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HMS, BM EQ-Bank 79

Evaluate how vitamin D status affects the interrelationship between endocrine function and calcium absorption during resistance training.   (8 marks)

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Sample Answer

Evaluation Statement:

  • Vitamin D status significantly affects the endocrine-calcium relationship during resistance training.
  • This evaluation examines hormonal effects and absorption efficiency.

Hormonal Effects:

  • Vitamin D acts as a hormone in the endocrine system controlling calcium regulation.
  • It triggers release of parathyroid hormone when blood calcium is low.
  • This hormone increases calcium release from bones for muscle contraction.
  • During resistance training, adequate vitamin D ensures proper hormone signalling.
  • Evidence shows athletes with optimal vitamin D have 25% better muscle strength.
  • This strongly meets the criterion of supporting hormonal function for performance.

Absorption Efficiency:

  • The digestive system requires vitamin D to absorb calcium from food.
  • Vitamin D enables special transport proteins in the intestine to capture calcium.
  • Without adequate vitamin D, only 10-15% of dietary calcium absorbs.
  • With sufficient vitamin D, absorption increases to 30-40%.
  • Resistance training increases calcium needs for bone strengthening and muscle function.
  • This partially fulfils efficiency needs as absorption still has limits.

Final Evaluation:

  • Vitamin D status proves highly effective in connecting endocrine and digestive functions.
  • Adequate levels optimise both hormone signalling and calcium absorption essential for resistance training.
  • However, even optimal vitamin D cannot overcome very low dietary calcium intake.
  • Athletes must maintain both sufficient vitamin D and calcium consumption for maximum training benefits.
Show Worked Solution

Sample Answer

Evaluation Statement:

  • Vitamin D status significantly affects the endocrine-calcium relationship during resistance training.
  • This evaluation examines hormonal effects and absorption efficiency.

Hormonal Effects:

  • Vitamin D acts as a hormone in the endocrine system controlling calcium regulation.
  • It triggers release of parathyroid hormone when blood calcium is low.
  • This hormone increases calcium release from bones for muscle contraction.
  • During resistance training, adequate vitamin D ensures proper hormone signalling.
  • Evidence shows athletes with optimal vitamin D have 25% better muscle strength.
  • This strongly meets the criterion of supporting hormonal function for performance.

Absorption Efficiency:

  • The digestive system requires vitamin D to absorb calcium from food.
  • Vitamin D enables special transport proteins in the intestine to capture calcium.
  • Without adequate vitamin D, only 10-15% of dietary calcium absorbs.
  • With sufficient vitamin D, absorption increases to 30-40%.
  • Resistance training increases calcium needs for bone strengthening and muscle function.
  • This partially fulfils efficiency needs as absorption still has limits.

Final Evaluation:

  • Vitamin D status proves highly effective in connecting endocrine and digestive functions.
  • Adequate levels optimise both hormone signalling and calcium absorption essential for resistance training.
  • However, even optimal vitamin D cannot overcome very low dietary calcium intake.
  • Athletes must maintain both sufficient vitamin D and calcium consumption for maximum training benefits.

HMS, BM EQ-Bank 74 MC

A badminton player in a five-set match shows declining performance. Which combination of factors most likely explains this?

  1. Increased insulin, reduced glucose uptake
  2. Elevated cortisol, impaired digestion
  3. Enhanced adrenaline, improved absorption
  4. Decreased glucagon, enhanced nutrient storage
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: Prolonged stress hormones (cortisol) impair digestive function and energy availability.

Other Options:

  • A is incorrect: Insulin decreases during exercise to maintain glucose availability.
  • C is incorrect: Absorption decreases, not improves, during intense exercise.
  • D is incorrect: Glucagon increases during exercise; nutrient storage doesn’t occur during activity.

HMS, BM EQ-Bank 67

Explain how the cardiovascular system adapts to exercise at altitude (2500 metres) over both short-term (24 - 48 hours) and long-term (3+ weeks) periods.   (5 marks)

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Sample Answer

  • Reduced oxygen pressure at altitude triggers immediate cardiovascular responses within 24-48 hours.
  • Heart rate increases because the body needs to circulate blood faster to compensate for lower oxygen content.
  • Cardiac output also rises through increased stroke volume, ensuring tissues receive adequate oxygen supply.
  • These short-term changes maintain oxygen delivery to vital organs despite the thinner air.
  • Breathing rate accelerates in response to chemoreceptors detecting lower blood oxygen levels.
  • After several days, low oxygen levels stimulate the kidneys to produce EPO (erythropoietin).
  • EPO signals bone marrow to increase red blood cell production, which begins the long-term adaptation process.
  • Over 3-4 weeks, red blood cell count rises significantly, enhancing the blood’s oxygen-carrying capacity.
  • Increased haemoglobin concentration results from these higher red blood cell numbers.
  • More haemoglobin molecules enable better oxygen binding from each breath of thin air.
  • Blood vessels in tissues also increase through capillarisation, improving oxygen delivery at the cellular level.
  • Long-term adaptations therefore compensate for reduced atmospheric oxygen, allowing sustained performance at altitude.
Show Worked Solution

Sample Answer

  • Reduced oxygen pressure at altitude triggers immediate cardiovascular responses within 24-48 hours.
  • Heart rate increases because the body needs to circulate blood faster to compensate for lower oxygen content.
  • Cardiac output also rises through increased stroke volume, ensuring tissues receive adequate oxygen supply.
  • These short-term changes maintain oxygen delivery to vital organs despite the thinner air.
  • Breathing rate accelerates in response to chemoreceptors detecting lower blood oxygen levels.
  • After several days, low oxygen levels stimulate the kidneys to produce EPO (erythropoietin).
  • EPO signals bone marrow to increase red blood cell production, which begins the long-term adaptation process.
  • Over 3-4 weeks, red blood cell count rises significantly, enhancing the blood’s oxygen-carrying capacity.
  • Increased haemoglobin concentration results from these higher red blood cell numbers.
  • More haemoglobin molecules enable better oxygen binding from each breath of thin air.
  • Blood vessels in tissues also increase through capillarisation, improving oxygen delivery at the cellular level.
  • Long-term adaptations therefore compensate for reduced atmospheric oxygen, allowing sustained performance at altitude.

HMS, BM EQ-Bank 63

Analyse how the interrelationship between the respiratory and circulatory systems can contribute to improved endurance performance in athletes.   (8 marks)

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Sample Answer

Overview Statement

  • The respiratory and circulatory systems demonstrate interconnected adaptations that enhance endurance performance.
  • Key components include lung capacity, oxygen transport, gas exchange efficiency, and cellular adaptations.
  • Performance improvements result from the synergistic relationship between both systems.

Respiratory Adaptations and Oxygen Uptake

  • Training increases vital capacity and breathing efficiency, enhancing oxygen uptake at the alveolar level.
  • Stronger respiratory muscles enable sustained ventilation during prolonged exercise.
  • Greater lung volumes allow more air to be processed with each breath.
  • Enhanced respiratory function provides the foundation for improved oxygen availability.

Circulatory Adaptations and Delivery

  • Increased stroke volume and capillarisation improve oxygen transport to working muscles.
  • Higher stroke volume means more blood pumped per heartbeat.
  • Denser capillary networks create greater surface area for oxygen delivery.
  • Circulatory improvements directly interact with respiratory gains for compound benefits.

System Integration and Efficiency

  • Ventilation-perfusion matching becomes more precise through training, optimising gas exchange.
  • Blood flow aligns with alveolar ventilation at the lung level.
  • Trained athletes extract more oxygen from each breath due to improved matching.
  • Such synchronisation demonstrates true system integration for performance enhancement.

Implications for Endurance Performance

  • Mitochondrial density increases in response to improved oxygen delivery.
  • Both systems demonstrate reciprocal enhancement through training adaptations.
  • Respiratory improvements enable greater circulatory adaptations and vice versa.
  • Interdependence between systems multiplies individual gains for superior endurance capacity.
Show Worked Solution

Sample Answer

Overview Statement

  • The respiratory and circulatory systems demonstrate interconnected adaptations that enhance endurance performance.
  • Key components include lung capacity, oxygen transport, gas exchange efficiency, and cellular adaptations.
  • Performance improvements result from the synergistic relationship between both systems.

Respiratory Adaptations and Oxygen Uptake

  • Training increases vital capacity and breathing efficiency, enhancing oxygen uptake at the alveolar level.
  • Stronger respiratory muscles enable sustained ventilation during prolonged exercise.
  • Greater lung volumes allow more air to be processed with each breath.
  • Enhanced respiratory function provides the foundation for improved oxygen availability.

Circulatory Adaptations and Delivery

  • Increased stroke volume and capillarisation improve oxygen transport to working muscles.
  • Higher stroke volume means more blood pumped per heartbeat.
  • Denser capillary networks create greater surface area for oxygen delivery.
  • Circulatory improvements directly interact with respiratory gains for compound benefits.

System Integration and Efficiency

  • Ventilation-perfusion matching becomes more precise through training, optimising gas exchange.
  • Blood flow aligns with alveolar ventilation at the lung level.
  • Trained athletes extract more oxygen from each breath due to improved matching.
  • Such synchronisation demonstrates true system integration for performance enhancement.

Implications for Endurance Performance

  • Mitochondrial density increases in response to improved oxygen delivery.
  • Both systems demonstrate reciprocal enhancement through training adaptations.
  • Respiratory improvements enable greater circulatory adaptations and vice versa.
  • Interdependence between systems multiplies individual gains for superior endurance capacity.

HMS, BM EQ-Bank 57

Analyse how the structure of the respiratory and circulatory systems work together to support performance in a rock climber during a difficult ascent.  (8 marks)

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Sample Answer

Overview Statement

  • Rock climbing demands unique respiratory and circulatory adaptations due to body positioning and sustained muscle contractions.
  • Key components include respiratory muscles, capillary networks, heart structure, and blood flow regulation.
  • Performance depends on how these systems adapt to climbing-specific challenges.

Respiratory Adaptations During Compression

  • The diaphragm and intercostal muscles must function despite chest compression against rock faces.
  • Enhanced respiratory muscle strength enables breathing in restricted positions.
  • Chest wall flexibility allows sufficient lung expansion even when compressed.
  • Such adaptations ensure adequate oxygen intake throughout challenging postures.

Capillary Networks and Grip Endurance

  • Extensive capillarisation in forearm muscles meets extreme grip demands during climbing.
  • Dense capillary networks deliver oxygen during sustained isometric contractions.
  • Blood flow increases dramatically in active forearm muscles during difficult holds.
  • Vascular density directly influences grip endurance and climbing duration.

Heart Structure and Positional Changes

  • The four-chamber heart structure coordinates with rapid positional changes during climbing.
  • One-way valves prevent blood pooling when transitioning to inverted positions.
  • Rapid cardiovascular adjustments maintain circulation from vertical to overhang positions.
  • Structural features ensure continuous oxygen delivery regardless of body orientation.

Integrated System Response

  • Pulmonary circulation adapts to varied thoracic pressures during climbing movements.
  • Systemic circulation prioritises blood flow through intermittent vessel dilation and constriction.
  • Recovery between moves allows repayment of oxygen debt from sustained holds.
  • Combined adaptations determine overall climbing performance and ascent sustainability.
Show Worked Solution

Sample Answer

Overview Statement

  • Rock climbing demands unique respiratory and circulatory adaptations due to body positioning and sustained muscle contractions.
  • Key components include respiratory muscles, capillary networks, heart structure, and blood flow regulation.
  • Performance depends on how these systems adapt to climbing-specific challenges.

Respiratory Adaptations During Compression

  • The diaphragm and intercostal muscles must function despite chest compression against rock faces.
  • Enhanced respiratory muscle strength enables breathing in restricted positions.
  • Chest wall flexibility allows sufficient lung expansion even when compressed.
  • Such adaptations ensure adequate oxygen intake throughout challenging postures.

Capillary Networks and Grip Endurance

  • Extensive capillarisation in forearm muscles meets extreme grip demands during climbing.
  • Dense capillary networks deliver oxygen during sustained isometric contractions.
  • Blood flow increases dramatically in active forearm muscles during difficult holds.
  • Vascular density directly influences grip endurance and climbing duration.

Heart Structure and Positional Changes

  • The four-chamber heart structure coordinates with rapid positional changes during climbing.
  • One-way valves prevent blood pooling when transitioning to inverted positions.
  • Rapid cardiovascular adjustments maintain circulation from vertical to overhang positions.
  • Structural features ensure continuous oxygen delivery regardless of body orientation.

