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CHEMISTRY, M8 2025 HSC 36

Use the data sheet provided and the information in the table to answer this question.
 

 

Consider the molecule shown.

For each of the following instrumental techniques, predict the expected features of the spectra produced.

Refer to the structural features of the molecule in your answer.

  • Infrared (IR)    (Ignore any absorptions due to \(\ce{C - C}\) or \(\ce{C - H}\) )
  • Carbon-13 NMR
  • Proton NMR
  • Mass spectrometry   (7 marks)

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Infrared (IR)

  • Strong broad \(\ce{OH}\) signal at approximately 2750 cm\(^{-1}\)
  • Strong sharp \(\ce{CO}\) signal at approximately 1700 cm\(^{-1}\)

\(^{13}\text{C NMR}\)

  • 3 distinct carbon environments 
  • 1 signal at 5−40 ppm \(\ce{(CH3)}\)
  • 1 signal at 20−50 ppm \(\ce{(CH2)}\)
  • 1 signal at 160−185 ppm \(\ce{(acid CO)}\)

\(^{1}\text{H NMR (Proton NMR)}\)

  • 3 distinct hydrogen environments
  • 1 signal at 0.7−2.1 ppm \(\ce{(CH3)}\), triplet splitting pattern, integration of 3
  • 1 signal at 2.1−4.5 ppm \(\ce{(CH2)}\), quartet splitting pattern, integration of 2
  • 1 signal at 9.0−13.0 ppm \(\ce{(COOH)}\), singlet, integration of 1

Mass Spectrometry

  • Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)
Show Worked Solution

Infrared (IR)

  • Strong broad \(\ce{OH}\) signal at approximately 2750 cm\(^{-1}\)
  • Strong sharp \(\ce{CO}\) signal at approximately 1700 cm\(^{-1}\)

\(^{13}\text{C NMR}\)

  • 3 distinct carbon environments 
  • 1 signal at 5−40 ppm \(\ce{(CH3)}\)
  • 1 signal at 20−50 ppm \(\ce{(CH2)}\)
  • 1 signal at 160−185 ppm \(\ce{(acid CO)}\)

\(^{1}\text{H NMR (Proton NMR)}\)

  • 3 distinct hydrogen environments
  • 1 signal at 0.7−2.1 ppm \(\ce{(CH3)}\), triplet splitting pattern, integration of 3
  • 1 signal at 2.1−4.5 ppm \(\ce{(CH2)}\), quartet splitting pattern, integration of 2
  • 1 signal at 9.0−13.0 ppm \(\ce{(COOH)}\), singlet, integration of 1

Mass Spectrometry

  • Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)

Filed Under: Organic Substances Tagged With: Band 5, Band 6, smc-3683-10-C NMR, smc-3683-20-H NMR, smc-3683-40-Mass Spectrometry, smc-3683-43-IR Spectroscopy

CHEMISTRY, M5 2025 HSC 35

A purple solution at 25°C contains a mixture of two different cobalt\(\text{(II)}\) complexes which are at equilibrium.

\(\underset{\text{(blue)}}{\ce{CoCl4^{2-}(aq)}} \ce{+ 6H2O(l)} \rightleftharpoons
\underset{\text{(pink)}}{\ce{Co(H2O)6^{2+}(aq)}} \ce{+ 4Cl^{-}(aq)}\)

The results of heating and cooling a sample of this solution are given in the table.

\begin{array}{|l|c|c|}
\hline\rule{0pt}{2.5ex} \textit{Temperature} \ \text{(°C)} \rule[-1ex]{0pt}{0pt}& 80 & 0 \\
\hline \rule{0pt}{2.5ex}\textit{Colour of solution} \rule[-1ex]{0pt}{0pt} & \quad \text{blue} \quad  & \quad  \text{pink} \quad \\
\hline
\end{array}

The energy profile diagram for this reaction is shown.
 

How do collision theory and Le Chatelier's principle account for the colour change to pink when the solution is cooled? Refer to the energy profile diagram in your answer.   (5 marks)

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Energy Profile Diagram 

  • The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy \((\text{E}_{a1})\) is smaller than the reverse activation energy \((\text{E}_{a2})\).

Le Chatelier’s Principle

  • Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
  • Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
  • This shifts the equilibrium to the right, increasing \(\left[ \ce{Co(H2O)6^{2+}}\right]\) and turning the solution pink.

Collision Theory

  • Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
  • When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
  • In this way, both forward and reverse reaction rates decrease.
  • However, the reverse reaction has a higher activation energy \((\text{E}_{a2})\) than the forward reaction \((\text{E}_{a1})\), as shown in the energy profile diagram.
  • Cooling causes a greater proportion of particles to fall below \((\text{E}_{a2})\) than \((\text{E}_{a1})\), so the reverse reaction rate decreases more significantly than the forward rate.
  • This causes a net shift toward products, increasing \(\left[ \ce{Co(H2O)6^{2+}}\right]\) and changing the solution colour to pink.
Show Worked Solution

Energy Profile Diagram 

  • The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy \((\text{E}_{a1})\) is smaller than the reverse activation energy \((\text{E}_{a2})\).

Le Chatelier’s Principle

  • Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
  • Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
  • This shifts the equilibrium to the right, increasing \(\left[ \ce{Co(H2O)6^{2+}}\right]\) and turning the solution pink.

Collision Theory

  • Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
  • When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
  • In this way, both forward and reverse reaction rates decrease.
  • However, the reverse reaction has a higher activation energy \((\text{E}_{a2})\) than the forward reaction \((\text{E}_{a1})\), as shown in the energy profile diagram.
  • Cooling causes a greater proportion of particles to fall below \((\text{E}_{a2})\) than \((\text{E}_{a1})\), so the reverse reaction rate decreases more significantly than the forward rate.
  • This causes a net shift toward products, increasing \(\left[ \ce{Co(H2O)6^{2+}}\right]\) and changing the solution colour to pink.

Filed Under: Factors that Affect Equilibrium Tagged With: Band 4, Band 5, smc-3670-10-Temperature, smc-3670-40-Collision theory

CHEMISTRY, M6 2025 HSC 34

A 0.010 L aliquot of an acid was titrated with 0.10 mol L\(^{-1} \ \ce{NaOH}\), resulting in the following titration curve.
 

  1. Calculate the \(K_a\) for the acid used.   (3 marks)

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  2. The concentration of the \(\ce{NaOH}\) was 0.10 mol L\(^{-1}\).
  3. Explain why the pH of the final solution never reached 13.   (2 marks)
  4. --- 6 WORK AREA LINES (style=lined) ---

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a.   \(\text{Strategy 1:}\)
 

\(\text{From the shape of the titration curve, the acid was weak.}\)

\(\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} \)

\(\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}\)

\(\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}\).

\(K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}\)

\(\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}\).
 

\(\text{Strategy 2:}\)

\(\text{Equivalence point is at} \ \ce{0.024 L NaOH added}\).

\(\text{Shape of curve shows acid is monoprotic.}\)

\(\ce{[HA] \times 0.010=0.1 \times 0.024}\)

\(\ce{[HA]=0.24 mol L^{-1}}\)

\(\text{pH at start is approximately 2.5}\)

\(\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}\)

\(K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}\)

\(\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}\)

\(K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}\)
 

b.   pH of the final solution < 13:

  • Some of the hydroxide was neutralised by the acid.
  • The 10 mL of acid also diluted the NaOH.
  • So the \(\ce{NaOH}\) concentration of the mixture will be less than 0.1 mol L\(^{-1}\) and the pH will be less than 13.
Show Worked Solution

a.   \(\text{Strategy 1:}\)
 

\(\text{From the shape of the titration curve, the acid was weak.}\)

\(\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} \)

\(\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}\)

\(\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}\).

\(K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}\)

\(\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}\).
 

\(\text{Strategy 2:}\)

\(\text{Equivalence point is at} \ \ce{0.024 L NaOH added}\).

\(\text{Shape of curve shows acid is monoprotic.}\)

\(\ce{[HA] \times 0.010=0.1 \times 0.024}\)

\(\ce{[HA]=0.24 mol L^{-1}}\)

\(\text{pH at start is approximately 2.5}\)

\(\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}\)

\(K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}\)

\(\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}\)

\(K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}\)
 

b.   pH of the final solution < 13:

  • Some of the hydroxide was neutralised by the acid.
  • The 10 mL of acid also diluted the NaOH.
  • So the \(\ce{NaOH}\) concentration of the mixture will be less than 0.1 mol L\(^{-1}\) and the pH will be less than 13.

