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HMS, BM EQ-Bank 62

Explain how the respiratory and circulatory systems work together to support an athlete during a 5 kilometre run.   (6 marks)

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  • During a 5km run, increased muscle activity creates higher oxygen demand, which triggers the respiratory system to increase breathing rate and depth.
  • This enhanced ventilation allows more oxygen to reach the alveoli for gas exchange.
  • Simultaneously, the circulatory system increases heart rate and stroke volume, resulting in greater cardiac output.
  • These cardiovascular changes facilitate more oxygenated blood to be delivered to working muscles.
  • Gas exchange intensifies at the alveoli, where oxygen diffuses from air to blood due to concentration gradients.
  • The increased blood flow creates optimal conditions for oxygen uptake in the lungs.
  • At muscle capillaries, oxygen diffuses from blood to tissues, while carbon dioxide moves in the opposite direction.
  • This continuous exchange cycle ensures sustained aerobic energy production throughout the run.
  • Both systems adjust breathing and heart rate proportionally to exercise intensity, maintaining adequate oxygen supply.
  • As a result, the coordinated response of both systems enables the athlete to sustain performance during the 5km run.
Show Worked Solution
  • During a 5km run, increased muscle activity creates higher oxygen demand, which triggers the respiratory system to increase breathing rate and depth.
  • This enhanced ventilation allows more oxygen to reach the alveoli for gas exchange.
  • Simultaneously, the circulatory system increases heart rate and stroke volume, resulting in greater cardiac output.
  • These cardiovascular changes facilitate more oxygenated blood to be delivered to working muscles.
  • Gas exchange intensifies at the alveoli, where oxygen diffuses from air to blood due to concentration gradients.
  • The increased blood flow creates optimal conditions for oxygen uptake in the lungs.
  • At muscle capillaries, oxygen diffuses from blood to tissues, while carbon dioxide moves in the opposite direction.
  • This continuous exchange cycle ensures sustained aerobic energy production throughout the run.
  • Both systems adjust breathing and heart rate proportionally to exercise intensity, maintaining adequate oxygen supply.
  • As a result, the coordinated response of both systems enables the athlete to sustain performance during the 5km run.

HMS, BM EQ-Bank 60 MC

During high-intensity exercise, what best explains why an athlete's breathing rate increases?

  1. To increase tidal volume in the pulmonary arteries
  2. To decrease carbon dioxide levels in the alveoli
  3. To meet increased oxygen demands of working muscles
  4. To reduce blood flow through the pulmonary circuit
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct. Breathing rate increases primarily to meet the increased oxygen demands of working muscles during intense exercise.

Other Options:

  • A is incorrect: Tidal volume refers to air volume, not blood in pulmonary arteries.
  • B is incorrect: While CO2 removal increases, this is a result not the primary reason.
  • D is incorrect: Blood flow through pulmonary circuit actually increases during exercise.

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 56

Explain how the heart's structure supports blood flow during a 400 metre sprint.   (6 marks)

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Sample Answer

  • The heart’s four-chamber structure separates oxygenated and deoxygenated blood, which ensures muscles receive only oxygen-rich blood during sprinting.
  • The left ventricle’s thick muscular walls enable powerful contractions, therefore generating high pressure to pump blood throughout the body.
  • During a 400m sprint, these thick walls allow stroke volumes to double, resulting in increased oxygen delivery to working muscles.
  • Four one-way valves slam shut between beats, which prevents backflow despite rapid heart rates during sprinting.
  • This valve function is crucial because it maintains forward blood flow even when heart rate increases dramatically.
  • Coronary arteries branch immediately from the aorta, consequently prioritising oxygen delivery to the heart muscle during extreme demand.
  • The aorta’s elastic nature allows it to stretch with each contraction then recoil, which maintains blood pressure between beats.
  • Atrial chambers act as primer pumps, ensuring ventricles fill completely despite shortened filling time.
  • As a result, this coordinated structure enables cardiac output to increase five-fold during maximal sprinting.
Show Worked Solution

Sample Answer

  • The heart’s four-chamber structure separates oxygenated and deoxygenated blood, which ensures muscles receive only oxygen-rich blood during sprinting.
  • The left ventricle’s thick muscular walls enable powerful contractions, therefore generating high pressure to pump blood throughout the body.
  • During a 400m sprint, these thick walls allow stroke volumes to double, resulting in increased oxygen delivery to working muscles.
  • Four one-way valves slam shut between beats, which prevents backflow despite rapid heart rates during sprinting.
  • This valve function is crucial because it maintains forward blood flow even when heart rate increases dramatically.
  • Coronary arteries branch immediately from the aorta, consequently prioritising oxygen delivery to the heart muscle during extreme demand.
  • The aorta’s elastic nature allows it to stretch with each contraction then recoil, which maintains blood pressure between beats.
  • Atrial chambers act as primer pumps, ensuring ventricles fill completely despite shortened filling time.
  • As a result, this coordinated structure enables cardiac output to increase five-fold during maximal sprinting.

HMS, BM EQ-Bank 54 MC

During a netball game, which sequence accurately shows how the respiratory and circulatory systems work together in the goal shooter's muscles?

  1. Decreased lung capacity → slower blood flow → reduced oxygen to muscles
  2. Steady breathing rate → reduced heart rate → increased muscle oxygen
  3. Increased breathing rate → enhanced blood flow → greater oxygen delivery
  4. Rapid breathing → decreased circulation → higher muscle oxygen
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: It shows correct relationship between systems during sport-specific movement.

Other options:

  • All other options contain physiologically incorrect relationships.

HMS, BM EQ-Bank 50

Explain how the biomechanical principles of force and fluid mechanics interrelate with the musculoskeletal system to enable safe diving entry into water.   (5 marks)

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Sample Answer

  • The musculoskeletal system generates force through coordinated muscle contractions in legs and core during springboard compression. The reason for this is that muscles work in sequence from larger leg muscles to smaller stabilisers. Such sequencing creates optimal force transfer through aligned joints for maximum upward propulsion.
  • Joint angles at takeoff directly influence force direction and body trajectory. Consequently, properly flexed knees and extended ankles enable force to travel through the skeletal system efficiently. At a deeper level, correct alignment produces the parabolic flight path needed for safe entry angles.
  • During flight, core muscles maintain rigid body alignment to prepare for water entry. More specifically, muscular tension transforms the body into a streamlined projectile. In turn, streamlining reduces surface area contacting water and minimises impact forces through fluid dynamics principles.
  • Arms positioned overhead with biceps covering ears create a wedge shape for initial water penetration. It functions through allowing hands to break water surface tension first. Following this, the wedge generates a cavity for the body to follow, which significantly reduces deceleration forces on spine and joints.
  • The musculoskeletal system absorbs remaining impact forces through controlled muscle tension and joint positioning. Hence, slightly flexed joints and engaged muscles distribute forces throughout the body rather than concentrating them. To put it simply, force distribution prevents injury while maintaining the streamlined position essential for safe entry.
Show Worked Solution

Sample Answer

  • The musculoskeletal system generates force through coordinated muscle contractions in legs and core during springboard compression. The reason for this is that muscles work in sequence from larger leg muscles to smaller stabilisers. Such sequencing creates optimal force transfer through aligned joints for maximum upward propulsion.
  • Joint angles at takeoff directly influence force direction and body trajectory. Consequently, properly flexed knees and extended ankles enable force to travel through the skeletal system efficiently. At a deeper level, correct alignment produces the parabolic flight path needed for safe entry angles.
  • During flight, core muscles maintain rigid body alignment to prepare for water entry. More specifically, muscular tension transforms the body into a streamlined projectile. In turn, streamlining reduces surface area contacting water and minimises impact forces through fluid dynamics principles.
  • Arms positioned overhead with biceps covering ears create a wedge shape for initial water penetration. It functions through allowing hands to break water surface tension first. Following this, the wedge generates a cavity for the body to follow, which significantly reduces deceleration forces on spine and joints.
  • The musculoskeletal system absorbs remaining impact forces through controlled muscle tension and joint positioning. Hence, slightly flexed joints and engaged muscles distribute forces throughout the body rather than concentrating them. To put it simply, force distribution prevents injury while maintaining the streamlined position essential for safe entry.

