Band 4

PHYSICS, M6 2016 HSC 23b

The diagram shows electrons travelling in a vacuum at `5.2 × 10^(4) \ text{m s}^(-1)` entering an electric field of `10 \ text {V m}^(-1)`.

A magnetic field is applied so that the electrons continue undeflected.

What is the magnitude and direction of the magnetic field? (3 marks)

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`0.0002 text{T Up the page.}`

Show Worked Solution

For the electrons to continue undeflected the force on the electrons due to the electric field must be equal in magnitude (but opposite in direction) to the force on them due to the magnetic field.

`F_(E)`	`=F_(b)`
`qE`	`=qvB`
`B`	`=(E)/(v)`
	`=(10)/(5.2 xx10^(4))`
	`=0.0002 text{T (up the page)}`

PHYSICS, M5 2016 HSC 21a

Why does orbital decay occur more rapidly for satellites in a low-Earth orbit than for satellites in other orbits? (2 marks)

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Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.

Show Worked Solution

Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.

Mean mark 59%.

BIOLOGY, M8 EQ-Bank 2

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

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

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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

BIOLOGY, M8 EQ-Bank 1

Outline the cause of a non-infectious nutritional disease in humans. In your answer, describe any possible effects that may be experienced by the a person living with the disease. (3 marks)

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Example Solution 1

Iron deficiency anaemia:

→ Iron deficiency anaemia develops when there is insufficient dietary iron intake or absorption from food sources.

→ Without adequate iron, the body cannot produce enough haemoglobin, the protein in red blood cells responsible for carrying oxygen throughout the body.

→ This results in symptoms such as chronic fatigue, weakness, shortness of breath and pale skin due to reduced oxygen transport capacity in the blood.

Example Solution 2

Vitamin D deficiency:

→ Vitamin D deficiency develops primarily due to insufficient sunlight exposure to the skin and inadequate dietary intake from foods like fatty fish or fortified products.

→ Without adequate Vitamin D, the body cannot properly absorb and regulate calcium in the intestines and maintain calcium homeostasis in the blood.

→ This results in soft bones, muscle weakness, and increased risk of fractures due to compromised bone mineralization.

Show Worked Solution

Example Solution 1

Iron deficiency anaemia:

→ Iron deficiency anaemia develops when there is insufficient dietary iron intake or absorption from food sources.

→ Without adequate iron, the body cannot produce enough haemoglobin, the protein in red blood cells responsible for carrying oxygen throughout the body.

→ This results in symptoms such as chronic fatigue, weakness, shortness of breath and pale skin due to reduced oxygen transport capacity in the blood.

Example Solution 2

Vitamin D deficiency:

→ Vitamin D deficiency develops primarily due to insufficient sunlight exposure to the skin and inadequate dietary intake from foods like fatty fish or fortified products.

→ Without adequate Vitamin D, the body cannot properly absorb and regulate calcium in the intestines and maintain calcium homeostasis in the blood.

→ This results in soft bones, muscle weakness, and increased risk of fractures due to compromised bone mineralization.

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.

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

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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)

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ii.

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♦ Mean mark (i) 48%.

ii.

Mean mark (ii) 55%.

ENGINEERING, TE 2024 HSC 26b

To avoid overloading a cellular network, a logic gate is used to control the activation of a signal booster.

The signal booster will only activate when both of the following conditions are met:

the signal strength falls below an acceptable level
the network traffic flow falls below an acceptable level.

Complete a truth table for this scenario and identify a suitable logic gate. (3 marks)

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Truth table:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Acceptable signal} \quad & \quad \textit{Acceptable traffic} \quad & \quad \textit{Activate booster} \quad \\
\textit{strength} \rule[-2ex]{0pt}{0pt} & \textit{flow} & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} & & \\
\hline
\end{array}

Logic gate: ....................

Show Answers Only

Truth table:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Acceptable signal} \quad & \quad \textit{Acceptable traffic} \quad & \quad \textit{Activate booster} \quad \\
\textit{strength} \rule[-2ex]{0pt}{0pt} & \textit{flow} & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 1 & 1 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 1 & 0 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 0 & 1 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 0 & 0 & 1 \\
\hline
\end{array}

Logic gate: NOR

Show Worked Solution

Truth table:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Acceptable signal} \quad & \quad \textit{Acceptable traffic} \quad & \quad \textit{Activate booster} \quad \\
\textit{strength} \rule[-2ex]{0pt}{0pt} & \textit{flow} & \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 1 & 1 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 1 & 0 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 0 & 1 & 0 \\
\hline
\rule{0pt}{2.5ex} \rule[-2ex]{0pt}{0pt} 0 & 0 & 1 \\
\hline
\end{array}

Logic gate: NOR

ENGINEERING, TE 2024 HSC 23d

A public telephone booth is assembled using a variety of parts.

Complete the table. (4 marks)

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\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Part} & \textit{Material} & \textit{Justification of } & \textit{Manufacturing } \\
\textit{} \rule[-1ex]{0pt}{0pt} & \textit{} & \textit{material} & \textit{method} \\
\hline
\\ \\
\text{}\text{} & &\text{Needs to be} & \\
\text{Screen} & \text{..................................} & \text{impact resistant} & \text{..................................}\\
\\ \\
\hline
\\ \\
\text{} &\text{} & \text{..................................} & \text{} \\
\text{Hand Receiver} \quad & \text{ABS} & \text{..................................} & \text{..................................}\\
\text{} & \text{} & \text{..................................} & \\
\\ \\
\hline
\\ \\
\text{} &\text{} & \text{..................................} & \text{Cold rolled and} \\
\text{Shelf} & \text{..................................} & \text{..................................} & \text{pressed}\\
\text{} & \text{} & \text{..................................} & \\
\\ \\
\hline
\end{array}

