HSC Physics 2024

(a) [2 marks]. Convert to SI units first: $r = 18\text{ cm} = 0.18\text{ m}$.

(b) [4 marks]. Torque in a DC motor follows $\tau = nIA_{\perp}B$.

1. Increase the current I (for example by raising the supply voltage). A larger current increases the force on each side of the coil, $F = BIL$, so the torque increases.
2. Increase the number of turns n in the coil. Each extra turn is another current-carrying loop in the field, so the force, and hence the torque, scales with the number of turns.

(Increasing the coil area A, or using a stronger or radial field B, are also acceptable.)

Marker's note. Convert centimetres to metres before substituting, and read the angle correctly from the stimulus. Direct the explanation at the DC motor, linking each change to the torque equation, not at the spanner.

(a) [2 marks]. The straight line shows that the more distant a galaxy, the faster it recedes (recessional velocity is directly proportional to distance). This is evidence that the universe is expanding, and, traced backward, supports the Big Bang model of an expanding universe with a common origin.

(b) [3 marks]. The recessional velocities were found from red shift. The light from each galaxy shows the hydrogen spectral lines, but shifted toward longer (redder) wavelengths compared with a stationary hydrogen source measured on Earth. The size of this shift gives the velocity: the greater the red shift, the faster the galaxy is receding, so comparing each galaxy's shifted spectrum with the laboratory hydrogen spectrum yields its recessional velocity.

Marker's note. In (a) state the relationship from the graph and link it specifically to the expansion of the universe (not stars or planets). In (b) outline a clear method, comparing the red-shifted galactic spectra to a stationary hydrogen reference rather than just naming "red shift" or "Doppler effect".

(a)(i) [2 marks]

Paraffin is rich in hydrogen, whose nuclei are single protons. When the unknown radiation from the beryllium struck the paraffin, it knocked these protons out, and the energy and momentum of the ejected protons could be measured. This let Chadwick work out the properties of the unknown radiation.

(a)(ii) [3 marks]

From the energy of the ejected protons, Chadwick deduced that the unknown radiation was not a gamma ray but a stream of neutral particles with about the same mass as a proton. Naming this particle the neutron changed the model of the atom: the nucleus was now understood to contain both protons and neutrons, which explained the difference between atomic number and mass number and the existence of isotopes.

(b) [4 marks]

The Bohr-Rutherford model had to assume, without explanation, that electrons orbit only in certain allowed shells and, contrary to classical electromagnetism, do not radiate energy and spiral into the nucleus. De Broglie proposed that electrons have a wavelength, $\lambda = h/(mv)$. An electron orbit is then stable only when its circumference fits a whole number of electron wavelengths, forming a standing wave, $2\pi r = n\lambda$. This explained why only particular orbits (energy levels) are allowed. Because a standing wave does not represent an accelerating point charge radiating energy, it also removed the problem of the electron continuously losing energy and collapsing into the nucleus.

Marker's note. In (a)(i) tie paraffin's hydrogen content to the ejection of protons for analysis. In (a)(ii) link the neutral, proton-mass particle to the revised nuclear model. In (b) connect the standing-wave (whole number of wavelengths) condition to both the allowed orbits and the radiation problem of the earlier model.

(a) [2 marks]. Read the peak of the underlying continuous (black body) curve, at about $\lambda_{max} = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$. By Wien's law:

(b) [3 marks]. Use the transition $n_f = 2$, $n_i = 6$ with $R = 1.097 \times 10^{7}\text{ m}^{-1}$:

(c) [3 marks]. The hot, dense surface of the star emits a continuous spectrum of all visible wavelengths. As this light passes outward through the cooler hydrogen atmosphere, photons whose energy ($E = hf$) exactly matches the gap between two electron energy levels are absorbed, lifting electrons to higher levels. Those particular wavelengths are removed from the beam reaching us (the re-emitted light goes in all directions), so the spectrum shows dark dips of reduced intensity at those wavelengths, which are the absorption lines.

Marker's note. In (a) take the peak from the continuous curve, not from a dip. In (b) show the substitution of every value and convert correctly from $1/\lambda$ to $\lambda$ to f. In (c) describe the full process: a continuous source, absorption at quantised energies matching electron transitions, and a resulting decrease in intensity at those wavelengths.

(a) [2 marks]. For a circular orbit, gravity provides the centripetal force:

Substitute the orbital speed $v = \dfrac{2\pi r}{T}$:

Rearranging gives $T^2 = \dfrac{4\pi^2}{GM}r^3$.

