Ruse_2022_Physics_sol

3 4 Two balls are released simultaneously from the same vertical position on two tracks which have the same start and end height. Track A is mostly straight, while Track B first curves down and then back upwards by the same amount, as shown below. Rolling resistance can be assumed to be negligible. Which ball reaches the end of the track first? A. The balls on Track A and Track B reach the end at the same time. B. The ball on Track A arrives first. C. The ball on Track B arrives first. D. It is not possible to determine the answer without knowing the detailed shape of each track. 5 When monochromatic light, above a threshold frequency, is incident on a metallic surface, electrons are emitted from the surface. Which changes will result in electrons of greater kinetic energy being emitted from the surface? Frequency of incident light Intensity of incident light Work function of metal A. decrease increase decrease B. increase increase decrease C. decrease decrease increase D. increase decrease increase

6 10 Calcite is a crystal which produces two images of an object when viewed from above. This phenomenon (double refraction) occurs because calcite has a different refractive index for different polarisations of light. No calcite With calcite Double refraction in calcite played an important role in the debate between Newton and Huygens over the nature of light. Newton explained it by hypothesizing that light particles had “sides” and the two types of light particles were separated by the crystal. Huygens, who assumed light was a wave like sound, was unable to explain this phenomenon with his model. Newton could also explain colour in his model by proposing that different colours corresonded to different sized partices, whereas Huygen’s original model had no explanation for colour. How did the wave model of light change from Huygens’ original model so that it could successfully explain the phenomena of polarisation and colour? Type of wave in new model Origin of colour in new model A. Longitudinal Wavelength B. Transverse Wavelength C. Longitudinal Amplitude D. Transverse Amplitude

9 14 An ideal transformer is used to convert 240 V AC to 12 V AC. A small electric motor draws a current of 2.0 A from the secondary coil of the transformer. Which row of the table shows the correct number of turns and current in the primary coil if the secondary coil has 50 turns? Number of turns Current (A) A. 500 0.1 B. 1000 0 .1 C. 500 0.2 D. 1000 0.2 15 A very long straight wire carries a DC current 𝐼𝐼 . A rectangular metallic wire loop moves with velocity 𝑣𝑣 in the same plane of the wire, as shown in the diagram below. (I) (II) (III) In which cases will the coil experience an induced current? A. I and II B. I and III C. II and III D. I, II and III 𝑣𝑣 𝐼𝐼 𝑣𝑣 𝐼𝐼 𝑣𝑣 𝐼𝐼 Top View

12 19 The arrangement shown consists of 3 parallel current-carrying wires, A, B and C, placed perpendicular to the plane of the paper and equidistant apart. The conductors are 𝑎𝑎 meters apart and carry equal currents, 𝐼𝐼 . Which row of the table shows the correct magnitude and direction of the net force per unit length acting on wire A? magnitude direction A. 𝜇𝜇 0 2 𝜋𝜋𝑎𝑎 𝐼𝐼 2 upwards B. 𝜇𝜇 0 2 𝜋𝜋𝑎𝑎 √ 3 𝐼𝐼 2 upwards C. 𝜇𝜇 0 2 𝜋𝜋𝑎𝑎 𝐼𝐼 2 downwards D. 𝜇𝜇 0 2 𝜋𝜋𝑎𝑎 √ 3 𝐼𝐼 2 downwards 𝑎𝑎 𝑎𝑎 𝑎𝑎 60° 60° B A C

15 Physics Section II Answer Booklet 80 marks Attempt Questions 21 – 36 Allow about 2 hours and 25 minutes for this part Instructions • Write your Student Number at the top of pages 15 and 39 • Answer the questions in the spaces provided. These spaces provide guidance for the expected length of response. • Show all relevant working in questions involving calculations • Extra writing space is provided at the back of this booklet. If you use this space, clearly indicate which questions you are answering. JAMES RUSE AGRICULTURAL HIGH SCHOOL 2022

18 Question 23 (4 Marks) Two devices that convert energy to other forms of energy are the DC generator and the AC generator. DC generator AC generator (a) Describe the energy transformations that occur in the AC generator. 1 Conversion of mechanical energy to electrical energy Comment: About 5% of candidature believed it was the other way around. (b) By identifying the component labelled A, describe how the DC generator 3 is able to perform its function. Criteria Mark Identifies the appropriate part and provides a good description of how the DC generator performs its function (Part A, its function and mentions fluctuating, unidirectional current/voltage (camel humps, external circuit) or shows graphically) 3 Identifies the appropriate part and provides a moderate description of how the DC generator performs its function (omits fluctuating unidirectional current/voltage) 2 Provides some relevant information 1 DC generator: By rotating the coil in an external magnetic field, the coil experiences a change in flux and by Faraday’s law an emf is induced in the coil. The coil being part of a complete circuit, a current is induced in the internal circuit (from x to y), in accordance with Lenz’s law. Half a turn later, the current should still be flowing from x to y which is in contravention of Lenz’s law. With the force down here, the field from N to S, the induced current has to be from y to x to preserve Lenz’s law. Thus part A, the split-ring commutator ensures for every half a turn the current in the internal circuit reverses its direction, while the brushes maintains its polarity all the time, producing unidirectional A

21 Question 25 (5 marks) Consider the following Hertzsprung-Russell diagram with stars A, B, C and D. (a) Complete the following table: 4 Marking criteria Marks Correctly filled table with each entry distinct 4 1-2 errors/omissions with each filled entry distinct 3 3-4 errors 2 5-6 errors 1 Label Evolutionary stage of star Main process(es) occurring inside star A Main Sequence Hydrogen fusion in core (p-p chain) ( CNO is minor ) B Red giant H fusion in shell ( CNO cycle ) Helium fusion in core (Triple-alpha) C Red supergiant Fusion of elements up to iron in a series of shells D White dwarf None (b) Annotate the diagram to show the expected evolutionary sequence for our Sun. 1 Marking criteria Marks Correct sequence (A  B  D, not going too close to C) 1

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23 Question 26 (3 marks) A conducting bar with mass 𝑚𝑚 , length 𝐿𝐿 , is at rest on two horizontal conducting rails. 3 A uniform magnetic field 𝐵𝐵 perpendicular to the rails fills the region between it. When the rails are connected to a DC power source which maintains the current at a constant 𝐼𝐼 , the bar experiences a force 𝐹𝐹 due to the motor effect. Ignore the friction between the rails and the bar. Show that the speed attained by the bar is 𝑣𝑣 = � 2𝐵𝐵 𝐿𝐿 𝐼𝐼 𝑑𝑑 𝑚𝑚 , where 𝑑𝑑 is the horizontal distance travelled by the bar. Criteria Mark Correctly derives the relationship (Uses 𝐹𝐹 = 𝐵𝐵𝐼𝐼𝐿𝐿 , 𝑊𝑊 = ∆𝐾𝐾 = 𝐹𝐹 . 𝑑𝑑 , 𝑖𝑖𝑛𝑛𝑖𝑖𝑙𝑙𝑖𝑖𝑑𝑑𝑖𝑖𝑛𝑛𝑖𝑖 𝑖𝑖 = 0 𝑜𝑜𝑜𝑜 𝐾𝐾 𝑖𝑖 = 0) 3 Shows some correct steps or reasoning (uses 2 of the above) 2 Provides some relevant information (uses one of the above) 1 By the motor effect, the bar experiences a force given by 𝐹𝐹 = 𝐵𝐵𝐼𝐼𝐿𝐿 . But the work done on the bar is given by 𝑊𝑊 = ∆𝐾𝐾 . i.e. 𝐹𝐹 . 𝑑𝑑 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 With 𝑖𝑖 = 0, 𝑣𝑣 = � 2𝐹𝐹𝑑𝑑 𝑚𝑚 = � 2𝐵𝐵𝐼𝐼𝐿𝐿𝑑𝑑 𝑚𝑚 by sub 𝐹𝐹 Comments: This was a show question and many students omitted vital steps in their derivation. Please note that work done is not equal to the kinetic energy but to the change in kinetic energy i.e. 𝑊𝑊 = ∆𝐾𝐾 . bar X X X X X X X X X X X X X X DC Power source 𝐼𝐼 𝐵𝐵 Rails 𝐿𝐿