Filed Under: Quantitative Analysis Tagged With: Band 4, Band 5, smc-3675-20-Titration Curves and Conductivity Graphs, smc-3675-30-Ka/Kb

CHEMISTRY, M6 2025 HSC 33

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

\begin{array}{|l|c|c|c|}
\hline \rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \ \ \textit{Brand X} \ \ & \ \ \textit{Brand Y} \ \ & \ \ \textit{Brand Z} \ \ \\
\hline \rule{0pt}{2.5ex}\ce{CaCO3(\%)} \rule[-1ex]{0pt}{0pt}& 85.5 & 83.9 & 82.4 \\
\hline
\end{array}

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

  • A sample of chalk was crushed in a mortar and pestle.
  • A 3.00 g sample of the crushed chalk was placed in a conical flask.
  • 100.0 mL of 0.550 mol L\(^{-1} \ \ce{HCl(aq)}\) was added to the sample and left to react completely, resulting in a clear solution.
  • Four 20 mL aliquots of this mixture were then titrated with 0.10 mol L\(^{-1} \ \ce{KOH}\) .

The results of the titrations are recorded.

\begin{array}{|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Burette volume}\text{(mL)} \rule[-1ex]{0pt}{0pt}& \textit{Trial 1} & \textit{Trial 2} & \textit{Trial 3} & \textit{Trial 4} \\
\hline
\rule{0pt}{2.5ex}\text{Final} \rule[-1ex]{0pt}{0pt}& 7.80 & 14.90 & 22.10 & 29.25 \\
\hline
\rule{0pt}{2.5ex}\text{Initial} \rule[-1ex]{0pt}{0pt}& 0.00 & 7.80 & 14.90 & 22.10 \\
\hline
\rule{0pt}{2.5ex}\text{Total used} \rule[-1ex]{0pt}{0pt}& 7.80 & 7.10 & 7.20 & 7.15 \\
\hline
\end{array}

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer.   (7 marks)

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\(\text{Exclude the outlier (Trial 1):}\)

\(\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}\)

\(\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}\)

\(\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }\)

\(\text{Ratio}\ \ \ce{HCl:KOH=1: 1}\)

   \(\ce{0.000715 mol HCl} \ \text{for each sample}\)

   \(\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}\)
 

\(\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}\)

\(\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}\)

\(\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}\)

\(\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}\)

\(\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}\)

\(\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09\)

\(\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}\)

\(\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%\)

\(\text{Chalk sample has to be Brand X.}\)

Show Worked Solution

\(\text{Exclude the outlier (Trial 1):}\)

\(\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}\)

\(\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}\)

\(\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }\)

\(\text{Ratio}\ \ \ce{HCl:KOH=1: 1}\)

   \(\ce{0.000715 mol HCl} \ \text{for each sample}\)

   \(\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}\)
 

\(\text{Calculate}\ \ce{HCl}\ \text{that reacted with}\ \ce{CaCO3}:\)

\(\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}\)

\(\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}\)

\(\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}\)

\(\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}\)

\(\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}\)

\(\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09\)

\(\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}\)
 

\(\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%\)

\(\text{Chalk sample has to be Brand X.}\)

Filed Under: Quantitative Analysis Tagged With: Band 4, Band 5, smc-3675-10-Titration

CHEMISTRY, M5 2025 HSC 32

The following three solids were added together to 1 litre of water:

  • \(\ce{0.006\ \text{mol}\ Mg(NO3)2}\)
  • \(\ce{0.010\ \text{mol}\ NaOH}\)
  • \(\ce{0.002\ \text{mol}\ Na2CO3}\).

Which precipitate(s), if any, will form? Justify your answer with appropriate calculations.   (5 marks)

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All sodium and nitrate salts are soluble  \(\Rightarrow\)  possible precipitates are \(\ce{Mg(OH)2}\) and \(\ce{MgCO3}\). 

\(\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}\)

\(\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}\)

\(\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad  K_{\textit{sp}}=6.82 \times 10^{-6}}\)
 

\(\ce{Mg(OH)2}\) is a lot less soluble than \(\ce{MgCO3}\) and will precipitate preferentially.

\(\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}\)

\(\Rightarrow \ce{Mg(OH)2}\) will precipitate.
 

\(\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}\)

Since \(K_{\textit{sp}}\) is small, assume reaction goes to completion.

\(\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}\)

Using stoichiometric ratio \((1:2)\)

\(0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}\) is in excess.

Concentration drops to: \(6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}\).
 

Check if \(\ce{MgCO3}\) will precipitate:

\(\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}\) becomes  \(2 \times 10^{-6} <K_{\textit{sp}}\).

\(\ce{\Rightarrow\ MgCO3}\) won’t precipitate.

Show Worked Solution

All sodium and nitrate salts are soluble  \(\Rightarrow\)  possible precipitates are \(\ce{Mg(OH)2}\) and \(\ce{MgCO3}\). 

\(\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}\)

\(\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}\)

\(\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad  K_{\textit{sp}}=6.82 \times 10^{-6}}\)
 

\(\ce{Mg(OH)2}\) is a lot less soluble than \(\ce{MgCO3}\) and will precipitate preferentially.

\(\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}\)

\(\Rightarrow \ce{Mg(OH)2}\) will precipitate.
 

\(\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}\)

Since \(K_{\textit{sp}}\) is small, assume reaction goes to completion.

\(\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}\)

Using stoichiometric ratio \((1:2)\)

\(0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}\) is in excess.

Concentration drops to: \(6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}\).
 

Check if \(\ce{MgCO3}\) will precipitate:

\(\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}\) becomes  \(2 \times 10^{-6} <K_{\textit{sp}}\).

\(\ce{\Rightarrow\ MgCO3}\) won’t precipitate.

Filed Under: Solution Equilibria Tagged With: Band 4, Band 5, smc-3672-20-Calcs given K(sp), smc-3672-70-Precipitate

PHYSICS, M7 2025 HSC 19 MC

A system consists of a sealed glass jar containing some oxygen and a small strip of magnesium.

The magnesium reacts with the oxygen to produce magnesium oxide as a product. Energy is released from the system in this reaction.

The mass of the system will

  1. increase because oxygen is added to the magnesium.
  2. decrease because energy is removed from the system.
  3. increase because energy is added to the system by the reaction.
  4. decrease because magnesium and oxygen are lost in the reaction.
Show Answers Only

\(B\)

Show Worked Solution

Option \(B\) is correct.

  • The system is sealed, so no matter enters or leaves – the magnesium and oxygen simply rearrange into magnesium oxide inside the jar.
  • Energy is released from the system and according to Einstein’s mass-energy equivalence \((E = mc^2)\), this lost energy corresponds to a decrease in mass.

Other options:

  • \(A\) is incorrect. No oxygen is added – it was already in the sealed jar.
  • \(C\) is incorrect. Energy is released (removed), not added to the system.
  • \(D\) is incorrect. No magnesium or oxygen is lost – they’re converted to magnesium oxide within the sealed system.

\(\Rightarrow B\)

Filed Under: Light and Special Relativity Tagged With: Band 5, smc-3699-80-E = mc^2

PHYSICS, M5 2025 HSC 18 MC

The escape velocity from the surface of a planet, which has no atmosphere, is \(v\). A mass is launched at 45° to the planet's surface at \(v\).

What will be the subsequent motion of the mass?

  1. A circular orbit around the planet
  2. An elliptical orbit around the planet
  3. A parabolic trajectory, returning to land with velocity \(v\)
  4. A trajectory reaching zero velocity at an infinite distance
Show Answers Only

\(D\)

Show Worked Solution
  • The mass is launched at escape velocity which means the object has just enough energy to escape the planet’s gravity.
  • As the mass travels away from the planet, kinetic energy converts into gravitational potential energy.
  • Since its initial velocity was exactly escape velocity, at infinite distance, all kinetic energy has been converted to potential energy. This means the velocity reaches zero exactly when the distance becomes infinite.
  • The 45° launch angle does not matter – escape velocity depends only on speed, not direction.

\(\Rightarrow D\)

Filed Under: Motion in Gravitational Fields Tagged With: Band 5, smc-3692-70-Escape Velocity

PHYSICS, M5 2025 HSC 36

A satellite with velocity \(v\), is in a geostationary orbit as shown in Figure 1.
 

At point \(Y\), the satellite explodes and splits into two pieces \(m_{ a }\) and \(m_{ b }\), of identical mass. As a result of the explosion, the velocity of one piece, \(m_{ a }\), changes from \(v\) to \(2 v\) as shown in Figure 2.

