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/1<1<ffD4"ff2007 VCE Physics Unit 3 Exam Solutions (updated 18th June)
Consequentials are indicated as Conseq on 1
Suggested Marking Scheme in italics
These suggested solutions have been prepared by the AIP Education Committee. Every effort has been made to double check the solutions for errors and typos.
Motion in one and two dimensions
1. 1.0 m t = 0.45, u = 0, a = 10, s = ?, use s = ut+ at2, s = 0 + x 10 x (0.45)2 (1)
s = 1.01 m (1). Note: The use of 9.8 m/s2 gives 0.99 m to two significant figures.
2. A parabolic path starting horizontal, but curving over (1) to meeting the ground at a point immediately below the bike seat in position B (1). The horizontal velocity of the tail light keeps its original value and travels forward with the bike as it falls. Note 1: This answer assumes that Mary is stationary relative to the ground. Note 2: The vertical drop and the horizontal displacement of the tail-light are no on the same scale.
3. By conservation of momentum, momentum before = momentum after.
2 x 3 + 1 x 0 = 2 x v + 1 x 4 (1), solve for v (1).
4. Elastic (1) KE before = x 2 x 32 = 9 J (1),
KE after = x 2 x 12 + x 1 x 42 = 1 + 9 = 9J (1)
5. 400 N Average force = Change of momentum / Time taken (1)
Average force = (1 x 4 1 x 0) / 0.01 (1) = 400N (1).
6. Zero At 400 m air resistance equals the weight (1), so net force is zero (1).
7. 3.3 m/s2 At 100m, AR = 400 N, so net force = 600 400 = 200 N (1).
Acceln = 200/60 (1) = 3.3 m/s2 (1)
8. 250,000 J Area under the graph from zero to 500m. 12.5 squares (1), each of 20,000 J gives 250,000 J (1). Note: The question is not well worded.
9. 0.40 kg Extension = 0.6 0.4 = 0.2 (1) Mg = kx, find M. M x 10 = 20 x 0.20 (1),
M = 0.40 kg (1).
10. 0.50 J Increase in PE = kx22 - kx12 (1) = x 20 x ((0.30)2 (0.20)2) (1)
Increase in PE = 0.50 J (1).
11. D Total energy is conserved. (2)
12. A Gravitational force = GMm/r2. G, M and m are same for both,
so ratio equals (RP/RE)2 = (6/10.5)2 = 0.33. (2)
13. 574 years Using Keplers 3rd Law: R3/T2 is the same for both objects,
then (TE)2 = (RE/RP)3 x (TP)2 = (10.5/6)3 x 2482 (1) = 329623= 574 years. (1)
14. 3.4 s u= 40sin250, v = -40sin250, a = -10, s = 0, t =?. Use v = u+ at
t = 2 x 40 x sin250 / 10 (1) = 3.38 s. (1)
Note 1: The use of 9.8 m/s2 gives the same answer to two significant figures. Note 2: The data in the question is inconsistent. The time of flight can also be determined from the range;
Time of flight = range / horizontal component of the speed
Time = 127 / 40cos250 which gives a different answer of 3.5 s.
15. 14.2 m u= 40sin250, v = 0, a = -10, s = ?, t = 3.38 / 2 s (1), Use s = (u + v)t/2
s = 40sin250 x 3.38 / 4 (1) = 14.28 m (1) Conseq on Qn 14.
16. D Weight is only force and is always down. (2)
17. A, C A: same max height so time to go up and down is unchanged (1). B: twice the horizontal distance in the same time, so horizontal component is doubled, but max height is the same so vertical component in unchanged. So the initial speed is greater. C: acceleration equals g and unchanged (1). D: Horizontal component is doubled, so angle is less.
Note: It could be argued that the question includes what happens in the gun where the acceleration of the paintball is greater, so only giving A.
Electronics and photonics
1. Peak voltage of 20 mV input becomes a 20 x 10-3 x 80 V peak output = 1.6 V (1). Graph is an inverted sine wave (1) centred on zero with a peak voltage of 1.6 V.
2. A peak voltage of 120 mV gives an output peak voltage of 9.6 V above and below the 3.0 volts at Y. This means the output will be clipped at 6.0 V and at 0 V, which on the other side of the capacitor will give a square looking wave (1) between +3 and -3 volts. (1)
Note 1: This answer assumes the value of the emitter resistor is zero and ignores the collector emitter voltage.
Note 2: The frequency of the input signal has doubled as seen in Figure 3a, but it is not mentioned in the text of the question.
3. Saturation means maximum collector current (1). At this point the voltage drop across the collector resistor is a maximum and so the voltage at the collector relative to the earth is a minimum and cannot go any lower (1).This means the output voltage is clipped at the
bottom (1).