24 Question 27 (5 Marks) A beam of singly charged positive ions in vacuum enters a region of uniform magnetic field strength 𝐵𝐵 , where it undergoes anticlockwise deflection in a semicircular arc as shown in the diagram below. The ions travelling with speed 1.20 × 10 5 𝑚𝑚 𝑠𝑠 −1 , are detected by a fixed detector when the diameter of the arc is 10.8 𝑚𝑚 . (a) If the ions have a mass of 20 𝑖𝑖 , calculate the strength of the magnetic field. 2 Criteria Marks Correctly calculates the strength of the magnetic field (including Tesla as the unit for B, Wb not accepted) 2 Provides some relevant information 1 𝑚𝑚𝑣𝑣 2 𝑜𝑜 = 𝑞𝑞𝑣𝑣𝐵𝐵 𝐵𝐵 = 𝑚𝑚𝑣𝑣 𝑞𝑞𝑞𝑞 = 20 × 1 . 661 × 10 −27 × 1 . 20 × 10 5 1 . 602 × 10 −19 × 5 . 4 = 4.6 𝑚𝑚𝑚𝑚 Comments: Generally well done except for a few students who interpretated “singly charged” as being 1 coulomb of charge. It is prudent to reflect on the reasonableness of your answer and this will prompt you to double check your calculation, and correct if necessary. detector Beam of positive ions Uniform magnetic field 10.8 𝑚𝑚

25 (b) Ions with heavier mass and same charge are also present in the beam. 1 Sketch qualitatively the path of these ions in the same region of magnetic field. (c) Explain the adjustment that needs to be made to the magnetic field to enable the 2 heavier ions with the same speed to reach the detector. Criteria Marks Provides the correct adjustment and a relevant explanation. (makes reference to 𝐵𝐵 = 𝑚𝑚𝑣𝑣 𝑞𝑞𝑞𝑞 or to the centripetal force ( 𝑞𝑞𝑣𝑣𝐵𝐵 ) , and should provide direction) 2 Provides the correct adjustment with a superficial explanation 1 For the larger ions to be detected (i.e. 𝑜𝑜 = 5.4 𝑚𝑚 ) , the centripetal force would need to be increased. This is achieved by increasing the strength of the magnetic field, from 𝑚𝑚𝑣𝑣 2 𝑞𝑞 = 𝑞𝑞𝑣𝑣𝐵𝐵 , while retaining its direction. Comments: An easy question to explain using an equation which many students could easily do. The equation helps to verify the relationship when some variables are kept constant. detector Beam of positive ions Uniform magnetic 10.8 𝑚𝑚

26 Question 28 (4 marks) An experiment was carried out to model the Millikan oil-drop experiment. An oil drop was selected for viewing through the microscope and the size of the applied voltage altered until the oil drop was stationary. The forces on the drop at this point can be represented in the following free-body diagram: (a) The top plate was 5085 V higher than the bottom plate, with a separation of 16 mm . 3 The mass of the oil drop was 8.8 × 10 −11 g . Calculate the number of excess electrons on this oil drop. Marking criteria Marks Calculates number of excess electrons 3 As above with 1 error 2 As above with 2 errors 1

27 F net = 0 = weight – electric = mg – qE qE = mg = 8.8 x 10 -14 x 9.8 = 8.624 x 10 -13 N E = V/d = 5085/0.016 = 3.18 x 10 5 V m -1 or N C -1 q = 8.624 x 10 -13 / 3.178 x 10 5 = 2.714 x 10 -18 C Charge on electron = -1.602 x 10 -19 Number of electrons = 2.679 x 10 -18 / 1.602 x 10 -19 = 16.96 (approximately 17 excess as the electrostatic force on the charged particle is towards the positive plate) Comment: most students were able to make some progress on solving this. A number seemed to be thrown by the word excess. An oil drop has a large number of electrons, due to the fact that it is made from carbon, hydrogen and oxygen atoms (which contain electrons). These are balanced by the nuclear charge, and do not help the drop to experience a net force in an electric field. However, the excess electrons (those above and beyond the ones which were originally there) can contribute to a net force in an electric field. (b) The mass of the drop was estimated with the voltage source off, which resulted in 1 the oil drops falling downwards at constant speed. Draw a free body diagram for an oil drop next to the free body diagram shown earlier, using the same scale. Marking criteria Marks Free body diagram drawn to scale for terminal velocity 1 Comment: use a ruler.

28 Question 29 (4 marks) A student performed a photoelectric experiment in which he exposed an unidentified metal 4 to various frequencies of incident electromagnetic radiation, 𝑓𝑓 , and measured the stopping voltage, 𝑉𝑉 𝑠𝑠 , of the emitted photoelectrons. The results are shown in the table below. 𝑓𝑓 (× 10 15 Hz) 1.3 1.4 1.5 1.6 1.7 𝑉𝑉 𝑠𝑠 (V) 0.6 1.1 1.5 1.9 2.4 By analysing the data graphically, and using the work functions of some metals given in the table below, identify the unknown metal used by the student. Support your answer with relevant calculations. Metal Sodium Cobalt Lead Zinc Iron Silver Platinum Work function (eV) 2.28 3.90 4.14 4.31 4.50 4.79 6.35 Criteria Marks Determines the correct work function using the line of best fit (1 for plotting, 1 for extrapolating, 1 for relevant equation, 1 for correct work function) 4 Any three of the above 3 Correctly plots some points or Correctly draws a line of best fit, (With incorrect working for the work function) 2 Provides some relevant information on the graph, Or provides some relevant calculation/information 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 𝑓𝑓 (× 10 15 Hz) 𝑉𝑉 𝑠𝑠 (V) 2.5 2.0 1.5 1.0 0.5 0

29 𝜙𝜙 = ℎ𝑓𝑓 0 = 6.626 × 10 −34 × 1.17 × 10 15 = 7.75 × 10 −19 𝐽𝐽 = 4.83 𝑒𝑒𝑉𝑉 Hence the unknown metal is Silver Comments: Generally well done, with the majority of the cohort gaining maximum marks. Please consider the number of marks and the space provided when deciding on a line of action to follow. Many students, although correct, used the equation connecting the stopping voltage to the frequency ( 𝑉𝑉 𝑠𝑠 = ℎ 𝑒𝑒 𝑓𝑓 − ∅ 𝑒𝑒 ) and substituted a point in the equation to obtain the work function. This really took up a lot of space and definitely a lot of time!

