2014 Physics Examination Report Victorian Curriculum And-Books Pdf

2014 Physics examination report Victorian Curriculum and
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Examination, SPECIFIC INFORMATION, This report provides sample answers or an indication of what answers may have included Unless otherwise stated these. are not intended to be exemplary or complete responses. The statistics in this report may be subject to rounding errors resulting in a total less than 100 per cent. Area of study Motion in one and two dimensions, Question 1a. Marks 0 1 2 Average, 16 3 82 1 7, This question was about constant acceleration Using x ut at2 the distance travelled was 2 5 m. A common error was not squaring the time, Question 1b. Marks 0 1 2 Average, 50 1 49 1 0, The tension in the coupling was the only force accelerating the two trucks Students needed to apply Newton s second.
law F ma 20 000 0 2 4000 N It was also possible to apply Newton s law to the engine by first determining the. driving force of the engine F ma 60 000 0 2 12 000 then Fnet 12 000 T 40 000 0 2 giving T 4000 N. Using the mass of the engine 40 000 kg was very common This assumed that the only force acting on the engine was. the tension in the coupling and neglected the driving force of the engine. Question 1c, Marks 0 1 2 Average, 19 3 78 1 6, Using the conservation of momentum the speed was 2 0 m s 1. Some students confused the masses while others assumed it was an elastic collision and used conservation of kinetic. Question 1d, Marks 0 1 2 3 Average, 24 6 8 63 2 1, Students were required to determine the kinetic energy before 320 000 J and after 160 000 J the collision Since. kinetic energy had been lost it was an inelastic collision. Some students evaluated the momentum before and after the collision while others did not square the velocity when. calculating the kinetic energy, Question 2a, Marks 0 1 2 Average. 38 1 61 1 2, Any of the three sets of data could have been used With three 50 g masses the stretching force was 1 5 N so the spring. constant was k F x 1 5 0 3 5 0 N m 1, Physics GA 3 Exam Published 30 March 2015 2.
Examination, Question 2b, Marks 0 1 2 3 Average, 71 13 2 14 0 6. Taking the zero of gravitational potential energy at the lowest point of the oscillation the gravitational potential energy at. the top of the oscillation will equal the spring potential energy at the lowest point So mgh k x2 thus. 0 2 10 h 0 5 5 0 h2 giving h 0 8 m, Alternatively some students knew it would oscillate equally either side of the equilibrium position They first determined. the equilibrium position using F kx which gave 0 4 m Since it was released from 0 4 m above this point and would. reach 0 4 m below the equilibrium position the total extension from the unstretched length would be 0 8 m. Many students worked out the equilibrium position and added on the unstretched length. Question 2c, Marks 0 1 2 Average, 60 12 27 0 7, The graphs shown did not include kinetic energy If this had been included they would have added to a constant value. Students often wrote about one graph being linear and the other quadratic as the reason they did not add to a constant. Question 2d, Marks 0 1 2 3 4 Average, 74 15 3 1 7 0 5. As the mass oscillated the total energy kinetic spring and gravitational remained constant Take the zero of. gravitational potential energy at the lowest point At the highest point the only energy was gravitational mgh 1 6 J. Alternatively at the lowest point the only energy was spring potential energy kx2 1 6 J The maximum speed. occurred at the equilibrium position centre of oscillation At that point the total energy 1 6 J was a combination of. kinetic gravitational and spring 1 6 mv2 mgh kx2 0 5 0 2 v2 0 2 10 0 4 0 5 5 0 0 4 2 so that. the speed was 2 0 m s 1, Very few students were able to progress far with this question Many assumed that the spring potential energy was equal.
