HSC Physics 2021

(a) [2 marks]. Maximum torque occurs when the plane of the coil is parallel to the field, so $\tau_{max} = nBIA$ with $n = 1$.

(b) [2 marks]. The position shown is the maximum-torque position (the coil normal is perpendicular to the field). As the loop turns, the torque decreases to zero after a quarter turn (90 degrees), when the coil normal is parallel to the field. Through the second quarter turn the torque rises back to its maximum at 180 degrees.

Marker's note. A half rotation is 180 degrees, not 90. The maximum torque is at the position drawn (normal perpendicular to the field), so the magnitude falls to zero at 90 degrees and returns to a maximum at 180 degrees.

[3 marks]. Every point on a rigid disc sweeps the same angle in the same time, so the angular velocities of X and Y are equal ($\omega_X = \omega_Y$).

Instantaneous (linear) speed is $v = r\omega$. Because X is twice as far from the centre ($r_X = 2 r_Y$), the instantaneous speed of X is twice that of Y. Both instantaneous velocities are tangential, and at the labelled instant they point in the same direction.

Marker's note. State that the angular velocities are equal, then use $v = r\omega$ to show the instantaneous speed of X is double that of Y. Support the comparison with the radius information from the stimulus rather than asserting it.

[4 marks].

Thomson passed a beam of cathode-ray electrons through crossed electric and magnetic fields. He adjusted the field strengths so the electric force and the magnetic force balanced and the beam was undeflected, giving the speed. Switching off one field and measuring the deflection let him calculate the charge-to-mass ratio ($q/m$) of the electron.

Millikan sprayed charged oil drops between two horizontal plates and used an electric field to balance the gravitational force on a drop so it was suspended. From the balancing voltage and the drop's weight he found the charge on each drop, and the charges were all multiples of a smallest value, the charge on the electron ($q_e$).

Marker's note. Treat the two scientists separately, not as a single 'they' response. Name the fields each used and the distinct property each found: Thomson the charge-to-mass ratio, Millikan the charge on the electron.

[3 marks]. The flux through one turn changes from $\Phi = BA$ to zero:

By Faraday's law, $\varepsilon = -n \dfrac{\Delta \Phi}{\Delta t}$:

The magnitude of the induced emf is 0.26 V.

Marker's note. Show the change in flux as a separate step and carry the negative sign through Faraday's law before taking the magnitude. Full working (formula, substitution, answer) is expected.

(a) [2 marks]. As the satellite rises from rest on the surface into orbit, its gravitational potential energy increases (it moves further from Mars, so U becomes less negative) and its kinetic energy increases (it speeds up from stationary to orbital speed). Chemical energy from the launch is transformed into both.

(b) [3 marks]. The orbit is geostationary, so the period equals one Martian day:

Using Kepler's third law, $\dfrac{r^3}{T^2} = \dfrac{GM}{4\pi^2}$:

Marker's note. Convert the period fully into seconds (24 h 40 min = 88 800 s) before substituting, and use Kepler's third law with the correct units throughout.

(a) [3 marks]. Plot V (y-axis) against f (x-axis), choosing a scale that spreads the four points across the gridlines (for example 0 to 8 x 10^14 Hz and 0 to 3 V), then draw a single straight line of best fit that balances the points rather than joining them dot to dot.

(b) [3 marks]. A better method is to find h from the gradient of the line of best fit, not from a single point. Rearranging $V = \dfrac{h}{q_e} f$ shows the gradient is $\dfrac{h}{q_e}$, so $h = \text{gradient} \times q_e$. Reading two well-separated points on the line and computing the gradient, then multiplying by $q_e = 1.602 \times 10^{-19}$ C, gives h.

This is justified because the line of best fit uses all four data points, which averages out random errors in individual readings and reduces the effect of any single anomalous point. Using only data point 1 relies on one measurement and any error in it passes straight into h.

Marker's note. The command word is 'justify', so give a reason: the gradient method uses all the data and reduces the effect of random (and systematic) error. Do not just calculate h from one point.

(a) [3 marks]. Lay the rod horizontally. The magnetic force $F = BIL$ must point straight up to oppose the weight, which points straight down. Using the right-hand rule, the current and the field must be perpendicular to each other and to the upward force. For example, with the field horizontal and pointing to the right, the current must flow into the page so that $I \times B$ gives an upward force.

(b) [3 marks]. Levitation needs the upward magnetic force to be at least the weight. Per unit length, the magnetic force is $BI$ and the weight is $\dfrac{m}{L} g$. The rod just floats when they are equal:

If the mass per unit length exceeds 0.282 kg m^-1, the weight per metre is larger than the largest magnetic force per metre the field and current can provide, so the net force is downward and the rod falls. Therefore 0.282 kg m^-1 is the maximum value for which levitation is possible.

