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2014 HSC TRIAL EXAMINATION PHYSICS – MAPPING GRID Exam Section Section I: Part A: Multiple Choice Section I: Part B: Extended Response Question Marks Syllabus/Course Outcomes Content 1 2 1 1 H6 H9 3 1 H1, H2, H3, H6 4 1 H6, H9 5 6 7 8 9 10 11 1 1 1 1 1 1 1 H1, H5, H9 H1, H2 H6 H9, H13 H3, H9 H9 H7, H9 12 1 H2, H3, H9 13 14 15 1 1 1 H7, H9 H7, H9, H10 H1, H3, H8 16 17 18 19 20 21 1 1 1 1 1 5 H1, H2, H10, H14 H3, H7, H10 H2, H3 H1, H10, H14 H2, H3 H5, H6, H7, H9, H16 22 23 24 25 5 4 4 8 H6, H7 H1, H3, H7, H9 H1, H6, H12, H10, H14 H2, H9, H12, H13 26 27 6 4 H3, H4, H7, H13, H16 H3, H7, H9, H14 28 5 H1, H4, H13, H16 9.2.1 9.2.1, 9.2.2 9.2.2, 9.4.4 9.2.2, 9.2.4 9.2.2 9.2.4 9.2.4 9.3.1 9.3.1 9.3.1 9.3.2, 9.3.3 9.3.2, 9.4.1 9.3.2 9.3.2 9.4.2, 9.4.3 9.4.1 9.4.2 9.4.4 9.4.3 9.4.4 9.2.1, 9.2.3 9.2.2 9.2.2 9.2.4 9.2.1, 9.3.1 9.3.4 9.3.3, 9.3.4 9.4.2 Targeted Performance Bands 1–2 3–4 Answer 3–4 A 4–5 B 2–3 5–6 4–5 2–3 1–2 3–4 5–6 D A A C A B B 3–4 D 2–3 5–6 1–2 B C C 3–4 5–6 1–2 5–6 2–3 1–4 A D D C A D B 1–4 1–5 2–6 1–6 1–6 1–5 1–6 Disclaimer Every effort has been made to prepare this Examination in accordance with the Board of Studies documents. No guarantee or warranty is made or implied that the Examination paper mirrors in every respect the actual HSC Examination question paper in this course. This paper does not constitute ‘advice’ nor can it be construed as an authoritative interpretation of Board of Studies intentions. No liability for any reliance, use or purpose related to this paper is taken. Advice on HSC examination issues is only to be obtained from the NSW Board of Studies. The publisher does not accept any responsibility for accuracy of papers which have been modified. PHYTR14B_GUIDELINES Page 1 Section II 29 30 31 32 6 4 4 25 33 25 34 25 PHYTR14B_GUIDELINES H1, H9, H10 H1, H2, H3, H10 H1, H3, H5, H10, H13 H1, H2, H3, H4, H7, H8, H10, H12, H13, H14 H1, H2, H3, H5, H7, H8, H9, H11, H13, H14, H1, H2, H3, H5, H10, H11, H13, H14, H15, H16 9.4.1 9.4.2 9.4.3 1–5 1–6 1-5 1-6 1-6 1-6 Page 2 2014 HSC TRIAL EXAMINATION PHYSICS – MARKING GUIDELINES Section I Part A – 20 marks Questions 1-20 (1 mark each) Question Correct Response Outcomes Assessed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 D B A B D A A C A B B D B C C A D D C A H6 H9 H1, H2, H3, H6 H6, H9 H1, H5, H9 H1, H2 H6 H9, H13 H3, H9 H9 H7, H9 H2, H3, H9 H7, H9 H7, H9, H10 H1, H3, H8 H1, H2, H10, H14 H3, H7, H10 H2, H3 H1, H10, H14 H2, H3 PHYTR14B_GUIDELINES Targeted Performance Bands 1–2 3–4 3–4 4–5 2–3 5–6 4–5 2–3 1–2 3–4 4–5 3–4 2–3 5–6 1–2 3–4 5–6 1–2 5–6 2–3 Page 3 Section I Part B – 55 marks Question 21 (5 marks) 21 (a) (1 mark) Outcomes Assessed: H6, H9 Targeted Performance Bands: 1-2 Criteria Correct substitution into a correct formula, yielding an answer Sample answer m m EP G 1 2 r EP Mark 1 6.67 10 11 6.0 10 24 12 1.2 10 8 J 40 000 10 3 21 (b) (2 marks) Outcomes Assessed: H6, H7, H9 Targeted Performance Bands: 2-4 Criteria Clear statement that an inertial frame of reference cannot be accelerating Clear statement that the rock’s frame of reference is not inertial Only one of the above Marks 2 1 Sample answer In order for a frame of reference to be inertial it is necessary that it not be accelerating. The rock is attracted by Earth’s gravity (and the Sun, Moon etc.). While it is true that every body in the Universe is similarly affected by gravity, so strictly none can be perfectly inertial, the frame of reference of this rock certainly cannot be considered to be inertial. 21 (c) (2 marks) Outcomes Assessed: H5, H9, H16 Targeted Performance Bands: 1-3 Criteria Not all the laws of physics hold in a non-inertial frame of reference The moving Earth or the air has kinetic energy and Interaction (or friction) changes the KE to heat One of the above or The rock actually did possess KE because its frame is not inertial Marks or 2 1 Sample answer Although the rock had no kinetic energy in its own reference frame, the Earth certainly did have kinetic energy as it approached the rock. When there was an interaction between the air (moving with the Earth) and the rock, the air provided the energy to rapidly heat the rock. PHYTR14B_GUIDELINES Page 4 Question 22 (5 marks) 22 (a) (2 marks) Outcomes Assessed: Targeted Performance Bands: 2-4 Criteria Correct substitution into the correct formula to yield an answer Incorrect substitution into a correct formula Marks 2 1 Sample answer Vertical motion: Δ y = ? uy = 16 sin 60 = 13.856 m s-1 ay = –9.8 m s-2 t = 1.5 s Δ y = uy t + ½ ay t2 Δ y = 13.856 1.5 – 0.5 9.8 1.52 = 9.759 [9.8 m ↑] 22 (b) (2 marks) Outcomes Assessed: Targeted Performance Bands: 1-3 Criteria Correct substitution into a correct formula to obtain the final velocity Correct method to find time for parachute to descend Only one of the above attempted correctly Marks 2 1 Sample answer Vertical motion: uy = 13.856 m s-1 vy = ? ay = –9.8 m s-2 t = 1.5 s vy = uy + ay t vy = 13.856 – 9.8 1.5 = – 0.844 m s-1↓ Time required to drop 9.759 m at 0.844 m s-1 = 11.56 s. 22 (c) (1 mark) Outcomes Assessed: Targeted Performance Bands: 1-2 Criteria Correct substitution into a correct formula to yield a displacement Mark 1 Sample answer Horizontal motion: ux = vx = 16 sin 60 = 8.0 m s-1 Δ x = ux t = 8.0 1.5 = 12 m → PHYTR14B_GUIDELINES Page 5 Question 23 (4 marks) 23 (a) (3 marks) Outcomes Assessed: Targeted Performance Bands: 2-5 Criteria Correct substitution into the correct equation to obtain an answer Correctly determining r3, but not r or Incorrectly substituting T = 96 into the correct equation, producing an answer Other incorrect substitution into the correct equation or Correctly identifying that T = 96 60 seconds Marks 3 2 1 Sample answer GM r3 2 T 4 2 r3 96 60 2 6.67 10 11 6.0 10 24 4 2 r 3 3.363 10 20 r = 6.954 10 m [= 7 000 km] 6 23 (b) (1 mark) Outcomes Assessed: H5, H14, H15 Targeted Performance Bands: 1-2 Criteria Correct substitution into an appropriate formula yielding an answer Sample answer 2r 2 6.954 10 6 v v 96 60 T v 7.586 10 3 m s 1 7.6 km s 1 Mark 1 Question 24 (4 marks) 24 (a) (2 marks) Outcomes Assessed: H1, H3, H7, H9 Targeted Performance Bands: 3-5 Criteria Both principles correctly stated One principle correctly stated Marks 2 1 Sample answer (The relativity principle): The laws of physics are the same in all inertial frames of reference. (Constant speed of light): Light travels through a vacuum at a fixed speed c, independent of the source or the observer. 