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Aussie Maths & Science Teachers: Save your time with SmarterEd

ENGINEERING, CS 2024 HSC 25c

A pin-jointed truss is shown. Member  \(AB\)  has been determined to be 250 N in compression.
 

Complete the table. You must use the method specified to solve each component.   (6 marks)

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \textit{}\\
\hline
\rule{0pt}{2.5ex} \text{External} & \text{Mathematical} & \\
\text{reaction at \(E\)} \rule[-1ex]{0pt}{0pt} & \text{} &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Direction: ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Methods of} & \\
\text{reaction of}  & \text{section} &\\
\text{member of \(CF\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Graphical} & \\
\text{reaction of}  &  &\\
\text{member \(BG\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

Show Answers Only

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\Large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} \)  
  \(=19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

Show Worked Solution

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} \)  
  \(=19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

BIOLOGY, M8 EQ-Bank 2

  1. Provide an example of a genetic non-infectious disease and how develops at the genetic level.   (1 mark)

  2. Describe TWO major effects of the disease on the human body and why these occur.   (2 marks)

Show Answers Only

a.   Answers could include one of the following:

Cystic Fibrosis

→ Cystic fibrosis develops when a person inherits two faulty copies of the CFTR gene (one from each parent), resulting in defective chloride channels in cell membranes.
  

Huntington’s Disease

→ Huntington’s disease is caused by inheriting a dominant mutated copy of the HTT gene, resulting in the production of abnormal huntingtin protein.
 

b.  Answers could include one of the following:

Cystic Fibrosis

→ The defective chloride channels cause mucus to build up in the lungs, leading to frequent chest infections and breathing difficulties because bacteria become trapped in the airways.

→ The same mucus also blocks pancreatic ducts, preventing digestive enzymes from reaching the intestines, which results in poor nutrient absorption and growth problems.
  

Huntington’s Disease

→ The abnormal protein accumulates in brain cells, causing progressive death of neurons that control movement, leading to uncontrolled jerking and twitching movements.

→ The protein buildup also affects regions of the brain controlling cognitive function, resulting in progressive memory loss and personality changes as these neural networks deteriorate

Show Worked Solution

a.   Answers could include one of the following:

Cystic Fibrosis

→ Cystic fibrosis develops when a person inherits two faulty copies of the CFTR gene (one from each parent), resulting in defective chloride channels in cell membranes.
 

Huntington’s Disease

→ Huntington’s disease is caused by inheriting a dominant mutated copy of the HTT gene, resulting in the production of abnormal huntingtin protein.
 

b.  Answers could include one of the following:

Cystic Fibrosis

→ The defective chloride channels cause mucus to build up in the lungs, leading to frequent chest infections and breathing difficulties because bacteria become trapped in the airways.

→ The same mucus also blocks pancreatic ducts, preventing digestive enzymes from reaching the intestines, which results in poor nutrient absorption and growth problems.
 

Huntington’s Disease

→ The abnormal protein accumulates in brain cells, causing progressive death of neurons that control movement, leading to uncontrolled jerking and twitching movements.

→ The protein buildup also affects regions of the brain controlling cognitive function, resulting in progressive memory loss and personality changes as these neural networks deteriorate

ENGINEERING, CS 2024 HSC 27b

A towbar hitch and shear pin is shown.
 

Explain why hot forging would be used to manufacture the shear pin.   (3 marks)

Show Answers Only

→ Hot forging creates components that exhibit superior strength compared to casting or machining methods.

→ This process enables the production of more intricate geometries while achieving an ideal combination of mechanical properties.

→ These properties include maximum yield strength, increased ductility, and reduced hardness.

Show Worked Solution

→ Hot forging creates components that exhibit superior strength compared to casting or machining methods.

→ This process enables the production of more intricate geometries while achieving an ideal combination of mechanical properties.

→ These properties include maximum yield strength, increased ductility, and reduced hardness.

♦ Mean mark 49%.

ENGINEERING, AE 2024 HSC 26c

A drawing of a scale model aircraft flying at constant velocity in level flight is shown.
 

Assume that the lift acts as a point load only on each wing and is located as shown in the drawing.

  1. In the box below, draw the free body diagram of the forces acting on the aircraft.   (1 mark)

 

  1. Draw both a shear force diagram and a bending moment diagram of the aircraft as the forces act across the wingspan of the aircraft.   (5 marks)

     

 

 

Show Answers Only

i.    
         

ii.    
       

Show Worked Solution

i.    
       
♦ Mean mark (i) 48%.
ii.    
     
Mean mark (ii) 55%.

ENGINEERING, PPT 2024 HSC 24d

Explain why a fully hardened steel would need to be tempered.

Support your answer with a labelled sketch of the resulting tempered microstructure.   (4 marks)


 

Show Answers Only

Exemplar solution 1:

→ Steel in its fully hardened state exhibits extreme brittleness, limiting its practical applications and making it vulnerable to failure.

→ When exposed to abrupt forces or stress concentrations, fully hardened steel can develop cracks or break catastrophically due to its inability to deform.

→ The tempering process plays a vital role in optimising steel’s mechanical properties, creating an essential balance between hardness and toughness.

→ Through careful tempering, the steel’s characteristics can be precisely tuned to match specific operational requirements, ensuring reliable performance across diverse applications.

→ Martensite by hardening:

Exemplar solution 2:

→ The tempering process serves as a crucial follow-up treatment after quenching, specifically designed to mitigate the steel’s brittle characteristics.

→ The procedure involves carefully reheating the hardened steel to a temperature below its critical point, followed by another cooling cycle.

→ This methodical heating and cooling sequence helps release internal stresses that developed during the initial quenching phase.

→ Through tempering, manufacturers can precisely adjust the balance between the steel’s hardness and toughness to achieve desired mechanical properties.

→ Ferrite and finely dispersed cementite:

Show Worked Solution

Exemplar solution 1:

→ Steel in its fully hardened state exhibits extreme brittleness, limiting its practical applications and making it vulnerable to failure.

→ When exposed to abrupt forces or stress concentrations, fully hardened steel can develop cracks or break catastrophically due to its inability to deform.

→ The tempering process plays a vital role in optimising steel’s mechanical properties, creating an essential balance between hardness and toughness.

→ Through careful tempering, the steel’s characteristics can be precisely tuned to match specific operational requirements, ensuring reliable performance across diverse applications.

→ Martensite by hardening:

♦♦ Mean mark 32%.

Exemplar solution 2:

→ The tempering process serves as a crucial follow-up treatment after quenching, specifically designed to mitigate the steel’s brittle characteristics.

→ The procedure involves carefully reheating the hardened steel to a temperature below its critical point, followed by another cooling cycle.

→ This methodical heating and cooling sequence helps release internal stresses that developed during the initial quenching phase.

→ Through tempering, manufacturers can precisely adjust the balance between the steel’s hardness and toughness to achieve desired mechanical properties.

→ Ferrite and finely dispersed cementite:

ENGINEERING, PPT 2024 HSC 24b

Brushless DC motors are used to power electric-powered bicycles.

Why are these types of motors so well suited to this application?   (3 marks)

Show Answers Only

→ Electronic commutation in brushless DC motors enables straightforward control and enhanced operational efficiency.

→ This precise control of speed and torque makes these motors particularly suitable for e-bike applications.

→ Since there are no brushes to wear down, these motors offer improved durability and longevity.

→ Their ability to deliver substantial torque even at lower speeds gives riders the power they need when starting from a stop or tackling inclines.
 

Answers could also include the following points:

→ Regenerative braking

→ Efficiency

→ Compact size and lightweight.

Show Worked Solution

→ Electronic commutation in brushless DC motors enables straightforward control and enhanced operational efficiency.

→ This precise control of speed and torque makes these motors particularly suitable for e-bike applications.

→ Since there are no brushes to wear down, these motors offer improved durability and longevity.

→ Their ability to deliver substantial torque even at lower speeds gives riders the power they need when starting from a stop or tackling inclines.
 

Answers could also include the following points:

→ Regenerative braking

→ Efficiency

→ Compact size and lightweight.

♦ Mean mark 47%.

ENGINEERING, PPT 2024 HSC 24a

How would impact testing be used during the design and development of a motorcycle helmet?   (2 marks)

Show Answers Only

→ Laboratory crash simulations assess how helmets will perform in real world accidents.

→ These evaluations measure the helmet’s ability to absorb forces, maintain its structure, and meet required safety certifications.

Show Worked Solution

→ Laboratory crash simulations assess how helmets will perform in real world accidents.

→ These evaluations measure the helmet’s ability to absorb forces, maintain its structure, and meet required safety certifications.

♦ Mean mark 50%.

PHYSICS, M7 2024 HSC 32

Many scientists have performed experiments to explore the interaction of light and matter.

Analyse how evidence from at least THREE such experiments has contributed to our understanding of physics.   (8 marks)

Show Answers Only

Students could include any of the following experiments:

  • Black body radiation experiments (M7 Quantum Nature of Light)
  • Photoelectric experiments (M7 Quantum Nature of Light)
  • Spectroscopy experiments (M8 Origins of Elements)
  • Polarisation experiments (M7 Wave Nature of Light)
  • Interference and diffraction (M7 Wave Nature of Light)
  • Cosmic gamma rays (M7 Special Relativity and/or M8 Deep Inside the Atom and standard model).
     

Young’s Double-Slit Experiment:

→ Young’s 1801 double slit experiment aimed to determine light’s wave-particle nature.

→ He passed coherent light through two slits and observed the pattern on a screen.

→ Instead of Newton’s predicted two bright bands, Young observed alternating bright and dark bands.

→ This interference pattern occurred due to light diffraction and interference, which  re wave properties.

→ The experiment provided strong evidence for light behaving as a wave at macroscopic scales.
 

Planck and the Blackbody Radiation Crisis:

→ Late 19th century scientists studied the relationship between black body radiation’s wavelength and intensity.

→ Experimental observations showed intensity peaked at a specific wavelength, contradicting classical physics predictions.

→ Classical physics led to the “ultraviolet catastrophe,” which violated energy conservation.

→ Planck’s thought experiment resolved this by proposing energy was transferred in discrete packets (quanta) where  \(E=hf\).

→ This revolutionary idea marked a shift from classical physics to quantum theory.
 

Einstein and the Photoelectric Effect:

→ In 1905, Einstein built upon Plank’s idea of quantised energy to propose that light was made up of quantised photons where \(E=hf\).

→ Einstein proposition explained why electrons are ejected from metal surfaces only when light exceeds a minimum frequency.

→ Previous to Einstein’s explanation of the photoelectric effect a high intensity of light corresponds to a high energy.

→ Einstein proposed that the KE of the emitted electrons was proportion to the frequency of the light rather than the intensity of the light. 

→ This development in the understanding of the interaction of light and matter at the atomic level shifted our understanding of light to a wave-particle duality model.
 

Cosmic Ray Experiments and the development of the Standard Model:

→ In 1912, Victor Hess discovered cosmic rays through high-altitude balloon experiments, finding that radiation increased with altitude rather than decreased as expected.

→ The study of cosmic rays led to the unexpected discovery of new particles, including the positron and muon, which couldn’t be explained by the known models of matter.

→ These discoveries from cosmic rays helped inspire the development of modern particle accelerators and contributed to the formulation of the quark model in the 1960s.

→ Eventually further studies on these newly discovered particles led to the development of the Standard Model of particle physics, which organises all known elementary particles and their interactions.

Show Worked Solution

Students could include any of the following experiments:

  • Black body radiation experiments (M7 Quantum Nature of Light)
  • Photoelectric experiments (M7 Quantum Nature of Light)
  • Spectroscopy experiments (M8 Origins of Elements)
  • Polarisation experiments (M7 Wave Nature of Light)
  • Interference and diffraction (M7 Wave Nature of Light)
  • Cosmic gamma rays (M7 Special Relativity and/or M8 Deep Inside the Atom and standard model).
♦ Mean mark 50%.

Young’s Double-Slit Experiment:

→ Young’s 1801 double slit experiment aimed to determine light’s wave-particle nature.

→ He passed coherent light through two slits and observed the pattern on a screen.

→ Instead of Newton’s predicted two bright bands, Young observed alternating bright and dark bands.

→ This interference pattern occurred due to light diffraction and interference, which  re wave properties.

→ The experiment provided strong evidence for light behaving as a wave at macroscopic scales.
 

Planck and the Blackbody Radiation Crisis:

→ Late 19th century scientists studied the relationship between black body radiation’s wavelength and intensity.

→ Experimental observations showed intensity peaked at a specific wavelength, contradicting classical physics predictions.

→ Classical physics led to the “ultraviolet catastrophe,” which violated energy conservation.

→ Planck’s thought experiment resolved this by proposing energy was transferred in discrete packets (quanta) where  \(E=hf\).

