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See how you performed in the 2024 VCE Physics exams with our full 2024 VCE Physics exam solutions.
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Want to know how you went in the 2024 VCE Physics exam? Compare your answers with our full 2024 VCE Physics exam solutions and explanations, expertly crafted by the Matrix Science Team!
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| Question | Answer | Explanation |
| 1 | C | Since Jo is travelling at constant speed, there must be zero net force on her. The upwards force exerted by the floor is equal and opposite to her weight, \(W=mg=75 \times 9.81=736 N \) |
| 2 | B | The orbital speed of a satellite does not depend on its mass. \(v=\sqrt{\frac{GM}{r}}\) When the mass of the satellite is doubled, the gravitational force exerted on it will automatically double. |
| 3 | A | There are two forces on the ball: its weight, which acts downwards, and the tension, which acts diagonally along the string. |
| 4 | C | The spring constant is equal to the gradient of the F vs. x graph: \(k = \frac{40}{0.1} = 400 Nm^{-1}\) The stored elastic energy in the spring is \(E_s = \frac{1}{2}k x^2\) \(E_s = \frac{1}{2} \times 400 \times 0.1^{2} = 2 J\) (Alternately: the area under the graph from 0-0.1 m gives the work done to compress the spring) When this energy is converted to kinetic energy, \(E_k = \frac{1}{2}m v^2\), this gives a speed of 8.9 m/s. |
| 5 | A | As the rock falls, its gravitational potential energy is converted to kinetic energy. In the absence of air resistance, no mechanical energy is lost to heat. |
| 6 | A | Magnetic field lines point towards the South pole of each magnet, so the resultant field at X is a combination of equal strength fields pointing left and right. They will cancel each other out. |
| 7 | D | The force on the conductor is given by \(F=IlB\) and remains constant because the current always travels perpendicular to the magnetic field lines. However, the resulting torque will be greater in Diagram A, as the force acts perpendicular to the loop’s area. |
| 8 | C | The force on the conducting wire is calculated using the formula: \(F=0.04 \times 5 \times 2.0 \times 10^{-3} = 4 \times 10^{-4} N\) |
| 9 | B | The force between two charges is governed by Coulomb’s Law, which states that the force is inversely proportional to the square of the distance, \(d^2\) For the force to be six times as large, \(d^2\) must be six times smaller. This happens if the new distance is \(\frac{d}{\sqrt{6}}=0.41 d\) |
| 10 | C | Induced current is proportional to the rate of change of flux. The flux must therefore always be changing, but at a higher rate in the middle time interval. |
| 11 | B | The best way to transmit power is at high voltage and low current, as this reduces the rate at which electrical energy is dissipated as heat. |
| 12 | C | As the upwards flux from the magnet reduces, the coil will compensate by creating its own upwards flux. From the eye’s perspective, this requires the current to travel anticlockwise around the coil (use the Right Hand Coil Rule). |
| 13 | A | This is a DC generator due to the presence of a split-ring commutator. The current output from this design will always be in the same direction but its magnitude will vary periodically. |
| 14 | D | Length contraction only occurs along the direction of relative motion, so the vertical dimension of the window will be unaffected. |
| 15 | B | The total energy of the particles increases proportional to the Lorentz factor, so they experience a large energy increase in the booster ring. At relativistic speeds, most of this energy serves to increase the mass of the particles and there is only a small increase in speed. |
| 16 | D | The larger the energy difference between the two levels, the higher the frequency of the photon emitted during a transition. The transitions \(E_5 \rightarrow E_1, E_4 \rightarrow E_1\) and \(E_3 \rightarrow E_1\) produce the three high-frequency lines shown in option D, with the other low-frequency transitions being \(E_5 \rightarrow E_2\) and \(E_5 \rightarrow E_3\) |
| 17 | B | Given a frequency of 50 Hz, the period of the standing wave is 0.02 seconds. At t = 0.01s only half a cycle has occurred and we will observe the pattern shown in B – troughs become crests and vice versa. |
| 18 | D | The separation between bright bands is given by the equation on the formula sheet: \(\Delta x = \frac{\lambda L}{d}\) When L is doubled and d is halved, the new separation between bands will be four times as large. |
| 19 | D | Squaring both sides of the formula gives: \(v^2 = GM \times \frac{1}{r}\) That is, a graph of \(v^2\) versus \(\frac{1}{r}\) will display a linear trend, with the gradient equal to the constant product GM. |
| 20 | B | “Precision” refers to the tendency of an experimental procedure to give similar results each time. If the data is precise, there is little variation between results. (Note that this does not mean the data are necessarily accurate!) |
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a) The acceleration of the two boats will be the same and is given by the net external force on the two boats divided by their mass:
\(a = \frac{F_{net}}{m_{total}} = \frac{(9.2 – (1.2+1.4)) \times 10^4}{(5+6) \times 10^5} = 0.06 ms^{-2}\)
The tension in the rope between boats 2 and 3 is an internal force when we consider them as a pair.