(b) [4 marks]. From the line of best fit, take the gradient as $\dfrac{r^3}{T^2}$. Reading the graph, the gradient is about $\dfrac{1 \text{ unit of } 10^{12}\text{ km}^3}{0.0008\text{ days}^2}$. Convert to SI units, with $1\text{ km}^3 = 10^{9}\text{ m}^3$ and $1\text{ day} = 86\,400\text{ s}$:

This gradient is $\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM}$, so:

Marker's note. In (a) start from "by considering gravitational force" and show the substitution and rearrangement in sequence. In (b) calculate the gradient from two points on the line of best fit (show the triangle), convert both km$^3$ to m$^3$ and days$^2$ to s$^2$, then take the inverse of the gradient before substituting into Kepler's third law.

[3 marks]. Both frames agree the muons arrive, but explain it with different relativistic effects.

• Earth's frame of reference: the muon is moving very fast, so its decay clock runs slow by time dilation. Its half-life, as measured by an Earth observer, is stretched, so far more muons survive the travel time to reach the ground than the rest-frame half-life would suggest.
• Muon's frame of reference: the muon is at rest and Earth rushes up to meet it, so the distance to the surface is shortened by length contraction. Because the muon only has to cover this much smaller distance, it reaches the ground within its own (un-dilated) half-life.

Marker's note. Name the effect for each frame correctly: time dilation in Earth's frame, length contraction in the muon's frame. Avoid undefined acronyms such as "FoR", and keep the observed effect tied to the frame doing the observing.

(a) [2 marks]. Each pion is a quark-antiquark pair giving the right charge:

• $\pi^-$ (charge $-1$): $d\bar{u}$, since $-\tfrac{1}{3} + (-\tfrac{2}{3}) = -1$.
• $\pi^+$ (charge $+1$): $u\bar{d}$, since $+\tfrac{2}{3} + \tfrac{1}{3} = +1$.
• $\pi^0$ (charge 0): $u\bar{u}$, since $+\tfrac{2}{3} + (-\tfrac{2}{3}) = 0$.

(b) [2 marks]. Total rest mass before and after:

The mass converted to energy is $1880 - 420 = 1460\text{ MeV/c}^2$, shared over three pions:

(c) [3 marks]. The problem is that nothing with mass can reach or exceed the speed of light ($3 \times 10^{8}$ m s$^{-1}$), so a calculation predicting this is invalid. The resolution is relativistic: as a pion gains energy its relativistic mass (and momentum) increases without limit as its speed approaches c, so progressively more of the released energy goes into mass rather than into extra speed. The classical assumption that all the energy becomes ordinary kinetic energy ($\tfrac{1}{2}mv^2$) is wrong; using the relativistic relation keeps each pion's speed below c.

Marker's note. Read (a) carefully and give the correct quark-antiquark pair for each pion. In (b) work with the mass change and keep units consistent (MeV/c$^2$ to MeV). In (c) explain how the problem is solved (energy going into increased relativistic mass and momentum), not merely that exceeding c is impossible.

(a) [2 marks]. The electric force on the electron is $F = qE$, and $F = ma$, so:

(b) [2 marks]. The horizontal speed is constant, so the time spent between the plates (width $5.0\text{ mm} = 5.0 \times 10^{-3}\text{ m}$) is:

Vertical displacement (starting from zero vertical speed):

(c) [3 marks]. Vertical speed on leaving the plates:

Time to cross the 30 mm gap to the screen at the constant horizontal speed:

Extra vertical drop over this gap:

Total distance from X = displacement inside the plates + drop after the plates:

Marker's note. Convert millimetres to metres before substituting. In (b) and (c) show the chosen equation and each substitution. Carry the answers from (a) and (b) forward correctly into (c) rather than restarting.

(a) [2 marks]. Connect the battery so that current flows through both rods in opposite directions (parallel currents in opposite directions repel, pushing the rods apart into position 2). With the rods made of different materials, completing the circuit so each rod carries current from the same battery gives different current magnitudes because the rods have different resistance. The figure below shows the required connection.

(b) [4 marks]. The statement is incorrect as a cause of the larger displacement. By Newton's third law (and Ampere's force law) the magnetic force between two parallel current-carrying rods is equal in magnitude on both:

This single expression gives the force on A and on B; rod A pushes B with exactly the force with which B pushes A. So even though A carries the larger current, B does not feel a larger force than A feels. The reason B moves further must be that B has the smaller mass: for equal forces, the lighter rod has the larger acceleration ($a = F/m$) and so the larger displacement in the same time ($s = \tfrac{1}{2}at^2$).

Marker's note. Recognise that the wires (not the rods) have negligible mass. Use Newton's third law to argue the forces are equal in magnitude, then attribute B's greater displacement to its smaller mass and hence larger acceleration, supporting the argument with the force and motion equations.