Analyse the subsequent motion of BOTH \(m_{ a }\) and \(m_{ b }\) after the explosion. Include reference to relevant conservation laws and formulae in your answer.   (8 marks)

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At the point in time of the explosion:

  • By the law of conservation of momentum, the momentum changes of \(m_a\) and \(m_b\) caused by the explosion must be equal in magnitude but opposite in direction.
  • Since their masses are equal, both pieces experience the same change in velocity, but in opposite directions.
  • Because the velocity of \(m_a\) increases by \(v\) in the direction it was already travelling, the velocity of \(m_b\) must also change by \(v\) but in the opposite direction to its original motion.
  • Hence, relative to Earth, the velocity of \(m_b\) becomes zero at that instant.

Motion after the explosion:

  • The satellite’s orbital velocity before the explosion is  \(v_{\text{orb}}=\sqrt{\dfrac{GM}{r}}\)  and the instantaneous velocity of \(m_a\) becomes twice this value.
  • Since  \(2v_{\text{orb}} = 2 \times \sqrt{\dfrac{GM}{r}} >\sqrt{\dfrac{2GM}{r}} \left(v_{\text{esc}}\right)\), the speed of \(m_a\) after the explosion exceeds escape velocity.
  • After the explosion, \(m_a\) continues moving away from Earth with a decreasing speed, but because its initial speed is greater than escape velocity from that point, it will never return. Its total mechanical energy (kinetic + potential) remains constant and positive.
  • Meanwhile, \(m_b\) begins accelerating from an initial velocity of zero toward Earth’s centre. Its initial acceleration is less than \(\text{ 9.8 m s}^{-2}\) because it is not at Earth’s surface. As gravitational force increases, its acceleration also increases, and it continues gaining speed at an increasing rate until it reaches Earth’s atmosphere.
  • In accordance with the conservation of energy, until \(m_b\) reaches the atmosphere, the sum of its kinetic and potential energy remains constant and equal to its initial potential energy immediately after the explosion.
Show Worked Solution

At the point in time of the explosion:

  • By the law of conservation of momentum, the momentum changes of \(m_a\) and \(m_b\) caused by the explosion must be equal in magnitude but opposite in direction.
  • Since their masses are equal, both pieces experience the same change in velocity, but in opposite directions.
  • Because the velocity of \(m_a\) increases by \(v\) in the direction it was already travelling, the velocity of \(m_b\) must also change by \(v\) but in the opposite direction to its original motion.
  • Hence, relative to Earth, the velocity of \(m_b\) becomes zero at that instant.

Motion after the explosion:

  • The satellite’s orbital velocity before the explosion is  \(v_{\text{orb}}=\sqrt{\dfrac{GM}{r}}\)  and the instantaneous velocity of \(m_a\) becomes twice this value.
  • Since  \(2v_{\text{orb}} = 2 \times \sqrt{\dfrac{GM}{r}} >\sqrt{\dfrac{2GM}{r}} \left(v_{\text{esc}}\right)\), the speed of \(m_a\) after the explosion exceeds escape velocity.
  • After the explosion, \(m_a\) continues moving away from Earth with a decreasing speed, but because its initial speed is greater than escape velocity from that point, it will never return. Its total mechanical energy (kinetic + potential) remains constant and positive.
  • Meanwhile, \(m_b\) begins accelerating from an initial velocity of zero toward Earth’s centre. Its initial acceleration is less than \(\text{ 9.8 m s}^{-2}\) because it is not at Earth’s surface. As gravitational force increases, its acceleration also increases, and it continues gaining speed at an increasing rate until it reaches Earth’s atmosphere.
  • In accordance with the conservation of energy, until \(m_b\) reaches the atmosphere, the sum of its kinetic and potential energy remains constant and equal to its initial potential energy immediately after the explosion.

Filed Under: Motion in Gravitational Fields Tagged With: Band 5, Band 6, smc-3692-40-Satellites

PHYSICS, M6 2025 HSC 35

A hollow copper pipe is placed upright on an electronic balance, which shows a reading of 300 g. A 50 g magnet is suspended inside the pipe and subsequently released.

 

It was observed that the readings on the balance began to increase after the magnet began to fall, and that the reading reached a constant maximum of 350 g before the magnet reached the bottom of the tube.

Explain these observations.   (4 marks)

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Show Answers Only

The balance reading changes in two stages:

Stage 1 – The Magnet Accelerates

  • As the magnet begins to fall, it speeds up, causing a growing change in magnetic flux through the copper pipe.
  • This induces eddy currents that produce an upward electromagnetic force on the magnet, opposing the constant gravitational force.
  • By Newton’s third law, the pipe experiences an equal downward force, so the balance reading increases as the magnet accelerates.

Stage 2 – The Magnet Reaches a Constant Speed

  • Eventually the magnet reaches a constant speed, where the upward electromagnetic force balances its weight.
  • At this point, the magnet pushes down on the pipe with a force equal to its own weight, so the balance reads the combined weight of the pipe (300 g) plus the magnet (50 g), giving a constant maximum reading of 350 g.
Show Worked Solution

The balance reading changes in two stages:

Stage 1 – The Magnet Accelerates

  • As the magnet begins to fall, it speeds up, causing a growing change in magnetic flux through the copper pipe.
  • This induces eddy currents that produce an upward electromagnetic force on the magnet, opposing the constant gravitational force.
  • By Newton’s third law, the pipe experiences an equal downward force, so the balance reading increases as the magnet accelerates.

Stage 2 – The Magnet Reaches a Constant Speed

  • Eventually the magnet reaches a constant speed, where the upward electromagnetic force balances its weight.
  • At this point, the magnet pushes down on the pipe with a force equal to its own weight, so the balance reads the combined weight of the pipe (300 g) plus the magnet (50 g), giving a constant maximum reading of 350 g.

Filed Under: Applications of the Motor Effect Tagged With: Band 5, smc-3695-20-Back emf

PHYSICS, M7 2025 HSC 34

The diagram shows a model of the orbits of Earth, Jupiter and Io, including their orbital direction and periods of orbit. In this model, it is assumed that the orbits of Earth, Jupiter and Io are circular.
 

A method to determine the speed of light using this model is described below.

When Earth was at position \(P\), the orbital period of Io was measured, and the time that Io was at position \(R\) was recorded.

Six months later, Io had orbited Jupiter 103 times, and Earth had reached position \(Q\). The orbital period of Io was used to predict when it would be at position \(R\). Assume that Jupiter has not moved significantly in its orbit around the Sun.

The time for Io to reach position \(R\) was measured to be  \(1.000 \times 10^3\) seconds later than predicted, due to the time it takes light to cross the diameter of Earth's orbit from \(P\) to \(Q\).

  1. Use the measurements provided in the model to calculate the speed of light.   (2 marks)

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  2. Consider a modification to this model in which the Earth's orbit is elliptical.
  3. Explain how this modification will affect the determination of the speed of light.   (3 marks)

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a.    \(d=2 \times 1.471 \times 10^{11} = 2.942 \times 10^{11}\ \text{metres} \)

\(v=\dfrac{d}{t}=\dfrac{2.942 \times 10^{11}}{1 \times 10^3} = 2.942 \times 10^8\ \text{m s}^{-1}\)
 

b.    Model modifications:

  • The model now assumes an elliptical orbit and (importantly) the time for Io to reach position \(R\) is assumed to again be  \(1.000 \times 10^3\) seconds later.

Consider the long axis of the ellipse in line with \(PQ:\)

  • Along this “longer” eliptical axis, light travels further than \(2.942 \times 10^{11}\ \text{m} \).
  • The longer distance travelled in the same time would result in a faster calculation of the speed of light.

Consider the short axis of the ellipse in line with \(PQ:\)

  • Along this axis, light travels less than \(2.942 \times 10^{11}\ \text{m} \).
  • The shorter distance travelled in the same time would result in a slower calculation of the speed of light.
Show Worked Solution

a.    \(d=2 \times 1.471 \times 10^{11} = 2.942 \times 10^{11}\ \text{metres} \)

\(v=\dfrac{d}{t}=\dfrac{2.942 \times 10^{11}}{1 \times 10^3} = 2.942 \times 10^8\ \text{m s}^{-1}\)
 

b.    Model modifications:

  • The model now assumes an elliptical orbit and (importantly) the time for Io to reach position \(R\) is assumed to again be  \(1.000 \times 10^3\) seconds later.

