HSC Physics 2022

(a) [2 marks]. Surface temperature: X is hotter than Y, because X lies in a hotter (bluer) spectral class on the horizontal axis. Luminosity: both lie high on the diagram, but Y is a cool, very luminous giant, so X is less luminous than Y.

(b) [2 marks]. X is a hot main-sequence star, so it is fusing hydrogen into helium in its core. Y is an evolved cool giant, so its core is fusing helium into heavier elements such as carbon.

Marker's note. Compare both temperature and luminosity for both stars rather than describing one in isolation. In (b) name the specific element fusing in each core, hydrogen for X and helium for Y.

[4 marks]. Two features:

1. Laminated core. The core is built from thin insulated sheets rather than a solid block. This breaks up the paths available to induced eddy currents, reducing their size, so less energy is wasted as heat in the core and efficiency rises.
2. Continuous (closed) iron core. A complete iron loop links the primary and secondary coils, maximising the magnetic flux that threads both coils (flux linkage). Little flux leaks away, so a larger fraction of the input power is transferred to the secondary, improving efficiency.

(A third acceptable feature: the low-resistance copper wire reduces resistive heating in the windings.)

Marker's note. Link each named feature explicitly to how it improves efficiency, and note that energy is otherwise lost as heat. Use the correct term flux linkage, and say laminations reduce the size of eddy currents rather than prevent them entirely.

[4 marks]. Method: direct a beam of light at a mirror placed a large, measured distance away. Start a timer at the instant the light is sent and stop it the instant the reflected beam returns. Measure the total path length (twice the distance to the mirror). The speed of light is then the total distance divided by the measured time:

A factor limiting accuracy is human reaction time in starting and stopping the timer; over the very short travel time this introduces a large relative error. Using a long baseline reduces, but does not remove, this limitation.

(Fizeau's rotating toothed cog, or methods using a rotating mirror, resonant cavity or Roemer's timing of Jupiter's moons, are equally acceptable.)

Marker's note. State the measurements actually taken, relate them to a calculation of c, and identify a genuine limiting factor. Distinguish experimental errors such as reaction time from human mistakes, and make the link from method to the value of c explicit.

(a) [2 marks]. From the graph the mass falls from 80 to 40 micrograms in 16 hours, so the half-life is $t_{1/2} = 16$ h. Then

(b) [2 marks]. Starting from $N_0 = 80$ micrograms and using $N = N_0 e^{-\lambda t}$:

(About 53 hours if the unrounded decay constant is used.)

Marker's note. Read the half-life off the graph rather than assuming it, then substitute correctly. In (b) show all substitution and solve using natural logs.

(a) [2 marks]. Change in kinetic energy:

The magnitude is $2.2 \times 10^9$ J (kinetic energy decreases as the rocket climbs).

(b) [3 marks]. Energy is conserved, so the kinetic energy lost equals the gravitational potential energy gained:

The bracket is $2.326 \times 10^{-7} - 4.0 \times 10^{-8} = 1.926 \times 10^{-7}$ m$^{-1}$. So

Marker's note. This is not an orbit-change question: just use $\tfrac{1}{2}mv^2$ at each point. In (b) apply conservation of energy and take care subtracting the two fractions with different denominators.

(a) [3 marks]. Using the photoelectric equation $K_{max} = hf - \phi$, with the work function converted to joules ($\phi = 2.9 \times 1.602 \times 10^{-19} = 4.646 \times 10^{-19}$ J):

(b) [3 marks]. The electron decelerates against the field and stops when the work done by the field equals its kinetic energy. With $W = qEd$:

Marker's note. Convert the work function from electron volts to joules before subtracting. In (b) keep energy, electric field and voltage distinct, and use $W = qEd$ to relate the field to the stopping distance.

(a) [2 marks]. Newton's particle (corpuscular) model predicts that light particles travel in straight lines through each slit, so only two bright bands appear on the screen, one directly behind each slit, with no interference fringes between them.