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\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Part} & \textit{Material} & \textit{Justification of } & \textit{Manufacturing } \\
\textit{} \rule[-1ex]{0pt}{0pt} & \textit{} & \textit{material} & \textit{method} \\
\hline
\text{Screen} & \text{Acrylic} & \text{Needs to be impact} & \text{Cast}\\
\text{}\text{} & &\text{resistant} & \\
\hline
\text{Hand Receiver} \quad & \text{ABS} & \text{Complex shaped,} & \text{Injection moulded}\\
\text{} & \text{} & \text{easily moulded} & \\
\hline
\text{} &\text{} & \text{Corrosion} & \\
\text{Shelf} & \text{Stainless steel} & \text{resistance,} & \text{Cold rolled and }\\
\text{} & \text{} & \text{hardness,} & \text{pressed}\\
\text{} & \text{} & \text{cleanability} & \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \textit{Part} & \textit{Material} & \textit{Justification of } & \textit{Manufacturing } \\
\textit{} \rule[-1ex]{0pt}{0pt} & \textit{} & \textit{material} & \textit{method} \\
\hline
\text{Screen} & \text{Acrylic} & \text{Needs to be impact} & \text{Cast}\\
\text{}\text{} & &\text{resistant} & \\
\hline
\text{Hand Receiver} \quad & \text{ABS} & \text{Complex shaped,} & \text{Injection moulded}\\
\text{} & \text{} & \text{easily moulded} & \\
\hline
\text{} &\text{} & \text{Corrosion} & \\
\text{Shelf} & \text{Stainless steel} & \text{resistance,} & \text{Cold rolled and }\\
\text{} & \text{} & \text{hardness,} & \text{pressed}\\
\text{} & \text{} & \text{cleanability} & \\
\hline
\end{array}

ENGINEERING, PPT 2024 HSC 25b

A circuit diagram is shown.

Use Ohm's Law to determine the current in the circuit. (3 marks)

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\(I= 0.57\ \text{Amps}\)

Show Worked Solution

\(E=IR\ \ \Rightarrow\ \ I=\dfrac{E}{R}\)

\(R_{\text{total}} = 10\ \Omega + \dfrac{1}{\dfrac{1}{20\ \Omega} + \dfrac{1}{25\ \Omega}} = 21.11\ \Omega \)

\(I=\dfrac{E}{R} = \dfrac{12\ \text{V}}{21.11\ \Omega} = 0.57\ \text{Amps}\)

ENGINEERING, PPT 2024 HSC 24c

An escalator is used to move people from the first floor to the second floor. The second floor is located 5.2 metres above the first floor. The average person's mass is 64.8 kg .

Determine the power requirement of the escalator in order to move 10 people from the first floor to the second floor in one minute. (3 marks)

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\(561.6\ \text{W}\)

Show Worked Solution

\(W(\text{1 passenger)}\ = mgh = 64.8 \text{kg}\ \times 10 \text{m/s}^{2}\ \times 5.2\ \text{m}\ = 3369.6\ \text{J} \)

\(W(\text{10 passengers)}\ = 10 \times 3369.6 = 33\ 696\ \text{J}\)

\(P=\dfrac{W}{t} = \dfrac{33\ 696\ \text{J}}{60\ \text{s}} = 561.6\ \text{W}\)

ENGINEERING, TE 2024 HSC 23a

Outline a use of semiconductors in the telecommunications industry. Include an example in your response. (3 marks)

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→ Semiconductor amplifiers play a key role in wireless communications by enhancing signal processing.

→ In mobile phones, they boost and strengthen signals to ensure clear and reliable transmission.

→ When used in integrated circuits (ICs), these semiconductors handle signal modulation and demodulation, which is essential for efficient wireless communication.

Show Worked Solution

→ Semiconductor amplifiers play a key role in wireless communications by enhancing signal processing.

→ In mobile phones, they boost and strengthen signals to ensure clear and reliable transmission.

→ When used in integrated circuits (ICs), these semiconductors handle signal modulation and demodulation, which is essential for efficient wireless communication.

ENGINEERING, AE 2024 HSC 22d

How does fibre metal laminate (FML) contribute to enhanced structural performance in aircraft? (3 marks)

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→ FML offers several advantages over traditional aluminium alloys in aircraft construction.

→ Its key benefits include enhanced impact and fatigue resistance, better corrosion protection, and reduced weight.

→ The material’s strength comes from its directional layered construction, which can be customised for specific aircraft sections.

→ This ability to tailor FML’s properties during manufacturing makes it adaptable for various aircraft applications.

Show Worked Solution

→ FML offers several advantages over traditional aluminium alloys in aircraft construction.

→ Its key benefits include enhanced impact and fatigue resistance, better corrosion protection, and reduced weight.

→ The material’s strength comes from its directional layered construction, which can be customised for specific aircraft sections.

→ This ability to tailor FML’s properties during manufacturing makes it adaptable for various aircraft applications.

Mean mark 56%.

ENGINEERING, AE 2024 HSC 22b

The mass of a single-engine propeller aeroplane, without pilot or passengers, is 906 kg . It has a maximum lift to drag ratio of 10.5. When fully loaded, the plane needs to produce 1050 N of thrust to maintain cruising speed and altitude.

What is the maximum total allowable mass for the pilot and passengers in kilograms? (3 marks)

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\(\text{Max allowable mass (pilot and passengers)}\ =196.5\ \text{kg}\)

Show Worked Solution

\(mg\ \text{(empty plane)}\ = 906 \times 10=9060\ \text{N}\)

\(\text{When plane cruising}\ \ \Rightarrow \ \ \text{Thrust = Drag = 1050 N}\)

\(\text{Max Lift to Drag ratio = 10.5}\)

Mean mark 54%.

\(\dfrac{\text{Max Lift}}{\text{Drag}} \)	\(=10.5\)
\(\text{Max Lift}\)	\(=1050\ \text{N}\ \times 10.5\)
	\(=11\ 025\ \text{N}\)

\(\text{Max}\ mg\ \text{(pilot and passengers)}\ = 11\ 025-9060=1965\ \text{N}\)

\(\therefore\ \text{Max allowable mass (pilot and passengers)}\ =196.5\ \text{kg}\)

ENGINEERING, CS 2024 HSC 21d

An orthogonal drawing of a strap brace tensioner is shown.

