HSC Physics 2020

(a) [2 marks]. Use the Rydberg equation with $n_f = 2$ and $n_i = 3$:

(b) [3 marks]. In the Bohr model electrons occupy fixed, discrete energy levels (stationary states) and do not radiate while in them. An electron moves to a higher level only by absorbing energy, and returns to a lower level by emitting a photon. Energy is conserved at every transition: the photon energy absorbed or emitted equals exactly the difference between the two energy levels, $E_{photon} = E_i - E_f$, so no energy is created or lost.

Marker's note. Recall the specific postulates (discrete stationary states, no radiation within a level, photon emitted on dropping down). Link conservation of energy directly to the photon energy matching the level difference.

(a) [2 marks]. Use Wien's displacement law:

(b) [3 marks].

1. The capsule's speed (about 12 900 m s$^{-1}$) is a tiny fraction of the speed of light, so the Lorentz factor is effectively 1 and relativistic effects such as time dilation and length contraction are negligible.
2. The capsule is decelerating sharply, so it is a non-inertial (accelerating) reference frame. Special relativity applies only to inertial frames, so it cannot properly analyse this motion.

Marker's note. In (a) round correctly and keep the unit (metres). In (b) give two clearly distinct points: the speed being far below c, and the frame being non-inertial. Vague single answers were not enough.

[3 marks]. From the graph, at $\tau = 2.95$ Nm the angular velocity is $\omega = 100$ rad s$^{-1}$. Rearrange the torque equation for the current:

Since 4.1 A is below the 5 A trip threshold, the circuit breaker does not cut the current and the motor continues to run.

Marker's note. Read the angular velocity off the graph precisely, substitute into the rearranged equation, and then state the explicit consequence for the motor (it keeps running because the current stays under 5 A).

[4 marks]. From the graph, the maximum height is about 44 m (where $v_y = 0$) and the total time of flight is 6 s.

Vertical component, using $v_y^2 = u_y^2 + 2a_y s_y$ at the top ($v_y = 0$):

Horizontal component, from the range and time:

Combine by Pythagoras:

Angle of launch:

So the initial velocity is about 36.5 m s$^{-1}$ at 54 degrees above the horizontal.

Marker's note. Read the maximum height and the time of flight precisely off the graph, find both components, then combine by vector sum and quote a magnitude and a direction. Do not confuse initial with average velocity.

[4 marks]. A hydrogen atom is a single proton nucleus orbited by one electron. The proton is a hadron (specifically a baryon) made of three quarks: two up quarks (charge $+\tfrac{2}{3}$ each) and one down quark (charge $-\tfrac{1}{3}$), which sum to a charge of $+1$. These quarks are bound by the strong nuclear force, carried by gluons. The electron is a lepton, a fundamental particle with charge $-1$ that cannot be subdivided. The electron is held to the proton by the electromagnetic force, carried by the photon, so the atom is overall neutral.

Marker's note. Apply the Standard Model to the whole atom: classify the proton (hadron of up and down quarks) and the electron (lepton), and correctly associate the bosons with their forces (gluon for the strong force, photon for the electromagnetic force).

(a) [2 marks]. An incandescent lamp produces a continuous emission spectrum containing all visible wavelengths blended together. A gas discharge tube produces a line emission spectrum, a few bright discrete wavelengths separated by dark gaps, characteristic of the gas's energy levels.

(b) [2 marks].

• Model X (the classical curve, rising without limit at short wavelengths) assumes energy is absorbed and emitted continuously, in any amount.
• Model Y (the curve with a finite peak) assumes energy is absorbed and emitted only in discrete packets (quanta), $E = hf$.

(c) [4 marks].

The 4000 K curve lies entirely below the 5000 K curve and peaks at a longer wavelength.

• The peak shifts to a longer wavelength because the peak wavelength is inversely proportional to absolute temperature (Wien's law, $\lambda_{max} \propto 1/T$), so a cooler body peaks further into the red and infrared.
• The whole curve is lower, and the area under it is smaller, because at every wavelength a cooler body radiates less intensity, so its total power output is less (consistent with the Stefan-Boltzmann relationship).

Marker's note. In (b) give an assumption of each model (continuous versus quantised emission), not a consequence of it. In (c) draw the 4000 K curve wholly under the 5000 K curve with a lower peak at a longer wavelength, and make the written explanation match the sketch.

[4 marks]. For the bright fringes, $d\sin\theta = m\lambda$, so for the first maxima $\sin\theta = \lambda/d$. The fringe spacing on the screen is approximately $\Delta x = \lambda y / d$. Reducing $\lambda$ on its own would shrink the spacing, so it must be compensated.

