks and neutrons were discovered by studying their deflection under an electric field. 23.3 The Unification of Forces 43. Which two forces were first combined, signifying the eventual desire for a Grand Unified Theory? a. electric force and magnetic forces b. electric force and weak nuclear force c. gravitational force and the weak nuclear force d. electroweak force and strong nuclear force Short Answer 23.1 The Four Fundamental Forces 47. Why do people tend to be more aware of the gravitational and electromagnetic forces than the strong and weak nuclear forces? a. The gravitational and electromagnetic forces act at short ranges, while strong and weak nuclear forces act at comparatively long range. b. The strong and weak nuclear forces act at short ranges, while gravitational and electromagnetic forces act at comparatively long range. c. The strong and weak nuclear forces act between all objects, while gravitational and electromagnetic forces act between smaller objects. d. The strong and weak nuclear forces exist in outer space, while gravitational and electromagnetic forces exist everywhere. 48. What fundamental force is responsible for the force of friction? a. b. c. the electromagnetic force the strong nuclear force the weak nuclear force 49. How do carrier particles relate to the concept of a force field? a. Carrier particles carry mass from one location to another within a force field. b. Carrier particles carry force from one location to another within a force field. Access for free at openstax.org. from the others? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 45. What is the name of the device used by scientists to check for proton decay? the cyclotron a. the Large Hadron Collider b. the Super-Kamiokande c. the synchrotron d. 46. How do Feynman diagrams suggest the Grand Unified Theory? a. The electromagnetic, weak, and strong nuclear forces all have similar Feynman diagrams. b. The electromagnetic, weak, and gravitational forces all
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have similar Feynman diagrams. c. The electromagnetic, weak, and strong forces all have different Feynman diagrams. c. Carrier particles carry charge from one location to another within a force field. d. Carrier particles carry volume from one location to another within a force field. 50. Which carrier particle is transmitted solely between nucleons? a. graviton b. photon c. pion d. W and Z bosons 51. Two particles of the same mass are traveling at the same speed but in opposite directions when they collide headon. What is the final kinetic energy of this two-particle system? a. b. c. zero d. infinite the sum of the kinetic energies of the two particles the product of the kinetic energies of the two particles 52. Why do colliding beams result in the location of smaller particles? a. Colliding beams create energy, allowing more energy to be used to separate the colliding particles. b. Colliding beams lower the energy of the system, so it requires less energy to separate the colliding particles. c. Colliding beams reduce energy loss, so less energy is required to separate colliding particles. c. The four fundamental forces are represented by d. Colliding beams reduce energy loss, allowing more their carrier particles, the leptons. energy to be used to separate the colliding particles. d. The four fundamental forces are represented by their carrier particles, the quarks. Chapter 23 • Test Prep 803 23.2 Quarks 53. What two features of quarks determine the structure of a particle? a. b. c. d. the color and charge of individual quarks the color and size of individual quarks the charge and size of individual quarks the charge and mass of individual quarks 54. What fundamental force does quantum chromodynamics describe? the weak nuclear force a. the strong nuclear force b. the electromagnetic force c. the gravitational force d. 55. Is it possible for a baryon to be constructed of two quarks and an antiquark? a. Yes, the color of the three particles would be able to sum to white. b. No, the color of the three particles would not be able to sum to white. 56. Can baryons be more massive than mesons? a. no b. yes 57. If antimatter exists, why is it so difficult to find? a. There is a smaller amount of antimatter than matter in the universe; antimatter is quickly annihilated by its matter analogue
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. b. There is a smaller amount of matter than antimatter in the universe; matter is annihilated by its antimatter analogue. c. There is a smaller amount of antimatter than matter in universe; antimatter and its matter analogue coexist. d. There is a smaller amount of matter than antimatter in the universe; matter and its antimatter analogue coexist. 58. Does a neutron have an antimatter counterpart? a. No, the antineutron does not exist. b. Yes, the antineutron does exist. 59. How are the four fundamental forces incorporated into the Standard Model of the atom? a. The four fundamental forces are represented by their carrier particles, the electrons. 60. Which particles in the Standard Model account for the majority of matter with which we are familiar? a. particles in fourth column of the Standard Model b. particles in third column of the Standard Model c. particles in the second column of the Standard Model d. particles in the first column of the Standard Model 61. How can a particle gain mass by traveling through the Higgs field? a. The Higgs field slows down passing particles; the decrease in kinetic energy is transferred to the particle’s mass. b. The Higgs field accelerates passing particles; the decrease in kinetic energy is transferred to the particle’s mass. c. The Higgs field slows down passing particles; the increase in kinetic energy is transferred to the particle’s mass. d. The Higgs field accelerates passing particles; the increase in kinetic energy is transferred to the particle’s mass. 62. How does mass-energy conservation relate to the Higgs field? a. The increase in a particle’s energy when traveling through the Higgs field is countered by its increase in mass. b. The decrease in a particle’s kinetic energy when traveling through the Higgs field is countered by its increase in mass. c. The decrease in a particle’s energy when traveling through the Higgs field is countered by its decrease in mass. d. The increase in a particle’s energy when traveling through the Higgs field is countered by its decrease in mass. 23.3 The Unification of Forces 63. Why do scientists believe that the strong nuclear force and the electroweak force will combine under high energies? a. The electroweak force will have greater strength. b. The strong nuclear force and electroweak force will achieve the same strength. c. The strong nuclear force
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will have greater strength. b. The four fundamental forces are represented by 64. At what energy will the strong nuclear force their carrier particles, the gauge bosons. theoretically unite with the electroweak force? 804 Chapter 23 • Test Prep a. b. c. d. 65. While we can demonstrate the unification of certain forces within the laboratory, for how long were the four forces naturally unified within the universe? a. b. c. d. 66. How does the search for the Grand Unified Theory help test the standard cosmological model? a. Scientists are increasing energy in the lab that models the energy in earlier, denser stages of the universe. b. Scientists are increasing energy in the lab that models the energy in earlier, less dense stages of the universe. c. Scientists are decreasing energy in the lab that models the energy in earlier, denser stages of the universe. d. Scientists are decreasing energy in the lab that Extended Response 23.1 The Four Fundamental Forces 69. If the strong attractive force is the greatest of the four fundamental forces, are all masses fated to combine together at some point in the future? Explain. a. No, the strong attractive force acts only at incredibly small distances. As a result, only masses close enough to be within its range will combine. b. No, the strong attractive force acts only at large distances. As a result, only masses far enough apart will combine. c. Yes, the strong attractive force acts at any distance. As a result, all masses are fated to combine together at some point in the future. d. Yes, the strong attractive force acts at large distances. As a result, all masses are fated tocombine together at some point in the future. 70. How does the discussion of carrier particles relate to the concept of relativity? a. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel more slowly than the speed of sound. b. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of light. Access for free at openstax.org. models the energy in earlier, less dense stages of the universe. 67. Why does finding proof that protons do not decay not disprove all GUTs? a. Proton decay is not a premise of all GUTs, and current GUTs can be amended in response to new findings. b. Proton decay is a premise of all GUT
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s, but current GUTs can be amended in response to new findings. 68. When accelerating elementary particles in a particle accelerator, they quickly achieve a speed approaching the speed of light. However, as time continues, the particles maintain this speed yet continue to increase their kinetic energy. How is this possible? a. The speed remains the same, but the masses of the particles increase. b. The speed remains the same, but the masses of the particles decrease. c. The speed remains the same, and the masses of the particles remain the same. d. The speed and masses will remain the same, but temperature will increase. c. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of sound. d. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel faster than the speed of light. 71. Why are synchrotrons constructed to be very large? a. By using a large radius, high particle velocities can be achieved using a large centripetal force created by large electromagnets. b. By using a large radius, high particle velocities can be achieved without a large centripetal force created by large electromagnets. c. By using a large radius, the velocities of particles can be reduced without a large centripetal force created by large electromagnets. d. By using a large radius, the acceleration of particles can be decreased without a large centripetal force created by large electromagnets. 23.2 Quarks 72. In this image, how does the emission of the gluon cause the down quark to change from a red color to a green color? Chapter 23 • Test Prep 805 a. b. c. d. If the velocity of the neutrino is known, then the upper limit on mass of the neutrino can be set. If only the kinetic energy of the neutrino is known, then the upper limit on mass of the neutrino can be set. If either the velocity or the kinetic energy is known, then the upper limit on the mass of the neutrino can be set. If both the kinetic energy and the velocity of the neutrino are known, then the upper limit on the mass of the neutrino can be set. 76. The term force carrier particleis shorthand for the scientific term vector gauge boson. From that perspective, can the H
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iggs boson truly be considered a force carrier particle? a. No, the mass quality provided by the Higgs boson is a scalar quantity. b. Yes, the mass quality provided by the Higgs boson results in a change of particle’s direction. 23.3 The Unification of Forces 77. If a Grand Unified Theory is proven and the four forces are unified, it will still be correct to say that the orbit of the Moon is determined by the gravitational force. Explain why. a. Gravity will not be a property of the unified force. b. Gravity will be one property of the unified force. c. Apart from gravity, no other force depends on the mass of the object. d. Apart from gravity, no other force can make an a. The emitted red gluon is made up of a green and a red color. As a result, the down quark changes from a red color to a green color. b. The emitted red gluon is made up of an anti-green and an anti-red color. As a result, the down quark changes from a red color to a green color. c. The emitted red gluon is made up of a green and an anti-red color. As a result, the down quark changes from a red color to a green color. d. The emitted red gluon is made up of an anti-green and a red color. As a result, the down quark changes from a red color to a green color. 73. Neutrinos are much more difficult for scientists to find when compared to other hadrons and leptons. Why is this? a. Neutrinos are hadrons, and they lack charge. b. Neutrinos are not hadrons, and they lack charge. c. Neutrinos are hadrons, and they have positive charge. d. Neutrinos are not hadrons, and they have a positive object move in a fixed orbit. charge. 74. What happens to the masses of a particle and its antiparticle when the two annihilate at low energies? a. The masses of the particle and antiparticle are transformed into energy in the form of photons. b. The masses of the particle and antiparticle are converted into kinetic energy of the particle and antiparticle respectively. c. The mass of the antiparticle is converted into kinetic energy of the particle. d. The mass of the particle is converted into radiation energy of the
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antiparticle. 75. When a star erupts in a supernova explosion, huge numbers of electron neutrinos are formed in nuclear reactions. Such neutrinos from the 1987A supernova in the relatively nearby Magellanic Cloud were observed within hours of the initial brightening, indicating that they traveled to earth at approximately the speed of light. Explain how this data can be used to set an upper limit on the mass of the neutrino. 78. As the universe expanded and temperatures dropped, the strong nuclear force separated from the electroweak force. Is it likely that under cooler conditions, the force of electricity will separate from the force of magnetism? a. No, the electric force relies on the magnetic force and vice versa. b. Yes, the electric and magnetic forces can be separated from each other. 79. Two pool balls collide head-on and stop. Their original kinetic energy is converted to heat and sound. Given that this is not possible for particles, what happens to their converted energy? a. The kinetic energy is converted into relativistic potential energy, governed by the equation. b. The kinetic energy is converted into relativistic mass, governed by the equation. c. The kinetic energy is converted into relativistic potential energy, governed by the equation. 806 Chapter 23 • Test Prep d. Their kinetic energy is converted into relativistic mass, governed by the equation. Access for free at openstax.org. Appendix A • Reference Tables 807 APPENDIX A Reference Tables Figure A1 Periodic Table of Elements Prefix Symbol Value Prefix Symbol Value tera giga T G mega M kilo hecto k h deka da 1012 deci d c centi milli m micro nano pico µ n p 109 106 103 102 101 10–1 10–2 10–3 10–6 10–9 10–12 Table A1 Metric Prefixes for Powers of Ten and Their Symbols 808 Appendix A • Reference Tables Prefix Symbol Value Prefix Symbol Value ___ 100 femto f 10–15 Table A1 Metric Prefixes for Powers of Ten and Their Symbols Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force m kg s A rad Energy Power Pressure Frequency Electronic potential Capacitance Charge Resistance Magnetic field meter kilogram second ampere radian newton joule watt pascal hertz volt farad coulomb ohm tesla Nuclear decay rate becquerel Table A2 SI
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Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Table A3 Selected British Units Access for free at openstax.org. Appendix A • Reference Tables 809 Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055 × 103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb/in2 = 6.895 × 103 Pa Table A3 Selected British Units Length 1 light year (ly) = 9.46 × 1015 m 1 astronomical unit (au) = 1.50 × 1011 m 1 nautical mile = 1.852 km 1 angstrom(Å) = 10-10 m Area 1 acre (ac) = 4.05 × 103 m2 1 square foot (ft2) 9.29 × 10-2 m3 1 barn (b) = 10-28 m2 Volume 1 liter (L) = 10-3 m3 1 U.S. gallon (gal) = 3.785 × 10-3 m3 Mass 1 solar mass = 1.99 × 1030 kg 1 metric ton = 103 kg 1 atomic mass unit (u) = 1.6605 × 10-27 kg Time 1 year (y) = 3.16 × 107 s 1 day (d) = 86,400 s Speed 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h Angle 1 degree (°) = 1.745x10-2 rad 1 minute of arc (') = 1 / 60 degree 1 second of arc ('') = 1 / 60 minute of arc 1 grad = 1.571 × 10-2 rad Table A4 Other Units 810 Appendix A • Reference Tables Energy 1 kiloton TNT (kT) = 4.2 × 1012 J 1 kilowatt hour (kW h) = 3.60 × 106J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J 1 electron volt (cV) = 1.60 × 10-19 J Pressure 1 atmosphere (atm) = 1.013 × 105 Pa 1 millimeter of mercury (mm Hg) = 133.3 Pa 1 torricelli (torr) = 1 mm Hg = 133.3
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Pa Nuclear decay rate 1 curie (Ci) = 3.70 × 1010 Bq Table A4 Other Units Circumference of a circle with radius ror diameter d Area of a circle with radius ror diameter d Area of a sphere with radius r Volume of a sphere with radius r Table A5 Useful formulae Symbol Meaning Best Value Approximate Value c G NA k R σ k Speed of light in vacuum Gravitational constant Avogadro’s number Boltzmann’s constant Gas constant Stefan-Boltzmann Constant Coulomb force constant Table A6 Important Constants Access for free at openstax.org. Symbol Meaning Best Value Approximate Value Appendix A • Reference Tables 811 qe ε0 µ0 h Charge on electron Permittivity of free space Permeability of free space Planck’s constant Table A6 Important Constants Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Table A7 The Greek Alphabet 812 Appendix A • Reference Tables Upsilon Phi Chi Psi Omega Table A7 The Greek Alphabet Sun mass average radius 1.99 × 1030 kg 6.96 × 108 m Earth-sun distance (average) 1.496 × 1011 m Earth mass 5.9736 × 1024 kg average radius orbital period Moon mass average radius orbital period (average) 6.376 × 106 m 3.16 × 107 s 7.35 × 1022 kg 1.74 × 106 s 2.36 × 106 s Earth-moon distance (average) 3.84 × 108 m Table A8 Solar System Data Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 n 1H 1.008 665 β– 1.007 825 99.985% 2H or D 2.014 102 0.015% 10.37 min 3H or T 3.016 050 β– 12.33 y 3He 3.016 030 1.38 × 10−4 % 0 1 2 neutron Hydrogen Deuterium Tritium Helium 1 1 2 3 3 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 8
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13 3 4 5 6 7 8 9 10 11 Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium 4 6 7 7 9 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 23 24 Table A9 Atomic Masses and Decay 4He 4.002 603 ≈100% 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 12N 13N 14N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 23Na 24Na 6.015 121 7.5% 7.016 003 92.5% 7.016 928 EC 53.29 d 9.012 182 100% 10.012 937 19.9% 11.009 305 80.1% 11.011 432 EC, β+ 12.000 000 98.90% 13.003 355 1.10% 14.003 241 13.005 738 β– β+ 14.003 074 99.63% 15.000 108 0.37% 5730 y 9.96 min 15.003 065 EC, β+ 122 s 15.994 915 99.76% 17.999 160 0.200% 18.000 937 EC, β+ 1.83 h 18.998 403 100% 19.992 435 90.51% 21.991 383 9.22% 21.994 434 β+ 22.989 767 100% 23.990 961 β– 2.602 y 14.96 h 814 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 24Mg 23.985 042 78.99% 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 55Mn 56Fe 59Co 60Co 58Ni 26.981 539 100% 27.976 927 92.23% 2.62h 30.975 362 β– 30.973 762 100% 31.973 907 β– 31.972 070 95.02% 34.969 031 β– 34.968 852 75.77% 36.965 903 24.23% 39.962 384 99.60% 38.963 707
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93.26% 14.28 d 87.4 d 39.963 999 0.0117%, EC, β– 1.28 × 109 y 39.962 591 96.94% 44.955 910 100% 47.947 947 73.8% 50.943 962 99.75% 51.940 509 83.79% 54.938 047 100% 55.934 939 91.72% 58.933 198 100% 59.933 819 β– 5.271 y 57.935 346 68.27% 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 815 60 63 64 66 69 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 59.930 788 26.10% 62.939 598 69.17% 64.927 793 30.83% 63.929 145 48.6% 65.926 034 27.9% 68.925 580 60.1% 71.922 079 27.4% 73.921 177 36.5% 74.921 594 100% 79.916 520 49.7% 78.918 336 50.69% 83.911 507 57.0% 84.911 794 72.17% 85.909 267 9.86% 87.905 619 82.58% 89.907 738 β– 88.905 849
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100% 89.907 152 β– 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 98Tc 97.907 215 β– Ruthenium 102 102Ru 101.904 348 31.6% 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Table A9 Atomic Masses and Decay 28.8 y 64.1 h 4.2 × 106 y 816 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Cesium Barium Lanthanum Cerium 103 106 107 109 114 115 120 121 130 127 131 132 136 133 134 137 138 139 140 Praseodymium 141 Neodymium Promethium Samarium 142 145 152 Table A9 Atomic Masses and Decay Access for free at openstax.org. 103Rh 106Pd 107Ag 109Ag 114Cd 102.905 500 100% 105.903 478 27.33% 106.905 092 51.84% 108.904 757 48.16% 113.903 357 28.73% 115In 114.903 880 95.7%, β– 120Sn 121Sb 119.902 200 32.59% 120.903 821 57.3% 130Te 129.906 229 33.8%, β– 4.4 × 1014 y 2.5 × 1021 y 127I 131I 132Xe 136Xe 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 126.904 473 100% 130.906 114 β– 8.040 d 131.904 144 26.9% 135.907 214 8.9% 132.905 429 100% 133.906 696 EC, β– 2.06 y 136.905 812 11.23% 137.905 232 71.70% 138.906 346 99.91% 139.905 433 88.48% 140.907 647 100% 141.907 719 27.13% 144.912 743 EC, α 17.