24 (b) (1 mark) Outcomes Assessed: H7, H10 Targeted Performance Bands: 2-3 Criteria A correct statement relating to either of Einstein’s postulates Mark 1 Sample answer For the speed of light in a vacuum to be unchangeable, length and time cannot also be fixed. As objects travel at relativistic speeds their length is observed shorter and their time dilated. PHYTR14B_GUIDELINES Page 6 24 (c) (1 mark) Outcomes Assessed: H2, H3, H8, H11 Targeted Performance Bands: 5-6 Criteria A correct definition involving the speed of light in a vacuum Mark 1 Sample answer One metre is now defined to be the distance light will travel through a vacuum in 1/c seconds, where c is the velocity of light in a vacuum, i.e. 2.99792 108 m s-1. Question 25 (8 marks) 25 (a) (4 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 1-5 Criteria A correct label for both axes, and Correct and uniform scaling for both axes, and Marking in all five points correctly (within 0.5 of a unit either direction), Drawing a neat, straight, appropriate line of best fit Successfully completing three of the required outcomes Successfully completing two of the required outcomes Successfully completing one of the required outcomes Marks 4 and 3 2 1 + 16 + + 12 + + 8 4 0 Reading on the balance [ 10-3 kg] Sample answer 0 0.50 1.00 1.50 2.00 Current through the wire [A] 25 (b) (1 mark) Outcomes Assessed: H9, H12 Targeted Performance Bands: 2-3 Criteria Correctly identifying the value of the drawn line of best fit when I = 0 Mark 1 Sample answer The wire’s.mass is 2.5 10-3 kg; I = 0 (no magnetic force) when the graph line is at that point. PHYTR14B_GUIDELINES Page 7 25 (c) (3 marks) Outcomes Assessed: H H9, H13 Targeted Performance Bands: 3-6 Criteria Determination of the slope of the graph as F/I, and attempting to find it Conversion of the weight force into newtons Identification of the direction of the magnetic field One of the above required outcomes is missing One of the above outcomes is present Marks 3 2 1 Sample answer The slope of the line of best fit is found as: m 15.3 10 3 2.5 10 3 6.4 10 3 kg A 1 I 2.00 0 This represents the rate of change of the apparent mass of the wire as the current varies, but since the scale is measured in kg it must be converted to newtons, i.e. FB 6.4 9.8 10 3 6.272 10 2 N A 1 I FB B 6.272 10 2 0.40 B B 0.1568 [ 0.16] T Since FB B I I At any time the scale must apply an upward force that balances the weight of the length of wire plus the magnetic force, which in this case is clearly downwards. The direction of the current is due north, so the direction of the magnetic field must be due east. Question 26 (6 marks) Outcomes Assessed: H3, H4, H7, H13, H16 Targeted Performance Bands: 1-6 Criteria Response includes correct comparisons of resistivity, density and stress/strain Response explains preferred use of copper or aluminium in distinct situations Response lacking one or two of the facets of the above outcomes Identifies two differences in the usage of copper and aluminium wires or cables Makes one correct statement about copper or aluminium based on the given data Marks 5–6 3–4 2 1 Sample answer The significantly lower resistivity of copper means that for any particular length and diameter of wire there is lower energy loss if it is used to carry electric current rather than aluminium. In homes, workplaces and commercial locations this is favourable because no other factors need consideration. However, electricity generators are not popular in residential areas as they are noisy, always lit up, polluting (due to fuel wastes and wasted heat), and work continuously 24 hours each day, 365 days per year. With vast amounts of electric power to be transferred over long distances from generators to cities for use, other factors do need to be considered. Since the power loss in long-distance cables is high, and PLOSS = I2 R it is essential to minimise the current carried by each cable, whilst maintaining the total power supplied. To permit this the cables’ potential is high, so the wires are strung between tall pylons, and heavily insulated against short-circuiting with the ground, the pylons, or other objects such as trees. Pylons are expensive to construct and maintain, so minimising their number is economically important. PHYTR14B_GUIDELINES Page 8 Copper’s density is 3.3 times that of aluminium, so it is much harder to lift, carry and work with, and more pylons are needed to sustain copper cables than otherwise identical aluminium ones. Furthermore, the stress:strain ratio of aluminium is about half that of copper, meaning it sags far less, also reducing the number of pylons required to support those cables. Question 27 (4 marks) 27 (a) (2 marks) Outcomes Assessed: H3, H7 Targeted Performance Bands: 2-4 Criteria An appropriate labelled diagram A correct description of the principle of mutual induction An unlabelled, but otherwise correct diagram of a two-coil transformer An outline of the principle of mutual induction Marks 2 or 1 Sample answer The principle of mutual induction for transformers is that for two solenoids that are linked, a change in the magnetic flux threading one coil induces an emf within the other coil (in accordance with Faraday’s law), and the induced emf is generated such that it opposes the initial flux change in the first coil (in accordance with Lenz’s law). Coil 1 Coil 2 Switch ~ AC source Galvanometer 27 (b) (1 mark) Outcomes Assessed: H7 Targeted Performance Bands: 1-2 Criteria Correct substitution into the correct