4. B All the answers have difficulties, but it is likely that B is the expected answer. A: Wrong, an insulating medium is an insulator!, no movement of charge through it, although it could be argued that there is a AC electric field passing through the insulator. B: This answer confuses the operation of a capacitor as a decoupler or smoother with that of a static filled capacitor. The capacitor is not connected straight to an AC source, rather it is connected to the collector which is biased and always positive. The capacitor allows small changes in the voltage at one plate to be quickly passed to the other plate allowing AC to pass, but the actual voltage drop across the capacitor is slows to change and so blocks DC. C: Magnetic flux is Unit 4 content, this explanation would require a magnetisable, conductive medium, which it is not. D: Wrong (2)
5. 500 ohms At 10 mA, the voltage drop across the diode, from Fig 4, is 1.0 V (1), so there is 5.0 volts across the resistor (1). Using Ohms law,
its resistance is 5 / (10 x 10-3) = 500 ohms. (1)
6. 6.0 V (1) Diode is reverse biased, so the left part of Fig 4 applies (1). The current is zero, so the voltage drop across R is also zero and all the voltage is across
the diode (1).
7. 10 lux From figure 6, when R = 3k ohms, so intensity = 10 lux. (2)
8. 750 ohms From the figure 6, at 20 lux, Resistance of LDR = 1500 ohms (1). Using voltage divider equation: 4 = 6 x (1500/(1500+R)), solve for R. R = 750 ohms. (1) Alternatively voltage across LDR = 4V, so voltage across R = 2V, so R is half the resistance of the LDR.
9. decrease (1) Need: Vout = 4V for a higher lux value. Greater lux means smaller LDR resistance, so R will also need to decrease to keep each resistors fraction of the voltage the same. (1)
10. D Only B and D refer to brightness. Brightness cannot go negative, so D. Note: This answer assumes that the value at the origin of the graphs is zero and also assumes that circuit for the laser diode is sufficiently biased so that the laser diode still emits light even when the voltage at W goes negative. (2)
11. C Only A and C refer to voltage. When the photodiode does not conduct the voltage at Y would be constant, so A is not possible. Note: This answer assumes that the value at the origin of the graphs is zero. (2)
Einsteins relativity
1. light/interference/no. It could be argued that since the experiment was investigating the velocity of light relative to the earth, that either or both Earth and light could be considered. (1)(1)(1). Note also that neither Michelson nor Morley (typo in the paper) thought the experiment disproved the aether hypothesis.
2. 10 s Speed of sound is reduced to 290 m/s (1). Distance equals 3.00km, so time equals 3000/290 (1) = 10.3 s (1).
3. B The distance as measured by the pilot is 90,000 km, but the speed of the signal is unaffected by the speed of the spacecraft and is c, so the time will be 9 x 107 / (3 x 108) = 0.300 sec. (2)
4. 1.005 Substitute v = 0.1000c into the formula for the Lorentz factor. (1) (1)
5. B The length of the ship as measured by Ann is contracted or shortened,
L = L0/g. (2)
6. No (1) Both Ann and Bob observe a moving object (1). The direction is irrelevant. Each will measure their moving object as shorter than its proper length. (1)
7. B Moving clocks run slow. (2)
8. A Moving clocks run slow. The life time of a moving pion will be longer that its life time at rest. So a rest lifetime will be 4.16 x 10-7 / 16 = 2.60 x 10-8 s. (2)
9. See Qn 4 substitute v = 0.998c into the formula for the Lorentz factor and show the answer is close to 16. Actually it is 15.8. (1) (1) (1)
10. 4.4 x 109 kg Use E = mc2 (1). m = 4.0 x 1026 / (3.0 x 108)2 (1) = 4.4 x 109 kg. (1)
Investigating materials and their use in structures
1. length/shear/elasticity (1) (1) (1)
2. Taking torques about T2: T1 x 3L/4 (1) = Mg x L/4 (1). Cancelling and rearranging (1)
T1 = Mg/3.