1. Reduce the slit separation $d$. Since $\sin\theta = \lambda/d$, decreasing $d$ increases $\theta$. Keeping the ratio $\lambda/d$ constant keeps the fringe angle, and so the distance between A and B, unchanged.
2. Increase the slit-to-screen distance $y$. The fringe spacing $\Delta x = \lambda y / d$ is proportional to $y$, so increasing $y$ enlarges the spacing to offset the reduction caused by the smaller wavelength.

Marker's note. Use the equation to show the proportional relationships, name the pronumerals you use ($d$, $y$, $\lambda$), and explain how each adjustment compensates for the smaller wavelength rather than just naming it.

(a) [2 marks]. The area swept is the rail width times the distance moved, $A = 0.20 \times 0.30 = 0.060$ m$^2$. The field is vertical and the swept area lies on the rails at 30 degrees, so the flux uses $\cos 30^\circ$:

(b) [2 marks]. Faraday's law over the 2.5 s of motion:

By Lenz's law the induced current opposes the increasing upward flux, so it flows anticlockwise as viewed from above.

(c) [3 marks]. Less force is needed without the field. With the field present, the induced current experiences a force opposing the rod's motion (Lenz's law), and the applied force must do extra work to supply the electrical energy dissipated in the circuit. Removing the field removes this opposing force and that energy demand. The force is not zero, though, because the rod must still be pushed up the incline against the component of gravity along the rails.

Marker's note. In (a) include the $\cos 30^\circ$ factor and the $B = 1$ T value even though it is unity. In (c) explain both that less force is needed (no induced opposing force, by Lenz's law) and that some force remains because the rod is still raised against gravity.

(a) [3 marks]. The hypothesis was proposed because the radiation was not deflected by an electric field, showing it carried no charge, which is consistent with gamma rays being uncharged. It was then rejected on two grounds. First, the radiation ejected fast protons from paraffin; gamma photons have far too little momentum to knock out a proton, but a neutral particle of about the proton's mass could transfer enough momentum. Second, the radiation could not produce the photoelectric effect, whereas high-energy gamma photons readily eject electrons. Both observations contradict the gamma hypothesis, so it was rejected.

(b) [2 marks]. The experiments revealed a new neutral particle, the neutron, with mass close to the proton. The nuclear model changed from a nucleus of protons alone to one containing both protons and neutrons, which also explained the existence of isotopes.

Marker's note. Link all three observations to the argument: the no-deflection result supports the hypothesis, while the proton ejection and the absence of the photoelectric effect reject it. In (b) name the neutron and say how it altered the nuclear model.

(a) [3 marks]. In deep inelastic scattering experiments, a linear accelerator gave electrons enough energy to penetrate protons. The electrons scattered at large angles far more often than a uniform proton would allow, showing the charge inside a proton is concentrated at three point-like centres. This was evidence that the proton is not elementary but is made of three quarks, supporting the Standard Model's quark structure of hadrons.

(b) (i) [2 marks]. Use the de Broglie relation with $v = 0.1c = 3 \times 10^7$ m s$^{-1}$:

(b) (ii) [2 marks]. At 0.95c the proton's momentum must be calculated relativistically, $p = \gamma mv$, with $\gamma$ much larger than 1. Since $\lambda = h/p$ and the relativistic momentum is greater than the classical $mv$, the actual wavelength is shorter than the classical de Broglie value would predict.

Marker's note. In (a) give a specific accelerator example and a clear cause-and-effect link from the scattering result to internal proton structure. In (b)(i) pick the right equation and substitute the data-sheet proton mass carefully; in (b)(ii) explain via the increased relativistic momentum giving a shorter wavelength.

(a) [3 marks]. The comet moves in an ellipse, so total mechanical energy is conserved. As it falls toward the Sun, gravitational potential energy is converted into kinetic energy, so its speed increases and is greatest at perihelion (closest approach). As it swings away from the Sun, kinetic energy is converted back into gravitational potential energy, so it slows down and is slowest at aphelion (farthest point), before returning to P. Equivalently, the component of gravity along the velocity speeds the comet up on the way in and slows it on the way out.

(b) [3 marks]. Because both speeds are constant and the masses are equal, the two stars orbit their common centre of mass, which lies midway between them, so each moves in a circle of radius $x/2$. The gravitational force provides the centripetal force on star B:

By symmetry star A has the same speed, consistent with the stated constant speed.

Marker's note. In (a) apply conservation of mechanical energy explicitly, linking the potential-to-kinetic exchange to the speed at perihelion and aphelion. In (b) analyse forces (not energy), use the orbital radius $x/2$, and set the centripetal force equation out cleanly to reach $v_B = \sqrt{Gm/2x}$.

[7 marks]

Several factors set an upper limit on the speed, and they interact.

