HSC Physics 2025

[3 marks]. Two methods, each with the expected observation:

1. Penetration test. Place a sheet of paper between each source and a Geiger counter. The source whose count rate drops almost to background is the alpha emitter (alpha is stopped by paper); the source whose count is barely reduced is the beta emitter (beta passes through paper and is stopped only by a few millimetres of aluminium).
2. Magnetic deflection. Pass each beam through a magnetic field at right angles to its path. The beams deflect in opposite directions because alpha is positive and beta is negative, and the less-massive beta particles deflect much more than the heavier alpha particles, identifying each source.

Marker's note. State a clear observational difference for each method (for example "alpha is stopped by paper, beta passes through"), not just "test with paper". Choose methods designed for radiation identification and avoid unrelated experiments such as Rutherford scattering or Chadwick's neutron discovery.

[4 marks]. In transformer A and transformer B, most of the electrical energy is transferred from the primary to the secondary coil by a changing magnetic flux, but some electrical energy is transformed into heat. This heating comes from eddy currents induced in the iron core, from magnetic hysteresis as the core repeatedly re-magnetises, and from resistive heating in the windings, so the energy output is slightly less than the input.

In the transmission line, some electrical energy is transformed into heat in the wires because of their resistance ($P_{loss} = I^2R$). Transmitting at high voltage keeps the current low and limits this loss, which is why transformer A steps the voltage up before transmission and transformer B steps it down again.

Marker's note. Keep the focus on energy transformations in both transformers and the line, not a general description of step-up and step-down action. Naming the specific loss mechanisms (eddy currents, hysteresis, resistive heating) shows deeper understanding.

(a) [3 marks]. Force on the current-carrying side in the field:

The pivot is 0.20 m from this side, so the torque is

The torque acts to rotate the loop about its axis (clockwise viewed from the side shown).

(b) [2 marks]. Torque is proportional to $B \times I$, and it must stay in the same direction, so neither $B$ nor $I$ may be reversed. To double the product while keeping both positive, for example quadruple the magnetic field and halve the current ($4B \times \tfrac{1}{2}I = 2BI$). Any factor pair multiplying to 2 with both unchanged in sign works.

Marker's note. The wire side that gives the torque is the one in the field; the loop area is not needed because the coil is not fully inside a uniform field. In (b), quantify both changes so they multiply to a factor of 2 without reversing direction.

[3 marks]. For a circular orbit the gravitational force provides the centripetal force:

The kinetic energy is therefore

Setting the kinetic energies equal:

With $r_A = 3r_B$,

Marker's note. Show every step from the circular-orbit condition to the kinetic-energy expression, then equate. Avoid bringing in unnecessary formulas; the clean route is $E_k = GMm/2r$.

(a) [3 marks]. Plot stopping voltage (y-axis) against frequency (x-axis) and draw a straight line of best fit through the points. The data are linear (a 0.5 V rise for every $0.1 \times 10^{15}$ Hz). Extrapolating the line back to where the stopping voltage is zero gives the threshold frequency:

(b) [3 marks]. In the particle model, light arrives as photons of energy $E = hf$. A photoelectron is ejected only when a single photon supplies at least the work function $\phi$ of potassium, which sets the threshold frequency $f_0 = \phi/h$ below which no current flows however bright the light. Above $f_0$, the surplus energy becomes the electron's maximum kinetic energy, $E_{k,\max} = hf - \phi$. The stopping voltage measures this kinetic energy ($qV_{stop} = E_{k,\max}$), so as the frequency rises the photon energy rises, the electrons leave with more kinetic energy, and a larger stopping voltage is needed - giving the straight-line increase observed.

Marker's note. Tie each graph feature to the model: the x-intercept to the threshold frequency, and the positive slope to $E_k = hf - \phi$. The gradient of the line equals $h/q$.

(a) [2 marks]. Average emf is the rate of change of flux, $\varepsilon = \Delta\phi/\Delta t$ with $\phi = BA$. The loop turns from maximum flux to zero flux through the 90 degrees, so $\Delta\phi = BA$:

(b) [3 marks]. The new curve should show three changes relative to the 10 rev/s curve:

• Double the frequency: twice as many cycles in the same time (the period halves from 0.1 s to 0.05 s).
• Double the peak voltage: peak emf is proportional to the rotational speed, so the amplitude doubles.
• Starts at maximum magnitude with opposite sign: at the given starting position the emf begins at its peak value, and reversing the rotation direction flips the sign, so the curve begins at the opposite-sign peak.

Marker's note. Mark in guiding points first (maxima, minima, intercepts). The increased speed changes both the number of cycles and the peak voltage, and at this starting position the emf begins at its greatest magnitude, not at zero.

[3 marks]. Two differences:

1. Probability clouds versus fixed paths. In Rutherford's model the electron is a particle on a definite orbit, and in Bohr's model it follows fixed circular orbits at set radii. Schrödinger replaced these with orbitals - three-dimensional regions describing only the probability of finding an electron, with no definite path.
2. Wave description. Following de Broglie, Schrödinger treated the electron as a standing matter wave described by a wavefunction, rather than as a classical particle as in the Rutherford and Bohr models.