With reference to the orthogonal drawing and modified title block, complete the table. (3 marks)

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\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{What does 'M' stand for in} \quad \quad & \text{What is the allowable range for the} \quad \\
\text{M5 \(\times\) 0.8?} & \text {R\(2.5 \pm 0.5\) dimension?} \\
\\
\\
\text{........................................................} & \text{........................................................}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{Which dimension is not to AS1100?} & \text{What does the 0.8 refer to in the} \quad \\
\text{} & \text {expression M5 \(\times 0.8\) ?} \\
\\
\\
\text{........................................................} & \text{........................................................}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{Name the tensioner material} & \text{What is the unit of measurement} \quad \\
\text{} & \text {used in this drawing} \\
\\
\\
\text{........................................................} & \text{........................................................}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{What is the name of the curve} & \text{State ONE radial dimension} \quad \\
\text{indicated by the letter 'A'?} & \text {} \\
\\
\\
\text{........................................................} & \text{........................................................}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{What is the angle of projection?} & \text{Which line type is missing from the} \quad \\
\text{} & \text {SIDE VIEW?} \\
\\
\\
\text{........................................................} & \text{........................................................}\rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}

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\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{What does ‘M’ stand for in} \quad \quad & \text{What is the allowable range for the} \quad \\
\text{M5 \(\times\) 0.8?} & \text {R\(2.5 \pm 0.5\) dimension?} \\
\\
\text{Metric} & \text{From 2 mm to 3 mm (or 1 mm)}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{Which dimension is not to AS1100?} & \text{What does the 0.8 refer to in the} \quad \\
\text{} & \text {expression M5 \(\times 0.8\) ?} \\
\\
\text{2.5 mm} & \text{Pitch of thread}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{Name the tensioner material} & \text{What is the unit of measurement} \quad \\
\text{} & \text {used in this drawing} \\
\\
\text{Galvanised Steel (GALZ275)} & \text{Millimetres (mm)}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{What is the name of the curve} & \text{State ONE radial dimension} \quad \\
\text{indicated by the letter ‘A’?} & \text {} \\
\\
\text{Fillet} & \text{R18 or R2.5}\rule[-1ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{What is the angle of projection?} & \text{Which line type is missing from the} \quad \\
\text{} & \text {SIDE VIEW?} \\
\\
\text{3rd} & \text{Hidden detail}\rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}

Show Worked Solution

ENGINEERING, CS 2024 HSC 21c

Strap bracing is used in the construction of many residential structures. It has holes punched along its length to allow tensioners to be applied as shown.

A 6 mm diameter hole is to be punched into mild steel strap bracing.

If the shear strength of the bracing is 380 MPa and the shear force required to punch the hole is 5.73 kN, what is the maximum allowable thickness of the strap bracing? (3 marks)

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\(0.8\ \text{mm}\)

Show Worked Solution

\(\text{Shear stress}\)	\(= \dfrac{\text{Shear force}}{\text{Shear area}} \)
\(380 \times 10^{6}\ \text{Pa}\)	\(= \dfrac{5730}{\text{Shear area}}\)
\(\text{Shear area}\)	\(= \dfrac{5730}{380 \times 10^{6}}\)
	\(=1.508 \times 10^{-5}\ \text{m}^2 \)

\(\text{Shear area}\)	\(= \pi \times\ \text{diameter}\ \times \ \text{thickness} \)
\(\text{Thickness}\)	\(= \dfrac{\text{Shear area}}{\pi \times\ \text{diameter}} \)
	\(= \dfrac{1.508 \times 10^{-5}}{\pi \times 0.006}\)
	\(=0.0008\ \text{m}\)
	\(=0.8\ \text{mm}\)

ENGINEERING, AE 2024 HSC 18 MC

A transition duct in an aircraft air conditioning system is shown.

Which is the correct shape and size for side \(1, a, b, 2\) of the transition piece?

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\(A\)

Show Worked Solution

\(\Rightarrow A\)

ENGINEERING, AE 2024 HSC 14 MC

Which propulsion system requires fuel and an oxidiser such as liquid oxygen to be carried on board?

Ramjet
Rocket
Turbofan
Turboprop

Show Answers Only

\(B\)

Show Worked Solution

→ A rocket is the only propulsion system option that carries both fuel and an oxidiser onboard, allowing it to operate in space where there is no atmospheric oxygen.

→ Air-breathing engines like ramjets, turbofans, and turboprops take in oxygen from the atmosphere and don’t need an oxidiser.

\(\Rightarrow B\)

ENGINEERING, AE 2024 HSC 12 MC

What is the effect of increasing the angle of attack on the lift and induced drag of an aircraft?

\begin{align*}
\begin{array}{c}
\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} \text{Lift} \rule[-1ex]{0pt}{0pt} & \text{Induced drag} \\
\hline
\rule{0pt}{2.5ex} \ \ \ \ \ \text{Decreases} \ \ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \text{Decreases} \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Decreases} \rule[-1ex]{0pt}{0pt} & \text{Increases} \\
\hline
\rule{0pt}{2.5ex} \text{Increases} \rule[-1ex]{0pt}{0pt} & \text{Decreases} \\
\hline
\rule{0pt}{2.5ex} \text{Increases} \rule[-1ex]{0pt}{0pt} & \text{Increases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution

As the angle of attack increases:

→ it creates greater lift due to increased airflow deflection under the wing

→ it creates greater induced drag due to stronger wingtip vortices and pressure differential effects.

\(\Rightarrow D\)

ENGINEERING, CS 2024 HSC 11 MC

Why are gap joints strategically placed in pedestrian footpaths?

To improve visual appeal
To improve wheelchair access
To allow for contraction and movement
To allow for movement of debris and water

Show Answers Only

\(C\)

Show Worked Solution

→ Expansion joints or gaps in footpaths are necessary to prevent cracking by accommodating the natural expansion and contraction of concrete due to temperature changes and ground movement.

\(\Rightarrow C\)

ENGINEERING, TE 2024 HSC 7 MC

Which row of the table correctly identifies the types of cables?

Show Answers Only

\(A\)

Show Worked Solution

→ 1st column is coaxial cable: cross-section has a central copper wire, surrounded by an insulating material, which is then encased in a aluminium braiding (Eliminate B and D).