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 45

Compare the biomechanical principles involved in safe pushing versus pulling movements.   (5 marks)

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Sample Answer

Force Direction and Body Position

Similarities

  • Both movements require staggered stances to create stable base of support

Differences

  • Pushing directs force away from body requiring forward lean, while pulling brings force toward body needing backward lean.
  • Foot positioning varies
    • pushing uses rear leg drive with forward stance
    • pulling anchors through front leg in backward stance.

Muscle Activation Patterns

Similarities

  • Both require core muscle engagement for spinal stability and protection.

Differences

  • Pushing engages anterior muscles (pectorals, anterior deltoids, triceps) while pulling activates posterior muscles (latissimus dorsi, rhomboids, biceps).
  • Joint stress varies
    • pushing loads anterior shoulder structures
    • pulling stresses posterior shoulder and elbow differently

Centre of Gravity and Balance Requirements

Similarities

  • Pushing shifts centre of gravity forward beyond base of support, while pulling positions it behind force application point.
  • Fall risks differ
    • pushing risks forward falls if force releases suddenly
    • pulling risks backward falls
  • Spinal protection varies
    • pushing prevents hyperextension
    • pulling guards against excessive flexion
Show Worked Solution

Sample Answer

Force Direction and Body Position

Similarities

  • Both movements require staggered stances to create stable base of support

Differences

  • Pushing directs force away from body requiring forward lean, while pulling brings force toward body needing backward lean.
  • Foot positioning varies
    • pushing uses rear leg drive with forward stance
    • pulling anchors through front leg in backward stance.

Muscle Activation Patterns

Similarities

  • Both require core muscle engagement for spinal stability and protection.

Differences

  • Pushing engages anterior muscles (pectorals, anterior deltoids, triceps) while pulling activates posterior muscles (latissimus dorsi, rhomboids, biceps).
  • Joint stress varies
    • pushing loads anterior shoulder structures
    • pulling stresses posterior shoulder and elbow differently

Centre of Gravity and Balance Requirements

Similarities

  • Pushing shifts centre of gravity forward beyond base of support, while pulling positions it behind force application point.
  • Fall risks differ
    • pushing risks forward falls if force releases suddenly
    • pulling risks backward falls
  • Spinal protection varies
    • pushing prevents hyperextension
    • pulling guards against excessive flexion

HMS, BM EQ-Bank 43

Describe how biomechanical principles influence the safe execution of a landing from a jump.   (4 marks)

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Sample Answer

Force absorption 

  • Quadriceps, hamstrings and calf muscles contract eccentrically during landing.
  • This controlled lengthening prevents sudden joint compression and distributes forces.

Joint flexion 

  • Ankles, knees and hips bend simultaneously upon ground contact.
  • This flexion increases absorption time and transforms peak forces into manageable loads.

Base of support

  • Feet positioned shoulder-width apart provide lateral stability during landing.
  • This wider stance prevents sideways falling and enables balanced force distribution through both legs.

Centre of gravity

  • Deep knee bend lowers the body’s centre of gravity toward ground.
  • Athletes maintain better equilibrium when mass is positioned lower.
  • Positioning the body in this way enhances balance control reducing fall risk.
Show Worked Solution

Sample Answer

Force absorption 

  • Quadriceps, hamstrings and calf muscles contract eccentrically during landing.
  • This controlled lengthening prevents sudden joint compression and distributes forces.

Joint flexion 

  • Ankles, knees and hips bend simultaneously upon ground contact.
  • This flexion increases absorption time and transforms peak forces into manageable loads.

Base of support

  • Feet positioned shoulder-width apart provide lateral stability during landing.
  • This wider stance prevents sideways falling and enables balanced force distribution through both legs.

Centre of gravity

  • Deep knee bend lowers the body’s centre of gravity toward ground.
  • Athletes maintain better equilibrium when mass is positioned lower.
  • Positioning the body in this way enhances balance control reducing fall risk.

HMS, BM EQ-Bank 32 MC

Which characteristic best describes Type IIb (fast twitch) muscle fibres?

  1. High mitochondrial density
  2. Slow contraction speed
  3. Rapid force production
  4. High fatigue resistance
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: Type IIb fibres produce rapid, powerful contractions.

Other Options:

  • A is incorrect: Characteristic of Type I (slow twitch) fibres.
  • B is incorrect: Type IIb fibres have fast contraction speeds
  • D is incorrect: Type IIb fibres fatigue quickly

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, BM EQ-Bank 22

Describe how the coordination of joint actions at the hip, knee and ankle contributes to the generation of power in a vertical jump. Use biomechanical principles in your response.   (3 marks)

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Sample Answer

Preparation phase:

  • Concurrent flexion at hips, knees and ankles lowers the centre of gravity. This creates a countermovement, initiating the stretch-shortening cycle in leg muscles to store elastic energy.

Propulsive phase:

  • Sequential triple extension follows a proximal-to-distal pattern – hips extend first, then knees, finally ankles. This kinetic chain transfers force upward, with each joint adding velocity to the movement.
  • Ground reaction forces peak when all joints extend together. Coordinated timing maximises vertical impulse at take-off.
Show Worked Solution

Sample Answer

Preparation phase:

  • Concurrent flexion at hips, knees and ankles lowers the centre of gravity. This creates a countermovement, initiating the stretch-shortening cycle in leg muscles to store elastic energy.

Propulsive phase:

  • Sequential triple extension follows a proximal-to-distal pattern – hips extend first, then knees, finally ankles. This kinetic chain transfers force upward, with each joint adding velocity to the movement.
  • Ground reaction forces peak when all joints extend together. Coordinated timing maximises vertical impulse at take-off.

HMS, BM EQ-Bank 19

Describe how the structural differences between the hip joint and shoulder joint reflect their contrasting functional requirements in human movement.   (3 marks)

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Sample Answer

Hip joint:

  • Features a deep acetabulum (socket) that encloses most of the femoral head, with short, strong ligaments providing robust support. This stable structure reflects its primary function of weight-bearing during standing and locomotion.

Shoulder joint:

  • Has a shallow glenoid cavity with loose joint capsule and longer ligaments. This flexible structure allows extensive multi-directional movement required for reaching, throwing, and overhead activities.

Structural Differences:

  • These directly match functional needs: stability for weight-bearing (hip) versus mobility for upper limb activities (shoulder).
Show Worked Solution

Sample Answer

Hip joint:

  • Features a deep acetabulum (socket) that encloses most of the femoral head, with short, strong ligaments providing robust support. This stable structure reflects its primary function of weight-bearing during standing and locomotion.