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7 y 151.919 729 26.7% Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 817 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 202 205 Table A9 Atomic Masses and Decay 153Eu 158Gd 159Tb 164Dy 165Ho 166Ho 152.921 225 52.2% 157.924 099 24.84% 158.925 342 100% 163.929 171 28.2% 164.930 319 100% 165.930 290 33.6% 169Tm 168.934 212 100% 174Yb 175Lu 180Hf 181Ta 184W 173.938 859 31.8% 174.940 770 97.41% 179.946 545 35.10% 180.947 992 99.98% 183.950 928 30.67% 187Re 186.955 744 62.6%, β– 190.960 920 β– 191.961 467 41.0% 190.960 584 37.3% 192.962 917 62.7% 194.964 766 33.8% 196.966 543 100% 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 202Hg 205Tl 4.6 × 1010y 15.4 d 197.968 217 β– 2.696 d 198.968 253 16.87% 201.970 617 29.86% 204.974 401 70.48% 818 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 82 Lead 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 206Pb 207Pb 208Pb 205.974 440 24.1% 206.975 872 22.1% 207.9
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257.099 480 EC, α Rutherfordium 261 261Rf 261.108 690 α Dubnium Seaborgium Bohrium Hassium Meitnerium 262 263 262 264 266 262Db 263Sg 262Bh 264Hs 266Mt 262.113 760 α, fission 263.11 86 α, fission 262.123 1 264.128 5 266.137 8 α α α 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 108 Table A9 Atomic Masses and Decay 4.47 × 109 y 23.5 min 2.355 d 2.41 × 104 y 7.37 × 103 y 8.50 × 103 y 1.38 × 103 y 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 mim 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms 820 Appendix A • Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 3H 14C 13N 12.33 y 5730 y 9.96 min 22Na 2.602 y 32P 35S 36Ci 40K 43K 45Ca 51Cr 14.28 d 87.4 d 3.00 × 105 y 1.28 × 109 y 22.3 h 165 d 27.70 d 52Mn 5.59d β– β– β+ β+ β– β– β– β– β– β– EC β+ 100% 100% 100% 90% γ 1.27 100% 0.0186 0.156 1.20 1.20 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% γs 0.373 0.618 γ 0.320 0.257 100% 3.69 28% γ s 1.33 52Fe 8.27 h β+ 1.80 43% 1.43 0.169 0.378 59Fe 44.6 d β– s 60Co 5.271 y β– 65Zn 67Ga 244.1 d 78.3 h EC EC 0.273 0.466 0.318 45% 55% γ s 1.10 1.29 100% γ s 1.17 1.33 γ 1.12 γ s 0.0933 0.185 0.300 Table A10 Selected Radioactive
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Isotopes Access for free at openstax.org. 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent Appendix A • Reference Tables 821 75Se 118.5 d EC 86Rb 18.8 d β– s 85Sr 90Sr 90Y 64.8 d 28.8 y 64.1 h 99mTc 6.02 h 113mIn 99.5 min 123I 131I 13.0 h 8.040 d EC β– β– IT IT EC β– s 0.69 1.77 0.546 2.28 0.248 0.607 others others γ s 0.121 0.136 0.265 0.280 others γ 1.08 20% 65% 68% 20% 9% γ 0.514 1 100% 9% 91% 100% 100% γ γ γ 0.142 0.392 0.159 7% 93% γ s 0.364 others 129Cs 32.3 h EC γ s 0.0400 0.372 0.411 others γ 0.662 95% 5% ≈100% γ s 0.030 137Cs 30.17 y β– s 140Ba 12.79 d β– Table A10 Selected Radioactive Isotopes 0.511 1.17 1.035 100% 100% ≈100% 85% 35% 32% 25% 95% 25% 822 Appendix A • Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 198Au 197Hg 210Po 226Ra 235U 238U 2.696 d 64.1 h 138.38 d 1.60 × 103 y β– EC α αs 7.038 × 108 y α 4.468 × 109 y αs 1.161 ≈100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 0.044 0.537 others 0.412 0.0733 γ γ 65% 24% ≈100% 100% γ 0.186 1 100% ≈100% γ s Numerous <0.400% γ 0.050 23% 23%
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77% 237Np 2.14 × 106 y αs numerous γ s numerous <0.250% 239Pu 2.41 × 104 y αs 4.96 (max.) 5.19 5.23 5.24 11% 15% 73% γ s 7.5 × 10-5 0.013 0.052 others 73% 15% 15% 243Am 7.37 × 103 y αs Max. 5.44 γ s 0.075 others 88% 11% 5.37 5.32 others Table A10 Selected Radioactive Isotopes Symbol Meaning Best Value Approximate Value me Electron mass 9.10938291(40) × 10–31 kg 9.11 × 10–31 kg Table A11 Submicroscopic masses Access for free at openstax.org. Appendix A • Reference Tables 823 Symbol Meaning Best Value Approximate Value mp mn u Proton mass 1.672621777(74) × 10–27 kg 1.6726 × 10–27 kg Neutron mass 1.674927351(74) × 10–27 kg 1.6749 × 10–27 kg Atomic mass unit 1.660538921(73) × 10–27 kg 1.6605 × 10–27 kg Table A11 Submicroscopic masses Substance p(kg/m3) Substance p(kg/m3) Air 1.29 Air (at 20°C and Atmospheric pressure) 1.20 Iron Lead 7.86 × 103 11.3 × 103 Aluminum Benzene Brass Copper Ethyl alcohol Fresh water Glycerin Gold Helium gas Hydrogen gas Ice 2.70 × 103 Mercury 13.6 × 103 0.879 × 103 Nitrogen gas 1.25 8.4 × 103 Oak 0.710 × 103 8.92 × 103 Osmium 22.6 × 103 0.806 × 103 Oxygen gas 1.43 1.00 × 103 Pine 0.373 × 103 1.26 × 103 Platinum 21.4 × 103 1.93 × 103 Seawater 1.03 × 103 1.79 × 10–1 Silver 10.5 × 103 8.99 × 10–2 Tin 7.30 × 103 0.917 × 103 Uranium 18.7 × 103 Table A12 Densities of common substances (including water at various temperatures) Substance Specific Heat (J/kg °C) Substance Specific Heat (J/kg °C)
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Elemental solids Other solids Aluminum Beryllium Cadmium 900 1830 230 Brass Glass 380 837 Ice (–5 °C) 2090 Table A13 Specific heats of common substances 824 Appendix A • Reference Tables Substance Specific Heat (J/kg °C) Substance Specific Heat (J/kg °C) Copper Germanium Gold Iron Lead Silicon Silver 387 322 129 448 128 703 234 860 1700 Marble Wood Liquids Alcohol (ethyl) 2400 Mercury 140 Water (15 °C) 4186 Gas Steam (100 °C) 2010 Note: To convert values to units of cal/g °C, divide by 4186 Table A13 Specific heats of common substances Substance Melting Point (°C) Latent Heat of Fusion (J/kg) Boiling Point (°C) Latent Heat of Vaporization (J/kg) Helium –272.2 Oxygen –218.79 Nitrogen –209.97 Ethyl Alcohol –114 Water 0.00 Sulfur 119 Lead 327.3 Aluminum 660 Silver 960.80 Gold 1063.00 Copper 1083 5.23 × 103 1.38 × 104 2.55 × 104 1.04 × 105 3.33 × 105 3.81 × 104 3.97 × 105 3.97 × 105 8.82 × 104 6.44 × 104 1.34 × 105 –268.93 –182.97 –195.81 78 100.00 444.60 1750 2516 2162 2856 2562 2.09 × 104 2.13 × 105 2.01 × 105 8.54 × 105 2.26 × 106 2.90 × 105 8.70 × 105 1.05 × 107 2.33 × 106 1.58 × 106 5.06 × 106 Table A14 Heats of fusion and vaporization for common substances Access for free at openstax.org. Appendix A • Reference Tables 825 Materials (Solids) Average Linear Expansion Coefficient (a)(°C)–1 Material (Liquids and Gases) Average Volume Expansion Coefficient (B)(°C)–1 Aluminum 24 × 10–6 Acetone 1.5 × 10–4 Brass and Bronze 19 × 10–6 Concrete 12 × 10–6 Copper 17 × 10–6 Glass (ordinary) 9 × 10–6 Glass (Pyrex) 3.2 × 10–6 Invar (Ni-Fe alloy) Lead Steel 1.3 × 10–6 29 × 10–
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6 13 × 10–6 Alcohol, ethyl 1.12 × 10–4 Benzene Gasoline Glycerin Mercury 1.24 × 10–4 9.6 × 10–4 4.85 × 10–4 1.82 × 10–4 Turpentine 9.0 × 10–4 Air* at 0°C Helium* 3.67 × 10–3 3.665 × 10–3 * The values given here assume the gases undergo expansion at constant pressure. However, the expansion of gases depends on the pressure applied to the gas. Therefore, gases do not have a specific value for the volume expansion coefficient. Table A15 Coefficients of thermal expansion for common substances Medium v(m/s) Medium v(m/s) Medium v(m/s) Gases Liquids at 25°C Hydrogen 1286 Glycerol Helium Air Air Oxygen 972 343 331 317 Seawater Water Mercury Kerosene Methyl Alcohol Carbon tetrachloride 1904 1533 1493 1450 1324 1143 926 Table A16 Speed of sound in various substances Solids* Pyrex glass Iron Aluminum Brass Copper Gold Lucite Lead Rubber 5640 5950 5100 4700 3560 3240 2680 1322 1600 826 Appendix A • Reference Tables Medium v(m/s) Medium v(m/s) Medium v(m/s) *Values given here are for propagation of longitudinal waves in bulk media. However, speeds for longitudinal waves in thin rods are slower, and speeds of transverse waves in bulk are even slower. Table A16 Speed of sound in various substances Source of Sound B(dB) Nearby jet airplane 150 Jackhammer machine gun 130 Siren; rock concert 120 Subway; power lawn mower 100 Busy traffic Vacuum cleaner Normal Conversation Mosquito buzzing whisper Rustling leaves Threshold of hearing 80 70 60 40 30 10 0 Table A17 Conversion of sound intensity to decibel level Wavelength Range (nm) Color Description 400-430 430-485 485-560 560-590 590-625 625-700 Violet Blue Green Yellow Orange Red Table A18 Wavelengths of visible light Access for free at openstax.org. Appendix A • Reference Tables 827 Substance Index of Refraction Substance Index of Refraction Solids at 20°C Cubic zirconia Diamond (C) Flourite (CaF2) 2.15 2.419 1.434 Liquids at 20°C Benz
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ene 1.501 Carbon disulfide 1.628 Carbon tetrachloride 1.461 Fused quartz (SiO2) 1.458 Ethyl alcohol Gallium phosphide 3.50 Glass, crown Glass, flint Ice (H2O) Polystyrene 1.52 1.66 1.309 1.49 Glycerin Water 1.361 1.473 1.333 Gases at 0°C, 1 atm Air 1.000 293 Sodium chloride (NaCl) 1.544 Carbon dioxide 1.000 45 Note: These values assume that light has a wavelength of 589 nm in vacuum. Table A19 Indices of refraction Hoop or thin cylindrical shell Hollow cylinder Solid cylinder or disk Rectangular plane Long, thin rod with rotation axis through center Long, thin rod with rotation axis through end Solid sphere Thin spherical shell Table A20 Moments of inertia for different shapes μs μk Rubber on dry concrete 1.0 0.8 Table A21 Coefficients of friction for common objects on other objects 828 Appendix A • Reference Tables Steel on steel Aluminum on steel Glass on glass Copper on steel Wood on wood Waxed wood on wet snow Waxed wood on dry snow Metal on metal (lubricated) Teflon on Teflon Ice on ice Synovial joints in humans μs μk 0.74 0.61 0.94 0.53 0.25-0.5 0.14 0.1 0.15 0.04 0.1 0.01 0.57 0.47 0.4 0.36 0.2 0.1 0.04 0.06 0.04 0.03 0.003 Note: All values are approximate. In some cases, the coefficient of friction can exceed 1.0. Table A21 Coefficients of friction for common objects on other objects Material Dielectric Constant ĸ Dielectric Strength* (106V/m) Air (dry) Bakelite Fused quartz Mylar Neoprene rubber Nylon Paper 1.000 59 4.9 4.3 3.2 6.7 3.4 3.7 Paraffin-impregnated paper 3.5 Polystyrene Polyvinyl chloride Porcelain 2.56 3.4 6 Table A22 Dielectric constants 3 24 8 7 12 14 16 11 24 40 8 Access for free at openstax.org. Appendix A • Reference Tables 829 Material Dielectric Constant ĸ Dielectric Strength
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motion does the spacing between dots represent? 4. Have you covered all possible motions in the runs you did? If not, which ones did you omit? Give reasons why you omitted some types of motion. Think About It 1. What is motion? 2. What types of motion can objects undergo? 3. What are some words used to describe motion? 4. How can you determine how far and how fast an object moves? 5. What does the term “falling” mean? 6. Describe the motion of falling objects. 7. Which falls faster: a heavy object or a light object? 8. How can a ball that’s moving upward still be falling? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 1 Graphs and equations describe motion in one dimension. 5 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 6 1.1 The Language of Motion When describing motions, such as the ones in Figures 1.1 and 1.2, you can use many different expressions and words. The English language is rich with them. Many sports broadcasters invent words or expressions to convey action and to excite the imagination. Phrases such as “a cannonating drive from the point” or “blistering speed” have become commonplace in our sports lingo. In physics, you must be precise in your language and use clearly defined terms to describe motion. Kinematics is the branch of physics that describes motion. kinematics: a branch of physics that describes motion Figure 1.2 How would you describe the motions shown in these photos? origin: a reference point position: the straight-line distance between the origin and an object’s location; includes magnitude and direction scalar quantity: a measurement that has magnitude only vector quantity: a measurement that has both magnitude and direction 6 Unit I Kinematics M I N D S O N How Do Objects Move? Study the photos in Figure 1.2. Describe the motion of the puck, the wheelchair, and the harpoon with respect to time. Jot down your descriptions and underline key words associated with motion. Compare your key words with those of a partner. Compile a class list on the chalkboard. When describing motion, certain words, such as speed, acceleration, and velocity, are common. These words
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have slightly different meanings in physics than they do in everyday speech, as you will learn in the following subsection. Physics Terms It’s Saturday night and the Edmonton Oilers are playing the Calgary Flames. In order to locate players on the ice, you need a reference system. In this case, select the centre of the ice as the reference point, or origin. You can then measure the straight-line distances, d, of players from the origin, such as 5.0 m. If you specify a direction from the origin along with the distance, then you define a player’s position, d, for example, 5.0 m [E] (Figure 1.3). The arrow over the variable indicates that the variable is a vector quantity. The number and unit are called the magnitude of the vector. Distance, which has a magnitude but no direction associated with it, is an example of a scalar quantity. Vector quantities have both magnitude and direction. Position is an example of a vector quantity. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 7 W N S E di 5.0 m origin 26.0 m Figure 1.3 The player’s position is 5.0 m [east of the origin] or simply 5.0 m [E]. The player is at a distance of 5.0 m from the origin. 60.0 m If the player, initially 5.0 m [east of the origin], skates to the east end of the rink to the goal area, his position changes. It is now 25.0 m [east of the origin] or 25.0 m [E] (Figure 1.4). You can state that he has travelled a straight-line distance of 20.0 m, and has a displacement of 20.0 m [E] relative to his initial position. W N S E 25.0 m df 26.0 m origin 60.0 m PHYSICS INSIGHT Technically, if you are standing away from the origin, you are displaced a certain distance and in a certain direction. However, the sign is not used with position unless the object you are referring to has moved from the origin to its current position, that is, unless the object has experienced a change in position. Figure 1.4 The player’s position has changed. A change in position is called displacement. Distance travelled is the length of the path taken to move from one position to another, regardless of direction
