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2020 HSC Physics Paper Solutions and Explanations

In this post, we reveal the solutions to the 2020 HSC Physics Exam Paper!

The Matrix 2020 HSC Physics Exam Paper Solutions are here! The Matrix Science team has been hard at work on them.

 

2020 HSC Physics Exam Paper Solutions

Check out our the solutions below and see how you went!

Section 1: Multiple Choice

Question Answer Solution
1  D Huygens proposed that light propagated via the formation of wavelets.
2 A An AC motor has no brushes as current is applied to the stator.

Strictly speaking, an AC induction motor does not require current to be supplied to the rotor instead, the current in the rotor is induced by the changing stator field.

3  C Maxwell was a theorist who showed that changing magnetic fields could induce electric fields and vice versa, resulting in electromagnetic waves.
4  B According to the graph, the half-life is about 100 years (time when the remaining mass is half of the starting value).

\(\lambda = \frac{\text{ln }2}{100} = 0.0069 \text{ years}^{-1}\)

5  B The initial vertical component of velocity  determines the time of flight according to the equation \(s_y = u_y t + \frac{1}{2}at^2\).
6  A The stars vary in colour as indicated by the x-axis. They also vary in luminosity, as indicated by their magnitude on the y-axis. These factors indicate that they also vary in mass.

Being part of a cluster suggests that they have approximately the same age. There are long-lived red main sequence stars and the larger stars have moved on to being giants or white dwarfs

7  D The output shown is that of a DC generator, which uses a split-ring commutator to convert alternating current in the coil to pulsed direct current in the external circuit.
8  C The process reduces both the mass number and atomic number by 2, so it must involve the emission of an alpha particle. The element with atomic number 90 is thorium.
9  A The Bohr model was the first to incorporate discrete energy levels, which explained why only certain wavelengths were produced in an element’s emission spectrum.
10 B The bottom plate must be positively charged to exert the electric force shown, so the electric field points towards the top of the page.

According to the RH palm rule, a magnetic field out of the page would be required to produce an opposing force upwards.

11  C The final mass is smaller than the initial mass, so mass has been converted to energy and released. This reaction is a transmutation rather than a fusion, as it does not combine two nuclei.
12  D The escape velocity does not depend on the mass of the satellite, but does depend on initial position as given by \(v_{esc} = \sqrt{\frac{2Gm}{r}}\).  The close the satellite is to Earth, the smaller the value of \(r\) and the greater the escape velocity.
13  A Reading from the graph tells us that electrons from metal Y will have a lower maximum kinetic energy. This corresponds to their maximum velocity also being lower.
14  C Reversing the current will result in the wires being attracted to each other. Doubling the current in Y will result in the magnitude of force doubling, according to Ampere’s Law.
15  D If the rocket is initially travelling horizontally at a constant speed of \(20 \text{ ms}^{-1}\), there must have been zero net force on it. At that time, the thrust must have been equal to the rocket’s weight: \(w = mg = 7800 \times 9.8 = 76440 \text{ N}\). When the thrust is increased to \(90000 \text{ N}\) there will be a net force upwards, and hence an acceleration upwards.
16  A The question indicates that the moderator increases the rate of fission, so its loss will cause the rate of fission to decrease.

Withdrawing the control rods will increase the rate of fission as fewer neutrons are absorbed.

17  C Observer Y sees the carriage as being length contracted, so the distance from the source to the sensor is reduced. Since both observers must agree on the speed of light, Y sees it take a shorter time than X to reach the sensor.
18  B When the Earth is travelling away from Jupiter, the light signal from the start of one orbit will take longer to reach Earth than the light signal from the start of the previous orbit,  as it needs to travel a longer distance to reach Earth. Therefore the period will appear to be longer.
19  C In the starting position shown, charges in the conductor are all moving to the right, parallel to the magnetic field. They do not experience a magnetic force, so there is no EMF.

After a quarter of a turn, when the conductor is horizontal, all the charges will be moving vertically (perpendicular to the field). They experience a magnetic force: positive charges are pushed into the page and negative charges are pushed out of the page. This does not result in a charge separation across P and Q, and does not produce an EMF between them.

20  A This question combines aspects of projectile and circular motion.

There is a constant downwards force due to the weight which is constant.