b) Using the sum of the forces on Boat 3,
\(F_{net} = T – (1.4 \times 10^4)\)
Hence \(T = F_{net} +1.4 \times 10^4\)
Using the acceleration from part (a), the net force on boat 3 is
\(F_{net}=ma=6.0 \times 10^5 \times 0.06\)
and hence \(T = 6.0\times 10^5 \times 0.06 + 1.4 \times 10^4 = 5.0 \times 10^4 N\)
a) By considering the vertical forces on the car, the vertical component of the normal force will be balanced by the weight force, giving \( N\cos \theta = mg\). Considering the horizontal forces on the car, the centripetal force is provided by the horizontal component of the normal force, giving \( N\sin\theta = \frac{mv^2}{r} \).
Solving these equations simultaneously by eliminating the normal force and mass gives the speed of the car to be:
\(v = \sqrt{gr \tan \theta}\)
Substituting in variables,
\(v = \sqrt{9.81 \times 1200 \times \tan 33 ^{\circ}} = 87 ms^{-1}\)
b) Since the normal force has a component that is horizontal when the track is banked, this horizontal component of the normal force points towards the centre of the circle, and thus provides the necessary centripetal force on the car.
a) Decomposing the initial velocity of the ball gives
\(u_x=u \cos \theta=48 \times \cos 35^{\circ}, u_y = u \sin \theta = 48 \times \sin 35 ^{\circ} \)
The height of the cliff is found using the vertical displacement equation:
\( h = u_y t +\frac{1}{2} at^2 = (48 \times \sin 35 ^{\circ}) \times 6.2 + \frac{1}{2} \times (-9.8) \times 6.2^2 = -18 m\)
Therefore h = 18 m.
b) Using the horizontal displacement equation:
\( s_x = u_x t = (48 \times \cos 35 ^{\circ}) \times 6.2 = 240 m \)
c) Air resistance always acts in the opposite direction to the velocity of the projectile, hence decreasing both vertical and horizontal velocity during flight. This would hence reduce the maximum height reached by the projectile as well as the time of flight and the horizontal distance travelled.
a) The impulse is found from the area under the force-time graph in SI units:
\(I = \frac{1}{2} \times 240 \times 10^{-3} \times 24000 = 2880 Ns\)
b) As the impulse \( I = \Delta p \) experienced by the dummy must be the same, doubling the time taken would halve the force experienced. Hence the new graph would be a triangle of double the base width but half the height of the previous scenario, with the maximum height of 12,000 N occurring at t = 240 ms.
a) The change in momentum is \(\Delta p = m \Delta v = 0.63 \times (10 – – 12) = 14 kgms^{-1} upwards\)
b) The collision is inelastic as the final speed is lower than the initial speed, indicating that the kinetic energy of the ball has decreased.
For Kwan’s suggestion to work, the centripetal force on the supply craft must exactly match the gravitational force acting on it at that orbital radius. Equating the two equations allows us to find the required speed at this height:
\(G \frac{m_1 m_2}{r^2} = \frac{m_1 v^2}{r} \)
This gives
\(v = \sqrt{\frac{Gm_2}{r}} = \sqrt{\frac{6.67\times 10^{-11}\times 4.5\times 10^5}{500}} = 2.5 \times 10^-4 ms^{-1} \)
which is much slower than the proposed speed and hence will not work. In other words, the gravitational force between the ISS and supply craft is too small to keep it in this orbit.