Consider the long axis of the ellipse in line with \(PQ:\)

  • Along this “longer” eliptical axis, light travels further than \(2.942 \times 10^{11}\ \text{m} \).
  • The longer distance travelled in the same time would result in a faster calculation of the speed of light.

Consider the short axis of the ellipse in line with \(PQ:\)

  • Along this axis, light travels less than \(2.942 \times 10^{11}\ \text{m} \).
  • The shorter distance travelled in the same time would result in a slower calculation of the speed of light.

Filed Under: Electromagnetic Spectrum Tagged With: Band 5, Band 6, smc-3696-20-Measuring c

Algebra, STD2 EQ-Bank 11

The amount of water (\(W\)) in litres used by a garden irrigation system varies directly with the time (\(t\)) in minutes that the system operates.

This relationship is modelled by the formula \(W=kt\), where \(k\) is a constant.

The irrigation system uses 96 litres of water when it operates for 24 minutes.

  1. Show that the value of \(k\) is 4.   (1 mark)

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  2. The water tank for the irrigation system contains 650 litres of water. Calculate how many minutes the irrigation system can operate before the tank is empty.   (2 marks)

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a.    \(W=kt\)

\(\text{When } W = 96 \text{ and } t = 24:\)

\(96\) \(=k \times 24\)
\(k\) \(=\dfrac{96}{24}\)
\(k\) \(=4\ \text{(as required)}\)

b.     \(162.5\ \text{minutes}\)

Show Worked Solution

a.    \(W=kt\)

\(\text{When } W = 96 \text{ and } t = 24:\)

\(96\) \(=k \times 24\)
\(k\) \(=\dfrac{96}{24}\)
\(k\) \(=4\ \text{(as required)}\)

  
b.    \(W = 4t\)

\(\text{When } W = 650:\)

\(650\) \(=4\times t\)
\(t\) \( =\dfrac{650}{4}\)
  \(=162.5\ \text{minutes}\)

    
\(\therefore\ \text{The irrigation system can operate for } 162.5 \text{ minutes.}\)

Filed Under: Direct Variation (Std1-2027), Direct Variation and Currency Conversion (Std2-2027) Tagged With: Band 4, Band 5, smc-6249-10-Find k, smc-6249-30-Graphical Solutions, smc-6513-30-Graphical Solutions, smc-6514-10-Find k, syllabus-2027

Algebra, STD2 EQ-Bank 10

The cost (\(C\)) of copper wire varies directly with the length (\(L\)) in metres of the wire.

This relationship is modelled by the formula \(C = kL\), where \(k\) is a constant.

A 250 metre roll of copper wire costs $87.50.

  1. Show that the value of \(k\) is 0.35.   (1 mark)

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  2. A builder has a budget of $140 for copper wire. Calculate the maximum length of wire that can be purchased.   (2 marks)

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a.    \(C=kL\)

\(\text{When } C = 87.50 \text{ and } L = 250:\)

\(87.50\) \(=k \times 250\)
\(k\) \(=\dfrac{87.50}{250}\)
\(k\) \(=0.35\ \text{(as required)}\)

b.     \(400\ \text{m}\)

Show Worked Solution

a.    \(C=kL\)

\(\text{When } C = 87.50 \text{ and } L = 250:\)

\(87.50\) \(=k \times 250\)
\(k\) \(=\dfrac{87.50}{250}\)
\(k\) \(=0.35\ \text{(as required)}\)

 
b.    \(C = 0.35L\)

\(\text{When } C = 140:\)

\(140\) \(=0.35\times L\)
\(L\) \( =\dfrac{140}{0.35}\)
  \(=400\ \text{m}\)

    
\(\therefore\ \text{The builder can purchase } 400 \text{ metres of wire.}\)

Filed Under: Direct Variation (Std1-2027), Direct Variation and Currency Conversion (Std2-2027) Tagged With: Band 4, Band 5, smc-6249-10-Find k, smc-6249-30-Graphical Solutions, smc-6513-30-Graphical Solutions, smc-6514-10-Find k, syllabus-2027

Algebra, STD2 EQ-Bank 09 MC

The amount of paint (\(P\)) needed to cover a wall varies directly with the area (\(A\)) of the wall.

A painter uses 3.5 litres of paint to cover a wall with an area of 28 square metres.

How much paint is needed to cover a wall with an area of 42 square metres?

  1. 4.2 L
  2. 4.75 L
  3. 5.25 L
  4. 5.75 L
Show Answers Only

\(C\)

Show Worked Solution

\(P \propto A\)

\(P=kA\)

\(\text{When } P = 3.5 \text{ and } A = 28:\)

\(3.5\) \(=k \times 28\)
\(k\) \(=\dfrac{3.5}{28}\)
\(k\) \(=0.125\)

  
\(\text{When } A = 42:\)  

\(P\) \(=0.125\times 42\)
  \(=5.25\)

  
\(\Rightarrow C\)

Filed Under: Direct Variation (Std1-2027), Direct Variation and Currency Conversion (Std2-2027) Tagged With: Band 5, smc-6249-10-Find k, smc-6249-20-Algebraic Solutions, smc-6514-10-Find k, smc-6514-20-Algebraic Solutions, syllabus-2027

PHYSICS, M6 2025 HSC 17 MC

A circular loop of wire is placed at position \(X\), next to a straight current-carrying wire with the current direction shown.

The loop is moved to position \(Y\) at a constant speed.
 

 

Which row in the table best describes the induced electromotive force (emf) in the loop as it moves from \(X\) to \(Y\) ?

Show Answers Only

\(C\)

Show Worked Solution
  • The downward current in the straight wire creates a magnetic field out of the page on the loop’s side (right-hand grip rule).
  • As the loop moves from \(X\) to \(Y\), this outward flux decreases. Lenz’s Law means that the loop induces an anticlockwise current to oppose the change (eliminate \(B\) and \(D\)).
  • Using  \(\epsilon = -\dfrac{\Delta \Phi}{\Delta t}\), the induced emf is equal to the negative slope of the flux-time graph.
  • Because the magnetic field gets weaker with distance \((B \propto \frac{1}{r})\) and the loop moves at constant speed, the flux drops in a curved (non-linear) way rather than straight.
  • This means the induced emf also decreases in a curved way, not as a straight-line fall.

\(\Rightarrow C\)

Filed Under: Electromagnetic Induction Tagged With: Band 5, smc-3705-20-Induced emf, smc-3705-40-Lenz and Faraday

PHYSICS, M8 2025 HSC 16 MC

A neutron is absorbed by a nucleus, \(X\).

The resulting nucleus undergoes alpha decay, producing lithium-7.

What is nucleus \(X\) ?

  1. Boron-10
  2. Boron-11
  3. Lithium-6
  4. Lithium-10
Show Answers Only

\(A\)

Show Worked Solution
  • The alpha decay described can be expresses as:
  •      \(\ce{^1_0n + ^{10}_5Bu \rightarrow ^7_3 Li + ^4_2 He}\)

\(\Rightarrow A\)

Filed Under: Properties of the Nucleus Tagged With: Band 5, smc-3703-40-Radioactive Decay

PHYSICS, M7 2025 HSC 12 MC

Which graph correctly represents Malus' Law?
 

 

Show Answers Only

\(A\)

Show Worked Solution
  • Malus’ Law: \(I=I_{\text{max}}\cos^{2}\theta\)
  • Light intensity is therefore directly proportional to \(\cos^{2}\theta\) (noting \(I_{\text{max}}\) is a constant).

\(\Rightarrow A\)

Filed Under: Light: Wave Model Tagged With: Band 5, smc-3697-40-Polarisation

PHYSICS, M6 2025 HSC 11 MC

A DC motor connected to a 12 V supply is maintaining a constant rotational speed. An ammeter in the circuit reads 0.5 A .
 

Some material falls into the running motor, causing it to slow down.

What is the subsequent ammeter reading likely to be?

  1. 0
  2. Between 0 and 0.5 A
  3. 0.5 A
  4. Greater than 0.5 A
Show Answers Only

\(D\)

Show Worked Solution
  • When the motor slows down, its back EMF decreases, meaning it opposes the supply voltage less.
  • With a lower back EMF, the net voltage across the motor’s windings increases, causing more current to flow.
  • Therefore the ammeter reading becomes greater than 0.5 A.

\(\Rightarrow D\)

Filed Under: Applications of the Motor Effect Tagged With: Band 5, smc-3695-20-Back emf

PHYSICS, M8 2025 HSC 10 MC

The diagram shows four lines, \(W, X, Y\) and \(Z\), depicting radioactivity varying with time.
 