Integrated System Response

  • Pulmonary circulation adapts to varied thoracic pressures during climbing movements.
  • Systemic circulation prioritises blood flow through intermittent vessel dilation and constriction.
  • Recovery between moves allows repayment of oxygen debt from sustained holds.
  • Combined adaptations determine overall climbing performance and ascent sustainability.

HMS, BM EQ-Bank 46

How does correct joint alignment help to prevent injury during weight-bearing activities.   (5 marks)

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Sample Answer

Force Distribution Through Joint Surfaces

  • Correct alignment positions bones so that weight-bearing forces spread evenly across entire joint surface.
  • This even distribution occurs because aligned bones create uniform contact between joint surfaces.
  • As a result, cartilage experiences balanced compression rather than concentrated pressure points, preventing localised wear and degradation of specific cartilage areas.
  • Misalignment creates high-stress zones which leads to damaged cartilage and eventual osteoarthritis.

Ligament and Tendon Protection

  • Proper joint positioning maintains ligaments and tendons within optimal length ranges by keeping anatomical relationships correct.
  • This positioning enables these structures to handle loads at appropriate angles.
  • Consequently, ligaments avoid overstretching which prevents tears and chronic laxity.
  • Correct alignment ensures tendons track smoothly through anatomical pathways by maintaining proper bone positions.
  • This smooth tracking prevents friction and inflammation from abnormal movement patterns.

Muscular Efficiency and Support

  • Joint alignment enables muscles to operate at ideal length-tension relationships through optimal positioning.
  • This positioning allows maximum force production while minimising energy expenditure.
  • As a result, efficient muscle function provides dynamic stabilisation during activities.
  • Well-aligned joints create balanced muscle activation where opposing groups share loads appropriately.
  • This balanced activation prevents single muscles from overworking which reduces strain injury risk.
  • Proper positioning eliminates compensatory movements thereby preventing cascade effects throughout kinetic chain.
Show Worked Solution

Sample Answer

Force Distribution Through Joint Surfaces

  • Correct alignment positions bones so that weight-bearing forces spread evenly across entire joint surface.
  • This even distribution occurs because aligned bones create uniform contact between joint surfaces.
  • As a result, cartilage experiences balanced compression rather than concentrated pressure points, preventing localised wear and degradation of specific cartilage areas.
  • Misalignment creates high-stress zones which leads to damaged cartilage and eventual osteoarthritis.

Ligament and Tendon Protection

  • Proper joint positioning maintains ligaments and tendons within optimal length ranges by keeping anatomical relationships correct.
  • This positioning enables these structures to handle loads at appropriate angles.
  • Consequently, ligaments avoid overstretching which prevents tears and chronic laxity.
  • Correct alignment ensures tendons track smoothly through anatomical pathways by maintaining proper bone positions.
  • This smooth tracking prevents friction and inflammation from abnormal movement patterns.

Muscular Efficiency and Support

  • Joint alignment enables muscles to operate at ideal length-tension relationships through optimal positioning.
  • This positioning allows maximum force production while minimising energy expenditure.
  • As a result, efficient muscle function provides dynamic stabilisation during activities.
  • Well-aligned joints create balanced muscle activation where opposing groups share loads appropriately.
  • This balanced activation prevents single muscles from overworking which reduces strain injury risk.
  • Proper positioning eliminates compensatory movements thereby preventing cascade effects throughout kinetic chain.

HMS, BM EQ-Bank 31

Explain the role of major muscles in performing a deadlift.
  

In your response, identify the types of muscle contractions occurring and explain how these muscles work together to execute the movement safely.   (5 marks)

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Sample Answer

  • During the lifting phase, multiple muscle groups work simultaneously. Erector spinae muscles contract isometrically, maintaining a rigid spine position. This prevents dangerous spinal flexion under load. Meanwhile, gluteus maximus and hamstrings perform concentric contractions to extend the hips. Similarly, quadriceps contract concentrically to extend the knees.
  • These coordinated actions create the upward force needed to lift the weight. The reason for simultaneous activation is load distribution – sharing the work prevents any single muscle group from overloading. Additionally, trapezius muscles contract isometrically to stabilise the shoulder girdle and maintain bar position.
  • In the lowering phase, the same muscles perform eccentric contractions. This controlled lengthening prevents the weight from dropping suddenly. Hamstrings and glutes gradually lengthen while maintaining tension, which protects the lower back from sudden loading.
  • Throughout both phases, core muscles (rectus abdominis, transverse abdominis) maintain isometric contraction. This continuous bracing protects the spine and enables efficient force transfer. Therefore, coordinated muscle contractions ensure both effective lifting and injury prevention.
Show Worked Solution

Sample Answer

  • During the lifting phase, multiple muscle groups work simultaneously. Erector spinae muscles contract isometrically, maintaining a rigid spine position. This prevents dangerous spinal flexion under load. Meanwhile, gluteus maximus and hamstrings perform concentric contractions to extend the hips. Similarly, quadriceps contract concentrically to extend the knees.
  • These coordinated actions create the upward force needed to lift the weight. The reason for simultaneous activation is load distribution – sharing the work prevents any single muscle group from overloading. Additionally, trapezius muscles contract isometrically to stabilise the shoulder girdle and maintain bar position.
  • In the lowering phase, the same muscles perform eccentric contractions. This controlled lengthening prevents the weight from dropping suddenly. Hamstrings and glutes gradually lengthen while maintaining tension, which protects the lower back from sudden loading.
  • Throughout both phases, core muscles (rectus abdominis, transverse abdominis) maintain isometric contraction. This continuous bracing protects the spine and enables efficient force transfer. Therefore, coordinated muscle contractions ensure both effective lifting and injury prevention.

HMS, TIP EQ-Bank 4 MC

A sprinter is working to improve their block start performance. Which biomechanical sequence would most effectively generate maximal horizontal force from the blocks?

\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}{|l|l|l|}
\hline
\rule{0pt}{2.5ex}\textbf{Rear Leg Action}\rule[-1ex]{0pt}{0pt}& \textbf{Front Leg Action}& \textbf{Trunk Position} \\
\hline
\rule{0pt}{2.5ex}\text{Concentric hip extension}\rule[-1ex]{0pt}{0pt}&\text{Isometric knee extension}&\text{Forward lean 45°}\\
\hline
\rule{0pt}{2.5ex}\text{Eccentric knee extension}\rule[-1ex]{0pt}{0pt}& \text{Concentric hip extension}&\text{Forward lean 30°}\\
\hline
\rule{0pt}{2.5ex}\text{Concentric knee extension}\rule[-1ex]{0pt}{0pt}& \text{Concentric hip extension}&\text{Forward lean 90°} \\
\hline
\rule{0pt}{2.5ex}\text{Concentric hip and knee extension}\rule[-1ex]{0pt}{0pt}& \text{Concentric hip and knee extension}&\text{Forward lean 60°} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: A forward lean of 60° optimises horizontal force production. Simultaneous concentric hip and knee extension in both legs maximises power. The combination creates the most effective angle of force application.

Other options:

  • A, B and C incorrect: Show incorrect sequence to achieve desired outcome

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)

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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).

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.
♦ Mean mark 50%.

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%.

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

  \(g^{\prime}(x)\) \(=\dfrac{1}{3}(x-k)^{-\frac{2}{3}}\)
  \(g^{\prime}(0)\) \(=\dfrac{1}{3}(0-k)^{-\frac{2}{3}}=\dfrac{1}{3}(-k)^{-\frac{2}{3}} =\dfrac{1}{3}(k)^{-\frac{2}{3}}\)
♦ Mean mark (b) 39%.
c.     \(\text{When }y=0:\)
     \(-\sqrt[3]{k}+m\) \(=0\)  
  \(\sqrt[3]{k}\) \(=m\)  
  \(k\) \(=m^3\)  
♦ Mean mark (c) 47%.
d.     \(g^{\prime}(x)\) \(=g^{\prime}(0)\)
  \(\dfrac{1}{3}(x-k)^{-\frac{2}{3}}\) \(=\dfrac{1}{3(-k)^{\frac{2}{3}}}\)
  \(\dfrac{1}{(x-k)^{\frac{2}{3}}}\) \(=\dfrac{1}{k^{\frac{2}{3}}}\)
  \((x-k)^{\frac{2}{3}}\) \(=k^{\frac{2}{3}}\)
  \((x-k)^2\) \(=k^2\)
  \(x-k\) \(=\pm k\)
  \(x\) \(=0, \ 2k\)
♦♦♦ Mean mark (d) 15%.

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

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

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

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

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

\(f^{\prime}(x)=0\ \text{at some point in the interval.}\)

\(\therefore\ \text{A stationary point must exist in the given range.}\)

c. 

Show Worked Solution

a.     \(A\) \(=\dfrac{\pi}{3}\times\dfrac{1}{2}\left(f(0)+2f\left(\dfrac{\pi}{3}\right)+2f\left(\dfrac{2\pi}{3}\right)+f(\pi)\right)\)
    \(=\dfrac{\pi}{6}\left(0+2\times \dfrac{\pi}{3}\sin\left(\dfrac{\pi}{3}\right)+2\times \dfrac{2\pi}{3}\sin\left(\dfrac{2\pi}{3}\right)+\pi\sin({\pi})\right)\)
    \(=\dfrac{\pi}{6}\left(2\times \dfrac{\pi}{3}\times\dfrac{\sqrt{3}}{2}+2\times\dfrac{2\pi}{3}\times \dfrac{\sqrt{3}}{2}+0\right)\)
    \(=\dfrac{\pi}{6}\left(\dfrac{2\pi\sqrt{3}}{6}+\dfrac{4\pi\sqrt{3}}{6}\right)\)
    \(=\dfrac{\pi}{6}\times \dfrac{6\pi\sqrt{3}}{6}\)
    \(=\dfrac{\sqrt{3}\pi^2}{6}\)
♦ Mean mark (a) 48%.
bi.    \(f(x)\) \(=x\sin(x)\)
  \(f^{\prime}(x)\) \(=x\cos(x)+\sin(x)\)

 

b.ii.  \(\text{Gradient in given range gradually decreases.}\)

\(\text{Range of}\ f^{\prime}(x)\ \text{will be defined by the endpoints.}\)

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

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

♦♦♦ Mean mark (b.ii.) 20%.
♦♦♦ Mean mark (b.iii.) 12%.

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

\(f^{\prime}(x)=0\ \text{at some point in the interval.}\)

\(\therefore\ \text{A stationary point must exist in the given range.}\)

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

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

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

♦♦ Mean mark (c) 36%.

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.
    2. Use  \(z=2\)  for this confidence interval.   (2 marks)
    3. A new sample of \(n\) people results in the same sample proportion.
    4. 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.   \(\text{Transformations:}\)

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

\(\text{2- Translation of 1500 units upwards}\)

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

♦♦♦ Mean mark (a) 17%.

\(\text{Given}\ \ h(0)=h_1(0):\)

\(\dfrac{1500}{0+1} +C= 3000\ \ \Rightarrow\ \ C=1500\)

\(h_1(t)=\dfrac{1500}{t+1}+1500\)
 

\(\text{Transformations:}\)

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

\(\text{2- Translation of 1500 units upwards}\)
 

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.   \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\dfrac{6}{25n}\) \(=\dfrac{2}{2500}\)
  \(\dfrac{25n}{6}\) \(=1250\)
  \(n\) \(=\dfrac{6}{25}\times 1250\)
    \(=300\)
♦♦ Mean mark (c.ii.) 33%.

PHYSICS, M6 2024 HSC 18 MC

The diagram shows a magnet moving towards a coil \(X\).
 

This action causes a current to be induced in the coil.

Which situation will induce a current in coil \(X\) that is in the same direction as the current induced by the movement of the magnet?
 

Show Answers Only

\(D\)

Show Worked Solution
  • As the north pole of the magnet moves towards coil \(X\), the magnetic flux through \(X\) is increased. By Lenz’s law, a current will be generated in the coil that produces a magnetic field to oppose the increase the flux. Using Lenz’s law, the current in \(X\) will generate a north pole to the left to repel the approaching magnet. The induced current will run up the coils as viewed from the front. 

  • The current will be induced in the same direction in \(D\). Using the right hand grip rule, the current though the white coil will produce magnetic field lines gong to the left. As the current is decreasing, the change of flux through \(X\) is decreasing. Therefore by Lenz’s law, the current induced in \(X\) will strengthen the magnetic field to oppose the decreasing magnetic flux. 
  • Therefore, the magnetic field lines produced by the current in \(X\) will also be to the left, and using the right hand grip rule, the current through \(X\) will run up the coils (front view).
  • In all the other options, the current will run down the coils (front view).

\(\Rightarrow D\)

♦♦♦ Mean mark 28%.