formula, giving an answer Mark 1 Sample answer VP n n 240 P P 12.24 12 VS nS nS 19.6 27 (c) (1 mark) Outcomes Assessed: H9, H14 Targeted Performance Bands: 4-5 Criteria Identification of the name rectification, or a description of the output Mark 1 Sample answer In order for the AC mains supply to become DC it is necessary to rectify the AC output of the transformer. A diode could be used for the purpose. PHYTR14B_GUIDELINES Page 9 Question 28 (5 marks) Outcomes Assessed: H1, H4, H13, H16 Targeted Performance Bands: 1-6 Criteria At least one example supporting the opinion of the article At least one example opposing the opinion of the article At least one clear reference to the Einstein-Planck ‘disagreement’ A stated assessment of the correctness of the article provided A substantial response, but lacking in one or two of the above One clear example either supporting or opposing the article’s comments or One statement in support of the article and one opposing it A statement relating the article’s contents to the Einstein-Planck ‘disagreement’ Marks 5 3–4 2 1 Sample answer Einstein was a renowned believer that human knowledge belongs equally to all people, and that all research discoveries should be available to anyone interested, especially to those who are intending research along similar lines. In this way the trend towards advancing knowledge is most likely to be maximised. There is a common saying about not needing to “reinvent the wheel”, but whenever the results of research are concealed others must independently follow the same path to rediscover similar outcomes, instead of being able to progress from there. The great inventor Nikola Tesla was determined that his genius should be useful for all, so he did not patent a number of his greatest inventions which could have made him an extremely wealthy man. Unfortunately, as we know from coursework, those whom he worked for were not of the same inclination. They became very rich, while Tesla died in poverty. The First World War began a century ago. At that time there was an intense patriotic feeling in Germany, that discoveries of all kinds belonged to the Fatherland, and Max Planck supported this, at least at the time. Nowadays, however, there is a considerable number of research journals available to readers, allowing those interested to find a treasure of articles associated with what they may intend to investigate. Intense efforts are made by researchers to have articles included in the prestigious journals, and it is similarly a requirement to include supporting material from a large array of articles found within the journals to support or challenge the findings of investigation in order to have an article accepted for publication. This would certainly appear to dispute the claims made in Mr Trounson’s article. However, in today’s Australia, there is a struggle for research funding. Government money is available for research of all kinds, but it is limited. Grants are supervised by independent Boards, but they are competitive. In 2013 applications were invited for grants to improve the interest in science education in schools right across Australia, due to the National Syllabus, and because the standard of science and maths education here was seen to be slipping – but it was competitive! Instead of cooperation between universities, sharing ideas and selecting the best of them so all researchers could work together, the secrecy certainly limited the potential. Research money is also available from industry as well as military budgets, but it is rarely in the interests of industry to share research information before exhausting every avenue to gain financial benefits from what they have paid for, while discoveries made via military research are almost inevitably top secret, to be revealed only during conflict. Just as in those days of World War 1, discoveries such as the Haber process to make nitrates, and hence gunpowder, from air, are very closely-guarded secrets. PHYTR14B_GUIDELINES Page 10 Mr Stone, and Mr Trounson as the author of the article, seem to be correct insofar as Australia is concerned, though whether we differ from other countries in this is questionable. Question 29 (6 marks) 29 (a) (1 mark) Outcomes Assessed: H9 Targeted Performance Bands: 2-3 Criteria Eight straight lines evenly separated, with arrows directed towards the 0 V plate Mark 1 Sample answer +30 V 0V 29 (b) (2 marks) Outcomes Assessed: H9, H10 Targeted Performance Bands: 1-3 Criteria Correct substitution into the correct formula yielding an answer Correctly identifying the direction of the force One of the above Sample answer qV FE q E d and Marks 2 1 1.602 10 19 30 FE 2.0 10 16 N 2 2.4 10 29 (c) (1 mark) Outcomes Assessed: H10 Targeted Performance Bands: 4-5 Criteria Correct substitution into the correct formula giving an answer, direction not needed Mark 1 Sample answer The impulse of the force is given by: Fd 2.0 10 16 4.5 10 2 Impulse F t Impulse 1.67 10 24 N s 6 v 5.4 10 PHYTR14B_GUIDELINES Page 11 29 (d) (2 marks) Outcomes Assessed: H1 Targeted Performance Bands: 2-4 Criteria Correct substitution into the correct formula yielding an answer Correctly identifying the direction of the magnetic field One of the above and Marks 2 1 Sample answer FB = B q v For the electron to not be deviated, FE ↑ = FB ↓ FB = 2.0 10-16 N ↓ 2.0 10-16 = B 1.602 10-19 5.4 106 = 8.65 10-13 B B = 2.312 10-4 T. It must be directed due south to balance the electric force Question 30 (4 marks) 30 (a) (1 mark) Outcomes Assessed: H1, H3, H10 Targeted Performance Bands: 1-2 Criteria Response that correctly identifies a blackbody radiation curve Mark 1 Sample answer The controversial graph is a blackbody radiation curve, showing the energy emitted by an ideal blackbody of a fixed temperature across all wavelengths of electromagnetic radiation. 