Which of the four lines is consistent with a decay graph with the smallest decay constant \((\lambda)\) ?

  1. \(W\)
  2. \(X\)
  3. \(Y\)
  4. \(Z\)
Show Answers Only

\(C\)

Show Worked Solution
  • A radioactive decay graph is shaped exponentially and is not linear (eliminate \(W\) and \(Z\).
  • Since the decay constant \(=\dfrac{\lambda}{t_{\frac{1}{2}}}\), and \(Y\) has a higher half-life than \(X\) (its activity graaph is higher), \(Y\) must have the smaller decay constant.

\(\Rightarrow C\)

Filed Under: Properties of the Nucleus Tagged With: Band 5, smc-3703-40-Radioactive Decay

PHYSICS, M5 2025 HSC 7 MC

A satellite in a circular orbit around Earth at an altitude of 500 km is moved to a new circular orbit at a higher altitude of 35 800 km .

Which statement correctly compares properties of the satellite in the higher orbit with its properties in the lower orbit?

  1. Its period is greater, and its acceleration is the same.
  2. Its kinetic energy is less, and its acceleration is less.
  3. Its orbital velocity is less, and its potential energy is positive.
  4. Its escape velocity is greater, and the centripetal force is less.
Show Answers Only

\(B\)

Show Worked Solution

Option \(B\) is correct:

  • Orbital velocity decreases when altitude increases \(\left( v=\sqrt{\frac{GM}{r}}\right) \)
  • Kinetic energy decreases \(\left(K=\frac{1}{2}mv^{2}\right)\)
  • Acceleration decreases  \(\left(a_c=\frac{v^{2}}{r}\right) \)

Other options:

  • \(A\) is incorrect. Acceleration does not stay the same.
  • \(C\) is incorrect. Gravitational potential energy is always negative.
  • \(D\) is incorrect. Escape velocity is lower \(\left( v_{esc}=\sqrt{\dfrac{2GM}{r}} \right)\)

\(\Rightarrow B\)

Filed Under: Motion in Gravitational Fields Tagged With: Band 5, smc-3692-30-Orbital Velocity/Period/Radius, smc-3692-40-Satellites

PHYSICS, M8 2025 HSC 30

A beam of electrons travelling at \(4 \times 10^3 \ \text{m s}^{-1}\) exits an electron gun and is directed toward two narrow slits with a separation, \(d\), of 1 \(\mu\text{m}\). The resulting interference pattern is detected on a screen 50 cm from the slits.

  1. Show that the wavelength of the electrons in this experiment is 182 nm.   (2 marks)

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  2. An interference fringe occurs on the screen where constructive interference takes place.
     

  1. Determine the distance between the central interference fringe \(A\) and the centre of the next bright fringe \(B\).   (2 marks)

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  2. Determine the potential difference acting in the electron gun to accelerate the electrons in the beam from rest to \(4 \times 10^3 \ \text{m s}^{-1}\).   (2 marks)

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a.    \(\text {Using} \ \ \lambda=\dfrac{h}{m v}:\)

\(\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}\)

b.    \(x=9.1 \ \text{cm}\)

c.    \(V=4.5 \times 10^{-5} \ \text{V}\)

Show Worked Solution

a.    \(\text {Using} \ \ \lambda=\dfrac{h}{m v}:\)

\(\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}\)
 

b.    \(\text {Using}\ \ x=\dfrac{\lambda m L}{d}\)

\(\text{where} \ \ x=\text{distance between middle of adjacent bright spots}\)

\(x=\dfrac{182 \times 10^{-9} \times 1 \times 0.5}{1 \times 10^{-6}}=0.091 \ \text{m}=9.1 \ \text{cm}\)
 

c.    \(\text{Work done by field}=\Delta K=K_f-K_i\)

\(qV\) \(=\dfrac{1}{2} m v^2-0\)
\(V\) \(=\dfrac{m v^2}{2 q}=\dfrac{9.109 \times 10^{-31} \times\left(4 \times 10^3\right)^2}{2 \times 1.602 \times 10^{-19}}=4.5 \times 10^{-5} \ \text{V}\)

Filed Under: Charged Particles, Conductors and Electric/Magnetic Fields, Light: Wave Model, Quantum Mechanical Nature of the Atom Tagged With: Band 4, Band 5, smc-3693-35-Kinematic/Work Calcs, smc-3697-30-Double Slit Calcs, smc-3702-40-De Broglie

PHYSICS, M8 2025 HSC 33

Analyse the role of experimental evidence and theoretical ideas in developing the Standard Model of matter.   (6 marks)

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Overview Statement

  • The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
  • These components interact with each other, where theory guides experiments and results validate or refine theory.

Particle Discovery

  • Theoretical predictions lead to targeted experimental searches for specific particles.
  • The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
  • Cloud chambers discovered antimatter after theory predicted its existence.
  • Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
  • This pattern shows theory provides the framework while experiments confirm reality.
  • Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

  • High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
  • These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
  • Unexpected experimental results sometimes cause theoretical modifications or new predictions.
  • The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

  • This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
  • Neither component alone could have produced the model.
  • Together, they form a self-correcting system advancing our understanding of fundamental matter.
Show Worked Solution

Overview Statement

  • The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
  • These components interact with each other, where theory guides experiments and results validate or refine theory.

Particle Discovery

  • Theoretical predictions lead to targeted experimental searches for specific particles.
  • The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
  • Cloud chambers discovered antimatter after theory predicted its existence.
  • Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
  • This pattern shows theory provides the framework while experiments confirm reality.
  • Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

  • High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
  • These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
  • Unexpected experimental results sometimes cause theoretical modifications or new predictions.
  • The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

  • This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
  • Neither component alone could have produced the model.
  • Together, they form a self-correcting system advancing our understanding of fundamental matter.

Filed Under: Deep Inside the Atom Tagged With: Band 4, Band 5, smc-3704-10-Standard Model

PHYSICS, M7 2025 HSC 32

Analyse the consequences of the theory of special relativity in relation to length, time and motion. Support your answer with reference to experimental evidence.   (8 marks)

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Overview Statement

  • Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
  • These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

  • Muon-decay experiments show how time dilation and length contraction depend on relative motion.
  • Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
  • In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
  • In the muon’s frame of reference, the atmosphere is length-contracted according to the equation  \(l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}\), so the distance they travel is much shorter.
  • Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

  • Relativistic momentum explains how objects behave as they approach light speed.
  • Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
  • Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
  • This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
  • Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

  • The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
  • At high velocities, motion fundamentally alters measurements of time, distance and momentum.
  • Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.
Show Worked Solution

Overview Statement

  • Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
  • These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

  • Muon-decay experiments show how time dilation and length contraction depend on relative motion.
  • Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
  • In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
  • In the muon’s frame of reference, the atmosphere is length-contracted according to the equation  \(l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}\), so the distance they travel is much shorter.
  • Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

  • Relativistic momentum explains how objects behave as they approach light speed.
  • Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
  • Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
  • This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
  • Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

  • The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
  • At high velocities, motion fundamentally alters measurements of time, distance and momentum.
  • Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

Filed Under: Light and Special Relativity Tagged With: Band 4, Band 5, Band 6, smc-3699-10-Time Dilation, smc-3699-20-Length Contraction, smc-3699-30-Momentum Dilation and calcs, smc-3699-50-Scientific Theories

PHYSICS, M8 2025 HSC 31

Experiments have been carried out by scientists to investigate cathode rays.

Assess the contribution of the results of these experiments in developing an understanding of the existence and properties of electrons.   (5 marks)

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Judgment Statement

  • The cathode ray experiments were highly valuable in establishing both the existence and properties of electrons through definitive experimental evidence and quantitative measurements.

Demonstrating Particle Nature

  • Electric field deflection experiments produced significant results by proving cathode rays were negatively charged particles rather than electromagnetic radiation.
  • This was highly effective because it eliminated the competing theory that cathode rays were electromagnetic waves, since waves are not deflected by electric fields.
  • The consistent deflection pattern across many experiments provided strong evidence for the particle nature of electrons.

Quantifying Electron Properties

  • By adjusting electric and magnetic field strengths within experiments, the charge-to-mass ratio of the electron was determined.
  • This measurement proved highly effective as it provided the first quantitative property of electrons.
  • The e/m ratio demonstrated considerable value by revealing electrons were much lighter than atoms, indicating subatomic particles existed.

Overall Assessment

  • Assessment reveals these experiments achieved major significance in atomic theory development.
  • The combined results produced measurable, reproducible data that definitively established electrons as fundamental charged particles with specific properties.
  • Overall, these contributions proved essential for understanding atomic structure.