a) The speed of the electron can be found from the equation of its radius in the magnetic field:
\( r = \frac{mv}{qB}\)
Rearranging,
\(v = \frac{qBr}{m} = \frac{(1.6 \times 10^{-19}) \times (5.00 \times 10^{-3})\times (1.5 \times 10^{-2})}{9.11 \times 10^{-31}} = 1.32 \times 10^7 ms^{-1} \)
b) From the energy transformation equation in the electric field:
\( \frac{1}{2} mv^2 = qV_0\)
Solving for voltage,
\(V_0 = \frac{(9.11 \times 10^{-31}) \times (1.32 \times 10^7)^2}{2\times (1.6\times 10^{-19})} = 496 V \)
c) Having a positive charge, the proton will be deflected upwards in the magnetic field rather than downwards. Since the mass m is much larger for the proton than the electron, then the radius according to \( r = \frac{mv}{qB} \) will be larger for the proton.
a) Electric field lines always point from positive to negative, hence there should be a uniform electric field between the clouds of equally-spaced parallel field lines pointing upwards.
b) \(E = \frac{V}{d} = \frac{1.2 \times 10^9}{850} = 1.4 \times 10^6 Vm^{-1} \)
c) The total power provided by the lightning bolt is given by
\( P = VI = (1.2\times 10^9) \times (30.0\times 10^3) = 3.6\times 10^{13} \) and hence the total energy transferred is
\( E = Pt = (3.6 \times 10^{13}) \times (60.0 \times 10^{-6}) = 2.16 \times 10^9 J \)
a) In the coil’s current orientation as pictured in Figure 11, the current passing through side FG is parallel to the magnetic field and hence there is no force on this side.
b) When the coil is vertical (a quarter of a rotation from its current position in Figure 11) it will not rotate from rest. The torque on sides EF and GH will be zero, as the force experienced on these sides by the magnetic field will be parallel to the displacement vector from the axis.
c) According to the restrictions, neither the current nor the number of windings can be increased, so the strength of the magnetic field must increase to increase the torque.
a) The slip rings provide a continuous electrical connection between the coil and the external circuit, feeding the alternating current generated by the coil to the oscilloscope.
b) The peak-to-peak voltage is measured from the crest to trough of the signal, which is a total of 8 V. The period of oscillation is 0.2 s from the graph, hence the frequency is \( f = \frac{1}{0.2} = 5 Hz \)
c) As the induced EMF has doubled in magnitude but not changed in frequency, this could be attributed to doubling the strength of the magnetic field or doubling the area of the coil used (or doubling the number of windings, as a third option). Any of these would double the change in flux from maximum to minimum and hence the EMF according to Faraday’s Law.
a) Using the transformer equation relating voltage to number of turns,
\(\frac{N_1}{N_2} = \frac{V_1}{V_2}\)
\(N_2 = N_1 \frac{V_2}{V_1} = 460\times \frac{36}{230} = 72 turns\)
b) Using the transformer equation relating current to number of turns,
\( \frac{N_1}{N_2} = \frac{I_2}{I_1}\)
\(I_1 = I_2 \frac{N_2}{N_1} = 0.80 \times \frac{72}{460} = 0.13 A \)
c) Transformers rely on the Principle of Electromagnetic Induction, where an EMF (voltage) can only be generated from a changing flux in the core of the windings (Faraday’s Law). Hence AC is required as the changing current creates a constantly changing magnetic flux. A constant DC current in the primary would produce a steady magnetic flux and hence no induction in the secondary.
a) Total power is given by \( P = 50 \times 600 = 3\times 10^4 W \)
This is equivalent to 30 kW.
b) Individual panels have a maximum output of 20 V and 600W, giving a current of \( I = \frac{P}{V} = \frac{600}{20} = 30 A \).
When panels are connected in series, the voltages add but the current stays the same. For a string of 10 panels, each string produces a maximum voltage output of \( 10 \times 20 = 200 V \) and I = 30 A.
c) As each string is connected in parallel, the total maximum output voltage will be 200 V and the total maximum current will be \( I = 30 \times 5 = 150 A \).
d) The solar panels work on the photovoltaic effect which produces DC, so the inverter is used to convert this current to AC for use in the home (or to feed back into the grid).
a) Using classical mechanics, the time taken is the distance travelled divided by the speed.