(b) [3 marks]. A is the second-order maximum, so $m = 2$. Using $d \sin\theta = m\lambda$:

(c) [2 marks]. Green light has a shorter wavelength (520 nm versus 655 nm). Since $\sin\theta = m\lambda/d$, a smaller $\lambda$ gives a smaller angle for each order, so the bright lines move closer together and the fringe spacing on the screen decreases.

Marker's note. In (a) show the two distinct bands of the particle model and label the diagram. In (b) identify $m = 2$, keep units consistent and show full working. In (c) make the cause and effect explicit: shorter wavelength gives a smaller angle and closer bands.

[3 marks]. The energy released equals the mass defect (mass lost) converted by $E = \Delta m c^2$. For Step Y:

Using the conversion 931.5 MeV per atomic mass unit:

Step Y releases about 6.5 MeV, much more than the 1.20 MeV of Step X, because a greater mass defect is converted to energy in Step Y. The larger mass lost means a larger energy release.

Marker's note. Keep units consistent and avoid rounding intermediate mass values, since small errors in atomic mass units distort the energy. Link the larger energy release directly to the larger mass defect.

[4 marks]. Horizontal: once released, no net horizontal force acts on the apple (ignoring air resistance), so its horizontal velocity is constant. That horizontal velocity is the vector sum of the apple's eastward velocity relative to the car and the car's northward velocity relative to the ground, so the apple travels in a straight horizontal line in a north-east direction at constant speed.

Vertical: the only vertical force is gravity, which gives a constant downward acceleration of $9.8$ m s$^{-2}$. The apple starts with zero vertical velocity and accelerates uniformly downward, so its vertical speed increases steadily throughout the flight.

The two components are independent: the constant horizontal motion and the accelerating vertical motion combine to give a curved three-dimensional path.

Marker's note. Use correct terminology and refer to the actual apple in the stimulus, not a generic projectile. Recognise the horizontal velocity as the vector sum of the car's and the apple's velocities, and tie each component to the force (or absence of force) acting on it.

(a) [3 marks]. Both observers must measure the same speed of light, a postulate of special relativity. The outside observer sees the light follow a longer, slanted path (X to Y to X plus the carriage's motion) than the straight up-and-down path seen on the train. Since the same speed covers a greater distance, $t = d/c$ means the outside observer measures a longer time for the same event. Time runs slow in the moving frame relative to the outside observer: this is time dilation.

(b) [3 marks]. The 15 ns measured on the train is the proper time $t_0$ (both events at the same place on the train). The outside observer measures the dilated time:

Marker's note. Use unambiguous language about which observer sees the longer path and longer time. In (b) recognise the train time as the proper time $t_0$, recall that 1 nanosecond is $1 \times 10^{-9}$ s, and apply the time dilation equation correctly.

[9 marks]

Accounting for the Geiger-Marsden results

In the model the atom is mostly empty space with a tiny, dense, positively charged nucleus. Most alpha particles pass far from the nucleus and travel almost straight through the foil, which explains why most were undeflected. The few alpha particles heading close to a nucleus feel a strong repulsive electrostatic (Coulomb) force from the like positive charge; because the nucleus is very massive compared with an alpha particle, it stays nearly stationary while the alpha particle is deflected through a large angle, even back towards the source. Particles passing at a greater distance feel a weaker force and are deflected less.

Accounting for the hydrogen (Balmer) spectrum

Bohr added that electrons occupy stable, quantised energy levels and do not radiate while in them. When an electron drops from a higher level $n_i$ to a lower level $n_f$, it emits a photon whose energy equals the difference between the levels:

The four visible Balmer lines correspond to electrons falling to the $n_f = 2$ level from $n_i = 3, 4, 5, 6$. The wavelengths follow:

Sample calculation ($n_f = 2$, $n_i = 3$, $R = 1.097 \times 10^7$ m$^{-1}$):

which is the red line at the far right of the Balmer series, matching the observed spectrum.