30 Question 30 (4 marks) The ideal transformer is a device capable of changing the voltage and current characteristics of an AC power supply without dissipating any energy according to the following model. 𝑉𝑉 𝑝𝑝 𝑉𝑉 𝑠𝑠 = 𝑁𝑁 𝑝𝑝 𝑁𝑁 𝑠𝑠 𝑉𝑉 𝑝𝑝 𝐼𝐼 𝑝𝑝 = 𝑉𝑉 𝑠𝑠 𝐼𝐼 𝑠𝑠 (a) Identify two limitations of the ideal transformer model when applied to real 2 transformers. Criteria mark identifies any two limitations of the ideal transformer model 2 Any one of the above 1 Any Two of the following limitations: • Incomplete flux linkage resulting in less flux from the primary being transferred to the secondary • Production of eddy currents in the core results in power losses in the core • Magnetic hysteresis in core resulting in loss of flux • Resistive heating effects in the conducting wires (b) Outline qualitatively two strategies used to improve transformer efficiency. 2 Criteria mark outlines any two strategies to increase efficiency 2 Any one of the above 1 Any two of the following strategies:  Increase flux linkage by tightly wrapping coils around core and in multiple orientations  Use of soft iron core made of ferrite material to reduce magnetic hysteresis  Laminate the core to reduce eddy current formation, thereby reducing power losses  Use of thicker resistance wires in core to reduce resistive heating effects Comments: Although generally well done, students believed that energy and power are synonymous. Furthermore loss of flux and loss of power seemed to be synonymous as well. Also, the pictorial model shown above is a single transformer and not two transformers! Each loop shown, depicts the primary and secondary coils.

31 Question 31 (8 marks) A coracle is a very small, light boat that can be used for fishing. It floats high so that it barely disturbs the water. Figure A (left): An example of a coracle. Figure B (right): Trajectory of the fish. A child of mass 30.0 kg sits in a coracle of mass 10.0 kg on the water at rest. The child throws a fish of mass 5.0 kg onto the riverbank. The fish leaves their hand with at a speed of 5.0 ms −1 at an angle of 10° above the horizontal. (a) Determine the initial vertical component of the velocity of the fish. 2 Criteria Marks Calculates the correct value for the initial vertical velocity, including units 2 Provides some relevant information. 1 𝑖𝑖 𝑦𝑦 = 𝑖𝑖 𝑠𝑠𝑖𝑖𝑛𝑛 θ = 5 𝑚𝑚𝑠𝑠 −1 × 𝑠𝑠𝑖𝑖𝑛𝑛 10 = 0.87 𝑚𝑚𝑠𝑠 −1 (b) The fish lands a horizontal distance of 1.2 m from the child’s hand. 3 Determine the vertical displacement, 𝑦𝑦 𝑓𝑓 , of the fish from the point it left the child’s hand to where it lands. Criteria Marks Calculates the correct vertical displacement, including sign. 3 Applies a correct process to calculate the vertical displacement, but makes a numerical error. 2 Provides some relevant information. 1 Use the range to find the time of flight via 𝑥𝑥 = 𝑖𝑖 𝑖𝑖𝑜𝑜𝑠𝑠 θ 𝑡𝑡 Rearranging: 𝑡𝑡 = 𝑥𝑥 𝑢𝑢 𝑐𝑐𝑜𝑜𝑠𝑠 θ = 1 . 2𝑚𝑚 5 . 0𝑚𝑚𝑠𝑠 −1 𝑐𝑐𝑜𝑜𝑠𝑠 10 = 0.244 𝑠𝑠 The change in vertical displacement is given by 𝑦𝑦 𝑓𝑓 = 𝑖𝑖 𝑦𝑦 𝑡𝑡 + 1 2 𝑎𝑎𝑡𝑡 2 = 0.87 𝑚𝑚𝑠𝑠 −1 × 0.244 𝑠𝑠 + 1 2 ( − 9.8 𝑚𝑚𝑠𝑠 −2 ) × (0.244 𝑠𝑠 ) 2 = − 7.9 × 10 −2 𝑚𝑚 The fish has a vertical displacement of 𝑦𝑦 𝑓𝑓 = − 7.9 𝑖𝑖𝑚𝑚 .

32 (c) How fast are the boat and child moving horizontally after the throw? 3 Identify any assumption/s you make in this calculation. Criteria Marks • Correctly applies conservation of momentum to calculate the speed of the boat and child after the fish leaves the child’s hand, showing all working. • Identifies and justifies any assumptions made Note: simply saying “assume momentum is conserved” is not sufficient – momentum is always conserved if you choose a system containing all interacting objects. By applying conservation of momentum to a system consisting of just the boat, boy and fish you must assume all external forces acting on this system are negligible compared to the force between the fish and boy. 3 Applies conservation of momentum, but makes a numerical error AND Correctly identifies an assumption made OR Correctly calculates the speed of the boat. 2 Provides some relevant information. 1 We will apply conservation of momentum to the throw. The assumption that we will make is that there is no external force/s acting on the system consisting of the boat, child and fish during the throw. If the coracle floats high so that it barely disturbs the water then this implies that the force between the boat and coracle is small, so it is a reasonable assumption. 𝑚𝑚 𝐵𝐵𝐵𝐵 𝑖𝑖 𝐵𝐵𝐵𝐵 + 𝑚𝑚 𝐹𝐹 𝑖𝑖 𝐹𝐹 = 𝑚𝑚 𝐵𝐵𝐵𝐵 𝑖𝑖 𝐵𝐵𝐵𝐵 + 𝑚𝑚 𝐹𝐹 𝑖𝑖 𝐹𝐹 As fish and coracle are initially at rest the left-hand side is zero. As we wish to find the final velocity of the coracle in the horizontal direction, we need to determine the final velocity of the fish in the horizontal direction. This is 𝑖𝑖 𝑥𝑥 = 𝑖𝑖 𝑖𝑖𝑜𝑜𝑠𝑠 θ = 5.0 𝑚𝑚𝑠𝑠 −1 × 𝑖𝑖𝑜𝑜𝑠𝑠 10 = 4.9 𝑚𝑚𝑠𝑠 −1 (As this is close to 5m/s, students can use this value, but would then need to list this as an assumption for full marks) Solving for the final velocity of the boat + child: 𝑣𝑣 𝐵𝐵𝐵𝐵 = − 𝑚𝑚 𝐹𝐹 𝑢𝑢 𝐹𝐹 𝑚𝑚 𝐵𝐵𝐵𝐵 = − 5𝑘𝑘𝑘𝑘 × 4 . 9𝑚𝑚𝑠𝑠 −1 40𝑘𝑘𝑘𝑘 = − 0.62 𝑚𝑚𝑠𝑠 −1 The boat and child are moving at a speed of 0.62m/s away from the bank.

33 Question 32 (5 marks) In your course you used a diffraction grating to observe spectra from discharge tubes. (a) Identify the purpose of the diffraction grating in this investigation. 1 Criteria Marks Identifies the purpose of the diffraction grating as separating the different colours/lines in the spectra (or equivalent) 1 The diffraction grating is used to separate the light from the discharge tube into its constituent colours. The image below is a photograph of the visible spectra from a hydrogen discharge tube after it passes through a diffraction grating which has 300 lines per mm. Central maxima cyan 1 red 1 cyan 2 red 2 The central maxima (containing all colours of light from the discharge tube) is visible on the far left in the photo. The first and second order cyan spectral lines (both with 𝜆𝜆 = 486 nm ), marked “cyan 1” and “cyan 2” are labelled above the spectrum. The first and second order red spectral lines (both with 𝜆𝜆 = 656 nm ) are marked “red 1” and “red 2”. The angular separation between the cyan 1 line and the central maxima is 𝜃𝜃 𝑐𝑐 = 8.4 𝑜𝑜 . (b) Find the angular separation 𝜃𝜃 𝑞𝑞 between the red 1 line and the central maxima. 2 Criteria Marks Correctly calculates the angular separation of the red lines in the spectrum. 2 Substitutes the correct value for 𝑑𝑑 or λ into the correct equation OR Makes an error in substitution but otherwise solves the problem using a correct process. 1 Using 𝑑𝑑 𝑠𝑠𝑖𝑖𝑛𝑛 𝜃𝜃 = 𝑛𝑛𝜆𝜆 , with 𝑑𝑑 = 1 × 10 −3 300 = 3.33 × 10 −6 𝑚𝑚 , 𝜆𝜆 = 656 × 10 −9 𝑚𝑚 and 𝑛𝑛 = 1 , we have 𝑠𝑠𝑖𝑖𝑛𝑛 θ = 656 × 10 −9 𝑚𝑚 3 . 33 × 10 −6 𝑚𝑚 = 0.197 So 𝜃𝜃 = 11.3 𝑜𝑜