Calculus, MET2 2024 VCAA 17 MC

Consider the algorithm below, which prints the roots of the cubic polynomial  \(f(x)=x^3-2 x^2-9 x+18\).

\begin{array} {l}
\rule{0pt}{2.5ex} \textbf{ define} \ \text{ f (x) }  \\
\rule{0pt}{2.5ex} \quad \quad \textbf{return} \ \text{(x} ^3 - 2  \text{x}^2 - 9 \text{x} + 18) \\
\rule{0pt}{2.5ex}  \text{c} \leftarrow \text{f} \ (0) \\
\rule{0pt}{2.5ex} \textbf{if}\  \ \text{c < 0} \ \textbf{then}\\
\rule{0pt}{2.5ex} \quad \quad \text{c} \ \leftarrow \ \text{(- c)} \\
\rule{0pt}{2.5ex} \textbf{end if} \\
\rule{0pt}{2.5ex} \textbf{while} \ \text{ c > 0 } \\
\rule{0pt}{2.5ex} \quad \quad \textbf{if} \ \ \text{f (c) = 0 } \ \textbf{then}  \\
\rule{0pt}{2.5ex} \quad \quad \quad \quad \textbf{print} \ \text{c }   \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{end if} \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{if} \ \ \text{f (-c) = 0} \ \textbf{then} \\
\rule{0pt}{2.5ex} \quad \quad \quad \quad \textbf{print} \ \text{-c }   \\
\rule{0pt}{2.5ex} \quad \quad  \textbf{end if} \\
\rule{0pt}{2.5ex} \quad \quad  \text{c} \ \leftarrow \ \text{c - 1} \\
\rule{0pt}{2.5ex} \textbf{ end while } \\
\end{array}

In order, the algorithm prints the values

  1. \(-3, 3, 2\)
  2. \(-3, 2, 3\)
  3. \(3, 2, -3\)
  4. \(3, -3, 2\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Solve}\ \ \ x^3-2 x^2-9 x+18=0\ \ \text{(using CAS):}\)

\(\text{Roots are:}\ \ \ x=3,\ 2, -3\)

\(\text{Using the algorithm the order is }\ c, -c,\ c-1\ \rightarrow \ 3, -3,\ 2\)

\(\Rightarrow D\)

♦♦♦ Mean mark 27%.

PHYSICS, M7 2024 HSC 20 MC

Three identical atomic clocks are made so that they tick at precisely the same rate. One is kept in a laboratory, \(X\), on Earth's equator. Another is placed on board a satellite, \(Y\), in a circular orbit with a period of 12 hours. A third is placed in a satellite, \(Z\), that is in a geostationary orbit. The satellites orbit Earth in the equatorial plane.
 

Assume that the satellites are inertial frames of reference and the clocks are affected ONLY by the predictions of special relativity.

Which statement correctly compares the rates at which the clocks tick, as determined by an observer at \(X\), when the satellites are in the positions shown in the diagram?

  1. The clock at \(Y\) ticks faster than either the clock at \(X\) or the clock at \(Z\).
  2. The clock at \(Y\) ticks slower than either the clock at \(X\) or the clock at \(Z\).
  3. The clocks tick at different rates, with \(X\) being the fastest and \(Y\) being the slowest.
  4. The clocks tick at different rates, with \(Z\) being the slowest and \(X\) being the fastest.
Show Answers Only

\(B\)

Show Worked Solution
  • Using Einstein’s special theory of relativity in relation to time dilation, the faster a clock travels relative to a stationary observer, the slower time moves for the clock. This means that the clock moving relative to the observer will tick more slowly.
  • The clock placed at \(X\) will be stationary relative to the observer at \(X\).
  • This is also true for the clock placed at \(Z\). As the clock being placed in a satellite which is in a geostationary orbit, the satellite will appear to be stationary in the sky. Therefore, the observer at \(X\) is in the same frame of reference as the clock at \(Z\) and no effects of time dilation will be observed.
  • The clock at \(Y\) has a period of 12 hours, hence it must have a smaller orbital radius and so a higher linear velocity than \(Z\). Thus the clock at \(Y\) would be moving faster as seen by the observer at \(X\) which is in the same frame of reference as \(Z\). Therefore, the observer at \(X\) would see the clock at \(Y\) tick slower than the clocks at \(X\) and \(Z\) due to the effects of time dilation.

\(\Rightarrow B\)

Note: As the question assumes that all satellites are in inertial frames of reference, students can discount the rotational velocities of the satellites and the centripetal forces of gravity on the satellites, effectively treating the Earth as flat.

♦♦♦ Mean mark 21%.

PHYSICS, M6 2024 HSC 19 MC

In a vacuum chamber there is a uniform electric field and a uniform magnetic field.

A proton having a velocity, \(v\), enters the chamber. Its velocity remains unchanged as it travels through the chamber.

A second proton having a velocity, \(2v\), in the same direction as the first proton, then enters the chamber at the same point as the first proton.

In the chamber, the acceleration of the second proton

  1. is zero.
  2. is constant in magnitude and direction.
  3. changes in both magnitude and direction.
  4. is constant in magnitude, but not direction.
Show Answers Only

\(C\)

Show Worked Solution
  • The first proton, with velocity \(v\), travels through with no deflection. The net force acting on the proton is zero.
  • However, as the second proton has a velocity of \(2v\), \(F_B > F_E\) and the initial direction of \(F_B\) will be in the opposite direction of \(F_E\). This will cause the second proton to undergo circular motion (force is perpendicular to the velocity). 
  • The acceleration of the second proton will change direction as the centripetal force/acceleration will act towards the centre of the circular path that the second proton undertakes.
  • The magnitude of the acceleration of the second proton will also change. Initially \(F_B\) opposes \(F_E\) and \(F_{\text{net}}\) is at a minimum. As the direction of \(F_B\) changes, \(F_{\text{net}}\) will increase as \(F_E\) will not directly oppose \(F_B\).
  • As the magnitude of the net force on the second proton increases, the magnitude of the acceleration on the second proton will also increase.

\(\Rightarrow C\)

♦♦♦ Mean mark 26%.

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, 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. 

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.
♦♦ Mean mark 48%.

CHEMISTRY, M5 2024 HSC 39

Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid \(\left(\ce{BrCH2COOH}\right)\) is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.

\(\ce{BrCH2COOH(aq) \rightleftharpoons BrCH2COOH}\textit{(octan-l-ol)}\)

The equilibrium constant expression for this system is

\(K_{e q}=\dfrac{\left[\ce{BrCH2COOH}\textit{(octan-l-ol)}\right]}{\left[\ce{BrCH2COOH}\textit{(aq)}\right]}\).

An aqueous solution of bromoacetic acid with an initial concentration of 0.1000 mol L \(^{-1}\) is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with \(K_a=1.29 \times 10^{-3}\). When the system has reached equilibrium, the \(\left[\ce{H+}\right]\) is \(9.18 \times 10^{-3} \text{ mol L}^{-1}\).

Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the \(K_{eq}\) for the octan-1-ol and water system.   (4 marks)

Show Answers Only

\(0.390\) 

Show Worked Solution

  • The ionisation of bromoacetic acid in water is:
  •    \(\ce{BrCH2COOH(aq) \rightleftharpoons BrCH2COO^-(aq) + H^+(aq)}\)
  • At equilibrium \(\ce{[BrCH2COO^-(aq)] = [H^+(aq)]} = 9.18 \times 10^{-3}\ \text{mol L}^{-1}\) as the are formed in a \(1:1\).
\(K_{a}\) \(=\dfrac{\ce{[H^+][BrCH2COO^-]}}{\ce{[BrCH2COOH]_{eq}}}\)  
\(\ce{[BrCH2COOH]_{eq}}\) \(=\dfrac{\ce{[H^+][BrCH2COO^-]}}{K_a}\)  
  \(=\dfrac{(9.18 \times 10^{-3})^2}{1.29 \times 10^{-3}}\)  
  \(=0.06533\ \text{mol L}^{-1}\)  

  

\(\ce{[BrCH2COOH]_{\text{total}}}=\ce{[BrCH2COOH(aq)]_{eq} + [BrCH2COO^-(aq)]}\)

\(\ce{+ [BrCH2COOH(octan-1-ol)]_{eq}}\)

\(\ce{[BrCH2COOH(octan-1-ol)]_{eq}}\) \(=0.1000-0.06533-9.18 \times 10^{-3}\)  
  \(=0.02549\ \text{mol L}^{-1}\)    
     
  • Since the volume of the aqueous solution of bromoacetic acid and octane is the same, the concentration values between the water and octane solutions can be added/subtracted in one equation and mole calculations are not required.
  •    \(K_{eq} = \dfrac{\ce{[BrCH2COOH(octan-1-ol)]_{eq}}}{\ce{[BrCH2COOH(aq)]_{eq}}}= \dfrac{0.02549}{0.06533}=0.390\ \text{(3 sig. fig.)}\)
♦♦ Mean mark 27%.
COMMENT: Students who identified the acid conc in the organic solvent often succeeded in this question.

Networks, GEN2 2024 VCAA 15

An upgrade to the supermarket requires the completion of 11 activities, \(A\) to \(K\).

The directed network below shows these activities and their completion time, in weeks.

The minimum completion time for the project is 29 weeks.
 

 

  1. Write down the critical path.   (1 mark)

  2. Which activity can be delayed for the longest time without affecting the minimum completion time of the project?   (1 mark)

Use the following information to answer parts c-e.

A change is made to the order of activities.

The table below shows the activities and their new latest starting times in weeks.

\begin{array}{|c|c|}
\hline
\textbf{Activity} & \textbf{Latest Starting}\\
&\textbf{time} \text{(weeks)}\\
\hline A & 0 \\
\hline B & 2 \\
\hline C & 10 \\
\hline D & 9 \\
\hline E & 13 \\
\hline F & 14 \\
\hline G & 18 \\
\hline H & 17 \\
\hline I & 19 \\
\hline J & 25 \\
\hline K & 22 \\
\hline
\end{array}

A dummy activity is now required in the network.

  1. On the directed network below, draw a directed edge to represent the dummy activity. Include a label.  (1 mark)

     

  1. What is the new minimum completion time of the project?  (1 mark)
  2. The owners of the supermarket want the project completed earlier.
  3. They will pay to reduce the time of some of the activities.
  4. A reduction in completion time of an activity will incur an additional cost of $10 000 per week.
  5. Activities can be reduced by a maximum of two weeks.
  6. The minimum number of weeks an activity can be reduced to is seven weeks.
  7. What is the minimum amount the owners of the supermarket will have to pay to reduce the completion time of the project as much as possible?   (1 mark)

Show Answers Only

a.    \(A, C, H, J\)

b.    \(\text{Activity E}\)

c.    

d.    \(\text{30 weeks}\)

e.    \($50\,000\)

Show Worked Solution

a.    \(\text{Critical path:  }A, C, H, J\)
 

 
b.    \(\text{Activity with the largest float time can be delayed the longest.}\)

\(\text{Consider Activity E:}\)

\(\text{EST = 11, LST}= 18-4=14\rightarrow\ \text{Float time = 3 weeks}\)

\(\therefore\ \text{Activity E can be delayed the longest.}\)
 

♦♦ Mean mark (b) 34%.

c.    

♦♦♦ Mean mark (c) 10%.

d.    \(\text{New minimum completion time is 30 weeks.}\)
 

♦♦♦ Mean mark (d) 27%.

e.    \(\text{Activities that can be reduced:}\)

\(-A\ \text{can be reduced by 2 weeks}\)

\(-B, D\ \text{can each be reduced by 1 week each}\)

\(-H\ \text{can be reduced by 1 week}\)

\(\text{Total reduction = 5 weeks}\)

\(\Rightarrow \ \text{Minimum payment}=$50\,000\)

♦♦♦ Mean mark (e) 7%.

Networks, GEN2 2024 VCAA 14

A manufacturer \((M)\) makes deliveries to the supermarket \((S)\) via a number of storage warehouses, \(L, N, O, P, Q\) and \(R\). These eight locations are represented as vertices in the network below.

The numbers on the edges represent the maximum number of deliveries that can be made between these locations each day.
 

  1. When considering the possible flow of deliveries through this network, many different cuts can be made.   
  2. Determine the capacity of Cut 1, shown above.   (1 mark)

  3. Determine the maximum number of deliveries that can be made each day from the manufacturer to the supermarket.   (1 mark)

  4. The manufacturer wants to increase the number of deliveries to the supermarket.
  5. This can be achieved by increasing the number of deliveries between one pair of locations.
  6. Complete the following sentence by writing the locations on the lines provided:
  7. To maximise this increase, the number of deliveries should be increased between
    locations ____ and  ____.       (1 mark)

    

Show Answers Only

a.    \(46\)

b.    \(37\)

c.    \(\text{R and S}\)

Show Worked Solution

a.    \(13+18+6+9=46\)

\(\text{(Reverse flow}\ Q → O\ \text{is not counted.)}\)
 

b.  