Marker's note. Keep the focus on electron behaviour (probability orbitals and wave nature), not on protons or neutrons. Compare Schrödinger separately against Rutherford and against Bohr, not against a merged "Rutherford-Bohr" model.

[4 marks]. Treat it as projectile motion launched at $\theta = 20$ degrees. The two ramp tops are at the same height, so the vertical displacement over the flight is $s_y = 0$. Resolve the launch speed $u$ into components $u_x = u\cos\theta$ and $u_y = u\sin\theta$.

Vertical (taking up as positive, $a = -9.8\ \text{m s}^{-2}$):

Horizontal: $s_x = u_x t = u\cos\theta \cdot t$. Substituting for $t$:

Rearranging for $u$ with $s_x = 16$ m and $\theta = 20$ degrees:

So the minimum launch speed is about 15.6 m s$^{-1}$.

Marker's note. Resolve the velocity into components before using the kinematics equations, and use $s_y = 0$ because the ramps are level. Working symbolically first and substituting numbers last avoids rounding errors.

(a) [2 marks]. At B the string tension is zero, so gravity alone supplies the centripetal force:

(b) [3 marks]. A is lower than B by a height $r$, and mechanical energy is conserved. Take A as the reference level:

Substituting $v_B^2 = rg$:

So $v_A = \sqrt{3}\,v_B \approx 1.7\,v_B$: the mass is faster at A. As it climbs from A to B it converts kinetic energy into gravitational potential energy, so it slows.

Marker's note. Recognise that the object has both kinetic and potential energy at A and at B, and use conservation of energy to relate the two speeds mathematically rather than just stating "A is faster".

(a) [2 marks]. Use the de Broglie relation $\lambda = h/(mv)$ with the electron mass $9.109 \times 10^{-31}$ kg:

(b) [2 marks]. For the first bright fringe ($m = 1$), the fringe spacing is $x = m\lambda L / d$:

(c) [2 marks]. The work done by the accelerating voltage equals the gain in kinetic energy, $qV = \tfrac{1}{2}mv^2$:

Marker's note. Show the substitution into de Broglie's equation, not just the final value. Watch the unit prefixes (mm, nm) in (b), and remember to square the velocity and halve it in (c).

[5 marks]. Cathode-ray experiments were essential to establishing both that electrons exist and what their properties are.

• Existence and particle nature. Crookes observed that cathode rays travelled in straight lines and cast sharp shadows, and that a paddle wheel placed in the beam turned - evidence the rays were streams of particles carrying momentum, not waves. Deflection by an electric field then showed the particles were negatively charged (electromagnetic waves would not be deflected).
• A constituent of all atoms. Thomson found the rays behaved identically regardless of the cathode material, so the particle was a building block common to every atom, not a product of a particular metal - strong evidence the electron is a fundamental constituent of matter.
• Quantitative properties. By balancing electric and magnetic deflections, Thomson measured the charge-to-mass ratio of the electron and found it far larger than that of a hydrogen ion, implying a very small mass.

Overall these results, building on one another, moved the electron from a hypothesis to an established sub-atomic particle with a measured charge sign and charge-to-mass ratio, so their contribution was decisive.

Marker's note. Address all parts: link specific experiments to the existence of the electron, not only its properties, and finish with an assessment (how decisive the contribution was) rather than a list.

[8 marks]

Special relativity rests on two postulates: the laws of physics are the same in all inertial frames, and the speed of light in a vacuum is constant for all observers. Three consequences follow.

Time (time dilation)

A clock moving relative to an observer runs slow, $t = t_0/\sqrt{1 - v^2/c^2}$. The proper time $t_0$ is measured in the frame where the events occur at one place.

Length (length contraction)

An object moving relative to an observer is contracted along the direction of motion, $L = L_0\sqrt{1 - v^2/c^2}$.

Motion (relativistic momentum and the speed limit)

Momentum becomes $p = mv/\sqrt{1 - v^2/c^2}$, which grows without limit as $v \to c$. A massive object therefore needs ever more energy to accelerate and can never reach $c$.

Experimental evidence - cosmic-ray muons

Muons created high in the atmosphere have a half-life so short that, classically, almost none should reach the ground, yet many do.

• From Earth's frame, the muons' internal clocks run slow (time dilation), so they survive long enough to arrive.
• From the muons' frame, the atmosphere is length-contracted, so they have a much shorter distance to travel.

Both descriptions agree on the measured muon flux, confirming time dilation and length contraction together.

Experimental evidence - particle accelerators. Particles such as electrons require increasing energy for diminishing speed gains as they approach $c$, exactly matching the relativistic-momentum prediction and confirming that $c$ is an unreachable limit. Together these results show the consequences for time, length and motion are real, not merely mathematical.