Shoulder joint:

  • Has a shallow glenoid cavity with loose joint capsule and longer ligaments. This flexible structure allows extensive multi-directional movement required for reaching, throwing, and overhead activities.

Structural Differences:

  • These directly match functional needs: stability for weight-bearing (hip) versus mobility for upper limb activities (shoulder).

HMS, HIC EQ-Bank 013

Explain how TWO sociological determinants can influence young people's participation in risky health behaviours. Provide examples in your response.   (4 marks)

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Answers could include TWO of the following:

Peer group influence

  • Significant sociological determinant as young people often modify their behaviour to gain acceptance from their peer group.
  • For example, a young person may begin vaping or drinking alcohol at social gatherings due to perceived peer pressure and desire to fit in with their social circle.

Religion

  • Religious restrictions can lead young people to rebel against perceived constraints.
  • This can result in participation in prohibited behaviours like underage drinking or drug taking when they feel an emotional need to act out.

Culture

  • Cultural norms within Western societies can normalise unhealthy practices.
  • For example, fast food consumption and tobacco smoking are more acceptable in certain cultures and contribute to obesity and respiratory conditions.
  • Young people often engage in these risky behaviours as they navigate identity formation within their cultural context, particularly when society presents these behaviours as desirable.
Show Worked Solution

Answers could include TWO of the following:

Peer group influence

  • Significant sociological determinant as young people often modify their behaviour to gain acceptance from their peer group.
  • For example, a young person may begin vaping or drinking alcohol at social gatherings due to perceived peer pressure and desire to fit in with their social circle.

Religion

  • Religious restrictions can lead young people to rebel against perceived constraints.
  • This can result in participation in prohibited behaviours like underage drinking or drug taking when they feel an emotional need to act out.

Culture

  • Cultural norms within Western societies can normalise unhealthy practices.
  • For example, fast food consumption and tobacco smoking are more acceptable in certain cultures and contribute to obesity and respiratory conditions.
  • Young people often engage in these risky behaviours as they navigate identity formation within their cultural context, particularly when society presents these behaviours as desirable.

HMS, HIC EQ-Bank 503

Outline how TWO social justice principles are demonstrated through a rural health service's mobile dental clinic that visits remote communities.   (3 marks)

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The social justice principles of equity and creating supportive environments are demonstrated in this mobile dental service.

Equity is shown through

  • bringing essential dental care directly to remote communities
  • reducing geographical barriers
  • ensuring rural residents have similar access to dental services as metropolitan areas

Creating supportive environments is demonstrated by

  • modifying health service delivery to suit the needs of remote communities
  • establishing temporary but fully equipped dental clinics within their local area rather than requiring long-distance travel
Show Worked Solution

The social justice principles of equity and creating supportive environments are demonstrated in this mobile dental service.

Equity is shown through

  • bringing essential dental care directly to remote communities
  • reducing geographical barriers
  • ensuring rural residents have similar access to dental services as metropolitan areas

Creating supportive environments is demonstrated by

  • modifying health service delivery to suit the needs of remote communities
  • establishing temporary but fully equipped dental clinics within their local area rather than requiring long-distance travel

HMS, HIC EQ-Bank 502

Outline how TWO social justice principles are demonstrated in a local council's initiative to provide free exercise classes for seniors in community centres.   (3 marks)

Show Answers Only
  • The social justice principles of equity and participation are demonstrated in this initiative.
  • Equity is shown through providing free exercise classes which removes financial barriers for seniors who may be on fixed incomes.
  • Ensures equal access to physical activity opportunities regardless of economic status.
  • Participation is demonstrated by actively engaging seniors in community-based physical activity programs.
  • Enables seniors to be involved in health-promoting activities while also fostering social connections within their local area.
Show Worked Solution
  • The social justice principles of equity and participation are demonstrated in this initiative.
  • Equity is shown through providing free exercise classes which removes financial barriers for seniors who may be on fixed incomes.
  • Ensures equal access to physical activity opportunities regardless of economic status.
  • Participation is demonstrated by actively engaging seniors in community-based physical activity programs.
  • Enables seniors to be involved in health-promoting activities while also fostering social connections within their local area.

HMS, HIC EQ-Bank 4 MC

A local council supported residents to establish community gardens and cooking programs to improve food security and nutrition in their neighborhood.

Which action area of the Ottawa Charter is this strategy an example of?

  1. Building healthy public policy
  2. Creating supportive environments
  3. Developing personal skills
  4. Strengthening community action
Show Answers Only

\( D\)

Show Worked Solution

Consider option D:  Strengthening community action

  • Initiative empowers residents to take collective action for their own health outcomes
  • Shows community ownership and participation in identifying and addressing local health needs
  • Demonstrates community capacity building through shared gardens and programs
  • Council acts as facilitator while community leads and implements the health initiative
  • Focus is on collective community action rather than individual skills, policy or environment changes

\(\Rightarrow D\)

HMS, HIC EQ-Bank 3 MC

A hospital introduced Aboriginal Health Workers and cultural awareness training for staff to improve health outcomes for First Nations patients.

Which action area of the Ottawa Charter is this strategy an example of?

  1. Building healthy public policy
  2. Reorienting health services
  3. Developing personal skills
  4. Strengthening community action
Show Answers Only

\( B\)

Show Worked Solution

Consider option B: Reorienting health services

  • Involves changing the way healthcare is organised and delivered to better meet population needs
  • The introduction of Aboriginal Health Workers represents a structural change to how hospital services are delivered
  • Cultural awareness training for staff is changing the way health professionals provide care to make it more culturally appropriate and responsive
  • The focus is on the healthcare organisation itself making changes to how it delivers services, which is a defining feature of reorienting health services

\(\Rightarrow B\)

HMS, BM 2022 HSC 9 MC

During a game of touch football, a skilled player successfully passes the ball in a high-pressure situation.

Which of the following identifies the types of feedback the player is most likely to have experienced?

  1. Intrinsic, delayed and knowledge of results
  2. Extrinsic, delayed and knowledge of results
  3. Intrinsic, concurrent and knowledge of performance
  4. Extrinsic, concurrent and knowledge of performance
Show Answers Only

\( C \)

Show Worked Solution
  • C is correct: Player feels ball contact and body position during performance.

Other Options:

  • A is incorrect: Feedback occurs during performance, not delayed.
  • B is incorrect: Internal body sensations, not external sources.
  • D is incorrect: Internal sensations from player, not external sources.

♦ Mean mark 50%.

ENGINEERING, CS 2024 HSC 25c

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

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

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

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

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

Show Answers Only

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

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

 

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

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

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

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

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

Show Worked Solution

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

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

 

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

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

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

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

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

BIOLOGY, M8 EQ-Bank 2

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

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

Show Answers Only

a.   Answers could include one of the following:

Cystic Fibrosis

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

Huntington’s Disease

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

b.  Answers could include one of the following:

Cystic Fibrosis

  • The defective chloride channels cause mucus to build up in the lungs, leading to frequent chest infections and breathing difficulties because bacteria become trapped in the airways.
  • The same mucus also blocks pancreatic ducts, preventing digestive enzymes from reaching the intestines, which results in poor nutrient absorption and growth problems.