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. Displacement, d, is the change in position. The player’s displacement is written as distance: the length of the path taken to move from one position to another d 20.0 m [E] where is the Greek letter delta that means “change in.” Calculate the change in a quantity by subtracting the initial quantity from the final quantity. In algebraic notation, R Rf Ri. You can calculate the displacement of the player in the following manner: d d f d i 25.0 m [E] 5.0 [E] 20.0 m [E] displacement: a straight line between initial and final positions; includes magnitude and direction info BIT Pilots use radar vectors when landing their aircraft. Radar vectors are instructions to fly in a particular direction and usually include altitude and speed restrictions. Chapter 1 Graphs and equations describe motion in one dimension. 7 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 8 Sign Conventions How would you determine your final distance and displacement if you moved from a position 5.0 m [W] to a position 10.0 m [E] (Figure 1.5)? W 10.0 m 8.0 m 6.0 m 4.0 m 2.0 m 2.0 m 4.0 m 6.0 m 8.0 m 0.0 m E 10.0 m Figure 1.5 The person travels a distance of 5.0 m 10.0 m 15.0 m. What is the person’s displacement? What is the person’s final position relative to the bus stop? To calculate the distance travelled in the above scenario, you need only add the magnitudes of the two position vectors. d 5.0 m 10.0 m 15.0 m To find displacement, you need to subtract the initial position, d i, f. Let d from the final position, d 5.0 m [W] and d f 10.0 m [E]. i d d f d i 10.0 m [E] 5.0 m [W] Note that subtracting a vector is the same as adding its opposite, so the negative west direction is the same as the positive east direction. d 10.0 m [E] 5.0 m [W] 10.0 m [E] 5.0 m [E] 15.0 m [E] PHYSICS INSIGHT When doing calculations with measured
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values, follow the rules on rounding and the number of significant digits. Refer to pages 876–877 in this book. W N up E S down L R Figure 1.6 Let east be positive and west negative. Similarly, north, up, and right are usually designated as positive. Another way of solving for displacement is to designate the east direction as positive and the west direction as negative (Figure 1.6). The two position 10.0 m [E] 10.0 m. vectors become d i Now calculate displacement: 5.0 m [W] 5.0 m and d f d d f d i 10.0 m (5.0 m) 15.0 m Since east is positive, the positive sign indicates that the person has moved 15.0 m east. Practise finding position and displacement in the next Skills Practice and example. info BIT On April 26, 2004, Stephane Gras of France did 445 chin-ups in one hour. If you consider up as positive, then Gras made 445 positive displacements and 445 negative displacements, meaning that his net displacement was zero! 8 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page Finding Position and Displacement 1. Create a scale using the dimensions of the hockey rink (Figure 1.7). Measuring from the centre of the player’s helmet, (a) find each player’s position relative to the north 26.0 m and south sides of the rink. (b) find each player’s position relative to the east and west sides of the rink. (c) If the player moves from position 2 to position 4 on the rink, what is his displacement? 1 2 5 origin 4 3 60.0 m Figure 1.7 Example 1.1 A traveller initially standing 1.5 m to the right of the inukshuk moves so that she is 3.5 m to the left of the inukshuk (Figure 1.8). Determine the traveller’s displacement algebraically (a) using directions (b) using plus and minus signs Given d i d f 1.5 m [right] 3.5 m [left] Required displacement (d ) inukshuk 3.5 m [left] origin 1.5 m [right] Figure 1.8 Analysis and Solution To find displacement, use the equation d (a) d d i d f 3.5 m [
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left] 1.5 m [right] 3.5 m [left] (1.5 m [left]) 3.5 m [left] 1.5 m [left] 5.0 m [left] d d i. f i d d f d (b) Consider right to be positive. 1.5 m [right] 1.5 m 3.5 m [left] 3.5 m d f 3.5 m (1.5 m) 3.5 m 1.5 m 5.0 m d i The answer is negative, so the direction is left. Paraphrase The traveller’s displacement is 5.0 m [left] of her initial position. Note that the direction of displacement is relative to initial position, whereas the direction of position is relative to the designated origin, in this case, the inukshuk. Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. What is the sprinter’s displacement from the initial position? 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. What is the player’s displacement from the starting position? 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, calculate the distance and displacement her hand travels during that time. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m Chapter 1 Graphs and equations describe motion in one dimension. 9 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 10 For all subsequent problems in this book, you will be using plus and minus signs to indicate direction. This method is more flexible for problem solving and easier to use. Like distance and displacement, speed and velocity is another scalar-vector pair. Speed is the rate at which an object moves. It is a scalar quantity, so it has magnitude only; for example, v 50 km/h (Figure 1.9). Velocity is a vector quantity, so it has both magnitude (speed) and direction. If you are travelling south from Fort McMurray to Lethbridge at 50 km/h
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, your velocity is written as v 50 km/h [S]. If you designate south as negative, then v 50 km/h. Acceleration is a vector quantity that represents the rate of change of velocity. You will study aspects of displacement, velocity, and acceleration, and their interrelationships, in the sections that follow. Figure 1.9 Scalar or vector? 1.1 Check and Reflect 1.1 Check and Reflect Knowledge 1. What two categories of terms are used to describe motion? Give an example of each. 2. Compare and contrast distance and displacement. 3. What is the significance of a reference point? Applications 4. Draw a seating plan using the statements below. (a) Chad is 2.0 m [left] of Dolores. (b) Ed is 4.5 m [right] of Chad. (c) Greg is 7.5 m [left] of Chad. (d) Hannah is 1.0 m [right] of Ed. (e) What is the displacement of a teacher who walks from Greg to Hannah? 5. A person’s displacement is 50.0 km [W]. What is his final position if he started at 5.0 km [E]? 6. Using an autuk (a type of sealskin racquet), two children play catch. Standing 3.0 m apart, the child on the right tosses the ball to the child on the left, and then moves 5.0 m [right] to catch the ball again. Determine the horizontal distance and displacement the ball travels from its initial position (ignore any vertical motion). 3.0 m 5.0 m 7. Below is a seating plan for the head table at a wedding reception. Relative to the bride, describe the positions of the groom, best man, maid of honour, and flower girl. 0.75 m 0.75 m 0.50 m 0.75 m 0.75 m Flower girl Best man Bride Groom Maid of Honour Ring boy e TEST To check your understanding of scalar and vector quantities, follow the eTest links at www.pearsoned.ca/school/physicssource. 10 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 11 1.2 Position-time Graphs and Uniform Motion You are competing to win the Masters Golf Tournament. The hole is 5.0 m away (Figure 1.10). You gently hit the ball with
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your club and hold your breath. Time seems to stop. Then, 5.0 s later, it rolls into the hole. You have won the tournament! From section 1.1, you know that displacement is the change in an object’s position. If you replay the sequence of motions of your winning putt in 1.0-s intervals, you can measure the displacements of the golf ball from you, the putter, to the hole (Figure 1.11). Figure 1.10 You can represent motion in sports using vectors and graphs. 0.0 m 0.0 s origin 1.0 m 1.0 s 2.0 m 2.0 s 3.0 m 3.0 s 4.0 m 4.0 s 5.0 m 5.0 s Figure 1.11 What is the golf ball’s displacement after each second? Table 1.1 displays the data from Figure 1.11 for the golf ball’s position from you at 1.0-s intervals. By graphing the data, you can visualize the motion of the golf ball more clearly (Figure 1.12). ▼ Table 1.1 Position-time data Time (s) Position (m [right]) 0.0 1.0 2.0 3.0 4.0 5.0 t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 Velocity Position vs. Time 6.0 5.0 4.0 3.0 2.0 1.0 0..0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.12 A position-time graph of the golf ball Notice that the graph in Figure 1.12 is a straight line. A straight line has a constant slope. What does constant slope tell you about the ball’s motion? To answer this question, calculate the slope and keep track of the units. Designate toward the hole, to the right, as the positive direction. Chapter 1 Graphs and equations describe motion in one dimension. 11 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 12 e SIM Practise calculating average speed and average velocity. Go to www.pearsoned.ca/ school/physicssource. Recall that slope rise run. For position–time graphs, this equation becomes slope change in position change in time A change
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in position is displacement. So, the equation for slope becomes PHYSICS INSIGHT Speed has magnitude only. Velocity has both magnitude and direction. velocity: rate of change in position slope d t d d i f ti tf 5.0 m 0.0 m 5.0 s 0.0 s 1.0 m/s The answer is positive, so the golf ball moves at a rate of 1.0 m/s [right]. Notice that the units are m/s (read metres per second). These units indicate speed or velocity. Since displacement is a vector quantity, the slope of the position-time graph in Figure 1.12 gives you the velocity, v, of the ball: the change in position per unit time. Because you have calculated velocity over a time interval rather than at an instant in time, it is the average velocity. d v t Speed and Velocity 4 m/s 4 m/s Figure 1.13 Objects with the same speed can have different velocities. Objects travelling at the same speed can have different velocities. For example, a tram carries passengers across a ravine at a constant speed. A passenger going to the observation deck has a velocity of 4 m/s [right] and a passenger leaving the deck has a velocity of 4 m/s [left] (Figure 1.13). Their speeds are the same, but because they are travelling in opposite directions, their velocities are different. 1-2 Decision-Making Analysis 1-2 Decision-Making Analysis Traffic Safety Is Everyone’s Business The Issue In an average year in Alberta, traffic accidents claim six times more lives than homicide, eight times more lives than AIDS, and 100 times more lives than meningitis. Collisions represent one of the greatest threats to public safety. Background Information In the Canadian 2002 Nerves of Steel: Aggressive Driving Study, speeding was identified as one of two common aggressive behaviours that contribute to a significant percentage of all crashes. The Alberta Motor Association’s Alberta Traffic Safety Progress Report has suggested that a province-wide speed management program could significantly improve levels of road safety, decreasing both speed and casualties. One suggested program is the implementation of the vehicle tachograph, a device required in Europe to improve road safety. 12 Unit I Kinematics e WEB To learn more about how speeding is a key contributing factor in casualty collisions in Alberta, follow the links at www.pearsoned.ca/school/ physicssource. 01-PearsonPhys20-
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Chap01 7/23/08 11:43 AM Page 13 Analysis Your group has been asked to research different traffic safety initiatives. The government will use the results of your research to make the most appropriate decision. 1. Research (a) how state- or province-wide speed management programs have influenced driver behaviour (b) the societal cost of vehicle crashes (c) driver attitudes toward enforcement of and education about traffic safety issues 2. Analyze your research and decide which management program should be used. 3. Once your group has completed a written report recommending a particular program, present the report to the rest of the class, who will act as representatives of the government and the community. So far, you have learned that the slope of a position-time graph represents a rate of change in position, or velocity. If an object moves at constant velocity (constant magnitude and direction), the object is undergoing uniform motion. A position-time graph for an object at rest is a horizontal line (Figure 1.14). An object at rest is still said to be undergoing uniform motion because its change in position remains constant over equal time intervals. Concept Check (a) Describe the position of dots on a ticker tape at rest. What is the slope of the graph in Figure 1.14? (b) Describe the shape of a position-time graph for an object travelling at a constant velocity. List three possibilities. Frame of Reference uniform motion: constant velocity (motion or rest) at rest: not moving; stationary ) ].0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (min) 4.0 5.0 Figure 1.14 A position-time graph for a stationary object If you were to designate the hole, rather than the putter, as the origin (starting point) in the golf tournament (Figure 1.15(a)), your data table would start at 5.0 m [left] at time 0.0 s, instead of at 0.0 m and 0.0 s (Table 1.2). The values to the left of the hole are positive. ▼ Table 1.2 Position-time data Time (s) Position (m [left]) t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 5.0 4.0 3.0 2.