30 (b) (1 mark) Outcomes Assessed: H1, H3, H10 Targeted Performance Bands: 3-4 Criteria Suitable correct answers for both axes required to gain this mark Mark 1 Sample answer The x-axis represents the wavelength of the e-m radiation being emitted. The y-axis represents the amount of energy emitted by the blackbody at the particular wavelength. (“Intensity” would also be acceptable.) PHYTR14B_GUIDELINES Page 12 30 (c) (2 marks) Outcomes Assessed: H1, H2, H3, H10 Targeted Performance Bands: 4-6 Criteria Response describes the major difference between experimental and theoretical curves and The significance of the theoretical curve A correct statement about a difference between the two graphs Marks 2 1 Sample answer The graph indicates the experimental relationship between the identified variables. It shows that for any ideal blackbody at a fixed temperature, the amount of radiation emitted increases steadily for shorter wavelengths up to a specific value, called the dominant wavelength, beyond which the energy released becomes less with decreasing wavelength, approaching zero emission as the wavelength approaches zero. The dominant wavelength depends only on the surface temperature of the radiating blackbody. It was controversial because no theory based on the Classical physics at that time was able to generate a curve matching the actual blackbody curve. There was no maximum, so the curve continued to rise exponentially as the wavelength became shorter, even in the case of very cold blackbodies, which apart from not matching experiment was also not logical, and did not obey the fundamental Law of Conservation of Energy. Question 31 (4 marks) 31 (a) (1 mark) Outcomes Assessed: H3 Targeted Performance Bands: 1-2 Criteria Response identifies germanium Mark 1 Sample answer The first semiconductor in common use was germanium. 31 (b) (2 marks) Outcomes Assessed: H1, H3, H5, H13 Targeted Performance Bands: 1-3 Criteria Outlines two correct reasons why silicon replaced germanium for common use Identifies one reason why silicon replaced germanium Marks 2 1 Sample answer Germanium is a very rare element, so much so that although when developing his periodic table of the elements Mendeleev predicted its existence and several of its physical and chemical properties, it had not been identified at that time. By contrast, silicon is abundant, the second most common element in the Earth’s crust. For this reason, silicon is far cheaper than germanium. Furthermore, germanium loses its semiconducting properties at fairly low temperatures, becoming a regular electrical conductor, whereas silicon retains its semiconducting identity to far higher temperatures. PHYTR14B_GUIDELINES Page 13 31 (c) (1 mark) Outcomes Assessed: H1, H3, H10, H13 Targeted Performance Bands: 4-5 Criteria Response describes the actual series of electron jumps making the hole ‘migrate’ Mark 1 Sample answer An electron hole is a location in a crystal where an electron is “missing”. Frequently they occur at places where an electron absorbs a photon, and gains sufficient energy to escape into the conduction band of the crystal. They also occur in crystals doped with a p-type impurity when an atom of the dopant shares electrons with surrounding atoms, but has insufficient to covalently bond with all its neighbours. An electron from a surrounding atom can jump into the vacant hole, in which case it leaves a new hole where it came from, which may be filled in its turn by another electron – which now creates another hole, and so on. Electrons are the actual moving charges, but it is the hole that “migrates” through the crystal in this manner. PHYTR14B_GUIDELINES Page 14 Section II Question 32 – Medical Physics (25 marks) 32 (a) (i) (1 mark) Outcomes Assessed: H2, H10 Targeted Performance Bands: 1-3 Criteria Four days Mark 1 Sample answer The graph shows a repeated decrease in quantity by half every four days. 32 (a) (ii) (2 marks) Outcomes Assessed: H1, H3, H7 Targeted Performance Bands: 2-3 Criteria Response concludes that a proton is likely to change into a neutron and This conclusion is based on correct and stated evidence A response stating that a positron is likely to be emitted Marks 2 1 Sample answer Since the only stable isotope of iodine is I-127, which has 53 protons and 74 neutrons, whilst unstable I-124 has 53 protons and only 71 neutrons, it clearly has insufficient neutrons to balance the number of protons in its nucleus. In such cases, one proton may change into a neutron by releasing a positron and a neutrino. 32 (a) (iii) (1 mark) Outcomes Assessed: H10 Targeted Performance Bands: 1-2 Criteria Positron Emission Tomography Mark 1 Sample answer Positron Emission Tomography uses positrons emitted as protons are converted to neutrons. 32 (b) (i) (2 marks) Outcomes Assessed: H1, H2, H3, H10 Targeted Performance Bands: 1-2 Criteria Two correct substitutions into the correct formula yielding an answer One correct substitution into the correct formula yielding an answer Marks 2 1 Sample answer Z = ρ v ZM = 1075 1590 = 1.71 106 kg m-2 s-1 (Rayls) ZB = 1908 4080 = 7.78 106 kg m-2 s-1 PHYTR14B_GUIDELINES Page 15 32 (b) (ii) (2 marks) Outcomes Assessed: H3, H10, H12, H14 Targeted Performance Bands: 3-6 Criteria Correct substitution into the correct formula yielding an answer and The correct analysis of the result leading to the correct conclusion One of the required outcomes correct Sample answer Z 2 Z1 IR I0 Z 2 Z1 2 Marks 2 1 2 IR 7.78 10 6 1.71 10 6 0.409 2 I0 7.78 10 6 1.71 10 6 0.387 Percentage variation from normal 100 94.6% Bone loss is not significant. 