HSC Physics 2023

(a) [2 marks]. The luminosity of a star is set by its surface temperature and its radius (surface area). From the Stefan-Boltzmann relation $L = 4\pi R^2 \sigma T^4$, a larger radius or a higher temperature both raise the luminosity. (Mass is also accepted, since it determines both of these.)

(b) [3 marks]. Star A is a main-sequence star: it is fusing hydrogen into helium in its core, is comparatively large and hot, and has a high luminosity (well above $1\,L_{sun}$). Star B is a white dwarf: it has no nuclear fusion occurring, is very small and dense (about Earth-sized), and despite a high surface temperature has a low luminosity (below $1\,L_{sun}$) because of its tiny surface area. So Star A is far more luminous and is still fusing fuel, whereas Star B is a stellar remnant.

Marker's note. Keep luminosity (total power output) distinct from brightness (as seen from Earth). For (b), read the H-R diagram carefully and compare real features such as fuel source and size, not just position on the chart.

[3 marks]. The proper time $t_0 = 3.1 \times 10^{-9}$ s is measured on the spacecraft (where the two pulses occur at the same place). The Earth observer measures the dilated time:

Marker's note. The time measured on the spacecraft is the proper (relativistic) time $t_0$. The common error was evaluating the denominator $\sqrt{1 - 0.9^2}$ incorrectly.

(a) [2 marks].

(b) [3 marks]. The minimum photon energy corresponds to the longest wavelength ($\lambda = 2.8 \times 10^{-5}$ m), since $E = hc/\lambda$:

(c) [2 marks]. Using Wien's displacement law $\lambda_{max} = b/T$ with $b = 2.898 \times 10^{-3}$ m K:

Marker's note. Handle scientific notation carefully (for example $6 \times 10^{-7}$ equals $0.0000006$). In (b) the lower energy goes with the longer wavelength. In (c) report the temperature in kelvin, the SI unit used in astronomy.

[3 marks]. The magnetic force supplies the centripetal force, so $qvB = \dfrac{mv^2}{r}$, giving:

The field must be perpendicular to the velocity (out of the page for the path shown) so that $F = qv \times B$ points toward the centre.

Marker's note. Equate the magnetic force to the centripetal force ($F = qvB = mv^2/r$). Magnetic field is a vector, so state a direction. Ampere's law (for straight conductors) does not apply to a single moving charge.

(a) [2 marks]. Part X is the split-ring commutator. Its function is to reverse the direction of the current in the coil every half turn, just as the coil passes through the plane perpendicular to the field. This keeps the force on each side of the coil acting the same rotational way, so the coil continues turning in one direction rather than reversing.

(b) [2 marks]. As the coil spins it cuts field lines and, by Lenz's law, induces a back emf that opposes the supply voltage. The back emf grows with rotational speed, so the net voltage and therefore the motor current fall. Since torque is proportional to current ($\tau \propto BIA$), the torque decreases as the speed increases.

Marker's note. In (a), answer the key word "describe" and name the commutator's role (reversing the current). In (b), show the cause-and-effect chain: faster rotation, larger back emf, smaller current, smaller torque. Eddy currents and back emf are not the same thing, and parts (a) and (b) are separate.

[3 marks]. Mass defect = (mass of products) - (mass of reactants):

The magnitude $0.056$ u is converted to energy. Using $1\ \text{u} = 931.5$ MeV:

So about 52 MeV is released.

Marker's note. Transpose the masses from the table carefully and set out the working logically. The table gives mass in atomic mass units, not energy. Avoid rounding too early, which loses significant figures. State the unit (MeV or joules).

(a) [4 marks]

Composition: each element absorbs and emits light at a unique set of wavelengths. The pattern of dark absorption lines in a star's spectrum is matched against the known line pattern of elements measured in the laboratory; the elements whose patterns appear are the elements present in the star's outer layers.

Temperature

a star radiates approximately as a black body, with a continuous curve whose peak intensity lies at a wavelength $\lambda_{max}$ that depends only on temperature (Wien's law, $\lambda_{max} = b/T$). Measuring the wavelength of peak intensity gives the surface temperature: a hotter star peaks at a shorter (bluer) wavelength. The relative strengths of certain lines also indicate temperature.