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0 1.0 0.0 5.0 m 0.0 s 4.0 m 1.0 s 3.0 m 2.0 s 2.0 m 3.0 s 1.0 m 4.0 s 0.0 m 5.0 s origin Figure 1.15(a) Designating an origin is arbitrary. In this example, the hole is the origin and all positions are measured relative to it. Chapter 1 Graphs and equations describe motion in one dimension. 13 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 14 info BIT On May 31, 2004 in Moscow, Ashrita Furman of the USA walked 1.6 km while continuously hulahooping in 14 min 25 s. He also holds the world record for the fastest time for pushing an orange with his nose. On August 12, 2004, he pushed an orange 1.6 km in 24 min 36 s. What was his speed, in km/h and m/s, for each case? (See Unit Conversions on page 878.) e WEB In November 2004, at an altitude of 33 000 m, the X-43A recorded a speed of Mach 9. Use the Internet or your local library to research the term “Mach” as used to describe the speed of an object. How did this term originate? What is the difference between Mach and ultrasonic? Write a brief summary of your findings. To learn more about Mach, follow the links at www.pearsoned.ca/school/ physicssource. 14 Unit I Kinematics The corresponding position-time graph is shown in Figure 1.15(b). ) ].0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.15(b) Compare this graph with the graph in Figure 1.12. If you change your reference frame, the position-time graph also changes. From the graph, slope v d t d d i f ti tf 0.0 m (5.0 m) 5.0 s 0.0 s 1.0 m/s The velocity of the golf ball is 1.0 m/s. What does the negative sign mean? It means the ball is travelling opposite to the direction to which the positions of the ball are measured. It does
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not mean that the golf ball is slowing down. Since positions, now measured to the left of the hole (the new origin) are designated positive, any motion directed to the right is described as being negative. In this case, you can also see that the ball is decreasing its position from the origin with increasing time. The ball travels to the right toward the hole, decreasing its position each second by 1.0 m — it travels at 1.0 m/s to the right or 1.0 m/s [left] as indicated by the downward slope on the graph. Concept Check Determine how the velocity of the golf ball can be positive if the hole is at the origin. Below is a summary of what you have learned: • The slope of a position-time graph represents velocity. • The velocity is the average velocity for the time interval. • Your choice of reference frame affects the direction (sign) of your answer. • A straight line on a position-time graph represents uniform motion. Comparing the Motion of Two or More Objects on a Position-time Graph You can represent the motions of two objects on one graph, as long as the origin is the same for both objects. You can then use the graph to compare their motions, as in the next example. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 15 Example 1.2 At the end of the school day, student A and student B say goodbye and head in opposite directions, walking at constant rates. Student B heads west to the bus stop while student A walks east to her house. After 3.0 min, student A is 300 m [E] and student B is 450 m [W] (Figure 1.16). (a) Graph the position of each student on one graph after 3.0 min. (b) Determine the velocity in m/s of each student algebraically. Given Choose east to be positive. dd A dd B t 3.0 min 300 m [E] 300 m 450 m [W] 450 m Required (a) position–time graph (b) velocity (v B) A and v BUS STOP bus stop school Lakeview School W N S E+ Student B’s position: 450 m [W] origin Student A’s position: 300 m [E] Figure 1.16 Analysis and Solution (a) Since east is the positive direction, plot student A’s position (3.0 min, 300 m)
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above the time axis and student B’s position (3.0 min, 450 m) below the time axis (Figure 1.17). Position vs. Time 1.0 2.0 3.0 Time (min 600 400 200 0 200 400 600 Figure 1.17 (b) Convert time in minutes to time in seconds. d t to find the velocity Then use the equation v of each student. 60 s 1 min t 3.0 min 180 s v A 300 m 180 s 1.7 m/s The sign is positive, so the direction is east. v B 450 m 180 s 2.5 m/s The sign is negative, so the direction is west. Paraphrase (b) Student A’s velocity is 1.7 m/s [E] and student B’s velocity is 2.5 m/s [W]. Practice Problems 1. A wildlife biologist measures how long it takes four animals to cover a displacement of 200 m [forward]. (a) Graph the data from the table below. (b) Determine each animal’s average velocity. Animal Time taken (s) Elk Coyote Grizzly bear Moose 10.0 10.4 18.0 12.9 Answers 1. (a 250.0 200.0 150.0 100.0 50.0 0.0 Position vs. Time 0 2 4 6 8 10 12 14 16 18 20 Time (s) (b) Elk: 20.0 m/s [forward] Coyote: 19.2 m/s [forward] Grizzly bear: 11.1 m/s [forward] Moose: 15.5 m/s [forward] Chapter 1 Graphs and equations describe motion in one dimension. 15 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 16 So far, you have learned that the slope of a position–time graph represents velocity. By comparing the slopes of two graphs, you can determine which object is moving faster. From the slopes of the graphs in Figure 1.17, which student is moving faster? When you represent the motions of two objects on the same graph, you can also tell whether the objects are approaching or moving apart by checking if the lines are converging or diverging. An important event occurs at the point where the two lines intersect. Both objects have the same position, so the objects meet at this point. Concept Check Describe the shape of a graph showing the motion of two
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objects approaching each other. In the next example, two objects start at different times and have different speeds. You will graphically find their meeting point. Example 1.3 Two rollerbladers, A and B, are having a race. B gives A a head start of 5.0 s (Figure 1.18). Each rollerblader moves with a constant velocity. Assume that the time taken to reach constant velocity is negligible. If A travels 100.0 m [right] in 20.0 s and B travels 112.5 m [right] in 15.0 s, (a) graph the motions of both rollerbladers on the same graph. (b) find the time, position, and displacement at which B catches up with A. B A Practice Problems 1. The two rollerbladers in Example 1.3 have a second race in which they each travel the original time and distance. In this race, they start at the same time, but B’s initial position is 10.0 m left of A. Take the starting position of A as the reference. (a) Graph the motions of the rollerbladers. (b) Find the time, position, and B’s displacement at which B catches up with A. Answers 1. (b) t 4.0 s 20.0 m [right] d 30.0 m [right] d 16 Unit I Kinematics distance travelled by A in 5.0 s Figure 1.18 Given Choose right to be positive. d A tA d B tB 100.0 m [right] 100.0 m 20.0 s 112.5 m [right] 112.5 m 15.0 s, started 5.0 s later Required (a) position-time graph (b) time (t), position (d catches up with A ), and displacement (d ) when B 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 17 Analysis and Solution (a) Assume that t 0.0 s at the start of A’s motion. Thus, the position-time graph of A’s motion starts at the origin. A’s final position is 100.0 m at 20.0 s. The position-time graph for B’s motion starts at 0.0 m and 5.0 s (because B started 5.0 s after A). B starts moving after 5.0 s for 15.0 s
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. Thus, at 20.0 s (5.0 s 15.0 s), B’s position is 112.5 m. Each rollerblader travels with a constant velocity, so the lines connecting their initial and final positions are straight (Figure 1.19(a)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(a) (b) On the graph in Figure 1.19(a), look for a point of intersection. At this point, both rollerbladers have the same final position. From the graph, you can see that this point occurs at t 15.0 s. The corresponding position is 75.0 m (Figure 1.19(b)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(b) To find B’s displacement, find the change in position: d Both A and B started from the same position, d both have the same final position at the point of intersection, d f d 75.0 m. 75.0 m 0.0 m i d d i. f 0. Since they 75.0 m The answer is positive, so the direction is to the right. Paraphrase (b) B catches up with A 15.0 s after A started. B’s position and displacement are 75.0 m [right] of the origin. Chapter 1 Graphs and equations describe motion in one dimension. 17 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 18 Example 1.4 From the graph in Example 1.3, find the velocities of the two rollerbladers. Given Choose right to be positive. At the point of intersection (Figure 1.19(b)), d A tA d B tB 75.0 m [right] 75.0 m 15.0 s 75.0 m [right] 75.0 m 15.0 s 5.0 s 10.0 s Required velocities of A and B (vv A, v B) Analysis and Solution To find the velocity of each
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rollerblader, remember that the slope of a position–time graph is velocity. Because the motions are uniform, the slopes will be constant for each rollerblader. vv vvv A d t 75.0 m 0.0 m 15.0 s 0.0 s 5.0 m/s vvv B 75.0 m 0.0 m 15.0 s 5.0 s 75.0 m 0.0 m 10.0 s 7.5 m/s The answers are both positive, so the direction is to the right. You can see that, in order for B to cover the same distance as A, B must move faster because B started later. Paraphrase A’s velocity is 5.0 m/s [right] and B’s velocity is 7.5 m/s [right]. Practice Problems 1. Suppose rollerblader B gives A a head start of 5.0 s and takes 10.0 s to catch up with A at 100.0 m [right]. Determine the velocities of rollerbladers A and B. Answers 1. A: 6.67 m/s [right] B: 10.0 m/s [right] 18 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 19 1-3 Inquiry Lab 1-3 Inquiry Lab Car Activity Question What are the speeds of two different toy cars? If one car is released 3.0 s after the other, where will they meet? Variables Identify the manipulated, responding, and controlled variables. Materials and Equipment two battery-operated toy cars ticker tape carbon disk spark timer (60 Hz) masking tape ruler graph paper Procedure 1 On a flat surface, such as the floor or lab bench, mark the initial starting position of car 1 with masking tape. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Analysis 1. Draw a line through the first dot on each ticker tape and label it t = 0 s. 2. Depending on the calibration of your ticker timer, count from the starting position, and place a mark after a fixed number of dots, e.g., 6, 12, 18, 24, etc. Label each mark t1, t2, etc. On a 60-Hz timer, every sixth dot represents 0.10 s. 3. Measure the distance from t = 0 to t1
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, t = 0 to t2, t = 0 to t3, etc. Record the data in a position–time table. 4. Using an appropriate scale, graph each set of data for each toy car, separately. 5. Determine the slope of the line of best fit for each graph. See pages 872–873 for explicit instruction on how to draw a line of best fit. 2 Using masking tape, attach 1.0 m of ticker tape to the 6. What is the speed of each toy car? end of car 1. 3 Thread the ticker tape through the spark timer (Figure 1.20). 4 Turn the car on. 7. How do the speeds of car 1 and car 2 compare? 8. Assuming uniform motion, how far would car 1 travel in 15 s? 9. Assuming uniform motion, how long would it take car 2 5 Turn the spark timer on as you release the car from its to travel 30 m? initial position. 6 Observe the path of car 1 until the ticker tape is used up. Label the ticker tape “car 1.” 7 Repeat steps 2–6 for car 2. recording timer recording tape toy car Figure 1.20 10. Imagine that you release the faster car 3.0 s after the slower car. Graphically determine the position where the two cars meet. Assume uniform motion. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. In summary, you can see how a position-time graph helps you visualize the event you are analyzing. Calculating the slope of a position-time graph provides new information about the motion, namely, the object’s velocity. In the next sections, you will expand on your graphing knowledge by analyzing motion using a velocity-time graph. Chapter 1 Graphs and equations describe motion in one dimension. 19 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 20 1.2 Check and Reflect 1.2 Check and Reflect Knowledge Applications 1. For an object at rest, what quantities of 8. Two children on racing bikes start from motion remain the same over equal time intervals? 2. For an object travelling at a constant velocity, what quantity of motion remains the same over equal time intervals? 3. Match each ticker tape below with the correct position-time graph. (i) (ii) (iii) (iv.0
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7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time A B C D 1.0 2.0 3.0 Time (s) 4.0 5.0 4. Two friends start walking on a football field in the same direction. Person A walks twice as fast as person B. However, person B has a head start of 20.0 m. If person A walks at 3.0 m/s, find the distance between the two friends after walking for 20.0 s and determine who is ahead at this time. Sketch a position-time graph for both people. 5. A camper kayaks 16 km [E] from a camping site, stops, and then paddles 23 km [W]. What is the camper’s final position with respect to the campsite? 6. Sketch a position-time graph for a bear starting 1.2 m from a reference point, walking slowly away at constant velocity for 3.0 s, stopping for 5.0 s, backing up at half the speed for 2.0 s, and finally stopping. 7. Sketch a position-time graph for a student (a) walking east to school with a constant velocity (b) stopping at the school, which is 5 km east of home (c) cycling home with a constant velocity 20 Unit I Kinematics the same reference point. Child A travels 5.0 m/s [right] and child B travels 4.5 m/s [right]. How much farther from the point of origin is child A than child B after 5.0 s? 9. Insect A moves 5.0 m/min and insect B moves 9.0 cm/s. Determine which insect is ahead and by how much after 3.0 min. Assume both insects are moving in the same direction. 10. Describe the motion in each lettered stage for the object depicted by the positiontime graph below 20 16 12 8 4 0 Position vs. Time B A C 0 4 8 12 Time (s) 16 20 11. A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, at what point (distance and time) will the mosquito hit your sunglasses if you are travelling toward the mosquito with a speed of 2.0 m/s and the mosquito is travelling in a straight path? 12
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. Spotting a friend 5.0 m directly in front of you, walking 2.0 m/s [N], you start walking 2.25 m/s [N] to catch up. How long will it take for you to intercept your friend and what will be your displacement? 13. Two vehicles, separated by a distance of 450 m, travel in opposite directions toward a traffic light. When will the vehicles pass one another if vehicle A is travelling 35 km/h and is 300 m [E] of the traffic light while vehicle B is travelling 40 km/h? When will each vehicle pass the traffic light, assuming the light remains green the entire time? e TEST To check your understanding of uniform motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 21 1.3 Velocity-time Graphs: Uniform and Nonuniform Motion Recently installed video screens in aircraft provide passengers with information about the aircraft’s velocity during the flight (Figure 1.21). x km 0.0 h x km 1.0 h x km 2.0 h x km 3.0 h x km 4.0 h x km 5.0 h Figure 1.22 A plane flies at a constant speed, so the distances within each time interval are equal. Break the plane’s motion into a series of snapshots. Record your data in a data table and then graph it. Figure 1.22 shows the data of the plane’s path. Like position-time graphs, velocity-time graphs provide useful information about the motion of an object. The shape of the velocity-time graph reveals whether the object is at rest, moving at constant speed, speeding up, or slowing down. Suppose an airplane has a cruising altitude of 10 600 m and travels at a constant velocity of 900 km/h [E] for 5.0 h. Table 1.3 shows the velocity-time data for the airplane. If you graph the data, you can determine the relationship between the two variables, velocity and time (Figure 1.23). ▼ Table 1.3 Time (h) Velocity (km/h) [E] Figure 1.21 Video screens are an example of an application of velocitytime graphs. 0.0 1.0 2.0 3.0 4.0 5. 900 800 700 600 500 400 300 200 100 0 0.0
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900 900 900 900 900 900 Velocity vs. Time for an Airplane 1.0 2.0 3.0 4.0 5.0 Time (h) Figure 1.23 A velocity-time graph for an airflight Chapter 1 Graphs and equations describe motion in one dimension. 21 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 22 Designating east as the positive direction, the slope of the velocity-time graph is: slope rise run v t v v i f ti tf m k k m 900 900 h h 5.0 h 1.0 h 0 km/h2 From the graph in Figure 1.23, there is no change in the plane’s velocity, so the slope of the velocity-time graph is zero. Notice the units of the slope of the velocity-time graph: km/h2. These units are units of acceleration. Because the plane is moving at a constant velocity, its acceleration is zero. In general, you can recognize acceleration values by their units, which are always distance divided by time squared. In physics, the standard units for acceleration are metres per second per second, which is generally abbreviated to m/s2 (read metres per second squared). e TECH Determine the velocity of an object based on the shape Concept Check of its position-time graph. Go to www.pearsoned.ca/ school/physicssource. (a) What does the slope of a position-time graph represent? (b) What does the slope of a velocity-time graph represent? Non-uniform Motion Although objects may experience constant velocity over short time intervals, even a car operating on cruise control has fluctuations in speed or direction (Figure 1.24). How can you describe and illustrate a change of velocity using the concepts of kinematics? Recall from section 1.2 that an object moving at a constant velocity is undergoing uniform motion. But is uniform motion the only type of motion? Perform the next QuickLab to find out. Figure 1.24 Consider the kinds of changes in velocity this car experiences during the trip. 22 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 23 1-4 QuickLab 1-4 QuickLab Match a Graph Problem What type of motion does each graph describe? 5 Print out the graphs from your experiment. For each graph, construct a table of values for position and time. Materials LM 1-