0.409 2 32 (b) (iii) (2 marks) Outcomes Assessed: H3, H11, H12, H13 Targeted Performance Bands: 1-2 Criteria The response describes the difference between A-scans and B-scans Identifies why there would have been an advantage using a B-scan Identifies a difference between ultrasound A-scans and B-scans and Marks 2 1 Sample answer A-mode ultrasound scans cannot produce images. Instead, the oscilloscope detector registers a “spike” as each echo returns after reflection from tissue boundaries. The intensity of the echo is displayed as the amplitude of the spike. The time delay is measured very accurately. In B-scans signals are sent in many different directions. The intensity and time information from the echoes are recorded as spots of different brightness on a TV screen. The location of the spot depends upon the direction the signal was sent and the time delay, so images are made as the information provided by each point combines. Using a B-scan images proceeding from within the bones of the cosmonauts would have provided more detailed information. PHYTR14B_GUIDELINES Page 16 32 (b) (iv) (3 marks) Outcomes Assessed: H1, H11, H13, H14 Targeted Performance Bands: 4-5 Criteria The acoustic impedances of both air and skin are calculated and The ratio of reflected U/S to the original signal is found and A correct conclusion is reached regarding the use of a coupling gel Two of the above outcomes are achieved Correct substitution into one correct formula to yield an answer Sample answer Z = ρ v ZAIR = 1.3 330 = 429 kg m-2 s-1 ZSK = 1040 1540 = 1.60164 106 kg m-2 s-1 Z 2 Z1 IR I0 Z 2 Z1 2 Marks 3 2 1 2 I 1.60164 10 6 429 R 99.9% 2 I0 1.60164 10 6 429 When ultrasound passing through air reaches the skin, almost all the signal is reflected, only 0.1% enters the skin. Also, when the reflected signal carrying data returns to the skin to cross to the transducer, for the same reason only 0.1% of the signal reaches the transducer. For this reason a coupling gel must always be placed between the transducer and the skin. 2 32 (c) (i) (1 mark) Outcomes Assessed: H10 Targeted Performance Bands: 1-2 Criteria Answer gives a clear description of the proton’s ability to act as a magnet Mark 1 Sample answer Since hydrogen has a single proton it has a net magnetic moment because that proton spins as the nucleus is spinning. This allows the nucleus to act as a tiny magnet. 32 (c) (ii) (4 marks) Outcomes Assessed: H1, H3. H8 Targeted Performance Bands: 3-6 Criteria Response describes the precession of H-nuclei in a strong magnetic field A radio signal at the resonant frequency transfers energy to bodily tissues This energy is released when the signal stops, and analysed by computer Movement of the external signal source creates a sequence of images One or two of the processes unclear or omitted One of the above processes outlined Marks 4 2–3 1 Sample answer MRI scans produce detailed, high-quality images, especially of soft body tissues. Patients are located on a horizontal platform which slides into a large cylinder within which a very strong magnetic field is present. When subjected to a very powerful external magnetic field many of the hydrogen nuclei in the body of the patient undergoing the MRI scan start to precess (wobble) around the field at their special resonant frequency, known as the Larmor frequency. An external radio signal, the gradient field, with precisely the Larmor frequency, are directed PHYTR14B_GUIDELINES Page 17 at the body of the patient. Though these signals are weak, because they are at exactly the same frequency they produce resonance, so the hydrogen nuclei gain a great deal of energy, causing their axis of rotation to change. Whenever the radio signal stops the nuclei return to their original axis of rotation. During this relaxation process they release energy as radio waves that are detected and converted into signals analysed by a computer. The hydrogen in different tissues relaxes at distinct rates, so different signals are transmitted, producing images of high clarity. Since hydrogen is so common throughout the body the scanning technique can be used everywhere. As the radio transmitter moves along the body it causes responses from the hydrogen in the body tissues in sequence, producing a series of 2-dimensional “slices”. 32 (c) (iii) (1 mark) Outcomes Assessed: H1, H4, H12, H14 Targeted Performance Bands: 2-3 Criteria A correct advantage of the MRI technique is identified Mark 1 Sample answer All the other investigative techniques mentioned require the use of very high frequency electromagnetic radiation; CAT scans use X-rays and PET uses γ-radiation. MRI does not use this, so it is intrinsically less hazardous, and will not damage patients unless they have certain types of metallic implants in their bodies, such as pace-makers. 32 (d) (6 marks) Outcomes Assessed: H1, H3, H4, H13 Targeted Performance Bands: 1-6 Criteria Describes advantages of computer processing of the many CAT images CAT scans work well for both hard and soft tissue Describes higher risks in CAT images, and other drawbacks Gives an assessment based on the identified impacts Good explanation of the MRI process, but misses identified feature(s) Correctly identifies two features of CAT scans as above Correctly identifies one feature of CAT scans Marks 5–6 3–4 2 1 Sample answer Following the discovery of X-rays in 1895 and the publication of their capacity to “see inside” bodies, they started being used throughout the world, even before their exact nature had been established. Initially they were used mainly to check for broken bones, but a large number of other uses have developed since then. Now we have CAT scans. Images produced by this modern branch of X-ray technology leave the old system in the dark, as seen on the X-ray image introducing this question. The normal X-ray shown in Image 1 was captured by a single exposure, but the CAT scan (Image 2) consists of multi-images captured by a rotating gantry camera. Although both techniques are used successfully in brain scans when searching for signs of tumours, a CAT scan is able to identify possible tiny new tumours by scanning the patient a second time a half-hour after being injected with an X-ray fluorescent dye that moves with the blood to the site of a developing tumour. CAT images can be computer-processed to form 2-D and 3-D, whereas the single image made by an X-ray cannot be further processed. There is no doubt that human society has benefited from the appearance of this technology. PHYTR14B_GUIDELINES Page 18 However, it is not