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2015 HIGHER SCHOOL CERTIFICATE MID-YEAR EXAMINATION PHYSICS – MAPPING GRID Exam Section Part A: Multiple Choice Part B: Free Response Question Marks Syllabus/Course Outcomes Content 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5 5 7 3 5 7 6 6 4 7 H6, H9 H9 H6, H9 H9 H11 H9 H1 H2, H9 H2, H9 H2, H9 H2, H9 H9 H7, H9 H7, H9 H9, H10 H9, H10 H10 H10 H10 H9 H9 H2, H9 H6 H6 H9 H1, H2, H9 H4, H7, H9 H2, H9, H10 H2, H9 H3, H4, H9 9.2.1 9.2.1 9.2.1 9.2.2 9.2.2 9.2.2 9.2.4 9.3.1 9.3.1 9.3.2 9.3.2 9.3.2 9.3.3 9.3.4 9.4.1 9.4.1 9.4.2 9.4.3, 9.4.2 9.4.3 9.4 .4 9.2 2 9.2.2 9.2.2 9.2.4 9.3.1 9.3.2 9.3.1, 9.3.5 9.4.1 9.4.3 9.4.4 Targeted Performance Bands 3-5 4-6 2-4 3-5 3–5 2–4 3–5 2–4 4–6 2–4 3-5 3–5 2–4 3–5 3–5 4–6 2–4 4–6 3–5 3–5 3–5 3–5 4–6 2–4 4–6 3–5 3–5 2–5 2–5 3–5 Answer B C D C B D A C B B B D A C B A C D A B 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. MIDPHY15_GUIDELINES Page 1 2015 HIGHER SCHOOL CERTIFICATE MID-YEAR EXAMINATION PHYSICS – MARKING GUIDELINES 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 B C D C B D A C B B B D A C B A C D A B H6, H9 H9 H6, H9 H9 H11 H9 H1 H2, H9 H2, H9 H2, H9 H2, H9 H9 H7, H9 H7, H9 H9, H10 H9, H10 H10 H10 H10 H9 MIDPHY15_GUIDELINES 2 Targeted Performance Bands 3-5 4-6 2-4 3-5 3-5 2-4 3-5 2-4 4-6 2-4 3-5 3-5 2-4 3-5 3-5 4-6 2-4 4-6 3-5 3-5 Part B – 55 marks Question 21 (7 marks) 21 (a) (2 marks) Outcomes Assessed: H9 Targeted Performance Bands: 3 – 5 Criteria Calculates the initial horizontal velocity and initial vertical velocity Partial calculation Marks 2 1 Sample answer ux = u cos = 42 x cos 50 = 27.0 ms-1 . uy = u sin = 42 x sin 50 = 32.2 ms-1 . 21 (b) (2 marks) Outcomes Assessed: H9 Targeted Performance Bands: 4 – 6 Criteria Calculates the height and gives the unit Partial calculation involving the equations of accelerated motion Marks 2 1 Sample answer At top vy = 0 v2 = u2 + 2as 0 = 32.22 + 2 x -9.8 x h h = 52.8 m 21 (c) (2 marks) Outcomes Assessed: H9 Targeted Performance Bands: 4 – 6 Criteria Calculates the velocity and gives the angle Calculates the velocity or the angle Sample answer vx = 27.0 ms-1 Marks 2 1 vy = uy + at = 32.2 + -9.8 x 3 = 2.8 ms-1 . V2 = 27.02 + 2.82 V = 27.1 = 5.9° 21 (d) (1 mark) Outcomes Assessed: H9 Targeted Performance Bands: 3 – 5 Criteria States that V is max at initial and final position Mark 1 Sample answer The velocity of the cannon ball is maximum at the initial position (out of the cannon) and at the final position just before impact. MIDPHY15_GUIDELINES 3 Question 22 (5 marks) 22 (a) (2 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 2 – 4 Criteria Describes how the two masses fall, and gives the reason Outlines how the two masses fall Marks 2 1 Sample answer The two masses, which were the same size but different of weight, fell to the ground in the same time. This is because gravity accelerates all masses, big and small, at the same rate. 22 (b) (3 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 3 – 5 Criteria Identifies that the foam ball will take longer, and gives an explanation involving increased air resistance on the foam due to its surface Identifies that the foam ball will take longer, and mentions air resistance Identifies that the foam ball will take longer OR identifies that air resistance slows the ball Marks 3 2 1 Sample answer The result above would differ because the foam ball would take longer to hit the ground. This is because it is slowed down by air resistance to a greater degree than the smooth metal ball, because it has a rougher surface. Question 23 (7 marks) 23 (a) (2 marks) Outcomes Assessed: H6 Targeted Performance Bands: 2 – 4 Criteria Shows how Kepler’s 3 law links period and radius and planet mass Outlines Kepler’s 3rd law OR links period and radius and planet mass rd Marks 2 1 Sample answer Kepler’s 3rd law states that R3 / T2 = GM/42 which shows that the period of satellite A depends on its orbital radius, and on the mass of the planet it is orbiting. MIDPHY15_GUIDELINES 4 23 (b) (2 marks) Outcomes Assessed: H6 Targeted Performance Bands: 4 – 6 Criteria Correct calculation for radius and orbital period Partial calculation Marks 2 1 Sample answer R3 / T2 = GM/42 R3 = T2 GM/42 R3 = (6 x 60 x 60)2 x 6.67x10-11 x 5.1x1024 /42 = 4.02 x 1021 R = 1.59 x 107 m Orbital speed v = 2r/T = 2 x 1.59 x 107 /(6 x 60 x 60) = 4 625 ms-1 23 (c) (2 marks) Outcomes Assessed: H6 Targeted Performance Bands: 3 – 5 Criteria Identifies that for B radius is less, and