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1 (provided by your teacher) ruler motion sensor masking tape Procedure 1 Study the different position-time graphs on LM 1-1. With a partner, decide what type of motion each graph illustrates. 2 Set up the motion sensor to plot position vs. time. 3 Label a starting position with masking tape approximately 1 m in front of the motion sensor. Move away from the motion sensor in such a way that the graph of the motion captured approximates the one on the LM. 4 Switch roles with your partner and repeat steps 1–3. Questions 1. Describe your motion when a horizontal line was being produced on the position–time graph. 2. What relationship exists between the type of motion and change in position? 3. Suggest two different ways in which you could classify the motion described by the four graphs. 4. What would the graph look like if you moved away from and then back toward the motion sensor? 5. What happens to the graph when you move away from your initial position and then move back toward and then beyond your initial position? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Concept Check Which ticker tape in Figure 1.25 represents accelerated motion? Explain. Figure 1.25 Consider an object, such as a drag racer (Figure 1.26), starting from rest and reaching a constant velocity over a time interval (Figure 1.27). During this time interval, the vehicle has to change its velocity from a value of zero to a final non-zero value. An object whose velocity changes (increases or decreases) over a time interval is undergoing acceleration, represented by the variable a. Acceleration is a vector quantity. It is also called non-uniform motion because the object’s speed or direction is changing. Figure 1.26 A drag racer accelerates from rest. acceleration: a vector quantity representing the change in velocity (magnitude or direction) per unit time scale 1.0 m 0.0 m 0.0 s 2.0 m 1.0 s 8.0 m 2.0 s 18.0 m 3.0 s Figure 1.27 This sequence illustrates a car undergoing non-uniform motion. Chapter 1 Graphs and equations describe motion in one dimension. 23 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 24 PHYSICS INSIGHT An object is accelerating if it is speeding up, slowing down, or changing
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direction. The following scenario illustrates acceleration. A drag race is a 402-m (quarter-mile) contest between two vehicles. Starting from rest, the vehicles leave the starting line at the same time, and the first vehicle to cross the finish line is the winner. A fan records the position of her favourite vehicle during the drag race. Her results are recorded in Table 1.4. The position-time graph for this data is shown in Figure 1.28. From the graph, note that the object is speeding up because the displacement between data points increases for each successive time interval. Which ticker tape in Figure 1.25 matches the graph in Figure 1.28? ▼ Table 1.4 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 2.0 8.0 18.0 32.0 50.0 Position vs. Time for a Dragster ) ] 50.0 40.0 30.0 20.0 10.0 0.0 0.0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.28 What does the slope of the graph indicate about the speed of the car? Instantaneous Velocity Instantaneous velocity is the moment-to-moment measure of an object’s velocity. Imagine recording the speed of your car once every second while driving north. These data form a series of instantaneous velocities that describe your trip in detail. Earlier in this section, you learned that determining the velocity of an object from a position-time graph requires calculating the slope of the position-time graph. But how can you obtain the slope of a curve? Remember that each point on the curve indicates the position of the object (in this case, the dragster) at an instant in time. To determine the velocity of an object at any instant, physicists use tangents. A tangent is a straight line that touches a curve at only one point (Figure 1.29(a)). Each tangent on a curve has a unique slope, which represents the velocity at that instant. In order for the object to be at that position, at that time, it must have an instantaneous velocity equal to the slope of the tangent at that point. Determining the slopes of the tangents at different points on a position-time curve gives the instantaneous velocities at different times. Consider forward to be the positive direction. PHYS
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ICS INSIGHT When you calculate the slope of a line or curve at a single point, you are finding an instantaneous value. When you calculate the slope between two points, you are finding an average value. tangent: a straight line that touches a curved-line graph at only one point 24 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 25 Position vs. Time for a Dragster Position vs. Time for a Dragster Position vs. Time for a Dragster 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0 5.0 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0 5.0 50.0 40.0 30.0 20.0 10..0 0.0 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.29(a) Figure 1.29(b) Figure 1.29(c) The slope of the tangent at 2.0 s is slope d t 14.0 m 0.0 m 3.0 s 1.0 s 14.0 m 2.0 s 7.0 m/s The slope of the tangent at 3.0 s is slope 30.0 m 0.0 m 4.0 s 1.75 s The slope of the tangent at 4.0 s is slope 47.0 m (15.0 m) 5.0 s 3.0 s 30.0 m 2.25 s 13 m/s 32.0 m 2.0 s 16 m/s The sign is positive, so at 2.0 s, the velocity of the dragster is 7.0 m/s [forward]. At 3.0 s, the velocity of the dragster is 13 m/s [forward]. At 4.0 s, the velocity of the dragster is 16 m/s [forward]. Using Slopes of Position-time Graphs to Draw Velocity-time Graphs You can now create a new table using the slopes of the position-time graphs in Figures 1.29(a), (b), and (c). See Table 1.5. Remember that the slope of a position-time graph is velocity. These slope values
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are actually instantaneous velocities at the given times. You can use these three velocities to draw a velocity-time graph (Figure 1.30). The resulting velocity-time graph is a straight line that goes through the origin when extended. This means that the dragster has started from rest (0 velocity). The graph has a positive slope. To find the meaning of slope, check the units of the slope of a velocity-time graph. They are (m/s)/s, which simplify to m/s2. These units are the units of acceleration. Since the velocity-time graph in this example is a straight line with non-zero slope, the acceleration of the object is constant, so the object must be undergoing uniformly accelerated motion. info BIT When jet fighters come in to land on an aircraft carrier, they stop so quickly that pilots sometimes lose consciousness for a few seconds. The same thing can happen when a pilot ejects from an aircraft, due to enormous acceleration. uniformly accelerated motion: constant change in velocity per unit time Chapter 1 Graphs and equations describe motion in one dimension. 25 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 26 ▼ Table 1.5 Time (s) Velocity (m/s [forward]) 2.0 3.0 4.0 7.0 13 16 ) ] 20 15 10 5 0 Velocity vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.30 This velocity-time graph represents an object undergoing uniformly accelerated motion. 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 Acceleration vs. Time Just as the slope of a position-time graph reveals the rate at which position changes (velocity), the slope of a velocity-time graph reveals the rate at which velocity changes (acceleration). Calculate the slope of the line in Figure 1.30 as follows, designating forward as positive: 1.0 2.0 Time (s) 3.0 4.0 slope rise run Figure 1.31 An accelerationtime graph for an object undergoing uniformly accelerated motion is a straight line with zero slope. PHYSICS INSIGHT If the acceleration-time graph has a non-zero slope, the acceleration is changing (is nonuniform). The slope of an acceleration-time graph is called jerk, with units m/s
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3. v v i f t t i f 10 m/s (4 m/s) 2.5 s 1.0 s 4 m/s2 The answer is positive, so the car is accelerating at 4 m/s2 [forward]. The resulting acceleration-time graph is shown in Figure 1.31. You know that the velocity-time graph for an object undergoing uniform motion is a horizontal line (with zero slope, as in Figure 1.23). Similarly, a horizontal line on an acceleration-time graph indicates uniform acceleration. Concept Check If the position-time graph for an object undergoing positive acceleration is a parabola, such as the one in Figure 1.28, what is the shape of the position-time graph for an object undergoing negative acceleration? What would a ticker tape of the motion of an object that is slowing down look like? After driving your all-terrain vehicle (ATV, Figure 1.32) through a field, you see a wide river just ahead, so you quickly bring the vehicle to a complete stop. Notice in Figure 1.33 that, as your ATV slows down, the displacement in each time interval decreases. Figure 1.32 ATVs can undergo a wide variety of motions. 26 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 27 scale 1.0 m 0.0 m 0.0 s 13.5 m 1.0 s 24.0 m 2.0 s 31.5 m 3.0 s 36.0 m 4.0 s 37.5 m 5.0 s Figure 1.33 This ATV is undergoing non-uniform motion. It is accelerating, in this case, slowing down. Example 1.5 shows the calculations and resulting velocity-time graph for an object that is slowing down uniformly. Example 1.5 The position-time data for an ATV approaching a river are given in Table 1.6. Using these data, (a) draw a position-time graph (b) draw a velocity-time graph (c) calculate acceleration Analysis and Solution Designate the forward direction as positive. (a) For the position-time graph, ▼ Table 1.6 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 13.5 24.0 31.5 36.0 37.5 plot the data in Table
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1.6 (Figure 1.34). Position vs. Time 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0. Practice Problems 1. Draw a position-time graph from the velocity-time graph given below. Velocity vs. Time ) ] 14 12 10 8 6 4 2 0 0 10 20 Time (s) 30 40 2. Calculate the acceleration using the graph below. Velocity vs. Time ) ] 12 10 Time (s) 10 12 1.0 2.0 3.0 4.0 5.0 Time (s) Answers 1. Position vs. Time Figure 1.34 (b) Since the position-time graph is non-linear, find the slope of the tangent at 2.0 s, 3.0 s, and 5.0 s (Figures 1.35(a), (b), and (c)). ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time d t 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(a) slope d t 32.5 m15.5 m 3.0 s1.0 s 17.0 m 2.0 s 8.5 m/s 250 200 150 100 50 ) ]. 1.0 m/s2 [N] 10 20 Time (s) 30 40 Chapter 1 Graphs and equations describe motion in one dimension. 27 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 28 Position vs. Time ( 40.0 30.0 20.0 10.0 0.0 0.0 1.0 3.0 2.0 Time (s) 4.0 5.0 e MATH Figure 1.35(b) For an alternative method to create a velocity-time graph from the position- time data points, visit www.pearsoned.ca/school/ physicssource ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(c) slope d t 37.0 m(26.0 m) 4.0 s2.0 s 1 1 m.0 s.0 2 5.5 m/s This tangent is a
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horizontal line, so its slope is zero. The slopes of the tangents give the instantaneous velocities (Table 1.7). Positive signs mean that the direction is forward. Plot the data on a velocity-time graph (Figure 1.36). ▼ Table 1.7 Time (s) Velocity (m/s [forward]) 2.0 3.0 5.0 8.5 5. 15.0 14.0 13.0 12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Velocity vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 6.0 (c) Find acceleration by calculating the slope of the Figure 1.36 velocity-time graph. v a t 0.0 m/s (8.5 m/s) 5.0 s 2.0 s 2.8 m/s2 The acceleration of the ATV is 2.8 m/s2. Because the forward direction was designated as positive, the negative sign means that the direction of acceleration is backward. Negative Acceleration Does Not Necessarily Mean Slowing Down In Example 1.5, the value for acceleration is negative. What is the meaning of negative acceleration? When interpreting the sign of acceleration, you need to compare it to the sign of velocity. For example, for the drag racer that is speeding up, the direction of its velocity is the same as the direction of its acceleration (see the calculation of the slope of the velocitytime graph for Figure 1.30). When the directions (signs) of velocity and acceleration are the same (positive or negative), the object is speeding up. PHYSICS INSIGHT v a v a v aa v a speeding up speeding up slowing down slowing down 28 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 29 For the ATV in Example 1.5, the direction of its velocity is opposite to the direction of its acceleration, so it is slowing down. When velocity and acceleration have opposite directions (signs), the object slows down. Concept Check (a) Think of two more examples of objects not mentioned in this text that are speeding up and slowing down. In each case, indicate the signs or directions of velocity and acceleration. (b) Under what circumstances can an object
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have a negative acceleration and be speeding up? (c) You are given a position-time graph that is a curve. How can you use the slope of the tangent to determine whether the object represented in the graph is speeding up or slowing down? (Hint: How does the slope of the tangent change as you move along the position-time curve?) THEN, NOW, AND FUTURE Biomechanics and the Role of the Crash Test Dummy Understanding how biological systems move is a branch of physics known as biomechanics. For automobile manufacturers, understanding how the human body moves during a car accident is very important. To study, collect, and analyze data on how the human body moves during a vehicular collision requires a test subject. Human cadavers were the first test subjects used. While live human testing was valuable, it was limited in its scope due to the physical discomfort required and injury potential for some of the tests. Despite the production of reliable applicable data, most automobile manufacturers discontinued live animal testing in 1993 for moral and ethical reasons. Clearly, a different type of test subject needed to be designed and built. It came in the form of the now recognizable crash test dummy. Sam W. Alderson created “Sierra Sam” in 1949 to test aircraft ejection seats and pilot restraint harnesses. Then came the VIP-50 series and Sierra Stan in the 1950s. Engineers combined the best features of these different models and debuted Hybrid I in 1971. Hybrid I was known as the “50th percentile male” dummy (meaning approximately 50% of men are larger and 50% of men are smaller), with a height of 168 cm and a mass of 77 kg. A year later, Hybrid II, with improved shoulder, spine, and knee responses, was produced to test lap and shoulder belts. Still crude, their use was limited, leading to the advent of the Hybrid III family of crash test dummies that include a 50th percentile male, a 50th percentile female, and two child dummies. This family of crash test dummies is designed to measure spine and rib acceleration, and demonstrate neck movement in rearend collisions. Equipped with a more humanlike spine and pelvis, THOR (Figure 1.37) is the successor of Hybrid III. Its face contains a number of sensors for facial impact analysis. Since front and side air bags have reduced upper body injury, lower extremity injury has become more prevalent. Therefore, THOR is built with an Achilles tendon to
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better mimic the side-to-side, up-and-down, and rotational movements of the ankle. Even with sophisticated crash test dummies, plastic and steel can only approximate how the human body will move. The study of soft tissue injury can only be accomplished with real-life subjects. Therefore, the future of crash testing will be in cre- ating detailed computer models of human systems. Even though it is slow and cumbersome for full body simulations, the computer has the advantage of repeatability and lower cost. The programmer has the ability to control every variable and repeat each and any event. 1. Why are crash test dummies used? 2. What are some of the advantages of THOR over his previous prototypes? 3. Will crash test dummies become obsolete? Explain. Figure 1.37 THOR Chapter 1 Graphs and equations describe motion in one dimension. 29 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 30 1.3 Check and Reflect 1.3 Check and Reflect Applications 3. Match each velocity-time graph below 1. A sprinter in a championship race accelerates to his top speed in a short time. The velocity-time data for part of the race are given in the table below. Use the data to find the (a) average acceleration from 0.00 s to 0.50 s (b) average acceleration from 0.50 s to 3.00 s (c) average acceleration from 5.00 s to 6.00 s (d) Describe what was happening to the acceleration and velocity over 6.00 s. Time (s) Velocity (m/s [forward]) 0.00 0.12 0.14 0.50 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 9.83 9.93 0.00 0.00 0.00 2.80 5.00 8.00 9.80 10.80 11.30 11.60 11.70 11.80 11.90 11.95 11.97 2. Describe the motion of the object as illustrated in the graph below. Position vs. Time Time (s) 2.0 4.0 6.0 8.0 10. 20.0 0.0 20.0 40.0 60.0 80.0 100.0 30 Unit I Kinematics with the correct statement. (i) negative acceleration (ii) positive acceleration (
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iii) moving with zero acceleration (iv) stationary object Extensions 4. In your notebook, complete the velocity- time data table for the graph below. Position vs. Time ) ] 36 33 30 27 24 21 18 15 12 9 6 3 0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) Time (s) Velocity (m/s [forward]) 2.0 4.0 6.0 8.0 e TEST To check your understanding of uniformly accelerated motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 31 1.4 Analyzing Velocity-time Graphs When a plane flies across Alberta with constant speed and direction, it is said to be undergoing uniform motion (Figure 1.38(a)). If you were sitting in the plane, you would experience a smooth ride. An all-terrain vehicle (ATV) bouncing and careening along a rough trail is constantly changing speed and direction in order to stay on the road. A ride in the ATV illustrates non-uniform motion, or acceleration (Figure 1.39(a)). You can distinguish between uniform and non-uniform motion by simple observation and gathering data from your observations (see Figures 1.38(b) and 1.39(b)). There are several ways to interpret the data. One way is to analyze graphs by determining their slopes to obtain further information about an object’s motion, as you did in section 1.3. In this section, you will develop this method further and learn another method of graphical analysis: how to find the area under a graph. First review the information you can obtain from the slopes of position-time and velocity-time graphs. Figure 1.38(a) Uniform motion Figure 1.39(a) Non-uniform motion Velocity vs. Time Velocity vs. Time ) ] Time (h) Figure 1.38(b) A graph representing uniform motion ) ] Time (s) Figure 1.39(b) A graph representing non-uniform motion Chapter 1 Graphs and equations describe motion in one dimension. 31 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 32 Slopes of Graphs Reveal How Objects Move Consider the three photos and velocity-time graphs in Figure 1.40. You can interpret each graph