all good news. The hazardous nature of X-rays has often been ignored, but it does exist, and CAT scan images are the result of numerous X-rays, the data from which is compiled by a computer, as there is a vast amount of data to store and many calculations to be performed to prepare the images. CAT scans are far more expensive than regular X-ray images, so some patients may be unable to afford them. Meanwhile, the government health care system is placed under increased pressure by subsidising many CAT scans when X-rays would have been sufficient. However, in spite of the increased potential hazards of CAT scans, in many situations they can provide comforting certainty to those fearing cancer, or at least earlier detection with an improved chance of survival, so in all the benefits far outweigh the drawbacks. PHYTR14B_GUIDELINES Page 19 Question 33 – Astrophysics (25 marks) 33 (a) (i) (1 mark) Outcomes Assessed: H1, H2, H3 Targeted Performance Bands: 1-3 Criteria Correct substitution into the correct formula yielding an answer Sample answer mE mH LH 100 5 LE L H 100 LE 6 20 5 100 2.8 Mark 1 . LH 2.512 10 6 LE The brightest objects that Hubble camera can observe without being “blinded by the light” have a luminosity just 2.5 millionths of that of the dimmest objects visible by human eyes. 33 (a) (ii) (2 marks) Outcomes Assessed: H7, H8, H11 Targeted Performance Bands: 3-5 Criteria Clear explanation why direct sunlight does not enter the camera Clear explanation why scattered light does not enter the camera One of the above outcomes and Marks 2 1 Sample answer Unlike ground-based telescopes, those above the atmosphere do not have problems with light being scattered off air molecules (unless they are observing Earth). For that reason, provided their target is in a direction sufficiently apart from the Sun, the Moon and the planets, once set in a particular direction the camera will continue to receive light from the sources it observes. Even though there is no “night” for an orbiting telescope, with the Sun always visible unless it is temporarily eclipsed by the Earth, while the camera is aimed in a different direction light from the sun will not enter it. 33 (b) (3 marks) Outcomes Assessed: H1, H3, H11, H14 Targeted Performance Bands: 2-5 Criteria Two correct reasons for telescopes on high mountains outlined One correct reason outlined or Two reasons identified One reason stated Marks 3 2 1 Sample answer There are several reasons why telescopes are frequently sited at the top of high mountains. One of these is light pollution; so much scattered light from buildings in large cities produces a “noise” reducing the sensitivity and resolution of telescopes. Air molecules do cause some scattering, but more is caused by exhaust gases, dust and other particles polluting the air, but if the telescope is located where there is less air and little pollution the problem is reduced. A second factor that improves “seeing” from the top of high mountains is that there is far less water vapour there. Water vapour also causes scattering, but more significantly with regard to telescopes, it causes cloud and rain. The air is colder at greater heights, and as shown in the photograph, the clouds (and rain) are usually at levels below, so viewing is not impeded. PHYTR14B_GUIDELINES Page 20 33 (c) (i) (2 marks) Outcomes Assessed: H1, H2, H3, H7 Targeted Performance Bands: 2-3 Criteria Determining the colour index of the star to be +0.29 Using the table to estimate the spectral class of Alderamin to be A6 or A7 Determining the colour index of Alderamin to be +0.29 or Incorrectly finding a CI, but correctly using that CI to find a spectral class Marks 2 1 Sample answer CI = mB – mV CI = 2.73 – 2.44 = 0.29. The CI of an A0 star is exactly 0; the CI of an F0 star is +0.4. Hence Alderamin is an A6 type. 33 (c) (ii) (2 marks) Outcomes Assessed: H1, H2, H3, H14 Targeted Performance Bands: 2-4 Criteria If the star is determined to be an A6 or A7, estimating from +1 to +1.5 or If the star was found to be anything else, estimating a range of ± 0.3 If the star is determined to be an A6 or A7, estimating from 0 to +2 If the star was found to be anything else, estimating a range of ± 1 Marks 2 1 Sample answer Using the centre line of the given H–R diagram, and estimating the location of an A6 star, the corresponding value of its absolute magnitude would be +1. 33 (c) (iii) (1 mark) Outcomes Assessed: H1, H2, H3 Targeted Performance Bands: 2-6 Criteria Substituting the appropriate values into the correct formula to get an answer Mark 1 Sample answer d d 1.44 d 1 2.44 5 log 10 log 10 10 10 5 10 The distance of Alderamin from Earth is therefore approximately 19.4 parsecs. M m 5 log 10 PHYTR14B_GUIDELINES Page 21 33 (c) (iv) (1 mark) Outcomes Assessed: H1, H2, H3, H14 Targeted Performance Bands: 2-4 Criteria Correctly naming the process spectroscopic parallax Mark 1 Sample answer The name of this process is spectroscopic parallax. 33 (d) (3 marks) Outcomes Assessed: H2, H8, H11 Targeted Performance Bands: 3-6 Criteria States that the F-star will be more luminous than the K-star and Its dominant wavelength will be bluer than that of the K-type and There will be differences in the absorption spectral lines of the stars Identifies two differences as above or Identifies all three, but without stating which star has which properties Identifies one difference between the star types Marks 3 2 1 Sample answer The spectrograms of stars are composed of their blackbody radiation curve superimposed by a series of dark absorption lines that appear like troughs of lower intensity. The blackbody curve depends purely on the surface temperature of the star. An F-type star is yellow-white, with a surface temperature of around 7500K, whereas a K-type is orange, with a surface temperature around 4000K. Consequently, the F-type has far greater luminosity (around 16 times greater) than that of the K-type. Furthermore, the dominant wavelength of the F-type will be located towards the centre of the visible spectrum, around green or blue, while the dominant wavelength of the K-type would be located at the red end of the spectrum. The other major difference will be in the dark absorption lines distinguishing the spectra. An F-type star will display clear hydrogen lines amongst others. The hydrogen lines in the K-type star’s spectrum will be far less prominent, whereas there will be clear lines of various metals such as calcium. 