2025 HSC Physics Exam Paper Solutions

2025 HSC Physics Exam Paper Solutions

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Section 1. Multiple Choice

 

Question 21

Place each source behind a piece of paper with the detector on the opposite side. The source which still gives a signal at the detector is the beta emitter, because alpha particles will be blocked by the paper.

Place each source in the same electric field. The beta particles will deflect more sharply than the alpha particles because their charge to mass ratio is much larger and \(a = \frac{qE}{m}\).

Note that the question does not specify whether the beta particles are positive or negative.

Question 22

Transformer A is a step up transformer. It uses electromagnetic induction to transfer electrical energy from the power station to another circuit at higher voltage. Transformer B also uses electromagnetic induction to reduce the voltage before it reaches the house.

There are small energy losses in the transformers if they are non ideal, because some amount of electrical energy is converted to heat.

There is some energy converted to heat in the transmission line due to its electrical resistance. The rate of energy loss is given by \(P = I^2 R\), where P is the power converted to heat in watts.

Question 23

a) First calculate the force exerted on the 5 cm length of wire:

\(F = LIB \sin \theta\) \(F = 0.05 \times 2 \times 3 \times 10^{-2} \times \sin 90^{\circ} = 3 \times 10^{-3} \text{ N (downwards)}\)

Then calculate the resulting torque:

\(\tau = r F \sin \theta\) \(\tau = 0.2 \times 3 \times 10^{-3} \times \sin 90^{\circ} = 6 \times 10^{-4} \text{ Nm (clockwise)}\)

b) In order to double the magnitude of the torque we must double the force exerted by the magnetic field.

This will occur if the product \(IB\) doubles, for example if both I and B increase by a factor \(\sqrt{2}\).

Question 24

The orbital velocity of satellites in circular orbits is given by \(v = \sqrt{\frac{GM}{r}}\), giving them a kinetic energy of \(K = \frac{1}{2}mv^2 = \frac{1}{2}\frac{GMm}{r}\).

If \(r_A = 3r_B\) then for their kinetic energies to be equal, satellite A must have three times the mass of satellite B, i.e. \(m_A = 3m_B\).

Question 25

When the points are plotted, \(0.9 \times 10^{15} \text{ Hz}\) appears to be an outlier. Without this point, the x intercept and hence the threshold frequency is \(0.7 \times 10^{15} \text{ Hz}\).

The threshold frequency is the minimum frequency that will cause photoelectron emission. This is because at this frequency, individual photons have enough energy \(E = hf\) such that they can release the least tightly bound electrons from the metal. As the frequency of light increases, the photon energy increases and so does the kinetic energy of electrons released from the metal; the stopping voltage also increases as more work is required to stop electrons reaching the anode.

Question 26

The average EMF will be equal to \(\varepsilon = – \frac{N \Delta \Phi}{\Delta t}\).

The flux through the coil in the starting position is zero. After a quarter of a turn, the flux through the loop will be

\(\Phi = BA \cos \theta = 0.5 \times 0.4 \times 0.3 \times \cos 0^{\circ} = 0.06 \text{ Wb}\)

Hence the average EMF is \(\varepsilon = – \frac{1 \times 0.06}{0.1} = -0.6 \text{ V}\). The magnitude of the average EMF is 0.6 V.

The new graph should have double the amplitude and half the period. It should also start from a trough instead of a crest, because rotating the opposite way will reverse the direction of the initial EMF.

Question 27

The Schrödinger model describes the electron as a quantum object with wave like properties instead of a classical particle.

The model describes the electrons existing in three dimensional orbitals instead of one dimensional orbits, with the orbitals representing the probability of finding the electron at each location.

Question 28

The initial velocity components as the rider leaves the first ramp will be \(u_x = u \cos 20^{\circ}\) and \(u_y = u \sin 20^{\circ}\).

The horizontal velocity must be enough to travel 16 metres in a time t. This gives \(t = \frac{16}{u \cos 20^{\circ}}\).

Meanwhile, from the time the bike leaves the ramp to the time it lands, the vertical displacement is zero. Starting from \(s = ut + \frac{1}{2}at^2\):

\(0 = u \sin 20^{\circ} \left(\frac{16}{u \cos 20^{\circ}}\right) – 4.9\left(\frac{16}{u \cos 20^{\circ}}\right)^2\)

Solving this equation for u gives \(u = 15.4 \text{ ms}^{-1}\) (to three significant figures).

Question 29

At the highest point, the only force acting on the mass is its weight. This must provide the entire centripetal force:

\(mg = \frac{mv^2}{r}\)

Solving for speed gives \(v = \sqrt{rg}\).

The question states that mechanical energy is conserved. If we allow the bottom of the circle to be \(h = 0\), then at the top of the circle:

\(U = mg(2r)\) and \(K = \frac{1}{2}mv^2 = \frac{1}{2}mgr\).

At point A on the circle \(U = mgr\). By conservation of energy, \(K = \frac{3}{2}mgr\) and hence \(v = \sqrt{3rg}\).

Question 30

The wavelength of the electrons can be calculated from de Broglie’s equation:

\(\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3} = 1.82 \times 10^{-7} \text{ m} = 182 \times 10^{-9} \text{ m}\)

The equation to calculate diffraction maxima is \(d \sin \theta = m \lambda\). Using the small angle approximation \(\sin \theta \approx \tan \theta\),

\(d \frac{h}{L} = m \lambda\)

We are solving for h. We set \(m = 1\) because B is the first order maximum.