→ 2nd column is fibre optic cable: cross-section has a central glass core (carrying light signals) surrounded by a glass cladding with a lower refractive index and an outer protective jacket (diagram shows multiple-fibre cable).

\(\Rightarrow A\)

PHYSICS, M5 2024 HSC 30

An object sits on the floor of a hollow cylinder rotating around an axis, as shown. The cylinder's rotation causes the object to undergo uniform circular motion.

Explain the effect on all of the forces acting on the object if the period of the cylinder's rotation is halved. Ignore the effects of friction. (4 marks)

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Forces acting on the object:

The gravitational force pulls the object straight down toward Earth
The floor provides an upward force that pushes back against gravity
A centripetal force acts inward from the cylinder’s wall, directing the object toward the centre of the cylinder

Consider the centripetal force:

\(F_C = \dfrac{mv^{2}}{r}=\dfrac{m}{v} \Big( \dfrac{2\pi r}{T} \Big)^{2} = \dfrac{4m \pi^{2} r}{T^{2}}\)

→ When the period is cut in half \((T \rightarrow \frac{T}{2})\), the centripetal force becomes four times stronger.

→ The other forces, gravity pulling down and the floor pushing up, stay the same since they aren’t affected by how quickly the object moves in its circular path (i.e. its period).

Show Worked Solution

Forces acting on the object:

The gravitational force pulls the object straight down toward Earth
The floor provides an upward force that pushes back against gravity
A centripetal force acts inward from the cylinder’s wall, directing the object toward the centre of the cylinder

Consider the centripetal force:

\(F_C = \dfrac{mv^{2}}{r}=\dfrac{m}{v} \Big( \dfrac{2\pi r}{T} \Big)^{2} = \dfrac{4m \pi^{2} r}{T^{2}}\)

→ When the period is cut in half \((T \rightarrow \frac{T}{2})\), the centripetal force becomes four times stronger.

→ The other forces, gravity pulling down and the floor pushing up, stay the same since they aren’t affected by how quickly the object moves in its circular path (i.e. its period).

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.

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

--- 4 WORK AREA LINES (style=lined) ---
Show that the vertical displacement of the electron beam at the end of the parallel plates is approximately 8.1 mm. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
How far from point \(X\) will the electron beam strike the screen? (3 marks)

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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}

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

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

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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\(^{-1}\).
Explain the problem with this prediction and how it can be resolved. (3 marks)

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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 \(y=g(x)\).

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

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

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

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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.
Given that the graph of \(y=g(x)\) passes through the origin, find the coordinates of \(Q\) in terms of \(m\). (3 marks)

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

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

c. \(k=m^3\)

d. \(Q(2m^3,2m)\)

Show Worked Solution

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

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

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

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

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

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

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

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

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

Calculus, MET1 2024 VCAA 7

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

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

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

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

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

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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.
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)

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a. \(\dfrac{\sqrt{3}\pi^2}{6}\)

bi. \(x\cos(x)+\sin(x)\)

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

Functions, MET1 2024 VCAA 5

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

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

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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.
State a sequence of two transformations that maps \(h\) to this new model \(h_1\). (2 marks)

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In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads. Use \(z=2\) for this confidence interval. (2 marks)
  
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2. A new sample of \(n\) people results in the same sample proportion.
3. Find the smallest value of \(n\) to achieve a standard deviation of \(\dfrac{\sqrt{2}}{50}\) for the sample proportion. (1 mark)
  
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a. \(600\)

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

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

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

cii. \(300\)

Show Worked Solution

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

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

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

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

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

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

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

Probability, MET1 2024 VCAA 4

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

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

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

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

b. \(\dfrac{37}{10\,000}\)

Show Worked Solution

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

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

Calculus, MET1 2024 VCAA 3

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

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)

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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)

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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.	\(-\displaystyle\int_{-2}^0 (x+3)^{-3}-2\,dx\)	\(=-\left[\dfrac{1}{-1}(x+3)^{-1}-2x\right]_{-2}^0\)
		\(=-\left[\dfrac{-1}{x+3}-2x\right]_{-2}^0\)
		\(=-\left[\dfrac{-1}{3}-\left(\dfrac{-1}{-2+3}-2(-2)\right)\right]\)
		\(=\dfrac{10}{3}\ \text{sq units}\)

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.

A cubic polynomial \(p ;(0,12] \rightarrow R, p(t)=a t^3+b t^2+c t+d\) can be used to model monthly online sales in 2021.
The graph of \(y=p(f)\) is shown as a dashed curve on the set of axes above.

It has a local minimum at (2,2500) and a local maximum at (11,4400).
1. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\). (3 marks)

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1. Let \(q:(12,24] \rightarrow R, q(t)=p(t-h)+k\) be a cubic function obtained by translating \(p\), which can be used to model monthly online sales in 2022.
  Find the values of \(h\) and \(k\) such that the graph of \(y=q(t)\) has a local maximum at \((23,4750)\). (2 marks)

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

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

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

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1. Find the derivative \(f^{\prime}(t)\) (1 mark)

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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)

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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.

Express the derivative \(f^{\prime}(t)\) as a hybrid function. (2 marks)

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Find the average rate of change in temperature predicted by the model between \(t=0\) and \(t=\dfrac{1}{2}\).
Give your answer in degrees Celsius per hour. (1 mark)

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

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1. Find the value of \(t\) for which \(g^{\prime}(t)=10\).
  Give your answer correct to three decimal places. (1 mark)

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Find the time \(t \in(0,1)\) when the temperatures predicted by the models \(f\) and \(g\) are equal.
Give your answer correct to two decimal places. (1 mark)

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Find the time \(t \in(0,1)\) when the difference between the temperatures predicted by the two models is the greatest.
Give your answer correct to two decimal places. (1 mark)

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

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

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

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

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1. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.
  
  Give your answer in hours. (1 mark)

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1. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.
  