\(\text{Max deliveries (min cut)}\ =13+5+11+8=37\)

♦ Mean mark (b) 29%.

 
c.   \(\text{The number of deliveries should be increased between}\)

\(\text{locations R and S.}\)

♦ Mean mark (c) 22%.

Matrices, GEN2 2024 VCAA 12

When the construction company established the construction site at the beginning of 2023, it employed 390 staff to work on the site.

The staff comprised 330 construction workers \((C)\), 50 foremen \((F)\) and 10 managers \((M)\).

At the beginning of each year, staff can choose to stay in the same job, move to a different job on the site, or leave the site \((L)\) and not return.

The transition diagram below shows the proportion of staff who are expected to change their job at the site each year.
 

This situation can be modelled by the recurrence relation

\(S_{n+1}=T S_n\), where

\(T\) is the transitional matrix, \(S_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L \end{aligned}\)  and \(n\) is the number of years after 2023.

  1. Calculate the predicted percentage decrease in the number of foremen \((F)\) on the site from 2023 to 2025.   (1 mark)

  2. Determine the total number of staff on the site in the long term.  (1 mark)

To encourage more construction workers \((C)\) to stay, the construction company has given workers an incentive to move into the job of foreman \((F)\).

Matrix \(R\) below shows the ways in which staff are expected to change their jobs from year to year with this new incentive in place.

\begin{aligned}
& \quad \quad \ \ \textit{this year} \\
& \quad  C \quad  \ \ F \quad  \ \  M \quad L\\
R = & \begin{bmatrix}
0.4 & 0.2 & 0 & 0 \\
0.4 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.2 & 0.4 & 0.3 & 1
\end{bmatrix}\begin{array}{l}
C\\
F\\
M\\
L
\end{array} \quad \textit{next year}
\end{aligned}

The site always requires at least 330 construction workers.

To ensure that this happens, the company hires an additional 190 construction workers \((C)\) at the beginning of 2024 and each year thereafter.

The matrix  \(V_{n+1}\)  will then be given by

\(V_{n+1}=R V_n+Z\), where

\(V_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \quad\quad\quad Z=\left[\begin{array}{c}190 \\ 0 \\ 0 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \ \ \) and \(n\) is the number of years after 2023.

  1. How many more staff are there on the site in 2024 than there were in 2023 ?    (1 mark)

  2. Based on this new model, the company has realised that in the long term there will be more than 200 foremen on site.
  3. In which year will the number of foremen first be above 200?   (1 mark)

Show Answers Only

a.    \(14\%\)

b.    \(\text{Zero}\)

c.    \(101\)

d.    \(2027\)

Show Worked Solution

a.    \(\text{Using CAS:}\)

\(\text{Transition matrix}(T):\ \begin{aligned}
& \quad \quad \quad \ \ \ \ \textit{from} \\
& \quad \ \ \ \  C \ \ \ \ \  F \ \ \ \ \ \  M \ \ \ \ \ \  L \ \ \\
\ \ \textit{to}\ \ \ & \begin{array}{l}
C \\
F \\
M \\
L
\end{array}\begin{bmatrix}
0.3 & 0.2 & 0 & 0 \\
0.2 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.5 & 0.4 & 0.3 & 1
\end{bmatrix}
\end{aligned}\)

♦♦♦ Mean mark (a) 25%.

\(S_{2023}=\begin{bmatrix}
330 \\
50 \\
10 \\
0 \\
\end{bmatrix}\ \ ,\ \  S_{2024}=T\times S_{2023}=\begin{bmatrix}
109 \\
80 \\
13 \\
188 \\
\end{bmatrix}\)

\( S_{2025}=T\times S_{2024}=\begin{bmatrix}
48.7 \\
43 \\
19.9 \\
278.4 \\
\end{bmatrix}\)

 
\(\text{% decrease in foremen}\ (F)=\dfrac{50-43}{50}\times 100\%=14\%\)
 

b.   \(\text{Test for 10 years:}\)

\(\text{Using CAS:}\)

\(S_{2024}=T^{10}\times S_{2023}=\begin{bmatrix}
0.490748 \\
0.734232 \\
0.488858 \\
388.286 \\
\end{bmatrix}\)

 
\(\text{In the long term, there will be zero employees on site.}\)
 

♦♦♦ Mean mark (b) 24%.

c.    \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\)
  

\(\text{Difference in staff from}\ 2023-2024\)

\(=(332+146+13)-390\)

\(=101\ \text{more staff.}\)
 

\(\text{NOTE: 89 not included in calculation as these are the staff}\)

\(\text{who have left the company during the year.}\)
 

♦♦♦ Mean mark (c) 12%.

d.   \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\ \ ,\ \  V_{2025}=R\times V_{2024}+Z=\begin{bmatrix}
352 \\
167.2 \\
33.1 \\
217.7 \\
\end{bmatrix}\ \\\)

♦♦♦ Mean mark (d) 26%.

\(V_{2026}=R\times V_{2025}+Z=\begin{bmatrix}
364.24\\
187.48 \\
43.37 \\
364.91 \\
\end{bmatrix}\ \ ,\ \  V_{2027}=R\times V_{2026}+Z=\begin{bmatrix}
373.192 \\
200.54 \\
50.507 \\
525.761 \\
\end{bmatrix}\ \\\)

 
\(\therefore\ \text{In 2027 the number of foremen will be over 200.}\)

Matrices, GEN2 2024 VCAA 11

A population of a native animal species lives near the construction site.

To ensure that the species is protected, information about the initial female population was collected at the beginning of 2023. The birth rates and the survival rates of the females in this population were also recorded.

This species has a life span of 4 years and the information collected has been categorised into four age groups: 0-1 year, 1-2 years, 2-3 years, and 3-4 years.

This information is displayed in the initial population matrix, \(R_0\), and the Leslie matrix, \(L\), below.

\(R_0=\left[\begin{array}{c}70 \\ 80 \\ 90 \\ 40\end{array}\right] \quad \quad L=\left[\begin{array}{cccc}0.4 & 0.75 & 0.4 & 0 \\ 0.4 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.5 & 0\end{array}\right]\)

  1. Using the information above
  2.  i. complete the following transition diagram.   (1 mark) 

  1. ii. complete the following table, showing the initial female population, and the predicted female population after one year, for each of the age groups.  (1 mark)  

  1. It is predicted that if this species is not protected, the female population of each of the four age groups will rapidly decrease within the next 10 years.
  2. After how many years is it predicted that the total female population of this species will first be half the initial female population?   (1 mark)

Show Answers Only

a.i. 

a.ii.

  \(\textbf{Age Group}\)
  \(\ 0-1\ \text{year}\ \) \(\ 1-2\ \text{years}\ \) \(\ 2-3\ \text{years}\ \) \(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \) \(70\) \(80\) \(90\) \(40\)
\(\ \text{Population after}\ \)
\(\ \text{one year}\)		 \(124\) \(28\) \(56\) \(45\)

 

b.    \(\text{5 years}\)

Show Worked Solution

a.i. 

♦♦♦ Mean mark (a) 24%.

a.ii.  \(\text{Population after 1 yr calculations}\)

\(0-1\ \text{year}\ =0.4\times 70+0.75\times 80+0.4\times 90=124\)

\(1-2\ \text{years}\ =0.4\times 70=28\)

\(2-3\ \text{years}\ =0.7\times 80=56\)

\(3-4\ \text{years}\ =0.5\times 90=45\)

 

  \(\textbf{Age Group}\)
  \(\ 0-1\ \text{year}\ \) \(\ 1-2\ \text{years}\ \) \(\ 2-3\ \text{years}\ \) \(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \) \(70\) \(80\) \(90\) \(40\)
\(\ \text{Population after}\ \)
\(\ \text{one year}\)		 \(124\) \(28\) \(56\) \(45\)

  

b.    \(\text{Using CAS:}\)

\(R_1=L\times R_0=\begin{bmatrix}
124  \\
28 \\
56  \\
45  \end{bmatrix}\ \ \text{Total = 253}\ ,\ \ R_2=L\times R_1=\begin{bmatrix}
93  \\
49.6 \\
19.6  \\
28  \end{bmatrix}\ \ \text{Total = 190.2}\)

 

\(R_3=L\times R_2=\begin{bmatrix}
82.24  \\
37.2 \\
34.72  \\
9.8  \end{bmatrix}\ \ \text{Total = 163.96}\ ,\ \ R_4=L\times R_3=\begin{bmatrix}
74.684  \\
32.896 \\
26.04  \\
17.36  \end{bmatrix}\ \ \text{Total =150.98}\)

 

\(R_5=L\times R_4=\begin{bmatrix}
64.9616 \\
29.8736 \\
23.0272 \\
13.02 \end{bmatrix}\ \ \text{Total = 130.8824}\)

 
\(\therefore\ \text{Total female population less than 140 after 5 years}\)

♦ Mean mark (b) 39%.

Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\), where \(O\) is the origin and the points \(A, B\) and \(C\) are all different from each other and the origin.

The point \(M\) is the point such that  \(\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}\).

  1. Show that \(M\) lies on the line passing through \(A\) and \(B\).   (1 mark)

  2. The point \(G\) is the point such that  \(\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}\).
  3. Show that \(G\) lies on the line passing through \(M\) and \(C\), and lies between \(M\) and \(C\).   (2 marks)

  4. The complex numbers \(x, w\) and \(z\) are all different and all have modulus 1.
  5. Using part (ii), or otherwise, show that  \(\dfrac{1}{3}(x+w+z)\) is never a cube root of \(x w z\).   (2 marks)

Show Answers Only

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).
 

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Show Worked Solution

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

♦ Mean mark (ii) 43%.

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

♦♦♦ Mean mark (iii) 10%.

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Data Analysis, GEN2 2024 VCAA 2

The boxplot below displays the distribution of all gold medal-winning heights for the women's high jump, \(\textit{Wgold}\), in metres, for the 19 Olympic Games held from 1948 to 2020.

  1. Describe the shape of this data distribution.   (1 mark)

  2. For this boxplot, what is the smallest possible number of \(\textit{Wgold}\) heights lower than 1.85 m?   (1 mark)

  3.  i. Using the boxplot, show that the lower fence is 1.565 m and the upper fence is 2.325 m.  (1 mark)

  4. ii. Referring to the boxplot, the lower fence and the upper fence, explain why no outliers exist.  (1 mark)

Show Answers Only

a.    \(\text{Negatively skewed}\)

b.    \(1\)

c.i.  \(Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19\)

\(\text{Lower Fence}\) \(=Q_1-1.5\times IQR\)
  \(=1.85-1.5\times 0.19\)
  \(=1.565\)
\(\text{Upper Fence}\) \(=Q_1+1.5\times IQR\)
  \(=2.04+1.5\times 0.19\)
  \(=2.325\)

c.ii. \(\text{No values exist below the lower fence or above the upper fence.}\)

\(\therefore\ \text{No outliers exist.}\)

Show Worked Solution

a.    \(\text{Negatively skewed.}\)
 

b.    \(\text{Only 1 value is needed to extend the whisker below the}\)

\(\text{range of the}\ IQR.\)

♦♦♦ Mean mark (b) 3%.

c.i.  \(Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19\)

\(\text{Lower Fence}\) \(=Q_1-1.5\times IQR\)
  \(=1.85-1.5\times 0.19\)
  \(=1.565\)
\(\text{Upper Fence}\) \(=Q_1+1.5\times IQR\)
  \(=2.04+1.5\times 0.19\)
  \(=2.325\)

   

c.ii. \(\text{No values exist below the lower fence or above the upper fence.}\)

\(\therefore\ \text{No outliers exist.}\)

Networks, GEN1 2024 VCAA 39 MC

Anush, Blake, Carly and Dexter are workers on a construction site. They are each allocated one task.