34 (c) Explain quantitatively why the distance between the cyan 2 and red 2 lines on the 2 photograph is double the distance between the cyan 1 and red 1 lines. You may assume sin 𝜃𝜃 ≈ tan 𝜃𝜃 ≈ θ . Criteria Marks Shows quantitatively that separation of the second order lines is double that of the first order lines. 2 Identifies that the separation of the lines is proportional to the difference in the angular separation of the red lines and cyan lines. 1 For small angles, the separation of the red 1 and cyan 1 lines is proportional to 𝜃𝜃 𝑞𝑞 − 𝜃𝜃 𝑐𝑐 . For small angles, 𝜃𝜃 2𝑞𝑞 = 2𝜆𝜆 𝑟𝑟 𝑑𝑑 = 2 𝜃𝜃 𝑞𝑞 and 𝜃𝜃 2𝑐𝑐 = 2𝜆𝜆 𝑐𝑐 𝑑𝑑 = 2 𝜃𝜃 𝑐𝑐 The separation of the red 2 and cyan 2 lines is therefore proportional to 2 𝜃𝜃 𝑞𝑞 − 2 𝜃𝜃 𝑐𝑐 = 2( 𝜃𝜃 𝑞𝑞 − 𝜃𝜃 𝑐𝑐 ) , and the separation of the red 2 and cyan 2 lines should be double the separation of the red 1 and cyan 1 lines.

35 Question 33 (5 marks) The Deep Space Climate Observatory (DSCOVR), is a space probe that is located at a position where it orbits the Sun with the same period as the Earth, maintaining a fixed distance from the Earth and the Sun. (a) Calculate the radius of the Earth’s orbit, 𝑅𝑅 . 2 Using Kepler’s 3 rd law relating the period of the earth to its radius, we have 𝑜𝑜 𝐸𝐸 3 𝑚𝑚 𝐸𝐸 2 = 𝐺𝐺𝐺𝐺 4 π 2 Which can be rearranged for the earth’s radius as 𝑜𝑜 𝐸𝐸 = � 𝐺𝐺𝐺𝐺𝑚𝑚 𝐸𝐸 2 4 π 2 � 1 / 3 = � 6.67 × 10 −11 × 2.0 × 10 3 0 × (3.16 × 10 7 ) 2 4 π 2 � 1 / 3 = 1.50 × 10 11 𝑚𝑚 Criteria Marks Correctly calculates the radius of the Earth 2 Provides some relevant information 1 Data: Mass of the DSCOVR: 𝑚𝑚 = 570 kg Mass of the Sun: 𝐺𝐺 𝑆𝑆 = 2.0 × 10 30 kg Period of the Earth’s orbit: 𝑚𝑚 𝐸𝐸 = 3.16 × 10 7 s .

36 (b) Identify the force/s acting on DSCOVR. 3 Write down an equation which could be solved using the information provided to determine the distance 𝑑𝑑 . You do not need to solve this equation. Criteria Marks Obtains a correct equation, which includes the information about the period of the orbit. 3 Attempts to apply Newton’s 2 nd law to DSCOVR, including identifying the centripetal force, and the gravitational forces due to the Sun and Earth. 2 Identifies at least one force acting on DSCOVR 1 There are two forces acting on DSCOVR, one due to the gravitational force of the Earth and one due to the gravitational force of the Sun. Together these forces provide the centripetal force required for DSCOVR to orbit the Sun in a circular orbit with the same period as the earth. We need to write down Newton’s 2nd law for DSCOVR as: Σ𝐹𝐹 = 𝑚𝑚𝑎𝑎 𝐺𝐺𝐺𝐺 𝑆𝑆 𝑚𝑚 ( 𝑅𝑅 − 𝑑𝑑 ) 2 − 𝐺𝐺𝐺𝐺 𝐸𝐸 𝑚𝑚 𝑑𝑑 2 = 𝑚𝑚𝑣𝑣 2 ( 𝑅𝑅 − 𝑑𝑑 ) To find the orbital speed of DSCOVR, we need to use the information that its period is the same as Earth’s orbital period, so 𝑣𝑣 = 2 π ( 𝑅𝑅 − 𝑑𝑑 ) 𝑚𝑚 𝐸𝐸 So our final equation is 𝐺𝐺𝐺𝐺 𝑆𝑆 ( 𝑅𝑅 − 𝑑𝑑 ) 2 − 𝐺𝐺𝐺𝐺 𝐸𝐸 𝑑𝑑 2 = 4 π 2 ( 𝑅𝑅 − 𝑑𝑑 ) 𝑚𝑚 𝐸𝐸 2 Which could be solved to find 𝑑𝑑 using the quadratic formula.

37 Question 34 (11 marks) Cobalt-60 is an isotope that is used to treat some type of cancers. One of the products of the decay is a β -particle. (a) Show that the rest mass of the β -particle is equivalent to 0.511 MeV. 1 Marking criteria Marks Full working 1 E=mc 2 m = 9.1 x 10 -31 kg, c = 3 x 10 8 ms -1 E = 9.1 x 10 -31 x (3 x 10 8 ) 2 = 8.19 x 10 -14 J 1 eV = 1.602 x 10 -19 J 1 J = 1/(1.602 x 10 -19 ) eV 8.19 x 10 -14 J = 8.19 x 10 -14 /(1.602 x 10 -19 ) = 5.11 x 10 5 eV = 0.511 MeV Cobalt-60 undergoes radioactive decay according to the following equation: 𝐶𝐶𝑜𝑜 → 𝑁𝑁𝑖𝑖 28 60 + 𝑒𝑒 + 2 𝛾𝛾 0 0 −1 0 27 60 This can be represented diagrammatically, as shown below: producing stable nickel- 60, a β - particle with a maximum 0.31 MeV kinetic energy, two γ photons and an antineutrino (not shown). The γ photons are responsible for killing rapidly dividing cells, specifically the tumour cells. (b) The mass of a cobalt-60 nucleus is 59.919015 u. Calculate the mass of the nickel-60 3 nucleus.

38 Marking criteria Marks Calculates energy change Calculates mass change Calculates mass of Ni nucleus Consistent units 3 2 of the following: Calculates energy change Calculates mass change Calculates mass of Ni nucleus OR 1 error in calculation 2 1 of the following: Calculates energy change Calculates mass change Calculates mass of Ni nucleus OR 2 errors in calculation 1 Change in energy = 0.511 + 0.31 + 1.17 + 1.33 = 3.321 MeV Change in mass (in u) = 3.321/931.5 = 0.003565 Mass of Ni nucleus = 59.919015 – 0.003565 = 59.91545 u Comments: a large number of students forgot to include the mass of the β -particle, despite calculating it in part (a), some forgot the KE or the mass converted to photons . There was also some vey bad setting out, please show your working in a neat and ordered manner, as part mark are possible if it clear to the marker what you were doing. A few students tried to add up the mass of the protons and neutrons, this does not account for binding energy, some also considered only the mass defect for a neutron, again if you made these error, please revise binding energy of nuclei. Please be careful when using significant figures, we keep the same number of decimal places when adding/subtracting. (c) Small pellets, about the size of a grain of rice, are injected into the tumour and left 2 for a set period of time before being removed. The cobalt-60 pellets typically have an activity of 18.5 GBq per pellet when fresh, where 1Bq = 1 decay per second. Each pellet can be used for many treatments, but they need to be replaced once the activity has decreased to 7.0 GBq per pellet. The half-life for cobalt-60 is 5.27 years. Assuming the activity is proportional to the amount of Cobalt-60 present, how frequently would the cobalt-60 pellets need to be replaced to maintain sufficient activity?

