Physics 30 Released Item January 2001

ii Multiple Choice • Decide which of the choices best completes the statement or answers the question. • Locate that question number on the separate answer sheet provided and fill in the circle that corresponds to your choice. Example This examination is for the subject of A. science B. physics C. biology D. chemistry Answer Sheet A B C D Numerical Response • Record your answer on the answer sheet provided by writing it in the boxes and then filling in the corresponding circles. • If an answer is a value between 0 and 1 (e.g., 0.25), then be sure to record the 0 before the decimal place. • Enter the first digit of your answer in the left-hand box and leave any unused boxes blank. Examples Calculation Question and Solution If a 121 N force is applied to a 77.7 kg mass at rest on a frictionless surface, the acceleration of the mass will be _________ m/s 2 . (Record your three-digit answer in the numerical-response section on the answer sheet.) m F a = 2 m/s 557 1 kg 77.7 N 121 . a = = 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 . . 1 . 5 6 Record 1.56 on the answer sheet Calculation Question and Solution A microwave of wavelength 16 cm has a frequency, expressed in scientific notation, of b × 10 w Hz. The value of b is _______ . (Record your two-digit answer in the numerical-response section on the answer sheet.) λ c f = Hz 10 875 . 1 m 0.16 m/s 10 00 . 3 9 8 × = × = f 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 . . 1 . 9 Record 1.9 on the answer sheet

2 2 Use the following information to answer the next question. At a coal mine, a train engine bumps an empty hopper car that has a mass of 1.00 × 10 4 kg such that it rolls at a constant speed of 2.00 m/s under a coal storage bin. When the hopper car triggers an electromagnetic switch on the track below the storage bin, the bin drops a load of 1.20 × 10 4 kg of coal into the hopper car. Hopper car m = 1.00 × 10 4 kg Cross section of coal storage bin v = 2.00 m/s 3. The speed of the hopper car immediately after receiving the load of coal, expressed in scientific notation, is b × 10 – w m/s. The value of b is __________. (Record your three-digit answer in the numerical-response section on the answer sheet.)

5 5 3. The statements that describe the motion of the charged particle in diagram 1 are A. I and III B. I and IV C. II and III D. II and IV 4. The statements that describe the motion of the charged particle in diagram 2 are A. I and III B. I and IV C. II and III D. II and IV 5. The direction of the uniform magnetic field in diagram 2 is A. toward the top of the page B. toward the bottom of the page C. to the left of the page D. to the right of the page

8 8 Use the following information to answer the next question. Three types of radiation pass through an electric field along the paths shown below. Negatively charged plate Naturally radioactive substance inside shielded box X Z Y Positively charged plate 10. The types of radiation taking paths X , Y , and Z are, respectively, A. beta, alpha, and gamma B. beta, gamma, and alpha C. gamma, alpha, and beta D. alpha, gamma, and beta

11 11 12. The minimum electron speed necessary to ionize xenon atoms is A. 2.66 × 10 31 m/s B. 5.15 × 10 15 m/s C. 4.25 × 10 12 m/s D. 2.06 × 10 6 m/s 13. The electric potential difference across the electrified grids that is required to accelerate a xenon ion from rest to its exit speed is A. 2.93 × 10 –5 V B. 1.26 × 10 –3 V C. 1.26 × 10 3 V D. 4.71 × 10 29 V 14. If all of the xenon propellant could be expelled in a single short burst, the change in the speed of the DS1 capsule after all the fuel has been exhausted would be A. 6.14 m/s B. 7.16 m/s C. 6.14 × 10 3 m/s D. 7.16 × 10 3 m/s 15. The physics principle that best describes the propulsion of the DS1 capsule is the Law of Conservation of A. Charge B. Energy C. Momentum D. Nucleon Number

14 14 18. A copper wire is connected to a battery so that it has a current in it. A segment of the wire is perpendicular to a horizontal 1.5 T magnetic field. The length of the wire in the magnetic field is 3.0 cm, and the mass of the wire affected by the magnetic field is 20 g. In order to suspend the segment of wire, the minimum current in the wire must be A. 0.044 A B. 0.23 A C. 4.4 A D. 44 A Use the following information to answer the next three questions. The AC adapter for a pocket calculator contains a transformer that converts 120 volts into 3.0 volts. The pocket calculator draws 450 mA of current from the transformer. Assume that the transformer is an ideal transformer. 19. If the transformer’s secondary coil has exactly 50 turns, then the number of turns in the primary coil is A. 7 turns B. 40 turns C. 50 turns D. 2.0 × 10 3 turns 20. The current in the primary coil of the adapter is A. 0.80 mA B. 1.3 mA C. 11 mA D. 18 A Numerical Response 0 8. The power supplied by the primary coil is __________ W. (Record your two-digit answer in the numerical-response section on the answer sheet.)