The time, in hours, it takes for each worker to complete each task is shown in the table below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& \textbf{Task 1} & \textbf{Task 2} & \textbf{Task 3} & \textbf{Task 4} \\
\hline
\rule{0pt}{2.5ex} \textbf{Anush} \rule[-1ex]{0pt}{0pt}& 12 & 8 & 16 & 9 \\
\hline
\rule{0pt}{2.5ex} \textbf{Blake} \rule[-1ex]{0pt}{0pt}& 10 & 7 & 15 & 10 \\
\hline
\rule{0pt}{2.5ex} \textbf{Carly} \rule[-1ex]{0pt}{0pt}& 11 & 10 & 18 & 12 \\
\hline
\rule{0pt}{2.5ex} \textbf{Dexter} \rule[-1ex]{0pt}{0pt}& 10 & 14 & 16 & 11 \\
\hline
\end{array}

The tasks must be completed sequentially and in numerical order: Task 1, Task 2, Task 3 and then Task 4.

Management makes an initial allocation of tasks to minimise the amount of time required, but then decides that it takes the workers too long.

Another worker, Edgar, is brought in to complete one of the tasks.

His completion times, in hours, are listed below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& \textbf{Task 1} & \textbf{Task 2} & \textbf{Task 3} & \textbf{Task 4} \\
\hline
\rule{0pt}{2.5ex} \textbf{Edgar} \rule[-1ex]{0pt}{0pt}& 9 & 5 & 14 & 8 \\
\hline
\end{array}

When a new allocation is made and Edgar takes over one of the tasks, the minimum total completion time compared to the initial allocation will be reduced by

  1. 1 hour.
  2. 2 hours.
  3. 3 hours.
  4. 4 hours.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using Hungarian Algorithm on CAS:}\)

\(\text{Original Allocation}\)

 

\(\text{Minimal Sum}\ =9+7+11+16=43\)

\(\text{Task Allocation: Anush – T4, Blake – T2, Carly – T1, Dexter – T3}\)

♦♦♦ Mean mark 19%.

\(\text{Allocation with Edgar included}\)

\(\text{Minimal Sum}\ =9+15+10+5=39\)

\(\text{Task Allocation: Anush – T4, Blake – T3, Carly – None, Dexter – T1, Edgar – T2}\)

\(\therefore\ \text{Minimum completion time reduces from 43 to 39 hours i.e. by 4 hours.}\)

\(\Rightarrow D\) 

Networks, GEN1 2024 VCAA 38 MC

A connected graph has six vertices and six edges.

How many of the following four statements must always be true?

  • the graph has no vertices of odd degree
  • the graph contains a Eulerian trail
  • the graph contains a Hamiltonian path
  • the sum of the degrees of the vertices is 12
  1. 1
  2. 2
  3. 3
  4. 4
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Solve by drawing simple hypothetical example(s):}\)
 

♦♦♦ Mean mark 18%.

\(\text{Considering each option}\)

\(\text{1.  Both Graph 1 & 2 above have vertices of odd degree }\rightarrow\ \text{Incorrect} \)

\(\text{2.  Graph 2 has 4 vertices of odd degree which means no Eulerian Trail}\rightarrow\ \text{Incorrect} \)

\(\text{3.  Path cannot be completed in Graph 2 without repeating a vertex}\)

\(\ \ \ \ \ \text{No Hamiltonian Path}\rightarrow\ \text{Incorrect} \)

\(\text{4. The sum of the degrees of the vertices is always 12}\ \rightarrow\  \text{Correct in both examples}\)

\(\text{Given the available options, only 1 statement can always be true.}\)

\(\Rightarrow A\)

Matrices, GEN1 2024 VCAA 32 MC

A large sporting event is held over a period of four consecutive days: Thursday, Friday, Saturday and Sunday.

People can watch the event at four different sites throughout the city: Botanical Gardens \((G)\), City Square \((C)\), Riverbank \((R)\) or Main Beach \((M)\).

Let \(S_n\) be the state matrix that shows the number of people at each location \(n\) days after Thursday.

The expected number of people at each location can be determined by the matrix recurrence rule

\(S_{n+1}=TS_n+A\)

\begin{aligned}
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \textit{this day} \\
& \quad \quad \quad \quad \quad \quad \quad \ G \quad \ \  C \quad \ \ R \quad \ \  M \\
& \text{where} \quad T=\begin{bmatrix}
0.4 & 0.2 & 0.4 & 0 \\
0.4 & 0.1 & 0.3 & 0.3 \\
0.1 & 0.4 & 0.1 & 0.2 \\
0.1 & 0.3 & 0.2 & 0.5
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array} \text { next day } \quad \text{and}& A=\begin{bmatrix}
300 \\
200\\
100 \\
300
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

 

\begin{aligned} \text{Given the state matrix}& \quad \quad S_3=\begin{bmatrix}
5620\\
6386\\
4892\\
6902
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

the number of people watching the event at the Botanical Gardens \((G)\) from Thursday to Sunday has

  1. decreased by 162
  2. decreased by 212
  3. increased by 124
  4. increased by 696
Show Answers Only

\(D\)

Show Worked Solution
\(S_{n+1}\) \(=TS_n+A\)
\(TS_n\) \(=S_{n+1}-A\)
\(S_n\)  \(=T^{-1}\times \left(S_{n+1}-A\right)\)
♦♦♦ Mean mark 33%.

  
\(\text{Using CAS:}\)
  
\(T =\begin{bmatrix}
\ 0.4 & 0.2 & 0.4 & 0 \\
\ 0.4 & 0.1 & 0.3 & 0.3 \\
\ 0.1 & 0.4 & 0.1 & 0.2 \\
\ 0.1 & 0.3 & 0.2 & 0.5 
\end{bmatrix}\ \rightarrow \ \ T^{-1}=
 \begin{bmatrix}
\ -2.6 & 5.4 & 3.4 & -4.6 \\
\ 0.2 & -0.8 & 3.2 & -0.8 \\
\ 5 & -5 & -5 & 5 \\
\ -1.6 & 1.4 & -0.6 & 1.4   
\end{bmatrix}\)

 

\(\text{Sunday} =S_3=\begin{bmatrix}
\ 5620 \\
\ 6386 \\
\ 4892  \\
\ 6902
\end{bmatrix} \)

\(\text{Saturday} =S_2=T^{-1}\times \left(S_{3}-A\right)=\begin{bmatrix}
\ 5496 \\
\ 6168 \\
\ 4720  \\
\ 6516
\end{bmatrix} \)

\(\text{Friday} =S_1=T^{-1}\times \left(S_{2}-A\right)=\begin{bmatrix}
\ 5832 \\
\ 6076 \\
\ 4120  \\
\ 5972
\end{bmatrix} \)

\(\text{Thursday} =S_0=T^{-1}\times \left(S_{1}-A\right)=\begin{bmatrix}
\ 4924 \\
\ 4732 \\
\ 6540  \\
\ 4904
\end{bmatrix} \)

  
\(5620-4924=696\)

\(\therefore\ \text{Botanical Gardens attendance has increased by 696 people.}\)

\(\Rightarrow D\)

Mechanics, EXT2 M1 2024 HSC 16c

Two particles, \(A\) and \(B\), each have mass 1 kg and are in a medium that exerts a resistance to motion equal to \(k v\), where  \(k>0\)  and \(v\) is the velocity of any particle. Both particles maintain vertical trajectories.

The acceleration due to gravity is \(g\) ms\(^{-2}\), where  \(g>0\).

The two particles are simultaneously projected towards each other with the same speed, \(v_0\) ms\(^{-1}\), where  \(0<v_0<\dfrac{g}{k}\).

The particle \(A\) is initially \(d\) metres directly above particle \(B\), where  \(d<\dfrac{2 v_0}{k}\).

Find the time taken for the particles to meet.   (4 marks)

Show Answers Only

\(t=-\dfrac{1}{k} \ln \left(\dfrac{2v_0-dk}{v_0}\right)\)

Show Worked Solution

\(\text{Particle A:}\)

\(\ddot{x}=\dfrac{dv_{\small{A}}}{dt}=-g-k v_{\small{A}}\)

\(\dfrac{dt}{dv_{\small{A}}}=\dfrac{1}{-g-kv_{\small{A}}}\)

\(t=-\displaystyle \int \dfrac{1}{g+k v_{\small{A}}} \, dv_{\small{A}}=-\dfrac{1}{k} \ln \left(g+k v_A\right)+c\)

\(\text{At} \ \ t=0, \ v_{\small{A}}=-v_0 \ \Rightarrow \ c=\dfrac{1}{k} \ln \left(g-k v_0\right)\)

\(t=\dfrac{1}{k} \ln \left(g-k v_0\right)-\dfrac{1}{k} \ln \left(g+kv_{\small{A}}\right)=\dfrac{1}{k} \ln \left(\dfrac{g-kv_0}{g+kv_{\small{A}}}\right)\)

♦♦♦ Mean mark 28%.

\(\text{Find} \ v_{\small{A}}:\)

  \(\dfrac{g-k v_0}{g+k v_{\small{A}}}\) \(=e^{kt}\)
  \(g-kv_0\) \(=e^{kt} \cdot g+e^{kt} \cdot kv_{\small{A}}\)
  \(kv_{\small{A}}\) \(=e^{-kt}\left(g-kv_0\right)-g\)
  \(v_{\small{A}}\) \(=\dfrac{1}{k}\left[e^{-kt}\left(g-kv_0\right)-g\right]\)

  \(x\) \(=\displaystyle \frac{1}{k} \int e^{-kt} \cdot g-e^{-kt} \cdot v_0-g \, dt\)
    \(=\dfrac{1}{k}\left[-\dfrac{g}{k}e^{-kt}+v_0 e^{-kt}-gt\right]+c\)
    \(=-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-kt}-gt+c\)

 
\(\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{g}{k^2}-\dfrac{v_0}{k}\)

\(x_{\small{A}}=d-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-kt}-gt+\dfrac{g}{k^2}-\dfrac{v_0}{k}\)
 

\(\text{Particle B:}\)

\(\ddot{x}=-g-k v_{\small{B}}\)

\(t=-\dfrac{1}{k} \ln \left(g+k v_{\small{B}}\right)+c\)
 

\(\text{When} \ \ t=0, v_B=v_0 \ \Rightarrow \ c=\dfrac{1}{k} \ln \left(g+kv_0\right)\)

  \(v_{\small{B}}\) \(=\dfrac{1}{k}\left[e^{-kt}\left(g+kv_0\right)-g\right]\)
  \(x_{\small{B}}\) \(=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-kt}-gt+c\)

 

\(\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{g}{k^2}+\dfrac{v_0}{k}\)

\(x_{\small{B}}=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-kt}-gt+\dfrac{g}{k^2}+\dfrac{v_0}{k}\)
 

\(\text{Find  \(t\)  when \(\ x_{\small{A}}=x_{\small{B}}\):}\)

\(d-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-k t}-g t+\dfrac{g}{k^2}-\dfrac{v_0}{k}=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-k t}-g t+\dfrac{g}{k^2}+\dfrac{v_0}{k}\)

  \(\dfrac{2 v_0}{k} \cdot e^{-k t}\) \(=\dfrac{2 v_0}{k}-d\)
  \(e^{-kt}\) \(=\left(\dfrac{2 v_0-d k}{k}\right) \cdot \dfrac{k}{v_0}\)
  \(-kt\) \(=\ln \left(\dfrac{2 v_0-d k}{v_0}\right)\)
  \(t\) \(=-\dfrac{1}{k} \ln \left(\dfrac{2v_0-dk}{v_0}\right)\)

Vectors, EXT2 V1 2024 HSC 10 MC

Three unit vectors \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\), in 3 dimensions, are to be chosen so that  \(\underset{\sim}{a} \perp \underset{\sim}{b}, \ \underset{\sim}{b} \perp \underset{\sim}{c}\)  and the angle \(\theta\) between \(\underset{\sim}{a}\) and  \(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\)  is as small as possible.

What is the value of \(\cos \theta\) ?

  1. \(0\)
  2. \(\dfrac{1}{\sqrt{3}}\)
  3. \(\dfrac{1}{\sqrt{2}}\)
  4. \(\dfrac{2}{\sqrt{5}}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}}\)

♦♦♦ Mean mark 25%.

\(\theta_{\text{min}} \ \Rightarrow \ \cos\,\theta_{\text{max}}:\)

\(\underset{\sim}{a} \cdot\left(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\right)=\abs{\underset{\sim}{a}}^2+\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{a} \cdot \underset{\sim}{c}=1+\underset{\sim}{a} \cdot \underset{\sim}{c}\)

\(\text{Find}\ \ \abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}:\)

  \(\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}^2\) \(=\abs{\underset{\sim}{a}}^2+\abs{\underset{\sim}{b}}^2+\abs{\underset{\sim}{c}}^2+2\left(\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{c}+\underset{\sim}{a} \cdot \underset{\sim}{c}\right)\)
    \(=3+2 \underset{\sim}{a} \cdot \underset{\sim}{c}\)

 
\(\cos \theta=\dfrac{1+a \cdot c}{\sqrt{3+2 a \cdot c}}\)

\(\underset{\sim}{a} \cdot\underset{\sim}{c}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{c}} \cos\, \alpha = \cos\,\alpha \)

\(0 \leqslant \cos\, \alpha \, \leqslant 1\)

\(\therefore \cos \theta_{\text{max}}=\dfrac{2}{\sqrt{5}}\)

\(\Rightarrow D\)

Complex Numbers, EXT2 N2 2024 HSC 16b

The number  \(w=e^{\small{\dfrac{2 \pi i}{3}}}\)  is a complex cube root of unity. The number \(\gamma\) is a cube root of \(w\).