 
Other answers could include:

  • By using electrodes made of different materials, Thomson was able to deduce that the cathode rays’ properties were independent of the source of the electrons and hence that they were a constituent of atoms themselves rather than being a product of the cathode ray.
  • Cathode rays were passed through thin metal foils and the analysis of this behaviour allowed scientists (Lenard/Hertz) to deduce that the electrons had mass.
  • Crookes’ observation that cathode rays travelled in straight lines and cast sharp shadows from which he deduced that the rays were particles and not waves (which would have shown diffraction effects).
Show Worked Solution

Judgment Statement

  • The cathode ray experiments were highly valuable in establishing both the existence and properties of electrons through definitive experimental evidence and quantitative measurements.

Demonstrating Particle Nature

  • Electric field deflection experiments produced significant results by proving cathode rays were negatively charged particles rather than electromagnetic radiation.
  • This was highly effective because it eliminated the competing theory that cathode rays were electromagnetic waves, since waves are not deflected by electric fields.
  • The consistent deflection pattern across many experiments provided strong evidence for the particle nature of electrons.

Quantifying Electron Properties

  • By adjusting electric and magnetic field strengths within experiments, the charge-to-mass ratio of the electron was determined.
  • This measurement proved highly effective as it provided the first quantitative property of electrons.
  • The e/m ratio demonstrated considerable value by revealing electrons were much lighter than atoms, indicating subatomic particles existed.

Overall Assessment

  • Assessment reveals these experiments achieved major significance in atomic theory development.
  • The combined results produced measurable, reproducible data that definitively established electrons as fundamental charged particles with specific properties.
  • Overall, these contributions proved essential for understanding atomic structure.

 
Other answers could include:

  • By using electrodes made of different materials, Thomson was able to deduce that the cathode rays’ properties were independent of the source of the electrons and hence that they were a constituent of atoms themselves rather than being a product of the cathode ray.
  • Cathode rays were passed through thin metal foils and the analysis of this behaviour allowed scientists (Lenard/Hertz) to deduce that the electrons had mass.
  • Crookes’ observation that cathode rays travelled in straight lines and cast sharp shadows from which he deduced that the rays were particles and not waves (which would have shown diffraction effects).

Filed Under: Structure of the Atom Tagged With: Band 4, Band 5, smc-3701-10-Cathode Rays and Thomson

PHYSICS, M5 2025 HSC 29

A mass moves around a vertical circular path of radius \(r\), in Earth's gravitational field, without loss of mechanical energy. A string of length \(r\) maintains the circular motion of the mass.

When the mass is at its highest point \(B\), the tension in the string is zero.
 

  1. Show that the speed of the mass at the highest point, \(B\), is given by  \(v=\sqrt{r g}\).   (2 marks)

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  2. Compare the speed of the mass at point \(A\) to that at point \(B\). Support your answer using appropriate mathematical relationships.   (3 marks)

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a.    \(\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}\)

\(\dfrac{mv_B^2}{r}\) \(=T+mg\)
\(v_B^2\) \(=rg \ \ (T=0)\)
\(v_B\) \(=\sqrt{r g}\)

 
b.
    \(\text {Total} \ ME=E_k+GPE\)

\(\text{At point} \ B :\)

 \(\text {Total} \ ME\) \(=\dfrac{1}{2} m v_B^2+mg(2r)\)
  \(=\dfrac{1}{2} m \times r g+2 mrg\)
  \(=\dfrac{5}{2} m r g\)

 
\(\text{At point} \ A :\)

\(\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg\)
 

\(\text{Since \(ME\) is conserved:}\)

\(\dfrac{5}{2} mrg\) \(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\) \(=\dfrac{3}{2} m r g\)
\(v_A^2\) \(=3 rg\)
\(v_A\) \(=\sqrt{3rg}=\sqrt{3} \times v_B\)
Show Worked Solution

a.    \(\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}\)

\(\dfrac{mv_B^2}{r}\) \(=T+mg\)
\(v_B^2\) \(=rg \ \ (T=0)\)
\(v_B\) \(=\sqrt{r g}\)

 
b.
    \(\text {Total} \ ME=E_k+GPE\)

\(\text{At point} \ B :\)

 \(\text {Total} \ ME\) \(=\dfrac{1}{2} m v_B^2+mg(2r)\)
  \(=\dfrac{1}{2} m \times r g+2 mrg\)
  \(=\dfrac{5}{2} m r g\)

 
\(\text{At point} \ A :\)

\(\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg\)
 

\(\text{Since \(ME\) is conserved:}\)

\(\dfrac{5}{2} mrg\) \(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\) \(=\dfrac{3}{2} m r g\)
\(v_A^2\) \(=3 rg\)
\(v_A\) \(=\sqrt{3rg}=\sqrt{3} \times v_B\)

Filed Under: Circular Motion Tagged With: Band 4, Band 5, smc-3691-10-Centripetal Forces

PHYSICS, M6 2025 HSC 26

The starting position of a simple AC generator is shown. It consists of a single rectangular loop of wire in a uniform magnetic field of 0.5 T. This loop is connected to two slip rings and the slip rings are connected via brushes to a voltmeter.
 

  1. The loop is rotated at a constant rate through an angle of 90 degrees from the starting position in the direction indicated, in 0.1 seconds.
  2. Calculate the magnitude of the average emf generated during this rotation.   (2 marks)

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  1. The same coil was then rotated at 10 revolutions per second from the starting position. The voltage varies with time, as shown in the graph.
     

  1. On the same axes, sketch a graph that shows the variation of voltage with time if the rotational speed is 20 revolutions per second in the opposite direction, beginning at the original starting position.   (3 marks)

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a.    \(\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}\)

\(\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}\)
 

b.    
           

Show Worked Solution

a.    \(\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}\)

\(\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}\)
 

b.    

Filed Under: Electromagnetic Induction Tagged With: Band 4, Band 5, smc-3705-10-Generated emf

BIOLOGY, M8 2025 HSC 24

The following flow chart represents the control of body temperature in humans.
 

  1. Complete the flow chart to give an example of mechanism A and an example of mechanism B.   (2 marks)

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  2. Outline how mechanism B maintains homeostasis.   (2 marks)

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a.    Mechanism A (Decreases temperature):

  • Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

  • Shivering → Vasoconstriction

b.    Maintaining homeostasis

  • When body temperature drops below normal range, thermoreceptors detect the change.
  • The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
  • Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
  • Body temperature increases back to normal range, restoring homeostasis.

Show Worked Solution

a.   Mechanism A (Decreases temperature):

  • Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

  • Shivering → Vasoconstriction

b.    Maintaining homeostasis

  • When body temperature drops below normal range, thermoreceptors detect the change.
  • The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
  • Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
  • Body temperature increases back to normal range, restoring homeostasis.

Filed Under: Homeostasis Tagged With: Band 4, Band 5, smc-3659-15-Temperature

BIOLOGY, M5 2025 HSC 31

Congenital amegakaryocytic thrombocytopenia (CAMT) is a rare, inherited disorder where bone marrow no longer makes platelets that are important for clotting and preventing bleeding. The pedigree below shows the inheritance of CAMT in a family.
 

  1. What type of inheritance is shown in the pedigree above? Justify your answer?   (3 marks)
Type of Inheritance:  

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  1. A CAMT mutation was found to produce the following amino acid sequence:
    1. Glutamine – Tyrosine – Isoleucine – Aspartic acid.
  2. The same DNA fragment has been sequenced from an unaffected individual.
  3. Template strand           GTC ATA CAG CTG.
  4. The following codon chart displays all the codons and corresponding amino acids. The chart translates mRNA sequences into amino acids.
      

  5. Use the codon chart shown to explain the type of mutation which causes CAMT.   (3 marks)

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a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.
Show Worked Solution

a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

Filed Under: Genetic Variation / Inheritance Tagged With: Band 4, Band 5, smc-3648-10-Pedigrees, smc-3648-75-Inheritance Patterns

BIOLOGY, M5 2025 HSC 33

The following diagram shows the cell division processes occurring in two related individuals.
 