  Give your answer in hours, correct to two decimal places. (2 marks)

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

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

\(=22-10e^{-6t}\)

\(\text{Using CAS:}\)

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

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. \(\text{Using CAS:}\)

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

Functions, MET2 2024 VCAA 1

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

State, in terms of \(a\) where required, the values of \(x\) for which \(f(x)=0\). (1 mark)

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Find the values of \(a\) for which the graph of \(y=f(x)\) has
1. exactly three \(x\)-intercepts. (2 marks)

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

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

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

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

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

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

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

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

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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 18 MC

Find the value of \(x\) which maximises the area of the trapezium below.

10
\(5 \sqrt{2}\)
7
\(\sqrt{10}\)

Show Answers Only

\(B\)

Show Worked Solution

\(A\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{4x}{2}\sqrt{100-x^2}\quad\text{(Using Pythagoras)}\)
	\(=2x\sqrt{100-x^2}\)
\(\dfrac{dA}{dx}\)	\(=\dfrac{4(50-x^2)}{\sqrt{100-x^2}}\quad\text{(Using product rule)}\)

\(\text{For maximum }\ \dfrac{dA}{dx}=0\)

\(\dfrac{4(50-x^2)}{\sqrt{100-x^2}}\)	\(=0\quad(x \neq 10)\)
\(50-x^2\)	\(=0\)
(\(\sqrt{50}-x)(\sqrt{50}+x)\)	\(=0\)
\(\therefore\ x\)	\(=\sqrt{50}=5\sqrt{2}\quad (0<x<10)\)

\(\text{Check gradient for max using table}\)

\(x\)	\(7\)	\(\sqrt{50}\)	\(7.2\)
\(A^{\prime}\)	0.56	0	-1.06
\(\text{Gradient}\)	+	0	–

\(\therefore\ x=5\sqrt{2}\ \text{maximises the area}\)

\(\Rightarrow B\)

Calculus, MET2 2024 VCAA 17 MC

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

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

In order, the algorithm prints the values

\(-3, 3, 2\)
\(-3, 2, 3\)
\(3, 2, -3\)
\(3, -3, 2\)

Show Answers Only

\(D\)

Show Worked Solution

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

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

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

\(\Rightarrow D\)

Calculus, MET2 2024 VCAA 16 MC

Suppose that a differentiable function \( f: R \rightarrow R \) and its derivative \(f^{\prime}: R \rightarrow R\) satisfy \(f(4)=25\) and \(f^{\prime}(4)=15\).

Determine the gradient of the tangent line to the graph of \( {\displaystyle y=\sqrt{f(x)} } \) at \( x=4 \).

\(\sqrt{15}\)
\(\dfrac{1}{10}\)
\(\dfrac{15}{2}\)
\(\dfrac{3}{2}\)

Show Answers Only

\(D\)

Show Worked Solution

\(y\)	\(=\sqrt{f(x)}={f(x)}^{\frac{1}{2}}\)
\(y^{\prime}\)	\(=\dfrac{1}{2}\left({f(x)}^{-\frac{1}{2}}\right)\times f^{\prime}(x)\)
	\(=\dfrac{f^{\prime}(x)}{2\sqrt{f(x)}}\)
\(\text{Given}\ \ f(4)=25,\ f^{\prime}(4)=15\)
\(\therefore\ y^{\prime}\)	\(=\dfrac{15}{2\sqrt{25}}\)
	\(=\dfrac{3}{2}\)

\(\Rightarrow D\)

PHYSICS, M5 2024 HSC 25

The mathematical model below shows the relationship between the orbital radius of a satellite and its period.

\(\dfrac{r^3}{T^2}=\dfrac{G M}{4 \pi^2}\)

By considering gravitational force, show how this model can be derived. (2 marks)

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A planet with five moons is discovered. The following graph is produced from observations of the orbital radius of the moons and their orbital periods, measured in Earth days.

Use the graph to calculate the mass of the planet. (4 marks)

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Show Answers Only

a. See worked solutions

b. \(9.9 \times 10^{25}\ \text{kg}\)

Show Worked Solution

a. Centripetal force = gravitational force

\(F_c\)	\(=F_g\)
\(\dfrac{mv^2}{r}\)	\(=\dfrac{GMm}{r^2}\)
\(v^2\)	\(=\dfrac{GM}{r}\ \ \text{where}\ \ v=\dfrac{2 \pi r}{T}\)
\( \left(\dfrac{2\pi r}{T}\right)^2\)	\(=\dfrac{GM}{r}\)
\(\dfrac{4 \pi^2 r^2}{T^2}\)	\(=\dfrac{GM}{r}\)
\(\dfrac{r^3}{T^2}\)	\(=\dfrac{GM}{4 \pi^2}\)

b. Find gradient of line through \((200, 0.16)\) and \((300, 0.24)\):

\(m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{0.24-0.16}{300 \times 10^{12}-200 \times 10^{12}} = 8 \times 10^{-16}\ \text{days}^{2}\text{\km}^{3}\)

\(\Rightarrow\ \) To use \(\dfrac{r^3}{T^2} = \dfrac{GM}{4 \pi^2}\), \(r\) and \(T\) must be in SI units:

\(8 \times 10^{-16} \times \dfrac{(24 \times 60 \times 60)^2}{1000^3} = 5.972 \times 10^{-15}\ \text{s}^{2}\text{\m}^{3}\)

\(\Rightarrow \ \dfrac{T^2}{r^3}= 5.972 \times 10^{-15}\ \text{s}^{2}\text{\m}^{3}\ \ \Rightarrow\ \ \dfrac{r^3}{T^2} = 1.6745 \times 10^{14}\)

\(\dfrac{r^3}{T^2}\)	\(=\dfrac{GM}{4 \pi^2}\)
\(M\)	\(=\dfrac{r^3}{T^2} \times \dfrac{4 \pi^2}{G}\)
	\(=1.6745 \times 10^{14} \times \dfrac{4 \pi^2}{6.67 \times 10^{-11}}\)
	\(=9.9 \times 10^{25}\ \text{kg}\)

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

\(\dfrac{1}{2}\)
\(\dfrac{15}{16}\)
\(\dfrac{3}{16}\)
\(\dfrac{49}{48}\)

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using CAS}:\)

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

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

\(\Rightarrow B\)

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.