  1. Show that  \(\gamma+\bar{\gamma}\)  is a real root of  \(z^3-3 z+1=0\).   (3 marks)

  2. By using part (i) to find the exact value of  \(\cos \dfrac{2 \pi}{9} \cos \dfrac{4 \pi}{9} \cos \dfrac{8 \pi}{9}\), deduce the value(s) of  \(\cos \dfrac{2^n \pi}{9} \cos \dfrac{2^{n+1} \pi}{9} \cos \dfrac{2^{n+2} \pi}{9}\)  for all integers  \(n \geq 1\). Justify your answer.   (3 marks)

Show Answers Only

i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 
\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Show Worked Solution

i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

♦♦ Mean mark (i) 36%.
  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 

\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

♦♦♦ Mean mark (ii) 18%.

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Vectors, EXT2 V1 2024 HSC 16a

Consider the function  \(y=\cos (k x)\), where  \(k>0\). The value of \(k\) has been chosen so that a circle can be drawn, centred at the origin, which has exactly two points of intersection with the graph of the function and so that the circle is never above the graph of the function. The point  \(P(a, b)\)  is the point of intersection in the first quadrant, so  \(a>0\)  and  \(b>0\),  as shown in the diagram.

The vector joining the origin to the point \(P(a, b)\) is perpendicular to the tangent to the graph of the function at that point. (Do NOT prove this.)

Show that  \(k>1\).   (4 marks)

Show Answers Only

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)

 
\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Show Worked Solution

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)
♦♦♦ Mean mark 8%.

\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Mechanics, EXT2 M1 2024 HSC 15c

A bar magnet is held vertically. An object that is repelled by the magnet is to be dropped from directly above the magnet and will maintain a vertical trajectory. Let \(x\) be the distance of the object above the magnet.
 

The object is subject to acceleration due to gravity, \(g\), and an acceleration due to the magnet \(\dfrac{27 g}{x^3}\), so that the total acceleration of the object is given by

 \(a=\dfrac{27 g}{x^3}-g\)

The object is released from rest at  \(x=6\).

  1. Show that  \(v^2=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)\).   (2 marks)

  2. Find where the object next comes to rest, giving your answer correct to 1 decimal place.  (2 marks)

Show Answers Only

 

Show Worked Solution

i.    \(a=\dfrac{27 g}{x^3}-g\)

\(\dfrac{d}{dx}(\frac{1}{2}v^{2})\) \(= \dfrac{27g}{x^3}-g\)  
\(\dfrac{1}{2} v^2\) \(=-\dfrac{27 g}{2 x^2}-g x+c\)  

 
\(\text{When}\ \ x=6, v=0:\)

\(0\) \(=-\dfrac{27g}{2 \times 6^2}-6g+c\)  
\(c\) \(=\dfrac{459 g}{72}=\dfrac{51 g}{8}\)  

 

  \(\dfrac{1}{2} v^2\) \(=-\dfrac{27 g}{2 x^2}-g x+\dfrac{51 g}{8}\)
  \(v^2\) \(=-\dfrac{27 g}{x^2}-2 g x+\dfrac{51g }{4}\)
    \(=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)\)

  

ii.    \(\text{Find \(x\) when  \(v=0\):}\)

\(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\) \(=0\)  
\(51 x^2-8 x^3-108\) \(=0\)  
\(8 x^3-51 x^2+108\) \(=0\)  

 
\(\text{Given  \(x=6\)  is a root:}\)

♦♦♦ Mean mark (ii) 24%.

\(8 x^3-51 x^2+108=(x-6)\left(8 x^2-3 x-18\right)\)

\(\text{Other roots:}\)

  \(x\) \(=\dfrac{3 \pm \sqrt{9-4 \cdot 8 \cdot 18}}{2 \times 8}\)
    \(=\dfrac{3 \pm \sqrt{585}}{16}\)
    \(=\dfrac{3+3 \sqrt{65}}{16} \quad(x>0)\)
    \(=1.7 \ \text{units (1 d.p.)}\)

 
\(\therefore \ \text{Object next comes to rest at  \(x=1.7\) units}\) 

Vectors, EXT1 V1 2024 HSC 14d

A particle is projected from the origin, with initial speed \(V\) at an angle of \(\theta\) to the horizontal. The position vector of the particle, \(\underset{\sim}{r}(t)\), where \(t\) is the time after projection and \(g\) is the acceleration due to gravity, is given by

\(\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right)\).   (Do NOT prove this.)

Let \(D(t)\) be the distance of the particle from the origin at time \(t\), so  \(D(t)=|\underset{\sim}{r}(t)|\).

Show that for  \(\theta<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)\)  the distance, \(D(t)\), is increasing for all  \(t>0\).   (4 marks)

Show Answers Only

\(\text{See Worked Solutions.}\)

Show Worked Solution

\(\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right), \ D(t)=\abs{\underset{\sim}{r}(t)}\)

  \(D(t)^2\) \(=(V t \cos \theta)^2+\left(V t \sin \theta-\dfrac{g t^2}{2}\right)^2\)
    \(=V^2 t^2 \cos ^2 \theta+V^2 t^2 \sin ^2 \theta-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
♦♦♦ Mean mark 25%.

\(\text{Let }F(t)=D(t)^2\)

  \(F^{\prime}(t)\) \(=2 V^2 t-3 V g t^2 \sin \theta+g^2 t^3\)
    \(=t\left(2 V^2-3 V g t \sin \theta+g^2 t^2\right)\)

\(\text {Monotonically increasing when } F^{\prime}(t)>0 \text { for all } t.\)

\(g^2 t^2-3 V g t\, \sin \theta+2 V^2>0\)
 

\(\text {Find } t \text { when } \Delta<0:\)

  \((-3 V g \sin \theta)^2-4 \cdot g^2 \cdot 2 V^2\) \(<0\)
  \(9 V^2 g^2 \sin ^2 \theta-8 V^2 g^2\) \(<0\)
  \(\sin ^2 \theta\) \(<\dfrac{8}{9}\)
  \(\sin \theta\) \(<\sqrt{\dfrac{8}{9}} \ \ \left(\theta \in\left(0, \dfrac{\pi}{2}\right) \Rightarrow \ \sin \theta \in(0,1)\right)\)
  \(\theta\) \(<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)\)

Trigonometry, EXT1 T3 2024 HSC 10 MC

For real numbers \(a\) and \(b\), where  \(a \neq 0\)  and  \(b \neq 0\), we can find numbers \(\alpha\), \(\beta\), \(\gamma\), \(\delta\) and \(R\) such that  \(a\,\cos x + b\,\sin x\)  can be written in the following 4 forms:

\(R\,\sin(x + \alpha)\)

\(R\,\sin(x-\beta)\)

\(R\,\cos(x + \gamma)\)

\(R\,\cos(x-\delta)\)

where  \(R \gt 0\)  and  \(0<\alpha, \beta, \gamma, \delta \lt 2\pi\).

What is the value of  \(\alpha + \beta + \gamma + \delta\)?

  1. \(0\)
  2. \(\pi\)
  3. \(2\pi\)
  4. \(4\pi\)
Show Answers Only

\(D\)

Show Worked Solution

\(a\,\sin\,x+b\,\cos\,x\ \ \text{can be written 4 ways.}\)

\(\text{Consider the case:}\)

\(\cos\,x+\sin\,x\ \ \text{where}\ a=b=1,\ \ R=\sqrt{2}\)

\(\sqrt{2}\,\sin(x+\alpha)\) \(= \sqrt{2}(\sin\,x\,\cos\,\alpha + \cos\,x\,\sin\,\alpha)\)  
  \(= \sqrt{2}\Big(\sin\,x \cdot \dfrac{1}{\sqrt2} + \cos\,x \cdot \dfrac{1}{\sqrt2}\Big)\)  

 
\(\cos\, \alpha^{+}, \sin\,\alpha^{+}\ \Rightarrow\ \ \alpha=\dfrac{\pi}{4} \)

♦♦♦ Mean mark 19%.

\(\text{Similarly for other 3 cases:}\)

\(\sqrt{2}\,\sin(x-\beta): \ \cos\, \beta^{+}, \sin\,\beta^{-}\ \Rightarrow\ \ \beta=\dfrac{7\pi}{4} \)

\(\sqrt{2}\,\cos(x+\gamma): \ \cos\, \gamma^{+}, \sin\,\gamma^{-}\ \Rightarrow\ \ \gamma=\dfrac{7\pi}{4} \)

\(\sqrt{2}\,\cos(x-\delta): \ \cos\, \delta^{+}, \sin\,\delta^{+}\ \Rightarrow\ \ \delta=\dfrac{\pi}{4} \)

\(\therefore \alpha + \beta + \gamma + \delta = 4\pi\)

\(\Rightarrow D\)

Trigonometry, EXT1 T3 2024 HSC 14c

  1. Explain why the equation  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta\), where  \(-\pi<\theta<\pi\), has exactly one solution.   (1 marks)

  2. Solve  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\).   (2 marks)

Show Answers Only

i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

 \(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.   \(x=\dfrac{1}{2}\)

Show Worked Solution

i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

♦♦♦ Mean mark (i) 11%.

\(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.    \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\)

\(\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)\)

♦♦ Mean mark (ii) 32%.

\(\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1\)

  \(\dfrac{3 x+10 x}{1-30 x^2}\) \(=-1\)
  \(13 x\) \(=30 x^2-1\)
  \(30 x^2-13 x-1\) \(=0\)
  \((15 x+1)(2 x-1)\) \(=0\)

 
\(x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}\)

\(\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)\)

\(\therefore x=\dfrac{1}{2}\)

CHEMISTRY, M2 EQ-Bank 10

  1. A student is asked to prepare 500.0 mL of a 0.150 mol L\(^{-1}\) standard solution of oxalic acid \(\ce{(C2H2O4.2H2O)}\), and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L\(^{-1}\) solution. Outline and explain each step in this process, including the calculations involved and choice of equipment.   (5 marks)
  1. Justify the procedure in part (a.) by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced.   (2 marks)
Show Answers Only

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.
Show Worked Solution

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Statistics, STD1 S1 2024 HSC 24

Students in two classes, Class \(A\) and Class \(B\), recorded the number of text messages they sent in a day. Each class has 18 students.

The results are shown in the dot plots.
 

 

Compare the two datasets by examining the skewness, median and spread of the distributions.   (3 marks)

Show Answers Only

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Range} & \quad Q_1\quad & \text{Median}&\quad Q_3\quad & IQR\\
\hline
\rule{0pt}{2.5ex} \text{Class } A\rule[-1ex]{0pt}{0pt} & 5-1=4&3&4&5&5-3=2\\
\hline
\rule{0pt}{2.5ex} \text{Class } B\rule[-1ex]{0pt}{0pt} & 6-2=4 &2&3&4&4-2=2\\
\hline
\end{array}

\(\text{Skewness:}\)

\(\text{Class A is slightly negatively skewed, while Class B is more symmetrical with}\)

\(\text{a slightly positive skew.}\)
 

\(\text{Median:}\)

\(\text{Class A has a higher median (4) compared to Class B (3).}\)
 

\(\text{Spread:}\)

\(\text{Both classes have the same range (4) and IQR (2), suggesting similar variability}\)

\(\text{in the data.}\)
 

\(\text{In conclusion, while the spread of data is similar for both classes, Class A
tends to}\)

\(\text{have a higher number of text messages sent (higher median and negative skew)}\)

\(\text{compared to Class B, which has a more symmetrical, slightly positive distribution.}\)

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Range} & \quad Q_1\quad & \text{Median}&\quad Q_3\quad & IQR\\
\hline
\rule{0pt}{2.5ex} \text{Class } A\rule[-1ex]{0pt}{0pt} & 5-1=4&3&4&5&5-3=2\\
\hline
\rule{0pt}{2.5ex} \text{Class } B\rule[-1ex]{0pt}{0pt} & 6-2=4 &2&3&4&4-2=2\\
\hline
\end{array}

♦♦♦ Mean mark 16%.