17 17 24. The energy gained by a proton that moves through a potential difference of 1.0 V is A. 1.0 J B. 1.0 eV C. 6.3 × 10 18 J D. 1.6 × 10 –19 eV 11. A metal has a work function of 2.91 × 10 –19 J. Light with a frequency of 8.26 × 10 14 Hz is incident on the metal. The stopping voltage is __________V. (Record your three-digit answer in the numerical-response section on the answer sheet.) 25. If a light with a wavelength of 3.25 × 10 –8 m illuminates a metal surface with a work function of 5.60 × 10 –19 J, the maximum kinetic energy of the emitted photoelectrons is A. 5.60 × 10 –19 J B. 5.56 × 10 –18 J C. 6.12 × 10 –18 J D. 6.68 × 10 –18 J

18 18 Use the following information to answer the next three questions. A graph of data obtained from a photoelectric effect experiment is shown below. Stopping Voltage as a Function of the Frequency of Incident Light on a Cesium Plate Frequency (10 14 Hz) Stopping voltage (V) 0 2 4 6 8 10 12 0 0.5 1.0 1.5 2.0 2.5 3.0 P Point P corresponds to a trial using light at the frequency indicated. 26. The type of light indicated by point P is A. visible B. infrared C. microwave D. ultraviolet 27. The energy of a photon of light indicated by point P is A. 4.1 eV B. 2.3 eV C. 1.7 eV D. 0.0 eV

19 19 28. Photons of light, as indicated by point P , bombard the cesium plate. The maximum kinetic energy of an emitted electron is A. 4.1 eV B. 2.3 eV C. 1.7 eV D. 0.0 eV 29. The Compton experiment was significant in that it demonstrated that photons have A. mass B. momentum C. wave properties D. a speed of 3.00 × 10 8 m/s 30. An experiment starts with 1.45 kg of iodine-131. After 32.2 days, 90.6 g are left. The half-life of iodine-131 is A. 32.2 days B. 16.1 days C. 8.05 days D. 4.04 days

20 20 Use the following information to answer the next four questions. Fusion Research Interest in nuclear fusion is growing because of the amount of energy available from nuclear reactions. A major difficulty in producing a nuclear fusion reaction is that in order for nuclei to fuse, the nuclei must possess a large amount of kinetic energy. Under most circumstances, 0.25 MeV per nucleus is sufficient. At such high energies, the nuclear fuel is called a plasma. The average kinetic energy of a nucleus within a plasma can be found using bT E 2 3 k = where T is the temperature of the plasma, in Kelvin, and b is a physical constant equal to 1.4 × 10 –23 J/K. One method of obtaining the temperatures necessary for fusion is to use a high-intensity laser to heat a small cluster of nuclei. One such laser emits a 1.0 × 10 15 W pulse of ultraviolet radiation that lasts for 1.0 × 10 –12 s. The wavelength of this laser is 280 nm. A Fusion Reaction Equation → + H H 3 1 2 1 X + neutron 31. The missing product, X , in the fusion reaction given above is A. He 5 2 B. He 4 2 C. H 4 1 D. He 3 2

21 21 32. The main reason that the nuclei need to have such large kinetic energies is that A. fusion releases large amounts of energy B. fission must occur before fusion can occur C. this kinetic energy is converted into nuclear energy D. the nuclei must overcome a strong electrostatic repulsion 33. When the average kinetic energy of the nuclei in a plasma is 0.25 MeV, then the temperature is A. 1.9 × 10 9 K B. 2.9 × 10 9 K C. 4.3 × 10 9 K D. 1.2 × 10 28 K 34. The energy of a single photon of the ultraviolet laser is A. 7.1 × 10 –19 J B. 1.0 × 10 –27 J C. 7.1 × 10 –28 J D. 1.9 × 10 –40 J