(b) [4 marks]

As the star rotates on its axis, one edge approaches Earth and the other recedes. Light from the approaching edge is blue-shifted to a slightly shorter wavelength, and light from the receding edge is red-shifted to a slightly longer wavelength; light from the centre is unshifted. Because the whole disc contributes a continuous range of these shifts, the single 656 nm line is broadened (smeared symmetrically about 656 nm) rather than split. A faster rotation gives a wider line.

Marker's note. In (a) be explicit that the wavelength of peak intensity (not "maximum wavelength") gives temperature via Wien's law, and link unique element line patterns to composition. In (b) tie blue-shift and red-shift to the approaching and receding sides of the rotation (not the star's revolution), giving a broadened line.

(a) [2 marks]. For an ideal transformer $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$:

With the switch open only globe X carries current, so the full 40 V of the secondary is across globe X.

(b) [3 marks]. Closing the switch puts globes X and Y in parallel across the secondary. Two equal resistances in parallel have a smaller total resistance than X alone, so by Ohm's law ($I = V/R$) the secondary current increases. An ideal transformer conserves energy, so $P_{primary} = P_{secondary}$; with the primary voltage fixed at 240 V, a larger secondary power draws a larger primary current. So the primary current is greater after the switch is closed.

Marker's note. In (a) show the substitution into the turns ratio. In (b) work through the chain: adding Y in parallel lowers the secondary resistance, raising the secondary current, and conservation of energy then raises the primary current.

[4 marks]. Light from an incandescent lamp is unpolarised: the electric-field oscillations of its electromagnetic waves point in all directions perpendicular to the direction of travel, with equal intensity in every plane.

A plane polarising filter transmits only the component of each wave's electric field that is parallel to its polarising axis, and absorbs the perpendicular component. For a wave at an angle $\theta$ to the axis, only the projected component (proportional to $\cos\theta$) passes. Averaged over all the random orientations in unpolarised light, only about half the energy gets through, and what emerges is plane-polarised. Because part of every wave's field is absorbed, the transmitted intensity is reduced.

Marker's note. State the key feature of the light before (unpolarised) and after (polarised) the filter, and explain that the filter absorbs the field components not parallel to its axis, reducing intensity. A labelled diagram strengthens the answer.

(a) [4 marks]. From 0 to 0.03 s the coil's magnetic field, and hence the flux through ring X, increases at a constant rate. By Faraday's law this changing flux induces an emf and therefore a current around the complete ring. By Lenz's law that induced current opposes the increase, creating a magnetic field that opposes the coil's field, so the ring is repelled from the coil with a steady force over this interval.

From 0.03 s to 0.05 s the field (and flux) is constant, so there is no change in flux, no induced emf, no induced current, and therefore no magnetic force on the ring.

(b)(i) [2 marks]. Ring Y has a gap, so it is not a complete circuit. A changing flux still induces an emf around it, but no current can flow, so there is no induced magnetic field and no force acts on the ring; it stays still.

(b)(ii) [2 marks]. The maximum emf occurs while the flux is changing (0 to 0.03 s). Using $\text{emf} = \dfrac{\Delta\Phi}{\Delta t} = \dfrac{(\Delta B)\,A}{\Delta t}$:

Marker's note. In (a) connect the rate of change of flux to the induced current and then to the repulsive force, and note there is no force when the field is constant. In (b)(i) the gap means no current and so no motion. In (b)(ii) substitute the correct values, including the powers of ten.

[5 marks]. Similarities. Both curves have the same overall shape. As the car enters the fin, more magnets engage and the rate of change of flux in the aluminium rises, so the induced eddy currents and the opposing (braking) force grow, giving an increasing deceleration that reaches a maximum at about 0.8 s. After that, the car is slowing, so the flux changes more slowly, the eddy currents weaken, and the braking deceleration falls back toward zero in both cases.

Differences. The faster (12 m/s) entry produces a greater maximum deceleration, because the higher speed gives a greater rate of change of flux and so larger eddy currents and a larger retarding force. The faster case also keeps a greater deceleration for most of the time (until about 3 s), since it remains faster and keeps inducing stronger eddy currents than the slower case throughout the overlap.