to the kinetic energy but did not give an explanation Others assumed that the kinetic energy was equal to the total. Question 3a, Marks 0 1 2 Average, 24 3 73 1 5, The vertical component of the initial velocity was 20sin30 10 m s 1 Then using a constant acceleration formula for the. vertical motion v2 u2 2ax gave 0 100 2 10 x and so x 5 0 m It was also possible to determine the time to the. top of its path and use it to get the height, Some students neglected the vector nature of the equations and others did not square values substituted into equations. Question 3b, Marks 0 1 2 3 Average, 41 9 4 46 1 6, Using the horizontal component of the velocity the time of flight was determined 26 20t cos30 giving t 1 5 s This. time was then used in a constant acceleration formula for the vertical motion x 20sin30 1 5 10 1 52 3 75 m. Some students used more steps They determined the time to reach the top of the flight path then the total time from the. horizontal motion By subtracting these they obtained the time from the top to the wall and used this to find the distance. down from the top This was then subtracted from the maximum height to get the required height. A number of students applied a derived formula for the range which did not apply to this situation Others neglected the. vector nature of the quantities by omitting negative signs There were also a number of mathematical errors. Physics GA 3 Exam Published 30 March 2015 3, Examination. Question 4a, Marks 0 1 2 Average, 27 4 69 1 4, To feel weightless there would be no normal reaction force Therefore the only force acting gravity would be providing.
the centripetal force So mg mv2 r hence 10 v2 20 and v 14 1 m s 1. Question 4b, Marks 0 1 2 3 Average, 30 31 28 11 1 2. They felt weightless because the normal reaction force was zero They were in freefall with their acceleration equal to. gravity They were not actually weightless because the strength of the gravitational field was not zero. Some students believed that the gravity force down was equal to the normal force up and therefore they cancelled out or. stated that the net force was zero, Question 4c, Marks 0 1 2 3 Average. 17 14 25 44 2 0, The two forces acting were the gravity force down and the normal force up Because they were moving in a circular. section of the track the resultant force had to be up. mg or weight or grav force, The quality of some students diagrams was quite poor The arrows needed to be drawn clearly so that they could be. distinguished They also needed arrowheads to indicate direction It was common to see a resultant force and a centripetal. force There were also arrows representing acceleration and velocity. Question 5a, Marks 0 1 2 3 4 Average, 26 9 10 12 43 2 4.
2 2 2 2 3 2, By transposing GMm R 4 Rm T to give M 4 R GT and substituting the appropriate values the mass of the star. was calculated to be 1 1 1031 kg, Common errors included not converting the period to seconds forgetting to square or cube values incorrect transposing. of the equation copying from notes and an inability to use the calculator correctly. Question 5b, Marks 0 1 2 Average, 37 48 16 0 8, It was not possible to determine the mass of the planet from the data provided because its mass appears on both sides of. the equation and thus cancels out of the equation, Area of study Electronics and photonics. Question 6a, Marks 0 1 Average, The voltmeter would read 0 V because there is no difference in electrical potential between the terminals of the meter.
Physics GA 3 Exam Published 30 March 2015 4, Examination. Question 6b, Marks 0 1 2 Average, 26 6 69 1 5, The voltmeter was now measuring the potential difference across the 300 resistor Treating it as a voltage divider. would give 300 1200 8 2 0 V Alternatively students could have determined the current in the circuit V R. 8 1200 0 00666 and then use Ohm s law to determine the voltage across the 300 IR 0 00666 300 2 0 V. Although they realised it could be treated as a voltage divider some students got the incorrect ratio Of those who first. calculated the current some rounded off this value before using it in the Ohm s law calculation Students should be. encouraged to carry the value in the calculator register. Question 7a, Marks 0 1 2 3 Average, 24 12 7 57 2 0. From the graph the voltage across the LED was 2 V By knowing the power of the LED the current through it was. calculated with P VI 0 3 2 I hence I 0 15 A The voltage across the resistor was 9 V and the current through it. was 0 15 A so using Ohm s law the resistance was 60. Some students applied the power formula P VI to the resistor perhaps believing that the 300 W referred to it instead of. Question 7b, Marks 0 1 2 Average, 32 5 63 1 3, Since the voltage across the LED was 3 V the voltage across the resistor was 9 V So the power dissipated by the resistor. was P VI 9 0 5 4 5 W, Some students seemed to assume that the LED was the same LED as in part a and thus determined that the voltage.