gives an explanation using Kepler’s Laws Identifies that for B radius is less Marks 2 1 Sample answer If satellite B has the same orbital period as satellite A, the orbital radius of satellite B will be less than the orbital radius of A. Kepler’s law shows that R3 / T2 = GM/42 so R3 / = T2GM/42 for A, RA 3 = T2GM/42 for B, RB 3 = T2G( ½ M )/42 approximately, giving RB 3 = ½ T2GM/42 which shows that the radius for B is less. 23 (d) (1 mark) Outcomes Assessed: H6 Targeted Performance Bands: 2 – 4 Criteria Outline identifies no change in speed Mark 1 Sample answer The orbital speed would not change, as there is only one value for orbital speed at a certain orbital radius. MIDPHY15_GUIDELINES 5 Question 24 (3 marks) 24 (a) (2 marks) Outcomes Assessed: H6 Targeted Performance Bands: 3 – 5 Criteria Outline identifies that no experiment is possible, and links this to relativity and the absence of an ether Identifies that no experiment is possible Marks 2 1 Sample answer According to Einstein’s special theory of relativity, there is no experiment that the on-board astronaut can do to be able to measure the rockets’ absolute velocity relative to a stationary ether. In fact, the ether has not been detected, which would also preclude such a measurement. 24 (b) (1 mark) Outcomes Assessed: H6 Targeted Performance Bands: 4 – 6 Criteria Mark 1 Correct substitution Sample answer Tv = t0 /[ 1 – v2/c2 ]0.5 = 10/[ 1 – 0.982 ] = 50.3 s Question 25 (6 marks) 25 (a) (4 marks) Outcomes Assessed: H9 Targeted Performance Bands: 3 – 5 Criteria Explanation involving four features; - no current with open circuit - copper is non-magnetic - a closed circuit bringing about a current and a force - the force on a current is BIL Explanation involving three features Outline involving two features Outline involving one feature Marks 4 3 2 1 Sample answer The copper wire will not move when the switch is open, because there is no current flow, and copper is not magnetic so it will not be attracted to the magnetic poles. The copper wire will move when the switch is closed, because there is a current flowing due to the circuit being compete, and a force acts on a current carrying wire in a magnetic field given by F = BIL. MIDPHY15_GUIDELINES 6 25 (b) (2 marks) Outcomes Assessed: H9 Targeted Performance Bands: 3 – 5 Criteria Marks 2 1 Outline involving two features Outline involving one feature Sample answer The copper wire has a magnetic field given by F = BIL = 55 x 10-3 x 3.9 x 0.12 = 0.026 N down the page. Question 26 (7 marks) 26 (a) (2 marks) Outcomes Assessed: H1 Targeted Performance Bands: 2 – 4 Criteria Identifies Michael Faraday and electromagnetic induction Identifies Michael Faraday OR electromagnetic induction Marks 2 1 Sample answer The early electricity pioneer who carries this out was Michael Faraday and the effect is electromagnetic induction. 26 (b) (3 marks) Outcomes Assessed: H1. H2, H9 Targeted Performance Bands: 3 – 5 Criteria Explanation with all three features; - broken current flow producing a changing flux - changing flux inducing an emf - a changing induced current produce a changing magnetic force and a flickering magnetic needle Outline with two features Identifies 1 feature Marks 3 2 1 Sample answer When the switch was constantly changed from on to off a broken current was made to flow in the 1st circuit. This meant that a changing magnetic field was produced. The changing magnetic field cut across the wire in the 2nd circuit and produced an emf, due to induction. The induced current was an alternating current, and this produced an alternating magnetic field in the solenoid coil with the compass needle, making the compass needle move and flicker. MIDPHY15_GUIDELINES 7 26 (c) (2 marks) Outcomes Assessed: H1. H2 Targeted Performance Bands: 2 – 4 Criteria Outline involves verifying results and consistency Outline involves verifying results OR consistency Marks 2 1 Sample answer In order for the results here to become accepted by the scientific community, the experimenter should repeat the experiment many more times and verify that the effect is real, and he should make sure that the results are consistent. Question 27 (6 marks) Outcomes Assessed: H4, H7, H9 Targeted Performance Bands: 3 – 6 Criteria Marks All six features present ; similarities given differences given details on structure details on benefits details on drawbacks appropriate table with headings Five features present Four features present Three features present Outline involving a comparison One fact given 6 5 4 3 2 1 Sample answer Structure Benefits Drawbacks MIDPHY15_GUIDELINES The AC coil motor The AC induction motor Similarity – both use an external voltage source, a magnetic field, and a rotor with a current carrying coil. Similarity – both convert electrical energy into mechanical energy. If the motor gets stuck the current will quickly rise which may cause a burn out / fire. Difference- the induction motor’s rotor coil is not connected to the external voltage supply and runs on induced current. Difference- the induction motor is more reliable since there is no wearing down of brushes. Is slow to get to optimal speed because of lower torque. 