→ This revolutionary idea marked a shift from classical physics to quantum theory.
 

Einstein and the Photoelectric Effect:

→ In 1905, Einstein built upon Plank’s idea of quantised energy to propose that light was made up of quantised photons where \(E=hf\).

→ Einstein proposition explained why electrons are ejected from metal surfaces only when light exceeds a minimum frequency.

→ Previous to Einstein’s explanation of the photoelectric effect a high intensity of light corresponds to a high energy.

→ Einstein proposed that the KE of the emitted electrons was proportion to the frequency of the light rather than the intensity of the light. 

→ This development in the understanding of the interaction of light and matter at the atomic level shifted our understanding of light to a wave-particle duality model.
 

Cosmic Ray Experiments and the development of the Standard Model:

→ In 1912, Victor Hess discovered cosmic rays through high-altitude balloon experiments, finding that radiation increased with altitude rather than decreased as expected.

→ The study of cosmic rays led to the unexpected discovery of new particles, including the positron and muon, which couldn’t be explained by the known models of matter.

→ These discoveries from cosmic rays helped inspire the development of modern particle accelerators and contributed to the formulation of the quark model in the 1960s.

→ Eventually further studies on these newly discovered particles led to the development of the Standard Model of particle physics, which organises all known elementary particles and their interactions.

ENGINEERING, CS 2024 HSC 20 MC

A cross-section of a beam is shown.

Which of the following is the maximum stress in the beam if the maximum bending moment is 200 kN m and \(I_{\text{xx}}\) is 168.75 × \(10^{-6} \text{ m}^4\)?

  1. 177.8 MPa
  2. 177.8 GPa
  3. 355.6 MPa
  4. 355.6 GPa
Show Answers Only

\(A\)

Show Worked Solution

\(M=200\ \text{kN m}\ = 200\ 000\ \text{N m}\)

\(y=\ \text{distance to neutral axis}\ = \dfrac{300}{2} = 150\ \text{mm}\ = 0.15\ \text{m} \)

\(\sigma = \dfrac{My}{I_{\text{XX}}} = \dfrac{200\ 000 \times 0.15}{168.75 \times 10^{-6}} = 177.8\ \text{MPa}\) 

\(\Rightarrow A\)

♦ Mean mark 44%.

ENGINEERING, CS 2024 HSC 21b

A geotextile used in the construction of a retaining wall is shown.
 

Why are geotextiles used in the construction of retaining walls?   (3 marks)

Show Answers Only

Answers could include three of the following:

→ They facilitate drainage in water-prone areas while simultaneously distributing structural loads more evenly across the soil base.

→ Additionally, these materials create a stable foundation that prevents soil erosion by holding soil particles in place against water movement.

→ Geotextiles enhance soil stability, by preventing mixing of different soil layers and
improving load-bearing capacity.

→ They reinforce the soil mass, by increasing the overall strength and longevity of the retaining wall system.

Show Worked Solution

Answers could include three of the following:

→ They facilitate drainage in water-prone areas while simultaneously distributing structural loads more evenly across the soil base.

→ Additionally, these materials create a stable foundation that prevents soil erosion by holding soil particles in place against water movement.

→ Geotextiles enhance soil stability, by preventing mixing of different soil layers and improving load-bearing capacity.

→ They reinforce the soil mass, by increasing the overall strength and longevity of the retaining wall system.

♦ Mean mark 50%.

ENGINEERING, TE 2024 HSC 17 MC

Why is single-mode cable used for long distance telecommunications rather than other fibre optic cables?

  1. It has lower power consumption.
  2. It has greater attenuation over distance.
  3. It has a higher transmission rate and bandwidth.
  4. It has a greater change to the index of refraction.
Show Answers Only

\(C\)

Show Worked Solution

→ Single-mode fibre optic cable only allows one path for light to travel, reducing signal dispersion and allowing for faster, clearer transmission over long distances.

→ This results in a higher transmission rate and bandwidth compared to multi-mode cables which allow multiple light paths.

\(\Rightarrow C\)

♦♦ Mean mark 35%.

ENGINEERING, CS 2024 HSC 16 MC

An \(\text{I}\)-beam is loaded as shown.
 

The load is then increased.

Which increase in dimension would provide the most resistance to bending?

  1. \( \textit{A} \)
  2. \( \textit{B} \)
  3. \( \textit{C} \)
  4. \( \textit{D} \)
Show Answers Only

\(D\)

Show Worked Solution

→ The most efficient way to increase bending resistance in an \(\text{I}\)-beam would be to increase the height/depth of the web \((D)\), since resistance to bending increases with the cube of the depth from neutral axis.

\(\Rightarrow D\)

♦ Mean mark 46%.

ENGINEERING, PPT 2024 HSC 15 MC

Which property is necessary in the manufacture of a sheet steel car door panel?

  1. Fatigue
  2. Ductility
  3. Elasticity
  4. Corrosion resistance
Show Answers Only

\(B\)

Show Worked Solution

→ Ductility is the ability of the steel to be plastically deformed without breaking.

→ This is essential for pressing and stamping sheet metal into the complex curves and shapes required for car door panels.

\(\Rightarrow B\)

♦♦ Mean mark 32%.

ENGINEERING, PPT 2024 HSC 10 MC

An inclined plane is 5 metres long and 2 metres high. A force of 50 N is required to push a box up along the length of the slope.
 

Assuming no friction and 100% efficiency, what is the mass of the box to the nearest kilogram?

  1. 5
  2. 13
  3. 20
  4. 125
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Energy moving box}\ = 50 \times 5 = 250\ \text{J}\)

\(\text{Energy used = increase in potential energy of box (100% efficiency)}\)

\(m \times g \times h\) \(=250\)  
\(m\) \(=\dfrac{250}{10 \times 2} \)  
  \(=12.5\ \text{kg}\)  

 
\(\Rightarrow B\)

♦ Mean mark 40%.

ENGINEERING, PPT 2024 HSC 9 MC

When producing glass, what molten material is the glass floated on in order to obtain a smooth surface?

  1. Aluminium
  2. Bronze
  3. Steel
  4. Tin
Show Answers Only

\(D\)

Show Worked Solution

→ In the glass manufacturing process, molten glass is poured onto a bed of molten tin.

→ Tin provides an extremely flat and smooth surface due to tin’s high density and surface tension, allowing the glass to float and spread evenly while cooling.

\(\Rightarrow D\)

♦♦ Mean mark 33%.

ENGINEERING, AE 2024 HSC 8 MC

The cross-section of a tapered tube is shown.
 

   

What are the pressure readings at both gauge \(A\) and at gauge \(B\) as the air flows through the tube?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex} \textbf{A.} \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex} \textbf{B.} \rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex} \textbf{C.}\rule[-1ex]{0pt}{0pt} \\
\rule{0pt}{2.5ex} \textbf{D.}\rule[-1ex]{0pt}{0pt} \\
\end{array}
\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Pressure at gauge A} \rule[-1ex]{0pt}{0pt} & \textit{Pressure at gauge B} \\
\hline
\rule{0pt}{2.5ex} \text{High} \rule[-1ex]{0pt}{0pt} & \text{High} \\
\hline
\rule{0pt}{2.5ex} \text{High} \rule[-1ex]{0pt}{0pt} & \text{Low} \\
\hline
\rule{0pt}{2.5ex} \text{Low} \rule[-1ex]{0pt}{0pt} & \text{Low} \\
\hline
\rule{0pt}{2.5ex} \text{Low} \rule[-1ex]{0pt}{0pt} & \text{High} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution

→ According to Bernoulli’s principle, as the air velocity increases through the tapered section, the pressure must decrease.

→ This results in high pressure at the wide entrance (gauge A) where velocity is low and low pressure at the narrow exit (gauge B) where velocity is high.

\(\Rightarrow B\)

♦ Mean mark 46%.

ENGINEERING, CS 2024 HSC 6 MC

A shear force diagram is shown.

Which of the following beams is represented in the shear force diagram?

  1. A cantilevered beam with a point load
  2. A simply supported beam with a point load
  3. A cantilevered beam with a uniformly distributed load
  4. A simply supported beam with a uniformly distributed load
Show Answers Only

\(D\)

Show Worked Solution

→ For a simply supported beam with a uniformly distributed load, the shear force diagram shows a linear variation that begins with a positive reaction at the left support and decreases linearly to a negative value at the right support.

→ This linear pattern occurs because the distributed load gradually transfers the shear force from one support to the other, with zero shear at the beam’s midspan where the direction of shear changes.

\(\Rightarrow D \)

♦ Mean mark 55%.

PHYSICS, M6 2024 HSC 33

A magnet is swinging as a pendulum. Close below it is an aluminium (non-ferromagnetic) can. The can is free to spin around a fixed axis as shown.
 

Analyse the motion and energy transformations of both the can and the magnet.   (7 marks)

Show Answers Only

→ When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.

→ As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.

→ The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).

→ This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.

→ The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.

→ Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.

Show Worked Solution

→ When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.

→ As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.

→ The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).

→ This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.

→ The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.

→ Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.

♦♦ Mean mark 47%.

PHYSICS, M5 2024 HSC 31

In a thought experiment, a projectile is launched vertically from Earth's surface. Its initial velocity is less than the escape velocity.

The behaviour of the projectile can be analysed by using two different models, Model \(A\) and Model \(B\) as shown.
  

The effects of Earth's atmosphere and Earth's rotational and orbital motions can be ignored.

Compare the maximum height reached by the projectile, using each model. In your answer, describe the energy changes of the projectile.   (4 marks)

Show Answers Only

→ Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.

→ In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.

→ In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.

→ The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.

Show Worked Solution

→ Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.

→ In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.

→ In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.

→ The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.

♦ Mean mark 47%.

PHYSICS, M6 2024 HSC 29

Two horizontal metal rods, \(A\) and \(B\), of different materials are resting on a frictionless table. Initially they are at rest in position 1.

Both rods are then connected to a battery using wires. After the switch is turned on, currents of different magnitude flow in each rod. The rods move to position 2 after time, \(t\). In position 2, \(B\) has a larger displacement than \(A\) from position 1. The masses of the wires are negligible.
 

 

  1. Position 1 is reproduced below. Draw wires to show how the battery must be connected to the ends of the two rods in order for the magnitude of the current in each rod to be different, and for position 2 to be reached. No components, other than the wires, are required.   (2 marks)
     
     


 
 
 

  1. When the switch is turned on, the current in rod \(A\) is greater than the current in rod \(B\).
  2. Consider this statement.
  3.     Position 2 results from the larger current in rod \(A\), causing a larger force to act on rod \(B\).
  4. Evaluate this statement with reference to relevant physics principles.   (4 marks)

Show Answers Only

a.   
           
  

b.   Position 2 results from the larger current in rod \(A\), causing a larger force to act on rod \(B\).

→ While rod \(B\) shows a greater displacement from its starting position compared to rod \(A\), this isn’t because of a larger current in rod \(A\).

→ Following Newton’s third law, the electromagnetic force that rod \(A\) exerts on rod \(B\) is equal in magnitude to the force that rod B exerts on rod \(A\).

→ The magnitude of the electromagnetic force between the rods can be calculated using the formula:  \(\dfrac{F}{l} = \dfrac{\mu_0}{2\pi} \, \dfrac{I_1 I_2}{r}\).

→ Rod \(B\) has a larger displacement over the same time \((t)\ \Rightarrow\) Rod \(B\) experiences greater acceleration.

→ Since the greater acceleration of Rod \(B\) isn’t due to a stronger electromagnetic force, we can conclude that Rod \(B\) has less mass than Rod \(A\) \(\Big(a=\dfrac{F}{m}\Big)\).

→ Since Rod \(B\) experiences more acceleration, its displacement is larger over a given time \((t)\) according to the formula  \(s=\dfrac{1}{2}at^{2} \).

Show Worked Solution

a.   
           
♦ Mean mark (a) 47%.

b.   Position 2 results from the larger current in rod \(A\), causing a larger force to act on rod \(B\).

→ While rod \(B\) shows a greater displacement from its starting position compared to rod \(A\), this isn’t because of a larger current in rod \(A\).

→ Following Newton’s third law, the electromagnetic force that rod \(A\) exerts on rod \(B\) is equal in magnitude to the force that rod B exerts on rod \(A\).

→ The magnitude of the electromagnetic force between the rods can be calculated using the formula:  \(\dfrac{F}{l} = \dfrac{\mu_0}{2\pi} \, \dfrac{I_1 I_2}{r}\).

→ Rod \(B\) has a larger displacement over the same time \((t)\ \Rightarrow\) Rod \(B\) experiences greater acceleration.

→ Since the greater acceleration of Rod \(B\) isn’t due to a stronger electromagnetic force, we can conclude that Rod \(B\) has less mass than Rod \(A\) \(\Big(a=\dfrac{F}{m}\Big)\).