Huntington’s Disease

  • The abnormal protein accumulates in brain cells, causing progressive death of neurons that control movement, leading to uncontrolled jerking and twitching movements.
  • The protein buildup also affects regions of the brain controlling cognitive function, resulting in progressive memory loss and personality changes as these neural networks deteriorate
Show Worked Solution

a.   Answers could include one of the following:

Cystic Fibrosis

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

Huntington’s Disease

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

b.  Answers could include one of the following:

Cystic Fibrosis

  • The defective chloride channels cause mucus to build up in the lungs, leading to frequent chest infections and breathing difficulties because bacteria become trapped in the airways.
  • The same mucus also blocks pancreatic ducts, preventing digestive enzymes from reaching the intestines, which results in poor nutrient absorption and growth problems.

Huntington’s Disease

  • The abnormal protein accumulates in brain cells, causing progressive death of neurons that control movement, leading to uncontrolled jerking and twitching movements.
  • The protein buildup also affects regions of the brain controlling cognitive function, resulting in progressive memory loss and personality changes as these neural networks deteriorate

ENGINEERING, PPT 2024 HSC 27b

A towbar hitch and shear pin is shown.
 

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

Show Answers Only

  • Hot forging creates components that exhibit superior strength compared to casting or machining methods.
  • This process enables the production of more intricate geometries while achieving an ideal combination of mechanical properties.
  • These properties include maximum yield strength, increased ductility, and reduced hardness.

Show Worked Solution

  • Hot forging creates components that exhibit superior strength compared to casting or machining methods.
  • This process enables the production of more intricate geometries while achieving an ideal combination of mechanical properties.
  • These properties include maximum yield strength, increased ductility, and reduced hardness.
♦ Mean mark 49%.

ENGINEERING, AE 2024 HSC 26c

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

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

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

 

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

     

 

 

Show Answers Only

i.    
         

ii.    
       

Show Worked Solution

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

ENGINEERING, PPT 2024 HSC 24d

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

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


 

Show Answers Only

Exemplar solution 1:

  • Steel in its fully hardened state exhibits extreme brittleness, limiting its practical applications and making it vulnerable to failure.
  • When exposed to abrupt forces or stress concentrations, fully hardened steel can develop cracks or break catastrophically due to its inability to deform.
  • The tempering process plays a vital role in optimising steel’s mechanical properties, creating an essential balance between hardness and toughness.
  • Through careful tempering, the steel’s characteristics can be precisely tuned to match specific operational requirements, ensuring reliable performance across diverse applications.
  •  Martensite by hardening:

Exemplar solution 2:

  • The tempering process serves as a crucial follow-up treatment after quenching, specifically designed to mitigate the steel’s brittle characteristics.
  • The procedure involves carefully reheating the hardened steel to a temperature below its critical point, followed by another cooling cycle.
  • This methodical heating and cooling sequence helps release internal stresses that developed during the initial quenching phase.
  • Through tempering, manufacturers can precisely adjust the balance between the steel’s hardness and toughness to achieve desired mechanical properties.
  • Ferrite and finely dispersed cementite:

Show Worked Solution

Exemplar solution 1:

  • Steel in its fully hardened state exhibits extreme brittleness, limiting its practical applications and making it vulnerable to failure.
  • When exposed to abrupt forces or stress concentrations, fully hardened steel can develop cracks or break catastrophically due to its inability to deform.
  • The tempering process plays a vital role in optimising steel’s mechanical properties, creating an essential balance between hardness and toughness.
  • Through careful tempering, the steel’s characteristics can be precisely tuned to match specific operational requirements, ensuring reliable performance across diverse applications.
  • Martensite by hardening:

♦♦ Mean mark 32%.

Exemplar solution 2:

  • The tempering process serves as a crucial follow-up treatment after quenching, specifically designed to mitigate the steel’s brittle characteristics.
  • The procedure involves carefully reheating the hardened steel to a temperature below its critical point, followed by another cooling cycle.
  • This methodical heating and cooling sequence helps release internal stresses that developed during the initial quenching phase.
  • Through tempering, manufacturers can precisely adjust the balance between the steel’s hardness and toughness to achieve desired mechanical properties.
  • Ferrite and finely dispersed cementite:

ENGINEERING, PPT 2024 HSC 24b

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

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

Show Answers Only

  • Electronic commutation in brushless DC motors enables straightforward control and enhanced operational efficiency.
  • This precise control of speed and torque makes these motors particularly suitable for e-bike applications.
  • Since there are no brushes to wear down, these motors offer improved durability and longevity.
  • Their ability to deliver substantial torque even at lower speeds gives riders the power they need when starting from a stop or tackling inclines.

Answers could also include the following points:

  • Regenerative braking
  • Efficiency
  • Compact size and lightweight.

Show Worked Solution

  • Electronic commutation in brushless DC motors enables straightforward control and enhanced operational efficiency.
  • This precise control of speed and torque makes these motors particularly suitable for e-bike applications.
  • Since there are no brushes to wear down, these motors offer improved durability and longevity.
  • Their ability to deliver substantial torque even at lower speeds gives riders the power they need when starting from a stop or tackling inclines.

Answers could also include the following points:

  • Regenerative braking
  • Efficiency
  • Compact size and lightweight.
♦ Mean mark 47%.

ENGINEERING, PPT 2024 HSC 24a

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

Show Answers Only

  • Laboratory crash simulations assess how helmets will perform in real world accidents.
  • These evaluations measure the helmet’s ability to absorb forces, maintain its structure, and meet required safety certifications.

Show Worked Solution

  • Laboratory crash simulations assess how helmets will perform in real world accidents.
  • These evaluations measure the helmet’s ability to absorb forces, maintain its structure, and meet required safety certifications.
♦ Mean mark 50%.

PHYSICS, M7 2024 HSC 32

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

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

Show Answers Only

Students could include any of the following experiments:

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

Young’s Double-Slit Experiment:

  • Young’s 1801 double slit experiment aimed to determine light’s wave-particle nature.
  • He passed coherent light through two slits and observed the pattern on a screen.
  • Instead of Newton’s predicted two bright bands, Young observed alternating bright and dark bands.
  • This interference pattern occurred due to light diffraction and interference, which  re wave properties.
  • The experiment provided strong evidence for light behaving as a wave at macroscopic scales. 

Planck and the Blackbody Radiation Crisis:

  • Late 19th century scientists studied the relationship between black body radiation’s wavelength and intensity.
  • Experimental observations showed intensity peaked at a specific wavelength, contradicting classical physics predictions.
  • Classical physics led to the “ultraviolet catastrophe,” which violated energy conservation.
  • Planck’s thought experiment resolved this by proposing energy was transferred in discrete packets (quanta) where  \(E=hf\).
  • This revolutionary idea marked a shift from classical physics to quantum theory. 

Einstein and the Photoelectric Effect:

  • In 1905, Einstein built upon Plank’s idea of quantised energy to propose that light was made up of quantised photons where \(E=hf\).
  • Einstein proposition explained why electrons are ejected from metal surfaces only when light exceeds a minimum frequency.
  • Previous to Einstein’s explanation of the photoelectric effect a high intensity of light corresponds to a high energy.
  • Einstein proposed that the KE of the emitted electrons was proportion to the frequency of the light rather than the intensity of the light. 
  • This development in the understanding of the interaction of light and matter at the atomic level shifted our understanding of light to a wave-particle duality model.

