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# 2023 HSC Physics Solutions Question 1

D – The mass of the Earth, spacecraft do not change. The density of the atmosphere is unrelated to the gravitational field strength. The distance of the spacecraft from the Earth’s centre will decrease gravitational field strength according to this equation:

Question 2

C – Voltage increases from the power station using a step-up transformer, then stepped-down before it reaches the home. The current travelling from the power station to the houses is low to conserve power loss which is proportional to current by the equation:

Question 3

C  The double slit experiment will produce multiple fringes with a central maximum at the centre. The only image which fits this description is C.

Question 4

B  Mathematical reasoning:

Question 5

D – Kepler’s Second Law states that a line segment connecting a planet and its sun will sweep out equal areas during equal time intervals. Because the area made by S and P is the largest, it will take the longest time to travel that distance.

Question 6

D – An electron will radiate an electromagnetic wave whenever it is accelerating. Since only D is a scenario with an accelerating electron (by centripetal acceleration) it will produce an EM wave.

Question 7

C – The de Broglie wavelength equation is

So, since the neutron’s mass is larger, it will have a smaller wavelength than the proton at the same speed.

Question 8

B – Increasing and the height of the platform will make the ball travel further. Increasing the mass of the ball will have no effect on the ball’s final position. Increasing the launch angle can reduce its final distance, especially if launched at a high, almost upwards angle.

Question 9

A –  As can be seen in this image, intensity at all wavelengths will be less for a lower temperature black body.

Question 10

D – The force acting on side will always remain the same since it follows the equation

Since the angle between the current and magnetic field is always 90°, force is constant. That is, for force: I = II.

The torque acting on the loops follows the equation

Since the angle between the area vector and magnetic field are parallel in Fig II, we have  θ = 0 meaning there is no torque in Fig II. That is, for torque, I > II

Question 11

A – Element and have atomic numbers 90 and 91 respectively meaning X is Th and Y is Pa.

X is achieved by losing two protons, and losing atomic mass of 4 which corresponds to alpha decay.

Y is achieved by gaining a proton with no change in atomic mass, which corresponds to beta decay.

Question 12

B – Mathematical reasoning:

In Fig I, work can be written as

In Fig II, we have a change in electric field due to a larger distance:

Question 13

C – The binding energy of a nucleon can appear as extra mass by the equation

So, nucleus having greater binding energy means it has a greater mass defect than . Options A and B are incorrect because stability is associated with the binding energy per nucleon, not solely the binding energy. For instance, Uranium is unstable despite having a high binding energy, but it exhibits a low binding energy per nucleon.

Question 14

C – Mathematical reasoning

So, for Planet X, increasing radius by 3 times, and mass by 4 times changes the escape velocity to become

Question 15

A – Einstein’s model of light suggests that light is a packet of energy inversely proportional to its wavelength. This is consistent with option A since if light has too large a wavelength, then it will not have enough energy to liberate electrons from a metal.

Option B is not true since increasing the intensity only influences the amount of photoelectrons emitted, not their kinetic energy. Option C is not true since the maximum kinetic energy is dependent on the type of metal used. Option D is not true either.

Question 16

C – If object X is at rest, the force exerted by the table on X is equal to the force acting on X from above. Likewise, if object Y is stationary, the repulsive force from the magnetic fields ( B) must equal the gravitational force (Fgy). That means, the total force acting on X from above is the weight of Y plus the regular weight force of X (Fgy + Fgx).

Question 17

B – Torque is the force acting on the ball perpendicular to the rod. If we consider position , the tension of the rod balances the gravitational force, meaning Y the force is zero (and thus the torque is zero).

This means we can rule out options A, and D since if there was never torque, we would get no rotation. We can also rule out C since we established torque at Y was zero. Our findings are consistent with option B.

Question 18

A – Mass does not change kinematics in gravitational fields. For a proof—if you don’t believe me—the acceleration of an object in a gravitational field is found by the following steps

which shows acceleration is independent of mass.

