2017 HSC Physics Paper Solutions

Have you seen the 2017 HSC Physics Paper, yet? Did you solve the questions? Did you find some of them particularly difficult? In this post, we will work our way through the 2017 HSC Physics Paper and give you the solutions to all of the questions.

Read on, to see how to answer all of the 2017 questions.

 

2017 HSC Physics Paper Solutions

This paper was broken into two sections.

• Section 1 is based on the core unit. This section has 2 parts:
  • Part A – 20 Multiple Choice Questions
  • Part B – 10 Short Answer Questions
• Section 2 has five questions:
  • • Each question is on an option topic and is split into several parts.
    • In this post we have included the answers to:
      • Medical Physics
      • Astrophysics
      • From Quanta to Quarks

Section 1 – 75 Marks

Part A – 20 Marks

 

Part B – 55 Marks

Question 21 (5 Marks)

 

(a)

\(f=\frac{c}{\lambda}=\frac{3.00 \times 10{^8}}{550 \times 10{^14}} = 5.45 \times 10{^14} Hz\)

(b)

According to Einstein’s explanation of the photoelectric effect, light of frequency f consists of photons of energy E=hf and one electron can only absorb the energy from one photon at a time. If the energy is insufficient to eject the electron, it will relax back down to a low energy before absorbing another photon.

The energy carried by each photon of 550 nm light is E= hf = 6.626 × 10^-34 × 5.45 × 10¹⁴ = 3.6 × 10^-19 J . This is less than the minimum energy required for electrons to absorb in order to be ejected from the surface so electrons will not be ejected.

 

Question 22 (5 marks)

(a)

τ = Fd cos θ, with θ = 0 here. To increase torque the magnitude of force can be increased, and the distance from the axis d at which the force is applied can be increased.

 

(b)

Force produced by BC is F = nBIL = 15 × 0.2 × 7.0 × 0.08 = 1.68 N

Torque produced is \(T=Fd=1.68 \times \frac{0.06}{2} = 0.0504 Nm\)

 

The current is clockwise so force is into the page at BC by Right Hand Palm Rule, so torque is clockwise as viewed from the bottom of the page.

 

Question 23 (5 marks)

To validate theories in physics, independent valid experiments must be performed to test the different predictions made by the theory. If the results of different independent experiments match the predictions of the theory and no valid experiments contradict the predictions then the theory becomes validated.

Special Relativity is based on two principles and makes several predictions. The two principles are that (1) the laws of physics are the same in all inertial frames. (2) The speed of light has the same value for any observer.

Principle (1) has been validated since before special relativity, in that experiments done in inertial frames give the same result as if they were at rest. Principle (2) was validated by many experiments measuring the speed of light, and particularly by the Michelson–Morley experiments where there was no change in the speed of light due to the motion of the Earth.

The predictions of special relativity include time dilation, length contraction, relativity of simultaneity, mass dilation and the speed limit of massive objects. Time dilation has been validated by comparing moving atomic clocks to stationary ones, and by needing to adjust GPS satellite times. Mass dilation is witnessed in particle accelerators where the kinetic energy of particles increases even though their velocities never reach c. The appearance of muons at the surface of the earth, created by cosmic rays in the upper atmosphere, are incompatible with classical physics but explained by length contraction and time dilation. Observations from many independent sources agree with a range of predictions of special relativity and no valid observations disagree, so this theory is validated.

 

Question 24 (5 marks)

(a)

\(v=\sqrt{\frac{2\times6.67\times10{^-11}\times6.39\times10^{23}}{3.39\times10{^6}}}=5014.5ms{^-1}\)

 

(b)

Using conservation of energy, the initial mechanical energy (kinetic + potential) must equal the final mechanical energy:

KE_i + GPE_i = KE_f + GPE_f

To escape from the planet the object must reach an infinite distance away, where potential energy is zero, so GPEf = 0. As it moves further its KE will reduce, so we assume it had just enough energy to escape and KE reduces to 0, so KE_f = 0.

The object must therefore be provided with enough kinetic energy on the planet (providing it with escape velocity v_esc at radius r_planet) to convert to potential energy at an infinite distance away. Its total energy must be zero:

\(\frac{1}{2}{mv{^2}{_esc}}-\frac{GMplanet{^m}}{r{_{planet}}}=0\)

Mass of the spacecraft, m, can be cancelled out from the equation, yielding \(v{_esc}= \sqrt{\frac{2GM{_{planet}}}{r{_{planet}}}}\) as provided in the question.