22 22 35. The absorption spectrum of hydrogen is produced when electrons A. emit radio frequency photons B. emit short wavelength photons C. jump from a higher orbital to a lower orbital D. jump from a lower orbital to a higher orbital 36. An accelerated electron with 8.77 eV of energy strikes a mercury atom and leaves the collision with 2.10 eV of energy. The maximum frequency of light that can be emitted by the mercury atom is A. 1.01 × 10 14 Hz B. 5.07 × 10 14 Hz C. 1.61 × 10 15 Hz D. 2.12 × 10 15 Hz

23 23 37. For a hydrogen atom, the difference in radii between the sixth Bohr orbital and the second Bohr orbital is A. 1.69 × 10 –9 m B. 8.46 × 10 –10 m C. 1.17 × 10 –11 m D. 1.32 × 10 –11 m 12. An electron in a hydrogen atom is in the fourth orbital and jumps down to the second orbital. The energy released is __________ eV. (Record your three-digit answer in the numerical-response section on the answer sheet.)

24 Use the following information to answer the next question. In a physics demonstration, a student inflates a balloon by blowing into it. The end of the balloon is then tied. The balloon is rubbed with fur and develops an electrostatic charge. The balloon is placed against the ceiling and released. It remains “stuck” to the ceiling. The teacher then presents the following challenges to the students: • explain how the balloon received the electrostatic charge • explain why the balloon is attracted to the ceiling • provide a procedure that would determine if the charge on the balloon is positive or negative. Include a list of any additional equipment needed. • provide a procedure that could be used to determine if there is a relationship between the amount of rubbing and the amount of charge developed on an inflated balloon. Include a list of any additional equipment needed. Written Response—15% 1. Using concepts from Physics 30, provide a response to each of the teacher’s challenges. Marks will be awarded for the physics used to solve this problem and for the effective communication of your response.

26 Use the following information to answer the next question. Falling Magnet Experiment Two hollow tubes, one made of glass and the other made of aluminum, are positioned vertically. A student holds identical cylindrical magnets against the outside of the tubes and observe that neither tube attracts a magnet. Based on this observation, the student predicts that each magnet will fall through its respective tube with an acceleration of 9.81 m/s 2 . The student and his lab partner then drop the magnets into the tubes from rest at position A, as shown below. Position A Position B Glass tube (electrical insulator) Aluminum tube (electrica conductor Magnet Ma The students make the following observations: The magnets do not touch the sides of the tubes as they fall. The time for the magnet to fall through the aluminum tube is much greater than is the time for the identical magnet to fall through the glass tube. Glass Tube Aluminum Tube Mass of magnet (kg) 0.150 0.150 Tube length (m) 0.95 0.95 Time for magnet to fall from position A to position B (s) 0.44 0.76

27 Written Response—15% 2. Analyze the students’ observations from the falling magnet experiment by • completing the chart below. Include calculations to support the values you write in the chart • explaining the results of this experiment in terms of Lenz’s Law Clearly communicate your understanding of the physics principles that you are using to solve this question. You may communicate this understanding mathematically, graphically, and/or with written statements. Glass Tube Aluminum Tube Potential Energy of the magnet at position A (J) Acceleration of the magnet through the tube (m/s 2 ) Kinetic Energy of the magnet at position B (J) Mechanical Energy of the magnet at position A (J) Mechanical Energy of the magnet at position B (J) Resisting Force on the magnet (N)