  1. Compare the cell division processes carried out by cells \(R\) and \(S\) in Individual 1.   (3 marks)

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  2. Explain the relationship between Individuals 1 and 2.   (2 marks)

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  3. \(A\) and \(B\) are two separate mutations. Analyse how mutations \(A\) and \(B\) affect the genetic information present in cells  \(U\), \(V\), \(W\) and \(X\).   (4 marks)

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a.    Similarities:

  • Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

  • Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
  • Cell S undergoes meiosis producing four genetically different haploid gametes.
  • Mitosis in R maintains chromosome number for growth and repair.
  • Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b.    Relationship between Individuals 1 and 2

  • Individual 2 is the offspring of Individual 1.
  • This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c.    Mutation’s affect on genetic information

  • Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
  • Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
  • Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
  • The significance is that only mutation A can be inherited by offspring, while mutation B cannot.
Show Worked Solution

a.    Similarities:

  • Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

  • Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
  • Cell S undergoes meiosis producing four genetically different haploid gametes.
  • Mitosis in R maintains chromosome number for growth and repair.
  • Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b.    Relationship between Individuals 1 and 2

  • Individual 2 is the offspring of Individual 1.
  • This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c.    Mutation’s affect on genetic information

  • Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
  • Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
  • Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
  • The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

Filed Under: Cell Replication, Mutation Tagged With: Band 5, Band 6, smc-3649-20-Meiosis, smc-3649-30-Mitosis, smc-3652-20-Somatic vs. Germ-Line

BIOLOGY, M6 2025 HSC 34

The following graph shows the changes in allele frequencies in two separate populations of the same species. Each line represents an introduced allele.

  1. Explain why the fluctuations of the allele frequencies are more pronounced in the small population, compared to the larger population.   (2 marks)

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  2. Evaluate the effects of gene flow on the gene pools of the two populations.   (4 marks)

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a.  Genetic Drift and Population Size

  • Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
  • Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b.    Evaluation Statement

  • Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

  • The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
  • This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
  • The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

  • The small population (20) shows limited benefit from gene flow.
  • Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
  • Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

  • Overall, gene flow proves highly effective in large populations for maintaining diversity,.
  • However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.
Show Worked Solution

a.  Genetic Drift and Population Size

  • Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
  • Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b.    Evaluation Statement

  • Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

  • The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
  • This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
  • The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

  • The small population (20) shows limited benefit from gene flow.
  • Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
  • Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

  • Overall, gene flow proves highly effective in large populations for maintaining diversity,.
  • However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

Filed Under: Mutation Tagged With: Band 4, Band 5, smc-3652-10-Mechanisms of Change/Variation

BIOLOGY, M8 2025 HSC 26

The diagram shows the steps in LASIK (Laser-Assisted In Situ Keratomileusis) surgery.
 

Compare the LASIK technology shown with ONE other technology that can be used to treat a named visual disorder.   (4 marks)

Visual disorder:  

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Visual disorder: Myopia (short-sightedness)

Similarities:

  • Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
  • Both technologies enable clear vision at distance for myopia patients.

Differences:

  • LASIK permanently reshapes the cornea using laser ablation to correct vision.
  • Spectacles use external convex or concave lenses to refract light without altering eye structure.
  • LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
  • LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

Show Worked Solution

Visual disorder: Myopia (short-sightedness)

Similarities:

  • Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
  • Both technologies enable clear vision at distance for myopia patients.

Differences:

  • LASIK permanently reshapes the cornea using laser ablation to correct vision.
  • Spectacles use external convex or concave lenses to refract light without altering eye structure.
  • LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
  • LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

Filed Under: Technologies and Disorders Tagged With: Band 4, Band 5, smc-3663-20-Eyes

BIOLOGY, M7 2025 HSC 28

Alpha-gal syndrome (AGS) is a tick-borne allergy to red meat caused by tick bites. Alpha-gal is a sugar molecule found in most mammals but not humans, and can also be found in the saliva of ticks. The diagram shows how a tick bite might cause a person to develop an allergic reaction to red meat.
 

 

  1. The flow chart shows the process of antibody production following exposure to alpha-gal. 
  2.   
  3. Describe the role of X, Y and Z in the process of antibody production.   (4 marks)

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  4. An allergic reaction to alpha-gal sugar is similar to a secondary immune response.
    1.    
  5. Describe the features of antibody production shown in the graph.   (2 marks)

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  6. Explain the role of memory cells in the immune response.   (3 marks)

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a.    Antibody Production Process

  • X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
  • Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
  • Y is a B-cell that has receptors specific to the alpha-gal antigen.
  • B-cells are activated by Helper T-cells and undergo clonal expansion.
  • Some B-cells differentiate into memory cells for long-term immunity.
  • Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
  • Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b.    Features of Antibody Production

  • Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
  • Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c.    Role of Memory Cells

  • Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
  • Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
  • This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.
Show Worked Solution

a.    Antibody Production Process

  • X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
  • Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
  • Y is a B-cell that has receptors specific to the alpha-gal antigen.
  • B-cells are activated by Helper T-cells and undergo clonal expansion.
  • Some B-cells differentiate into memory cells for long-term immunity.
  • Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
  • Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b.    Features of Antibody Production

  • Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
  • Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c.    Role of Memory Cells

  • Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
  • Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
  • This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

Filed Under: Immunity and Responses to Pathogens Tagged With: Band 4, Band 5, smc-3657-10-Antibody Mediated, smc-3657-25-Antibodies

Financial Maths, STD2 EQ-Bank 01 MC

Jacinta buys several items at the supermarket. The docket for her purchases is shown below.

What is the amount of GST included in the total? 

  1. $1.15
  2. $1.27
  3. $1.55
  4. $1.71
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Total price of taxable items}\ +\ 10\%\ \text{ GST}\ =1.29+7.23+4.13=$12.65\)

\(\therefore\ 110\%\ \text{of original price}\) \(=$12.65\)
\(\therefore\ \text{GST}\) \(=$12.65\times\dfrac{100}{110}\)
  \(=$1.15\)

\(\Rightarrow A\)

Filed Under: Purchasing Goods (Std1-2027), Purchasing Goods (Std2-2027), Tax and Percentage Increase/Decrease (Std 1) Tagged With: Band 5, smc-1125-20-GST, smc-6278-20-GST/VAT, smc-6517-20-GST/VAT, syllabus-2027

Measurement, STD2 EQ-Bank 01 MC

The sheets of paper Jenny uses in her photocopier are 21 cm by 30 cm. The paper is 80 gsm, which means that one square metre of this paper has a mass of 80 grams. Jenny has a pile of this paper weighing 25.2 kg.

How many sheets of paper are in the pile?

  1. 500
  2. 2000
  3. 2500
  4. 5000
Show Answers Only

\(D\)

Show Worked Solution

\(1\ \text{square metre} = 100\ \text{cm}\times 100\ \text{cm}=10\,000\ \text{cm}^2\)

\(\text{Area of paper sheet}\ = \ 21\times 30=630\ \text{cm}^2\)

\(\text{Number of 80 gsm sheets in 25.2 kg}\ =\dfrac{25.2\times 1000}{80}=315\)

\(\therefore\ \text{Sheets in pile}\) \(=315\times\dfrac{10\,000}{630}\)
  \(=5000\)

  

Filed Under: Identify and Convert Between Units (Std2-2027) Tagged With: Band 5, smc-6303-30-Capacity/Volume/Mass, syllabus-2027

CHEMISTRY, M6 2025 HSC 31

Hydrazine is a compound of hydrogen and nitrogen. The complete combustion of 1.0 L of gaseous hydrazine requires 3.0 L of oxygen, producing 2.0 L of nitrogen dioxide gas and 2.0 L of water vapour. All volumes are measured at 400°C.

  1. Use the chemical equation for the combustion of hydrazine to show that the molecular formula for hydrazine is \(\ce{N2H4}\).   (2 marks)

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  2. The relationship between the acid equilibrium constant \(\left(K_a\right)\) and the corresponding conjugate base equilibrium constant \(\left(K_b\right)\) is shown.
      1. \(K_a \times K_b=K_w\)
  3. Use a relevant chemical equation to calculate the pH of a  0.20 mol L\(^{-1}\) solution of \(\ce{N2H5+}\) using the following data:
    • the \(K_b\) of hydrazine is \(1.7 \times 10^{-6}\) at 25°C
    • \(\ce{N2H5+}\) is the conjugate acid of \(\ce{N2H4}\).   (4 marks)
Show Answers Only

a.    Using Avogadro’s law:

  • At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
  •    \(\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}\)
  • Mole ratio  \(\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ \) matches the volume ratios given.
  • Therefore \(\ce{N2H4}\) is the correct molecular formula for hydrazine.

b.    \(\text{pH} = 4.46\)

Show Worked Solution

a.    Using Avogadro’s law:

  • At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
  •    \(\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}\)
  • Mole ratio  \(\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ \) matches the volume ratios given.
  • Therefore \(\ce{N2H4}\) is the correct molecular formula for hydrazine.