39 Marking criteria Marks Calculates time of replacement for Co-60 2 Calculates decay constant OR Calculates replacement time based on incorrect decay constant 1 λ=ln2/t 1/2 t 1/2 = 5.27 years λ = ln2/5.27 = 0.1315 years -1 N t = N o e^(- λt) N t = 7.0 GBq, N 0 = 18.5GBq 7 = 18.5 e^(- 0.1315t) Ln(7/18.5)=-0.1315t t = 7.4 years Comments: please be careful with rounding when you are doing calculations. Keep extra significant figures until the final number, particularly when using exponentials. (d) In one treatment regime, a fresh pellet is inserted into the tumour and removed after 10 minutes. (i) Calculate the change in temperature of a 60 kg patient caused by the ejected 3 particles. Assume: • the specific heat capacity of the patient is 3.5 × 10 3 J kg −1 K −1 • the activity remains constant during the 10 minutes. • 1 GBq = 10 9 Bq • 𝑄𝑄 = 𝑚𝑚 𝑖𝑖 ∆𝑚𝑚 Marking criteria Marks Calculates change in temperature 3 As above with 1 error 2 As above with 2 errors 1 Number of decays in 10 minutes = 60 x 10 x 18.5 x 10 9 = 1.11 x 10 13 Energy per decay = 0.31 + 1.17 + 1.33 MeV = 2.81 MeV = 2.81 x 10 6 x 1.602 x 10 -19 = 4.5 x 10 -13 J Energy released in 10 minutes = 1.11 x 10 13 x 4.5 x 10 -13 = 4.997 J m = 60 kg c = 3.5 × 10 3 J kg -1 K -1 Q = mcΔT 4.997 = 60 x 3.5 x 10 3 x ΔT ΔT = 4.997/(60 x 3.5 x 10 3 ) = 0.0000238 = 2 x 10 -5 K (an increase) Comments: A number of students did not carefully read the assumptions that you were instructed to make, in particular that the activity remained constant for 10 minutes (not unreasonable given the half- life of 7.4 years. The mass of the β -particle is not included in the calculation here (as it would remain an electron), but its KE should be as it will likely be captured by the tissue and converted to heat. Again, setting out is important, as is writing units as you go so that you can see that you haven’t forgotten to use a conversion factor.

40 (ii) Identify TWO other significant physical assumptions that you made in your 2 calculation above. Marking criteria Marks Two suitable significant assumptions 2 One suitable significant assumptions 1 All of the energy is converted to heat energy/absorbed by the body. All of the beta particles have close to the max KE. The temperature change is uniformly distributed throughout the body. Comment: the question was about significant assumptions made about the energy due to the ejected particles. Significant in this sense means an assumption that would change your answer by at least 10% (given that the mass was only to 1 sig fig)

41 Question 35 (5 marks) A carousel (below, left) is a children’s ride that rotates at a constant angular velocity. A mass 𝑚𝑚 is attached by a string to one of the vertical bars at a distance 𝑅𝑅 from the center of the carousel (below, right). The mass hangs at an angle 𝜃𝜃 when the ride is rotating with a constant angular velocity of ω . a) Draw a free body diagram showing the forces acting on the mass, to scale. 2 Indicate the net force with a dotted arrow. Criteria Marks Draws: - a tension force acting in the direction of the string - a weight force acting directly downwards such that: - the vertical component of the net force is zero - the horizontal component of the net force is equal to the horizontal component of the tension force, and points to the center of the carousel. 2 Correctly identifies at least one of the forces acting on the mass 1

42 b) Explain, qualitatively and quantitatively, how the angle 𝜃𝜃 would change if the 3 string was attached to a pole at a distance 1 2 𝑅𝑅 from the center of the carousel while it is still rotating at the same constant angular velocity ω . You may assume the angle 𝜃𝜃 is small. Criteria Marks A correct qualitative and quantitative explanation. 3 A correct qualitative explanation (e.g. 𝜃𝜃 is smaller as required 𝐹𝐹 𝐵𝐵 is smaller) OR A general algebraic expression that is not evaluated to answer the question OR Derives an expression relating 𝜃𝜃 and 𝑅𝑅 (eliminating 𝑣𝑣 ) but makes the incorrect substitution 𝜔𝜔 = 𝑣𝑣𝑜𝑜 (to obtain 𝜃𝜃 doubles instead of halves) 2 Provides some relevant information, e.g. formula for centripetal force. 1 Qualitative explanation: The string hangs at a smaller angle as a smaller centripetal force (provided by the horizontal component of tension) is required to move it in uniform circular motion when the radius is smaller and the angular speed remains constant. Quantitative explanation: Applying Newton’s 2 nd law to the mass in the vertical direction, where the acceleration is zero, we have Σ𝐹𝐹 𝑥𝑥 = 𝑚𝑚𝑎𝑎 𝑥𝑥 𝑚𝑚 𝑖𝑖𝑜𝑜𝑠𝑠 θ − 𝑚𝑚𝑖𝑖 = 0 So 𝑚𝑚 𝑖𝑖𝑜𝑜𝑠𝑠 θ = 𝑚𝑚𝑖𝑖 Applying Newton’s 2 nd law in the vertical direction and assuming θ is small enough that the radius of the circular motion equals 𝑅𝑅 , we have Σ𝐹𝐹 𝑦𝑦 = 𝑚𝑚𝑎𝑎 𝑦𝑦 𝑚𝑚 𝑠𝑠𝑖𝑖𝑛𝑛 θ = 𝑚𝑚 𝑣𝑣 2 𝑅𝑅 = 𝑚𝑚𝑅𝑅ω 2 Eliminating the tension, we have: 𝑡𝑡𝑎𝑎𝑛𝑛 θ = 𝑚𝑚𝑅𝑅ω 2 If θ is small, then this approximately equals θ = 𝑚𝑚𝑅𝑅ω 2 So halving the radius (for the same angular speed) will halve the angle.

43 Question 36 (9 Marks) A 90 g mass is attached via a light string to the shaft of a DC motor. An investigation is performed consisting of three stages: Stage 1: The power supply is switched on and the mass moves up. Stage 2: The string is fully wound up and the shaft is stationary. Stage 3: The power supply is switched off and the mass moves down. The vertical velocity of the mass during each stage is measured using video analysis and is shown below. Question 36 continues on page 41 The current through the motor and the potential difference (voltage) across the motor were also measured during each stage and are shown below. 2 nd stage 3 rd stage Figure 1: Vertical velocity of 90 g mass versus time 1 st stage

44 Analyse the physics involved, and as well as how the law of conservation of energy applies 9 to the system consisting of the 90 g mass and the motor during each of the three stages of the investigation. 1 st stage 2 nd stage 3 rd stage Figure 2: Current versus time Figure 3: Potential difference versus time