To calculate the acceleration of a charged object in an electric field, we have

Therefore, if we increase the mass, we decrease the acceleration downwards. Therefore, the particle should travel further.

Question 19

B – If we are looking from the frame of Wire , then the current should remain the same as the initial case, which leaves only options A and B.

Since the currents are travelling in the opposite direction, from the perspective of we would see charged particles passing through the same position in at a faster rate.

(Or by analogy, think of it this way): if you’re in a moving car on the left side of the road, you will observe more cars passing you each second on the opposite side of the road compared to when you’re sitting in a stationary car.

Question 20

D – If the universe were not accelerating, we would expect a straight line in the graph. However, if the universe were accelerating, assuming a non-accelerating universe would lead to an overestimation of the recessional velocity of galaxies at earlier points in time. Since the light we receive from galaxies farther away originates from a more distant past in the universe, this implies that the recessional velocities we observe in these more distant galaxies should be less than what would be predicted by a linear model, as indicated in option D.

Question 21

• Surface area
• Temperature

b. Describe differences between Stars A & B.

Star Type

Reading from the HR Diagram, Star A is likely a main sequence star, while Star B is likely a White Dwarf.

Star Sizes

Since both stars have roughly the same temperature, yet Star A is more luminous, it suggests that Star B is much smaller than Star A.

Age and Evolutionary path

Additionally, the stage in the star’s evolution is different. Star A is in its main sequence and is therefore younger than Star B. It will transition into a red giant star once hydrogen has been depleted and begin to fuse helium and heavier elements.

Star B, however, is older, has exhausted its nuclear fuel and reached the end of its main sequence phase. Over an extremely long period of time (billions to trillions of years), the star will slowly cool down and fade into a brown, then black dwarf.

Question 22

The time between pulses is given by the time-dilation equation:

Question 23

(a)

The magnitude of gravitational force the Sun exerts on the JWST is calculated as

(b)

The minimum photon energy the JWST can detect will be the energy of largest wavelength it can detect, which is 2.8 x 10 -5 m.

Photons of this wavelength will have energy:

(c)

Temperature is found using Wien’s law

Question 24

The magnetic force should equal the centripetal force. We can then solve for B.

Using the right-hand grip rule, the magnetic field must be directed out of the page to produce a counterclockwise rotation.

Question 25

(a)

Part X is a split-ring commutator, a metal cylinder divided into two halves that establishes contact with the brushes and the loop. Its primary role is to reverse the direction of current flow within the loop every 180 degrees, thereby ensuring torque remains in the same direction. Without this component, when the DC motor is activated, the loop would reverse in torque after the upright position, causing it to remain stationary.

(b)

The torque of a DC motor decreases as its rotational speed increases due to the back EMF that is generated when the motor rotates. This back EMF opposes the applied voltage and reduces the effective voltage across the armature, which in turn decreases the armature current. Since torque is directly proportional to armature current, a decrease in armature current results in a decrease in torque.

Question 26

The energy released is found by calculating the mass defect in the reaction, and using Einstein’s mass–energy equivalence relation.

Mass of Reactants: 12.064 + 1.008 = 13.072 u

Mass of Products: 9.013 + 4.003 = 13.016 u

Mass defect given by the difference between these two masses:

Mass defect is converted into released energy by the mass–energy equivalence relation:

Question 27

(a) Composition

Each element absorbs or emits light at specific wavelengths, creating a unique pattern of spectral lines. By comparing the emission or absorption lines of stars with laboratory measurements of elements, we can identify the elements present in the star, revealing its composition.

Example Spectrum – Star has similar spectral lines to hydrogen, suggesting the star has hydrogen in it’s composition.

Temperature

A star’s temperature is inferred from the peak of the star’s continuous spectrum (the star’s intensity at different wavelengths). Mathematically, the the temperature of a star is found as a function of it’s peak wavelength:

Hotter stars emit more light at shorter wavelengths, so they appear bluer, while cooler stars emit more light at longer wavelengths and appear redder.

