2023 HSC Physics Exam Paper Solutions

Congratulations on completing your HSC exams! See how you went with our 2023 HSC Physics Exam Paper Solutions.

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Matrix Science Team
2023 HSC Physics Exam Paper Solutions

2023 HSC Physics Exam Paper Solutions

The 2023 HSC Physics exam was held on 2 November. How did you go? Check your answers with the 2023 HSC Physics Exam Paper Solutions written by the Matrix Science Team!

Find the full 2023 HSC Physics Exam Paper Questions, here.

 

Section 1. Multiple Choice

QuestionAnswerExplanation
1DThe gravitational field strength of Earth is given by g=GM/r2  where r=rEarth +h and the altitude above ground is h. As h increases, the denominator increases and hence g decreases.
2CA step-up transformer must be placed at the power station to increase the transmission voltage and hence decrease the transmission current to minimise power loss Ploss =I2R. A step-down transformer must be used at the house to decrease voltage for use and safety.
3CThe diffraction pattern seen from a double slit has a bright central maximum, and has symmetric bright maxima on either side.
4BIn 90 years, three half-lives will have passed for caesium-137, meaning that (1/2)3=1/8 of the original 120 g will be left, which is 15 g.
5DBy Kepler’s 2nd Law, the exoplanet will travel the lowest velocity at the furthest point from the star, and hence will have the longest travel time between S and P.
6DBy Maxwell’s equations, accelerating charges will produce electromagnetic radiation. This can be achieved by an electron moving in a circular path at constant speed as it is accelerating while doing so, with a centripetal acceleration.

Electrons in stable orbits within atoms do not produce electromagnetic waves (Bohr-de Broglie model).

7CNeutrons are slightly heavier than protons. The de Broglie wavelength is given by λ=h/mv. If both the proton and the neutron have the same velocity, then their wavelength differs only due to their mass. The wavelength is inversely proportional to the mass, so the neutron would have a smaller wavelength due to its greater mass.
8BTo land at B, the range must decrease. As range is given by sx=uxt, range can be decreased by either decreasing the initial horizontal speed or the time of flight. The time of flight for the projectile increases as the platform height h increases or the launch angle is made higher, so it is only possible to decrease the range by decreasing the initial horizontal velocity, which can be done by increasing the launch angle.

The trajectory of a projectile is independent of its mass.

9AThree changes occur when the blackbody decreases in temperature:

  • The peak wavelength shifts to a longer wavelength
  • The peak intensity decreases
  • The intensity at all wavelengths decreases.
10DIn a uniform magnetic field, the force on side XY remains constant throughout the rotation, because the angle between the current and B field remains the same. 

The torque however is maximum in Figure I (force is down and r is horizontal on side XY) and zero in Figure II (force is down and r is vertical on side XY).

11AEach diagonal decay represents an alpha decay (mass number decreases by 4, atomic number decreases by 2), while each horizontal decay represents a beta minus decay (mass number stays the same, atomic number increases by 1). 
12BThe work done on the charge is W = qV. In Figure I, the charge passes through the full 100 V. However when the distance between the plates is doubled, the charge only traverses through a 50 V difference in the same distance, hence the work done is halved.
13CGreater binding energy indicates a greater mass defect via E=mc2, not to be confused with binding energy per nucleon which indicates stability.
14CThe escape velocity formula for Earth is: vesc=√2GME/rE

MX=4ME and rX=3rE , and

vesc,X =√2GMX/rX = √(2G4×ME/3rE) = √4/3 × √2GME/rE = 2/3 × 11.2 = 12.9 kms-1

15AEmission of photoelectrons required a threshold frequency f0, in which emission required a frequency of light to be greater than f0 and hence lower than a specific wavelength as they are inversely proportional.

The maximum kinetic energy of the emitted photoelectrons is given by Kmax= hf – ϕ, where the work function depends on the metal.

Increasing intensity increases the photocurrent, not the kinetic energy.

The maximum kinetic energy depends on the work function and hence depends on the metal.

The release of photoelectrons is instantaneous, and independent of the duration of exposure to the light.

16DBoth rods will exert an equal and opposite force on each other.

Since rod Y remains stationary, the magnetic force must be equal and opposite to its weight. This means the rods must be repelling each other.

The force that X exerts on the table is a combination of its own weight and the downwards magnetic force, with a total equal to the combined weight of X and Y. 

By Newton’s 3rd Law the table will exert the same amount of reaction force on X.