45 Your analysis should include both qualitative and quantitative information. Criteria Marks - Provides a comprehensive, correct, qualitative and quantitative analysis of: o the physics involved in at least two stages (substantially correct for third) o how energy conservation applies in at least two stages (substantially correct for third). - Communicates scientific understanding logically, succinctly, and effectively using correct and precise scientific terms. 9 - Provides a thorough, correct or substantially correct, qualitative and quantitative analysis of: o the physics involved in each of the three stages o how energy conservation applies in each of the three stages. - Communicates scientific understanding logically, succinctly, and effectively using correct and precise scientific terms. 8 - Applies the law of conservation of energy in a substantially correct way to at least one of the stages - Provides a substantially correct analysis of the physics involved in at least one the stages - Some quantitative analysis (not simply transcribing info from graphs) - Communicates scientific understanding logically and effectively using correct scientific terms. 6-7 - Relates the law of conservation of energy in a substantially correct way to at least one of the stages AND/OR - Provides a correct description of some aspects of the physics involved in at least one of the stages AND/OR - Includes quantitative information 4-5 Shows some understanding of conservation of energy and/or the physics involved in the investigation 2-3 Provides some relevant information 1

46

47 Physics: Correct application of Σ𝐹𝐹 = 𝑚𝑚𝑎𝑎 Correct application of Conservation of energy Units used correctly Communication: Logical structure Succinct, precise use of scientific language Effective/efficient communication via diagrams/annotations/equations Stg Qualitative observations Qualitative analysis Quantitative observations Quantitative analysis/calculations 1 - Velocity is positive and increasing - Acceleration is positive and decreasing - Current is initially high then decreases - Two forces act on mass, tension (up) and weight force (down) - The current flowing through the rotor of the motor produces a torque due to the motor effect - KE of the mass increases - GPE of mass increases (if mass and earth the system) OR gravity does negative work on the mass (if mass is the system) - Electrical power is supplied to the motor - back emf is generated, opposing supply - Acceleration upwards means that T>mg - By COE, total change in KE of mass + work done against gravity (i.e. change in GPE) + total heat generated in coils/bearings = total energy supplied Draws links between: o Decreasing acceleration due to o reduction in torque, due to o decrease in current, due to o increased back emf due to o increased angular velocity and so o increased rate of change of B flux due to velocity increasing over time - Gradient of v-t graph ≈ 1 𝑚𝑚𝑠𝑠 −2 (spread of values possible here) - Area under graph 𝐴𝐴 ≈ 0.4 𝑚𝑚 - N’s 2 nd law for mass 𝑚𝑚 − 𝑚𝑚𝑖𝑖 = 𝑚𝑚𝑎𝑎 - Torque due to ME is 𝜏𝜏 = 𝑛𝑛𝐵𝐵𝐼𝐼𝐴𝐴 - The tension and torque due to ME are connected via 𝜏𝜏 = 𝑜𝑜𝑚𝑚 - There is a back emf in the motor due to Faraday’s law 𝜖𝜖 𝑏𝑏𝑏𝑏𝑐𝑐𝑘𝑘 = −𝑁𝑁 ΔΦ Δ𝑡𝑡 which opposes the supply voltage. - Electrical energy 𝑊𝑊 𝐸𝐸 = 𝑃𝑃𝑡𝑡 = 𝑉𝑉𝐼𝐼𝑡𝑡 - Change in Kinetic energy of mass Δ𝐾𝐾𝐾𝐾 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 - Work done by gravity 𝑊𝑊 𝐺𝐺 = 𝑚𝑚𝑖𝑖Δ y - Acc of mass 𝑎𝑎 ≈ 1 𝑚𝑚𝑠𝑠 −2 - Area under graph is displacement of mass 𝐴𝐴 ≈ 0.4 𝑚𝑚 - Tension 𝑚𝑚 = 0.09 𝑘𝑘𝑖𝑖 (9.8 + 1) 𝑚𝑚𝑠𝑠 −2 ≈ 1 𝑁𝑁 - Electrical energy 𝑊𝑊 𝐸𝐸 = 𝑃𝑃𝑡𝑡 = 𝑉𝑉𝐼𝐼𝑡𝑡 ≈ 2 𝑉𝑉 × 2.1 𝐴𝐴 × 0.7 𝑠𝑠 ≈ 3.0 𝐽𝐽 - Change in KE of mass Δ𝐾𝐾𝐾𝐾 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 = 0.5 × 0.09 𝑘𝑘𝑖𝑖 × (1 𝑚𝑚𝑠𝑠 −1 ) 2 ≈ 0.05 𝐽𝐽 - Work done by gravity 𝑊𝑊 𝐺𝐺 = 𝑚𝑚𝑖𝑖Δ𝑦𝑦 = 0.09 𝑘𝑘𝑖𝑖 × 9.8 𝑚𝑚𝑠𝑠 −2 × 0.4 𝑚𝑚 ≈ − 0.4 𝐽𝐽 (negative as mass moves up) - 𝐼𝐼 2 𝑅𝑅 heat during stage 1 is Δ𝑚𝑚 𝑡𝑡ℎ = 𝑊𝑊 𝐸𝐸 + 𝑊𝑊 𝐺𝐺 − Δ𝐾𝐾𝐾𝐾 = 3 𝐽𝐽 − 0.4 𝐽𝐽 − 0.05 𝐽𝐽 ≈ 2.5 𝐽𝐽 2 - The net force on the mass is zero - There is a normal force from the casing of the motor downwards ( HARD ) (This normal force down + weight down = tension from string up) - Current and Voltage constant - Power supplied to the motor is constant as voltage and current are constant. Draws links between: o Large constant current due to o Zero back emf, due to o Constant B flux - As there is no change in KE or GPE of the mass, all the electrical power is dissipated as heat in the coils. - Resistance should increase as motor gets hot and 𝑅𝑅 ∝ 𝑚𝑚 - 𝑃𝑃 = 𝑉𝑉𝐼𝐼 - 𝐾𝐾 = 𝑃𝑃𝑡𝑡 - 𝑅𝑅 𝑐𝑐𝑜𝑜𝑖𝑖𝑐𝑐𝑠𝑠 = 𝑉𝑉 𝐼𝐼 - 𝑃𝑃 𝑐𝑐𝑜𝑜𝑖𝑖𝑐𝑐𝑠𝑠 = 𝐼𝐼 2 𝑅𝑅 - 𝑃𝑃 = 𝑉𝑉𝐼𝐼 = 1.9 𝑉𝑉 × 2.7 𝐴𝐴 ≈ 5.1 𝑊𝑊 - 𝐾𝐾 = 𝑃𝑃𝑡𝑡 = 5.1 𝑊𝑊 × 0.7 𝑠𝑠 = 3.6 𝐽𝐽 - 𝑅𝑅 𝑐𝑐𝑜𝑜𝑖𝑖𝑐𝑐𝑠𝑠 = 𝑉𝑉 𝐼𝐼 = 1 . 9 2 . 7 = 0.70 Ω 3 - The motor acts as a generator. - Mass moves with an increasing negative velocity - Constant negative acceleration of mass - Tension is less than the weight force - Current flows in the same direction as in stages 1 and 2 - Voltage across the motor is in the opposite direction to stages 1 and 2 - Work done by gravity converted to KE and electrical energy, which is converted to heat - The tension force in the string exerts a torque 𝜏𝜏 = 𝑜𝑜𝑚𝑚 on rotor, so 𝜔𝜔 id increasing - Increasing speed produces increasing emf - By N’s 3 rd law, the reaction force is the back torque acting due to F = BIL force acting on the generated current. ( HARD ) Draws links between: o shaft rotating in the opposite direction to stage 1 means that o the generated emf is in the opposite direction to the back emf in stage 1, accounting for: o Why current is same direction as S1 ( HARD ) o Why voltage is negative in S3 ( HARD ) - Gradient v-t graph ≈ −2−0 1 . 7−1 . 4 = − 6.7 𝑚𝑚𝑠𝑠 −2 - The induced emf is given by Faraday’s law 𝜖𝜖 = −𝑁𝑁 ΔΦ Δ𝑡𝑡 - Distance mass falls is area under v-t graph - 𝑊𝑊 𝐺𝐺 = 𝑚𝑚𝑖𝑖Δ𝑦𝑦 - Δ𝐾𝐾𝐾𝐾 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 - Acceleration of mass is 𝑎𝑎 = − 6.7 𝑚𝑚𝑠𝑠 −2 - Σ𝐹𝐹 = 𝑚𝑚𝑎𝑎 = 0.09 𝑘𝑘𝑖𝑖 × ( − 6.7) = 0.60 𝑁𝑁 downwards. - Σ𝐹𝐹 = 𝑚𝑚 − 𝑚𝑚𝑖𝑖 , so 𝑚𝑚 = − 0.6 𝑁𝑁 + 0.09 × 9.8 = 0.28 𝑁𝑁 upwards. - Δ𝑦𝑦 = 1 2 2.6 𝑚𝑚𝑠𝑠 −1 × 0.4 𝑠𝑠 = 0.52 𝑚𝑚 - 𝑊𝑊 𝐺𝐺 = 𝑚𝑚𝑖𝑖Δ𝑦𝑦 = 0.09 𝑘𝑘𝑖𝑖 × ( − 9.8 𝑚𝑚𝑠𝑠 −2 ) × ( − 0.52 𝑚𝑚 ) = 0.46 𝐽𝐽 - Δ𝐾𝐾𝐾𝐾 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 = 1 2 × 0.09 𝑘𝑘𝑖𝑖 × (2.6 𝑚𝑚𝑠𝑠 −1 ) 2 = 0.30 𝐽𝐽 - By conservation of energy 𝑊𝑊 𝐺𝐺 = Δ𝐾𝐾𝐾𝐾 + Δ𝐾𝐾 𝐸𝐸 , so Δ𝐾𝐾 𝐸𝐸 = 0.46 𝐽𝐽 − 0.3 𝐽𝐽 = 0.16 𝐽𝐽 - Thermal energy = electrical energy = 0.16J