8 Question 28 (7 marks) 28 (a) (2 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 2 – 4 Criteria Outline links large voltage to acceleration of cathode rays Mentions large voltage OR acceleration of cathode rays Marks 2 1 Sample answer He used an induction coil to produce a large voltage across the ends of the tube: this was needed to accelerate the cathode rays across the tube. 28 (b) (3 marks) Outcomes Assessed: H2. H9 Targeted Performance Bands: 3 – 5 Criteria Explanation involving the cathode rays bending downward, the electric field E = V/d, and the force Explanation involving the cathode rays bending downward, and the electric field OR the force One true fact about the cathode rays or the plates Marks 3 2 1 Sample answer The cathode rays, when made to pass through electric plates, will bend downward and produce a curved path. This is because there is an electric field E = V/d between the plates which will create a force on the negatively charged cathode rays, attracting them to the positive plate. 28 (c) (2 marks) Outcomes Assessed: H2. H9 Targeted Performance Bands: 3 – 5 Criteria Two inferences about the canal rays One inference about the canal rays Marks 2 1 Sample answer The canal rays are charged particles, sub- atomic in size, and oppositely charged to the cathode rays. They too are formed in the cathode ray tube but travel in the opposite direction of the electric field that drives the cathode rays from cathode to anode. MIDPHY15_GUIDELINES 9 Question 29 (4 marks) 29 (a) (2 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 2 – 4 Criteria Outline shows that the n-type silicon formed from impurities of an identified element and the p-type silicon formed from impurities of an identified element Outline mentions the n-type silicon formed from impurities of an identified element OR the p-type silicon formed from impurities of an identified element Marks 2 1 Sample answer To make p-type silicon, atoms of an element with fewer valence electrons are added as impurities to the silicon. E.g. aluminium. To make n-type silicon, atoms of an element with more valence electrons are added as impurities to the silicon. E.g. phosphorus. 29 (b) (2 marks) Outcomes Assessed: H2, H9 Targeted Performance Bands: 3 – 5 Criteria Explanation mentions the junction field and the small positive voltage producing free conduction for conventional current Outline mentions the junction field OR the small positive voltage producing free conduction for conventional current Marks 2 1 Sample answer At the p-n junction electrons on the n-type jump over to the holes on the p-type, forming an electric field and depletion region in between. This junction field will push negative charges back towards the n-type or right on the diagram, and positive charges are pushed back towards the p-type or left. When a small positive voltage is applied to the p-type, the junction field is nullified and a positive charge (conventional current) will move freely across from left to right, so the diode is effectively a conductor. MIDPHY15_GUIDELINES 10 Question 30 (3 marks) Outcomes Assessed: H3, H4, H9 Targeted Performance Bands: 3 – 5 Criteria Positives and negatives are both outlined, and an overall conclusion is given Positives are outlined, and an overall conclusion is given OR negatives are outlined, and an overall conclusion is given Outline mentions positives or negatives Sample answer High temperature superconductors can be used in electric motors. Positives; - stronger magnetic fields can be produced - higher torque - more efficient use of energy as far less heat is generated Negatives; - more costly to manufacture - needs extra systems and cryogenics to keep temperatures cold Conclusion; Beneficial overall as efficiency and speed are increased, but very costly. MIDPHY15_GUIDELINES 11 Marks 3 2 1