2010 QAT HSC Physics Exam Solutions PDF

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2010 TRIAL HIGHER SCHOOL CERTIFICATE EXAMINATION PHYSICS – MAPPING GRID

Marks

Syllabus/Course Outcomes

Content

Targeted Performance Bands

Answer

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 3 7 6

9.2.1 9.2.2 9.2.3 9.2.2 9.2.2 9.2.4 9.2.2 9.3.1 9.3.1, 9.3.5 9.3.1 9.3.2 9.3.2 9.4.1 9.4.2 9.4.2 9.4.3 9.4.4 9.4.1 9.4.1 9.1.12, 9.4.3 9.2.1, 9.2.2 9.2.4 9.2.2 9.3.4

3-4 5-6 2-3 4-5 3-4 5-6 1-2 2-3 3-4 4-5 5-6 4-5 5-6 2-3 4-5 1-2 3-4 3-4 4-5 3-4 3-6 2-6 2-6 1-5

D B A A C D A D C B D C B C D B C A A B

25 26 27 28 29 30

7 4 6 8 6 25

H2, H6, H9 H6, H9 H3, H9 H6, H9 H6, H9 H5, H6 H6, H9 H7, H9 H2, H14 H2 H2, H11 H2, H14 H9, H10 H7 H2, H10 H10, H11 H2, H14, H15 H9 H1, H10 H1, H2, H10 H2, H6, H7, H9 H1, H2, H8 H6, H9 H3, H4, H7, H9, H13, H16 H2, H3, H5, H7, H10 H1, H3, H7, H9 H2, H3, H4, H9, H13 H2, H7, H8, H10, H12 H1, H3, H7, H8, H10 H3, H4, H6, H9, H10

1-5 2-5 1-6 2-6 1-6 1-5

31

25

H1, H3, H6, H7, H8, H10, H12, H13, H14

32

25

H2, H3, H5, H7, H8, H9, H10, H11, H14, H15, H16

9.3.1, 9.3.3 9.3.1, 9.3.5 9.4.1 9.4.2 9.4.3 9.6.1, 9.6.2, 9.6.3, 9.6.4 9.7.1, 9.7.2, 9.7.3, 9.7.4, 9.7.5, 9.7.6 9.4.1, 9.8.1, 9.8.2, 9.8.3, 9.8.4

Exam Section Question

Section I: Part A: Multiple Choice

Section I: Part B: Extended Response

Section II: Options

1-6

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.

PHYSICS_TR_10B_Guidelines_2010

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2010 TRIAL HIGHER SCHOOL CERTIFICATE EXAMINATION PHYSICS – MARKING GUIDELINES

Section I Part A – 20 marks Questions 1-20 (1 mark each) Question

Correct Response

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

D B A A C D A

D C B D C B C D B C A A B

Outcomes Assessed H2, H6, H9 H6, H9 H9 H6, H9 H6, H9 H6 H6, H9

H7, H9 H2, H14 H2 H2, H11 H2, H14 H9, H10 H7 H2, H10 H10, H11 H2, H14, H15 H9 H1, H10 H1, H2, H10

Targeted Performance Bands 3-4 5-6 2-3 4-5 3-4 5-6 1-2

2-3 3-4 4-5 5-6 4-5 5-6 2-3 4-5 1-2 3-4 3-4 4-5 3-4

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.

PHYSICS_TR_10B_Guidelines_2010

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Section I Part B – 55 marks Question 21 (8 marks) 21 (a) (3 marks) Outcomes Assessed: H2, H6, H9 Targeted Performance Bands: 3-5   

Criteria Describes the effect of freefall, the Earth’s surface curving away, and gravitational attraction Describes the effect of freefall and gravitational attraction Describes the effect of freefall or gravitational attraction

Marks 3 2 1

Sample answer While astronauts in the process of orbiting still experience gravitational force from the Earth, and are falling towards the Earth, they get no closer because the Earth’s surface curves away at the same rate. Because the astronauts and craft are in free fall, the craft does not push back against them, and the lack of that reaction force makes astronauts feel they are weightless.

21 (b) (2 marks) Outcomes Assessed: H6, H7, H9 Targeted Performance Bands: 3-5 Criteria

Marks 2 1

 Correct calculation and working  Partially correct calculation Sample answer GPE is given by Ep = – G M m/r = -6.67  10-11  6  1024  100 000 / [6380000 + 320000]  Ep = –5.97  1012 J

21 (c) (3 marks) Outcomes Assessed: H7, H9 Targeted Performance Bands: 3-6    

Criteria Correct calculation of the orbital velocity and kinetic energy and correct units Correct calculation of the orbital velocity and correct unit used or Correct calculation of orbital velocity and correct calculation of kinetic energy Partially correct calculation

Sample answer In circular orbit Fc = Fg

so m v2/ r = G M m/r2

giving

Marks 3 2 1

v = (G M/r)

So v = (G M/r) = (6.67  10-11  6  1024 / [6380000+320000])  v = 7.73  103 ms-1 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.