17BTorque is given by τ =rFsinθ

The force that provides the torque on the mass is the weight force, which always acts vertically downwards. Torque is zero at Y as r and F are parallel. Torque is at a maximum at X and Z as the angle between r and F is a maximum at these points.

18AThe acceleration due to gravity is independent of the mass of the object in the field, so the object’s trajectory in a gravitational field will be unchanged and the mass will land exactly on X.

The acceleration of a particle in an electric field is given by a=qE/m and hence acceleration decreases as mass increases. This will cause the mass to land further away than Y.

19BIn the frame of reference of a charge in Y, the charges in Y will look unaffected as they are stationary within this frame. In comparison, a charge in Y would view the charges in X as flowing, and would measure a contracted length between the charges in X.
20DThe light that we detect from the most distant galaxies has taken the longest amount of time to reach the Earth. Therefore it was released the greatest amount of time ago (when the expansion rate was lower). The gradient of the graph, which determines Hubble’s constant, will appear to be lower.

Question 21

a)

Luminosity depends on:

  • Surface area, or radius of the star
  • Temperature of the star

b)

Star A is a main sequence star, in the main part of its life and fusing hydrogen to helium in the core, whereas star B is a white dwarf star and is the remnant core of a low-mass star that has died and is no longer undergoing fusion reactions. Star A and B have the same surface temperature, however star A has a much greater luminosity than star B, arising from a larger surface area or radius.

 

Question 22

The crew on the spacecraft measures the proper time between the emitted pulses, hence the people on Earth measure the dilated time:

tv = t0/1-v2/c2 = (3.1 × 10-9)/1-0.92 = 7.1 × 10-9s

 

Question 23

a)

F=GMm/r2 = (6.67 × 10-11  × 1.99 × 1030 × 6.1 × 103)/(1.52 × 1011)2 = 35 N

 

b)

Minimum photon energy corresponds to maximum wavelength:

E = hf = hc/λ = (6.626 × 10-34 × 3 × 108)/2.8 × 10-5 = 7.1 × 10-21J

 

c)

From Wien’s displacement law: λ =b/T hence

T = b/λ = 2.89 × 10-3/1.14 × 10-5 = 253.5 K

 

Question 24

The force from the magnetic field provides the centripetal force on the electron:

qvB = mv2/r → B=mv/qr = (9.109 × 10-31 × 3.0 × 106)/(1.602 × 10-19 × 10) = 1.71 × 10-6T

 

Question 25

a)

Part X is the split ring commutator. It reverses the direction of the current in the rotor every half turn to ensure that torque stays in the same direction, and the motor continues to rotate in the same direction.

b)

The torque in a DC motor is given by τ =nIABsinθ, hence maximum torque depends directly on the current in the coil. As the rotational speed increases, induction causes the back EMF in the coil to increase and hence the total voltage in the coil decreases. As resistance remains constant, the current decreases by Ohm’s law V=IR and thus the maximum torque decreases.

 

Question 26

Mass defect = (12.064 + 1.008) – (9.013 + 4.003) = 0.056 u

Energy released = 0.056 x 931.5 = 52.164 MeV

 

Question 27

a)

A star’s spectrum resembles a black body curve with added absorption lines.

The temperature of a star can be obtained from the peak emission wavelength of its spectrum, and is calculated using Wien’s displacement law: λpeak = b/T

The chemical composition of the star can be determined from the wavelengths of spectral lines in the star’s absorption spectrum. These spectral lines are unique for each element, corresponding to the energy transitions occurring in the atom, and can hence be compared with known spectra of elements to determine the elements present.

Additionally, broader spectral features are a common indicator of molecules being present in the star, as the energies of the elements in the molecule will overlap and create broader features rather than fine spectral lines.

b)

A star that is rotating will have one side redshifted (the side that is receding) and the other side will be blueshifted (the side that is moving towards Earth). This results in a simultaneous red and blue shift of the spectral lines, and hence results in a broadening of the line. The diagram should be modified by thickening the line, where a thicker line results from a faster rotational velocity. The line should be fainter towards the edges, as there are fewer particles with the most extreme rotational velocities.

 

Question 28

a)

Using the transformer equation VS/VP=NS/N→ VS = 240 × 50/300 = 40 V

b)

When the switch is open, globe Y is disconnected from the circuit. Hence only the resistance of globe X contributes to the current in the secondary coil. 