Marker's note. Engage with all parts of the graph (rise to the peak and the decay), and use cause and effect: speed sets the rate of flux change, which sets the eddy currents and so the braking force.

[7 marks]. The ball leaves with two independent motions: a vertical launch velocity and a horizontal velocity equal to the rim speed of the disc.

Horizontal (tangential) speed of the rim: the period of one revolution is $t_{rev} = 1/3 = 0.334$ s, so

Time of flight (vertical): the ball returns to ground level, so with $u_y = 5.72$ m/s and $a = -9.8$ m/s^2,

Range of the ball: travelling in a straight line tangentially,

Where the launcher is now: in 1.17 s the disc turns $3 \times 1.17 = 3.50$ revolutions, that is half a revolution past a whole number, so the launcher sits diametrically opposite its firing point, 2 m on the far side of the centre (4 m back along the line of flight from the firing point).

Position of the ball relative to the launcher's new position:

So the ball lands about 44.1 m from the launcher's new position, at about 5.2 degrees to the line of flight.

Marker's note. The key insight is that the ball keeps a constant horizontal (tangential) velocity while the launcher keeps rotating. Lay out the working clearly, label positions and angles, and avoid rounding too early (small differences give answers up to about 44.2 m and angles down to about 5.1 degrees).

[9 marks]. Both particle-field and particle-particle investigations have driven our understanding forward. At least two of each, explained, justify the statement.

Particle-field interactions.

• Thomson's experiment. A beam of cathode rays was deflected by electric and magnetic fields. Balancing the deflections let Thomson measure the charge-to-mass ratio of the electron, showing cathode rays were tiny negative particles common to all matter. This revealed the electron and overturned the idea of an indivisible atom.
• Millikan's oil-drop experiment. Charged oil drops were suspended in an electric field by balancing the electric force against gravity. This measured the fundamental charge on the electron, showing charge is quantised in units of $e$.
• Photoelectric effect. Light ejecting electrons from a metal, with a threshold frequency and kinetic energies independent of intensity, showed light interacts with electrons as quanta (photons), supporting the quantum model of light.

Particle-particle interactions.

• Geiger-Marsden experiment. Alpha particles fired at thin gold foil mostly passed through, but a few scattered through large angles. This scattering showed the atom is mostly empty space with a tiny, dense, positively charged nucleus.
• Chadwick's experiment. Bombarding beryllium produced a neutral, penetrating radiation that knocked protons from paraffin. Applying conservation of momentum and energy identified the neutron, completing the proton-neutron picture of the nucleus.
• Particle accelerators. High-energy collisions between particles produced showers of new particles, revealing quarks and building the Standard Model, deepening our understanding of the fundamental constituents of matter.

Across these, deflecting particles in fields measured their basic properties (charge, mass, quantised charge), while colliding particles with matter exposed the structure of the atom and nucleus and uncovered new particles - directly supporting the statement.

Marker's note. Deconstruct the question and address every part: at least two observations and at least two experiments, covering both particle-field and particle-particle interactions. Know the methods and outcomes of the key experiments and use diagrams where they help.

(a) [2 marks]. After the burn at P, the engine is off, so only gravity acts and total mechanical energy is conserved. As the satellite moves out from P to Q its orbital radius increases, so its gravitational potential energy increases (becomes less negative). By conservation of energy its kinetic energy decreases by the same amount. So $K$ is converted into $U$ as the satellite climbs from P to Q.

(b) [4 marks]. The total energy just after the burn is the original total energy plus the kinetic energy added:

Energy is conserved from P to Q, so $E = U_Q + K_Q$. The potential energy at Q is

Therefore

(c) [3 marks]. Compare the satellite's speed at Q with the speed needed for a circular orbit there. From $K_Q = \tfrac{1}{2}mv_Q^2$:

The speed required for a circular orbit at $r_Q$ is

Since $v_Q > v_{circ}$, the satellite is moving faster than the circular-orbit speed at Q, so gravity cannot hold it in a circle there. It will continue on its trajectory, increasing its distance from Earth (the orbit is now an ellipse with Q below its apogee).

Marker's note. Use conservation of energy throughout, with clear subscripts for the P and Q values, and remember an increase in radius makes $U$ less negative (an increase). In (c) support the explanation with the comparison of the actual speed to the required circular-orbit speed.