→ Since Rod \(B\) experiences more acceleration, its displacement is larger over a given time \((t)\) according to the formula  \(s=\dfrac{1}{2}at^{2} \).

♦ Mean mark (b) 47%.

PHYSICS, M6 2024 HSC 28

An electron gun fires a beam of electrons at 2.0 × 10\(^6\) m s\(^{-1}\) through a pair of parallel charged plates towards a screen that is 30 mm from the end of the plates as shown.

There is a uniform electric field between the plates of 1.5 × 10\(^4\) N C\(^{-1}\). The plates are 5.0 mm wide and 20 mm apart. The electron beam enters mid-way between the plates. \(X\) marks the spot on the screen where an undeflected beam would strike.

Ignore gravitational effects on the electron beam.
 

 

  1. Show that the acceleration of an electron between the parallel plates is 2.6 × 10\(^{15}\) m s\(^{-2}\).   (2 marks)

  2. Show that the vertical displacement of the electron beam at the end of the parallel plates is approximately 8.1 mm.   (2 marks)

  3. How far from point \(X\) will the electron beam strike the screen?   (3 marks)

Show Answers Only

a.   \(\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:\)

\(a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} \)
 

b.  \(\text{Time of beam between the plates:}\)

\(\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}\)

\(t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} \)

\(\text{Find vertical displacement at end of plates:}\)

\(s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} \)
 

c.   \(\text{Find vertical velocity of beam when leaving the plates:}\)

\(v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Time for beam to hit screen:}\)

\(\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}\)

\(t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} \)

\(\text{Vertical displacement (from end of plates):}\)

\(s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} \)

\(\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} \)

Show Worked Solution

a.   \(\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:\)

\(a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} \)
 

b.  \(\text{Time of beam between the plates:}\)

\(\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}\)

\(t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} \)

\(\text{Find vertical displacement at end of plates:}\)

\(s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} \)
 

c.   \(\text{Find vertical velocity of beam when leaving the plates:}\)

\(v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Time for beam to hit screen:}\)

\(\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}\)

\(t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} \)

\(\text{Vertical displacement (from end of plates):}\)

\(s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} \)

\(\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} \)

♦ Mean mark (c) 40%.

PHYSICS, M7 2024 HSC 27

The simplified model below shows the reactants and products of a proton-antiproton reaction which produces three particles called pions, each having a different charge.

\(\text{p}+\overline{\text{p}} \rightarrow \pi^{+}+\pi^0+\pi^{-}\)

There are no other products in this process, which involves only the rearrangement of quarks. No electromagnetic radiation is produced. Assume that the initial kinetic energy of the proton and antiproton is negligible.

Protons consist of two up quarks \(\text{(u)}\) and a down quark \(\text{(d)}\) . Antiprotons consist of two up antiquarks \((\overline{\text{u}})\) and a down antiquark \((\overline{\text{d}})\). Each of the pions consists of two quarks.

The following tables provide information about hadrons and quarks.

Table 1: Hadron Information

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} & \ \ \textit{Rest mass} \ \ & \quad \textit{Charge} \quad \\
& \left(\text{MeV/c}^2\right)&\\
\hline
\rule{0pt}{2.5ex} \text {proton (p)} \rule[-1ex]{0pt}{0pt} & 940 &  +1 \\
\hline
\rule{0pt}{2.5ex} \text {antiproton}(\overline{\text{p}}) \rule[-1ex]{0pt}{0pt} & 940 & -1  \\
\hline
\rule{0pt}{2.5ex} \text {neutral pion }\left(\pi^0\right) \rule[-1ex]{0pt}{0pt} & 140 & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{positive pion }\left(\pi^{+}\right) \rule[-1ex]{0pt}{0pt} & 140 & +1 \\
\hline
\rule{0pt}{2.5ex}\text {negative pion }\left(\pi^{-}\right) \rule[-1ex]{0pt}{0pt} & 140 &  -1\\
\hline
\end{array}

 
Table 2: Quark charges

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} \rule[-1ex]{0pt}{0pt} & \quad \textit{Charge} \quad \\
\hline
\rule{0pt}{2.5ex} \text {down quark (d)} \rule[-1ex]{0pt}{0pt} & -\dfrac{1}{3}  \\
\hline
\rule{0pt}{2.5ex} \text {up quark (u)} \rule[-1ex]{0pt}{0pt} & +\dfrac{2}{3}\\
\hline
\rule{0pt}{2.5ex} \text {down antiquark}(\overline{\text{d}}) \rule[-1ex]{0pt}{0pt} & +\dfrac{1}{3}\\
\hline
\rule{0pt}{2.5ex} \text{up antiquark }(\overline{\text{u}}) \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3}  \\
\hline
\end{array}

  1. Identify the quarks present in the \(\pi^{-}, \pi^{+}\)and the \(\pi^0\) particles.   (2 marks)

  2. The energy released in the reaction is shared equally between the pions.
  3. Calculate the energy released per pion in this reaction.   (2 marks)

  4. Calculation of the pions' velocities using classical physics predicts that each pion has a velocity, relative to the point at which the proton-antiproton reaction occurred, which exceeds 3 × 10\(^8\) m s\(^{-1}\).
  5. Explain the problem with this prediction and how it can be resolved.   (3 marks)

Show Answers Only

a.   \(\pi^{-}:\ \overline{\text{u}}\text{d}\)

\(\pi^{+}:\ \text{u}\overline{\text{d}}\)

\(\pi^{0}:\ \text{u}\overline{\text{d}}\)
 

b.   \(\text{Initial mass}\ = 940 + 940 =  1880\ \text{MeV/c}^{2} \)

\(\text{Final mass}\ = 3 \times 140 = 420\ \text{MeV/c}^{2} \)

\(\Delta \text{Mass (per pion)}\ = \dfrac{1460}{3} = 487\ \text{MeV/c}^{2} \)

\(\text{Using}\ \ E=mc^2:\)

\(\text{Energy released (per pion)}\ = 487\ \text{MeV}\)
 

c.   Problem with prediction:

→ The calculation shows the pions moving faster than light speed (3 × 10\(^{8}\) m/s), which can’t happen in reality.

Resolving the problem:

→ Since these pions are moving at extremely high speeds, we need to account for relativity.

→ Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.

→ Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

Show Worked Solution

a.   \(\pi^{-}:\ \overline{\text{u}}\text{d}\)

\(\pi^{+}:\ \text{u}\overline{\text{d}}\)

\(\pi^{0}:\ \text{u}\overline{\text{d}}\)
 

b.   \(\text{Initial mass}\ = 940 + 940 =  1880\ \text{MeV/c}^{2} \)

\(\text{Final mass}\ = 3 \times 140 = 420\ \text{MeV/c}^{2} \)

\(\Delta \text{Mass (per pion)}\ = \dfrac{1460}{3} = 487\ \text{MeV/c}^{2} \)

\(\text{Using}\ \ E=mc^2:\)

\(\text{Energy released (per pion)}\ = 487\ \text{MeV}\)
 

♦ Mean mark (b) 41%.

c.   Problem with prediction:

→ The calculation shows the pions moving faster than light speed (3 × 10\(^{8}\) m/s), which can’t happen in reality.

Resolving the problem:

→ Since these pions are moving at extremely high speeds, we need to account for relativity.

→ Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.

→ Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

♦ Mean mark (c) 43%.

Calculus, MET1 2024 VCAA 8

Let  \(g: R \rightarrow R, g(x)=\sqrt[3]{x-k}+m\),  where  \(k \in R \backslash\{0\}\)  and  \(m \in R\).

Let the point \(P\) be the \(y\)-intercept of the graph of  \(y=g(x)\).

  1. Find the coordinates of \(P\), in terms of \(k\) and \(m\).   (1 mark)

  2. Find the gradient of \(g\) at \(P\), in terms of \(k\).  (2 marks)

  3. Given that the graph of  \(y=g(x)\)  passes through the origin, express \(k\) in terms of \(m\).  (1 mark)

  4. Let the point \(Q\) be a point different from the point \(P\), such that the gradient of \(g\) at points \(P\) and \(Q\) are equal.
  5. Given that the graph of  \(y=g(x)\)  passes through the origin, find the coordinates of \(Q\) in terms of \(m\).  (3 marks)

Show Answers Only

a.    \(\bigg(0, -\sqrt[3]{k}+m\bigg)\)

b.    \(\dfrac{1}{3}(-k)^{-\frac{2}{3}}\)

c.    \(k=m^3\)

d.    \(Q(2m^3,2m)\)

Show Worked Solution

a.     \(g(0)\) \(=\sqrt[3]{0-k}+m\)
    \(=-\sqrt[3]{k}+m\)

\(P(0, -\sqrt[3]{k}+m)\)

b.     \(g(x)\) \(=\sqrt[3]{x-k}+m\)
    \(=(x-k)^{\frac{1}{3}}+m\)
  \(g^{\prime}(x)\) \(=\dfrac{1}{3}(x-k)^{-\frac{2}{3}}\)
  \(g^{\prime}(0)\) \(=\dfrac{1}{3}(0-k)^{-\frac{2}{3}}\)
    \(=\dfrac{1}{3}(-k)^{-\frac{2}{3}}\)

 

c.     \(\text{When }y=0\)
     \(-\sqrt[3]{k}+m\) \(=0\)  
  \(\sqrt[3]{k}\) \(=m\)  
  \(k\) \(=m^3\)  

 

d.     \(g^{\prime}(x)\) \(=g^{\prime}(0)\)
  \(\dfrac{1}{3}(x-k)^{-\frac{2}{3}}\) \(=\dfrac{1}{3(-k)^{\frac{2}{3}}}\)
  \(\dfrac{1}{(x-k)^{\frac{2}{3}}}\) \(=\dfrac{1}{k^{\frac{2}{3}}}\)
  \((x-k)^{\frac{2}{3}}\) \(=k^{\frac{2}{3}}\)
  \((x-k)^2\) \(=k^2\)
  \(x-k\) \(=\pm k\)

\(\therefore\ x=0, x=2k\)

\(\text{When }x=2k\)

\(y=(2k-k)^{\frac{1}{3}}+m=k^{\frac{1}{3}}+m=2m\ \text{(from (c))}\)

\(\text{Coordinates of are}\ Q(2k, 2m)\Rightarrow\ Q(2m^3,2m)\)

Calculus, MET1 2024 VCAA 7

Part of the graph of  \(f:[-\pi, \pi] \rightarrow R, f(x)=x \sin (x)\)  is shown below.

  1. Use the trapezium rule with a step size of  \(\frac{\pi}{3}\)  to determine an approximation of the total area between the graph of  \(y=f(x)\) and the \(x\)-axis over the interval  \(x \in[0, \pi]\).   (3 marks)

  2.   i. Find  \(f^{\prime}(x)\).  (1 mark)

  3.  ii. Determine the range of  \(f^{\prime}(x)\)  over the interval  \(\left[\dfrac{\pi}{2}, \dfrac{2 \pi}{3}\right]\).  (1 mark)

  4. iii. Hence, verify that \(f(x)\) has a stationary point for \(x \in\left[\dfrac{\pi}{2}, \frac{2 \pi}{3}\right]\).  (1 mark)

  5. On the set of axes below, sketch the graph of \(y=f^{\prime}(x)\) on the domain \([-\pi, \pi]\), labelling the endpoints with their coordinates.
  6. You may use the fact that the graph of \(y=f^{\prime}(x)\) has a local minimum at approximately \((-1.1,-1.4)\) and a local maximum at approximately \((1.1,1.4)\).  (3 marks)

Show Answers Only

a.    \(\dfrac{\sqrt{3}\pi^2}{6}\)

bi.   \(x\cos(x)+\sin(x)\)

bii.  \(\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2},\ 1\right]\)

biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\)

\(\therefore\ f^{\prime}(x)=0\ \text{at some point in the interval, and hence a stationary point exists.}\)

c. 