Power of the motor

The motor delivers a finite power $P = VI$. The mechanical power it supplies at the rope is $P = Fv$, so for a given useful force the maximum steady speed is $v = P/F$. Higher power allows a higher speed.

Friction on the mass

The mass slides against the surface, so kinetic friction $F_f = \mu N = \mu mg$ opposes the motion. At constant (maximum) speed the rope tension only has to overcome friction, $F = \mu mg$, and the larger the mass or the coefficient of friction, the larger this opposing force and the lower the achievable speed.

Torque and the pulley

The motor delivers a torque $\tau$, and the rope tension is $F = \tau / r$ for a pulley of radius $r$ (from $\tau = rF$). A pulley converts the motor's torque into rope tension; the available torque and pulley radius together cap the force, and hence the speed.

Back emf

As the motor spins faster the back emf rises, reducing the net voltage and the current, so $P = VI$ and the available torque fall as the speed climbs. This naturally limits the motor to a maximum speed where the supplied force just balances friction.

Combining $P = Fv$ with $F = \mu mg$, the top speed is about $v_{max} = P/(\mu mg)$: it rises with motor power and falls with the mass and the friction coefficient.

Marker's note. Give more than one factor (power, friction, torque, back emf) and support each with a genuine mathematical model ($P = VI$, $P = Fv$, $F = \mu N$, $\tau = rF$). Use cause and effect to link the motor's action to the mass's motion, and distinguish force from torque.

[9 marks]. The motion has three phases.

Phase 1, free fall X to Y (about the first 0.4 s). Only gravity acts, so the magnet accelerates downward at $g = 9.8$ m s$^{-2}$, shown by the curved (non-linear) part of the displacement-time graph. Gravitational potential energy converts into kinetic energy. Falling 0.78 m, the entry speed at Y is

Phase 2, through the cylinder Y to Z (0.5 s). As the magnet enters, its changing flux induces eddy currents in the copper cylinder. By Lenz's law these currents oppose the motion, producing a retarding force. The magnet quickly reaches a terminal (constant) velocity, shown by the straight, constant-gradient section of the graph:

At constant velocity the net force is zero: the magnetic retarding force balances gravity. Here gravitational potential energy is no longer becoming kinetic energy (speed is constant); instead it is dissipated as heat in the resistance of the copper by the eddy currents (and a small amount as the change in kinetic energy from 3.9 to 0.4 m s$^{-1}$ on entry). The kinetic energy lost on entry is

which, together with the lost potential energy through the cylinder, becomes heat in the copper.

Phase 3, free fall below Z. Once clear of the cylinder the induced currents stop, gravity again acts alone, and the magnet accelerates, so the graph curves once more.

Throughout, total energy is conserved: gravitational potential energy becomes kinetic energy in free fall, and becomes heat in the copper while passing through the cylinder, with no energy lost overall.

Marker's note. Read the three regions of the displacement-time graph correctly (curved free fall, straight constant-velocity section, curved free fall again). Apply conservation of energy quantitatively, referring to gravity and to the eddy-current heating in the copper. Use correct terms, for example accelerates due to gravity rather than just falls.

(a) [3 marks]. Work done equals force times the vertical distance moved toward plate Y. The field is $E = V/d$, the force is $qE$, and each particle starts midway and is deflected to the far plate.

For $q_1$ (separation d, starting d/2 from each plate), it moves $d/2$ to reach plate Y:

For $q_2$ (separation 2d, field $E_2 = V/2d$, starting at the same distance from plate X as $q_1$, that is $d/2$ from X). Plate Y is now $3d/2$ away, so it moves $3d/2$:

Therefore $W_2 / W_1 = \tfrac{3}{2}$: the work done on $q_2$ is 1.5 times that done on $q_1$.

(b) [3 marks]. The horizontal distance is $x = v_x t$, with $v_x$ the same for both, so it depends only on the time spent in the field, set by the vertical motion. The vertical acceleration is $a = qE/m$, and the vertical distance to reach plate Y is $\tfrac{1}{2}at^2$, so $t = \sqrt{2(\text{vertical distance})/a}$.

The field is halved in Figure 2 ($E_2 = V/2d = \tfrac{1}{2}E_1$), so $a_2 = \tfrac{1}{2}a_1$. The vertical distance also changes from $d/2$ to $3d/2$, three times larger.

Since horizontal distance is proportional to the time in the field, $q_2$ travels $\sqrt{6}$ (about 2.4) times the horizontal distance of $q_1$.

Marker's note. Label vertical and horizontal quantities systematically to avoid confusion. Recognise that doubling the separation halves the field, and that the deflection is projectile-style motion (constant horizontal velocity, uniform vertical acceleration), giving the $\sqrt{6}$ ratio.