Marker's note. Address each of time, length and motion, with mathematical detail or a thought experiment for each, and tie the consequences to specific evidence (the muon flux, accelerator energy requirements). A clear diagram or sketch helps.

[6 marks]. The Standard Model grew from a continual interplay of prediction and experimental confirmation across its three pillars - the matter particles (quarks and leptons), the force-carrying bosons, and the Higgs mechanism.

• Theory proposing, experiment confirming. Quarks were first proposed theoretically to explain patterns among the many particles being discovered. Deep inelastic scattering in particle accelerators then showed that protons and neutrons have internal point-like structure, confirming quarks exist.
• Force carriers (bosons). The theory predicts forces are mediated by gauge bosons - for example the photon for electromagnetism and the W and Z bosons for the weak interaction. Electroweak theory unified the electromagnetic and weak forces, and accelerator experiments later detected the W and Z bosons with the predicted masses, validating it.
• The Higgs boson. The Higgs mechanism was proposed decades in advance to explain how particles acquire mass; the discovery of a Higgs boson at the Large Hadron Collider confirmed this final piece. (The Higgs is not a force carrier.)
• Earlier tools. Simpler experiments such as the cloud chamber confirmed antimatter after its theoretical prediction, an early example of the same theory-then-experiment cycle.

So experimental evidence repeatedly tested and confirmed theoretical proposals, and each confirmation enabled the next prediction, which experiment then verified - the loop that built the Standard Model.

Marker's note. Cover all three components (fundamental particles, force-carrying bosons, the Higgs), describe a suitable accelerator or cloud-chamber experiment, and show the Higgs boson is not a gauge boson.

(a) [2 marks]. Light crosses the diameter of Earth's orbit, twice the radius:

(b) [3 marks]. With a circular orbit the path length is fixed at the diameter. If the orbit is elliptical, the distance light must travel between the two observing positions depends on where in the ellipse Earth lies relative to the line of sight to Jupiter. If the measurement is taken along the major (long) axis, the path is longer than $2.942 \times 10^{11}$ m, so dividing by the same measured delay gives a larger calculated speed of light. If taken along the minor (short) axis, the path is shorter and the calculated speed is smaller. The circular-orbit assumption therefore introduces an error whose sign depends on Earth's position in the ellipse.

Marker's note. Remember to double the radius in (a). In (b), relate the change in path length to the change in the calculated speed, and note that an annotated diagram of the long versus short axis helps.

[4 marks]

The falling magnet changes the magnetic flux through the copper pipe, inducing eddy currents in the pipe (Faraday's law). By Lenz's law these currents oppose the change, so they exert an upward retarding force on the magnet.

Phase 1 (accelerating, reading rises)

Just after release the magnet speeds up, so the rate of change of flux and the induced eddy currents grow, increasing the upward braking force on the magnet. By Newton's third law the magnet pushes down on the pipe with an equal and opposite force, which adds to the pipe's weight on the balance, so the reading climbs above 300 g.

Phase 2 (terminal velocity, constant 350 g)

When the upward magnetic force equals the magnet's weight (50 g of force), the magnet stops accelerating and falls at constant terminal velocity. The downward reaction force on the pipe is then steady and equal to the magnet's weight, so the balance reads the pipe plus the magnet, $300 + 50 = 350$ g, and stays there.

Marker's note. Split the answer into the two phases and connect each to the balance reading. Use Newton's third law to explain why the force on the magnet appears as an added downward force on the pipe.

[8 marks]

Conservation of momentum at the explosion

The two fragments have equal mass, so by conservation of momentum their changes in momentum are equal and opposite. Piece $m_a$ gains an extra $v$ in its direction of travel (going from $v$ to $2v$), so $m_b$ must lose $v$, that is its velocity changes by $-v$. Since $m_b$ was moving at $v$, it becomes stationary relative to Earth at that instant.

Motion of $m_a$

For a circular orbit, $v_{orbit} = \sqrt{GM/r}$, and the escape velocity from that radius is $v_{esc} = \sqrt{2GM/r} = \sqrt{2}\,v_{orbit}$. The new speed of $m_a$ is $2v = 2v_{orbit}$, which exceeds $v_{esc}$ (because $2 > \sqrt{2}$). Its total mechanical energy ($E_k + E_p$) is therefore positive, so $m_a$ follows an open (hyperbolic) path: it travels away from Earth, slowing as gravity does negative work on it, but never stops and never returns.

Motion of $m_b$

With zero velocity at point Y, $m_b$ has only gravitational potential energy. It falls straight toward Earth's centre. Its acceleration starts below $9.8\ \text{m s}^{-2}$ (it is far above the surface) and increases as it approaches Earth, so it speeds up at an increasing rate. By conservation of energy, the sum of its kinetic and potential energy stays constant and equal to its initial potential energy until it reaches the atmosphere.

Marker's note. Use conservation of momentum to fix $m_b$'s initial velocity at zero, then conservation of energy plus the escape-velocity formula to show $m_a$ escapes while $m_b$ falls back. Focus on the subsequent motion of both pieces, supported by the relevant formulae.