33 (e) (i) (2 marks) Outcomes Assessed: H2, H13, H14 Targeted Performance Bands: 3-5 Criteria Correctly converts AU into metres and days into seconds and Correctly substitutes into the correct formula leading to an answer One of the above outcomes Sample answer The combined mass of the two stars is found using: M1 M 2 4 2 r 3 GT 2 M1 M 2 4 2 0.12 1.5 1011 Marks 2 1 3 6.67 10 4.015 24 3600 11 2 = 2.868 1031 kg PHYTR14B_GUIDELINES Page 22 33 (e) (ii) (2 marks) Outcomes Assessed: H1, H3, H14 Targeted Performance Bands: 1-4 Criteria A correct light-curve, having 2 complete cycles Appropriately labelled axes and eclipses, and correct time scale An appropriate light-curve or Appropriately labelled axes and correct time scale Marks 2 1 Luminosity Sample answer Seen from the edge the light-curve of the largest stars of the Spica binary would appear as: Larger star eclipses smaller 0 2 Smaller star eclipses larger 4 6 8 Time (days) 33 (f) (6 marks) Outcomes Assessed: H1, H3, H5, H9, H11, H12, H13, H14 Targeted Performance Bands: 1-4 Criteria A response that describes at least two assumptions about stars in clusters The logical basis for the identified assumptions How at least one general property of stars was developed from these How star evolution or evolutionary paths were inferred from these A response that follows the above, but lacks details linking assumptions to observations to inferences in two or three places Two of the required outcomes clearly stated One property of star clusters identified Marks 5–6 3–4 2 1 Sample answer All the stars in a cluster are assumed to have formed from the same nebula. Consequently, they are considered to be composed of the usual ratio of hydrogen and helium, as well as the same mixture of “metals” in the same proportions, and to have been “born” at approximately the same time, as well as being effectively the same distance from Earth. These assumptions provide numerous advantages for astronomers when determining the properties of stars, as demonstrated below. If all the stars in a cluster are the same distance away the distance factor plays no part in their observed brightness, so astronomers were able to relate the brightness of a star to its spectral class, and therefore to its colour, allowing the arrangement of stars into O, B, A, F, G, K and M classes, not only because of the intensity of the hydrogen absorption lines as was originally applied, but also in terms of their luminosity. In this way it was also found that a few K- and M-type stars appeared to be far too luminous relative to others having the same colour, so the concept of red giant stars was developed. In a similar manner, white dwarfs were recognised because a few B-, A- and F-types located in clusters were found to have much lower luminosity than their normal counterparts. PHYTR14B_GUIDELINES Page 23 Assuming all the stars in a cluster have effectively the same material in the same proportions means that the differing absorption spectral lines observed in their individual spectra must be caused by factors such as surface temperature, adding a further factor into the arrangement of stars into their spectral classes. Assuming that all the stars in a cluster were “born” at roughly the same time allowed the very significant interpretation to be made that the brightest stars also exhaust their fuel much faster than their dimmer siblings, because there are far fewer O- and B-type stars present in older clusters relative to those present in newer clusters, while there are also far more red giants and white dwarfs present in older clusters. From the data gathered the evolutionary path of stars of the different spectral classes was gradually understood. PHYTR14B_GUIDELINES Page 24 Question 34 – From Quanta to Quarks (25 marks) 34 (a) (i) (2 marks) Outcomes Assessed: H1, H2 Targeted Performance Bands: 2-4 Criteria Correct substitution into Rydberg’s formula to find an answer for 1/λ and Subsequent processing to find the photon energy in either J or eV One correct process leading to an answer Marks 2 1 Sample answer 1 1 1 R 2 2 ni n f 1 1 1 1.097 10 7 2 2 1 5 1.053 10 7 m 1 c f 3.0 108 1.053 107 3.16 1015 Hz E h f 6.63 1034 3.16 1015 2.095 1018 J [ 13.08 eV] f 34 (a) (ii) (2 marks) Outcomes Assessed: H1, H2, H14 Targeted Performance Bands: 2-5 Criteria Identifying that nf must be 2 Correctly substituting into the correct formula to yield an answer Substituting incorrect data into the correct formula Marks 2 1 Sample answer For the photon to be visible it must belong to the Balmer series, i.e. nf = 2. 1 1 1 1 1 1 R 2 2 1.097 107 2 2 2.303 106 m 1 ni 5 2 n f λ = 434 nm 34 (b) (3 marks) Outcomes Assessed: H1, H3, H13 Targeted Performance Bands: 3-6 Criteria Original concepts of electron orbits and orbitals are identified Heisenberg’s uncertainty principle is stated and/or explained in this context A diagram representing some feature of the probability concept included One of the above steps either not present, or inappropriately considered An attempt made to relate the principal to orbits by diagram or writing Marks 3 2 1 Sample answer Heisenberg’s uncertainty principle had the effect of changing the concept of particle-electrons moving in circular (Bohr) or elliptical orbits (Pauli) with integral numbers of wavelengths (de Broglie), becoming almost like cloud-electrons. The electrons’ mass is so small and their speed so great their location becomes impossible to ascertain, becoming instead like a shadow picture representing the probability of finding the electron at a point at any instant, as represented in 2-D in this diagram. PHYTR14B_GUIDELINES Page 25 34 (c) (i) (1 mark) Outcomes Assessed: H3, H10, H13, H14 Targeted Performance Bands: 1-5 Criteria The difficulty in determining the chemical identity of product nuclei explained Mark 1 Sample answer Bombardment of nuclei with α-particles, or even neutrons, does not always lead to a change in the target nucleus. Even if it were 100% successful every time, there is such an enormous number of nuclei of any substance present in even a microscopic sample that it would take years for the number of successfully transmutated nuclei to be detectable by chemical means. 