PHYSICS_TR_10B_Guidelines_2010

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and the kinetic energy is Ek = ½ mv2 = ½  100000  (7.73 103 )2 = 2.99  1012 J Question 22 (3 marks) Outcomes Assessed: H1, H2, H6 Targeted Performance Bands: 2-6 Criteria Marks  Outlines what simultaneous events are, describes two events which are seen as 3 simultaneous by one observer but are not simultaneous for another observer, and uses a labelled diagram to support the description 2  Outlines what a simultaneous event is and partially describes an event which is seen differently by stationary / moving observers and uses a diagram to support their description 1  Partially correct description of simultaneous events or  Use of an appropriate diagram Sample answer The relativity of simultaneity means that there is no such thing as two events occurring at the same time for all observers. Consider Einstein’s thought experiment; a train is racing along to the right on a track with a (moving) observer standing in the middle of the carriage. Another observer (stationary) standing on the platform sees two lightning bolts strike both ends of the carriage simultaneously. The inside observer will see the bolt in front strike first since they are rushing towards this signal, and the bolt in the rear strike second, so they will say the bolts did not strike simultaneously. moving observer train

Light signal Stationary observer

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.

PHYSICS_TR_10B_Guidelines_2010

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Question 23 (6 marks) 23 (a) (1 mark) Outcomes Assessed: H6, H9 Targeted Performance Bands: 2-3 Criteria 

Correctly calculates vx and vy

Mark 1

Sample answer horizontal component vx = v cos  = 5.1 cos 68 = 1.91 m s-1 vertical component vy = v sin  = 5.1 sin 68 = 4.73 m s-1

23 (b) (2 marks) Outcomes Assessed: H6, H9 Targeted Performance Bands: 3-5  

Criteria Correct substitution into correct formula, yielding an answer Incorrect substitution into correct formula, or no final answer

Marks 2 1

Sample answer y = u t + ½ a t2 becomes 0.80 = 4.73 t + ½  –9.8 t2  0.80 = 4.73 t – 4.9 t2  t = 0.75, 0.22 seconds

23 (c) (1 mark) Outcomes Assessed: H6, H9 Targeted Performance Bands: 2-3 

Criteria Correctly identifies rising and falling moments

Mark 1

Sample answer There are two times at which the ball achieves a height of 0.8m, once when going upwards (0.22 s) and once when it is dropping (at 0.75 s).   23 (d) (2 marks) Outcomes Assessed: H6, H9 Targeted Performance Bands: 4-6 Criteria Marks  Correctly identifies longer time, and gives a valid reason for this 2  Correctly identifies longer time 1 Sample answer The ball must be dropping down in order to go through the hoop, so the longer time is correct. 23 (e) (1 mark) Outcomes Assessed: H6, H9 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.

PHYSICS_TR_10B_Guidelines_2010

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Targeted Performance Bands: 3-4 Criteria 

Correct calculation

Mark 1

Sample answer Horizontally R = vx t = 1.91  0.75 = 1.43 m

Question 24 (6 marks) 24 (a) (3 marks) Outcomes Assessed: H7, H9, H13 Targeted Performance Bands: 1-5     

Criteria Identification of two coils, a primary and a secondary Clearly describing that AC causes change in the magnetic flux linking them Including the functioning principle – mutual induction, Faraday’s or Lenz’s law Omitting one of the above A correct statement of what a transformer can do

Marks 3

2 1

Sample answer Transformers function according to a principle called ‘mutual induction’, which is an application of Faraday’s Law. AC current is applied to one coil, called the primary coil, causing a rapidly varying magnetic flux to be produced. The secondary coil is located within that changing magnetic flux, so an emf is induced within the secondary coil. The strength of that emf relative to that in the primary depends upon the ratio of the relative number of loops in the two coils. To improve the energy efficiency of the device, in most cases the two coils are linked by a soft iron core, which channels the magnetic flux from primary to secondary.

24 (b) (1 mark) Outcomes Assessed: H3, H4, H16 Targeted Performance Bands: 1-2 

Criteria Identifying both the step-up and the step-down processes

Mark 1

Sample answer Transformers are important in the grid since they allow the electric potential to be increased to minimise power loss between generator and city, and decreased to reduce hazards when used.

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.