Example Spectrum – Star has peak wavelength of roughly 300 nm, which can be used to find temperature of the star.

(b)

A rotating star will have one side that moves away from the observer while the other side moves towards the observer. Consequently, the side approaching the Earth will experience a Doppler shift towards the red end of the spectrum, while the side moving away from Earth will undergo a Doppler shift towards the blue end. This effect leads to a broadening of the star’s spectrum.

The hydrogen emission line would be modified as such (assuming the star’s translation velocity is zero, and there is no Doppler shift).

Question 28

(a)

The step down transformer will reduce the voltage to become:

Since the switch is open, we can ignore Globe Y, and the voltage drop across Globe X is simply 40 V.

(b)

When the switch is closed, the resistance in the secondary circuit decreases, resulting in a significant decrease in the current flowing through each branch. This reduction in current and total resistance has a substantial impact on minimising power loss, as indicated by the power loss equation:

Ploss = 12R

As a result, closing the switch maximises the total power in the secondary circuit. The current in the primary coil can be expressed in terms of the total power in the secondary coil as:

Therefore, closing the switch leads to an increase in the total power in the secondary coil, which, in turn, results in an increased current in the primary coil.

Question 29

Light from an incandescent lamp is unpolarised, meaning it oscillates in all directions perpendicular to the direction of propagation. When this light passes through a plane polarising filter, only the components of the light waves vibrating in the direction of the filter can pass through. This effectively blocks half of the light, thus reducing its intensity. The remaining light is said to be polarised because it now vibrates in only one direction.

Question 30

(a)

The electromotive force (EMF) acting on the ring is given by Faraday’s law of induction:

change in flux in Ring X, meaning an EMF and current will be induced. This results in the ring generating a magnetic field in the direction which opposes the magnetic field of the coil (by Lenz’s law).

Specifically, the solenoid will produce a north pole oriented towards Ring X, causing Ring X to produce a magnetic field with a north pole facing the coil. The interaction of these two magnetic fields produces a repulsive force which pushes the ring away from the solenoid.

For times 0.03 < t < 0.05

Between times 0.03 and 0.05 seconds, there is a zero change in magnetic field strength over time, meaning the coil does not experience a change in flux and thus does not generate a magnetic field. Consequently, there is no force acting on the coil.

(b) (i)

In this scenario, an EMF is induced in Ring Y, as predicted by Faraday’s law of electromagnetic induction which states that a changing magnetic field over time in a conductor will induce a voltage in that conductor.

However, due to the ring’s discontinuity, it fails to form a closed circuit, preventing the flow of current across the gap. Consequently, Ring Y is unable to generate a magnetic field and, as a result, no magnetic field (and thus no force) is produced to counteract the change in magnetic flux through the ring.

Question 31

Similarities:

• Both acceleration-time graphs show a comparable shape with varying speeds, with decceleration rapidly reaching a peak and then slowly increasing again to zero.

This phenomenon occurs when magnets pass through the braking fin, resulting in a change in magnetic flux. According to Lenz’s law, eddy currents are generated in the fin in a direction that opposes this change in flux. At the front of the cart, the magnetic fields of the cart and the eddy currents repel each other, while at the end of the cart, they attract each other. These interactions produce a force that opposes the car’s motion.

Differences:

• At higher speeds, the peak deceleration is notably greater (with a difference of ~ 2 m/s²) and occurs sooner (with a difference of ~ 0.1 seconds).

This variation in peak deceleration suggests that the braking system’s strength or effectiveness is speed-dependent. This makes sense, since the rate of change in flux over time within the braking fin is proportional to the speed of the cart. Therefore, faster-moving cars will generate a greater EMF and, consequently, experience a stronger opposing force.

Question 32

Before starting, we will need the time of flight, and the horizontal speed of the ball after release.