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by reading values from it. To gain new information, you must analyze the graph by calculating its slope. The slope describes the object’s motion. Velocity vs. Time Velocity vs. Time Velocity vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.40(a) A velocity-time graph for an object at rest Figure 1.40(b) A velocity-time graph for an object undergoing uniform motion Figure 1.40(c) A velocity-time graph for an object undergoing uniformly accelerated motion Concept Check 1. Sketch position-time graphs for the following: (a) two possibilities for stopped motion (b) two possibilities for uniform motion (c) four possibilities for uniformly accelerated motion Describe each graph in terms of direction of travel and whether the object is speeding up or slowing down. 2. Sketch the corresponding velocity-time graph for each position-time graph in question 1. For each graph, describe how you determined the graph’s shape. By analyzing the units for the slope of a velocity-time graph, m/s2, you know from section 1.3 that the slope of a velocity-time graph represents the acceleration of the object. Concept Check Sketch all the types of acceleration-time graphs you have encountered thus far. Describe the kind of motion that each graph represents. 32 Unit I Kinematics 01-PearsonPhys20-Chap01 7/25/08 7:50 AM Page 33 1-5 Design a Lab 1-5 Design a Lab Tortoise or Hare? The Question In your class, who has the fastest acceleration and the fastest average speed in the 50-m dash? Design and Conduct Your Investigation Make a list of variables that you think are likely to influence the acceleration of each participant. For each variable on your list, write a hypothesis that predicts how changes in that variable will affect the participants’ acceleration. Write a procedure for an investigation that will test the effect of one of these variables on acceleration. Clearly outline all the steps that you will follow to complete your investigation. Identify the responding and manipulated variables. List all the materials and equipment you will need, as well as all safety precautions. Compare your experimental design and procedure with those of your classmates. Identify any strengths and weaknesses. With your teacher’s approval, conduct your investigation. State any problems or questions that you found during your investigation or analysis that would need additional investigation to answer. The Area Under a Velocity-time Graph Represents Displ
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acement Occasionally, due to a medical or other emergency, a pilot must turn the aircraft and land at the same or alternate airport. Consider the graph for the uniform motion of a plane travelling east at 300 km/h for 2.0 h only to turn back west for 0.5 h to make an emergency landing (Figure 1.41). What is the plane’s displacement for this time interval? Unit analysis indicates that the area under a velocity-time graph equals displacement. To calculate displacement using a velocity-time graph, look at the units on the axes. To end up with a unit of displacement (km) from the units km/h and h, you need to multiply: km h h km The shapes in Figure 1.41 are rectangles, so the area under the velocitytime graph is l w (length times width). In this case, find the sum of the areas above and below the time axis. Consider east to be positive. For eastward displacement, the area is above the time axis, so it is positive. For westward displacement, the area is below the time axis, so it is negative. For eastward displacement (above the time axis), Velocity vs. Time Time (h) 1.0 2.0 3. 400 200 0 200 400 Figure 1.41 To calculate net displacement, add the areas above and below the time axis. area dd 300 vt km h 600 km (2.0 h) Chapter 1 Graphs and equations describe motion in one dimension. 33 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 34 For westward displacement (below the time axis), dd vt 300 km h 150 km (0.5 h) PHYSICS INSIGHT Note that the answer has one significant digit because there is only one significant digit in 0.5 h. To find the plane’s net displacement, add the two areas. area ddd 600 km (150 km) 600 km 150 km 450 km 5 102 km [E] Because the net area is positive, the plane’s displacement is 5 102 km [E]. Unlike position-time graphs, where you can only calculate the slope to determine velocity, you can use velocity-time graphs to determine both acceleration and displacement, as in the next example. Example 1.6 From the graph in Figure 1.42, calculate (a) displacement (b) acceleration Velocity vs. Time Practice Problems 1. Calculate the displacement and acceleration from the
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graph. Velocity vs. Time ) ].4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 4.0 3.0 2.0 1..0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.42 0 2 4 6 Time (s) 8 10 Analysis and Solution (a) For displacement, find the sum of the areas under the velocity-time graph (Figure 1.43). Designate east (above the time axis) as the positive direction. Convert minutes to seconds. 2. Use the graph below to determine the displacement of the object. Velocity vs. Time Time (s) 2 4 6 8 10 12 ) ] 10 5 0 5 10 15 20 25 Answers 1. 22 m [N], 0 m/s2 [N] 2. 1.0 102 m [E] or 1.0 102 m [W] 34 Unit I Kinematics 4.0 3.0 2.0 1. Velocity vs. Time A B C 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 35 d Region A: vt 3.0 A 4.0 min m s 60 s 1 min (240 s) 3.0 m s 720 m Region B: d B (5.0 min 4.0 min) 1 2 60 s 1 min 3.0 m s 1.0 1.0 m s m s (5.0 min 4.0 min) 60 s 1 min (1.0 min) 60 s 1 min 2.0 1.0 m s m s (1.0 min) 60 s 1 min 1 2 1 2 (120 m) 60 m PHYSICS INSIGHT When calculating total displacement from a velocity-time graph, remember to keep track of whether the area is positive or negative. PHYSICS INSIGHT Check your answer by looking at the units. Do the units reflect the answer that you are asked to find? 120 m Region C: d C 1.0 m s 1.0 m s 300 m d A d B d d C (10.0 min 5.0 min) 60 s 1 min (5.0 min) 60
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s 1 min 720 m 120 m 300 m 1140 m 1.1 103 m The answer is positive, so the direction of the displacement is east. (b) For acceleration, find the slope of each section of the graph. In region A: v a t m m 3.0 3.0 s s 240 s 0.0 m/s2 This answer makes sense because the velocity-time graph is a horizontal line. In region B: v a t m m 3.0 1.0 s s 60 s 0.033 m/s2 or 0.033 m/s2[W] Chapter 1 Graphs and equations describe motion in one dimension. 35 01-PearsonPhys20-Chap01 7/25/08 7:54 AM Page 36 Since the third part of the graph is also a horizontal line, its slope is also zero. Paraphrase (a) The displacement is 1.1 103 m [E]. (b) The acceleration is zero in regions A and C and 0.033 m/s2 [W] in region B. Concept Check For the velocity-time graph of a ball thrown up in the air (Figure 1.44), what is the net displacement of the ball? 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 ) ] 10.00 15.00 Figure 1.44 Average Velocity from Velocity-time Graphs Objects rarely travel at constant velocity. Think of your journey to school today. Whether you travelled by car or bus, rode a bike, or walked, stop signs, traffic lights, corners, and obstacles caused a variation in your velocity, or rate of travel. If you describe your motion to a friend, you can use a series of instantaneous velocities. The more time instances you use to record your motion, the more details about your trip you can communicate to your friend (Figure 1.45(a)). However, if you were to use the equation v and substitute your total displacement d t for d and your total time of travel for t, you would lose most of the details of your journey. You would obtain a value for your average velocity, v ave (Figure 1.45(b)). Velocity vs. Time Velocity vs. Time ) ] Time (s) v ave Time (s Figure 1.45(a) By using a series of instantaneous velocities at
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the given times, you can precisely describe your journey. Figure 1.45(b) average velocity of the journey. It describes your journey but the detail of the motions is lost. The straight line represents the If you need to obtain an average velocity value from a velocity-time graph, recall that displacement, d, is the area under the graph. 36 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 37 To find average velocity, determine the area under the velocity-time graph and divide it by the total time. To calculate average velocity when given different displacements over different time intervals, simply add the total displacement and divide by the total time, as shown in the next example. Example 1.7 Find the average velocity of a student who jogs 750 m [E] in 5.0 min, does static stretches for 10.0 min, and then runs another 3.0 km [E] in 30.0 min. Given Choose east to be positive. Convert kilometres to metres. d 1 t1 d 2 t2 d 3 750 m [E] 750 m 5.0 min 0 m 10.0 min 3.0 km [E] 3.0 km 1000 m 1 km 3000 m t3 30.0 min Required average velocity (v ave) Analysis and Solution First add the displacement values. d d 2 d 1 750 m 0 m 3000 m 3750 m d 3 total Then add the time intervals and convert to seconds. The total time elapsed is ttotal t3 t2 t1 5.0 min 10.0 min 30.0 min (45.0 min)60 2700 s s min Average velocity equals total displacement divided by total time elapsed: v ave d t 3750 m 2700 s 1.4 m/s Since the answer is positive, the direction is east. Paraphrase The student’s average velocity is, therefore, 1.4 m/s [E]. Practice Problems 1. A person runs 10.0 m [E] in 2.0 s, then 5.0 m [E] in 1.5 s, and finally 30.0 m [W] in 5.0 s. Find the person’s average velocity. 2. Person A runs the 100-m dash in 9.84 s and then tags person B, who runs 200 m in 19.32 s. Person B then tags an out-of-shape person C,
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who runs 400 m in 1.90 min. Find the average velocity for the trio. Compare it to each individual’s average velocity. Assume they are all running in a straight line. Answers 1. 1.8 m/s [W] 2. 4.89 m/s [forward] A: 10.2 m/s [forward]; faster than the average velocity for the trio B: 10.4 m/s [forward]; faster than the average velocity for the trio C: 3.51 m/s [forward]; slower than the average velocity for the trio Chapter 1 Graphs and equations describe motion in one dimension. 37 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 38 As you have seen, velocity-time graphs are very useful. They provide the following information: – Reading the velocity-time graph gives you instantaneous velocity values. – Finding the slope of a velocity-time graph gives you an object’s acceleration. – The area under a velocity-time graph gives you the object’s displacement. – You can also determine the average velocity of an object over a time interval from a velocity-time graph. Example 1.8 shows you how to obtain information about an object’s velocity and acceleration. Example 1.8 A bird starts flying south. Its motion is described in the velocity-time graph in Figure 1.46.0 6.0 4.0 2.0 0.0 Velocity vs. Time F E 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Time (s) –2.0 A –4.0 –6.0 –8.0 B C D From the graph, determine (a) whether acceleration is positive, negative, or zero for each section (b) the value of the acceleration where it is not zero (c) when the bird changes direction Analysis and Solution Consider north to be the positive direction. (a) Acceleration is the slope of each section of the graph. A: Final velocity is more negative than the initial velocity, as the bird is speeding up in the south direction. So the slope of this line is negative. The bird’s acceleration is negative. 1 2 3 4 5 6 7 8 10 11 9 12 Time (s) Figure 1.46 Practice Problems 1. Position vs. Time ) ]
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10 1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 (a) Describe the motion of the object from the graph above. (b) Draw the corresponding velocity-time graph. (c) Determine the object’s displacement. (d) When is the object stopped? 38 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 39 B: Acceleration is zero because the slope is zero (the graph Answers 1. (a) 3 m/s for 2 s, rest for 3 s, 3 m/s for 4 s, 6 m/s for 1 s (b) Velocity vs. Time ) ] 2 –4 –6 –8 5 10 15 Time (s) (c) 12 m (d) 2–5 s is a horizontal line). C: Acceleration is negative because the slope of the line is negative (as in section A). D: Acceleration is zero because the slope of the line is zero (as in section B). E: Final velocity is positive because the bird is now flying north. So the slope of this line is positive. The bird’s acceleration is positive. F: Acceleration is zero because the slope of the line is zero. (b) Acceleration is not zero for sections A, C, and E. v A.0 4.0 s s 1.0 s 0.0 s 2.0 m/s2 2.0 m/s2 [S] v C.0 6.0 s s 4.0 s 3.0 s 2.0 m/s2 2.0 m/s2 [S] v E.0 6.0 s s 10.0 s 7.0 s m 12.0 s 3.0 s 4.0 m/s2 4.0 m/s2 [N] (c) The bird changes direction at 8.5 s — it crosses the time axis at this instant. Chapter 1 Graphs and equations describe motion in one dimension. 39 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 40 The next example shows you how to use areas to find the displacement of the bird and its average velocity from a velocity-time graph. Example 1.9 From the graph in Figure 1.46, determine (a) the displacement for each section (b) the displacement
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for the entire flight (c) the average velocity of the flight Analysis and Solution (a) Displacement is the area between the graph and the time axis. Practice Problems 1. In your notebook, redraw the graph in Example 1.8 Practice Problem 1, but label the vertical axis “velocity (m/s [N])”. (a) Find the displacements for 0–2 s, 2–5 s, 5–7 s, 7–9 s, and 9–10 s. (b) Find the object’s total displacement. (c) Find the average velocity of the object. Answers 1. (a) 6 m [N], 18 m [N], 6 m [N], 6 m [N], 9 m [N] (b) 15 m [N] (c) 1.5 m/s [N] A: A l w bh 1 2 d (2.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 3.0 m B: A l w d (4.0 m s 8.0 m )(3.0 s 1.0 s) C: A l w bh 1 2 d (4.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 5.0 m D: A l w d (6.0 m s 18 m )(7.0 s 4.0 s) E: A 1 2 1 bh bh 2 (1.5 s) 6.0 d 1 2 m s (1.5 s) 6.0 1 2 m s 0.0 m F: A l w (6.0 d m s )(3.0 s) 18 m (b) Add all the displacements calculated in (a). The displacement over the entire flight is –16 m. Since north is positive, the displacement is 16 m [S]. (c) v ave d T t m 6 1. s 0 13 1.2 m/s 40 Unit I Kinematics North is positive, so the average velocity for the flight is 1.2 m/s [S]. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 41 Drawing Position-time and Acceleration-time Graphs from Velocity-time Graphs In this section, you have learned how to
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use a velocity-time graph to calculate displacement. It is also useful to know how to draw position-time and acceleration-time graphs when given a velocity-time graph. Consider the following trip. A family travelling from Calgary to go camping in Banff National Park moves at 18.0 m/s [forward] in a camper van. The van accelerates for 4.0 s until it reaches a velocity of 30.0 m/s [forward]. It continues to travel at this velocity for 25.0 s. When approaching a check stop, the driver brakes, bringing the vehicle to a complete stop in 15.0 s. The velocity-time graph for the trip is given in Figure 1.47. Velocity vs. Time I II III Figure 1.47 The complete graph of the van’s motion 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 The next example shows you how to create an acceleration-time graph from a velocity-time graph. Example 1.10 Use the velocity-time graph in Figure 1.47 to draw the corresponding acceleration-time graph. Analysis and Solution To find acceleration, calculate the slope for each section of the graph. The velocity-time graph has three distinct sections. The slope in each section is constant. Consider forward to be positive. Velocity vs. Time Section l Time: 0.0 s to 4.0 s ti vi tf vf 0.0 s 18.0 m/s 4.0 s 30.0 m/s 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(a) Chapter 1 Graphs and equations describe motion in one dimension. 41 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 42 Practice Problems 1. For each velocity-time graph below, draw the corresponding acceleration-time graph. Velocity vs. Time (a 25 20 15 10 5 0 0 2.0 6.0 10.0 14.0 vf Time (s) (b) Velocity vs. Time ) ] 25 20 15 10 5
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0 0 2.0 6.0 Time (s) 10.0 14.0 Answers 1. (a) Acceleration vs. Time ( 15 10 5 0 0 2.0 6.0 Time (s) 10.0 14.0 Acceleration vs. Time 10 5 0 5 Time (s) 2.0 6.0 10.0 14.0 (b) slope a v v i f ti tf 30.0 m/s (18.0 m/s) 4.0 s 0.0 s 3.0 m/s2 Velocity vs. Time Section ll Time: 4.0 s to 29.0 s ti vi tf 4.0 s 30.0 m/s 4.0 s 25.0 s 29.0 s 30.0 m/s v v i f ti tf slope 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 30.0 m/s (30.0 m/s) 29.0 s 4.0 s 0.0 m/s2 Figure 1.48(b) Velocity vs. Time Section lll Time: 29.0 s to 44.0 s ti vi tf 29.0 s 30.0 m/s 29.0 s 15.0 s 44.0 s 0.0 m/s vf slope a v v i f ti tf ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(c) 0.0 m/s (30.0 m/s) 44.0 s 29.0 s 2.0 m/s2 Now plot the values on the acceleration-time graph (Figure 1.49). Each section of the graph is a horizontal line because acceleration is constant (uniform). Acceleration vs. Time ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 30.0 35.0 40.0 45.0 5.0 10.0 15.0