Cosmic Ray Experiments and the development of the Standard Model:

  • In 1912, Victor Hess discovered cosmic rays through high-altitude balloon experiments, finding that radiation increased with altitude rather than decreased as expected.
  • The study of cosmic rays led to the unexpected discovery of new particles, including the positron and muon, which couldn’t be explained by the known models of matter.
  • These discoveries from cosmic rays helped inspire the development of modern particle accelerators and contributed to the formulation of the quark model in the 1960s.
  • Eventually further studies on these newly discovered particles led to the development of the Standard Model of particle physics, which organises all known elementary particles and their interactions.

Show Worked Solution

Students could include any of the following experiments:

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

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%.

ENGINEERING, CS 2024 HSC 20 MC

A cross-section of a beam is shown.

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

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

\(A\)

Show Worked Solution

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

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

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

\(\Rightarrow A\)

♦ Mean mark 44%.

ENGINEERING, CS 2024 HSC 21b

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

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

Show Answers Only

Answers could include three of the following:

  • They facilitate drainage in water-prone areas while simultaneously distributing structural loads more evenly across the soil base.
  • Additionally, these materials create a stable foundation that prevents soil erosion by holding soil particles in place against water movement.
  • Geotextiles enhance soil stability, by preventing mixing of different soil layers and
    improving load-bearing capacity.
  • They reinforce the soil mass, by increasing the overall strength and longevity of the retaining wall system.

Show Worked Solution

Answers could include three of the following:

  • They facilitate drainage in water-prone areas while simultaneously distributing structural loads more evenly across the soil base.
  • Additionally, these materials create a stable foundation that prevents soil erosion by holding soil particles in place against water movement.
  • Geotextiles enhance soil stability, by preventing mixing of different soil layers and improving load-bearing capacity.
  • They reinforce the soil mass, by increasing the overall strength and longevity of the retaining wall system.
♦ Mean mark 50%.

ENGINEERING, TE 2024 HSC 17 MC

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

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

\(C\)

Show Worked Solution
  • Single-mode fibre optic cable only allows one path for light to travel, reducing signal dispersion and allowing for faster, clearer transmission over long distances.
  • This results in a higher transmission rate and bandwidth compared to multi-mode cables which allow multiple light paths.

\(\Rightarrow C\)

♦♦ Mean mark 35%.

ENGINEERING, CS 2024 HSC 16 MC

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

The load is then increased.

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

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

\(D\)

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

\(\Rightarrow D\)

♦ Mean mark 46%.

ENGINEERING, PPT 2024 HSC 15 MC

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

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

\(B\)

Show Worked Solution
  • Ductility is the ability of the steel to be plastically deformed without breaking.
  • This is essential for pressing and stamping sheet metal into the complex curves and shapes required for car door panels.

\(\Rightarrow B\)

♦♦ Mean mark 32%.

ENGINEERING, PPT 2024 HSC 10 MC

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

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

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

\(B\)

Show Worked Solution

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

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

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

 
\(\Rightarrow B\)

♦ Mean mark 40%.

ENGINEERING, PPT 2024 HSC 9 MC

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

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

\(D\)

Show Worked Solution
  • In the glass manufacturing process, molten glass is poured onto a bed of molten tin.
  • Tin provides an extremely flat and smooth surface due to tin’s high density and surface tension, allowing the glass to float and spread evenly while cooling.

\(\Rightarrow D\)

♦♦ Mean mark 33%.

ENGINEERING, AE 2024 HSC 8 MC

The cross-section of a tapered tube is shown.
 

   

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

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

Show Answers Only

\(B\)

Show Worked Solution
  • According to Bernoulli’s principle, as the air velocity increases through the tapered section, the pressure must decrease.
  • This results in high pressure at the wide entrance (gauge A) where velocity is low and low pressure at the narrow exit (gauge B) where velocity is high.

\(\Rightarrow B\)

♦ Mean mark 46%.

ENGINEERING, CS 2024 HSC 6 MC

A shear force diagram is shown.

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

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

\(D\)

Show Worked Solution
  • For a simply supported beam with a uniformly distributed load, the shear force diagram shows a linear variation that begins with a positive reaction at the left support and decreases linearly to a negative value at the right support.
  • This linear pattern occurs because the distributed load gradually transfers the shear force from one support to the other, with zero shear at the beam’s midspan where the direction of shear changes.

\(\Rightarrow D \)

♦ Mean mark 55%.

PHYSICS, M6 2024 HSC 33

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

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

Show Answers Only

  • When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.
  • As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.
  • The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).
  • This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.
  • The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.
  • Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.

Show Worked Solution

  • When the magnet swings down from its high position toward the can, its gravitational potential energy transforms into kinetic energy.
  • As the magnet moves, it creates changing magnetic flux through the aluminium can. This flux change is strongest when there’s the fastest relative motion between the magnet and can.
  • The induced emf is described in the equation  \({\large{\varepsilon}} = -N \dfrac{\Delta \phi}{\Delta t} \).
  • This emf creates eddy currents in the can, which produce both heat and a magnetic field. Following Lenz’s law, this magnetic field opposes the magnet’s motion.
  • The magnetic fields from both the magnet and the eddy currents interact, causing the can to initially rotate clockwise.
  • Eventually, this interaction dampens the magnet’s swing. The magnetic interaction between the eddy currents and the magnet causes the can to rotate back and forth with decreasing amplitude, as the system’s energy gradually converts to heat.
♦♦ Mean mark 47%.

PHYSICS, M5 2024 HSC 31

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

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

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

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

Show Answers Only

  • Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.
  • In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.
  • In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.
  • The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.

Show Worked Solution

  • Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.
  • In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.
  • In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.
  • The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.
♦ Mean mark 47%.

PHYSICS, M6 2024 HSC 29

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

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

 

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


 
 
 

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

Show Answers Only

a.   
           
  

b.   Evaluate statement:

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

  • While rod \(B\) shows a greater displacement from its starting position compared to rod \(A\), this isn’t because of a larger current in rod \(A\).
  • Following Newton’s third law, the electromagnetic force that rod \(A\) exerts on rod \(B\) is equal in magnitude to the force that rod B exerts on rod \(A\).
  • The magnitude of the electromagnetic force between the rods can be calculated using the formula:  \(\dfrac{F}{l} = \dfrac{\mu_0}{2\pi} \, \dfrac{I_1 I_2}{r}\).
  • Rod \(B\) has a larger displacement over the same time \((t)\ \Rightarrow\) Rod \(B\) experiences greater acceleration.
  • Since the greater acceleration of Rod \(B\) isn’t due to a stronger electromagnetic force, we can conclude that Rod \(B\) has less mass than Rod \(A\) \(\Big(a=\dfrac{F}{m}\Big)\).
  • Since Rod \(B\) experiences more acceleration, its displacement is larger over a given time \((t)\) according to the formula  \(s=\dfrac{1}{2}at^{2} \).

Show Worked Solution

a.   
           
♦ Mean mark (a) 47%.

b.   Evaluate statement:

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

  • While rod \(B\) shows a greater displacement from its starting position compared to rod \(A\), this isn’t because of a larger current in rod \(A\).
  • Following Newton’s third law, the electromagnetic force that rod \(A\) exerts on rod \(B\) is equal in magnitude to the force that rod B exerts on rod \(A\).
  • The magnitude of the electromagnetic force between the rods can be calculated using the formula:  \(\dfrac{F}{l} = \dfrac{\mu_0}{2\pi} \, \dfrac{I_1 I_2}{r}\).
  • Rod \(B\) has a larger displacement over the same time \((t)\ \Rightarrow\) Rod \(B\) experiences greater acceleration.
  • Since the greater acceleration of Rod \(B\) isn’t due to a stronger electromagnetic force, we can conclude that Rod \(B\) has less mass than Rod \(A\) \(\Big(a=\dfrac{F}{m}\Big)\).
  • Since Rod \(B\) experiences more acceleration, its displacement is larger over a given time \((t)\) according to the formula  \(s=\dfrac{1}{2}at^{2} \).
♦ Mean mark (b) 47%.