So the empty-space nucleus model explains the scattering, while quantised energy levels and photon emission explain the discrete hydrogen spectrum, drawing together force, energy and photon evidence into one model.

Marker's note. Use the stimulus directly: link the empty space and dense nucleus to the scattering pattern, and the energy levels to the Balmer lines. Include a worked Balmer calculation and a labelled energy-level diagram, and keep the response targeted rather than reciting everything known.

(a) [3 marks]. As the aluminium flywheel spins, each region of it moves through the magnets' field, so the magnetic flux through that region changes. By Faraday's law this changing flux induces eddy currents in the flywheel. By Lenz's law these eddy currents flow so that their own magnetic field opposes the change producing them, creating a force that opposes the flywheel's motion. This drag acts at the rim, producing a torque opposing the pedalling, so the rider must work harder.

(b) [3 marks]. Move the magnets closer to the flywheel surface. A smaller gap means a stronger field through the aluminium, so for the same spin speed the rate of change of flux is greater, inducing larger eddy currents. Larger eddy currents produce a stronger opposing force and therefore a greater opposing torque, making it harder to pedal. (Alternatively, moving the magnets outwards towards the rim increases the lever arm of the force, also raising the torque.)

Marker's note. Give the cause-and-effect sequence of electromagnetic induction in order, and distinguish Faraday's law (induced current from changing flux) from Lenz's law (the current opposes the change). In (b) describe a practical adjustment and combine torque with induction in the justification.

(a) [2 marks]. The launch speed equals the speed in circular motion:

The vertical component at 45 degrees:

(b) [4 marks]. The horizontal component is $v_x = 20.1 \cos 45^\circ = 14.2$ m s$^{-1}$. Take up as positive, $a = -9.8$ m s$^{-2}$.

Time to the top ($v_y = 0$): $0 = 14.2 - 9.8t$, so $t_{up} = 1.45$ s. Height risen above P:

Height above the ground at the top: $10.3 + 1.2 = 11.5$ m. Time to fall this distance from rest:

Total flight time: $t = 1.45 + 1.53 = 2.98$ s. Horizontal range:

Marker's note. Link the circular motion to the launch velocity and draw a labelled vector diagram. Recognise that the hammer lands 1.2 m below its launch height, so the trajectory is not symmetrical: do not assume zero net vertical displacement.

[7 marks]. In a magnetic field a moving charge feels a force $F = qvB$ acting perpendicular to its velocity. A constant perpendicular force makes each particle move in a circular arc, with the magnetic force providing the centripetal force:

Direction of curve. X and Y curve in the same direction, so they carry the same sign of charge (deflected the same way by the right-hand rule). Z curves in the opposite direction, so Z must carry the opposite sign of charge.

Radius of curve. With the same speed $v$ and the same field $B$, the radius depends on $m/q$: $r = (m/q)(v/B)$. X has a smaller radius than Y, so X has a smaller mass-to-charge ratio (smaller mass or greater charge) than Y. Z has the same radius as X but curves the opposite way, so Z has the same magnitude of mass-to-charge ratio as X but the opposite sign of charge.

In summary: the direction of curving reveals the sign of charge, and the tightness of the curve reveals the mass-to-charge ratio, with the magnetic force always supplying the centripetal force.

Marker's note. Explain (not just describe) each feature: the circular nature, the different radii of X, Y and Z, and the opposite curve direction of Z. Use $r = mv/qB$ to support the analysis and write clearly so the comparison is unambiguous.

[5 marks]. The centripetal force needed to keep the capsule on its circular path:

The gravitational force between the two craft:

The required centripetal force (3.3 N) is about a million times larger than the gravitational force between the capsule and the ISS ($8.4 \times 10^{-6}$ N). Gravity between the two craft is far too weak to supply the centripetal force, so the hypothesis is rejected: the circular motion must be maintained by some other force, such as thrust from the capsule.

Marker's note. Calculate both the required centripetal force and the gravitational force, then compare them. A hypothesis can be true or false on the evidence: here the calculations are the evidence that refutes it, so state a clear conclusion.