 

 

Question 25 (7 marks)

(a)

The Braggs discovered and developed X-ray crystallography. A collimated beam of X-rays are fired at a material. They diffract off the cations, producing an X-ray diffraction pattern captured by photographic film. The regular array of maxima in the diffraction pattern indicates the regular lattice (crystal lattice) structure of the material, with spacing of the maxima relating to lattice spacing and the shape of the pattern relating to shape of the crystal lattice. This was the first method of directly showing that some materials have a crystal lattice structure consisting of a repeating regular array of cations, allowing the dimensions of the lattice to be measured.

 

(b)

Doping pure silicon with boron will produce a p-type doped semiconductor. The valence of boron is 3, so it only forms 3 bonds with nearby silicon atoms. An empty acceptor level forms near the valence band of the silicon that can easily accept valence electrons at room temperature. This results in many mobile holes in the valence band of the silicon at room temperature. This increases the number of mobile charge carriers and significantly improves conductivity.

 

(c)

Maglev trains require strong magnetic force but low weight in order to be levitated. The magnets need to be compact in order to fit on board and be cooled below Tc more easily. Superconductors achieve strong field with low weight and size. They don’t require an ongoing large power source to maintain the magnetic field, so the Maglev train can run on batteries rather than need to be fed large power from the track.

 

Question 26 (6 marks)

(a)

At a re-entry angle of greater than 10° the craft enters into dense atmosphere at too high a speed and experiences a very large drag force. This results in large g-force on the craft that can destroy the craft or directly injure the crew, and converts KE into heat too rapidly leading to intense heating which could cause the craft to burn up. The consequence is destruction of the craft and crew.

 

(b)

A spacecraft converts large gravitational potential energy and kinetic energy to heat via friction and shock heating of the gases around the craft. This slows the craft down so that it can safely land, but causes heating to the craft which must be kept within safe limits. This also heats the air around the craft to 5000 K so it visibly glows, according to the graph. This radiation dissipates some of the heat to surrounding regions but the glow is accompanied by ionisation blackout whereby the craft cannot send or receive radio signals to the control station. This means the spacecraft is flying blind, unable to navigate to its safe landing location during ionization blackout.

 

Question 27 (5 marks)

 

(a)

Flux Φ = BA. Maximum B occurs at maximum flux, so \( B= \frac{\phi}{A}\)\( = \frac{0.6}{\pi \times 0.30{^2}}\)\( =2.12T\).

 

(b)
Faraday’s Law: \(\varepsilon  = -\frac{change in flux}{time}= -\frac{\Delta \phi}{\Delta t}\). I.e largest voltage is produced by the largest rate of change of flux. This corresponds to the part of the graph with steepest slope: from 10 to 12 seconds where the flux changed by -0.6 Wb.

\(V= \varepsilon  = – \frac{-0.6}{2}=0.3V\). The direction is determined by Lenz’s law. The flux into the page is decreasing so emf produces current to produce magnetic flux to oppose the change, requiring current clockwise to produce flux into the page. The clockwise current causes positive charges to accumulate at P, thus P is positive, Q is negative.

 

Question 28 (6 marks)

Eddy currents in both devices result in heat and magnetic force.

Transformers: Eddy currents have undesirable effects. They cause flux interference which cancels out some of the flux being produced by the primary before it reaches the secondary, reducing the efficiency of the transformer. They cause dissipation of electrical energy to heat, heating the transformer and increasing the temperature of the solenoid wires, increasing their electrical resistance and further reducing efficiency. The cores of transformers are laminated to reduce eddy currents.

Magnetic braking: Eddy currents produce a magnetic force which slows the disc or moving part of the brake, and also dissipate the kinetic energy to heat. These are the desirable functional effects of magnetic braking, hence the disc or moving part are solid metal so that large eddy currents can form.