Show Worked Solution

a.     \(A\) \(=\dfrac{\pi}{3}\times\dfrac{1}{2}\left(f(0)+2f\left(\dfrac{\pi}{3}\right)+2f\left(\dfrac{2\pi}{3}\right)+f(\pi)\right)\)
    \(=\dfrac{\pi}{3}\left(0+2.\dfrac{\pi}{3}\sin\left(\dfrac{\pi}{3}\right)+2.\dfrac{2\pi}{3}\sin\left(\dfrac{2\pi}{3}\right)+\pi\sin({\pi})\right)\)
    \(=\dfrac{\pi}{6}\left(2\times \dfrac{\pi}{3}\times\dfrac{\sqrt{3}}{2}+2\times\dfrac{2\pi}{3}\times \dfrac{\sqrt{3}}{2}+0\right)\)
    \(=\dfrac{\pi}{6}\left(\dfrac{2\pi\sqrt{3}}{6}+\dfrac{4\pi\sqrt{3}}{6}\right)\)
    \(=\dfrac{\pi}{6}\times \dfrac{6\pi\sqrt{3}}{6}\)
    \(=\dfrac{\sqrt{3}\pi^2}{6}\)

  

bi.    \(f(x)\) \(=x\sin(x)\)
  \(f^{\prime}(x)\) \(=x\cos(x)+\sin(x)\)

 

bii.  \(f^{\prime}\left(\dfrac{\pi}{2}\right)\) \(=1\)
  \(f^{\prime}\left(\dfrac{2\pi}{3}\right)\) \(=\dfrac{2\pi}{3}\left(\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}\right)\)
    \(=-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}\)

\(\therefore\ \text{Range of }\ f^{\prime}(x)\ \text{is}\quad\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2},\ 1\right]\)

biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\)

\(\therefore\ f^{\prime}(x)=0\ \text{at some point in the interval, and hence a stationary point exists.}\)

c.    \(f^{\prime}(\pi)=\pi\cos(\pi)+\sin(\pi)=-\pi\)

\(f^{\prime}(-\pi)=-\pi\cos(-\pi)+\sin(-\pi)=\pi\)

\(\therefore\ \text{Endpoints are }\ (-\pi,\ \pi)\ \text{and}\ (\pi,\ -\pi).\)

Functions, MET1 2024 VCAA 6

Solve  \(2 \log _3(x-4)+\log _3(x)=2\)  for \(x\).   (4 marks)

Show Answers Only

\(\dfrac{7 + \sqrt{13}}{2}\)

Show Worked Solution

\(2\log_3(x-4)+\log_3(x)\) \(=2\)
\(\log_3x(x-4)^2\) \(=2\)
\(3^2\) \(=x(x-4)^2\)
\(x(x^2-8x+16)-9\) \(=0\)
\(x^3-8x^2+16x-9\) \(=0\)
\(\text{Test }x=1\)  
\(1^3-8(1)^2+16(1)-9\) \(=0\)
\(\therefore\ x-1\ \text{is a factor} \)  

  

\((x-1)(x^2-7x+9)=0\)
  

\(\text{Using quadratic formula to solve}\ x^2-7x+9=0\)

\(x\) \(=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(1)(9)}}{2(1)}\)
  \(=\dfrac{7\pm \sqrt{49-36}}{2}\)
  \(=\dfrac{7\pm \sqrt{13}}{2}\)

\( x=1, \dfrac{7- \sqrt{13}}{2}, \dfrac{7 + \sqrt{13}}{2}\)

  
\(\text{For }\log_3(x-4)\ \text{to exist}\ x>4\)

\(\therefore\ \dfrac{7 + \sqrt{13}}{2}\ \text{ is the only possible solution.}\)

Functions, MET1 2024 VCAA 5

The function  \(h:[0, \infty) \rightarrow R, h(t)=\dfrac{3000}{t+1}\)  models the population of a town after \(t\) years.

  1. Use the model \(h(t)\) to predict the population of the town after four years.   (1 mark)

  2. A new function, \(h_1\), models a population where  \(h_1(0)=h(0)\) but \(h_1\)  decreases at half the rate of \(h\) at any point in time.
  3. State a sequence of two transformations that maps \(h\) to this new model \(h_1\).  (2 marks)

  4. In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
    1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads. Use  \(z=2\) for this confidence interval.  (2 marks)

    2. A new sample of \(n\) people results in the same sample proportion.
    3. Find the smallest value of \(n\) to achieve a standard deviation of  \(\dfrac{\sqrt{2}}{50}\)  for the sample proportion.  (1 mark)

Show Answers Only

a.    \(600\)

b.    \(\text{Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{followed by a translation of 1500 units upwards}\)

ci.   \(\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\ \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

cii.  \(300\)

Show Worked Solution

a.     \(h(4)\) \(=\dfrac{3000}{4+1}\)
    \(=600\)

    

b.     \(h(t)\) \(=3000(t+1)^{-1}\)
  \(h^{\prime}(t)\) \(=\dfrac{-3000}{(t+1)^2}\)
  \(h_1^{\prime}(t)\) \(=\dfrac{1}{2}h^{\prime}(t)\)
    \(=\dfrac{-1500}{(t+1)^2}\)
  \(\therefore\ h_1(t)\) \(=\dfrac{1500}{t+1}\)

\(\text{Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{followed by a translation of 1500 units upwards}\)

ci.    \(\hat{p}=\dfrac{60}{100}=\dfrac{3}{5},\quad 1-\hat{p}=\dfrac{2}{5},\quad z=2\)

\(\text{Approx CI}\) \(=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

  

cii.   \(\sqrt{\dfrac{\hat{P}(1-\hat{P}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\dfrac{\dfrac{6}{25}}{n}\) \(=\dfrac{2}{2500}\)
  \(n\) \(=\dfrac{6}{25}\times\dfrac{2500}{2}\)
    \(=\dfrac{600}{2}=300\)

Probability, MET1 2024 VCAA 4

Let \(X\) be a binomial random variable where  \(X \sim \operatorname{Bi}\left(4, \dfrac{9}{10}\right)\).

  1. Find the standard deviation of \(X\).   (1 mark)

  2. Find  \(\operatorname{Pr}(X<2)\).  (2 marks)

Show Answers Only

a.    \(\operatorname{sd}(X)=\dfrac{3}{5}\)

b.    \(\dfrac{37}{10\,000}\)

Show Worked Solution

a.     \(\operatorname{sd}(X)\) \(=\sqrt{n.p(1-p)}\)
    \(=\sqrt{4\times\dfrac{9}{10}\times\dfrac{1}{10}}\)
    \(=\sqrt{\dfrac{36}{100}}=\dfrac{6}{10}=\dfrac{3}{5}\)

 

b.     \(\operatorname{Pr}(X<2)\) \(=\operatorname{Pr}(X=0)+\operatorname{Pr}(X=1)\)
    \(=^4C\ _0\left(\dfrac{9}{10}\right)^0\left(\dfrac{1}{10}\right)^4+\ ^4C\ _1\left(\dfrac{9}{10}\right)^1\left(\dfrac{1}{10}\right)^3\)
    \(=\dfrac{37}{10\,000}\)

Functions, MET1 2024 VCAA 2

Consider the simultaneous linear equations

\(\begin{aligned} 3 k x-2 y & =k+4 \\ (k-4) x+k y & =-k\end{aligned}\)

where  \(x, y \in R\) and \(k\) is a real constant.

Determine the value of \(k\) for which the system of equations has no real solution.   (3 marks)

Show Answers Only

\(k=\dfrac{4}{3}\)

Show Worked Solution

\(\text{For no solutions lines must be parallel }(m_1=m_2)\)

\(\text{and y-intercepts not equal}\ (c_1\neq c_2).\)

\(\text{Eqn 1}:\quad\) \(3kx-2y\) \(=k+4\)
  \(2y\) \(=3kx-(k+4)\)
  \(y\) \(=\dfrac{3k}{2}x-\dfrac{k+4}{2}\)
  \(\therefore\ \) \(m_1=\dfrac{3k}{2},\ c_1=-\dfrac{k+4}{2}\)
     
\(\text{Eqn 2}:\quad\) \((k-4)x+ky\) \(=-k\)
  \(ky\) \(=-(k-4)x-k\)
  \(y\) \(=\dfrac{-(k-4)}{k}x-1\)
  \(\therefore\ \) \(m_2=\dfrac{-(k-4)}{k},\ c_2=-1\)

\(\text{Equating gradients}\)

\(\dfrac{3k}{2}\) \(=\dfrac{-(k-4)}{k}\)
\(3k^2\) \(=-2(k-4)\)\(\therefore\)\(\therefore\)
\(3k^2\) \(=-2k+8\)

\(3k^2+2k-8=0\)

\(3k-4)(k+2)=0\)

\(\therefore\ k=\dfrac{4}{3},\ k=-2\)

\(\text{Substituting to test }y-\text{intercepts}\)

\(\text{Case 1:}\ \ k=-2\quad\) \(c_1\) \(=-\dfrac{k+4}{2}\)
    \(=-\dfrac{-2+4}{2}=-1\)
  \(c_2\) \(=-1\)

\(\therefore\ k=-2 \text{, not a solution as intercepts cannot be equal.}\)

\(\text{Case 2:}\ \ \dfrac{4}{3}\quad\) \(c_1\) \(=-\dfrac{k+4}{2}\)
    \(=-\dfrac{\dfrac{4}{3}+4}{2}=-\dfrac{8}{3}\)
  \(c_2\) \(=-1\)

\(\text{Hence}\ \ k=\dfrac{4}{3} \text{, is the only possible solution as intercepts are not}\)

\(\text{equal and gradients are equal.}\)

Calculus, MET2 2024 VCAA 3

The points shown on the chart below represent monthly online sales in Australia.

The variable \(y\) represents sales in millions of dollars.

The variable \(t\) represents the month when the sales were made, where \(t=1\) corresponds to January 2021, \(t=2\) corresponds to February 2021 and so on.

  1. A cubic polynomial \(p ;(0,12] \rightarrow R, p(t)=a t^3+b t^2+c t+d\) can be used to model monthly online sales in 2021.

    The graph of \(y=p(f)\) is shown as a dashed curve on the set of axes above.

    It has a local minimum at (2,2500) and a local maximum at (11,4400).

    1. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\).  (3 marks)
    1. Let \(q:(12,24] \rightarrow R, q(t)=p(t-h)+k\) be a cubic function obtained by translating \(p\), which can be used to model monthly online sales in 2022.

      Find the values of \(h\) and \(k\) such that the graph of \(y=q(t)\) has a local maximum at \((23,4750)\).  (2 marks)

  1. Another function \(f\) can be used to model monthly online sales, where

    \(f:(0,36] \rightarrow R, f(t)=3000+30 t+700 \cos \left(\dfrac{\pi t}{6}\right)+400 \cos \left(\dfrac{\pi t}{3}\right)\)

    Part of the graph of \(f\) is shown on the axes below.

    1. Complete the graph of \(f\) on the set of axes above until December 2023, that is, for \(t \in(24,36]\).Label the endpoint at \(t=36\) with its coordinates.  (2 marks)
    2. The function \(f\) predicts that every 12 months, monthly online sales increase by \(n\) million dollars.

      Find the value of \(n\).  (1 mark)

    1. Find the derivative \(f^{\prime}(t)\)  (1 mark)
    1. Hence, find the maximum instantaneous rate of change for the function \(f\), correct to the nearest million dollars per month, and the values of \(t\) in the interval \((0,36]\) when this maximum rate occurs, correct to one decimal place.  (2 marks)
Show Answers Only

ai.   \(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii.  \(h=12, k=350\)

bi.  

bii.  \( n=360\)

biii. \(f^{\prime}(t)=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

biv. \(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate}\ \approx 725\ \text{million/month}\)

Show Worked Solution
ai.   \(\text{Given}\ p(2)=2500, p(11)=4400, p^{\prime}(2)=0\ \text{and}\ p^{\prime}(11)=0\) 
  
\(p(t)=at^3+bt^2+ct+d\ \Rightarrow\ p^{\prime}(t)=3at^2+2bt+c\)
  
\(\text{Using CAS:}\)
    
\(\text{Solve}
\begin{cases}
8a+4b+2c+d=2500 \\
1331a+121b+11c+d=4400 \\
12a+4b+c=0  \\
363a+22b+c=0
\end{cases}\)
  

\(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii. \(\text{Local maximim }p(t)\ \text{is}\ (11, 4400)\)

\(\therefore\ h\) \(=23-11=12\)
\(k\) \(=4750-4400=350\)

 

bi.   \(\text{Plotting points from CAS:}\)

\((24,4820), (26, 3930), (28, 3290), (30, 3600), (32, 3410), (34, 4170), (36, 5180)\)

bii.  \(\text{Using CAS: }\)

\(f(12)-f(0)\) \(=4460-4100=360\)
\(f(24)-f(12)\) \(=4820-4460=360\)
\(f(36)-f(24)\) \(=5180-4820=360\) 

\(\therefore\ n=360\)

biii.  \(f(t)\) \(=3000+30t+700\cos\left(\dfrac{\pi t}{6}\right)+400\cos\left(\dfrac{\pi t}{3}\right)\)
  \(f^{\prime}(t)\) \(=30-\dfrac{700\pi}{6}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)
    \(=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

  
biv.  \(\text{Max instantaneous rate of change occurs when }f^{\prime\prime}(t)=0\)

\(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate using CAS:}\)

\(f{\prime}(10.2)=f{\prime}(22.2)=f{\prime}(34.2)=725.396\approx 725\ \text{million/month}\)

Calculus, MET2 2024 VCAA 2

A model for the temperature in a room, in degrees Celsius, is given by

\(f(t)=\left\{
\begin{array}{cc}12+30 t & \quad \quad 0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.\)

where \(t\) represents time in hours after a heater is switched on.