PHYSICS_TR_10B_Guidelines_2010

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24 (c) (2 marks) Outcomes Assessed: H7, H9 Targeted Performance Bands: 2-4    

Criteria Diagram showing a primary coil, secondary coil, AC input/output and iron core Correct labelling of at least three of these Either a diagram showing a primary and a secondary coil or Correct labelling of at least one correct part

Marks 2 1

Sample answer

Question 25 (7 marks) 25 (a) (1 mark) Outcomes Assessed: H2, H10 Targeted Performance Bands: 2-3 

Criteria Identifies that the current would flow from x to y

Mark 1

Sample answer When the upper face of the coil is turned clockwise current will flow in the direction x to y

25 (b) (2 marks) Outcomes Assessed: H3, H5, H7 Targeted Performance Bands: 1-4  

Criteria Successfully links energy provided by the wind to electrical energy output Makes one correct statement about windmills, generators or energy

Marks 2 1

Sample answer The coil could be attached directly to an axle of a wind turbine, so it turns because the blades are forced to rotate as the wind blows on them. Generating electricity always requires energy to be put into the system (conservation of energy) – in this case the wind provides that energy.

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.

PHYSICS_TR_10B_Guidelines_2010

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25 (c) (2 marks) Outcomes Assessed: H2 Targeted Performance Bands: 2-4   

Criteria Identifies two likely improvements, and justifies one of these proposals Identifies two possible improvements or Attempts to justify a given proposal

Marks 2 1

Sample answer To improve the efficiency of this device the rotor could be forced to rotate in a radial magnetic field created by curved pole-pieces, because maximum emf is induced for more time as the rotor is turned through the field. Another improvement would be to increase the number of loops upwards from 54.

25 (d) (2 marks) Outcomes Assessed: H9 Targeted Performance Bands: 1-5    

Criteria Correct formula to find current, with correct substitution Correct formula to determine force per loop, with correct substitution Identification of the 54 sides One correct substitution into one correct formula

Marks 2

1

Sample answer V = I R  18 = 450  I  I = 4.0  10-2 A FB = B I ℓ sin θ  FB = 0.22  4.0  10-2  12.0  10-2 sin 90 newtons (per wire). There are 54 identical loops, each having a side x y, so FTOTAL = 5.7  10-2 m

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.

PHYSICS_TR_10B_Guidelines_2010

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Question 26 (4 marks) 26 (a) (1 mark) Outcomes Assessed: H1, H7, H9 Targeted Performance Bands: 3-4 

Criteria Identifies the AC induction motor

Mark 1

Sample answer This is a model of the principle of an AC Induction motor.

26 (b) (3 marks) Outcomes Assessed: H2, H3, H9 Targeted Performance Bands: 2-5     

Criteria Changing magnetic flux is clearly identified Induction of emf and consequent current flow in the metal Identification of Lenz’s law One of the above missing One of the above stated

Marks 3

2 1

Sample answer The aluminium pie pan is an electrical conductor, which is free to rotate on the needle axis. The rotating magnet creates a magnetic field that is rotating. As it is lowered until it is just above the pie pan, its changing magnetic flux induces emf in the aluminium opposing the changing flux, in accordance with Lenz’s law. Since the pan is a bulk metal, current flows in the pan, in an attempt to stop the changing flux. The rotational momentum of the spinning magnet is sufficient to force the pie pan to rotate (slower) in the same direction for some time.

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.

PHYSICS_TR_10B_Guidelines_2010

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Question 27 (6 marks) Outcomes Assessed: H2, H3, H4, H9, H10, H13 Targeted Performance Bands: 1-6 Criteria  Discovery of fluorescence caused by cathode-rays  Another relevant property of Crookes’ cathode-ray tube discoveries  Two relevant discoveries by Thomson (not the q/m ratio)  Successful linkage of all these features to one another  One other feature (e.g. the grid, or e-m signal) or a general explanation of how it combines to produce a picture  Missing one of the above, or failure to link the discoveries  Reasonable recognition of the experiments, but inability to link to TV  Statements, or identification of two relevant discoveries  One feature of cathode-rays, Crookes’ tubes or Thomson’s experiments

Mark 6

5 3-4 2 1

Sample answer [N.B. Labelled diagram(s) might be used to identify and/or explain several of these features.] Following the discovery by Geissler of cathode-rays in vacuum tubes containing electrodes, eventually Crookes decided to investigate their properties, using a variety of special cathoderay tubes he invented. In particular, the Maltese Cross experiment showed that cathode-rays move in straight lines, and that they are emitted from the cathode, the paddle-wheel tube was able to prove they possess momentum/kinetic energy, whilst another tube fitted with a coated metal strip demonstrated that cathode-rays can be manipulated using magnetic fields. It was already...