 

Question 29 (7 marks)

(a)

From line of best fit on graph, \( K= \frac {24-3}{0.08  – 0.01}=300Nm{^-1}\)

Spring initial energy is \(E{_p}= \frac{1}{2}kx{^2}=\frac{1}{2} \times 300 \times 0.08{^2}=0.96 J\)

The spring decompresses, converting spring energy to kinetic energy: E_k= 0.96 J. Thus, \(v= \sqrt {\frac{2E}{m}}=\sqrt{\frac{2 \times 0.96}{0.04}}=6.93ms{^-1}\)  . Direction of launch velocity not specified.

 

(b)

\( u{_x} = u \cos \theta = 10 \cos 60\deg= 5.0ms{^-1}\) \(u{_y} = u \sin \theta = 10 \sin60\deg = 5.0ms{^-1}\)

Time to maximum height is v_y = u_y + a_yt  with v_y = 0. Assume a_y = -9.8ms^-2.

\( t = \frac{-8.67}{-9.8}=0.88 s\)

Trajectory is symmetric about maximum height so time of flight is

t_total = 0.88 × 2 = 1.77 s.

Range is Δx = u_xt = 5.0 × 1.77 = 8.85 m.

 

Question 30 (4 marks)

The electric field results in acceleration of the proton parallel to electric and magnetic fields, hence its velocity parallel to the fields increases at a constant rate.

The magnetic force results in acceleration of the proton perpendicular to the fields, the proton follows uniform circular motion, moving anticlockwise when viewed from the right of the page (speed remains constant perpendicular to the fields).

The horizontal motion (parallel to the fields) is affected by the electric field only, and is independent from the motion perpendicular to the fields which is affected by the magnetic field only.

Thus the proton traces out a spiral motion with the spiral having a constant radius and an increasing pitch (length of each spiral).

 

2. Section 2 – 25 marks

 

Question 32 – Medical Physics (25 marks)

(a)  (i)

In coherent fibre bundles, the arrangement – the relative positions – of the individual optical fibres are the same at both ends of the bundle.

In incoherent bundles the relative positions are not the same from one end to the other.

 

(ii)

Light travels along an optical fibre by total internal reflection. The core has higher refractive index to the cladding so when light is input into the end of the core of the fibre, it totally internally reflects when it is incident on the core-cladding boundary and continues to reflect, bound in the core, until it exits the other side. The incoherent bundle is used to carry illuminating light from the outside of the body to the surface inside the body. A lens forms an image on the end of the coherent bundle, which accepts this light and carries it out of the body, thus transferring the image outside the body.

 

(b)  (i)

An ultrasound densitometer can determine the density of bone. It measures the time of transit and the attenuation of ultrasound through bone to determine velocity and acoustic impedance, allowing the system to calculate the density of the bone by Z = ρv.

(ii)

B scans have a wide array of transducers and produce a rectangular image so are more suited to imaging close to the skin. The sector scan has a narrow image near the skin and widens deeper in the body, so is suited to deeper imaging. High frequencies of ultrasound have higher resolution but are attenuated by tissue over shorter distances than low frequencies, so high frequencies are appropriate for B scans while low are required for deeper sector scans.

 

(c)

Nuclei with net spin in a strong external magnetic field align parallel or antiparallel with the magnetic field, with a slight majority parallel. Applying a radio frequency pulse at the Larmor frequency to the nuclei rotates the spin. Usually, a 90 degree pulse is applied, rotating the spins to the transverse plane, perpendicular to the magnetic field. Because these spins have a nonzero angle to a strong magnetic field, they precess at the Larmor frequency. This precession is in phase where the originally parallel spins point in the same direction, likewise for the originally antiparallel spins. This results in a rotating magnetic field, which emits radio waves at the Larmor frequency. This is called resonance. After some time the nuclei’s precession stops being in phase, and they also return to being parallel/antiparallel with the field.

 

(d)  (i)

[K-shell or Bremsstrahlung]

 

Electrons are accelerated by potential difference in a cathode ray tube. They hit the anode with high kinetic energy. The electron can knock out a ground state (“K-shell”) electron from an atom in the anode, leaving a hole. The hole is filled by a higher energy electron of the atom falling into it, releasing an X-ray photon.

 

OR:

The electron is accelerated (the direction and speed is changed) by a nucleus in the anode. This acceleration results in emission of electromagnetic radiation by Maxwell’s theory. Due to the high change in energy, the radiation is an X-ray photon.