PHYSICS, M6 2024 HSC 28

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

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

Ignore gravitational effects on the electron beam.
 

 

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

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

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

Show Answers Only

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

♦ Mean mark (c) 40%.

PHYSICS, M7 2024 HSC 27

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

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

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

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

The following tables provide information about hadrons and quarks.

Table 1: Hadron Information

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

 
Table 2: Quark charges

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

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

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

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

Show Answers Only

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

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

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

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

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

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

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

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

c.   Problem with prediction:

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

Resolving the problem:

  • Since these pions are moving at extremely high speeds, we need to account for relativity.
  • Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.
  • Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

Show Worked Solution

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

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

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

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

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

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

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

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

♦ Mean mark (b) 41%.

c.   Problem with prediction:

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

Resolving the problem:

  • Since these pions are moving at extremely high speeds, we need to account for relativity.
  • Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.
  • Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.
♦ Mean mark (c) 43%.

Calculus, MET1 2024 VCAA 8

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

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

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

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

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

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

Show Answers Only

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

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

c.    \(k=m^3\)

d.    \(Q(2m^3,2m)\)

Show Worked Solution

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

\(P(0, -\sqrt[3]{k}+m)\)
 

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

  \(g^{\prime}(x)\) \(=\dfrac{1}{3}(x-k)^{-\frac{2}{3}}\)
  \(g^{\prime}(0)\) \(=\dfrac{1}{3}(0-k)^{-\frac{2}{3}}=\dfrac{1}{3}(-k)^{-\frac{2}{3}} =\dfrac{1}{3}(k)^{-\frac{2}{3}}\)
♦ 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 6

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

Show Answers Only

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

Show Worked Solution

\(2\log_3(x-4)+\log_3(x)\) \(=2\)
\(\log_3x(x-4)^2\) \(=2\)
\(x(x-4)^2\) \(=3^2\)
\(x(x^2-8x+16)-9\) \(=0\)
\(x^3-8x^2+16x-9\) \(=0\)

 
\(\text{Find a factor}\ \ \Rightarrow\ \ \text{Test}\ \ x=1:\)

\(1^3-8(1)^2+16(1)-9=0\)

\(\therefore\ x-1\ \text{is a factor} \)

♦♦ Mean mark 36%.

\((x-1)(x^2-7x+9)=0\)
  

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

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

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

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

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

Functions, MET1 2024 VCAA 5

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

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

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

  4. In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
    1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads.
    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%.

Probability, MET1 2024 VCAA 4

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

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

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

Show Answers Only

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

b.    \(\dfrac{37}{10\,000}\)

Show Worked Solution

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

 

b.     \(\operatorname{Pr}(X<2)\) \(=\operatorname{Pr}(X=0)+\operatorname{Pr}(X=1)\)
    \(=\ ^4C _0\left(\dfrac{9}{10}\right)^0\left(\dfrac{1}{10}\right)^4+\ ^4C_1\left(\dfrac{9}{10}\right)^1\left(\dfrac{1}{10}\right)^3\)
    \(= 1 \times \dfrac {1}{10\,000} + 4 \times \dfrac{9}{10} \times \dfrac{1}{1000}\)
    \(=\dfrac{37}{10\,000}\)
Mean mark (b) 51%.

Calculus, MET1 2024 VCAA 3

Let  \(g: R \backslash\{-3\} \rightarrow R, \ g(x)=\dfrac{1}{(x+3)^2}-2\).

  1. On the axes below, sketch the graph of  \(y=g(x)\),  labelling all asymptotes with their equations and axis intercepts with their coordinates.   (3 marks)


  2. Determine the area of the region bounded by the line  \(x=-2\),  the \(x\)-axis, the \(y\)-axis and the graph of \(y=g(x)\).   (2 marks)

Show Answers Only

a.

b.    \(\dfrac{10}{3}\ \text{sq units}\)

Show Worked Solution

a.    \(y\text{-intercept:}\ x=0\)

\(y=\dfrac{1}{(0+3)^2}-2=-\dfrac{17}{9}\)

\(x\text{-intercepts:}\ y=0\)

\(\dfrac{1}{(x+3)^2}-2\) \(=0\)
\((x+3)^2\) \(=\dfrac{1}{2}\)
\(x+3\) \(=\pm\dfrac{1}{\sqrt{2}}\)
\(x\) \(=-3\pm\dfrac{1}{\sqrt{2}}\)

b.   \(\text{Area is below}\ x\text{-axis:}\)

  \(\text{Area}\) \(=-\displaystyle\int_{-2}^0 (x+3)^{-3}-2\,dx\)
    \(=-\left[\dfrac{1}{-1}(x+3)^{-1}-2x\right]_{-2}^0\)
    \(=-\left[\dfrac{-1}{x+3}-2x\right]_{-2}^0\)
    \(=-\left[\dfrac{-1}{3}-\left(\dfrac{-1}{-2+3}-2(-2)\right)\right]\)
    \(=-\Big[\dfrac{-1}{3}-(-1+4)\Big] \)
    \(=\dfrac{10}{3}\ \text{u}^{2}\)
♦ Mean mark (b) 40%.

Calculus, MET2 2024 VCAA 3

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

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

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

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

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

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

     i. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\).   (3 mark)

    ii. Let \(q:(12,24] \rightarrow R, q(t)=p(t-h)+k\) be a cubic function obtained by translating \(p\), which can be used to model monthly online sales in 2022.

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

  2. Another function \(f\) can be used to model monthly online sales, where
     
    \(f:(0,36] \rightarrow R, f(t)=3000+30 t+700 \cos \left(\dfrac{\pi t}{6}\right)+400 \cos \left(\dfrac{\pi t}{3}\right)\)

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

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

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

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

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

aii.  \(h=12, k=350\)

bi.  

bii.  \( n=360\)

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

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

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

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

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

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

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

 

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

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

bii.  \(\text{Using CAS: }\)

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

\(\therefore\ n=360\)

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

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

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

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

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

Calculus, MET2 2024 VCAA 2

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

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

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

  1. Express the derivative \(f^{\prime}(t)\) as a hybrid function.   (2 marks)
  1. Find the average rate of change in temperature predicted by the model between \(t=0\) and \(t=\dfrac{1}{2}\).
  2. Give your answer in degrees Celsius per hour.   (1 mark)
  1. Another model for the temperature in the room is given by \(g(t)=22-10 e^{-6 t}, t \geq 0\).
  2.  i. Find the derivative \(g^{\prime}(t)\).   (1 mark)
  3. ii. Find the value of \(t\) for which \(g^{\prime}(t)=10\).
  4.     Give your answer correct to three decimal places.   (1 mark)
  1. Find the time \(t \in(0,1)\) when the temperatures predicted by the models \(f\) and \(g\) are equal.
  2. Give your answer correct to two decimal places.   (1 mark)
  1. Find the time \(t \in(0,1)\) when the difference between the temperatures predicted by the two models is the greatest.
  2. Give your answer correct to two decimal places.   (1 mark)
  1. The amount of power, in kilowatts, used by the heater \(t\) hours after it is switched on, can be modelled by the continuous function \(p\), whose graph is shown below.