In the horizontal plane the ball is in uniform circular motion. The horizontal, centripetal force from the wall \(F_c = \frac{mv^2}{r}\) remains constant since the ball’s horizontal speed remains constant.

These two forces remain perpendicular to each other, so the magnitude of their sum remains constant.

 

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Section 2: Short Response

Question 21 ( 5 marks)

(a) Use the Rydberg equation:

\(\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right) = 1.097\times 10^7 \times \left(\frac{1}{2^2} – \frac{1}{3^2}\right)\) which gives \(\lambda = 656 \text{ nm}\)

(b) Electrons in the Bohr model exist in one of a number of stable energy states. They can only transition to a higher energy state by absorbing the exact energy needed from a photon, and can only transition to a lower energy state by releasing the excess energy through the emission of a photon. The total energy before and after the transition must be the same as stated by the Law of Conservation of Energy.

 

Question 22  (5 marks)

(a)  The peak wavelength is calculated using Wien’s Displacement Law.

\( \lambda = \frac{b}{T} = \frac{2.898\times 10^{-3}}{3200} = 9.056 \times 10^{-7} \text{ m} = 906 \text{ nm} \)

 

(b)  Special Relativity only has measurable effects when the capsule is moving close to the speed of light. Even its maximum speed is only a tiny fraction of the speed of light and the effects will be undetectable.

The capsule is also accelerating, so it is not an inertial frame of reference and cannot be analysed using Special Relativity.

 

 

Question 23 (3 marks)

When the motor produces a torque of \(2.95 \text{ Nm}\), we can read from the graph that the angular velocity is approximately \(100 \text{ rads}^{-1}\). Using these values, we can rearrange the torque equation and calculate the current:

\( I = \frac{\omega \tau}{\eta V} = \frac{100 \times 2.95}{0.3 \times 240} = 4.10 \text{ A}\)

 

This current is below the maximum allowed value, so the circuit breaker is not needed and the motor can operate.

 

 

Question 24 (4 marks)

In order to calculate the projectile’s initial velocity, we need both the horizontal and vertical components of the velocity.

We can calculate the horizontal velocity using the stated range and the time of flight.

\(u_x = \frac{s_x}{t} = \frac{130}{6} = 21.7 \text{ ms}^{-1}\)

 

In order to find the initial vertical velocity, we can use the fact that the maximum vertical displacement was reached after three seconds, at which point the vertical velocity would be zero.

\( v_y = u_y + at\)

 

\( 0 = u_y + (-9.8) \times 3 \)

 

\( u_y = 29.4 \text{ ms}^{-1} \)

 

The final velocity is the vector sum of the components, which is \( 36.5 \text{ ms}^{-1}\) at \( 54^\circ \) above the horizontal.

 

 

Question 25 (4 marks)

The hydrogen atom is composed of one proton and one electron.

The electron is a lepton, an elementary particle.

The proton is composed of three quarks, two up and one down. The quarks are held together by the strong nuclear force, which is mediated by gauge bosons known as gluons.

The electron orbits the proton due to the electromagnetic force which results in attraction between then, which is mediated by virtual photons.

 

 

Question 26 (8 marks

(a) A gas discharge tube will produce a spectrum composed of lines at discrete wavelengths (which will depend on the energy levels of the gas in the tube). An incandescent lamp produces a continuum spectrum, a black body spectrum, with the peak emission wavelength dependent on its temperature.

 

(b) Model X assumes that a black body radiates equally at all frequencies.

Model Y assumes that the radiation is emitted in energy packets, where the frequency of emission is proportional to the energy of a transition (i.e. to the energy carried by the packet being emitted). Some transitions are impossible and others occur with varying probability.

 

(c)  The curve for the \( 4000 \text{ K} \) black body will have a lower intensity at all wavelengths, because the total power emitted by a body decreases as temperature decreases. The curve will also have its peak at a longer wavelength (which should be calculated by Wien’s Displacement Law: \( \lambda = 725 \text{ nm} \)). This is because the probability of certain energy transitions occurring varies with temperature. Smaller energy transitions are more common at lower temperatures, where the average energy per particle is lower, and these correspond to the emission of longer wavelength photons as \( E = hf = \frac{hc}{\lambda}\).