  1. Express the derivative \(f^{\prime}(t)\) as a hybrid function.  (2 marks)
  1. Find the average rate of change in temperature predicted by the model between \(t=0\) and \(t=\dfrac{1}{2}\).

    Give your answer in degrees Celsius per hour.  (1 mark)

  1. Another model for the temperature in the room is given by \(g(t)=22-10 e^{-6 t}, t \geq 0\).
    1. Find the derivative \(g^{\prime}(t)\).  (1 mark)
    1. Find the value of \(t\) for which \(g^{\prime}(t)=10\).

      Give your answer correct to three decimal places.  (1 mark)

  1. Find the time \(t \in(0,1)\) when the temperatures predicted by the models \(f\) and \(g\) are equal.

    Give your answer correct to two decimal places.  (1 mark)

  1. Find the time \(t \in(0,1)\) when the difference between the temperatures predicted by the two models is the greatest.

    Give your answer correct to two decimal places.  (1 mark)

  1. The amount of power, in kilowatts, used by the heater \(t\) hours after it is switched on, can be modelled by the continuous function \(p\), whose graph is shown below.

\(p(t)=\left\{
\begin{array}{cl}1.5 & 0 \leq t \leq 0.4 \\
0.3+A e^{-10 t} & t>0.4
\end{array}\right.\)

The amount of energy used by the heater, in kilowatt hours, can be estimated by evaluating the area between the graph of \(y=p(t)\) and the \(t\)-axis.

    1. Given that \(p(t)\) is continuous for \(t \geq 0\), show that \(A=1.2 e^4\).  (1 mark)

    1. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.

      Give your answer in hours.  (1 mark)

    1. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.

      Give your answer in hours, correct to two decimal places.  (2 marks)

Show Answers Only

a.    \(f^{\prime}(t)=\left\{
\begin{array}{cc}30  & \quad \quad 0 \leq t <\dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.\)

b.    \(20^{\circ}\text{C/h}\)

ci.    \(g^{\prime}(t)=60e^{-6t}\)

cii.   \(0.299\ \text{(3 d.p.)}\)

d.    \(0.27\ \text{(2 d.p.)}\)

e.    \(0.12\ \text{(2 d.p.)}\)

fi.   \(\text{Because function is continuous}\)

\(0.3+Ae^{-10t}\) \(=1.5\)
\(Ae^{-10\times 0.4}\) \(=1.2\)
\(A\) \(=\dfrac{1.2}{e^{-10\times 0.4}}\)
  \(=1.2e^4\)

fii.  \(\dfrac{1}{3}\ \text{hours}\)

fiii. \(1.33\ \text{(2 d.p.)}\)

Show Worked Solution

a.    \(f^{\prime}(t)=\left\{
\begin{array}{cc}30  & \quad \quad 0 \leq t < \dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.\)

b.    \(\text{When }\ t=0, f(t)=12\ \ \text{and when }\ t=\dfrac{1}{2}, f(t)=22\)

\(\therefore\ \text{Average rate of change}\) \(=\dfrac{22-12}{\frac{1}{2}}\)
  \(=20^{\circ}\text{C/h}\)

 

ci.    \(g(t)\) \(=22-10 e^{-6 t}\)
  \(g^{\prime}(t)\) \(=60e^{-6t},\ \ t\geq 0\)

 

cii.   \(60e^{-6t}\) \(=10\)
  \(-6t\,\ln{e}\) \(=\ln{\dfrac{1}{6}}\)
  \(t\) \(=\dfrac{\ln{\frac{1}{6}}}{-6}\)
      \(=0.2986…\approx 0.299\ \text{(3 d.p.)}\)

 

d.     \(\left\{
\begin{array}{cc}12+30 t & \ \  0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.\)		 \(=22-10e^{-6t}\)

\(\text{Using CAS:}\)

\(\text{Temps equal when}\ t\approx 0.27\ \text{(2 d.p.)}\)

 

e.     \(\text{Difference }(D)\) \(=|g(t)-f(t)|\)
    \(=\left(22-10e^{-6t}\right)-(12+30t)\)
  \(\dfrac{dD}{dt}\) \(=60e^{-6t}-30\)

\(\text{Max time diff when}\ \dfrac{dD}{dt}=0\)

\(\therefore\ 60e^{-6t}-30\) \(=0\)
\(e^{-6t}\) \(=0.5\)
\(-6t\) \(=\ln{0.5}\)
\(t\) \(=0.1155\dots\)
  \(\approx 0.12\ \text{(2 d.p.)}\)

 

fi.   \(\text{Because function is continuous}\)

\(0.3+Ae^{-10t}\) \(=1.5\)
\(Ae^{-10\times 0.4}\) \(=1.2\)
\(A\) \(=\dfrac{1.2}{e^{-10\times 0.4}}\)
  \(=1.2e^4\)

  
fii.  \(\text{Using CAS:}\)

\(\text{Or, considering the graph, the area from 0 to 0.4 }=0.6\ \rightarrow t<0.4\)

\(\therefore\ \text{Solving}\ 1.5t=0.5\ \rightarrow\ t=\dfrac{1}{3}\)

\(\therefore\ \text{It takes }\dfrac{1}{3}\ \text{hours for heater to use 0.5  kilowatts.}\)
  

fiii. \(\text{Using CAS:}\)

\(1.5\times 4+\displaystyle\int_{0.4}^{t}0.3+1.2e^4.e^{-10t}dt\) \(=1\)
\(\Bigg[0.3t+0.12e^4.e^{-10t}\Bigg]_{0.4}^{t}\) \(=0.4\)
\(t\) \(=1.3333\dots\)
  \(\approx 1.33\ \text{(2 d.p.)}\)

 

PHYSICS, M7 2024 HSC 26

Muons are unstable particles produced when cosmic rays strike atoms high in the atmosphere. The muons travel downward, perpendicular to Earth's surface, at almost the speed of light.

Classical physics predicts that these muons will decay before they have time to reach Earth's surface.

Explain qualitatively why these muons can reach Earth's surface, regardless of whether their motion is considered from either the muon's frame of reference or the Earth's frame of reference.   (3 marks)

Show Answers Only

→ The muon’s are able to reach the Earth’s surface due to Einstein’s special theory of relativity in relation to length contraction and time dilation.

Muon’s frame of reference:

→ The distance to the Earth’s surface is contracted according to  \(l=l_o\sqrt{1-\frac{v^2}{c^2}}\). Muons see the Earth’s surface move towards them at speeds close to \(c\).

→ Since the muons have to travel a shorter distance than the proper length, they will have time to reach the Earth’s surface before they decay.
 

Earth’s frame of reference:

→ The time that it takes the muon to decay will be dilated according to  \(t=\dfrac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\)  as the muon’s are moving close to the speed of light.

→ Therefore, the muons have a longer half-life and lifespan than predicted by classical physics and will be able to reach the Earth’s surface before they decay.

→ In this way, muons can reach the surface of the Earth from either frame of reference. 

Show Worked Solution

→ The muon’s are able to reach the Earth’s surface due to Einstein’s special theory of relativity in relation to length contraction and time dilation.

Muon’s frame of reference:

→ The distance to the Earth’s surface is contracted according to  \(l=l_o\sqrt{1-\frac{v^2}{c^2}}\). Muons see the Earth’s surface move towards them at speeds close to \(c\).

→ Since the muons have to travel a shorter distance than the proper length, they will have time to reach the Earth’s surface before they decay.
 

Earth’s frame of reference:

→ The time that it takes the muon to decay will be dilated according to  \(t=\dfrac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\)  as the muon’s are moving close to the speed of light.

→ Therefore, the muons have a longer half-life and lifespan than predicted by classical physics and will be able to reach the Earth’s surface before they decay.

→ In this way, muons can reach the surface of the Earth from either frame of reference. 

♦ Mean mark 46%.

Functions, MET2 2024 VCAA 1

Consider the function  \( f: R \rightarrow R, f(x)=(x+1)(x+a)(x-2)(x-2 a) \text { where } a \in R \text {. } \)

  1. State, in terms of \(a\) where required, the values of \(x\) for which \(f(x)=0\).  (1 mark)
  1. Find the values of \(a\) for which the graph of \(y=f(x)\) has
    1. exactly three \(x\)-intercepts.  (2 marks)
    1. exactly four \(x\)-intercepts.  (1 mark)
  1. Let \(g\) be the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).
    1. Find \(g^{\prime}(x)\)  (1 mark)
    1. Find the coordinates of the local maximum of \(g\).  (1 mark)
    1. Find the values of \(x\) for which \(g^{\prime}(x)>0\).  (1 mark)

    1. Consider the two tangent lines to the graph of \(y=g(x)\) at the points where

      \(x=\dfrac{-\sqrt{3}+1}{2}\) and \(x=\dfrac{\sqrt{3}+1}{2}\).

      Determine the coordinates of the point of intersection of these two tangent lines.  (2 marks)

  1. Let \(g\) remain as the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).

    Let \(h\) be the function \(h: R \rightarrow R, h(x)=(x+1)(x-1)(x+2)(x-2)\), which is the function \(f\) where \(a=-1\).

    1. Using translations only, describe a sequence of transformations of \(h\), for which its image would have a local maximum at the same coordinates as that of \(g\).   (1 mark)
    1. Using a dilation and translations, describe a different sequence of transformations of \(h\), for which its image would have both local minimums at the same coordinates as that of \(g\).  (2 marks)

Show Answers Only

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.  \(g^{\prime}(x)=2(x-2)(x+1)(2x-1)\)

cii. \(\left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ. \(\left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.   \(\text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii.  \(\text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units}\)

Show Worked Solution

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(\text{The solution must be all }R\ \text{except those that give 3 or less solutions.}\)

\(\therefore\ R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.    \(g^{\prime}(x)\) \(=2(2-x)(x+1)^2+(x-2)^22(x+1)\)
    \(=2(x-2)(x+1)(x+1+x+2)\)
    \(=2(x-2)(x+1)(2x-1)\)

  
cii.  \(\text{When  }g^{\prime}(x)=0, x=2, -1, \dfrac{1}{2}\)

\(\text{From graph local maximum occurs when }x=\dfrac{1}{2}\)

\(g\left(\dfrac{1}{2}\right)\) \(=\left(\dfrac{1}{2}+1\right)^2\left(\dfrac{1}{2}-2\right)^2\)
  \(=\dfrac{9}{4}\times \dfrac{9}{4}=\dfrac{81}{16}\)

  
\(\therefore\ \text{Local maximum at}\ \left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(\text{From graph}\ g^{\prime}(x)>0\ \text{when }x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ.  \(\text{Use CAS to find tangent lines and solve to find intersection.}\)

\(\text{Point of intersection of tangent lines}\ \left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.  \(\text{Local maximum of }g(x)\ \rightarrow\left(\dfrac{1}{2}, \dfrac{81}{16}\right)\)

\(\text{From CAS local maximum of }h(x)\ \rightarrow \left(0, 4\right)\)

\(\therefore\ \text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii. \(\text{Using CAS to solve }g^{\prime}(x)=0\ \text{and }h^{\prime}(x)=0\)

\(\text{Local Minimums for }g(x)\ \text{at }(-1, 0)\ \text{and }(2, 0)\ \text{which are 3 apart.}\)

\(\text{Local minimums for at }h(x)\ \text{at }\left(-\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\ \text{and }\left(\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\)

\(\therefore\ \text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units.}\)

 

Calculus, MET2 2024 VCAA 20 MC

The function \(f: R \rightarrow R\) has an average value \(k\) on the interval \([0,2]\) and satisfies \(f(x)=f(x+2)\) for all \(x \in R\). The value of the definite integral \( {\displaystyle \int_2^6 f(x) d x } \) is

  1. \(2k\)
  2. \(3k\)
  3. \(4k\)
  4. \(6k\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{The function }f(x)\ \text{is a periodic function}\ \rightarrow\ \text{Period}=2\)

\(\text{Average value}\ \rightarrow\ k\) \(=\dfrac{1}{2}\displaystyle\int_0^2 f(x)\, dx\)
\(2k\) \(=\displaystyle\int_0^2 f(x)\, dx\)

\(\text{As function is periodic the average value remains the same}\ \rightarrow\ k,\text{for each period.}\)

\(\therefore\ \displaystyle\int_2^6 f(x)\, dx\) \(=2\times\displaystyle\int_0^2 f(x)\, dx\)
  \(=2\times 2k=4k\)

  
\(\Rightarrow C\)

Probability, MET2 2024 VCAA 19 MC

Consider the normal random variable \(X\) that satisfies \( \text{Pr (X < 10) = 0.2 and Pr(X > 18) = 0.2} \).