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20.0 25.0 Time (s) Figure 1.49 The next example shows you how to use a velocity-time graph to generate a position-time graph. 42 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 43 Example 1.11 Sketch a position-time graph from the velocity-time graph in Figure 1.47. Analysis and Solution To sketch the position-time graph, find the area under the velocitytime graph. Consider forward to be positive. In the first part of the velocity-time graph (0.0–4.0 s), area (displacement) is a rectangle and a triangle. The displacement is positive because the areas are above the time-axis. A l w bh 1 2 d 1 (18.0 m/s)(4.0 s) (4.0 s)(30.0 m/s 18.0 m/s) 2 96 m Since the velocity-time graph in this section has a positive slope, the car has positive acceleration, so the corresponding positiontime graph is a parabola that curves upward. On the positiontime graph, sketch a curve from the origin to the point t 4.0 s 96 m (Figure 1.50(a)). and d In the second part of the velocity-time graph (4.0–29.0 s), displacement is a rectangle. It is positive since the area is above the time-axis. A l w (30.0 m/s)(29.0 s 4.0 s) d 750 m Since the velocity-time graph has zero slope in this section, the car moves with constant velocity and the position-time graph is a straight line with a positive slope that extends from t 4.0 s and d t 29.0 s and d 96 m to 96 m 750 100 80 60 40 20 0 0 Position vs. Time 4 Time (s) Figure 1.50(a) Position vs. Time ) ] 1000 900 800 700 600 500 400 300 200 100 0 0 84 24 28 32 12 16 20 Time (s) Figure 1.50(b) Practice Problems 1. Velocity vs. Time ) ] 2 –4 –6 3 6 9 12 15 18 Time (s) (a) Describe the motion of the object illustrated above. Calculate its total displacement. (b) Draw the corresponding position-time graph. Answers 1. (a) travels with uniform motion, changes
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direction at 10 s, and travels with uniform motion; total displacement: 10 m [forward] (b) Position vs. Time ) ] 60 50 40 30 20 10 0 0 5 15 20 10 Time (s) 846 m (See Figure 1.50(b).) In the third part of the velocity-time graph (29.0– 44.0 s), displacement is a triangle. It is positive since the area is above the time-axis. 1 A bh 2 d (44.0 s 29.0 s)(30.0 m/s) 1 2 225 m Chapter 1 Graphs and equations describe motion in one dimension. 43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 44 ) ] 1200 1100 1000 900 800 700 600 0 Position vs. Time 0 24 28 32 36 40 44 48 Time (s) Figure 1.50(c) Since the velocity-time graph has a negative slope, the car undergoes negative acceleration, so the slopes of the tangents of the position-time graph decrease (approach zero). The position-time graph is a parabola that curves down, from t 29.0 s and 846 m to d t 44.0 s and 846 m 225 m d 1071 m Position vs. Time (See Figure 1.50(c).) The resulting position-time graph is shown in Figure 1.51 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 0 84 12 16 24 28 32 20 Time (s) 36 40 44 48 Figure 1.51 Concept Check If north is positive, sketch position-time, velocity-time, and acceleration-time graphs for an object (a) speeding up and going north (b) slowing down and going north (c) speeding up and going south (d) slowing down and going south 1.4 Check and Reflect 1.4 Check and Reflect Knowledge 1. On a ticker tape, how can you distinguish between uniform and uniformly accelerated motion? 2. Use the terms “displacement” and “velocity” to describe how uniformly accelerated motion differs from uniform motion. 3. What is the relationship between the slope of a position-time graph and velocity? 4. Compare the shape of a position-time graph for uniform motion with a positiontime graph representing uniformly accelerated motion. 5. What is the relationship between the slope of a velocity-time graph and acceleration? 6. If a velocity-
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time graph is a straight line with a non-zero slope, what kind of motion is the object undergoing? 44 Unit I Kinematics 7. Determine the displacement of the object whose motion is described by the following graph. Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 6.0 4.0 Time (s) 8.0 10.0 8. Calculate displacement from the velocity-time graph below.0 4.0 3.0 2.0 1.0 0.0 Velocity vs. Time 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 45 9. Describe the velocity-time graph for an object undergoing negative acceleration. 10. What quantity of motion can be determined from the area under a velocity-time graph? (b) Create a graph for the question and check your answer using graphing techniques. 17. Determine acceleration from the velocity- time graph given below. Velocity vs. Time ) ] 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 10.0 20.0 Time (s) 30.0 40.0 18. A truck travelling forward at 14.0 m/s accelerates at 1.85 m/s2 for 6.00 s. It then travels at the new speed for 35.0 s, when a construction zone forces the driver to push on the brakes, providing an acceleration of 2.65 m/s2 for 7.0 s. Draw the resulting velocity-time and position-time graphs for this motion. Extension 19. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 10.0 5.0 0.0 5.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s 10.0 e TEST To check your understanding of velocity-time graphs follow the eTest links at www.pearsoned.ca/school/physicssource. 11. Compare and contrast the shape of a velocity-time graph for an object experiencing uniform motion with one experiencing uniformly accelerated motion. 12. Describe the acceleration-time graph of a car travelling forward and applying its brakes. 13. Calculate the acceleration of an object using
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the velocity-time graph below. Velocity vs. Time 2.0 4.0 6.0 8.0 10.0 Time (s 100 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 14. Construct an acceleration-time graph using the graph given below. Velocity vs. Time ) ] 12 10 8 6 4 2 0 0 5 10 15 Time (s) Applications 15. A motorbike increases its velocity from 20.0 m/s [W] to 30.0 m/s [W] over a distance of 200 m. Find the acceleration and the time it takes to travel this distance. 16. (a) While driving north from Lake Louise to Jasper, you travel 75 min at a velocity of 70 km/h [N] and another 96 min at 90 km/h [N]. Calculate your average velocity. Chapter 1 Graphs and equations describe motion in one dimension. 45 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 46 1.5 The Kinematics Equations A cylindrical piston the length of a football field controls the launch of a fighter plane from the deck of a carrier ship (Figure 1.52). Too much pressure and the nose gear is ripped off; too little pressure and the plane crashes into the ocean. This propulsion system accelerates a 20 000-kg plane from rest to 74 m/s (266 km/h) in just 2.0 s! e TECH Study the physics of jet takeoffs by visiting www.pearsoned.ca/school/ physicssource and viewing the simulation. Figure 1.52 Analyzing complex motions requires many calculations, some of which involve using the kinematics equations you will study in this section. To determine the crucial values required for launching a plane, such as flight deck length, final velocity, and acceleration, physicists and engineers use kinematics equations similar to the ones you will know by the end of this section. In this section, you will practise your analytical skills by learning how to derive the kinematics equations from your current knowledge of graphs and then apply these equations to analyze complex motions such as airplane launches. Velocity vs. Time Concept Check Create a summary chart for the information you can gather by analyzing position-time, velocity-time, and acceleration-time graphs. Use the headings “Graph Type”, “Reading the Graph”, “Slope”, and “Area”
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. Consider the airplane taking off from a moving aircraft carrier (Figure 1.52). The plane must reach its takeoff speed before it comes to the end of the carrier’s runway. If the plane starts from rest, the velocitytime graph representing the plane’s motion is shown in Figure 1.53. Notice that the slope of the graph is constant. By checking the units on the graph, you know that the slope represents acceleration: rise run Therefore, the velocity-time graph is a straight line. m/s2. In this case, the acceleration is constant (uniform). m/s s 80 70 60 50 40 30 20 10 ) ].0 1.0 Time (s) 2.0 Figure 1.53 The slope of this velocity-time graph represents the plane’s acceleration. 46 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 47 From Figure 1.53, you can derive the first kinematics equation This equation can also be written as v a t The next example shows you how to apply this equation to solve a problem. Example 1.12 A hybrid car with an initial velocity of 10.0 m/s [E] accelerates at 3.0 m/s2 [E]. How long will it take the car to acquire a final velocity of 25.0 m/s [E]? Given Designate east as the positive direction. v 10.0 m/s [E] 10.0 m/s v 25.0 m/s [E] 25.0 m/s a 3.0 m/s2 [E] 3.0 m/s2 i f Required time (t) Analysis and Solution Use the equation v v v a i t t f Since you are dividing by a vector, and initial and final velocities and acceleration are in the same direction, use the scalar form of the equation. Isolate t and solve. t vi vf a 25 m/s 10 m/s 3.0 m/s2 m 15 s m 3.0 s2 5.0 s Practice Problems 1. A motorcycle with an initial velocity of 6.0 m/s [E] accelerates at 4.0 m/s2 [E]. How long will it take the motorcycle to reach a final velocity of 36.0 m/s [E]? 2. An elk moving at a velocity of 20 km/
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h [N] accelerates at 1.5 m/s2 [N] for 9.3 s until it reaches its maximum velocity. Calculate its maximum velocity, in km/h. Answers 1. 7.5 s 2. 70 km/h [N] PHYSICS INSIGHT The mathematics of multiplying vectors is beyond this text and division of vectors is not defined. So, when multiplying and dividing vectors, use the scalar versions of the kinematics equations. Paraphrase It will take the car 5.0 s to reach a velocity of 25.0 m/s [E]. As you know from section 1.4, you can calculate the area under a velocitytime graph. By checking the units, you can verify that the area represents displacement: l w m s s m Chapter 1 Graphs and equations describe motion in one dimension. 47 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 48 PHYSICS INSIGHT The area of a trapezoid 1 2 is given by l2)w. (l1 To calculate the displacement (area) from the velocity-time graph in Figure 1.54, you can use the formula for the area of a trapezoid, 1 A (l1 2 l2)w, which is simply the average of the parallel sides multiplied by the base. Velocity vs. Time 1 Area d (vf vi ) t 2 l2 l1 w vf vi ) ] ti t Time (s) tf Figure 1.54 graph to derive the equation d v t. ave Use the area under the velocity-time The second kinematics equation is v i (v d f)t 1 2 v where l1 to apply this equation. v i, l2 f, and w t. The next example shows you how Example 1.13 A cattle train travelling west at 16.0 m/s is brought to rest in 8.0 s. Find the displacement of the cattle train while it is coming to a stop. Assume uniform acceleration. Given Designate west as the positive direction. v v t 8.0 s 16.0 m/s [W] 16.0 m/s 0 m/s [W] 0 m/s i f Required displacement (d ) Analysis and Solution Use the equation d 1 2 (v i vv f)t and solve for d. Practice Problems 1. A hound running at a velocity of 16 m/s
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[S] slows down uniformly to a velocity of 4.0 m/s [S] in 4.0 s. What is the displacement of the hound during this time? 2. A ball moves up a hill with an initial velocity of 3.0 m/s. Four seconds later, it is moving down the hill at 9.0 m/s. Find the displacement of the ball from its initial point of release. Answers 1. 40 m [S] 2. 12 m 48 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 49 d (16.0 m/s 0 m/s)(8.0 s) 1 2 8.0 64 m (8.0 s) m s The sign is positive, so the train’s direction is west. Paraphrase The cattle train travels 64 m [W] before it stops. You can also calculate the area under a velocity-time graph by considering that the total area under the graph is made up of a triangle and a rectangle (Figure 1.55). vf v at vi vi area area of rectangle l w t vi t 1 2 bh 1 t 2 1 t 2 a1 2 v t a )2 t ( Figure 1.55 You can divide the area under the velocity-time graph into a triangle and a rectangle. info BIT The fastest time of covering 1.6 km while flipping tiddly winks — 52 min 10 s — was achieved by E. Wynn and J. Culliongham (UK) on August 31, 2002. Their speed was 1.8 km/h. Using this data and the displacement equation below, verify that they did indeed travel a distance of 1.6 km. In Figure 1.55, the area of a rectangle represents the displacement of an object travelling with a constant velocity, v i. The height of the rectangle is v i and the base is t. Therefore, the area of the rectangle is equal to v t. The area of the triangle represents the additional displacement resulting from the change in velocity. The height of the triangle is v and the base is t. The area of the triangle is equal to v v i f i 1 2 t(v). But vat. Therefore, the area of the triangle is equal to 1 2 (t)(at) a(t)2. Add both displacements to obtain 1 2 PHYSICS INSIGHT 2 ti (t2)
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tf ti)2. (t)2 (tf 2, whereas d v 1 t a(t)2 2 i The next example shows you how to apply the third kinematics equation. Chapter 1 Graphs and equations describe motion in one dimension. 49 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 50 Example 1.14 A golf ball that is initially travelling at 25 m/s hits a sand trap and slows down with an acceleration of 20 m/s2. Find its displacement after 2.0 s. Given Assign a positive direction for forward and a negative direction for backward. 25 m/s vi a 20 m/s2 t 2.0 s Practice Problems 1. A skier is moving down a uniform slope at 3.0 m/s. If the acceleration down the hill is 4.0 m/s2, find the skier’s displacement after 5.0 s. 2. A motorcycle travelling at 100 km/h on a flat road applies the brakes at 0.80 m/s2 for 1.0 min. How far did the motorcycle travel during this time? Answers 1. 65 m [down] 2. 2.3 102 m Required displacement (d ) Analysis and Solution Use the equation d v i 25 d m s (2.0 s) 1 2 50 m (40 m) 10 m t a(t)2 to solve for d. 1 2 20 m s2 (2.0 s)2 The sign is positive, so the direction is forward. Paraphrase The displacement of the golf ball is 10 m [forward]. To obtain the fourth kinematics equation, derive the value of a required variable in one equation, substitute the derived value into the second equation, and simplify. v i t f Start with a v Isolate v i. at v v Then substitute v (v v d f i f i 1 2 at for v i into the equation f)t. The equation becomes d 1 2 (v f at v f)t. This equation simplifies to v d 1 t a(t)2 2 f Apply this equation in the next example. info BIT The fastest lava flow ever recorded was 60 km/h in Nyiragongo (Democratic Republic of Congo) on January 10, 1977. At this speed, how far would the lava travel in 2 h 30 min? 50 Unit I Kinematics 01-PearsonPhys20-Chap01 7
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/23/08 11:43 AM Page 51 Example 1.15 Figure 1.56 A speedboat slows down at a rate of 5.0 m/s2 and comes to a stop (Figure 1.56). If the process took 15 s, find the displacement of the boat. f 0.0 m/s (because the boat comes to rest) Given Let forward be the positive direction. v t 15 s a 5.0 m/s2 (Acceleration is negative because the boat is slowing down, so its sign must be opposite to that of velocity (positive).) Practice Problems 1. If the arresting device on an aircraft carrier stops a plane in 150 m with an acceleration of 15 m/s2, find the time the plane takes to stop. 2. The 1968 Corvette took 6.2 s to accelerate to 160 km/h [N]. If it travelled 220 m [N], find its acceleration. Answers 1. 4.5 s 2. 2.9 m/s2 [N] Required displacement (d ) Analysis and Solution Use the equation d vv f t a(t)2 to solve for d 1 2. d (0.0 m/s)(15 s) 1 2 (5.0 m/s2)(15 s)2 562.5 m 5.6 102 m The sign is positive, so the direction of displacement is forward. Paraphrase The displacement of the speedboat is 5.6 102 m [forward]. Deriving the fifth and last kinematics equation involves using the difference of squares, another math technique. Isolate t in the equation a v f v i. Remember that, when multiplying t or dividing vectors, use the scalar form of the equation: t vi vf a PHYSICS INSIGHT Recall that the difference of squares is (a b)(a b) a2 b2 Then substitute the expression for t into d (vi d (vi vi vf a 1 2 1 2 vf)t: vf) 2 vi 2 vf a 1 d 2 Chapter 1 Graphs and equations describe motion in one dimension. 51 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 52 info BIT A tortoise covered 5.48 m in 43.7 s at the National Tortoise Championships in Tickhill, UK, to set a world record on July 2, 1977. It was moving at 0.45 km/h. The more standard form of
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the fifth kinematics equation is vf 2 vi 2 2ad This equation is applied in the next example. Example 1.16 Practice Problems 1. A jetliner lands on a runway at 70 m/s, reverses its engines to provide braking, and comes to a halt 29 s later. (a) What is the jet’s acceleration? (b) What length of runway did the jet require to come safely to a complete stop? 2. On-ramps are designed so that motorists can move seamlessly into highway traffic. If a car needs to increase its speed from 50 km/h to 100 km/h and the engine can provide a maximum acceleration of magnitude 3.8 m/s2, find the minimum length of the on-ramp. Answers 1. (a) 2.4 m/s2 [forward] (b) 1.0 km 2. 76 m A bullet accelerates the length of the barrel of a gun (0.750 m) with a magnitude of 5.35 105 m/s2. With what speed does the bullet exit the barrel? Given a 5.35 105 m/s2 d 0.750 m Required final speed (vf) Analysis and Solution Use the equation vf from rest, vi 0 m/s. 2 vi 2 2ad. Since the bullet starts vf vf 2 (0 m/s)2 2(5.35 105 m/s2)(0.750 m) 802 500 m2/s2 802 500 m2/s2 896 m/s Paraphrase The bullet leaves the barrel of the gun with a speed of 896 m/s. It is important to note that the velocity-time graph used to derive the kinematics equations has a constant slope (see Figure 1.54), so the equations derived from it are for objects undergoing uniformly accelerated motion (constant acceleration). General Method of Solving Kinematics Problems Now that you know five kinematics equations, how do you know which one to use to solve a problem? To answer this question, notice that each of the five kinematics equations has four variables. Each kinematics problem will provide you with three of these variables, as given values. The fourth variable represents the unknown value. When choosing your equation, make sure that all three known variables and the one unknown variable are represented in the equation (see Table 1.8). You may need to rearrange the equation to solve for the unknown variable.