[4 marks]. Three forces act on the object: gravity (down), the normal force from the floor (up) and the centripetal force from the cylinder wall (toward the centre).

The centripetal force relates to the period through $v = \dfrac{2\pi r}{T}$:

Because $F_c \propto \dfrac{1}{T^2}$, halving the period (T to T/2) increases the centripetal force by a factor of 4 (the wall pushes the object inward four times as hard).

The gravitational force is $mg$ and is unchanged, since g and m do not depend on the period. With no vertical acceleration, the floor's normal force still balances gravity, so it is also unchanged. Only the horizontal (centripetal) force changes.

Marker's note. Address every force, not just the centripetal one. Show the $F_c \propto 1/T^2$ relationship to justify the factor of 4, and explain that the vertical forces (gravity and the normal reaction) are unaffected by the period.

[4 marks]. In both models the projectile starts with the same kinetic energy, which is fully converted to gravitational potential energy at the maximum height (where the speed is zero), so $\tfrac{1}{2}mv^2 = \Delta U$.

• Model A (uniform field): g is constant, so the potential energy gained per metre of height, $mg\,\Delta h$, is the same all the way up. The kinetic energy is "spent" at a constant rate with height.
• Model B (radial field): the field strength decreases with height ($g \propto 1/r^2$), so the potential energy gained per metre of height shrinks as the projectile rises. The kinetic energy is spent more slowly the higher it goes.

Because Model B converts kinetic energy to potential energy more slowly with height, the projectile travels further before all its kinetic energy is used up, so it reaches a greater maximum height in Model B than in Model A.

Marker's note. Address both parts: compare the maximum heights and describe the energy conversion (kinetic to gravitational potential). Link the greater height in Model B explicitly to the weakening field with altitude, and avoid irrelevant detail.

[8 marks]

Three experiments on light-matter interaction reshaped physics.

1. Black body radiation (Planck, from the spectra of heated bodies)

The measured intensity-wavelength curves could not be explained by classical wave theory (the "ultraviolet catastrophe"). Planck fitted the data only by assuming energy is emitted in discrete packets, $E = hf$. This introduced quantisation and began quantum physics.

2. The photoelectric effect (Hertz and Lenard, explained by Einstein)

Electrons are ejected from a metal only above a threshold frequency, and their maximum kinetic energy depends on frequency, not intensity ($K_{max} = hf - \phi$). Wave theory could not explain the threshold or the instant emission. Einstein's photon model, treating light as quanta each carrying energy $hf$, explained every feature and gave strong evidence for the particle nature of light.

3. Spectroscopy (emission and absorption spectra of gases)

Each element produces a unique set of discrete spectral lines whose frequencies match transitions between quantised electron energy levels (the Balmer series, $\tfrac{1}{\lambda} = R(\tfrac{1}{n_f^2} - \tfrac{1}{n_i^2})$). This evidence supported the Bohr model of the atom and let astronomers identify the composition, temperature and motion (via red shift) of stars, advancing astrophysics.

Together this evidence drove the shift from a purely classical, continuous picture of light and energy to a quantised one, underpinning quantum physics and the modern atomic model.

Marker's note. Choose at least three genuine light-and-matter experiments, present the key observation from each, and link each clearly to the physics it advanced (quantisation, the photon, the atomic model). Structure the response and, where possible, use equations to support the analysis rather than just listing experiments.

[7 marks]

The magnet's motion and energy. As the magnet swings down from its high point it converts gravitational potential energy into kinetic energy, moving fastest at the bottom of the swing, where the relative motion between magnet and can is greatest.

Induction in the can

The moving magnet produces a changing magnetic flux through the nearby aluminium can. By Faraday's law, $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$, this induces an emf and, because aluminium conducts, eddy currents flow in the can. The induced current is largest when the rate of flux change is largest, near the bottom of the swing.

The can's motion (Lenz's law)

By Lenz's law the eddy currents create a magnetic field that opposes the change causing them, so they drag on the magnet and the can experiences a reaction. The can is pulled to follow the magnet, rotating about its axis first one way then the other, with diminishing amplitude as the magnet's swing decays.

Overall energy flow

The eddy currents dissipate energy as resistive (ohmic) heating in the can. So the magnet's mechanical energy is progressively transferred, via electromagnetic induction, into kinetic energy of the can and ultimately into heat. Both the magnet's swing and the can's rotation damp out over time, until effectively all the original gravitational potential energy of the magnet has become thermal energy.

Marker's note. Describe the interaction in terms of energy transformations (gravitational to kinetic to electrical to heat), not only forces. Track the motion of both the magnet and the can over a full cycle, and use induction (Faraday and Lenz) to predict the can's motion and the damping of the magnet.