Marker's note. Show the full substitution to reach 0.282 kg m^-1, then give the cause and effect: above this value the gravitational force per unit length exceeds the magnetic force, so levitation is no longer possible.

(a) [3 marks]. The astronaut's 9.4 years is the proper time $t_0$; the Earth time is $t = 15$ years. From time dilation $t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}}$, the Lorentz factor is

Solving for v, $\dfrac{v}{c} = \sqrt{1 - \dfrac{1}{\gamma^2}} = \sqrt{1 - \dfrac{1}{1.596^2}} = \sqrt{0.607} = 0.779$, so $v = 0.779c$.

The Earth observer sees the ship travel at v for 15 years:

This is about 11.7 light years.

(b) [2 marks]. As v approaches c, the relativistic momentum $p = \dfrac{mv}{\sqrt{1 - v^2/c^2}}$ grows without bound. Increasing the speed further would require ever larger force, and reaching c would need infinite force (and infinite energy), which is impossible. So no spaceship with mass can reach or exceed the speed of light; c is the limit.

Marker's note. In (a) identify $t_0 = 9.4$ years as the proper time, keep units consistent, and check the resulting speed is below c. In (b) link the rising relativistic momentum (or mass-energy) explicitly to the unattainable speed c.

[5 marks].

Bohr treated the electron as a negatively charged particle orbiting the nucleus in fixed circular paths. Only certain orbits, with quantised angular momentum and discrete energy levels, are allowed, and the electron does not radiate while in an allowed orbit.

de Broglie proposed that the electron has a wave nature as well as a particle nature (matter waves). The allowed Bohr orbits are exactly those whose circumference fits a whole number of electron wavelengths, forming a stable standing wave; this explained why orbits are quantised.

Schrodinger removed definite orbits altogether and described the electron by a wavefunction. The square of the wavefunction gives the probability of finding the electron in a region, so the electron is spread out as a probability cloud (orbital) rather than following a path.

The models therefore move from a localised particle in a fixed orbit (Bohr), to a wave-particle standing wave (de Broglie), to a purely probabilistic wave description (Schrodinger).

Marker's note. Link each description to the correct scientist and bring out the differences: a particle in quantised orbits, a standing matter wave, and a probabilistic wavefunction. Structure the response model by model rather than blending them.

(a) [2 marks]. By conservation of energy, the kinetic energy gained equals the electric potential energy lost. The proton starts from rest, so KE starts at zero, and its KE curve is the mirror image of the falling potential-energy graph (their sum stays constant). Where the potential-energy line falls by a certain amount, the KE line rises by the same amount, so the KE graph is the potential-energy graph reflected about a horizontal line, starting at zero.

(b) [3 marks]. The electron carries the same magnitude of charge as the proton but the opposite sign, so it feels a force of the same size in the opposite direction and accelerates towards the other plate. Because the electron's mass is about 1800 times smaller, from $a = F/m$ it has a much larger acceleration and reaches the plate in a much shorter time. It still gains the same kinetic energy over the same potential difference (charge magnitudes are equal), but it does so far more quickly.

Marker's note. Give clear cause and effect: opposite charge means opposite direction of motion; equal-magnitude charge means equal force; smaller mass means greater acceleration (and shorter time).

[7 marks]

Galvanometer on cart 1

closing the switch drives a steady direct current from the battery through solenoid 1, so G1 shows a constant deflection. This current makes solenoid 1 an electromagnet with, say, a south pole facing cart 2.

Galvanometer on cart 2

as solenoid 1's field grows from zero, the flux through solenoid 2 changes. By Faraday's law this changing flux induces an emf and a current in solenoid 2, so G2 shows a brief deflection. By Lenz's law the induced current opposes the change, so solenoid 2 forms a south pole facing cart 1. Once the current in solenoid 1 becomes steady, the flux through solenoid 2 stops changing and the current in G2 falls back to zero, so G2 deflects only momentarily.

Initial movement

the two solenoids present like poles to each other (south facing south), so they repel. The carts are pushed apart in opposite directions. The magnetic forces on the two carts are equal and opposite (Newton's third law), so momentum is conserved: $m_1 v_1 = m_2 v_2$. Since cart 1 has twice the mass of cart 2, cart 1 moves off at half the speed of cart 2.

Marker's note. Use the battery terminals and the right-hand grip rule to fix the field directions, link the induced current to Faraday's and Lenz's laws, and apply conservation of momentum to get the speed ratio (cart 1 at half the speed of cart 2). Opposing fields means like poles facing, not opposite poles facing.

[5 marks]. The model works well in its central feature. When the elastic is stretched, each marking moves away from the fixed end by an amount proportional to its original distance: the marking at d3 moves three times as far as the marking at d1 in the same time, so its recession speed is proportional to distance. This is exactly Hubble's evidence, that more distant galaxies recede faster, with recession speed proportional to distance ($v = H_0 d$). The model also captures the key idea that the space (the elastic) between the markings expands, rather than the markings travelling through a fixed space, just as space itself expands between galaxies.