 

(ii)

A plain X-ray image is a flat projection (shadow) of all radiopaque tissue in between the X-ray source and the detector. Deep soft tissues within hard tissues are not visible. A CAT scan is a full reconstruction of each cross-section of the body. It can create a 3D reconstruction of the position of hard and soft tissue at all depths in the body, even soft tissue within a shell of hard tissue. It also has higher contrast and resolution. It can be used for diagnosing soft tissue problems in the brain and chest whereas radiographs are used for diagnosing fractures and dense foreign objects.

 

(e)

Functional medical imaging uses radioisotopes introduced into a patient’s body to produce images of metabolic function. The radioisotope is an element or is included into a chemical that has some biological function (radiopharmaceutical).  This is introduced to the patient and binds to or accumulates in regions of interest. It decays and radiates from its location inside the body and the radiation is detected outside the body. The amount and location of the radiation produces the image, indicating ‘hot spots’ of high activity.

For example, Technetium-99m can label red blood cells. The blood travels through the heart muscles. If the heart muscle is dead, it will show up as dull on the image, diagnosing the function of the heart muscle whereas the structure of the heart is unchanged.

Tc-99m can also label a radiopharmaceutical that attaches to receptors on a type of cancer tumour. Then hot spots on the scan image will show the location of tumours in the body, even though the density of the tumour is similar to soft tissue so won’t show up on structural scans.

 

 

Question 33 – Astrophysics (25 marks)

(a)  (i)

Earth-based optical telescopes are affected by absorption and distortions of light by atmosphere, reducing the resolution and sensitivity of the telescope. A space-based telescope is free from these problems and also can take longer exposure times, increasing effective sensitivity in the information collected.

 

(ii)

\( \frac{I{_A}}{I{_B}} = 100{^{m{_B}-m{_A}/5}}=\frac{I{_P}}{I{_Q}}=100{^{\frac{8.5-10.3}{5}}}=0.43\)

 

P is 43% the brightness of Q.

 

(b)  (i)

X is B0 type. It is very hot, blue, with not many absorption features, just from neutral Helium and Hydrogen.

Y is M0 type. It is a cool red star with many absorption features including molecules such as TiO.

 

(ii)

Light is collected from the star by an optical telescope. The light just from the star is input into a spectrometer. This has a diffraction grating which splits the light across a screen into its wavelengths. Instead of an observation screen, an array of digital light sensors detects the light and measures the intensity at each wavelength. This histogram of intensity versus wavelength is plotted, producing the spectrum.

 

(c)

Colour index is the difference in absolute or relative magnitudes of a star through two different filters or bands. For example, blue magnitude minus visual magnitude. The redder magnitude is subtracted from the bluer magnitude, for example CI = B – V. The colour filters have a characteristic wavelength but a relatively wide band, about 100nm. Due to lower magnitude indicating larger brightness, a negative colour index indicates a bluer star and positive colour index indicates a redder star. The colour index of a star (once any red or blue shift is corrected for) determines the surface temperature of the star and the spectral class.

 

 

(d)   (i)

 

 

(ii)

In an open cluster the stars are young blue stars. There are a low number of stars (50-1000) loosely gravitationally bound.

In a globular cluster the stars are very old red stars. They are low metallicity so are extremely early generation stars. No new stars being formed. There is a large number of stars (1000-1000000) strongly gravitationally bound.

 

 

(e)

Starting from the main sequence stars:

Stars of about 1 solar mass become red giants as the core runs out of hydrogen and compresses as gravitational forces dominate over radiation pressure. This increases temperature causing the shell to fuse hydrogen. As the shell runs out of hydrogen the gravitational pressure is not enough to cause any further fusion. The radiation pressure overcomes gravity pushing the envelope out into a planetary nebula but the core remains behind as a white dwarf.

Large stars about 8 solar masses and larger become red giants in the same way, but are larger so are red supergiants. They can have core helium fusion and perhaps larger element fusion. When the core stops fusing the radiation pressure in the core ceases and the enormous gravitational force on the star collapses the core, generating large heat and causing a supernova, blasting off the atmosphere of the star. The core remains. For the ~8 solar mass stars the remnant becomes a neutron star. A neutron star is where gravitational force overcomes the pressure of electrons pushing against each other across atoms. The atoms are compressed into a huge mass of neutrons. For very large stars >25 solar masses, the remnant collapses into a black hole.