PHYSICS DATA SHEET CONSTANTS Acceleration Due to Gravity or Gravitational Field Near Earth ............ a g or g = 9.81 m/s 2 or 9.81 N/kg Gravitational Constant ........................ G = 6.67 × 10 –11 N . m 2 /kg 2 Mass of Earth ...................................... M e = 5.98 × 10 24 kg Radius of Earth ................................... R e = 6.37 × 10 6 m Coulomb’s Law Constant .................... k = 8.99 × 10 9 N . m 2 /C 2 Electron Volt ....................................... 1 eV = 1.60 × 10 –19 J Elementary Charge .............................. e = 1.60 × 10 –19 C Index of Refraction of Air ................... n = 1.00 Speed of Light in Vacuum .................. c = 3.00 × 10 8 m/s Energy of an Electron in the 1st Bohr Orbit of Hydrogen ..................... E 1 = –2.18 × 10 –18 J or –13.6 eV Planck’s Constant ............................... h = 6.63 × 10 –34 J . s or 4.14 × 10 –15 eV . s Radius of 1st Bohr Orbit of Hydrogen r 1 = 5.29 × 10 –11 m Rydberg’s Constant for Hydrogen ...... R H = 1.10 × 10 7 m 1 Rest Mass Charge Alpha Particle ............... kg 10 65 . 6 27 – × = α m α 2 + Electron ........................ m e = 9.11 × 10 –31 kg e – Neutron ......................... m n = 1.67 × 10 –27 kg n 0 Proton ........................... m p = 1.67 × 10 –27 kg p + hypotenuse opposite = sin θ hypotenuse adjacent = cos θ adjacent opposite = tan θ C c B b A a sin sin sin = = C ab b a c cos 2 – 2 2 2 + = For any Vector R 2 2 y x R R R + = x y R R = θ tan θ cos R R x = θ sin R R y = x : [ x min , x max , x scl ] y : [ y min , y max , y scl ] Exponential Prefix Symbol Value pico ............. p .................. 10 –12 nano ............ n .................. 10 –9 micro ........... µ .................. 10 –6 milli ............. m ................. 10 –3 centi ............ c .................. 10 –2 deci ............. d .................. 10 –1 Exponential Prefix Symbol Value tera ............. T ................. 10 12 giga ............. G ................. 10 9 mega ........... M ................ 10 6 kilo ............. k ................. 10 3 hecto ........... h ................. 10 2 deka ............ da ................ 10 1

EQUATIONS t d v G G = ave t v v a i f – G G G = 2 i 2 1 t a t v d G G G + = T r v π 2 = 2 f 2 1 t a – t v d G G G = t v v d         + = 2 i f G G G ad v v 2 2 i 2 f + = r v a 2 c = a m F G G = v m t F G G ∆ = ∆ g m F G G = g N f F F µ = x k F G G – s = 2 2 1 g r m Gm F = 2 1 r Gm g = r mv F 2 c = 2 2 c 4 T mr F π = v m p G G = Fd W = θ cos Fd E W = ∆ = t E t W P ∆ = = 2 k 2 1 mv E = mgh E = p 2 p 2 1 kx E = k m T π 2 = g l T π 2 = f T 1 = λ f v = l ; l = λ = λ 4 2 1 1 1 2 2 1 2 1 2 1 sin sin n n v v = = = λ λ θ θ nl xd = λ n d θ λ sin = 0 i 0 i – d d h h m = = i 0 1 1 1 d d f + = W E hf + = max k W hf = 0 stop max k qV E = λ hc hf E = =         = 2 i 2 f H 1 – 1 1 n n R λ 1 2 1 E n E n = 1 2 r n r n = n N N       = 2 1 0 2 mc E = λ h p = pc E c hf p = = ; 2 2 1 e r q kq F = 2 1 r kq E = G q F E e G G = d V E = G q E V ∆ = 3 2 1 R R R R + + = 3 2 1 1 1 1 1 R R R R + + = max eff 707 . 0 I I = IR V = IV P = t q I = ⊥ = IlB F m ⊥ = qvB F m ⊥ = lvB V p s s p s p I I V V N N = = max eff 707 . 0 V V = Tear-out Page Fold and tear along perforation.