After the switch is closed, two globes of identical resistance are connected in parallel, reducing the resistance in the secondary circuit to a half of its original value. By Ohm’s law, this will increase the current in the secondary circuit. 

Since the transformer is ideal, the power in the primary coil must be equal to the power in the secondary coil: VpIp = VsIs. Thus the increase in current in the secondary coil will result in changing the current in the primary coil.

Question 29

Light from an incandescent lamp is emitted as electromagnetic waves with random polarisation and hence its output is unpolarised. When a plane polarising filter is held in front of the light, only the component of the electric field that is parallel to the polarising filter plane will be able to pass through. Any component of a wave that is polarised perpendicular to the filter will be blocked, reducing the intensity of the transmitted light compared to the original emission.

 

Question 30

a)

As the magnetic field increases when the current is switched on, there will be a change in flux through ring X from 0s to 0.03s. This change in flux will induce an EMF in ring X by Faraday’s law for this duration of time, and thus an induced current. By Lenz’s law, the current will flow in ring X to oppose the change in flux experienced, and thus will produce a magnetic field in the ring that will create an opposing force between the ring and the solenoid (ie. creating a magnetic north nearest to the solenoid, such that the ring is repelled by the solenoid). Between 0.03 – 0.05s, there is no change in magnetic field and therefore no change in flux, which will reduce the current in ring X to zero and the repulsive force between the ring and the solenoid will be zero.

b)

(i) Ring Y will experience an induced EMF between t = 0 s and t = 0.03 s due to the changing flux through the centre. However, because it does not form a closed loop, no current can flow around it and hence no force is exerted between the ring and the solenoid.

(ii) From the graph:

ε =Δϕt = ΔBA/Δt = A × ΔB/Δt = 4 × 10-4 (6 × 10-3)/0.03 = 8 × 10-5 V

 

Question 31

The change in flux experienced by the braking fin (Faraday’s Law) induces eddy currents in the fin which create an opposing force on the rollercoaster (Lenz’s law). This leads to negative acceleration and the slowing down of the roller coaster at both initial speeds.

The induced current, and hence the braking force, is proportional to the rate of change of flux. Hence, braking force is proportional to the speed of the rollercoaster.

For this reason, the magnitude of acceleration of the higher speed (12 m/s) is larger than for the lower speed (10m/s) at all times due to the greater braking force from the greater rate of change of flux.

The shape of each data set is similar as each one experiences the same physics principles of electromagnetic braking with a braking force proportional to the speed of the rollercoaster. Each has a maximum braking force when maximum rate of change of flux is achieved, and the force decreases as the speed of the rollercoaster decreases (and the rate of change of flux decreases), thus giving the same shape for each dataset.

 

Question 32

First consider the motion of the ball: it will be launched into the air as a projectile, with a vertical velocity of 5.72 m/s and a horizontal velocity equal to the rotational velocity of the disc at the edge. 

The frequency of rotation is 3 Hz, and hence the angular velocity of the disc is ω = 2πf = 6π rad/s. Thus, the linear velocity at the edge of the disc is v = rω = 2 × 6π = 12π m/s.
Hence, we have a projectile with ux=12π m/s and uy = 5.72 m/s

Time taken to reach the maximum height occurs when vy = 0 → vy = uy + at → t = uy/g. For a launch from the ground, the total time of flight is twice this amount: t = 2uy/g.

Hence the range of the projectile is sx= uxt =2uxuy/g = (2 × 12π × 5.72)/9.8 = 44.01m from its launch position in the direction of the horizontal launch velocity.

Meanwhile the disc continues to rotate. The period of rotation of the disc is ⅓ = 0.33 s. While the projectile is in the air for t=2uy/g=25.729.8=1.167 s, the launcher has has undergone 1.167/0.33=3.5 rotations, placing it vertically opposite the launch position X. Since the radius of the disc is 2 m, the position of the launched is 4 m away from X. 

Finding the hypotenuse of the triangle with sides 44.01 m and 4 m gives the relative position of the projectile from the launcher’s new position as 44.19 m.

2023 HSC Physics Exam Paper Solutions

The angle relative to the horizontal in the diagram above is 5.2 degrees.

 

Question 33

This question is very open and lends itself to a broad range of possible answers, allowing students to showcase their knowledge and ability to link several aspects of physics to form a complete argument.

