HSC Physics 2019

[2 marks]. De Broglie proposed that matter has a wave nature: any particle with momentum has an associated wavelength, so electrons and other particles can show wave behaviour such as diffraction. The wavelength is given by

where $h$ is Planck's constant and $mv$ is the particle's momentum.

Marker's note. State the wave nature of matter and include the de Broglie wavelength equation. Do not confuse de Broglie's matter waves with Bohr's earlier energy-level model.

[3 marks]. Beyond composition and temperature, stellar spectra reveal:

• Velocity towards or away from Earth. A shift of the spectral lines to longer wavelengths (red shift) shows the star is receding; a shift to shorter wavelengths (blue shift) shows it is approaching. The size of the shift gives the radial speed.
• Rotational velocity. One edge of the star's disc moves towards us and the other away, so the lines are broadened by opposing red and blue shifts; broader lines mean faster rotation.
• Density. Denser stars produce broader, darker absorption lines through pressure broadening, while low-density stars give narrow, faint lines.

Marker's note. Relate a specific observation in the spectrum (line shift, line broadening) to a specific feature of the star (motion, rotation, density), not just restate composition and temperature.

[3 marks]. Plot maximum kinetic energy (eV) on the y-axis against frequency ($\times 10^{14}$ Hz) on the x-axis and draw a straight line of best fit that bisects the points evenly.

The photoelectric equation is $KE_{max} = hf - \phi$, so the work function $\phi$ is the magnitude of the y-intercept when the line is extrapolated. Equivalently, $\phi = hf_0$ at the threshold frequency $f_0$ where the line crosses $KE_{max} = 0$.

Reading the line of best fit, the threshold frequency is about $9.5 \times 10^{14}$ Hz, so

The work function is about 4 eV.

Marker's note. Draw a single straight line of best fit (the line that bisects the points evenly) and extrapolate it; the work function is the extrapolated $KE_{max}$ where frequency equals zero, read as a positive value.

(a) [3 marks]. Power into the primary is

The stated output is 500 J s$^{-1}$, that is 500 W. Energy cannot be created, so the output power can never exceed the input power; the figures show the input is the larger quantity and an ideal transformer would deliver at most 450 W. In a real transformer the useful output is in fact less than 450 W, because some energy is converted to heat in the windings and core. So the device is consistent with conservation of energy: input power equals useful output power plus the power lost as heat.

(b) [4 marks]. Two modifications:

1. Laminate the iron core. The solid core lets large eddy currents circulate, dissipating energy as heat. Building the core from thin, insulated laminations breaks up these current loops, reducing eddy-current losses and so improving efficiency.
2. Use the thicker copper wire for the primary coil. The primary carries the larger current (9 A), and power lost as heat is $I^2R$. Putting the low-resistance thick wire where the current is highest minimises resistive heating, improving efficiency.

Marker's note. In (a) actually compute $V_pI_p = 450$ W and link the difference to energy lost as heat. In (b) name two specific modifications (lamination, thicker primary winding) and explain how each reduces a particular loss.

[4 marks]. The model shows an electric field E and a magnetic field B oscillating perpendicular to each other and to the direction of travel. This matches Maxwell's central prediction that a changing electric field generates a changing magnetic field and a changing magnetic field generates a changing electric field, so the two regenerate each other and a self-sustaining wave travels through space.

The model also shows the disturbance originating from an oscillating charge, consistent with Maxwell's prediction that an accelerating charge radiates an electromagnetic wave. Finally, the single propagation speed v reflects Maxwell's result that all such waves travel through a vacuum at one speed, $c \approx 3 \times 10^8$ m s$^{-1}$, set by the electric and magnetic constants, and that light is one member of a whole spectrum of these waves.

Marker's note. Name E as the electric field and B as the magnetic field (not just waves), and tie features of the diagram directly to specific Maxwell predictions rather than restating the question.

(a) [3 marks]. Over the measured range (0 to about 25 degrees) the graph is a straight line through the origin, which is exactly the form $\tau = k\theta$ predicts. Its gradient is

matching the quoted $k$. Because the plotted points sit on this straight line, the model predicts the measured torque accurately within this range, so it is valid for small angles up to about 25 degrees.

(b) [3 marks]. The accuracy gets steadily worse beyond 25 degrees. Torque is more correctly modelled by

For small angles $\sin\theta \approx \theta$ (in radians), so the two models nearly agree; but as $\theta$ grows, $\sin\theta$ curves below the straight line. For example at 90 degrees the true torque is $100\sin 90 = 100$ Nm, whereas the linear model predicts $1.7 \times 90 = 153$ Nm, a large over-estimate. So the linear model increasingly over-predicts the torque as the angle rises beyond 25 degrees.