48 Sample answer Note: This sample answer is thorough , as it is designed to illustrate the variety of material that can be discussed in response to the question. Stage 1 Physics Involved : Current flowing through the coils in the DC motor experiences a force due to the motor effect 𝐹𝐹 = 𝐼𝐼𝐿𝐿𝐵𝐵 𝑠𝑠𝑖𝑖𝑛𝑛 𝜃𝜃 1 . This force produces a torque τ = 𝑜𝑜𝐹𝐹 𝑠𝑠𝑖𝑖𝑛𝑛 𝜃𝜃 2 (where 𝜃𝜃 2 is the angle between r and F, and r is the radius of the rotor) on the motor shaft. This torque exerts a tension force on the string of 𝑚𝑚 = τ / 𝑜𝑜 𝑠𝑠ℎ𝑏𝑏𝑓𝑓𝑡𝑡 . Figure 1 shows that the average acceleration of the mass during stage 1 is: 𝑎𝑎 𝑏𝑏𝑣𝑣1 = ( 1−0 . 5 ) 𝑚𝑚 / 𝑠𝑠 ( 0 . 6−0 . 1 ) 𝑠𝑠 = 1.0 𝑚𝑚𝑠𝑠 −2 Drawing a free body diagram from the 90g mass (right) and applying Newton’s 2 nd law ( Σ𝐹𝐹 = 𝑚𝑚𝑎𝑎 ) in the vertical direction we have: 𝑚𝑚 − 𝑚𝑚𝑖𝑖 = 𝑚𝑚𝑎𝑎 So the tension in the string is (on average) 𝑚𝑚 = 𝑚𝑚 ( 𝑎𝑎 + 𝑖𝑖 ) = 0.09 × (9.8 + 1.0) = 0.97 𝑁𝑁 . The average net force acting on the 90g mass during stage 1 is then Σ𝐹𝐹 = 0.09 𝑘𝑘𝑖𝑖 × 1.0 𝑚𝑚𝑠𝑠 −2 = 0.09 𝑁𝑁 A back emf is produced in the motor due to Faraday’s law ϵ 𝑏𝑏𝑏𝑏𝑐𝑐𝑘𝑘 = −𝑁𝑁 𝑑𝑑ϕ 𝐵𝐵 𝑑𝑑𝑡𝑡 in a direction which opposes the supply voltage (due to Lenz’s law) in order to satisfy conservation of energy. As the rate of rotation of the motor increases the magnitude of the back emf increases, so the current flowing through the motor decreases, as seen in figure 2 from 𝑡𝑡 = 0 𝑠𝑠 to 𝑡𝑡 = 0.7 𝑠𝑠 . The current is given by 𝐼𝐼 𝑞𝑞𝑜𝑜𝑡𝑡 = ϵ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 −ϵ 𝑏𝑏𝑏𝑏𝑐𝑐𝑏𝑏 𝑅𝑅 𝑐𝑐𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠 . As the torque is proportional to the current, this means that the tension force decreases over time, so the net force and therefore the acceleration decreases. Conservation of Energy : One way to organize our thinking about conservation of energy is to use an energy flow diagram and energy bar chart before writing down an equation for conservation of energy, just as we would draw a free body diagram before we apply Newton’s 2 nd law. Here the system has been chosen to be the mass and the motor (alternatively we could include the earth in the system as well). It is not an isolated system as energy is transferred in to the system in the form of electrical energy and transferred out of the system in the form of work done against the external force of gravity (if the system was chosen to include earth this would be an increase in gravitational potential energy). Energy transformations include increasing kinetic energy of the mass and increasing thermal energy in the system (primarily due to 𝐼𝐼 2 𝑅𝑅 heating in the coils of the motor).

JRAHS 2022 Physics HSC Trial Exam Page 49 The information in our energy bar chart expresses conservation of energy: The sum of the initial energy + work done (by power supply and gravity) = Final energy of the system 𝐾𝐾 𝑡𝑡ℎ1 + 𝐾𝐾 𝑘𝑘1 + 𝑊𝑊 𝐸𝐸 − 𝑊𝑊 𝐺𝐺 = 𝐾𝐾 𝑡𝑡ℎ2 + 𝐾𝐾 𝑘𝑘2 Expressing this in terms of changes: Δ E th + Δ E k = 𝑊𝑊 𝐸𝐸 − W G Alternatively, if earth was included in the system, we could express this as: Δ E th + Δ E k + Δ E G = 𝑊𝑊 𝐸𝐸 When the rotor is turning, the power supply does electrical work to drive current against the back emf and dissipates electrical energy as heat in the coils of the motor. The work done against the back emf provides the energy to raise the mass against the external force of gravity as well as to increase the kinetic energy of the mass. The electrical energy provided by the power supply in the first stage is the product of the potential, current and time (the area under the current-time graph times average voltage). This is approximately: 𝑃𝑃𝑡𝑡 = 𝑉𝑉𝐼𝐼𝑡𝑡 = 2 𝑉𝑉 × 2.1 𝐴𝐴 × 0.7 𝑠𝑠 = 2.94 𝐽𝐽 The work done against the external force of gravity is the force due to gravity times the displacement (area under the v-t graph): 𝑊𝑊 𝐺𝐺 = 𝐹𝐹𝑠𝑠 𝑖𝑖𝑜𝑜𝑠𝑠 θ = 𝐹𝐹𝑣𝑣𝑡𝑡 = 0.09 × 9.8 × (0.7 𝑠𝑠 × 0.4 𝑚𝑚 / 𝑠𝑠 + 0.5 × 0.6 𝑚𝑚 / 𝑠𝑠 × 0.7 𝑠𝑠 ) = 0.43 𝐽𝐽 The change in kinetic energy is Δ E k = 1 2 𝑚𝑚𝑣𝑣 2 = 0.5 × 0.09 × 1 2 = 0.05 𝐽𝐽 It can be seen from conservation of energy that the bulk of the electrical energy supplied by the power supply is dissipated as heat in the coils. Δ E th = 𝑊𝑊 𝐸𝐸 − W G − Δ E k = 2.94 − 0.05 − 0.43 = 2.5J Stage 2