 

 

Question 27 (4 marks)

The distance between maxima is determined by the wavelength of light but also by the distance between the slits and the distance to the screen.

When the wavelength \( \lambda \) is reduced, the spacing of the maxima reduces according to  \(d \text{sin} \theta = m \lambda \). There is a decrease in the angular separation. In order to bring the spacing back to its original value, we can decrease the separation between the slits \( d \), causing \( \theta \) to increase, or we can place the screen further away from the slits by increasing \( y \), causing their linear separation to increase.

 

 

Question 28 (7 marks)

(a) The magnetic flux through the loop is \( \Phi = BA \text{cos} \theta\).

The field strength is constant at \(1 \text{ T}\), the angle between the field and the area normal vector is \(30^\circ \), and the area will shrink by \(0.3 \text{ m} \times 0.2 \text{ m}\) when the rod is moved. This gives the change in flux:

\( \Delta \Phi = B \text{cos} \theta \times \Delta A = 1 \times \text{cos}30^\circ \times 0.3 \times 0.2 = 0.052 \text{ Wb} \)

 

(b) Using Faraday’s Law gives:

\(\varepsilon = – \frac{\Delta \Phi}{\Delta t} = – \frac{-0.052}{2.5} = 0.021 \text{ V} \)

 

 

(The change in flux is taken to be negative, since it is a decrease of flux.)

The current will flow anticlockwise as viewed from above. (This is because the upwards flux through the loop is decreasing, so the loop tries to generate additional flux upwards, according to Lenz’s Law.)

 

(c)  There is always a gravitational force pulling downwards on the rod. The downhill component of that must be overcome to move the rod as described in the question. Additionally, when the magnetic field is present, the induced current tends to produce a force opposing the motion according to Lenz’s Law. This makes it more difficult to move the rod as you must overcome both forces.

Without the magnetic field, the opposing force due to Lenz’s Law is not present and so less force is required to move the rod, as only the downhill component of the gravitational force must be overcome.

 

 

Question 29 (5 marks)

(a)  Gamma rays were known to have zero electric charge, which would explain their lack of deflection in an electric field. They were also known to carry energy, which could be transferred to protons in the paraffin to eject them. However, gamma rays would be capable of producing the photoelectric effect as they are high energy photons, so this disagrees with observations.

Scientists also found that the energy associated with gamma rays was not sufficient to eject the protons with the velocities observed, apparently violating conservation of energy.

 

(b) These experiments showed the existence of the neutron, which were the unknown radiation mentioned in the question. This allowed scientists to correctly identify the composition of the atomic nucleus – that it was composed of protons and neutrons in specific ratios.

 

 

Question 30 (7 marks)

(a) Energy can be converted to mass according to Special Relativity. By accelerating particles to high kinetic energies and colliding them, scientists at particle accelerators were able to create heavy, unstable particles that were not normally observed on Earth. The existence of these particles, their physical properties, and their interactions and decay were used as evidence for the Standard Model of matter.

For example, deep inelastic scattering experiments showed that particles could be removed from protons. This indicated protons are not fundamental and lead to the discovery of quarks. (Many examples would be acceptable here.)

 

(b) (i) Using de Broglie’s equation:

\(\lambda = \frac{h}{mv} = \frac{6.626 \times  10^{-34}}{1.673 \times 10^{-27} \times (0.1 \times 3 \times 10^8)} = 1.32 \times 10^{-14} \text{ m} \)

 

(ii) The relativistic effects would result in its wavelength being shorter. The wavelength is inversely proportional to momentum, \(\lambda = \frac{h}{p}\) , and, at these speeds, the proton’s relativistic momentum should be used which is larger than its non-relativistic momentum.


Question 31 (6 marks)

(a)  Kepler’s Second Law states that the vector joining the comet to the Sun will sweep out equal areas in equal times. As a consequence, the comet must travel at a higher speed when it is close to the Sun and a lower speed when it is further away.

Additionally, when the comet is far away from the Sun it has greater potential energy, and lower kinetic energy and hence speed.

 

(b)  The only way for \(A\) to travel with constant speed is if its orbit is circular. This will only occur if \(A\) and \(B\) are both orbiting in a circle around a common centre of mass, and are at opposite sides of the circle.

The gravitational force on \(B\) from \(A\) must be equal to the centripetal force keeping it in circular motion. Since \(x\) is the distance between the stars, the radius of the circular motion is \(\frac{x}{2}\).