Question 34 – From Quanta to Quarks (25 marks)

(a)  (i)

A helium-3 nucleus consists of two protons and one neutron. Each proton consists of two up quarks and one down quark, and the neutron consists of one up quark and two down quarks. Gluons transferred between the quarks bind the quarks into each nucleon.

 

(ii)

The strong nuclear force is the residual force from the strong force carried by gluons. The strong nuclear force binds nucleons together in the nucleus. It binds protons and neutrons by the same amount. It is repulsive below 0.5 femtometres, attractive from 0.5-2 femtometres, and becomes effectively zero strength above 2 femtometres. Its maximum strength is very large – some 10 000 N.

 

(b)  (i)

Nuclear transmutation is when a nucleus changes into a different element and releases radiation. Nuclear fission is when a nucleus splits into 2 or more daughter nuclei (and usually free neutrons).

 

(ii)

The fuel elements contain fissionable material. Any slow stray neutrons can result in fission, releasing fast neutrons.

The moderator material absorbs the kinetic energy from these fast neutrons, heating up and allowing the energy to be extracted as heat. It also results in slow neutrons which can then cause further fission in a chain reaction.

The control rods absorb neutrons and are used to control the reproduction constant of the chain reaction. It can cause the reaction rate to increase when removed or to decrease or stop when inserted in order to control the rate of the nuclear chain reaction.

 

(c)

Bohr postulated that electrons can orbit the nucleus only in particular allowed orbits. The electron could only change orbits by a discrete jump from one allowed orbit to another where it has to absorb a photon, or emit a photon, of energy equal to the difference in energy of the orbits. The allowed orbits and energies are determined by quantisation of the angular momentum of the orbiting electron \(L=mvr=\frac{nh}{2\pi}\), where n is a positive integer. The wavelength of light emitted or absorbed follows Rydberg’s equation \(\frac{1}{\lambda} = R{_H}(\frac{1}{n{^2_f}}-\frac{1}{n{^2_1}}\). The emission line in question is the third line of the Balmer series, where n_f = 2 , and thus n_i = 5 . Then \(\frac{1}{\lambda}=R{_H}(\frac{1}{4}-\frac{1}{25})\) and so \(\lambda= \frac{100}{21R{_H}}=\frac{100}{21 \times 1.097 \times 10{^7}} = 434.0nm\).

 

 

(d)  (i)

Orbiting electrons are in circular motion undergoing acceleration and thus should be consistently radiating according to Maxwell’s theory of electromagnetism. They would lose energy and be unstable, spiralling into the nucleus. Also, any orbit should be possible in classical physics.  However, in Bohr’s model only certain orbits are allowed and these are stable and do not radiate.

 

(ii)

De Broglie hypothesised that particles, including electrons, exhibit wave properties with a wavelength \( \lambda {_dB} = \frac{h}{mv}\) .

Under de Broglie’s hypothesis, electrons in atoms can occupy orbits only when an integer number of electron wavelengths fit around the circumference of the circular orbit: nλ = 2πr  in order to form a standing wave. Combining the two last equations reproduces Bohr’s quantisation of angular momentum.

Davisson and Germer observed electron diffraction when scattering electrons off a nickel crystal. This indicated electrons have wave properties. The wavelength of the electrons indicated by the position of the diffraction maximum matched the de Broglie wavelength predicted for the electrons of that velocity.

De Broglie’s hypothesis, supported by Davisson and Germer’s experiment, applied to electron orbits explained the quantisation of the orbits in Bohr’s model.

 

(e)

Protons are used in particle colliders to investigate higher energy particles that don’t exist in everyday life. Protons are often used because they are easily accessible from ionised hydrogen. They are massive so they can be accelerated and their kinetic energy increased. They are charged so they respond to magnetic and electric fields to trap them and increase their kinetic energy to extremely large values. When collided their energy is converted into matter to produce a wide variety of new particles because they feel all forces. In particle colliders the strong, weak and electromagnetic forces are the relevant forces. The properties of the product particles are studied to update theories and understanding of matter, i.e. the Standard Model.

Neutrons are used at low energy in neutron scattering and diffraction. They are neutral so they interact well with light nuclei rather than with electrons around heavy nuclei as X-rays do. They are massive so have short de Broglie wavelength allowing for probing on very short length scales, determining the structure of solid matter on the nuclear level.

 

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