Answers could include the following key points that address different parts of the question:

  • The experiments that lead to the discovery of electrons (observations and experiments)
  • The experiments that lead to the discovery of protons & neutrons in the nucleus (observations and experiments)
  • De Broglie’s matter wave hypothesis and a relevant experiment that confirmed it (observations and experiments, subatomic particle properties)
  • Schrodinger’s quantum model of the behaviour of electrons (subatomic particle properties)
  • Particle colliders and the Standard Model (observations and experiments, subatomic particle properties, interactions with fields)
  • Nuclear fission and fusion processes (interactions with fields, processes in the physical world)
  • Nuclear decay processes (interactions with fields, processes in the physical world)

 

Sample answer:

The discovery of the subatomic particles themselves lay the foundations for further observation, beginning with the discovery of the electron, the first subatomic particle to be discovered, by Thomson’s charge to mass ratio experiment, which saw the discovery of the electron through its interaction with electric and magnetic fields in a discharge tube. Its properties of fundamental charge and light mass were then discovered by Milikan’s oil drop experiment, with the interaction of electrons with electric and gravitational fields.

While the Geiger-Marsden experiment confirmed the existence of the nucleus, it was Rutherford’s discovery of the proton, and Chadwick’s discovery of neutrons via their interaction with paraffin wax, that completed the discovery of the stable subatomic particles that make the atom.

De Broglie’s matter wave hypothesis furthered our understanding of the wave nature of these subatomic particles, with the Davisson-Germer experiment demonstrating the wave properties of diffraction and interference of electrons through their interaction with nickel. Schrodinger completed our understanding of the wave nature of electrons through the wave mechanics of his equation. The wavefunction fully describes the wave properties of subatomic particles and hence describes their interactions with other subatomic particles.

Particle colliders use electric and magnetic fields to accelerate particles to near light speed, making use of the relativistic relationship E=mc2 to convert the particle’s kinetic energy to matter. Through this process, the fundamental particles of the Standard Model could be created and studied. These include the quarks which interact via all four fundamental forces, leptons which interact with the weak, EM and gravitational forces but not the strong force, and bosons which are the particles responsible for mediating the weak, strong, and EM forces. The interaction of the matter particles with the EM, weak and strong forces can hence be studied in particle colliders, with physical processes such as the decay of unstable fundamental particles proceeding via the W+, W- or Z bosons that mediate the weak force.

From the knowledge of the particles and their interactions with the Standard Model, other physical processes can be understood such as nuclear fission and fusion. These processes involve the release or absorption of the binding energy of the nuclei, directly related to the strong force binding the nucleus together. Nuclear fusion is observed to occur naturally within stars, fusing light elements to heavier ones up to iron.

In a nucleus, the strong force overcomes the electric repulsion between protons to stabilise the nucleus, however the nucleus will decay via the weak force to increase stability. One such example is the beta minus decay of a neutron-rich isotope, which releases an electron and antineutrino via a W- interaction. 

Through analysing the interaction of subatomic particles with fields and other particles, scientists have increased our understanding of crucial physical processes in the world as well as a gaining a thorough understanding of these particles themselves.

Question 34

a)

As the satellite moves from P to Q, its total energy remains constant once the engine is shut off. Its gravitational potential energy increases as r increases, and kinetic energy decreases by the same amount.

 

b)

The total energy of the satellite at point P after the engine is fired and turned off is Etotal= U + K = -1.195 × 1010 + 5.232 × 108= -1.14268 × 1010J. As stated in the previous answer, this total remains constant while travelling from P to Q.

The gravitational potential energy at point Q is

U=-GMm/r = -(6.67 × 10-11 × 6 × 1024 × 400)/(6.850 × 106)=-2.34 × 1010 J

hence the kinetic energy is

K = Etotal – U = -1.14268 × 1010 – (-2.34 × 1010) = 1.194 × 1010J

 

c)

According to Kepler’s First Law, the satellite will travel in an elliptical orbit, passing through Q and P, with the centre of the Earth at one of the focal points of the ellipse. 

According to Kepler’s Second Law, the satellite is moving the slowest at the furthest point from the Earth (the apogee, point Q). As it moves towards the closest point to Earth (the perigee, point P), it will decrease in altitude and speed up, reaching its maximum speed at P. Its total energy will remain constant, with kinetic energy increasing as gravitational potential energy decreases.

 

Written by Matrix Science Team

The Matrix Science Team are teachers and tutors with a passion for Science and a dedication to seeing Matrix Students achieving their academic goals.

© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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