JRAHS 2022 Physics HSC Trial Exam Page 50 Physics Involved : The rotor is stationary as the torque acting on the motor shaft due to the tension (from the string) and the motor effect (as discussed in the first stage) is equal (zero net torque). Alternatively, we can consider the forces acting on the mass. These are the weight force, the tension force due to the string upwards and the normal force on the mass due to the motor casing pushing on the mass once they make contact. Using Newton’s 2 nd law, and the fact that the mass has zero acceleration, we can reason that the sum of these forces is zero (see FBD, right). As the coil is not rotating in stage 2, there is no back emf (there is no change in B flux through the coil, so no emf as a result of Faraday’s law). When the shaft is stationary the current is given by 𝐼𝐼 𝑠𝑠𝑡𝑡𝑏𝑏𝑡𝑡 = ϵ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑅𝑅 𝑐𝑐𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠 which is larger than when the coil was rotating. As the current flowing through the coils determines the torque provided by the motor, via 𝜏𝜏 = 𝑛𝑛𝐵𝐵𝐼𝐼𝐴𝐴 , this means that the torque due to the motor effect is also larger in stage 2 than it was in stage 1, and so the tension in the string is larger as 𝜏𝜏 𝑚𝑚𝑏𝑏𝑥𝑥 𝑜𝑜 𝑠𝑠ℎ𝑏𝑏𝑓𝑓𝑡𝑡 = 𝑚𝑚 . Using information from figures 2 and 3, the resistance of the coils (+ contacts) is 𝑅𝑅 𝑐𝑐𝑜𝑜𝑖𝑖𝑐𝑐𝑠𝑠 = 1.9 𝑉𝑉 2.7 𝐴𝐴 = 0.70 Ω Conservation of Energy : Electrical energy continuously flows into the mass & motor system from the power supply at an (almost) constant rate in stage 2. This electrical energy is transformed into heat in the coils of the rotor during stage 2. The total electrical energy transformed to thermal energy in the motor during stage 2 is 𝑃𝑃𝑡𝑡 = 𝑉𝑉𝐼𝐼𝑡𝑡 = 1.9 𝑉𝑉 × 2.7 𝐴𝐴 × 0.7 𝑠𝑠 = 3.6 𝐽𝐽 . Stage 3

JRAHS 2022 Physics HSC Trial Exam Page 51 Physics Involved : During this stage the motor acts as a generator. Using figure 1, the average acceleration of the 90g mass is 𝑎𝑎 𝑏𝑏𝑣𝑣3 = ( −2−0 ) 𝑚𝑚 / 𝑠𝑠 0 . 3𝑠𝑠 = − 6.7 𝑚𝑚𝑠𝑠 −2 A net force downwards acts on the 90g mass equal to Σ𝐹𝐹 = 0.09 𝑘𝑘𝑖𝑖 × ( − 6.7 𝑚𝑚𝑠𝑠 −2 ) = − 0.6 𝑁𝑁 The tension in the string is then: 𝑚𝑚 = 𝑚𝑚 ( 𝑖𝑖 + 𝑎𝑎 ) = 0.09 × (9.8 − 6.7) = 0.28 𝑁𝑁 upwards. This tension exerts a torque on the shaft of the DC motor, causing an angular acceleration (increasing angular speed). The rotation of the shaft causes a change in magnetic flux and so an induced emf according to Faraday’s law ϵ = −𝑁𝑁 𝑑𝑑ϕ 𝐵𝐵 𝑑𝑑𝑡𝑡 , which drives a current. A “back torque” is produced by the F = BIL force which acts on the generated current. This back torque produces a force on the string that is the reaction force to the tension force by Newton’s 3 rd law. If motor was not connected to a complete circuit, the mass would accelerate at close to g as the tension would be close to zero (equal to zero if there was no friction in the shaft), as there would be no back torque. As the rotation of the shaft is in the opposite direction to stage 1, this means that the induced emf is in the opposite direction to the induced emf during stage 1 , and is increasing in magnitude (as the rate of rotation is increasing) to a maximum value of 0.3V (from figure 3). As the back emf during stage 1 would have driven current in the opposite direction to the supply voltage, this means that the induced current during stage 3 has to flow in the same direction as current in stages 1 and 2, which is what is observed in figure 2. The measured potential difference across the motor during stage 3 is negative. To understand why this is the case (i.e. the measured potential difference is negative but the current is positive), consider the back emf during stage 1 (below left): The back emf that appears across each coil in the motor during stage 1 must be in such a direction that it would act as a battery that “opposes” the supply (imagine connecting two batteries with the

JRAHS 2022 Physics HSC Trial Exam Page 52 positive terminals together – current flows in the direction that the battery with the larger potential is trying to drive it – the current flows against the back emf, which is how electrical work is done). As the supply voltage is larger than the back emf, current flows out of the positive terminal of the power supply (anticlockwise in the circuit shown). In stage 3, the “generator” is acting as a source of emf, so current flows out of its positive terminal, resulting in a current in the same direction as in stage 1 (see figure above right) even though the potential difference is opposite in sign to the situation in stage 1. Conservation of Energy : During stage 3, external work done by gravity is transferred to the system (if earth is included in the system this would instead be described as a change in gravitational potential energy). Some of this work is converted to kinetic energy of the mass, and the remainder is converted to electrical energy, which is then dissipated as heat in the coils of the generator (as these are the only resistance in the circuit in stage 3, the power supply when switched off has no resistance). Applying conservation of energy: 𝑊𝑊 𝐺𝐺 = Δ𝐾𝐾 𝑘𝑘 + Δ𝐾𝐾 𝑡𝑡ℎ The work done by gravity is the external force of gravity times the displacement of the 90g mass, which is the weight times the area under the velocity-time graph during stage 3. The mass moves through a distance equal to the area under the velocity-time graph Δ𝑦𝑦 = 1 2 2.6 𝑚𝑚𝑠𝑠 −1 × 0.4 𝑠𝑠 = 0.52 𝑚𝑚 The work done by the force of gravity is then 𝑊𝑊 𝐺𝐺 = 𝑚𝑚𝑖𝑖Δ𝑦𝑦 = 0.09 𝑘𝑘𝑖𝑖 × ( − 9.8 𝑚𝑚𝑠𝑠 −2 ) × ( − 0.52 𝑚𝑚 ) = 0.46 𝐽𝐽 The change in kinetic energy of the mass is Δ𝐾𝐾𝐾𝐾 = 1 2 𝑚𝑚𝑣𝑣 2 − 1 2 𝑚𝑚𝑖𝑖 2 = 1 2 × 0.09 𝑘𝑘𝑖𝑖 × (2.6 𝑚𝑚𝑠𝑠 −1 ) 2 = 0.30 𝐽𝐽 By conservation of energy 𝑊𝑊 𝐺𝐺 = Δ𝐾𝐾𝐾𝐾 + Δ𝐾𝐾 𝐸𝐸 , so the electrical energy produced (which is dissipated as heat in the coils of the motor) is:

JRAHS 2022 Physics HSC Trial Exam Page 53 Δ𝐾𝐾 𝐸𝐸 = 0.46 𝐽𝐽 − 0.3 𝐽𝐽 = 0.16 𝐽𝐽 Most of the work done by gravity is therefore converted into kinetic energy of the mass, and only a small fraction produces electrical work in the generator. END OF PAPER