3. 2 Substitute Mg = 3T1 (1) into the supplied equation T2 = Mg T1
gives T2 = 3T1 T1 =2T1 (1)
4. D Between the cables the bottom is under tension, to the right of T2, the top is under tension. (2)
5. B Both materials return to the same shape. (2)
6. C Cats iron is brittle, mild steel is ductile. (2)
7. 1.9 x 104 N Stress = Force/ Area (1). So Force = 240 x 106 x p x (10/2 x 10-3)2 (1)
Force = 1.9 x 104 N (1)
8. 3.8 J Strain energy = Area under graph x Volume (1)
Strain energy = x 240 x 106 x 0.4 x 10-3 x p x (10/2 x 10-3)2 x 1.0 (1) = 3.8 J (1)
9. D At B, weight of lamp is down, which is balanced by the upward component of the force within the rod, which must therefore be under compression. The horizontal and outward component of the compression force in the rod is balances by the inward component of the force in the cable, which must therefore be under tension. (2)
10. 120 N upward component of the force in the rod = Weight. CB x sin600 = Mg (1)
Outward component of the force in the rod = Force in AB. CB x cos600 = AB (1)
Eliminating CB, AB = Mg / tan600 = 20 x 10 / tan 600 = 115.5 N (1)
Further electronics
1. oscilloscope/undervoltage/capacitor. Although an audible hum could be argued for. (1) (1) (1)
2. C ratio of turns = 6.0/240, which is the same as 30/1200. (2)
3. C A: Wrong, current can go two ways at the top. B: Wrong, current cannot go either way at the top. D: Wrong, current returns to the supply without going through Vout. (2)
4. B full wave rectification. (2)
5. 400 mF Time constant (t = CR) is time to lose 63% of the voltage, which is about 40 milliseconds (1). R = 100 ohms, so C = 40 x 10-3 / 100 = 4.0 x 10-4
C = 400 x 10-6 F. (1)
6. C From Figure 1 in 20 ms the voltage across the capacitor will drop to 7.0 V, before it rises again on the next half wave peak. C fits this description. (2)
7. 6.0 V Zener diode is used in reverse bias mode, which is the graph in
the third quadrant. (2)
8. 3.0 V Supply voltage of 9.0 V minus 6.0 V across Zener (1) equals 3.0 V across R. (1) Conseq on Qn 7
9. 30 mA Using V = IR, I = 3 / 100 = 30/1000 (1) = 30 mA. (1) Conseq on Qn 8
10. With the whole class connected in parallel to the supply, the effective resistance is reduced (1). This reduces the Time constant, increases the ripple (1) and decreases the effective DC voltage.
11. Increase the power supply voltage and put a 12V zener diode with a resistor in series across the bridge rectifier (1). Increase the size of the capacitor. (1)
12. Trace should be a sine wave. The period, or time for one cycle, is 20 ms, which is 4 divisions across the graph (1). The peak voltage is 4 x sqrt of 2 = 5.6 volts, which is just under three divisions up the graph from the middle line, then the same amount below the middle line. (1)
Dot points not covered:
Motion is one and two dimensions
analyse uniform circular motion of an object in a horizontal plane;
analyse relative velocity of objects along a straight line and in two dimensions;
distinguish between stationary (inertial) frames of reference and frames of reference that are moving at constant speed relative to the stationary frame, including Galilean transformations in one dimension between frames of reference;
analyse transfers of energy between kinetic energy, potential energy and other forms of energy for objects that
undergo elastic and inelastic collisions
move from position to position in a changing gravitational field, using only areas under force-distance and field-distance graphs;
analyse planetary and satellite motion modelled as uniform circular orbital motion in a universal gravitation field, using
a = v2/r = 4p2r/T2
Electronics and photonics
" describe energy transfers and transformations in electrical optical, and optical electrical conversion systems using opto-electronic converters;
Einstein s relativity
" describe Maxwell s prediction that the speed of light depend sonly on the electrical and magnetic properties of the medium it is passing through and not on the speed of the source or the speed of the medium;
contrast Maxwells prediction with the principles of Galilean relativity (no absolute frame of reference; all velocity measurements are relative to the frame of reference);
compare Einsteins postulates and the assumptions of the Newtonian model;
use simple thought experiments to show that
the elapse of time occurs at different rates depending &24:;i+ , - N Q 8
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spatial measurements are different when measured in different frames of reference;
explain the concepts of proper time and proper length as quantities that are measured in the frame of reference in which objects are at rest;
explain movement at speeds approaching the speed of light in terms of the postulates of Einsteins theory of relativity;
explain the equivalence of work done into increased mass energy according to Einsteins equation, E = mc2,
Investigating materials and their use in structures
compare the tensile and compressive strengths and the stiffness or flexibility of different materials under load to determine their suitability for use in structures such as columns, beams and arches;
model the behaviour of materials under load in terms of extension and compression, graphically and algebraically using Youngs modulus;
use data to describe and predict the performance of a simple structure under load;
Further electronics
select measuring devices for circuit analysis and faults diagnosis;
select measurements of voltage and current (from the use of a multimeter and an oscilloscope) in the DC power supply circuit to evaluate the operation of the circuit in terms of its design brief;
calculate power dissipation in circuit elements;
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