Tear-out Page Periodic Table of the Elements Fold and tear along perforation. 7 VIIB 6 VIB 5 VB 4 IVB 3 IIIB 2 IIA 1 IA 9 VIIIB 8 57 138.91 La Ac 89 (277.03) lanthanum 58 140.12 Ce Th 90 (232.04) thorium cerium 59 140.91 Pr 91 (231.04) Pa protactinium 60 144.24 Nd 92 238.03 U uranium neodymium 61 (144.91) Pm 93 (237.05) Np neptunium promethium 62 150.35 Sm 94 (244.06 ) Pu plutonium samarium praseodymium actinium cesium rubidium potassium sodium lithium hydrogen 1 1.01 H 3 6.94 Li 11 22.99 Na 19 39.10 K 37 85.47 Rb 55 132.91 Cs Fr barium strontium calcium magnesium beryllium yttrium scandium hafnium zirconium titanium 22 47.90 Ti 40 91.22 Zr 72 178.49 Hf 104 (266.11) Unq tantalum niobium vanadium 23 50.94 V 41 92.91 Nb 73 180.95 Ta 105 (262.11) Unp tungsten molybdenum chromium 24 52.00 Cr 42 95.94 Mo 74 183.85 W 106 (263.12) Unh rhenium technetium manganese 25 54.94 Mn 43 (98.91) Tc 75 186.21 Re 107 (262.12) Uns osmium ruthenium iron 26 55.85 Fe 44 101.07 Ru 76 190.20 Os 108 (265) Uno iridium rhodium cobalt 27 58.93 Co 45 102.91 Rh 77 192.22 I r 109 (266) Une francium radium 4 9.01 Be 12 24.31 Mg 20 40.08 Ca 38 87.62 Sr 56 137.33 Ba 88 (226.03) Ra 21 44.96 Sc 39 88.91 Y 57-71 89-103 87 (223.02) unnilquadium unnilpentium unnilhexium unnilseptium unniloctium unnilennium americium europium platinum palladium nickel 28 58.71 Ni 46 106.40 Pd 78 195.09 Pt 64 157.25 Gd 96 (247.07) Cm curium gadolinium gold silver copper 29 63.55 Cu 47 107.87 Ag 79 196.97 Au 65 158.93 Tb 97 (247.07) Bk berkelium mercury cadmium zinc 30 65.38 Zn 48 112.41 Cd 80 200.59 Hg 66 162.50 Dy 98 (242.06) Cf californium thallium indium gallium aluminum 5 10.81 B 13 26.98 Al 31 69.72 Ga 49 114.82 I n 81 204.37 Tl 67 164.93 Ho 99 (252.08) Es einsteinium holmium lead tin germanium 6 12.01 C Si 32 72.59 Ge 50 118.69 Sn 82 207.19 Pb 68 167.26 Er 100 (257.10) Fm fermium erbium bismuth antimony arsenic phosphorus nitrogen 7 14.01 N 15 30.97 P 33 74.92 As 51 121.75 Sb 83 208.98 Bi 69 168.93 Tm 101 (258.10) Md thulium polonium tellurium selenium sulphur oxygen 8 16.00 O 16 32.06 S 34 78.96 Se 52 127.60 Te 84 (208.98) Po 70 173.04 Yb 102 (259.10) No nobelium ytterbium astatine iodine bromine chlorine fluorine 9 19.00 F 17 35.45 Cl 35 79.90 Br 53 126.90 I 85 (209.98) At 71 174.97 Lu 103 (260.11) Lr lawrencium lutetium 2 4.00 He 10 20.17 Ne 18 39.95 Ar 36 83.80 Kr Xe 86 (222.02) Rn 63 151.96 Eu 95 (243.06) Am 18 VIIIA or O 17 VIIA 16 VIA 15 VA 14 IVA 13 IIIA 12 IIB 11 IB carbon boron 54 131.30 argon helium xenon radon neon krypton 10 VIIIB mendelevium terbium dysprosium silicon 14 28.09 lithium Name 3 6.94 Li Symbol Atomic number Atomic molar mass Key Based on 12 6 C ( ) Indicates mass of the most stable isotope

Physics 30 January 2001 Diploma Examination Multiple Choice and Numerical Response Keys 1. B 20. C 2. B 21. C 3. C 22. A 4. B 23. B 5. B 24. B 6. D 25. B 7. A 26. D 8. B 27. A 9. C 28. B 10. B 29. B 11. B 30. C 12. D 31. B 13. C 32. D 14. D 33. A 15. C 34. A 16. C 35. D 17. C 36. C 18. C 37. A 19. D 1.08 2.63 9.09 3857 2.92 2.55 3.87 1.4 2.08 1.21 1.60 2.55 or 2.56