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

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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)

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

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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 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.

♦ 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.

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

1. What was the role of paraffin in Chadwick's experiment? (2 marks)
  
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2. How did Chadwick's experiment change the model of the atom? (3 marks)
  
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Explain how de Broglie's hypothesis regarding the nature of electrons addressed limitations in the Bohr-Rutherford model of the atom. (4 marks)

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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:

De Broglie’s hypothesis:

♦ Mean mark (b) 44%.

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

\(\text{Transformation is a combination of a dilation of }\dfrac{1}{2}\ \text{parallel to the }\)

\(x-\text{axis, followed by a translation of }\dfrac{1}{2}\ \text{to the left.}\)

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

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

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

\(\Rightarrow A\)

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

\(\dfrac{7}{44}\)
\(\dfrac{21}{44}\)
\(\dfrac{5}{22}\)
\(\dfrac{245}{576}\)

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Possibilities are: (R1, R1, R2), (R1, R2, R1), (R2, R1, R1)}\)

\(\text{Pr(Exactly 2 from R1)}\)	\(=\dfrac{7}{12}\times \dfrac{6}{11}\times \dfrac{5}{10}+\dfrac{7}{12}\times \dfrac{5}{11}\times \dfrac{6}{10}+\dfrac{5}{12}\times \dfrac{7}{11}\times \dfrac{6}{10}\)
	\(=\dfrac{21}{44}\)

\(\Rightarrow B\)

Calculus, MET2 2024 VCAA 10 MC

Suppose a function \(f:[0,5] \rightarrow R\) and its derivative \(f^{\prime}:[0,5] \rightarrow R\) are defined and continuous on their domains. If \(f^{\prime}(2)<0\) and \(f^{\prime}(4)>0\), which one of these statements must be true?

\(f\) is strictly decreasing on \([0,2]\).
\(f\) does not have an inverse function.
\(f\) is positive on \([4,5]\).
\(f\) has a local minimum at \(x=3\).

Show Answers Only

\(B\)

Show Worked Solution

\(\text{As }f^{\prime}(2)<0\ \text{and }f^{\prime}(4)>0\ \text{the function must have} \)

\(\text{a turning point between }x=2\ \text{and }x=4.\)

\(\therefore\ f(x)\ \text{will not have an inverse function, as it is many-to-one.}\)

\(\Rightarrow B\)

Probability, MET2 2024 VCAA 9 MC

At a Year 12 formal, 45% of the students travelled to the event in a hired limousine, while the remaining 55% were driven to the event by a parent.

Of the students who travelled in a hired limousine, 30% had a professional photo taken.

Of the students who were driven by a parent, 60% had a professional photo taken.

Given that a student had a professional photo taken, what is the probability that the student travelled to the event in a hired limousine?

\(\dfrac{1}{8}\)
\(\dfrac{27}{200}\)
\(\dfrac{9}{31}\)
\(\dfrac{22}{31}\)

Show Answers Only

\(C\)

Show Worked Solution

\(\text{Pr(Limo\|PP)}\)	\(=\dfrac{\text{Pr(Limo)}\ \cap\ \text{Pr(PP)}}{\text{Pr(PP)}}\)
	\(=\dfrac{0.45\times 0.30}{(0.45\times 0.30)+(0.55\times 0.6)}\)
	\(=\dfrac{0.135}{0.465}\)
	\(=\dfrac{9}{31}\)

\(\Rightarrow C\)

Probability, MET2 2024 VCAA 7 MC

A fair six-sided die is repeatedly rolled. What is the minimum number of rolls required so that the probability of rolling a six at least once is greater than 0.95 ?

Show Answers Only

\(C\)

Show Worked Solution

\(X\sim \text{Bi} \left(n, \dfrac{1}{6}\right)\)

\(\text{Pr}(X\geq 1)\)	\(>0.95\)
\(1-\text{Pr}(X=0)\)	\(>1\)
\(-\text{Pr}(X=0)\)	\(\geq -0.05\)
\(\text{Pr}(X=0)\)	\(< 0.05\quad \left[\text{CAS}:\text{invBinomN}\left(0.05,\dfrac{1}{6},0\right)\right]\)
	\(=17\)

\(\Rightarrow C\)

Functions, MET2 2024 VCAA 6 MC

Consider the function \(f(x)=\dfrac{2 x+1}{3-x}\) with domain \(x \in R \backslash\{3\}\)

The inverse of \(f\) is

\(f^{-1}(x)=\dfrac{3 x-1}{x+2}\) with domain \(x \in R \backslash\{3\}\)
\(f^{-1}(x)=3-\dfrac{7}{x+2}\) with domain \(x \in R \backslash\{-2\}\)
\(f^{-1}(x)=3+\dfrac{5}{x+2}\) with domain \(x \in R \backslash\{-2\}\)
\(f^{-1}(x)=\dfrac{1-3 x}{x+2}\) with domain \(x \in R \backslash\{-2\}\)

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Let}\ y=f(x)\)

\(\text{For inverse swap }x\leftrightarrow y\)

\(x\)	\(=\dfrac{2y+1}{3-y}\)
\(3x-xy\)	\(=2y+1\)
\(3x-1\)	\(=2y+xy\)
\(y(2+x)\)	\(=3x-1\)
\(y\)	\(=\dfrac{3x-1}{2+x}\)

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

\(y\)	\(=3-\dfrac{7}{x+2}\)
\(y\)	\(=\dfrac{3(x+2)-7}{x+2}\)
\(y\)	\(=\dfrac{3x-1}{2+x}\)

\(\Rightarrow B\)

Functions, MET2 2024 VCAA 5 MC

Consider the functions \(f:(1, \infty) \rightarrow R, f(x)=x^2-4 x\) and \(g: R \rightarrow R, g(x)=e^{-x}\).