\(t = \frac{s}{v} = \frac{9700 – 6500}{0.985\times (3\times 10^8)} = 1.08\times 10^{-5} = 10.8 \mu s\)
b) It is unlikely that the muons would reach the detector according to classical physics with a lifetime of 2.20 \( \mu s \) as \( \frac{10.8}{2.2} = 4.9 \approx 5 \) half lives would have passed in this time, leaving only \( 1 \div 2^5 = 3.125 \%\) of muons reaching the detector.
c) The Lorentz factor for the muons is
\(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-0.985^2}} = 5.8 \)
d) The mean half-life of the muon as measured by the physicists can be found from the time dilation equation, where the muons measure proper time.
\(t = \gamma t_0 = 5.8 \times 2.2 = 12.7 \mu s = 13 \mu s \) (to 2 sf).
e) Einstein’s special theory of relativity states that the muon lifetime will be dilated when measured by the physicist. As shown in part (d), the mean half-life of the muon in the physicist’s frame is longer than the time taken for the muons to travel to the detector in this frame, and thus will not decay before arriving and will be detected.
a) Annihilation reactions convert all of the particle mass into energy using Einstein’s \( E = mc^2\)
\(E = 2 \times (9.11\times 10^{-31}) \times (3/times 10^8)^2 = 1.64 \times 10^{-13} J \)
There is no kinetic energy as the particles are moving at negligible speed.
b) If the electron and positron are moving at negligible speed when they annihilate, the total initial momentum must be zero. As each gamma ray has a momentum according to its wavelength given by \(p = \frac{h}{\lambda}\), the gamma rays must move in opposite directions to also give a final momentum of zero to satisfy conservation of momentum.
a) The de Broglie wavelength gives a momentum of
\( p = \frac{h}{\lambda} \)
Substituting in values,
\(p = \frac{6.63\times 10^{-34}}{2\times 10^{-10}} = 3.32\times 10^{-24} kgms^{-1} \)
b) Considerable effects of diffraction only occur when the width (interatomic spacing, in this case) is less than or equal to the wavelength. Since \( \frac{\lambda}{w} < 1 \), they will not obtain a useful diffraction pattern and hence Max is correct.
c) Since the electrons and the X-rays have approximately the same wavelength and the interatomic spacing is the same for both, the maxima of the diffraction patterns will have the same spacing and hence will produce the same patterns.
d) The X-rays must have the same wavelength as the electrons, hence their energy (using the eV form of Planck’s constant) must be
\( E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{5.01 \times 10^{-11}} = 2.48 \times 10^4 eV \)
a) Controlled variables include (but are not limited to): same intensity of light, photocell, electrode, metal plate, electrical equipment
Dependent variable: stopping voltage
Independent variable: frequency of the incident light
b) 
c) i) From the graph, the gradient of the line is \( grad = \frac{1.5}{4.5\times 10^{14}} = 3.33\times 10^{-15} eVs \)
ii) From the x-intercept, the threshold frequency is \( 5.5\times 10^{14} Hz \)
iii) From the y-intercept, the work function is approximately 1.8 eV.
d) As the dependent variable is the stopping potential, the voltage measurements will give the maximum kinetic energy of the photoelectrons using the equation \( \frac{1}{2}mv^2 = qV \).
e) A different photocell would mean a different work function and hence a different threshold frequency, resulting in the line shifting up or down on the graph (different x and y intercepts) but no change to the gradient (Planck’s constant).
f) Changing a green filter to a blue one increases the frequency of the incident light, and hence would increase the stopping potential in magnitude (but still negative). Hence, the position of P will move to the left.
g) As the power of the light beam is unchanged, but the frequency of the photons is higher the number of photons in the beam is decreased and therefore the number of photoelectrons emitted will remain the same. Hence the horizontal line would be lowIer than previously.
h) The wave model of light could not explain why the stopping potential was dependent on the frequency of light, as it was predicted that the kinetic energy of the photoelectrons would be dependent on the intensity of the light.
However as seen in the experimental results, the frequency of light determines the stopping potential as the frequency determines the photoelectron kinetic energy.
This is explained by the particle model of light where the photoelectron kinetic energy is given by \(E_K = hf – \phi \), as the photon energy \( E = hf \) depends on the frequency and therefore directly affects the kinetic energy.
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