The value of \(\text{Pr}(X<12)\) is closest to

  1. 0.134
  2. 0.297
  3. 0.337
  4. 0.365
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Pr}\left(z<\dfrac{10-\mu}{\sigma}\right)=0.2\)

\(\text{Pr}\left(z\leq\dfrac{18-\mu}{\sigma}\right)=0.8\)

\(\text{Next use CAS to solve the simultaneous eqns below to find}\)

\(\text{the mean and SD.}\)

\(z_1=\dfrac{10-\mu}{\sigma}\ , z_2=\dfrac{18-\mu}{\sigma}\)

\(\text{Using CAS:}\)

\(\therefore\ \text{Pr}(X<12)\approx 0.336947\)

\(\Rightarrow C\)

Calculus, MET2 2024 VCAA 15 MC

The points of inflection of the graph of \(y=2-\tan \left(\pi\left(x-\dfrac{1}{4}\right)\right)\) are

  1. \(\left(k+\dfrac{1}{4}, 2\right), k \in Z\)
  2. \(\left(k-\dfrac{1}{4}, 2\right), k \in Z\)
  3. \(\left(k+\dfrac{1}{4},-2\right), k \in Z\)
  4. \(\left(k-\dfrac{3}{4},-2\right), k \in Z\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{The graph of}\ \ y=-\tan(\pi x)\ \text{is translated 2 units upwards and then}\)

\(\text{translated }\dfrac{1}{4}\ \text{units to the right to become the graph shown below.}\)

\(\text{Hence all points of inflection lie of the line}\ y=2\)

\(\rightarrow\ \text{Eliminate Options C and D}\)

\(\text{Consider Option A:}\)

\(\text{When}\ \ k=0, x=0+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\text{then}\ \ y=2-\tan \left(\pi\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right)=2\)

\(\text{Similarly for any}\ k \in Z\ \text{a point of inflection will be found at }\ \left(k+\dfrac{1}{4}, 2\right)\)

\(\Rightarrow A\)

PHYSICS, M8 2024 HSC 24

An absorption spectrum resulting from the passage of visible light from a star's surface through its hydrogen atmosphere is shown. Absorption lines are labelled \(W\) to \(Z\) in the diagram.
 

  1. Determine the surface temperature of the star.   (2 marks)

  2. Absorption line \(W\) originates from an electron transition between the second and sixth energy levels. Use  \(\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_{ f }^2}-\dfrac{1}{n_{ i }^2}\right)\)  to calculate the frequency of light absorbed to produce absorption line \(W\).   (3 marks)

  3. Explain the physical processes that produce an absorption spectrum.   (3 marks)

Show Answers Only

a.    \(5796\ \text{K}\)

b.    \(7.31 \times 10^{14}\ \text{Hz}\)

c.    An absorption spectra is produced when:

→ A continuous spectrum of light  from a black body such as a star passes through cooler and lower density gas in the outer atmosphere of the star.

→ As the light passes through the gas, electrons in the atoms that make up the cooler gas clouds absorb distinct wavelengths/energy levels of light equal to the difference in energy levels between the electron shells where \(E_i-E_f=hf=\dfrac{hc}{\lambda}\). 

→ As the electrons in the atoms fall back into their ground state, they emit the photon of light that they absorb and the photon is then scattered out of the continuous spectrum.

→ The light that remains is then passed through a prism to separate the wavelengths and record the intensities. The black or darkened lines in the absorption spectra is the result of the scattered wavelengths of light. 

Show Worked Solution

a.    Determined temperature using the peak wavelength:

\(T\) \(=\dfrac{b}{\lambda_{\text{max}}}\)  
  \(=\dfrac{2.898 \times 10^{-3}}{500 \times 10^{-9}}\)  
  \(=5796\ \text{K}\)  

 

b.     \(\dfrac{1}{\lambda}\) \(=R\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)\)
    \(=1.097 \times 10^7 \times \left(\dfrac{1}{2^2}-\dfrac{1}{6^2}\right)\)
    \(=2.438 \times 10^6\ \text{m}^{-1}\)
  \(\lambda\) \(=\dfrac{1}{2.438 \times 10^6}\)
    \(=410.2\ \text{nm}\)

 

\(\therefore f=\dfrac{c}{\lambda} = \dfrac{3 \times 10^8}{410.2 \times 10^{-9}} = 7.31 \times 10^{14}\ \text{Hz}\)
 

c.    Absorption spectra:

→ Produced when a continuous spectrum of light  from a black body such as a star passes through cooler and lower density gas in the outer atmosphere of the star.

→ As the light passes through the gas, electrons in the atoms that make up the cooler gas clouds absorb distinct wavelengths/energy levels of light equal to the difference in energy levels between the electron shells where \(E_i-E_f=hf=\dfrac{hc}{\lambda}\). 

→ As the electrons in the atoms fall back into their ground state, they emit the photon of light that they absorb and the photon is then scattered out of the continuous spectrum.

→ The light that remains is then passed through a prism to separate the wavelengths and record the intensities. The black or darkened lines in the absorption spectra is the result of the scattered wavelengths of light. 

♦ Mean mark (c) 51%.

PHYSICS, M8 2024 HSC 23

Development of models of the atom has resulted from both experimental investigations and hypotheses based on theoretical considerations.

  1. A key piece of experimental evidence supporting the nuclear model of the atom was a discovery by Chadwick in 1932.
  2. An aspect of the experimental design is shown.
     

    1. What was the role of paraffin in Chadwick's experiment?   (2 marks)

    2. How did Chadwick's experiment change the model of the atom?  (3 marks)

  2. Explain how de Broglie's hypothesis regarding the nature of electrons addressed limitations in the Bohr-Rutherford model of the atom.   (4 marks)

Show Answers Only

a.i.  Role of paraffin wax:

→ Paraffin wax is a rich source of protons.

→ When the paraffin was placed in front of the unknown radiation, the transfer of momentum from the radiation caused protons to be emitted from the paraffin wax.

→ The emitted protons could then be detected and analysed.

→ From studying the protons ejected from the paraffin wax, Chadwick proposed the existence of the neutron.
 

a.ii. Changes to the model of the atom:

→ Previous to Chadwick’s experiment, the model of the atom proposed by Rutherford consisted of a dense positive charge in the nucleus which was orbited by electrons.

→ In this model however, the protons did not account for the total mass of the nucleus.

→ Through using the conservation of momentum and energy in his experiment, Chadwick proposed the existence of the neutron particle which was slightly heavier than the proton.

→ The model of the atom was updated to include both protons and neutrons in the nucleus which then fully accounted for the mass of the nucleus.
 

b.   Limitations in the Bohr-Rutherford model:

→ Rutherford’s model of the atom stated that electrons orbited the nucleus and were electrostatically attracted to the positive nucleus. This meant that the electrons were in circular motion and were constantly under centripetal acceleration.

→ However, Maxwell predicted that an accelerating charge would emit electro-magnetic radiation and in Rutherford’s model, all atoms should have been unstable as the electrons would emit EMR, lose energy and spiral into the nucleus.

→ Bohr proposed that electrons orbited the nucleus in stationary states at fixed energies with no intermediate states possible where they would not emit EMR but provided no theoretical explanation for this.

De Broglie’s hypothesis:

→ De Broglie proposed that electrons could exhibit a wave nature and could act as matter-waves. The electrons would form standing waves around the nucleus and would no longer be an accelerating particle which addressed the limitation of all atoms being unstable.

→ Further, De Broglie proposed that the standing waves would occur at integer wavelengths where the circumference of the electron orbit would be equal to an integer electron wavelength, \(2\pi r=n\lambda\)  where  \(\lambda = \dfrac{h}{mv}\). At any other radii other than this, deconstructive interference would occur and a standing electron wave would not form. This addressed why electrons could only be present at fixed radii/energy levels in the atom.

Show Worked Solution

a.i.  Role of paraffin wax:

→ Paraffin wax is a rich source of protons.

→ When the paraffin was placed in front of the unknown radiation, the transfer of momentum from the radiation caused protons to be emitted from the paraffin wax.

→ The emitted protons could then be detected and analysed.

→ From studying the protons ejected from the paraffin wax, Chadwick proposed the existence of the neutron.
 

Mean mark (a)(i) 52%.

a.ii. Changes to the model of the atom:

→ Previous to Chadwick’s experiment, the model of the atom proposed by Rutherford consisted of a dense positive charge in the nucleus which was orbited by electrons.

→ In this model however, the protons did not account for the total mass of the nucleus.

→ Through using the conservation of momentum and energy in his experiment, Chadwick proposed the existence of the neutron particle which was slightly heavier than the proton.

→ The model of the atom was updated to include both protons and neutrons in the nucleus which then fully accounted for the mass of the nucleus.
 

b.   Limitations in the Bohr-Rutherford model:

→ Rutherford’s model of the atom stated that electrons orbited the nucleus and were electrostatically attracted to the positive nucleus. This meant that the electrons were in circular motion and were constantly under centripetal acceleration.

→ However, Maxwell predicted that an accelerating charge would emit electro-magnetic radiation and in Rutherford’s model, all atoms should have been unstable as the electrons would emit EMR, lose energy and spiral into the nucleus.

→ Bohr proposed that electrons orbited the nucleus in stationary states at fixed energies with no intermediate states possible where they would not emit EMR but provided no theoretical explanation for this.

De Broglie’s hypothesis:

→ De Broglie proposed that electrons could exhibit a wave nature and could act as matter-waves. The electrons would form standing waves around the nucleus and would no longer be an accelerating particle which addressed the limitation of all atoms being unstable.

→ Further, De Broglie proposed that the standing waves would occur at integer wavelengths where the circumference of the electron orbit would be equal to an integer electron wavelength, \(2\pi r=n\lambda\)  where  \(\lambda = \dfrac{h}{mv}\). At any other radii other than this, deconstructive interference would occur and a standing electron wave would not form. This addressed why electrons could only be present at fixed radii/energy levels in the atom.

♦ Mean mark (b) 44%.

Functions, MET2 2024 VCAA 13 MC

The function \(f:(0, \infty) \rightarrow R, f(x)=\dfrac{x}{2}+\dfrac{2}{x}\) is mapped to the function \(g\) with the following sequence of transformations:

  1. dilation by a factor of 3 from the \(y\)-axis
  2. translation by 1 unit in the negative direction of the \(y\)-axis.

The function \(g\) has a local minimum at the point with the coordinates

  1. \((6,1)\)
  2. \(\left(\dfrac{2}{3}, 1\right)\)
  3. \((2,5)\)
  4. \(\left(2,-\dfrac{1}{3}\right)\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Combination transformation}\rightarrow\ f\left(\dfrac{x}{3}\right)-1\)

\(\therefore f(x)\rightarrow g(x)\) \(=\dfrac{\frac{x}{3}}{2}+\dfrac{2}{\frac{x}{3}}-1\)
  \(=\dfrac{x}{6}+\dfrac{6}{x} -1, (0, \infty)\)
\(\text{and}\ g'(x)\) \(=\dfrac{1}{6}-6x^{-2}\)

\(\Rightarrow A\)

PHYSICS, M8 2024 HSC 22

The following graph, based on the data gathered by Hubble, shows the relationship between the recessional velocity of galaxies and their distance from Earth.
 

     

  1. Describe the significance of the graph to our understanding of the universe.   (2 marks)

  2. How were the recessional velocities of galaxies determined?  (3 marks)

Show Answers Only

a.   Recessional velocity vs distance from Earth graph:

→ The graph shows that the further away galaxies are from Earth, the faster these galaxies are moving away from Earth.

→ This relationships depicts Hubble’s law:  \(v=H_0 D\). 

→ This graph provides evidence that the universe is constantly expanding as predicted by the big bang theory.
 

b.   Determining recessional velocities:

→ The recessional velocities of the galaxies were determined by analysing their absorption spectras.

→ Light waves that are moving away from the Earth will appear to be stretched (wavelength increased) according to the doppler effect.

→ The absorption spectra of galaxies were compared with the spectra of the same elements on Earth, revealing that the galaxies’ spectra were redshifted.

→ The greater the extent of the red shift in the spectra, the greater the recessional velocity of the galaxy. 

Show Worked Solution

a.   Recessional velocity vs distance from Earth graph:

→ The graph shows that the further away galaxies are from Earth, the faster these galaxies are moving away from Earth.

→ This relationships depicts Hubble’s law:  \(v=H_0 D\). 

→ This graph provides evidence that the universe is constantly expanding as predicted by the big bang theory.
 

b.   Determining recessional velocities:

→ The recessional velocities of the galaxies were determined by analysing their absorption spectras.

→ Light waves that are moving away from the Earth will appear to be stretched (wavelength increased) according to the doppler effect.