\( F_G = F_C \)

 

\( \frac{Gmm}{x^2} = \frac{mv^2}{\frac{x}{2}} \)

 

\( v = \sqrt{\frac{Gm}{2x}} \)

 

Question 32 (7 marks)

Ultimately the factors that limit the speed are the power of the motor and the coefficient of kinetic friction between the mass and the surface it is sliding on.

The mass will experience friction given by \(f = \mu N\). Since the mass is not accelerating vertically, the normal force will be equal to its weight, hence \(f = \mu mg\). The friction must be balanced by the tension coming from the string (and the motor) for the mass to travel at constant speed.

The maximum speed will be determined by the power of the motor and the force required: \(P = Fv = \mu mg v\).

The force produced by the motor will depend on its torque and the radius of the pulley, \(\tau = Fr = \mu mg r\).

Expressing the mass’s speed in terms of the motor’s angular velocity, \(v = r \omega\) gives:

\(v = \frac{P}{\mu mg} = \frac{\tau \omega}{\mu mg}\).

The above shows that the speed is proportional to the power of the motor and inversely proportional to the coefficient of friction.

Possible additional discussion:

The torque in electric motors is proportional to the current through its coils. This will depend on both the external voltage and the rotation speed (through the back EMF induced by the rotating coils).

The motor’s maximum speed would be that for which the net voltage on the coil allows for sufficient current and torque to balance the torque on the pulley produced by the friction.

 

 

Question 33 ( marks)

The magnet is stationary at \(X\), so it has an energy of \(U = mgh = 0.04 \times 9.8 \times 0.98 = 0.38416 \text{ J}\) relative to the bottom of the cylinder.

As the magnet falls under the influence of gravity its potential energy is converted to kinetic energy. When it reaches the top of the cylinder, its potential energy reduces to \(U = 0.04 \times 9.8 \times 0.2 = 0.0784 \text{ J}\), and its kinetic energy is \(K = 0.30576 \text{ j}\), such that the total energy remains the same at \(0.38416 \text{ J}\).

(Another option: calculate its speed and hence kinetic energy from the graph, noting that it was falling for \(0.4 \text{ s}\).)

The cylinder experiences a change in flux when the magnet begins to pass through it, resulting in an induced EMF according to Faraday’s Law. A current flows around the circumference of the cylinder because it forms a closed circuit.

The current produces a magnetic field which interacts with the magnet to produce an opposing (upwards) force, according to Lenz’s Law.

The opposing force causes the magnet to slow down, and then fall at constant speed as seen in the graph (the force must be balancing its weight). From the graph, the speed of the magnet is \( 0.4 \text{ ms}^{-1}\), calculated by taking the gradient of the distance vs. time graph.

When the magnet reaches the bottom of the cylinder its potential energy is 0 and its kinetic energy is \( K = \frac{1}{2} mv^2 = 0.0032 \text{ J}\).

Its initial energy was \(0.38416 \text{ J}\) meaning that \(0.38416 – 0.0032 = 0.38096 \text{ J}\) were converted into the electrical energy of the induced current in the cylinder, and then to heat through the copper’s resistance.

The force due to Lenz’s Law did negative work on the falling magnet in order to reduce its kinetic energy.

 

 

Question 34 ( marks)

(a)  For \(q_1\):

\(W = qE \times s = q\frac{V}{d} \times  \frac{d}{2} = \frac{qV}{2} \)

 

For \(q_2\):

\(W = q\frac{V}{2d} \times \frac{3}{4}2d = \frac{3qV}{4} \)

 

Hence the work is greater for \(q_2\).

 

(b) The electric field is \(\frac{V}{d}\) for \(q_1\) compared to \(\frac{V}{2d}\) for \(q_2\) so the force \(F = qE\) on \(q_1\) would be stronger, and its acceleration upwards would be larger.

It also has a shorter distance to travel to reach plate \(Y\), so its time of flight and range will be shorter compared to \(q_2\).

 

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Written by Matrix Science Team

The Matrix Science Team are teachers and tutors with a passion for Science and a dedication to seeing Matrix Students achieving their academic goals.

 

© Matrix Education and www.matrix.edu.au, 2018. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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