Limitations weaken the analogy. The elastic stretches in only one dimension, whereas the real expansion is three-dimensional and has no fixed end or centre. The markings themselves stretch slightly, but real galaxies are held together by gravity and do not expand. The investigation also measures only positions, while Hubble's evidence came from redshift in galactic spectra, which measures recession speed directly.

Overall the elastic models the proportionality at the heart of Hubble's discovery well, but oversimplifies the dimensionality and the source of the evidence.

Marker's note. The question is about modelling, not just Hubble's discovery, so justify using the stimulus (compare the stretch of the near and far markings) and weigh both how the model succeeds and its limitations.

[9 marks].

Experiment A supports the wave model. Two coherent sources produce a pattern of bright and dark bands on the screen. The band spacing is

Bright bands occur where the path difference is a whole number of wavelengths (constructive interference) and dark bands where it is a half-odd number of wavelengths (destructive interference). Only waves can interfere to add and cancel like this, so the regularly spaced fringes are direct evidence for the wave model of light.

Experiment B supports the particle (photon) model. The energy of one 400 nm photon is

This single-photon energy is greater than the work function of calcium ($4.60 \times 10^{-19}$ J) but less than that of nickel ($8.25 \times 10^{-19}$ J), which is exactly why photoelectrons are freed from calcium but not from nickel. Each photon acts as one particle delivering all its energy to one electron, so increasing the intensity would not help nickel; only a higher photon energy would. The maximum kinetic energy of the calcium photoelectrons is

A wave model predicts that any colour should eventually free electrons if bright enough, which contradicts the result, so the photoelectric effect supports the particle model of light.

Together, the interference pattern (Experiment A) and the photoelectric effect (Experiment B) show that light behaves as a wave in some situations and as particles in others, the basis of wave-particle duality.

Marker's note. Address every part: calculate the fringe spacing for A, the photon energy (and the calcium photoelectron kinetic energy) for B, and link constructive and destructive interference to the wave model and the one-photon-one-electron threshold to the particle model.

(a) [4 marks]. The mass is launched upward at an angle, so it has both horizontal and vertical velocity. The horizontal component is constant (no horizontal force), but the vertical component is changed by gravity.

• At t0 (launch): KE is 864 J from the full horizontal and vertical components.
• At t1: KE is a minimum (284 J) but not zero. This is the top of the path, where the vertical velocity is momentarily zero; the remaining KE is entirely from the unchanged horizontal component.
• At t2 (impact): KE is the largest (1393 J). After the top, gravity speeds the mass up again as it falls, and because it lands at the base of the cliff (below the launch height) it has lost gravitational potential energy relative to t0, which has been converted to kinetic energy, so its KE exceeds the launch value.

The pattern (high, then minimum, then highest) reflects KE being lost to gravitational potential energy on the way up and regained, with extra, on the way down past the launch height.

(b) [3 marks]. Horizontal KE is constant, so the minimum KE at t1 is all horizontal:

At t0 the vertical KE is $864 - 284 = 580$ J, so $v_{y0} = \sqrt{\dfrac{2 \times 580}{3.0}} = 19.7 \text{ m s}^{-1}$ (upward).

At t2 the vertical KE is $1393 - 284 = 1109$ J, so $v_{y2} = \sqrt{\dfrac{2 \times 1109}{3.0}} = 27.2 \text{ m s}^{-1}$ (downward).

Taking up as positive and using $v_{y2} = v_{y0} - g t$ with $v_{y2} = -27.2$ m s^-1:

Marker's note. In (a) explain the minimum at the top (vertical velocity zero, horizontal KE remains) and why the impact KE exceeds the launch KE (lower final height). In (b) split the KE into horizontal and vertical parts, assign the correct signs to the vertical velocity, and add the times carefully.

(a) [3 marks]. The mass defect is

Converting to kilograms (1 u = 1.661 x 10^-27 kg) and using $E = \Delta m c^2$:

(b) [3 marks]. First find how many atoms decay in ten years. The decay constant is

The number remaining after 10 years is $N = N_0 e^{-\lambda t} = 9.0 \times 10^{24} \times e^{-0.0790} = 9.0 \times 10^{24} \times 0.9240 = 8.32 \times 10^{24}$.

The number that have decayed is

Each decay releases 9.0 x 10^-13 J, so the total energy is

Marker's note. In (a) calculate the mass defect, convert u to kg (or use 931.5 MeV per u), then apply $E = \Delta m c^2$ with all steps shown. In (b) plan the path: decay constant, atoms decayed in ten years, then multiply by the per-decay energy from (a).