across the resistor was 10 V, Question 8a, Marks 0 1 Average. Reading from the graph when the potential difference was zero the current was 3 6 mA. Question 8b, Marks 0 1 Average, With the switch open the current was zero Reading from the graph the potential difference across the photodiode was. Question 8c, Marks 0 1 2 3 Average, 10 21 12 57 2 2. From the graph operating the photodiode at 1 V meant a current of 1 6 mA Therefore the voltage across the resistor and. the current through it were also 1 V and 1 6 mA respectively Thus using Ohm s law the resistance was 625. Some students neglected to convert or incorrectly converted the current to ampere when substituting into Ohm s law. Others assumed that the voltage across the resistor was the maximum that the photodiode could provide 1 3 minus the. voltage at which it was now operating 1 0, Physics GA 3 Exam Published 30 March 2015 5. Examination, Question 9, Marks 0 1 2 Average, 66 9 25 0 6.
The modulated signal was, Some students drew the correct general shape but did not have the correct ratio of the amplitudes Others included a sine. wave within the rectangular blocks Many mirrored the signal under the time axis. Question 10a, Marks 0 1 2 3 Average, 19 10 18 53 2 1. The expected graph is shown below however graphs with biasing shown were also acceptable Common errors included. drawing the characteristics for an inverting amplifier not showing scales on the axes and having clipping commencing at. the wrong point or not shown at all Some students gave the incorrect amplification. Physics GA 3 Exam Published 30 March 2015 6, Examination. Question 10b, Marks 0 1 2 3 Average, 19 18 7 55 2 0. The output signal is shown below If the student showed biasing on the characteristics in part a other possibilities were. acceptable The main errors were inverting the signal or not showing clipping Some students sketches did not show the. same period as the original, Question 11, Marks 0 1 2 3 4 Average.
17 15 12 4 53 2 6, At 300 lux the resistance of the LDR was 500 The control unit switched when VC was 4 V Using the voltage divider. relationship VC RV RLDR RV 9 4 RV 500 RV 9 RV 400 Alternatively knowing that the resistance. of the LDR was 500 and that the voltage across it would have to be 5 V students could determine the current through it. to be 0 01 A using Ohm s law Then applying this to the resistor with a voltage of 4 V and the current 0 01 A the. resistance was 400 Another method was to use the fact that the ratio of the voltages would be the same as the ratio of. the resistances, Students commonly determined the resistance of the LDR as 500 but had the wrong voltage divider ratio or the voltage. required by the control unit, Area of study Electric power. Question 12a, Marks 0 1 Average, The current was going up the front of the coil and down the back so the magnetic field resulting went from right to left. At X it was pointing to the left so option A was correct. Question 12b, Marks 0 1 Average, At point X the magnetic field was to the left the current was up the page so the force applied was out of the page option.
Question 13a, Marks 0 1 Average, Physics GA 3 Exam Published 30 March 2015 7. Examination, The desired effect required a change in flux through the loop This could be achieved by moving it to the right so that it. moves out of the magnetic field or by rotating it So the correct answer was C moving the loop sideways to the right so. that it moves out of the magnetic field and D rotating the loop about a horizontal axis. This question required students to select one or more options Therefore it was important to check all possibilities. Question 13b, Marks 0 1 2 3 Average, 33 10 7 50 1 8. The EMF generated in the loop was determined using Faraday s law n B t 0 08 0 050 0 010 0 4 V This. value was then substituted into Ohm s law with the current to determine the resistance as 20. A common error involved students converting the time from milliseconds to seconds Others neglected the area of the. loop all together, Question 13c, Marks 0 1 2 3 Average. 43 24 13 21 1 1, Initially the flux was downwards and decreasing According to Lenz s law the induced current would oppose the change.
in flux Therefore the current induced in the loop would be clockwise. Students needed to refer to the change in flux not the change in magnetic field Some wrote about opposing the change. but it was unclear what change this referred to It was important to explain the initial conditions and how they were. changing Some students appeared not to realise that the loop in Figure 25 was th. 2014 Examination Report 2014 Physics GA 3 Examination GENERAL COMMENTS scientific mode and that it does not truncate answers after one or two decimal places The rounding off of calculations should be done only at the end not progressively after each step Answers should be simplified to decimal form Where values of constants are provided in the stem of the question or on the formula

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