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

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

  1.   i. Given that \(p(t)\) is continuous for \(t \geq 0\), show that \(A=1.2 e^4\).   (1 mark)
  2.  ii. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.
  3.     Give your answer in hours.   (1 mark)
  4. iii. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.
  5.     Give your answer in hours, correct to two decimal places.   (2 marks)

Show Answers Only

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

b.    \(20^{\circ}\text{C/h}\)

ci.    \(g^{\prime}(t)=60e^{-6t}\)

cii.   \(0.299\ \text{(3 d.p.)}\)

d.    \(0.27\ \text{(2 d.p.)}\)

e.    \(0.12\ \text{(2 d.p.)}\)

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

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

fii.  \(\dfrac{1}{3}\ \text{hours}\)

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

Show Worked Solution

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

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

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

 

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

 

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

 

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

\(\text{Using CAS:}\)

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

 

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

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

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

 

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

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

  
fii.  \(\text{Using CAS:}\)

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

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

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

fiii. \(\text{Using CAS:}\)

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

 

PHYSICS, M7 2024 HSC 26

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

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

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

Show Answers Only

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

Muon’s frame of reference:

  • The distance to the Earth’s surface is contracted according to  \(l=l_o\sqrt{1-\frac{v^2}{c^2}}\). Muons see the Earth’s surface move towards them at speeds close to \(c\).
  • Since the muons have to travel a shorter distance than the proper length, they will have time to reach the Earth’s surface before they decay. 

Earth’s frame of reference:

  • The time that it takes the muon to decay will be dilated according to  \(t=\dfrac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\)  as the muon’s are moving close to the speed of light.
  • Therefore, the muons have a longer half-life and lifespan than predicted by classical physics and will be able to reach the Earth’s surface before they decay.
  • In this way, muons can reach the surface of the Earth from either frame of reference. 

Show Worked Solution

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

Muon’s frame of reference:

  • The distance to the Earth’s surface is contracted according to  \(l=l_o\sqrt{1-\frac{v^2}{c^2}}\). Muons see the Earth’s surface move towards them at speeds close to \(c\).
  • Since the muons have to travel a shorter distance than the proper length, they will have time to reach the Earth’s surface before they decay.

Earth’s frame of reference:

  • The time that it takes the muon to decay will be dilated according to  \(t=\dfrac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\)  as the muon’s are moving close to the speed of light.
  • Therefore, the muons have a longer half-life and lifespan than predicted by classical physics and will be able to reach the Earth’s surface before they decay.
  • In this way, muons can reach the surface of the Earth from either frame of reference. 
♦ Mean mark 46%.

Functions, MET2 2024 VCAA 1

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

  1. State, in terms of \(a\) where required, the values of \(x\) for which \(f(x)=0\).  (1 mark
  1. Find the values of \(a\) for which the graph of \(y=f(x)\) has
      
     i. exactly three \(x\)-intercepts.   (2 marks)

    ii. exactly four \(x\)-intercepts.   (1 mark)

  1. Let \(g\) be the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).
      
      i. Find \(g^{\prime}(x)\)   (1 mark)

     ii. Find the coordinates of the local maximum of \(g\).   (1 mark)

    iii. Find the values of \(x\) for which \(g^{\prime}(x)>0\).   (1 mark)

     iv. Consider the two tangent lines to the graph of \(y=g(x)\) at the points where
    \(x=\dfrac{-\sqrt{3}+1}{2}\) and \(x=\dfrac{\sqrt{3}+1}{2}\). Determine the coordinates of the point of intersection of these two tangent lines.   (2 marks)

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

    Let \(h\) be the function \(h: R \rightarrow R, h(x)=(x+1)(x-1)(x+2)(x-2)\), which is the function \(f\) where \(a=-1\).
      
     i. Using translations only, describe a sequence of transformations of \(h\), for which its image would have a local maximum at the same coordinates as that of \(g\).   (1 mark)

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

Show Answers Only

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

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

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

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

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

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

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

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

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

Show Worked Solution

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

Calculus, MET2 2024 VCAA 20 MC

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

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

\(C\)

Show Worked Solution

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

\(\dfrac{1}{2}\displaystyle\int_0^2 f(x)\, dx = k\ \ \Rightarrow\ \ \displaystyle\int_0^2 f(x)\, dx = 2k\)

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

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

  
\(\Rightarrow C\)

Calculus, MET2 2024 VCAA 15 MC

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

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

\(A\)

Show Worked Solution

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

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

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

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

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

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

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

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

\(\Rightarrow A\)

Probability, MET2 2024 VCAA 14 MC

Let \(h\) be the probability density function for a continuous random variable \(X\), where

\(h(x)=\left\{
\begin{array} {c}
\rule{0pt}{2.5ex} \ \ \ \ \ \dfrac{x}{6}+k \rule[-1ex]{0pt}{0pt} & -3 \leq x<0 \\
\rule{0pt}{2.5ex} \ \ -\dfrac{x}{2}+k \rule[-1ex]{0pt}{0pt} & 0 \leq x \leq 1 \\
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & \text { elsewhere } \\
\end{array}\right.\)

and \(k\) is a positive real number.

The value of  \(\text{Pr}(X<0.5)\)  is

  1. \(\dfrac{1}{2}\)
  2. \(\dfrac{15}{16}\)
  3. \(\dfrac{3}{16}\)
  4. \(\dfrac{49}{48}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using CAS}:\)

♦ Mean mark 53%.

\(\text{Alternatively, given that}\ \ k=\dfrac{1}{2}\ \text{by CAS:}\)

\(\text{Pr}\left(X<\dfrac{1}{2}\right)\) \(=\displaystyle \int_{-3}^0 \dfrac{x}{6}+\dfrac{1}{2}\,dx +\int_0^{0.5} \dfrac{1}{2}-\dfrac{x}{2}\,dx\)
  \(=\dfrac{1}{2}{\left[\dfrac{x^2}{6}+x\right] _{-3}^0} +\dfrac{1}{2}{\left[x-\dfrac{x^2}{2}\right] _0^{0.5}}\) 
  \(=\dfrac{1}{2}\left[0-\left(\dfrac{9}{6}-3\right)\right]+\dfrac{1}{2}\left[\left(0.5-\dfrac{0.5^2}{2}\right)-0\right]\)
  \(=\dfrac{15}{16}\)

  
\(\Rightarrow B\)

PHYSICS, M8 2024 HSC 24

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

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

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

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

Show Answers Only

a.    \(5796\ \text{K}\)

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

c.    An absorption spectra is produced when:

  • A continuous spectrum of light  from a black body such as a star passes through cooler and lower density gas in the outer atmosphere of the star.
  • As the light passes through the gas, electrons in the atoms that make up the cooler gas clouds absorb distinct wavelengths/energy levels of light equal to the difference in energy levels between the electron shells where \(E_i-E_f=hf=\dfrac{hc}{\lambda}\). 
  • As the electrons in the atoms fall back into their ground state, they emit the photon of light that they absorb and the photon is then scattered out of the continuous spectrum.
  • The light that remains is then passed through a prism to separate the wavelengths and record the intensities. The black or darkened lines in the absorption spectra is the result of the scattered wavelengths of light. 
Show Worked Solution

a.    Determined temperature using the peak wavelength:

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

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

 

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

c.    Absorption spectra:

  • Produced when a continuous spectrum of light  from a black body such as a star passes through cooler and lower density gas in the outer atmosphere of the star.
  • As the light passes through the gas, electrons in the atoms that make up the cooler gas clouds absorb distinct wavelengths/energy levels of light equal to the difference in energy levels between the electron shells where \(E_i-E_f=hf=\dfrac{hc}{\lambda}\). 
  • As the electrons in the atoms fall back into their ground state, they emit the photon of light that they absorb and the photon is then scattered out of the continuous spectrum.
  • The light that remains is then passed through a prism to separate the wavelengths and record the intensities. The black or darkened lines in the absorption spectra is the result of the scattered wavelengths of light. 
♦ Mean mark (c) 51%.