→ The absorption spectra of galaxies were compared with the spectra of the same elements on Earth, revealing that the galaxies’ spectra were redshifted.

→ The greater the extent of the red shift in the spectra, the greater the recessional velocity of the galaxy. 

♦ Mean mark (b) 49%.

PHYSICS, M6 2024 HSC 17 MC

The diagram shows a type of particle accelerator called a cyclotron.

Cyclotrons accelerate charged particles, following the path as shown.
 

   

An electric field acts on a charged particle as it moves through the gap between the dees. A strong magnetic field is also in place.

Once a charged particle has the required velocity, it exits the accelerator towards a target.

Which of the following is true about a charged particle in a cyclotron?

  1. It increases speed while inside the dees.
  2. It only accelerates while between the dees.
  3. It undergoes acceleration inside and between the dees.
  4. It slows down inside the dees and speeds up between the dees.
Show Answers Only

\(C\)

Show Worked Solution

→ When the charged particle is between the dees, it will experience an acceleration due to the electric field present where \(a = \dfrac{qE}{m}\).

→ When the charged particle is inside of the dees, the particle undergoes uniform circular motion due to the strong magnetic field in place from the electromagnets.

→ While the magnitude of the velocity of the charged particle does not change, the direction of the velocity does. Hence, there is a change in velocity of the particle so it is experiencing an acceleration.

→ A charged particle will experience a centripetal force/acceleration when moving perpendicular to a magnetic field.

\(\Rightarrow C\)

♦ Mean mark 45%.

PHYSICS, M7 2024 HSC 16 MC

The graph shows the relationship between the maximum kinetic energy of emitted photoelectrons and the incident photon energy for four different metal surfaces.
 

Light of frequency \(7 \times 10^{14}\ \text{Hz}\) is incident on the metals.

From which metals are photoelectrons emitted?

  1. \(\ce{K}\), \(\ce{Li}\) only
  2. \(\ce{Mg}\), \(\ce{Ag}\) only
  3. All of the metals
  4. None of the metals
Show Answers Only

\(A\)

Show Worked Solution

→ The photon energy for a light frequency of \(7 \times 10^{14}\ \text{Hz}\) is:

\(E=hf = 6.626 \times 10^{-34} \times 7 \times 10^{14} = 4.64 \times 10^{-19}\ \text{J}\)

→ The energy of the light frequency in electron volts\(=\dfrac{4.64 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.9\ \text{eV}\).

→ Photoelectrons will only be emitted if the energy of the incident photons is enough to overcome the work function of the metals. On the graph above, the work function of each metal is the same value as the x-intercepts of their respective graphs.

→ Therefore photoelectrons will be emitted from potassium and lithium as the energy of the incident photons \((2.9\ \text{eV})\) is greater than the work functions of these metals.

\(\Rightarrow A\)

♦ Mean mark 47%.

PHYSICS, M6 2024 HSC 15 MC

A uniform magnetic field is directed into the page. A conductor \(P Q\) rotates about the end \(P\) at a constant rate.
 

 
 

Which graph shows the emf induced between the ends of the conductor, \(P\) and \(Q\), as it rotates one revolution from the position shown?
 


 

Show Answers Only

\(D\)

Show Worked Solution

→ As the conductor rotates in the magnetic field, all of the charges in the conductor will have a velocity.

→ This will produce a force on each positive and negative charge in the rod according to the equation \(F=qvB\).

→ Using the right hand rule, the direction of the force applied to each positive charge in the rod during the rotation will be towards \(P\). Similarly, the direction of force on each negative charge will be towards \(Q\). 

→ This separation of positive and negative charges to opposite ends of the conductor generates the emf.

→ As the velocity of the charges during the circular motion will remain constant, the force on the charges will be constant. Therefore, a constant emf will be produced in the conductor.

\(\Rightarrow D\)

♦♦ Mean mark 35%.

PHYSICS, M8 2024 HSC 14 MC

The velocity of a proton \( {\displaystyle \left({ }_1^1 \text{H}\right) } \) is twice the velocity of an alpha particle \( { \displaystyle \left({ }_2^4 \text{He}\right) } \). The proton has a de Broglie wavelength of \(\lambda\).

What is the de Broglie wavelength of the alpha particle?

  1. \(\dfrac{\lambda}{8}\)
  2. \(\dfrac{\lambda}{2}\)
  3. \(2 \lambda\)
  4. \(8 \lambda\)
Show Answers Only

\(B\)

Show Worked Solution

→ Let the mass of the proton be \(m\) and the velocity of the proton be \(v\).

→ Therefore the mass of the alpha particle will be \(4m\) and the velocity will be \(\dfrac{v}{2}\).

→ The de Broglie wavelength of the proton, \(\lambda = \dfrac{h}{mv}\).

→ Hence the de Broglie wavelngth of the alpha particle \(=\dfrac{h}{4m \times \frac{v}{2}}=\dfrac{h}{2mv} = \dfrac{\lambda}{2}\)

\(\Rightarrow B\)

♦ Mean mark 49%.

PHYSICS, M6 2024 HSC 10 MC

A rod carrying a current, \(I\), placed in a uniform magnetic field as shown, experiences a force \(F\).
 

How many degrees must the rod be rotated clockwise so that it experiences a force \(\dfrac{F}{2}\)?

  1. 30°
  2. 45°
  3. 60°
  4. 90°
Show Answers Only

\(C\)

Show Worked Solution

→ The force experienced by a current carrying conductor in a magnetic field is given by \(F=lIB\sin \theta\), where \(\sin \theta\) refers to the angle between the direction of the current through the conductor and the magnetic field lines.

→ For the rod to experience a force of \(\dfrac{F}{2}\), the value of  \(\sin \theta = \dfrac{1}{2}\).

\(\sin \theta=\dfrac{1}{2}\ \ \Rightarrow \ \theta=\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=30^{\circ}\) 

→ As the angle between the current carrying conductor and magnetic field lines originally is \(90^{\circ}\), the rod must be rotated by \(60^{\circ}\).

\(\Rightarrow C\) 

♦♦ Mean mark 31%.

PHYSICS, M8 2024 HSC 7 MC

A pure sample of polonium-210 undergoes alpha emission to produce the stable isotope lead-206.

The half-life of polonium-210 is 138 days.

At the end of 276 days, what is the ratio of polonium-210 atoms to lead-206 atoms in the sample?

  1. \(1 : 4\)
  2. \(1 : 3\)
  3. \(1 : 2\)
  4. \(1 : 1\)
Show Answers Only

\(B\)

Show Worked Solution

→ After two half-life’s of the polonium only one quarter of the sample of polonium-210 atoms will be left.

→ Therefore, three quarters of the polonium-210 atoms would have decayed into lead-206 atoms.

→ The ratio of polonium atoms to lead atoms is \(\dfrac{1}{4}:\dfrac{3}{4} = 1:3\)

\(\Rightarrow B\)

♦ Mean mark 45%.

PHYSICS, M8 2024 HSC 3 MC

Which of the following is a fundamental particle in the Standard Model of matter?

  1. Hadron
  2. Neutron
  3. Photon
  4. Proton
Show Answers Only

\(C\)

Show Worked Solution

→ Hadrons are subatomic particles which are composed of two or more quarks. Protons and Neutrons are categorised as hadrons as they are both composed of 3 quarks. 

→ Hadrons, protons and neutrons are not fundamental particles as they are all composed of quarks.

→ The photon is a fundamental particle, classified as a gauge boson and mediates the electromagnetic force.

\(\Rightarrow C\)

♦ Mean mark 55%.

BIOLOGY, M8 2024 HSC 35

The graph shows the results of a survey conducted to determine if children changed their method of communication after cochlear implantation.
 

With reference to the data, describe how cochlear implants work, and how they affect communication in children.   (5 marks)

Show Answers Only

→ Cochlear implants are surgical electronic devices that help restore hearing in patients with cochlear damage.

→ The implants are inserted directly into the cochlea and stimulate the auditory nerve by sending sound signals straight to the brain.

→ The graph demonstrates that the age of implantation significantly affects communication outcomes.

→ Early implantation (under 3 years): Dramatic decrease in sign language use, with only 10% (approximately) still signing after 5 years.

→ Middle age implantation (3-5 years): Moderate decrease in sign language use.

→ Late implantation (over 5 years): Little to no change in sign language use

→ The data clearly shows that earlier cochlear implantation leads to greater shifts away from sign language as the primary mode of communication.

Show Worked Solution

→ Cochlear implants are surgical electronic devices that help restore hearing in patients with cochlear damage.

→ The implants are inserted directly into the cochlea and stimulate the auditory nerve by sending sound signals straight to the brain.

→ The graph demonstrates that the age of implantation significantly affects communication outcomes.

→ Early implantation (under 3 years): Dramatic decrease in sign language use, with only 10% (approximately) still signing after 5 years.

→ Middle age implantation (3-5 years): Moderate decrease in sign language use.

→ Late implantation (over 5 years): Little to no change in sign language use

→ The data clearly shows that earlier cochlear implantation leads to greater shifts away from sign language as the primary mode of communication.

♦♦ Mean mark 52%.

BIOLOGY, M6 2024 HSC 34

Discuss the ethical implications and impacts on society of the use of TWO biotechnologies.   (7 marks)

Show Answers Only

Social impacts of recombinant DNA technology (plant biotechnology)

→ Recombinant DNA has created beneficial products like Bt corn and Bt cotton.

→ Bt crops require fewer pesticide applications since they produce their own insecticidal proteins.

→ This reduces chemical pesticide costs for farmers and decreases environmental impact from pesticide spraying.

Associated ethical concerns:

→ Farmers growing Bt corn and cotton must purchase, at a significant cost, new GM seeds each season. Traditional farmers can reuse their seeds for the following year’s crop planting.

→ This difference creates economic disparities in access to GM crops and the related market opportunities.

Social impacts of selective breeding/hybridisation (animal biotechnology)

→ Selective breeding/hybridisation has produced, for example, dairy cows capable of increased milk production.

→ This results in higher yields and greater food availability.

→ Improved profits and living standards for farmers who can access the technology.

→ Increased food production to support population growth.

Associated ethical issues:

→ High-yield dairy cows show decreased fertility.

→ May compromise animal welfare and quality of life.

Show Worked Solution

Social impacts of recombinant DNA technology (plant biotechnology)

→ Recombinant DNA has created beneficial products like Bt corn and Bt cotton.

→ Bt crops require fewer pesticide applications since they produce their own insecticidal proteins.

→ This reduces chemical pesticide costs for farmers and decreases environmental impact from pesticide spraying.

Associated ethical concerns:

→ Farmers growing Bt corn and cotton must purchase, at a significant cost, new GM seeds each season. Traditional farmers can reuse their seeds for the following year’s crop planting.

→ This difference creates economic disparities in access to GM crops and the related market opportunities.

♦ Mean mark 56%.

Social impacts of selective breeding/hybridisation (animal biotechnology)

→ Selective breeding/hybridisation has produced, for example, dairy cows capable of increased milk production.

→ This results in higher yields and greater food availability.

→ Improved profits and living standards for farmers who can access the technology.

→ Increased food production to support population growth.

Associated ethical issues:

→ High-yield dairy cows show decreased fertility.

→ May compromise animal welfare and quality of life.

BIOLOGY, M5 2024 HSC 33

Female Jack Jumper ants (Myrmecia pilosula) have a single pair of chromosomes. During meiosis, crossing over occurs. The diagram shows the crossing over and the position of three genes on the chromosomes.
 

  1. Outline the significance of crossing over for the Jack Jumper ants.   (2 marks)

  2. Draw the chromosomes of the four possible gametes after crossing over for the Jack Jumper ants occurs. Include the alleles for each gene.   (2 marks)
     

Show Answers Only

a.   Significance of crossing over:

→ Genetic variation increases in the Jack Jumper ant population through recombination.

→ This enhanced genetic diversity improves the species’ chances of survival when faced with environmental changes, as some ants may carry beneficial adaptations.
 

b.   

Show Worked Solution

a.   Significance of crossing over:

→ Genetic variation increases in the Jack Jumper ant population through recombination.

→ This enhanced genetic diversity improves the species’ chances of survival when faced with environmental changes, as some ants may carry beneficial adaptations.
 

b.   

♦ Mean mark (b) 54%.

BIOLOGY, M7 2024 HSC 32

Helicobacter pylori is a bacterium that invades the gut lining and can cause damage to the stomach as shown in the diagram.
 

With reference to innate and adaptive immunity, explain how the body responds after exposure to Helicobacter pylori.   (7 marks)

Show Answers Only

→ Damaged cells release chemicals that trigger inflammation as an initial response.

→ The inflammatory response causes blood vessels to dilate, increasing blood flow and allowing phagocytes (macrophages and neutrophils) to move into the infected area.