Marker's note. In (a) read the gradient from the graph and show it equals 1.7. In (b) use the alternative model $\tau = rF\sin\theta$ from the formula sheet and show the discrepancy grows with angle; treat the formula as a model that makes predictions.

(a) [3 marks]. Imagine a light clock on a train moving at close to the speed of light: a pulse bounces straight up and down between two mirrors, and one tick is the time for a round trip. For a passenger on the train the pulse travels straight up and back. For an observer on the platform the train has moved on, so the same pulse traces a longer, diagonal zig-zag path. The speed of light is the same for both observers, so the platform observer, seeing the light cover a longer path at the same speed, measures a longer time for one tick. The moving clock therefore runs slow as seen from the ground, which is time dilation.

(b) [3 marks]. Cosmic-ray muons provide the evidence. Muons created high in the atmosphere travel downward at more than 0.99c but have a very short half-life. Without relativity, almost all should decay before reaching the ground, so only a few should be detected at sea level compared with the number counted on a mountain top. In fact far more muons reach sea level than the non-relativistic prediction. The measured survival rate matches the prediction once time dilation stretches the muons' half-life in the ground frame, confirming time dilation. (Atomic clocks flown on aircraft give the same conclusion.)

Marker's note. In (a) make the link explicit: constant speed of light plus a longer light path means a longer measured time. The twin paradox alone is not accepted. In (b) name a real experiment (muon decay, flown atomic clocks) and state how its result supports time dilation.

[3 marks]. Side WX runs parallel to the rotation axis, so it stays perpendicular to the field throughout the turn. The force on it, $F = BIL$, keeps the same magnitude and the same direction before and after the rotation.

Side XY runs along the field direction at the start, so initially it experiences no force (current parallel to B). After the 90 degree rotation XY lies perpendicular to the field, so it now experiences a force of magnitude $BIL$, directed to the right.

Marker's note. Address the correct sides. State that the force on WX is unchanged in magnitude and direction, and that XY goes from zero force to a force to the right. Use unambiguous direction language; this question is about forces, not torque.

[3 marks]. The work done by the field on the charge equals the gain in kinetic energy:

The work done moving charge $q$ through potential difference $V$ is $W = qV$, and the particle starts from rest, so

Rearranging for $v$:

Marker's note. Show each step from $W = qV = \tfrac{1}{2}mv^2$ to the result. Keep the symbols distinct, in particular $V$ for potential difference versus $v$ for velocity.

(a) [3 marks]. The track is frictionless, so the loss in gravitational potential energy from P to Q equals the gain in kinetic energy:

The mass cancels:

The height difference is about 0.11 m.

(b) [3 marks]. Take downward as negative and resolve the launch velocity. The initial vertical velocity is

Using $s = u_y t + \tfrac{1}{2}a t^2$ with $u_y = +1.15$ m s$^{-1}$, $a = -9.8$ m s$^{-2}$, $t = 0.5$ s:

The displacement is 1.65 m below Q, so the height of Q above the floor is about 1.7 m.

Marker's note. In (a) use energy conservation and show $\Delta h = v^2/2g$. In (b) resolve the velocity into its vertical component first, then substitute into a constant-acceleration equation with consistent signs.

(a) [4 marks]. As the blades spin they push air downward. By Newton's third law the air pushes up on the fan with an equal and opposite force. This upward thrust adds to the support, so the net downward pull on the spring balance is reduced and the reading falls.

The thrust grows as the blades speed up, so the reading keeps decreasing for the first 10 seconds. At 10 seconds the fan reaches its maximum constant speed, so the thrust becomes constant; from 10 to 15 seconds the forces are balanced and the spring balance reading stays at this lower, constant value.

(b) [4 marks]. The prediction is partly wrong and partly right.

• 0 to 10 s (inaccurate). The student shows the current rising. In fact the current is largest at start-up and then falls. As the motor speeds up, the coils cut more flux per second, so the back emf grows; the back emf opposes the supply voltage, so the net driving voltage and therefore the current decrease as the motor speeds up.
• 10 to 15 s (accurate). Here the prediction shows a constant current, which is correct. At maximum constant speed the back emf is constant, so the net voltage and the current settle to a steady value.

Marker's note. In (a) explain more than one change and state the direction explicitly (the reading decreases, then holds). In (b) assess means judge correctness: identify the rising-current section as wrong using back emf, and the constant section as right.