\(\text{Skewness:}\)

\(\text{Class A is slightly negatively skewed, while Class B is more symmetrical with}\)

\(\text{a slightly positive skew.}\)
 

\(\text{Median:}\)

\(\text{Class A has a higher median (4) compared to Class B (3).}\)
 

\(\text{Spread:}\)

\(\text{Both classes have the same range (4) and IQR (2), suggesting similar variability}\)

\(\text{in the data.}\)
 

\(\text{In conclusion, while the spread of data is similar for both classes, Class A
tends to}\)

\(\text{have a higher number of text messages sent (higher median and negative skew)}\)

\(\text{compared to Class B, which has a more symmetrical, slightly positive distribution.}\)

Measurement, STD1 M5 2024 HSC 32

A scale diagram is shown with locations \(A, B\) and \(C\) marked (assume grid squares are 1 cm × 1 cm).

Jo takes 24 minutes to walk from \(A\) to \(B\) (in a straight line) when walking at 3 km per hour.
 

  1. What is the scale used in the diagram?   (3 marks)
  2. What is the distance from \(B\) to \(C\), in kilometres?   (2 marks)
Show Answers Only

a.    \(1:20\,000\)

b.    \(\text{1.4 km}\)

Show Worked Solution

a.   \(\text{Find distance }A\rightarrow B:\)

\(D=S \times T = 3 \times \dfrac{24}{60} = 1.2\ \text{km}\)
  

  \(\text{Scale }\rightarrow\ 6\ \text{cm }\) \(:\ 1.2\ \text{km}\)
  \(6\ \text{cm }\) \(:\ 1200\ \text{m}\)
  \(6\ \text{cm }\) \(:\ 1200\times 100\ \text{cm}\)
  \(6\ \) \(:\ 120\,000\)
  \(1\ \) \(:\ 20\,000\)
♦♦♦ Mean mark 15%.

b.   \(\text{Distance }B\rightarrow C = 7\ \text{grid units}\)

\(\text{Real distance }B\rightarrow C\) \(=7 \times 20\,000\)
  \(=140\,000\ \text{cm}\)
  \(=1400\ \text{m}\)
  \(=1.4\ \text{km}\)
♦♦♦ Mean mark 28%.

Measurement, STD1 M3 2024 HSC 14

A hotel is located 186 m north and 50 m west of a train station.
 

  1. What is the straight line distance from the hotel to the train station? Round your answer to the nearest metre.   (2 marks)

  2. What is the bearing of the hotel from the train station? Round your answer to the nearest degree.   (2 marks)

Show Answers Only

a.    \(193\ \text{m}\)

b.    \(345^\circ\ \text{(nearest degree)}\)

Show Worked Solution

a.   \(\text{By Pythagoras:}\)

  \(d^2\) \(=50^2+186^2\)
  \(d^2\) \(=37\,096\)
  \(d\) \(=\sqrt{37\,096}\)
    \(=192.603\dots\)
    \(=193\ \text{m (nearest metre)}\)
♦♦ Mean mark (a) 37%.
b.     \(\tan\theta\) \(=\dfrac{50}{186}\)
  \(\theta\) \(=15.046\dots^\circ\)
    \(\approx 15^\circ\ \text{(nearest degree)}\)

 

\(\text{Bearing}\ H\ \text{from}\ T =360-15=345^\circ\)

♦♦♦ Mean mark (b) 12%.

Statistics, STD1 S1 2024 HSC 13

Consider the following dataset.

\(1, \ 1, \ 2, \ 3, \ 5, \ 7, \ 15\)

  1. What is the interquartile range?   (1 mark)
  2. By using the outlier formula, determine whether 15 is an outlier.   (2 marks)
Show Answers Only

a.    \(\text{IQR = 6}\)

b.    \(15\ \text{is not an outlier as it is not greater than 16.}\)

Show Worked Solution

a.    \(Q_2=3, \ Q_1=1, \ Q_3=7\)

\(\therefore\ IQR=7-1=6\)
 

♦♦♦ Mean mark (a) 25%.

b.   \(\text{Find upper fence:}\)

\(Q_3+1.5\times IQR=7 + 1.5\times 6=16\)

\(\therefore\ \text{15 is not an outlier (15 < 16)}\)

♦♦ Mean mark (b) 32%.

Networks, STD1 N1 2024 HSC 20

The diagram shows a network with weighted edges.
 

  1. Draw a minimum spanning tree for this network and determine its weight.   (2 marks)
     


  1. Is it possible to find another spanning tree with the same weight? Give a reason for your answer.   (1 mark)

Show Answers Only

a.
         

b.    \(\text{Yes.}\)

\(\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\)

\(\text{to create a second MST (with equivalent weight = 24)}\)

Show Worked Solution

a.
         

♦ Mean mark 53%.

 
b.    \(\text{Yes.}\)

\(\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\)

\(\text{to create a second MST (with equivalent weight = 24)}\)

♦♦♦ Mean mark 16%.

Probability, 2ADV S1 2024 HSC 9 MC

A bag contains 2 red and 3 white marbles. Jovan randomly selects two marbles at the same time from this bag. The probability tree diagram shows the probabilities for each of the outcomes.
 

Given that one of the marbles that Jovan has selected is red, what is the probability that the other marble that he has selected is also red?

  1. \(\dfrac{1}{10}\)
  2. \(\dfrac{1}{7}\)
  3. \(\dfrac{1}{4}\)
  4. \(\dfrac{7}{10}\)
Show Answers Only

\(B\)

Show Worked Solution
\(P(R_2|_{\text{other is red}})\) \(=\dfrac{P(R_1 \cap R_2)}{P\text{(at least 1 red)}}\)  
  \(=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-P(WW)}\)  
  \(=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-(\frac{3}{5} \times \frac{2}{4})}\)  
  \(=\dfrac{1}{7}\)  

 
\(\Rightarrow B\)

♦♦♦ Mean mark 14%.

Calculus, 2ADV C3 2024 HSC 31

Two circles have the same centre \(O\). The smaller circle has radius 1 cm, while the larger circle has radius \((1 + x)\) cm. The circles enclose a region \(QRST\), which is subtended by an angle \(\theta\) at \(O\), as shaded.

The area of \(QRST\) is \(A\) cm\(^{2}\), where \(A\) is a constant and \(A \gt 0\).
 

Let \(P\) cm be the perimeter of \(QRST\).

  1. By finding expressions for the area and perimeter of \(QRST\), show that  \(P(x)=2x+\dfrac{2A}{x}\).   (3 marks)
  2. Show that if the perimeter, \(P(x)\), is minimised, then \(\theta\) must be less than 2.   (3 marks)
Show Answers Only

a.   \(\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS\)

\(A\) \(=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2\)  
\(A\) \(=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}\)  
\(A\) \(=\dfrac{\theta}{2}(1+2x+x^2-1)\)  
\(A\) \(=\dfrac{\theta}{2}x(x+2)\)  
\(\theta\) \(=\dfrac{2A}{x(x+2)}\)  

 

\(\text{Perimeter}\ QRST\) \(=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x \)  
  \(=\theta(x+1)+\theta+2x\)  
  \(=\theta(x+2)+2x\)  
  \(=\dfrac{2A}{x(x+2)}(x+2)+2x\)  
  \(=2x+\dfrac{2A}{x}\)  

 
b.   \(P(x)=2x+\dfrac{2A}{x}\)

\(P^{′}(x) = 2-\dfrac{2A}{x^2}\)

\(P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)\)

\(\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:\)

\(2-\dfrac{2A}{x^2}\) \(=0\)  
\(2x^2\) \(=2A\)  
\(x^2\) \(=A\)  

 
\(\text{Using part (a):}\)

\(\theta\) \(=\dfrac{2A}{x(x-2)}\)  
  \(=\dfrac{2x^2}{x(x+2)}\)  
  \(=\dfrac{2x}{x+2}\)  
  \(=2 \times \dfrac{x}{x+2} \)  
  \( \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) \)  

 
\(\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. \)

Show Worked Solution

a.   \(\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS\)

\(A\) \(=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2\)  
\(A\) \(=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}\)  
\(A\) \(=\dfrac{\theta}{2}(1+2x+x^2-1)\)  
\(A\) \(=\dfrac{\theta}{2}x(x+2)\)  
\(\theta\) \(=\dfrac{2A}{x(x+2)}\)  

 

\(\text{Perimeter}\ QRST\) \(=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x \)  
  \(=\theta(x+1)+\theta+2x\)  
  \(=\theta(x+2)+2x\)  
  \(=\dfrac{2A}{x(x+2)}(x+2)+2x\)  
  \(=2x+\dfrac{2A}{x}\)  
♦ Mean mark (a) 41%.
COMMENT: Sector/arc calculations used in solution are for those who don’t want to remember formulas.

b.   \(P(x)=2x+\dfrac{2A}{x}\)

\(P^{′}(x) = 2-\dfrac{2A}{x^2}\)

\(P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)\)

\(\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:\)

\(2-\dfrac{2A}{x^2}\) \(=0\)  
\(2x^2\) \(=2A\)  
\(x^2\) \(=A\)  

 
\(\text{Using part (a):}\)

\(\theta\) \(=\dfrac{2A}{x(x-2)}\)  
  \(=\dfrac{2x^2}{x(x+2)}\)  
  \(=\dfrac{2x}{x+2}\)  
  \(=2 \times \dfrac{x}{x+2} \)  
  \( \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) \)  

 
\(\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. \)

♦♦♦ Mean mark (b) 12%.
 

Measurement, STD2 M6 2024 HSC 40

A compass radial survey is shown.
 

Given that \(AC\) is a straight line, find the bearing of \(C\) from \(O\), correct to the nearest degree.   (3 marks)

Show Answers Only

\(\text{Bearing of}\ C\ \text{from}\ O =220^{\circ}\)

Show Worked Solution

\(\text{In}\ \Delta ODA,\ \text{using cosine rule:}\)

\(\cos \angle DOA\) \(=\dfrac{38^2+42^2-67.6^2}{2 \times 38 \times 42}\)  
  \(=-0.4223…\)  
\(\angle DOA\) \(=115.0^{\circ}\)  

 
\(\text{Let}\ X\ \text{be a point directly north of point}\ O:\)

\(\angle DOX=360-285=75^{\circ}\)

\(\angle XOA=115-75=40^{\circ}\)

\(\angle AOB=110-40=70^{\circ}\)

\(\angle COB=180-70=110^{\circ}\ \ (AOC\ \text{is a straight line)}\)

\(\therefore\ \text{Bearing of}\ C\ \text{from}\ O = 110+110=220^{\circ}\)

♦♦♦ Mean mark 19%
STRATEGY:  The information in \(\Delta ODA\) strongly suggests finding \(\angle DOA\) is a good strategy to explore.

Networks, STD2 N3 2024 HSC 39

A project involving nine activities is shown in the network diagram.

The duration of each activity is not yet known.
 

The following table gives the earliest start time (EST) and latest start time (LST) for three of the activities. All times are in hours.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Activity} \rule[-1ex]{0pt}{0pt} & EST & LST \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ 0\ \ \ \ \ \  & \ \ \ \ \ \ 2\ \ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 0 & 1 \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 12 & 12 \\
\hline
\end{array}

  1. What is the critical path?   (1 mark)
  2. The minimum time required for this project to be completed is 19 hours.
  3. What is the duration of activity \(I\)?   (1 mark)

  4. The duration of activity \(C\) is 3 hours.
  5. What is the maximum amount of time that could occur between the start of activity \(F\) and the end of activity \(H\)?   (1 mark)

Show Answers Only

a.   \(\text{Critical Path:}\ BEGI\)

b.   \(\text{Duration of}\ I =7\ \text{hours}\)

c.  \(\text{Max time}\ =8\ \text{hours}\)

Show Worked Solution

a.   \(\text{Activity}\ A\ \text{and}\ C: \ LST \gt EST\)

\(\Rightarrow\ \text{Activity}\ A\ \text{and}\ C\ \text{not on critical path.}\)

\(\text{Critical Path:}\ BEGI\)
 

♦ Mean mark (a) 43%.

b.   \(\text{Duration of}\ I = 19-12=7\ \text{hours}\)
 

c.   \(\text{Since}\ C + F + H + I\ \text{is not a critical path:}\)

\(C + F + H + I = 18\ \text{or less (C.P. = 19 hours)}\)

\(3+F+H+7 = 18\ \text{or less}\)

\(\Rightarrow\ F+H = 8\ \text{or less}\)

\(\therefore\ \text{Max time from start of}\ F\ \text{to end of}\ H = 8\ \text{hours}\)

♦♦♦ Mean mark (c) 10%.