34 (c) (ii) (2 marks) Outcomes Assessed: H1, H3, H10, H11 Targeted Performance Bands: 2-3 Criteria Diagram showing a collision of two particles, with two different particles afterwards and At least 2 particles correctly identified One of the outcomes identified above Marks 2 1 Sample answer Nitrogen nucleus Oxygen-17 nucleus Combined nucleus α-particle proton 34 (c) (iii) (2 marks) Outcomes Assessed: H10, H13 Targeted Performance Bands: 2-4 Criteria Response gives a completely balanced nuclear equation Response gives a nuclear equation that is not balanced Marks 2 1 Sample answer 4 2 14 7 N 18 9 PHYTR14B_GUIDELINES F 11 p 17 8 O Page 26 34 (d) (i) (1 mark) Outcomes Assessed: H2, H10, H14 Targeted Performance Bands: 2-3 Criteria Mark 1 Identifies the antineutrino Sample answer The missing reaction product is an antineutrino. 34 (d) (ii) (2 marks) Outcomes Assessed: H1, H7, H19 Targeted Performance Bands: 3-5 Criteria Correctly determines the mass defect and Uses a correct method to evaluate the energy released (in joules) Correctly applies one of the above outcomes Marks 2 1 Sample answer The mass of the reactant (one neutron) is 1.008665 amu. The mass of the products is 1.007276 + 0.000549 = 1.007825 (ignoring the antineutrino). Hence the mass defect is 8.4 10-4 amu. The energy released in this reaction is therefore 8.4 10-4 931.5 = 0.7825 MeV 1MeV = 1.602 10-13 J, so the energy release is 0.7825 1.602 10-13 = 1.25 10-13 J. 34 (d) (iii) (2 marks) Outcomes Assessed: H1, H2, H10 Targeted Performance Bands: 2-4 Criteria Response correctly identifies two properties of an anti-proton Correctly identifies one property of an anti-proton Marks 2 1 Sample answer An anti-proton will have precisely the same mass as a proton. Its charge will be negative, identical to that of an electron and opposite to that of a proton. d 34 (d) (iv) (1 mark) Outcomes Assessed: H1, H2, H5, H10 Targeted Performance Bands: 3-4 Criteria d u d Diagram shows that the anti-proton has 2 anti-up and 1 anti-down quarks u Sample answer An anti-proton: d u represents an anti-up quark d represents an anti-down quark Mark 1 u u u d PHYTR14B_GUIDELINES u u Page 27 34 (e) (7 marks) Outcomes Assessed: H1, H2, H3, H10, H11, H13, H15, H16 Targeted Performance Bands: 1-6 Criteria Historical development of the atomic model Problems with the Rutherford atomic model Discussion of Balmer’s (and others’) emission spectral lines Development of the quantum theory Description of the Bohr atomic model What the Bohr model explained successfully What the model was unable to successfully explain A sound consideration of the issue, lacking one or two of the listed issues A fair consideration of the issue, using statements rather than descriptions Identifying any two of the issues listed above A correct statement referring to the issue Marks 7 5–6 3–4 2 1 Sample answer Following Thomson’s experimental proof of the existence of electrons as part of atoms, it took little time before Rutherford’s gold leaf experiment led him to propose a new model of an atom. It defied Maxwell’s laws of electromagnetism, because electrons were postulated to orbit around a central positively charged nucleus, kept in orbit by their electrostatic attraction, but since they were accelerating charges they had to radiate e-m energy, causing the atom to collapse. Balmer, a Swedish maths teacher, noticed a mathematical relationship for the wavelengths of the visible spectral lines of excited hydrogen. Lyman and Paschen discovered similar series relationships for hydrogen’s U-V and infra-red lines too, and Rydberg united all three with a single formula to predict them, but it seemed an interesting coincidence that worked only for hydrogen atoms. Planck announced his quantum theory in 1900. Although received with great scepticism, with Einstein’s support it had become accepted in 1905. Bohr explained away the problem with Rutherford’s atomic model by applying the quantum theory, postulating that electrons could not gradually lose energy – rather, they could lose (or gain) only fixed quanta. They were normally located within stable energy shells, and could only jump to alternate shells either outwards by absorbing a photon of energy, or inwards by emitting the appropriate photon. Not only did this supply a reason allowing Rutherford’s atom to exist, it explained the spectral lines of Balmer, etc. Bohr used Classical theory wherever possible to explain phenomena, with quantum theory to explain away the difficult points. Being a hybrid theory, it is surprising how long it survived; in fact it is still taught in school chemistry! However, it was unable to explain 4 phenomena. In spite of much time and effort, Bohr’s atom could explain only hydrogen and other singleelectron species, He+, Li2+, Be3+ and so on. It could not explain the differences in intensity and width of spectral lines (called “sharp” – s, “primary” – p and “diffuse” – d). It was unable to explain “hyperfine splitting” of some spectral lines, and although it could explain the Zeeman effect (the splitting of single spectral lines into three separate polarised lines when light passes through a strong magnetic field, it could not explain the ‘anomalous’ Zeeman effect, where the lines could be split into up to 15 separate lines. Eventually, therefore, it was superseded. PHYTR14B_GUIDELINES Page 28