The range of the composite function \(g(f(x))\) is

\( (0, e^3) \)
\( (0, e^3 ] \)
\( (0, e^4) \)
\( (0, e^4 ]\)

Show Answers Only

\(D\)

Show Worked Solution

\(\text{dom}\left(g(f)\right)=\text{dom}(f)\)

\(g(f(x))=e^{-(x^2-4x)}\)

\(\text{From Graph }\rightarrow \text{range}\ = (0,\infty]\)

\(\Rightarrow D\)

PHYSICS, M8 2024 HSC 22

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

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

--- 4 WORK AREA LINES (style=lined) ---
How were the recessional velocities of galaxies determined? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. Recessional velocity vs distance from Earth graph:

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

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

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

b. Determining recessional velocities:

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

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

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

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

Show Worked Solution

a. Recessional velocity vs distance from Earth graph:

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

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

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

b. Determining recessional velocities:

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

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

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

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

♦ Mean mark (b) 49%.

PHYSICS, M5 2024 HSC 13 MC

The diagram shows two identical satellites, \(A\) and \(B\), orbiting a planet.

Which row in the table correctly compares the potential energy, \(U\), and kinetic energy, \(K\), of the satellites?

\begin{align*}
\begin{array}{l}
\ \rule{0pt}{2.5ex} \textbf{} \rule[-1.5ex]{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{Potential energy} \rule[-1ex]{0pt}{0pt} & \textit{Kinetic energy} \\
\hline
\rule{0pt}{2.5ex} U_\text{A} > U_\text{B} \rule[-1ex]{0pt}{0pt} & K_\text{A} < K_\text{B} \\
\hline
\rule{0pt}{2.5ex} U_\text{A} < U_\text{B} \rule[-1ex]{0pt}{0pt} & K_\text{A} > K_\text{B} \\
\hline
\rule{0pt}{2.5ex} U_\text{A} > U_\text{B} \rule[-1ex]{0pt}{0pt} & K_\text{A} > K_\text{B} \\
\hline
\rule{0pt}{2.5ex} U_\text{A} < U_\text{B} \rule[-1ex]{0pt}{0pt} & K_\text{A} < K_\text{B} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ As shown in the graph below, as the radius of a satellite increases the kinetic energy decreases and the gravitational potential energy increases.

→ This can also be concluded using the energy formulas for kinetic energy and gravitational potential energy:

\(K=\dfrac{GMm}{2r}\) and \(U=\dfrac{GMm}{r}\)

\(\Rightarrow A\)

Probability, MET2 2024 VCAA 3 MC

A discrete random variable \(X\) is defined using the probability distribution below, where \(k\) is a positive real number.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \text{0} \rule[-1ex]{0pt}{0pt} & \text{1} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{3} \rule[-1ex]{0pt}{0pt} & \text{4} \\
\hline
\rule{0pt}{2.5ex} \text{Pr} \ (X = x) \rule[-1ex]{0pt}{0pt} & 2k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 5k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 2k \\
\hline
\end{array}

Find \(\operatorname{Pr}(X<4 \mid X>1)\).

\(\dfrac{13}{15}\)
\(\dfrac{11}{13}\)
\(\dfrac{4}{5}\)
\(\dfrac{8}{15}\)

Show Answers Only

\(C\)

Show Worked Solution

\(2k+3k+5k+3k+2k=1\ \Rightarrow\ k=\dfrac{1}{15}\)

\(\operatorname{Pr}(X<4 \mid X>1)\)	\(=\dfrac{\operatorname{Pr}(1<X<4)}{\operatorname{Pr}(X>1)}\)
	\(=\dfrac{\frac{5}{15}+\frac{3}{15}}{\frac{5}{15}+\frac{3}{15}+\frac{2}{15}}\)
	\(=\dfrac{\frac{8}{15}}{\frac{10}{15}}\)
	\(=\dfrac{4}{5}\)

\(\Rightarrow C\)

PHYSICS, M7 2024 HSC 12 MC

A rod has a length, \(\mathrm{L}_0\), when measured in its own frame of reference.

The rod travels past a stationary observer at speed, \(v\), as shown in the diagram.

Which option represents the relationship between the speed of the rod, \(v\), and the length of the rod as measured by the stationary observer?

\( W \)
\( X \)
\( Y \)
\( Z \)

Show Answers Only

\(B\)

Show Worked Solution

→ Length contraction occurs when objects travelling at high speeds are observed from a stationary observer.

\(L=L_o\sqrt{1-\frac{v^2}{c^2}}\)

→ At low speeds, e.g. \((0.4c)\), the contraction factor is only small, \(\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-\frac{(0.4c)^2}{c^2}} = 0.92\)

→ As the speed of the rod increases, the contraction factor increases exponentially, e.g. \((0.9c)\), \(\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-\frac{(0.9c)^2}{c^2}} = 0.44\)

→ This exponential increase in the length contraction of an object as the object approaches \(c\) is demonstrated by the curve \(X\).

\(\Rightarrow B\)

PHYSICS, M5 2024 HSC 11 MC

A satellite is in a circular orbit.

What is the relationship between its orbital velocity, \(v\), and its orbital radius, \(r\)?

\(v\) is directly proportional to the square of \(r\).
\(v\) is inversely proportional to the square of \(r\).
\(v\) is directly proportional to the square root of \(r\).
\(v\) is inversely proportional to the square root of \(r\).

Show Answers Only

\(D\)

Show Worked Solution

→ The orbital velocity of a satellite is given by the equation: \(v=\sqrt{\dfrac{GM}{r}}\)

→ Hence \(v \propto \dfrac{1}{\sqrt{r}}\)

→ The orbital velocity is inversely proportional to the square root of the orbital radius.

\(\Rightarrow D\)

PHYSICS, M5 2024 HSC 9 MC

Object \(P\) is dropped from rest, and object \(Q\) is launched horizontally from the same height.

Which option correctly compares the projectile motion of \(P\) and \(Q\) ?

The acceleration of \(P\) is less than the acceleration of \(Q\).
The final velocity of \(Q\) is greater than the final velocity of \(P\).
The time of flight of \(Q\) is greater than the time of flight of \(P\).
The initial vertical velocity of \(P\) is less than the initial vertical velocity of \(Q\).