Time of Flight

At X, the ball will be launched to the east. The time of flight of the ball is calculated as

Horizontal Speed of Ball During Release

To find the velocity of the disc at position , we should consider that it does 3 revolutions per second, which can be rewritten as an angular velocity of

Therefore, at a radius of 2 metres, the disc’s tangential velocity is

This will also be the horizontal velocity of the ball before release.

Position of the Ball

Using our findings above, the ball’s eastward displacement is

Position of the Launcher

After t = 11661 s, the ball would have traversed an angle of

This means the launcher will be located exactly above its initial position.

Position of the Ball Relative to Launcher

A diagram of the final position of the ball and launcher is shown below.

The position of the ball relative to the launcher’s new position is thus found using Pythagoras’s theorem:

Therefore, the ball is 44 metres away from the final position of the launcher at an angle of S E.

Question 33

By examining how subatomic particles interact with fields, we gain insights into the forces that influence these particles. Furthermore, by studying their interactions with other particles, we learn about their inherent properties and the rules governing their behaviour. This argument will be justified by considering the two examples of:

• The discovery of electron characteristics from interactions with electric and magnetic fields, and
• The discovery of quarks and their characteristics from interactions with other particles.

Discovery of Electron Characteristics from Interactions with Fields.

The interaction of “cathode rays” with electric fields has deepened our understanding of the characteristics of the electron. Namely, in the late 19th century, physicist J.J. Thomson began experimenting with cathode ray tubes, which are sealed glass tubes from which most of the air has been evacuated. Then, a high voltage is applied across two electrodes at one end of the tube, which causes a beam of particles to flow from the cathode (the negatively-charged electrode) to the anode (the positively-charged electrode). To test the properties of these particles, Thomson placed two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. The amount by which the ray was deflected by a magnetic field helped Thomson determine the mass-to-charge ratio of the particles. This experiment led to the discovery that all atoms contain tiny negatively charged subatomic particles or electrons, which was a groundbreaking discovery in the study of atomic physics.

Discovery of Quarks from Other Particles.

Additionally, linear particle accelerators have been used to increase our understanding of the characteristics of quarks. In 1968 a series of electron-proton scattering experiments conducted at the Stanford Linear Accelerator Center (SLAC) revealed that nucleons have an inner structure. This works since the de Broglie wavelength of an electron is given as

so at high speeds, the electrons would have a small enough wavelength to probe a proton and neutron to interfere with its interior. By analysing the scattering patterns, it was identified that there exist point-like particles inside the proton called quarks. By combining these results with other experiments (such as neutrino-scattering in the Gargamelle bubble chamber at CERN), it became clear that quarks in the proton have charges of -1/3 and 2/3. Therefore, experiments involving the interaction of subatomic particles with other particles have increased our understanding of the existence of quarks and their properties, which are now a key part of the Standard Model.

Question 34

(a)

When the satellite’s engines are activated, it shifts to a higher orbit, resulting in an overall increase in its total energy. This leads to a reduction in kinetic energy, which is equivalent to the energy acquired by the satellite (indicated by a decrease in orbital velocity). Simultaneously, the potential energy of the satellite increases, and this increase is twice the energy gained by the satellite.

(b)

To do this, we will note that the total mechanical energy is conserved as the satellite transitions from P to Q. Therefore, by finding the initial kinetic energies, and the potential energies at P and Q we can then find the final kinetic energy.

Potential Energies at P and Q

The potential energy at P and Q is solely dependent on the satellites distance from Earth. Therefore, at P, we have

Initial Kinetic Energy at P

The initial kinetic energy at P is given as

After the added energy, we have

Calculating Kinetic Energy at Q

Here, we can take advantage of the conservation of energy, meaning the total energy at P equals the total energy at Q.

(c)

Total energy of the satellite is

This corresponds to a radius of

This suggests that the satellite will continue past Q until it reaches an orbital radius of 7.005 x 106m from the Earth’s centre. As the satellite rises, it will begin to lose kinetic energy as its potential energy increases, meaning the satellite will become slower. Eventually, it will reach a stable orbit around Earth unless an external force is applied to the satellite.

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