i Holistic Scoring Guide Draft Major Concepts Charging by contact, induced charge separation, experimental design Score Criteria 5 Excellent • The student provides a complete solution covering the full scope of the question. – The reader has no difficulty following the strategy or solution presented by the student. – Statements made in the response are supported explicitly but may contain minor errors or have minor omissions. 4 Good • The student provides a solution to the significant parts of the question. – The reader may have some difficulty following the strategy or solution presented by the student. – Statements made in the response are supported, but the support may be implicit. 3 Satisfactory • The student provides a solution in which he/she has made significant progress toward answering the question. – The reader has difficulty following the strategy or solution presented by the student. – Statements made in the response may be open to interpretation and may lack support. 2 Limited • The student provides a solution in which he/she has made some progress toward answering the question. − Statements made in the response lack support. 1 Poor • The student provides a solution that begins to answer the question. 0 Insufficient • The student provides a solution that is invalid for the major concepts addressed by the question. NR No response is given.

ii Use the following information to answer the next question. In a physics demonstration, a student inflates a balloon by blowing into it. The end of the balloon is then tied. The balloon is rubbed with fur and develops an electrostatic charge. The balloon is placed against the ceiling and released. It remains “stuck” to the ceiling. The teacher then presents the following challenges to the students: • explain how the balloon received the electrostatic charge • explain why the balloon is attracted to the ceiling • provide a procedure that would determine if the charge on the balloon is positive or negative. Include a list of any additional equipment needed. • provide a procedure that could be used to determine if there is a relationship between the amount of rubbing and the amount of charge developed on an inflated balloon. Include a list of any additional equipment needed. Written Response—15% 1. Using concepts from Physics 30, provide a response to each of the teacher’s challenges. Marks will be awarded for the physics used to solve this problem and for the effective communication of your response.

iii A complete response should include the following content. The clarity of the response is considered in assigning a mark. Expected Content: This question requires both a theoretical and an empirical understanding of electrostatics. Theoretical To completely address the theoretical aspect of the question, the student needs to address the movement of charges. The response should include the following: • A statement that the balloon receives its charge by friction or by contact. • A statement that electrons move between the balloon and the fur. (Students do not need to specify that the balloon will become negatively charged, or identify the charge that remains on the fur after contact.) • A description of the charge separation that occurs on the ceiling. • An explanation of the process of inducing a charge. • A statement that opposite charges attract. Empirical Two procedures are required to completely address the empirical aspect of the question. • A procedure is required to determine the type of charge on the balloon. The response must include instructions and equipment needed. The procedure must result in the determination of the nature of the charge. • A procedure is required to determine if there is a relationship between the amount of rubbing and the amount of charge developed on an inflated balloon. The response must include instructions and equipment needed. Instructions must specify repeated trials, and identify both the manipulated and responding variables. (Examples of the manipulated variable include: time of rubbing, number of strokes, or amount of pressure applied. Examples of the responding variable include: the time that the balloon remains stuck to the ceiling, divergence of the leaves of an electroscope, or force of attraction of the balloon to the ceiling.)

iv Scoring Guide for Anaholistic Questions Major Concepts: Potential energy; Acceleration and kinetic energy; Resisting force; Lenz’s Law Score Criteria 5 In the response, the student • uses an appropriate method that reflects an excellent understanding of all major concepts • provides a complete description of the method used and shows a complete solution for the problem • states formulas explicitly • may make a minor error, omission, or inconsistency; however, this does not hinder the understanding of the physics content • draws diagrams that are appropriate, correct, and complete • may have an error in significant digits or rounding 4 In the response, the student • uses an appropriate method that reflects a good understanding of all major concepts or that reflects an excellent understanding of three of the major concepts • provides explanations that are correct and detailed • states most formulas explicitly and applies them correctly • makes minor errors, omissions, or inconsistencies in calculations and/or substitutions; however, these do not hinder the understanding of the physics content • draws most diagrams appropriately, correctly, and completely • may have errors in units, significant digits, rounding, or graphing 3 In the response, the student • uses an appropriate method that reflects a basic understanding of all four of the major concepts or that reflects a good understanding of three of the major concepts • uses an appropriate method that reflects an excellent understanding of two of the major concepts and that reflects a basic understanding of one of the two remaining concepts • uses formulas and/or diagrams that may be implicit, and these are applied correctly; however, errors in calculations and/or substitutions that hinder the understanding of the physics content are present • provides explanations that are correct but lack detail • has a major omission or inconsistency 2 In the response, the student • uses an appropriate method that reflects a basic understanding of three of the four major concepts or that reflects a good understanding of two of the major concepts • gives formulas and/or diagrams that are implicitly correct; however, they are not applied to determine the final solution or errors in the application of equations are present, and the answer is consistent with calculated results 1 In the response, the student • attempts at least two of the major concepts or uses an appropriate method that reflects a good understanding of one of the major concepts • makes errors in the formulas, diagrams, and/or explanations, and the answer is not consistent with calculated results 0 In the response, the student • identifies an area of physics that does not apply to the major concepts • uses inappropriate formulas, diagrams, and/or explanations NR No response is given.