Show Answers Only

\(B\)

Show Worked Solution

→ Both objects start with the same initial vertical velocity and experience the same acceleration due to gravity. Therefore both objects will hit the ground at the same time with the same final vertical velocity.

→ Object \(Q\) however, has both vertical and horizontal velocity while object \(P\) only has vertical velocity. Therefore object \(Q\) will have a greater total final velocity when compared with \(P\).

\(\Rightarrow B\)

PHYSICS, M6 2024 HSC 8 MC

An ideal transformer produces an output of 6 volts when an input of 240 volts is applied.

What change would be needed to produce an output of 12 volts, using the same input voltage?

Increase the number of turns on the primary coil
Decrease the number of turns on the primary coil
Increase the resistance connected to the secondary coil
Decrease the resistance connected to the secondary coil

Show Answers Only

\(B\)

Show Worked Solution

→ Using the ideal transformer equation:

\(\dfrac{V_p}{V_s}\)	\(=\dfrac{N_p}{N_s}\)
\(V_s\)	\(=V_p \times \dfrac{N_s}{N_p}\)

→ To increase the output voltage while the input voltage remains the same, the value of \(\dfrac{N_s}{N_p}\) must increase.

→ Given the options, this can only be done by decreasing the number of turns on the primary coil.

\(\Rightarrow B\)

PHYSICS, M6 2024 HSC 4 MC

A conducting coil is mounted on an axle and placed in a uniform magnetic field. The diagram shows different ways of connecting the coil to a power source.

Which setup allows the conducting coil to rotate continuously?

Show Answers Only

\(D\)

Show Worked Solution

→ The split ring commutator is used in a DC motor to reverse the current in the arms every 180\(^{\circ}\) to ensure unidirectional torque.

→ Initially, the split ring commutator must be in contact with the carbon brushes to produce an electric current through the coil and the coil must lie in the horizontal plane so torque in produced.

→ Additionally, the split ring commutator should disconnect with the carbon brushes as the coil is vertically oriented.

\(\Rightarrow D\)

CHEMISTRY, M2 EQ-Bank 3

Balance the following chemical equations:

\(\ce{HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\) (1 mark)
\(\ce{CuSO4(aq) + AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\) (1 mark)

Show Answers Only

a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)

b. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)

Show Worked Solution

a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)

b. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)

BIOLOGY, M6 2024 HSC 34

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

--- 23 WORK AREA LINES (style=lined) ---

Show Answers Only

Social impacts of recombinant DNA technology (plant biotechnology)

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

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

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

Associated ethical concerns:

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

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

Social impacts of selective breeding/hybridisation (animal biotechnology)

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

→ This results in higher yields and greater food availability.

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

→ Increased food production to support population growth.

Associated ethical issues:

→ High-yield dairy cows show decreased fertility.

→ May compromise animal welfare and quality of life.

Show Worked Solution

Social impacts of recombinant DNA technology (plant biotechnology)

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

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

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

Associated ethical concerns:

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

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

♦ Mean mark 56%.

Social impacts of selective breeding/hybridisation (animal biotechnology)

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

→ This results in higher yields and greater food availability.

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

→ Increased food production to support population growth.

Associated ethical issues:

→ High-yield dairy cows show decreased fertility.

→ May compromise animal welfare and quality of life.

BIOLOGY, M5 2024 HSC 33

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

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

--- 4 WORK AREA LINES (style=lined) ---
Draw the chromosomes of the four possible gametes after crossing over for the Jack Jumper ants occurs. Include the alleles for each gene. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---

Show Answers Only

a. Significance of crossing over:

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

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

Show Worked Solution

a. Significance of crossing over:

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

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

♦ Mean mark (b) 54%.

BIOLOGY, M8 2024 HSC 29

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

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

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

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

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

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

b. 630 000 000 (630 million)

Show Worked Solution

♦ Mean mark (a) 56%.

b. Using the LOBF on the graph:

Population (%) with diabetes in 2040 = 7%

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

BIOLOGY, M8 2024 HSC 31

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

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 7 WORK AREA LINES (style=lined) ---

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. 4 am

b. Signs of torpor:

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

→ In contrast, the kookaburra exhibited classic torpor behaviour.

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

c. Kookaburra adaptation:

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

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

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

Show Worked Solution

a. 4 am

b. Signs of torpor:

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

→ In contrast, the kookaburra exhibited classic torpor behaviour.

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

Mean mark (b) 56%.

c. Kookaburra adaptation:

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

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

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

♦ Mean mark (c) 47%.

BIOLOGY, M5 2024 HSC 30

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

Compare Process \(A\) with DNA replication. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Explain the importance of mRNA and tRNA in polypeptide synthesis. (5 marks)

--- 13 WORK AREA LINES (style=lined) ---

Show Answers Only

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

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

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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

♦ Mean mark (b) 60%.

BIOLOGY, M7 2024 HSC 27

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

BIOLOGY, M7 2024 HSC 26

Describe a plant disease and its effect on agricultural production. (4 marks)

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Two possible examples (among many) are described below.

Plant disease: Wheat rust

→ Wheat rust is a fungal disease that produces reddish-brown pustules on wheat stems and leaves, weakening the plant and reducing grain development.

→ This disease can devastate entire wheat crops, significantly reducing grain yields and causing major economic losses for Australian grain producers.

Plant disease: Stone fruit scab

→ Caused by fungi, produces dark spots on plums, peaches and nectarines that develop into scabs, potentially cracking and destroying the fruit.

→ This disease significantly impacts crop quality and yield, resulting in economic losses for farmers.

Show Worked Solution

Two possible examples (among many) are described below.

Plant disease: Wheat rust

→ Wheat rust is a fungal disease that produces reddish-brown pustules on wheat stems and leaves, weakening the plant and reducing grain development.

→ This disease can devastate entire wheat crops, significantly reducing grain yields and causing major economic losses for Australian grain producers.

Plant disease: Stone fruit scab

→ Caused by fungi, produces dark spots on plums, peaches and nectarines that develop into scabs, potentially cracking and destroying the fruit.

→ This disease significantly impacts crop quality and yield, resulting in economic losses for farmers.