→ Phagocytes process H.pylori antigens and present them to helper T-cells, which launch the adaptive immune response by releasing cytokines.

→ This cytokine release activates both T and B cells to mount multiple specific defences.

→ Cytotoxic T-cells directly attack H.pylori while memory T-cells remain for secondary rapid responses.

→ Suppressor T-cells regulate the immune response and plasma B-cells produce H.pylori-specific antibodies. Memory B-cells persist for responding to future (secondary) infections.

→ Antibodies work in two ways – direct neutralisation of antigens and tagging antigens for destruction by phagocytes.
 

The immune response involves both innate and adaptive immunity systems working together:

→ Innate immunity provides rapid, immediate defence.

→ Adaptive immunity develops more slowly but offers long-term protection through memory cells.

Show Worked Solution

→ Damaged cells release chemicals that trigger inflammation as an initial response.

→ The inflammatory response causes blood vessels to dilate, increasing blood flow and allowing phagocytes (macrophages and neutrophils) to move into the infected area.

→ Phagocytes process H.pylori antigens and present them to helper T-cells, which launch the adaptive immune response by releasing cytokines.

→ This cytokine release activates both T and B cells to mount multiple specific defences.

→ Cytotoxic T-cells directly attack H.pylori while memory T-cells remain for secondary rapid responses.

→ Suppressor T-cells regulate the immune response and plasma B-cells produce H.pylori-specific antibodies. Memory B-cells persist for responding to future (secondary) infections.

→ Antibodies work in two ways – direct neutralisation of antigens and tagging antigens for destruction by phagocytes.

♦♦ Mean mark 48%.

The immune response involves both innate and adaptive immunity systems working together:

→ Innate immunity provides rapid, immediate defence.

→ Adaptive immunity develops more slowly but offers long-term protection through memory cells.

BIOLOGY, M8 2024 HSC 29

An epidemiological study was conducted to help model how many people will be affected by Type 2 diabetes globally in the future. Continuous data were collected from 1990 to 2020. From that data, the following data points were chosen to demonstrate the trend.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Year} \rule[-1ex]{0pt}{0pt} & \textit{Percentage of population affected} \\
\rule{0pt}{2.5ex} \textit{} \rule[-1ex]{0pt}{0pt} & \textit{by Type 2 diabetes (%)} \\
\hline
\rule{0pt}{2.5ex} \text{1990} \rule[-1ex]{0pt}{0pt} & \text{3.1} \\
\hline
\rule{0pt}{2.5ex} \text{2000} \rule[-1ex]{0pt}{0pt} & \text{3.7} \\
\hline
\rule{0pt}{2.5ex} \text{2010} \rule[-1ex]{0pt}{0pt} & \text{4.3} \\
\hline
\rule{0pt}{2.5ex} \text{2010} \rule[-1ex]{0pt}{0pt} & \text{5.6} \\
\hline
\end{array}

  1. Plot the data on the grid provided and include the line of best fit.   (2 marks)
     

  1. A prediction of the global population numbers suggests there will be about 9 billion \((9\ 000\ 000\ 000)\) people on the planet by 2040.

    Predict the number of people that will be affected by diabetes in 2040. Show working on your graph on the previous page and your calculations.   (3 marks)

Show Answers Only

a.   

b.   630 000 000 (630 million)

Show Worked Solution

a.   

♦ Mean mark (a) 56%.

b.   Using the LOBF on the graph:

Population (%) with diabetes in 2040 = 7%

People with diabetes in 2040 = \(\dfrac{7}{100} \times 9\ 000\ 000\ 000 = 630\ 000\ 000\)

BIOLOGY, M8 2024 HSC 31

A study monitored the changes in the body temperature of a kookaburra (an Australian bird) and a human over a 24-hour period. The results of the study are shown in the graph.
 

  1. At what time was the kookaburra's body temperature the lowest?   (1 mark)

  1. Some endothermic organisms can display torpor (a significant decrease in physiological activity).

    With reference to the graph, explain whether the human or the kookaburra was displaying torpor and if so, state the time this occurred.   (3 marks)

  1. Outline an adaptation that may lead to an increase in the kookaburra's body temperature during the inactive period.   (2 marks)

Show Answers Only

a.   4 am

b.   Signs of torpor:

→ The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity

→ In contrast, the kookaburra exhibited classic torpor behaviour.

→ Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.
 

c.   Kookaburra adaptation:

→ Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.

→ This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier

→ The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

Show Worked Solution

a.   4 am
 

b.   Signs of torpor:

→ The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity

→ In contrast, the kookaburra exhibited classic torpor behaviour.

→ Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.

Mean mark (b) 56%.

c.   Kookaburra adaptation:

→ Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.

→ This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier

→ The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

♦ Mean mark (c) 47%.

BIOLOGY, M5 2024 HSC 30

The diagram shows a simplified version of the process of polypeptide synthesis.
 

  1. Compare Process \(A\) with DNA replication.   (3 marks)
  1. Explain the importance of mRNA and tRNA in polypeptide synthesis.   (5 marks)

Show Answers Only

a.   Process \(A\) vs DNA replication:

→ Both DNA replication and transcription (Process \(A\)) begin with unwinding the DNA double helix.

→ DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.

→ In contrast, transcription copies just one DNA strand to produce a single mRNA strand.
 

b.   mRNA and tRNA’s role in polypeptide synthesis:

→ mRNA is created in the nucleus by copying a DNA template during transcription.

→ This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.

→ At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.

→ tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.

→ The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.

Show Worked Solution

a.   Process \(A\) vs DNA replication:

→ Both DNA replication and transcription (Process \(A\)) begin with unwinding the DNA double helix.

→ DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.

→ In contrast, transcription copies just one DNA strand to produce a single mRNA strand.
 

b.   mRNA and tRNA’s role in polypeptide synthesis:

→ mRNA is created in the nucleus by copying a DNA template during transcription.

→ This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.

→ At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.

→ tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.

→ The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.

♦ Mean mark (b) 60%.

BIOLOGY, M5 2024 HSC 28

Cystic fibrosis is an inherited disorder that causes damage to the lungs, digestive system and other organs in the body. A person with cystic fibrosis will have two faulty recessive alleles for the cystic fibrosis gene (CFTR) on chromosome 7.

  1. Two healthy parents, heterozygous for cystic fibrosis, have a child that does not have cystic fibrosis. They are planning to have a second child.

    Using a Punnett square, determine the probability of their second child being born with the condition. Use \(R\) for the normal CFTR allele, and \(r\) for the faulty CFTR allele.   (3 marks)

The defect that creates the faulty CFTR allele is often caused by the deletion of three nucleotides. The following diagram illustrates a small section of the CFTR gene and the corresponding amino acid sequence of the CFTR protein.
 
 
The following codon chart displays all the codons and the corresponding amino acids. The chart translates mRNA sequences into amino acids.
 
  1. Explain how the deletion of nucleotides in the CFTR gene removes only one amino acid. Include reference to the nucleotides that code for isoleucine and phenylalanine amino acids.   (4 marks)

Show Answers Only

a.    Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.
 

b.    Deletion of nucleotides in CFTR gene:

→ Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.

→ At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.

→ In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.

→ Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

Show Worked Solution

a.    Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.
 

b.    Deletion of nucleotides in CFTR gene:

→ Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.

→ At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.

→ In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.

→ Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

♦ Mean mark (b) 56%.

BIOLOGY, M7 2024 HSC 27

Milk pasteurisation (heating to approximately 70°C) was gradually introduced in America from the early 1900s. The graph shows the number of disease outbreaks in relation to raw (unpasteurised) and pasteurised milk in America from 1900-1975.
 

Explain the trends observed in the graph. In your response, refer to the role of Pasteur's work in pasteurisation.   (5 marks)

Show Answers Only

→ Louis Pasteur’s research was pivotal in debunking the theory of spontaneous generation and establishing our understanding of microorganisms.

→ His work revealed that microbes present in milk could be responsible for disease outbreaks.

→ He demonstrated that exposing substances to high temperatures effectively kills microorganisms, which is why heating milk to 70°C eliminates many harmful bacteria.

→ This scientific foundation – the presence of microbes in milk and their vulnerability to heat – explains the effectiveness of milk pasteurisation in preventing disease outbreaks.

→ The historical data presented in the graph supports this, showing significantly fewer disease outbreaks linked to pasteurised milk compared to raw milk.

→ A notable decline in raw milk-related outbreaks occurred after 1945, though this may also be attributed to decreased raw milk consumption during that period.

→ While pasteurised milk has generally proven safer, some disease outbreaks have still occurred with pasteurised products. These cases typically result from issues in the pasteurisation process itself or problems during subsequent storage and transportation of the milk.

Show Worked Solution

→ Louis Pasteur’s research was pivotal in debunking the theory of spontaneous generation and establishing our understanding of microorganisms.

→ His work revealed that microbes present in milk could be responsible for disease outbreaks.

→ He demonstrated that exposing substances to high temperatures effectively kills microorganisms, which is why heating milk to 70°C eliminates many harmful bacteria.

→ This scientific foundation – the presence of microbes in milk and their vulnerability to heat – explains the effectiveness of milk pasteurisation in preventing disease outbreaks.

→ The historical data presented in the graph supports this, showing significantly fewer disease outbreaks linked to pasteurised milk compared to raw milk.

→ A notable decline in raw milk-related outbreaks occurred after 1945, though this may also be attributed to decreased raw milk consumption during that period.

→ While pasteurised milk has generally proven safer, some disease outbreaks have still occurred with pasteurised products. These cases typically result from issues in the pasteurisation process itself or problems during subsequent storage and transportation of the milk.

BIOLOGY, M6 2024 HSC 19 MC

The diagram represents some experimental steps used in the production of large amounts of human growth hormone.
 

What makes this technology successful?

  1. DNA ligase can cut human growth hormone genes from human cells.
  2. Human growth hormone can cause E. coli to grow and mature rapidly.
  3. Human plasmids containing the gene of interest can be inserted into bacteria.
  4. Restriction enzymes can produce sticky ends on both bacterial and human DNA.
Show Answers Only

\(D\)

Show Worked Solution

→ The technology is possible because restriction enzymes create compatible sticky ends on both the bacterial plasmid and human DNA containing the growth hormone gene.

→ This allows them to be joined together via complementary base pairing before DNA ligase permanently connects them, which is essential for successful gene transfer.

\(\Rightarrow D\)

♦ Mean mark 53%.

BIOLOGY, M6 2024 HSC 18 MC

The following diagram models a population of glowworms in an isolated cave. The letter \(B\) and \(b\) represent the alleles for a gene in an individual.
 

Which row of the table correctly identifies the process and reason for the change in the gene pool?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Process} \rule[-1ex]{0pt}{0pt}& \textit{Reason} \\
\hline
\rule{0pt}{2.5ex}\text{Gene flow} & \text{Change in frequency} \\
& \text{of the \(b\) allele due to a mutation} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Genetic drift} & \text{Change in frequency} \\
& \text{of the \(b\) allele due to random events} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Gene flow} & \text{Introduction of more \(b\) alleles due } \\
& \text{to new members moving into the population} \rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex}\text{Genetic drift} & \text{Introduction of more \(b\) alleles due } \\
& \text{to this allele being more advantageous} \rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution

→ The diagram shows genetic drift, where the \(b\) allele becomes fixed in the population over time through random chance events in a small, isolated population (as indicated by the cave setting).

→ The context of an isolated cave means that the changes could not have been the result of natural selection or gene flow from outside populations.

\(\Rightarrow B\)

♦♦ Mean mark 33%.

BIOLOGY, M8 2024 HSC 17 MC

Over 12 months, the prevalence of a non-infectious disease will increase in a population if

  1. the total population increases.
  2. disease recovery time decreases.
  3. the incidence rate of the disease decreases.
  4. the survival time of individuals with the disease increases.
Show Answers Only

\(D\)

Show Worked Solution

→ Prevalence = number of existing cases in a population at a given time.

→ Factors affecting prevalence include new cases (incidence), duration of disease, death/recovery rates and population changes.

→ If people with the disease live longer, they remain in the “prevalent cases” count for a longer time, increasing overall prevalence.

\(\Rightarrow D\)

♦ Mean mark 51%.

BIOLOGY, M8 2024 HSC 16 MC

In a person with a particular visual disorder, light from a distant object focuses in front of the retina.
 

How can this disorder be corrected?

  1. Laser surgery to reshape the retina
  2. Use of a diverging lens in front of the eye
  3. Use of a converging lens in front of the eye
  4. Laser surgery to make the cornea more curved
Show Answers Only

\(B\)

Show Worked Solution

→ A diverging (concave) lens will spread light rays slightly before they enter the eye, moving the focal point back to the retina.

\(\Rightarrow B\)

♦ Mean mark 53%.