v Use the following information to answer Written Response 2. Falling Magnet Experiment Two hollow tubes, one made of glass and the other made of aluminum, are positioned vertically. A student holds identical cylindrical magnets against the outside of the tubes and observe that neither tube attracts a magnet. Based on this observation, the student predicts that each magnet will fall through its respective tube with an acceleration of 9.81 m/s 2 . The student and his lab partner then drop the magnets into the tubes from rest at position A, as shown below. Position A Position B Glass tube (electrical insulator) Aluminum tube (electrica conductor Magnet Ma The students make the following observations: The magnets do not touch the sides of the tubes as they fall. The time for the magnet to fall through the aluminum tube is much greater than is the time for the identical magnet to fall through the glass tube. Glass Tube Aluminum Tube Mass of magnet (kg) 0.150 0.150 Tube length (m) 0.95 0.95 Time for magnet to fall from position A to position B (s) 0.44 0.76

vi Written Response—15% 2. Analyze the students’ observations from the falling magnet experiment by • completing the chart below. Include calculations to support the values you write in the chart • explaining the results of this experiment in terms of Lenz’s Law Clearly communicate your understanding of the physics principles that you are using to solve this question. You may communicate this understanding mathematically, graphically, and/or with written statements. Sample Solution: Glass Tube Aluminum Tube Potential Energy of the magnet at position A (J) 1.4 1.4 Acceleration of the magnet through the tube (m/s 2 ) 9.8 3.3 Kinetic Energy of the magnet at position B (J) 1.4 0.47 Mechanical Energy of the magnet at position A (J) 1.4 1.4 Mechanical Energy of the magnet at position B (J) 1.4 0.47 Resisting Force on the magnet (N) 0 0.98

vii • Potential Energy E p A = mgh = (0.150 kg)(9.81 m/s 2 )(0.95 m) E p A = 1.398 J • Acceleration: from 2 2 1 at t v d i + = a = 2 2 d t a glass = 2 ) s 44 . 0 ( ) m 95 . 0 ( 2 a aluminum = 2 ) s 76 . 0 ( ) m 95 . 0 ( 2 a glass = 9.814 m/s 2 a aluminum = 3.289 m/s 2 • Kinetic Energy v f 2 = v i 2 + 2 ad v f 2 = 2 ad since v i = 0 E k = 2 f 2 1 mv E k = ) 2 ( net 2 1 d a m = ma net d E glass = ma net d E aluminum = ma net d = ) m 95 . 0 )( s m 8 . 9 )( kg 150 . 0 ( 2 / = ) m 95 . 0 )( s m 3 . 3 )( kg 150 . 0 ( 2 / E glass = 1.397 J E aluminum = 0.470 J Students can calculate v f from t v v d 2 f i + = . Since v i = 0, t d v 2 f = . As a result: v f ( g l a s s ) = 4.32 m/s, and v f ( a l u m i n u m ) = 2.50 m/s

viii • Resisting Force Method 1 ∆ E = Work done = Fd F glass = ∆ E d F aluminum = ∆ E d = m 95 . 0 J 4 . 1 J 4 . 1 − = m 95 . 0 J 47 . 0 J 4 . 1 − F glass = 0 N F aluminum = 0.979 N Method 2 F net = ma F g – F A = ma (0.150 kg)(9.81 m/s 2 ) – F A = (0.150 kg)(3.29 m/s 2 ) F A = 0.98 N • Lenz’s Law Explanation By Lenz’s Law, a changing magnetic field will cause an induced magnetic field in the aluminum tube that opposes the original changing magnetic field. Since the glass tube is an electrical insulator, there is no induced magnetic field.