PHYSICS, M8 2024 HSC 23

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

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

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

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

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

Show Answers Only

a.i.  Role of paraffin wax:

  • Paraffin wax is a rich source of protons.
  • When the paraffin was placed in front of the unknown radiation, the transfer of momentum from the radiation caused protons to be emitted from the paraffin wax.
  • The emitted protons could then be detected and analysed.
  • From studying the protons ejected from the paraffin wax, Chadwick proposed the existence of the neutron.
     

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

  • Previous to Chadwick’s experiment, the model of the atom proposed by Rutherford consisted of a dense positive charge in the nucleus which was orbited by electrons.
  • In this model however, the protons did not account for the total mass of the nucleus.
  • Through using the conservation of momentum and energy in his experiment, Chadwick proposed the existence of the neutron particle which was slightly heavier than the proton.
  • The model of the atom was updated to include both protons and neutrons in the nucleus which then fully accounted for the mass of the nucleus.
     

b.   Limitations in the Bohr-Rutherford model:

  • Rutherford’s model of the atom stated that electrons orbited the nucleus and were electrostatically attracted to the positive nucleus. This meant that the electrons were in circular motion and were constantly under centripetal acceleration.
  • However, Maxwell predicted that an accelerating charge would emit electro-magnetic radiation and in Rutherford’s model, all atoms should have been unstable as the electrons would emit EMR, lose energy and spiral into the nucleus.
  • Bohr proposed that electrons orbited the nucleus in stationary states at fixed energies with no intermediate states possible where they would not emit EMR but provided no theoretical explanation for this.

De Broglie’s hypothesis:

  • De Broglie proposed that electrons could exhibit a wave nature and could act as matter-waves. The electrons would form standing waves around the nucleus and would no longer be an accelerating particle which addressed the limitation of all atoms being unstable.
  • Further, De Broglie proposed that the standing waves would occur at integer wavelengths where the circumference of the electron orbit would be equal to an integer electron wavelength, \(2\pi r=n\lambda\)  where  \(\lambda = \dfrac{h}{mv}\). At any other radii other than this, deconstructive interference would occur and a standing electron wave would not form. This addressed why electrons could only be present at fixed radii/energy levels in the atom.

Show Worked Solution

a.i.  Role of paraffin wax:

  • Paraffin wax is a rich source of protons.
  • When the paraffin was placed in front of the unknown radiation, the transfer of momentum from the radiation caused protons to be emitted from the paraffin wax.
  • The emitted protons could then be detected and analysed.
  • From studying the protons ejected from the paraffin wax, Chadwick proposed the existence of the neutron. 
Mean mark (a)(i) 52%.

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

  • Previous to Chadwick’s experiment, the model of the atom proposed by Rutherford consisted of a dense positive charge in the nucleus which was orbited by electrons.
  • In this model however, the protons did not account for the total mass of the nucleus.
  • Through using the conservation of momentum and energy in his experiment, Chadwick proposed the existence of the neutron particle which was slightly heavier than the proton.
  • The model of the atom was updated to include both protons and neutrons in the nucleus which then fully accounted for the mass of the nucleus. 

b.   Limitations in the Bohr-Rutherford model:

  • Rutherford’s model of the atom stated that electrons orbited the nucleus and were electrostatically attracted to the positive nucleus. This meant that the electrons were in circular motion and were constantly under centripetal acceleration.
  • However, Maxwell predicted that an accelerating charge would emit electro-magnetic radiation and in Rutherford’s model, all atoms should have been unstable as the electrons would emit EMR, lose energy and spiral into the nucleus.
  • Bohr proposed that electrons orbited the nucleus in stationary states at fixed energies with no intermediate states possible where they would not emit EMR but provided no theoretical explanation for this.

De Broglie’s hypothesis:

  • De Broglie proposed that electrons could exhibit a wave nature and could act as matter-waves. The electrons would form standing waves around the nucleus and would no longer be an accelerating particle which addressed the limitation of all atoms being unstable.
  • Further, De Broglie proposed that the standing waves would occur at integer wavelengths where the circumference of the electron orbit would be equal to an integer electron wavelength, \(2\pi r=n\lambda\)  where  \(\lambda = \dfrac{h}{mv}\). At any other radii other than this, deconstructive interference would occur and a standing electron wave would not form. This addressed why electrons could only be present at fixed radii/energy levels in the atom.
♦ Mean mark (b) 44%.

Calculus, MET2 2024 VCAA 13 MC

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

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

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

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

Show Worked Solution

\(\text{Dilate by a factor of 3 from the}\ y\text{-axis:}\)

\(f(x) \rightarrow f_1(x)=\dfrac{\frac{x}{3}}{2}+\dfrac{2}{\frac{x}{3}}=\dfrac{x}{6}+\dfrac{6}{x}\)

\(\text{Translate 1 unit down:}\)

\(f_1(x) \rightarrow g(x)=\dfrac{x}{6}+\dfrac{6}{x}-1\)

\(g'(x)=\dfrac{1}{6}-6x^{-2}\)

♦ Mean mark 45%.
 

\(\Rightarrow A\)

Functions, MET2 2024 VCAA 12 MC

The graph of \(y=f(x)\) is shown below.

Which of the following options best represents the graph of \(y=f(2 x+1)\) ?
 

Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Graph has been dilated by a factor of}\ \dfrac{1}{2}\ \text{from}\ y\text{axis.}\)

→ \(\text{Eliminate C and D.}\)

\(\text{Graph is then translated}\ \dfrac{1}{2}\ \text{unit to the left.}\)

\(\text{Consider the turning point}\ (2, 1)\ \text{after translation:}\)

\(\left(2, 1\right)\ \rightarrow \ \left(2\times \dfrac{1}{2}-\dfrac{1}{2}, 1\right)=\left(\dfrac{1}{2}, 1\right)\)

\(\therefore\ \text{Option A is the only possible solution.}\)

\(\Rightarrow A\)

♦ Mean mark 47%.

Probability, MET2 2024 VCAA 11 MC

Twelve students sit in a classroom, with seven students in the first row and the other five students in the second row. Three students are chosen randomly from the class.

The probability that exactly two of the three students chosen are in the first row is

  1. \(\dfrac{7}{44}\)
  2. \(\dfrac{21}{44}\)
  3. \(\dfrac{5}{22}\)
  4. \(\dfrac{245}{576}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Pr(Exactly 2 from R1)}\ =\dfrac{\displaystyle \binom{7}{2}\binom{5}{1}}{\displaystyle \binom{12}{3}}=\dfrac{21}{44}\)

\(\Rightarrow B\)

♦ Mean mark 40%.