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52 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 53 ▼ Table 1.8 The Variables in the Five Kinematics Equations Equation v i v f)t v d 1 t a(t)2 2 i v d 1 t a(t)2 2 f 2 vi vf 2 2ad PHYSICS INSIGHT Remember these implied given values. – If the object starts from 0. – If the object comes to rest, v i f a stop, v 0. – If the object experiences uniform motion, a 0. 1.5 Check and Reflect 1.5 Check and Reflect Applications 1. How far will a humanoid robot travel in 3.0 s, accelerating at 1.0 cm/s2 [forward], if its initial velocity is 5.0 cm/s [forward]? 2. What is the displacement of a logging truck accelerating from 10 m/s [right] to 20 m/s [right] in 5.0 s? 3. How far will a car travel if it starts from rest and experiences an acceleration of magnitude 3.75 m/s2 [forward] for 5.65 s? 4. Determine the acceleration of a bullet starting from rest and leaving the muzzle 2.75 103 s later with a velocity of 460 m/s [forward]. 5. Some aircraft are capable of accelerations of magnitude 42.5 m/s2. If an aircraft starts from rest, how long will it take the aircraft to travel down the 2.6-km runway? 6. If a cyclist travelling at 14.0 m/s skids to a stop in 5.60 s, determine the skidding distance. Assume uniform acceleration. 7. Approaching a flashing pedestrianactivated traffic light, a driver must slow down to a speed of 30 km/h. If the crosswalk is 150 m away and the vehicle’s initial speed is 50 km/h, what must be the magnitude of the car’s acceleration to reach this speed limit? 8. A train’s stopping distance, even when full emergency brakes are engaged, is 1.3 km. If the train was travelling at an initial velocity of 90 km/h [forward], determine its acceleration under full emergency braking. 9. A rocket starts from rest and accelerates uniformly for 2.00 s over a displacement of 150 m [W]. Determine the rocket’s acceleration.
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10. A jet starting from rest reaches a speed of 241 km/h on 96.0 m of runway. Determine the magnitude of the jet’s acceleration. 11. What is a motorcycle’s acceleration if it starts from rest and travels 350.0 m [S] in 14.1 s? 12. Determine the magnitude of a car’s acceleration if its stopping distance is 39.0 m for an initial speed of 97.0 km/h. 13. A typical person can tolerate an acceleration of about 49 m/s2 [forward]. If you are in a car travelling at 110 km/h and have a collision with a solid immovable object, over what minimum distance must you stop so as to not exceed this acceleration? 14. Determine a submarine’s acceleration if its initial velocity is 9.0 m/s [N] and it travels 1.54 km [N] in 2.0 min. e TEST To check your understanding of the kinematics equations, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 53 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 54 1.6 Acceleration due to Gravity Figure 1.57 Amusement park rides are an application of physics. projectile motion: motion in a vertical plane projectile: an object released or thrown into the air Many amusement parks and midways showcase a ride based solely on acceleration due to gravity. The ride transports thrill seekers up to a dizzying height, allows them to come to rest, and then, without warning, releases them downward before coming to a controlled stop (Figure 1.57). In the previous sections, you learned about objects that move in a horizontal plane. Many objects move in a vertical plane. Flipping a coin, punting a football, and a free throw in basketball are all examples of objects experiencing motion in a vertical plane (Figure 1.58). This type of motion is called projectile motion. A projectile is any object thrown into the air. Projectiles include objects dropped from rest; objects thrown downward or vertically upward, such as a tennis ball for service; and objects moving upward at an angle, such as a punted football. First let’s consider projectile motion of an object moving straight up or down. What is the relationship between an object’s mass and the speed of its fall?
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Do the next QuickLab to find out. x (horizontal) y (vertical) Figure 1.58 A plane has two dimensions, x and y. 1-6 QuickLab 1-6 QuickLab The Bigger They Are... Problem Does mass affect how quickly an object falls? Materials two objects of similar size and shape but different mass, such as a marble and a ball bearing, a die and a sugar cube, a golf ball and a table tennis ball two pans chair Procedure 1 Place a pan on either side of the chair. 2 Standing on the chair, release each pair of objects from the same height at the same time. 3 Listen for the objects hitting the pans. 4 Repeat steps 2 and 3 for other pairs of objects. 5 Repeat steps 2 and 3 from a higher and a lower height. Questions 1. Did the pair of objects land at the same time? 2. How did a change in height affect how long it took the objects to drop? 3. How did a change in the objects’ shape affect how long it took each pair to drop? In the 16th century, Galileo conducted experiments that clearly demonstrated that objects falling near Earth’s surface have a constant acceleration, neglecting air resistance, called the acceleration due to gravity. You can determine the value of the acceleration due to Earth’s gravity by performing the following experiment. info BIT In 1971, astronaut David Scott tested Galileo’s theory with a feather and a hammer. With no air on the Moon, both objects hit the ground at the same time. 54 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 55 1-7 Inquiry Lab 1-7 Inquiry Lab Determining the Magnitude of the Acceleration due to Gravity Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How can position-time and velocity-time graphs be used to determine the acceleration due to gravity? 7 Using a ruler, measure the position of the object at each time interval and record it in your data table. 8 Plot your collected data on a position-time graph. Materials and Equipment 60-Hz spark timer ticker tape carbon disk power supply small mass metre-stick or ruler Procedure masking tape C-clamp retort stand graph paper cushion 9 With a sweeping motion, practise connecting the dots in a smooth curve that best fits the data. 10 Construct a data table in your notebook for recording instantaneous
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velocity and time. 11 Draw three tangents on the position-time graph. 12 Calculate the instantaneous velocities at these points by determining the slopes of the tangents. Record the data in your table. 1 Construct a data table in your notebook for recording 13 Plot a velocity-time graph of your collected data. time and position. 14 Draw a line of best fit. 2 Set up materials as shown in Figure 1.59(a), ensuring that the timer is 1.5 m above the floor. 15 Calculate the acceleration experienced by the object, in m/s2, by finding the slope of the velocity-time graph. recording timer Analysis ticker tape mass cushion 1. Determine the experimental value of the magnitude of acceleration due to gravity by averaging your group’s results. 2. Determine the percent error for your experimental value. Assume the theoretical magnitude of a is 9.81 m/s2. 3. Describe the shape of the position-time graph you drew in step 9. Figure 1.59(a) 4. From your graph, describe the relationship between time and displacement for an accelerating object. 3 Attach a 1.5-m strip of ticker tape to the mass and thread the ticker tape through the spark timer. 4 Turn on the spark timer just before your partner releases the mass. 5 Repeat steps 3 and 4 for each person in your group. 6 Analyze the ticker tape by drawing a line through the first distinct dot on the tape. Label it “start”. (On a 60-Hz timer, every sixth dot represents 0.10 s.) Continue labelling your ticker tape as shown in Figure 1.59(b). t 0.10 s start t 0.20 s t 0.30 s d 1 d 2 Figure 1.59(b) e LAB d 3 For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 55 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 56 e TECH Use graphical analysis to determine acceleration due to Earth’s gravity. Go to www.pearsoned.ca/school/ physicssource. vi 0 a 9.81 m/s2 [down] dy Figure 1.60 The time it takes the golf ball to hit the ground depends on the height from which it drops and
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on the acceleration due to gravity. PHYSICS INSIGHT The equations of parabolas are quadratic equations because they include a power of two, for example, y x2. The equation for the displacement of a vertical projectile is d v 1 t a(t)2. 2 i Gravity Causes Objects to Accelerate Downward Recall the kinematics equations for accelerated motion from section 1.5(t)2 2 1 t a(t)2 2 vf 2 vi 2 2ad You can also apply these equations to motion in a vertical plane. is the acceleration due Because vertical acceleration is due to gravity, a to gravity, or 9.81 m/s2 [down]. If you drop a golf ball from a height of 1.25 m, how long will it take for the ball to reach the ground (Figure 1.60)? Because the ball is moving in only one direction, down, choose down to be positive for simplicity. Since the golf ball is accelerating due to gravity starting from rest, v i 0 and a 9.81 m/s2 [down] 9.81 m/s2 The ball’s displacement can be expressed as 1.25 m [down], or 1.25 m. The equation that includes all the given variables and the unknown variable is d v i t a(t)2. The displacement and acceleration vectors 1 2 are both in the same direction, so use the scalar form of the equation to solve for time. Since vi 0, 1 d at2 2 2d t a 2(1.25 m) m 9.81 s2 0.505 s info BIT Without a parachute, Vesna Vulovic, a flight attendant, survived a fall of 10 160 m when the DC-9 airplane she was travelling in exploded. The golf ball takes 0.505 s to reach the ground when released from a rest height of 1.25 m. Note that the time it takes for an object to fall is directly proportional to the square root of the height it is dropped from: t. If there is 2d a no air resistance, the time it takes for a falling object to reach the ground depends only on the height from which it was dropped. The time does not depend on any other property of the object. 56 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 57 1-8 QuickLab 1-8 QuickLab Could You Be a Goalie for the
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NHL? Problem What is your reaction time? Materials and Equipment long ruler (30 cm or more) flat surface Procedure 1 Rest your arm on a flat surface with your wrist at the edge. 2 Ask your partner to hold the ruler vertically so that the ruler’s end is just above your hand. 3 Curl your fingers so that the space between your thumb and index finger is large enough for the ruler to pass through easily. 4 Without watching your partner, ask your partner to let go of the ruler without warning. 5 Try to close your thumb and index finger on the ruler as quickly as possible. 6 Record where your hand is on the ruler. 7 Repeat steps 1–6 several times. Questions 1. Determine the average distance the ruler falls in each of your trials. 2. Using the average distance, calculate the time. 3. An average NHL goalie has a reaction time of 0.15 s. How does your reaction time compare with your partner’s? 4. Certain drugs impair reaction time. What would you expect your results in this lab to be if your reaction time were increased? Instead of dropping an object such as a golf ball, what if you threw an object down? By throwing an object straight down, you give the object an initial vertical velocity downward. What effect does an initial velocity have on the motion of the object? The next example will show you. Example 1.17 While cliff diving in Mexico, a diver throws a smooth, round rock straight down with an initial speed of 4.00 m/s. If the rock takes 2.50 s to land in the water, how high is the cliff? Given For convenience, choose down to be positive because down is the only direction of the ball’s motion. v 4.00 m/s [down] 4.00 m/s t 2.50 s a 9.81 m/s2 [down] 9.81 m/s2 i Required height of cliff (d) Analysis and Solution The initial velocity and acceleration vectors are both in the same direction, so use the scalar form of the equation d v 1 t a(t)2. 2 i Chapter 1 Graphs and equations describe motion in one dimension. 57 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 58 d (4.00 m/s)(2.50 s) (9.81 m/s2)(2.50 s)2 1 2 10.0 m 30.
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7 m 40.7 m Paraphrase The cliff is 40.7 m high. Practice Problems 1. If a rock takes 0.750 s to hit the ground after being thrown down from a height of 4.80 m, determine the rock’s initial velocity. 2. Having scored a touchdown, a football player spikes the ball in the end zone. If the ball was thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long it is in the air. 3. An elevator moving downward at 4.00 m/s experiences an upward acceleration of 2.00 m/s2 for 1.80 s. What is its velocity at the end of the acceleration and how far has it travelled? Answers 1. 2.72 m/s [down] 2. 0.43 s 3. 0.400 m/s [down], 3.96 m What Goes Up Must Come Down Circus clowns are often accomplished jugglers (Figure 1.61). If a juggler throws a ball upward, giving it an initial velocity, what happens to the ball (Figure 1.62)? vi a dy Figure 1.61 Juggling is an example of projectile motion. Figure 1.62 The ball’s motion is called vertical projectile motion. When you throw an object up, its height (displacement) increases while its velocity decreases. The decrease in velocity occurs because the object experiences acceleration downward due to gravity (Figure 1.63(a)). The ball reaches its maximum height when its vertical velocity equals zero. In other words, it stops for an instant at the top of its path (Figure 1.63(b)). When the object falls back toward the ground, it speeds up because of the acceleration due to gravity (Figure 1.63(c)). 58 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 59 a vi v 0 a vf a Figure 1.63(a) Stage 1: Velocity and acceleration are in opposite directions, so the ball slows down. Figure 1.63(b) Stage 2: The ball has momentarily stopped, but its acceleration is still 9.81 m/s2 [down], which causes the ball to change direction. Figure 1.63(c) Stage 3: Velocity and acceleration are in the same direction, so the ball speeds up. The next two examples analyze different stages of the same object’s motion
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