\(h = \frac{mL \lambda}{d} = \frac{0.5 \times 182 \times 10^{-9}}{1 \times 10^{-6}} = 0.091 \text{ m}\)

When the electrons pass through the electron gun, work done on them gives them kinetic energy such that \(qV = \frac{1}{2}mv^2\):

\(V = \frac{mv^2}{2q} = \frac{9.109 \times 10^{-31} \times (4 \times 10^3)^2}{2 \times 1.602 \times 10^{-19}} = 4.55 \times 10^{-5} \text{ V}\)

Question 31

Atoms were previously considered to be fundamental particles before the discovery of the electron. When cathode rays were discovered, various experiments were performed to determine their properties. Because the rays deflected in the presence of electric and magnetic fields, scientists concluded that cathode rays were composed of an unknown, negatively charged particle.

Thomson’s experiment with cathode rays allowed him to determine the charge to mass ratio of the electron. From this value he concluded that it was smaller than an atom. He also found that it was present in all matter, and was therefore a fundamental subatomic particle.

Cathode ray experiments were instrumental in proving the existence of the subatomic electron, but did not completely determine its properties. Eventually the Millikan experiment, which did not use cathode rays, was able to determine the charge on the electron after which it was fully characterised. Cathode ray experiments also did not demonstrate the wave properties of electrons, which were discovered later.

Question 32

The two postulates of Special Relativity state that:

1. The laws of physics are the same in all inertial frames of reference.

2. The speed of light in a vacuum is constant for all observers.

Taken together, these postulates predict that two observers in different inertial frames can disagree on the time passing between two events as well as the length of an object, in order for them to agree on the speed of light.

One prediction of Special Relativity is that “moving clocks run slow” that is, when a frame is moving relative to an observer, time in that frame appears to run more slowly. This is supported by the Hafele Keating experiment, in which highly accurate atomic clocks travelled at different velocities with respect to the centre of the Earth. The result of this experiment was that the plane with the highest velocity recorded a shorter time than clocks that moved more slowly. Because the velocities were small with respect to the speed of light, the effects were on the order of nanoseconds.

Time dilation is also observed for high velocity muons travelling from the upper atmosphere to the surface of the Earth. Without relativistic effects, they would not be able to travel this distance before decaying. Since they are observed at the surface of the Earth, scientists must be observing their dilated lifetime. Muon experiments also support the prediction of length contraction; in the muons’ frame of reference, the distance they travel to the surface of the Earth is contracted.

The effects of Special Relativity can also be seen in particle accelerators, where time dilation extends the lifetime of unstable particles in the laboratory frame and the particles can convert relativistic kinetic energy into other types.

The predictions of Special Relativity correspond to measurable effects on length and time measurements for objects in motion.

Question 33

The Standard Model describes the fundamental particles that make up all matter as well as mediating the interactions between them. The majority of these particles were discovered with particle accelerators.

The electron was discovered in cathode ray tube experiments, which can be considered a rudimentary linear accelerator. By subjecting a beam of cathode rays to electric and magnetic fields, Thomson was able to determine that the electron was more fundamental than the atom, and was present in all atoms.

Einstein’s Special Theory of Relativity predicted that mass was a form of energy. Particle accelerators increase the energy of light, stable particles by factors of several thousand and collide them to form heavier, unstable particles. After a large number of different products were observed, scientists proposed they were made up of even more fundamental particles called quarks. This was supported by deep inelastic scattering, where electrons were used to eject quarks from a proton. Eventually, six quarks and six leptons were discovered.

Force carrier particles were proposed for the weak, strong and electromagnetic forces and eventually discovered in particle accelerators. Ultimately the Standard Model was developed from a series of unexpected observations, followed by theories that were proposed, and later experimentally confirmed.

Question 34

When Earth is at Q, it is \(2.942 \times 10^{11} \text{ m}\) further away from the point where Io emits light. Given that it takes \(1.000 \times 10^{3} \text{ s}\) more for the image of Io to reach Earth, the speed of light must be

\(v = \frac{s}{t} = \frac{2.942 \times 10^{11}}{1.000 \times 10^3} = 2.942 \times 10^8 \text{ ms}^{-1}\)

If the Earth’s orbit is elliptical, the extra distance travelled by light to reach it on the opposite side of the orbit after 6 months will depend on the alignment of Earth and Jupiter in their orbits. Hence the time for Io to reach position R will vary by different amounts depending on the positions of the objects. The calculation will have to account for the changing distances as well as the changing times. The speed of light will be the same.

Question 35

When the magnet first enters the copper pipe, changing magnetic flux induces eddy currents in the walls of the pipe. These exert a force upwards on the magnet according to Lenz’s Law. There is a reaction force according to Newton’s Third Law, and the magnet pushes down on the pipe so the scale reading goes up.

The magnet slows down until it reaches terminal velocity, where the magnetic force upwards equals the weight downwards. The additional reaction force on the pipe therefore reaches a value equal to the weight of the magnet, which the scale converts to an additional mass of 50 g.

Question 36

According to the Law of Conservation of Momentum, the total momentum of the two pieces must equal the total momentum of the original satellite.

Initial momentum \(= (m_a + m_b)v\)

Final momentum \(= m_a \times 2v + m_b v_b\)

Since the masses are equal, the piece \(m_b\) must have a velocity of zero after the explosion.

Piece B will experience a gravitational force towards the centre of the Earth and will begin to fall towards the surface in a straight line. As it approaches the Earth, the gravitational field strength and acceleration will increase until it hits the surface.

Meanwhile, piece A is travelling at a speed twice that of the orbital speed: \(v = 2\sqrt{\frac{GM}{r}}\) which is greater than the escape velocity \(v = \sqrt{\frac{2GM}{r}}\). Hence piece A will move in an escape path that gets further from Earth and eventually reaches an infinite distance away.