[5 marks]. The electron - Millikan's oil-drop experiment. Charged oil droplets were suspended between charged plates, balancing their weight against the electric force, and the same droplets were timed falling under gravity. The droplet charges were always whole-number multiples of one smallest value, showing charge is quantised and giving the charge on a single electron. Combined with Thomson's earlier charge-to-mass ratio from deflecting cathode rays in electric and magnetic fields, this fixed the mass of the electron and established it as a fundamental particle.

Another fundamental particle - the quark. In deep inelastic scattering experiments, a high-energy beam of electrons was fired at protons. The way the electrons scattered at large angles showed the proton is not a point but contains three smaller, point-like charged objects inside it. These were identified as quarks, confirming that protons are made of fundamental quarks while the electron itself has no internal structure.

Marker's note. Pick a fundamental particle correctly: protons and neutrons are not fundamental, but quarks are. For each particle name the experiment, its key observation, and what it told us.

[4 marks]. Similarity. Each particle feels a magnetic force $F = qvB$ that is always perpendicular to its velocity. A constant-magnitude force at right angles to the motion provides a centripetal force, so both particles travel in circular arcs (uniform circular motion) rather than simply being deflected once.

Difference. The radius of the circle is

The alpha particle has about four times the mass of the proton and twice the charge, so its $m/q$ ratio is about twice as large. With the same speed and field, the alpha particle follows a circle of about twice the radius of the proton's, so it curves more gently.

Marker's note. Describe the trajectories (circular paths from a perpendicular force), not just the forces. Explain the radius difference by equating the magnetic and centripetal forces, and note that mass and charge together set the radius.

[9 marks]

Production of energy. The sun produces almost all its energy by nuclear fusion in its core, mainly through the proton-proton chain. Hydrogen nuclei fuse through intermediate steps (forming deuterium, then helium-3) to build helium-4. The helium nucleus has slightly less mass than the four protons that formed it; this lost mass is converted to energy according to $E = mc^2$, which is the source of the sun's output.

Radiation of energy

The sun's surface behaves as a black body. Its emitted radiation follows a black-body curve that peaks at a particular wavelength, and that peak wavelength fixes the surface temperature through Wien's law.

Surface temperature (quantitative)

Reading the peak of the curve at $\lambda_{max} = 5.0 \times 10^{-7}$ m and using Wien's law with $b = 2.898 \times 10^{-3}$ m K:

Power output (quantitative). The sun radiates equally in all directions, so the 1360 W m$^{-2}$ measured at Earth is spread over a sphere of radius $r = 1.5 \times 10^{11}$ m. The total power is intensity times the area of that sphere:

So the sun fuses hydrogen to helium, converting mass to energy, and radiates roughly $3.9 \times 10^{26}$ W from a surface near 5800 K.

Marker's note. Address every part: name the proton-proton chain and mass-to-energy conversion, use the graph peak with Wien's law for the temperature, and use $P = I \times 4\pi r^2$ for the power. Carry units through and show full working.

[4 marks]. A string that stays deflected to a constant angle while the speed is constant means the mass has a constant horizontal acceleration. Since the car's speed is constant, that acceleration must be centripetal, so the car is turning in uniform circular motion. The deflection points outward from the turn, and a larger $\theta$ means a larger acceleration and so a tighter (smaller-radius) turn.

Resolve the string tension $T$ on the mass. Vertically the tension balances gravity, and horizontally it provides the centripetal force:

Dividing (2) by (1) cancels $T$ and $m$:

So measuring $\theta$ (and knowing $v$) gives the radius of the car's circular path.

Marker's note. Recognise the motion as uniform circular motion and use vector analysis: resolve the tension, then divide the two equations to eliminate $T$ and $m$. Show every step of the derivation.

[7 marks]. Mass defect. The mass lost in the decay is

Converting to kilograms ($1 \text{ u} = 1.661 \times 10^{-27}$ kg):

Total energy released. Using $E = \Delta m\,c^2$:

Kinetic energy of the alpha particle. This released energy is shared as the kinetic energy of the two products, so

Why the alpha's kinetic energy is so much larger. The radon atom is at rest, so by conservation of momentum the two products fly apart with equal and opposite momenta: $m_\alpha v_\alpha = m_{Po} v_{Po}$. The alpha particle has far less mass than the polonium, so it must have a far greater speed. Kinetic energy can be written $KE = p^2/(2m)$; with the same magnitude of momentum, the lighter alpha particle (smaller $m$) carries the larger kinetic energy. So although the alpha is much lighter, its much higher velocity wins out and it ends up with most of the released energy.

Marker's note. Convert the mass defect to energy with correct units before subtracting the polonium kinetic energy. Then apply conservation of momentum to the at-rest parent and use $KE = p^2/2m$ (or $KE = \tfrac{1}{2}mv^2$ with equal momenta) to explain why the lighter alpha takes the greater share.