Statistics, STD2 S5 2024 HSC 35

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the}\)

\(\text{percentage of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

♦♦ Mean mark (a) 36%.

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the}\)

\(\text{percentage of scores in this range will be twice the answer in part (a).}\)
 

♦♦♦ Mean mark (b) 19%.

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

♦♦ Mean mark (c) 30%.

Statistics, 2ADV S2 2024 HSC 8 MC

Some data are used to create a box plot shown.
 

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?
 


 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}\)

\(\text{ln options A, B and C (given 16 data points):}\)

\(Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}\)

\(Q_3\ \text{(13th data point)}\  \rightarrow \text{6th column}\)

\(\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}\)

♦♦♦ Mean mark 11%.

\(\text{ln option D:}\)

\(Q_1  \rightarrow \text{3rd column}\)

\(Q_3 \rightarrow \text{5th column}\)

\(\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}\)

\(\Rightarrow D\)

Statistics, STD2 S1 2024 HSC 15 MC

Some data are used to create a box plot shown.
 

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?
 


 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}\)

\(\text{ln options A, B and C (given 16 data points):}\)

\(Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}\)

\(Q_3\ \text{(13th data point)}\  \rightarrow \text{6th column}\)

\(\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}\)
 

\(\text{ln option D:}\)

\(Q_1  \rightarrow \text{3rd column}\)

\(Q_3 \rightarrow \text{5th column}\)

\(\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}\)

\(\Rightarrow D\)

♦♦♦ Mean mark 9%.

CHEMISTRY, M2 EQ-Bank 8

A gas at a temperature of \(9.0 \times 10^2\ \text{K}\) in a container with a volume of \(30.0\ \text{L}\) has a pressure of \(5.0 \times 10^2\ \text{kPa}\).

  1. How many moles of gas are in the container?   (3 marks)
  1. If the empty container weighs \(450.0\ \text{g}\) and the container with the gas weighs \(526.0\ \text{g}\), what is the gaseous element in the container?   (2 marks)
Show Answers Only

a.    \(2.0\ \text{mol}\)

b.    The unknown gas is fluorine.

Show Worked Solution

a.    Using the Ideal Gas Law:

\(PV=nRT \Rightarrow n=\dfrac{PV}{RT}\)

\(n=\dfrac{500 \times 30}{8.314 \times 900}=2.0\ \text{mol (2 sig.fig)}\) 

 

b.    Mass of the gas \(=526.0-450.0=76\ \text{g}\)

Molar Mass of the gas \(=\dfrac{m}{n}=\dfrac{76}{2.0}=38\ \text{g mol}^{-1}\)

  • This is equal to the MM of fluorine gas \(\ce{(F2)}\).
  • The unknown gas is fluorine.

CHEMISTRY, M3 EQ-Bank 8

A galvanic cell has been set up as illustrated in the diagram below.

  1. The standard potential for this reaction is 0.78 V. Use half equations to determine the unknown electrode.   (2 marks)
  1. The unknown solution is light green in colour. Explain what will happen to the colour of the unknown solution as the reaction proceeds.   (2 marks)
  1. After some time, a solid deposit formed on the copper electrode was removed and dried. The mass of the deposit was 0.150 g. Determine the final concentration of the copper nitrate solution.   (3 marks)
Show Answers Only

a.    \(\ce{Fe(s) -> Fe^{2+} + 2e^-}\)

b.    As the reaction progresses:

  • The solution will darken, taking on a more intense green colour.
  • This change occurs because nickel undergoes oxidation to form \(\ce{Fe^{2+}}\) ions, which are released into the solution, thereby increasing its colour intensity.

c.    \(0.137\ \text{mol L}^{-1}\)

Show Worked Solution
a.     \(E^{\circ}_{\text{cell}}\) \(=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}\)
  \(0.78\) \(=0.34-E^{\circ}_{\text{anode}}\)
  \(E^{\circ}_{\text{anode}}\) \(=0.34-0.78\)
  \(E^{\circ}_{\text{anode}}\) \(=-0.44\)
     
  • \(\ce{Fe^{2+} + 2e^- -> Fe(s)} \qquad -0.44\ \text{V}\)
  • \(\ce{Fe^{2+}}\) is undergoing oxidation. The correct half equation is: \(\ce{Fe(s) -> Fe^{2+} + 2e^-}\)
     

b.    As the reaction progresses:

  • The solution will darken, taking on a more intense green colour.
  • This change occurs because nickel undergoes oxidation to form \(\ce{Fe^{2+}}\) ions, which are released into the solution, thereby increasing its colour intensity.
     

c.    Moles of solid copper formed on electrode: \(\dfrac{m}{MM}=\dfrac{0.175}{63.55}=2.36 \times 10^{-3}\ \text{mol}\)

Moles of copper taken out of solution: \(2.36 \times 10^{-3}\ \text{mol}\)

Moles of copper remaining in solution: \((0.15 \times 0.18)-2.36 \times 10^{-3}= 0.0246\ \text{mol}\)

Final concentration: \(c=\dfrac{n}{V}=\dfrac{0.0246}{0.18}=0.137\ \text{mol L}^{-1}\)

CHEMISTRY, M3 EQ-Bank 20

During a laboratory investigation, a student mixed two solutions and observed a sudden colour change, an increase in temperature, and the formation of bubbles.

  1. Explain why these observations indicate a chemical change.   (3 marks)
  1. Describe two types of chemical reactions that could cause at least two of these observations each.   (2 marks)
Show Answers Only

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

 

Show Worked Solution

a.   Colour Change:

  • This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

  • The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

  • The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.
     

b.   Acid-Base Reaction:

  • When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

  • Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

CHEMISTRY, M3 EQ-Bank 16

Describe how activation energy, collision frequency, and molecular orientation work together to determine the rate of a chemical reaction. In your answer, define what each term refers to and relate these factors to collision theory.   (5 marks)

Show Answers Only
  • Activation Energy: For a reaction to occur, the colliding molecules must have enough energy to overcome the activation energy barrier, which is the minimum energy required to break the bonds in the reactants and initiate the reaction. 
  • Collision Frequency: The rate of a reaction is also influenced by how frequently reactant molecules collide.
  • Molecular Orientation: In addition to having enough energy, molecules must collide with the correct orientation for a reaction to take place. Reactant molecules need to align in a way that allows bonds to break and new bonds to form. 
  • Increasing collision frequency increases the number of opportunities for molecules to collide, but only those collisions with enough energy and the correct orientation will lead to successful bond rearrangements.
  • For the maximum rate of reaction there needs to be a lower activation energy which makes it easier for collisions to result in a reaction, the proper orientation that ensures when collisions occur, they lead to the formation of products and a high collision frequency.
Show Worked Solution
  • Activation Energy: For a reaction to occur, the colliding molecules must have enough energy to overcome the activation energy barrier, which is the minimum energy required to break the bonds in the reactants and initiate the reaction. 
  • Collision Frequency: The rate of a reaction is also influenced by how frequently reactant molecules collide.
  • Molecular Orientation: In addition to having enough energy, molecules must collide with the correct orientation for a reaction to take place. Reactant molecules need to align in a way that allows bonds to break and new bonds to form. 
  • Increasing collision frequency increases the number of opportunities for molecules to collide, but only those collisions with enough energy and the correct orientation will lead to successful bond rearrangements.
  • For the maximum rate of reaction there needs to be a lower activation energy which makes it easier for collisions to result in a reaction, the proper orientation that ensures when collisions occur, they lead to the formation of products and a high collision frequency.

CHEMISTRY, M3 EQ-Bank 28v4

A student stirs 2.50 g of silver (I) nitrate powder into 100.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)
  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)
Show Answers Only

a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(2.11 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\)

b.    \(\ce{n(AgNO3) = \frac{m}{M} = \frac{2.50}{107.9 + 14.01 + 16.00 \times 3} = \frac{2.50}{169.91} = 0.01471 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.100 = 0.150 \text{ mol}}\)

\(\ce{AgNO3} \text{ is the limiting reagent}\)

\(\ce{n(AgCl) = n(AgNO3) = 0.01471 \text{ mol}}\)

\(\ce{m(AgCl) = n \times M = 0.01471 \times (107.9 + 35.45) = 0.01471 \times 143.35 = 2.11 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

CHEMISTRY, M3 EQ-Bank 28v2

A student stirs 3.50 g of copper (II) nitrate powder into 150.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)
  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)
Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(2.51 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(\ce{n(Cu(NO3)2) = \frac{m}{M} = \frac{3.50}{63.55 + 2 \times (14.01 + 16.00 \times 3)} = \frac{3.50}{187.57} = 0.01866 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.150 = 0.225 \text{ mol}}\)

\(\ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCl2) = n(Cu(NO3)2) = 0.01866 \text{ mol}}\)

\(\ce{m(CuCl2) = n \times M = 0.01866 \times (63.55 + 2 \times 35.45) = 0.01866 \times 134.45 = 2.51 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

CHEMISTRY, M4 EQ-Bank 6

Explain why using bond energies is not an accurate method of calculating enthalpy changes.   (2 marks)

Show Answers Only
  • Enthalpy changes calculated using bond energies are not entirely accurate because bond energies represent averages of similar bond dissociations.
  • The actual energy of a specific bond can vary depending on the electrochemical environment around it, such as the presence of nearby electronegative atoms, adjacent double bonds, or the overall size of the molecule.
  • The energy of a specific bond depends also on the pressure, temperature, and state of the molecule. Bond energies assume standard laboratory conditions and all compounds already being in gas form and so are inaccurate. 
Show Worked Solution
  • Enthalpy changes calculated using bond energies are not entirely accurate because bond energies represent averages of similar bond dissociations.
  • The actual energy of a specific bond can vary depending on the electrochemical environment around it, such as the presence of nearby electronegative atoms, adjacent double bonds, or the overall size of the molecule.
  • The energy of a specific bond depends also on the pressure, temperature, and state of the molecule. Bond energies assume standard laboratory conditions and all compounds already being in gas form and so are inaccurate. 

CHEMISTRY, M4 EQ-Bank 5

The chemical equation for the complete combustion of ethane  \(\ce{(C2H6)}\)  is given below:

\(\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)}\)

The structural formula for ethane and standard bond energies for provided for you.

\begin{array} {|c|c|}
\hline Bond & Enthalpy \text{ (kJ mol}^{-1})  \\     
\hline \ce{C-C} & 348 \\
\hline \ce{O-O} & 146 \\
\hline \ce{O=O} & 495 \\
\hline \ce{C-O} & 358 \\
\hline \ce{C=O} & 799 \\
\hline \ce{C-H} & 413 \\
\hline \ce{H-O} & 463 \\
\hline \end{array}

Using the bond energies provided, calculate the enthalpy for the complete combustion of one mole of ethane.   (3 marks)

Show Answers Only

\(-1415.5\ \text{kJ mol}^{-1}\)

Show Worked Solution
\(\Delta H\) \(=\Sigma\,{\text{bonds broken}}-\Sigma\,{\text{bonds formed}}\)  
  \(=((12 \times 413) + (2 \times 348) + (7 \times 495))-((8 \times 799) + (12 \times 463))\)  
  \(=9117-11948\)  
  \(=-2831\)  (for two moles of ethane, as per the equation)  
     
  • \(\Delta H\) for the combustion of one mole of ethane is \(-1415.5 \text{ kJ mol}^{-1}\)

CHEMISTRY, M4 EQ-Bank 4

The chemical equation for the combustion of butane \(\ce{(C4H10)}\) is given below:

\(\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)} \qquad \Delta H = -5754\ \text{kJ mol}^{-1}\)

Given that the standard enthalpy of formation of \(\ce{CO2(g)}\) is –393 kJ mol\(^{-1}\) and \(\ce{H2O(g)}\) is –241 kJ mol \(^{-1}\), calculate the standard enthalpy of formation of butane.   (3 marks)

Show Answers Only

\(-126\ \text{kJ mol}^{-1}\)

Show Worked Solution
  • The enthalpy change for the combustion of 2 moles of butane is -5754 kJ.
\(\Delta H\) \(=\Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}\)  
\(-5754\) \(=(8 \times -393 + 10 \times -241)-(2 \times \Delta H_f \text{ (butane)})\)  
\(-5754\) \(=-4334-2 \times \Delta H_f \text{ (butane)}\)  
\(\Delta H_f \text{ (butane)}\) \(=\dfrac{-4334 + 5754}{2}\)  
  \(=-126\ \text{kJ mol}^{-1}\)  
     
  • The standard enthalpy of formation of butane is \(-126\ \text{kJ mol}^{-1}\).