b.    \(K_a(\ce{N2H5+}) = \dfrac{K_w}{K_b(\ce{N2H4})} = \dfrac{1 \times 10^{-14}}{1.7 \times 10^{-6}} = 5.88235 \times 10^{-9}\)

  • The ionisation of \(\ce{N2H5+}\) is given the chemical equation below:
  •    \(\ce{N2H5+(aq) + H2O(l) \leftrightharpoons N2H4(aq) + H3O+(aq)}\)
  •    \(K_a = \dfrac{\ce{[H3O+][N2H4]}}{\ce{[N2H5+]}}\)
     
  • Using an Ice Table where all numbers are in mol L\(^{-1}\).

\begin{array} {|c|c|c|c|}
\hline  & \ce{[N2H5+]} & \ce{[N2H4]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20 -x & x & x \\
\hline \end{array}

 

  • Substituting into the \(K_a\) expression:
   \(\dfrac{x^2}{0.20-x}\) \(=5.88235 \times 10^{-9}\)  
\(\dfrac{x^2}{0.20}\) \(=5.88235 \times 10^{-9}\), as \(x\) is really small  
\(x\) \(=3.42997 \times 10^{-5}\)  

 

   \(\text{pH}\) \(=-\log_{10}(\ce{[H3O+]})\)  
  \(=-\log_{10}(3.42997 \times 10^{-5})\)  
  \(=4.46\)  

Filed Under: Bronsted-Lowry Theory, Quantitative Analysis Tagged With: Band 4, Band 5, smc-3674-10-Calculations Involving pH, smc-3675-30-Ka/Kb

CHEMISTRY, M5 2025 HSC 29

Consider the following reaction.

\(\ce{2NO2(g) \rightleftharpoons N2O4(g) \quad \quad \Delta H= -57.2 \ \text{kJ mol}^{-1}}\)

A sealed reaction vessel of fixed volume contains a mixture of \(\ce{NO2}\) and \(\ce{N2O4}\) gases at equilibrium.

Explain the impact of the addition of argon, an inert gas, on the temperature of the system.   (4 marks)

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Show Answers Only
  • The equilibrium expression for the reaction is:
  •    \(Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}​\)
  • When argon is added at constant volume, the total pressure increases, but the concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of \(Q\) remains equal to \(K_{eq}\), so the equilibrium position does not shift.
  • Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.
Show Worked Solution
  • The equilibrium expression for the reaction is:
  •    \(Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}​\)
  • When argon is added at constant volume, the total pressure increases, but the concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of \(Q\) remains equal to \(K_{eq}\), so the equilibrium position does not shift.
  • Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.

Filed Under: Factors that Affect Equilibrium Tagged With: Band 5, smc-3670-29-Pressure, smc-3670-35-Volume

CHEMISTRY, M7 2025 HSC 28

Kevlar and polystyrene are two common polymers.

A section of their structures is shown.
 

     

  1. Kevlar is produced through a reaction of two different monomers, one of which is shown. Draw the missing monomer in the box provided.   (1 mark)

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  1. Kevlar chains are hard to pull apart, whereas polystyrene chains are not.
  2. With reference to intermolecular forces, explain the difference in the physical properties of the two polymers.   (3 marks)

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Show Answers Only

a.    
           

b.    The physical differences between the two polymers are:

  • Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
  • The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
  • Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
  • These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.
Show Worked Solution

a.    
           

b.    The physical differences between the two polymers are:

  • Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
  • The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
  • Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
  • These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

Filed Under: Polymers Tagged With: Band 4, Band 5, smc-3681-10-Addition Polymers, smc-3681-20-Condensation Polymers, smc-3681-30-Monomers

CHEMISTRY, M7 2025 HSC 27

Mixtures of hydrocarbons can be obtained from crude oil by the process of fractional distillation. Examples include petrol, diesel and natural gas.

  1. Outline an environmental implication for a use of a named hydrocarbon mixture that is obtained from crude oil.   (2 marks)

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  2. Ethene is a simple hydrocarbon obtained from crude oil.
  3. Using structural formulae, write the chemical equation for the conversion of ethene to ethanol, including any other necessary reagents.   (3 marks)

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  1. When ethanol is reacted with ethanoic acid, ethyl ethanoate is formed, as shown by the equation.
      1. \(\text{ethanol} \ + \ \text{ethanoic acid } \ \rightleftharpoons \ \text{ethyl ethanoate} \ +\ \text{water}\)
  2. The graph below shows the concentration of ethanol from the start of the reaction, \(t_0\), up to a time \(t_1\).
  3. At time \(t_1\), an additional amount of ethanol is added to the system.
  4. Sketch on the graph the changes that occur in the concentration of ethanol between time \(t_1\), and when the system reaches a new equilibrium before time \(t_2\).   (3 marks)

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a.    Environmental implication:

  • The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
  • Increased \(\ce{CO2}\) levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
  • A typical component of petrol, octane, burns according to:
  •    \(\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}\)

b.    
           

c.    
       

Show Worked Solution

a.    Environmental implication:

  • The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
  • Increased \(\ce{CO2}\) levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
  • A typical component of petrol, octane, burns according to:
  •    \(\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}\)

b.    
           

c.    
       

  • There will be a sudden increase at \(t_1\) and then the concentration of ethanol will decrease smoothly until a new equlibrium concentration (greater than the original equilibrium concentration) is reached.

Filed Under: Factors that Affect Equilibrium, Hydrocarbons and Associated Reactions Tagged With: Band 4, Band 5, smc-3670-30-Concentration, smc-3670-60-Concentration diagrams, smc-3677-70-Reactions of Unsaturated Hydrocarbons, smc-3677-90-Environment

CHEMISTRY, M5 2025 HSC 20 MC

The solubility constant for silver\(\text{(I)}\) oxalate \(\ce{(Ag2C₂O4)}\) was determined using the following method.

  • 2.0 g of solid \(\ce{Ag2C2O4}\) was added to 100 mL of distilled water.
  • A sample of the saturated solution above the undissolved \(\ce{Ag₂C₂O}\) was diluted by a factor of 2000, using distilled water.
  • This diluted solution was analysed using atomic absorption spectroscopy (AAS).

The calibration curve for the AAS is provided below.
 

The absorbance of the diluted sample was 0.055.

What is the \(K_{s p}\) for silver oxalate?

  1. \(8.8 \times 10^{-14}\)
  2. \(5.3 \times 10^{-12}\)
  3. \(1.1 \times 10^{-11}\)
  4. \(2.1 \times 10^{-11}\)
Show Answers Only

\(B\)

Show Worked Solution
  • By interpolation, the observed concentration of silver ions for an absorbance of \(0.055\)  is  \(0.11 \times 10^{-6}\ \text{mol L}^{-1}\).
  • Since this sample was diluted by a factor of 2000:
  •    \(\ce{[Ag+]_{\text{saturated}} = 2000 \times [Ag+]_{\text{diluted}}} = 2000 \times 0.11 \times 10^{-6} = 2.2 \times 10^{-4}\ \text{mol L}^{-1}\)
  • The equation for the dissolution of \(\ce{Ag2C₂O4}\) is
  •    \(\ce{Ag2C₂O4 \leftrightharpoons 2Ag+(aq) + C2O4^{2-}(aq)}\)
  • Hence \(\ce{[C2O4^{2-}]_{\text{saturated}} = 0.5 \times [Ag+]_{\text{saturated}}} = 0.5 \times 2.2 \times 10^{-4} = 1.1 \times 10^{-4}\)
  •    \(K_{sp} = \ce{[Ag+]^2[C2O4^{2-}]} = (2.2 \times 10^{-4})^2(1.1 \times 10^{-4}) = 5.3 \times 10^{-12}\)

\(\Rightarrow B\)

Filed Under: Inorganic Substances, Solution Equilibria Tagged With: Band 5, smc-3672-15-Find K(sp), smc-3682-60-Colourimetry UV-Vis and AAS

CHEMISTRY, M8 2025 HSC 18 MC

The concentration of silver ions in a solution is determined by titrating it with aqueous sodium chloride, using yellow potassium chromate as the indicator.

Which row of the table correctly identifies the colour change at the endpoint and the more soluble salt?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \text{Colour change} \ \ & \text{More soluble salt} \\
\text{at endpoint}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • The two key reactions that take place in this titration are
  •    \(\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}\)
  •    \(\ce{2Ag+(aq) + CrO4^{2-}(aq) -> Ag2CrO4(s)}\)
  • For the titration to occur effectively all of the \(\ce{Cl-}\) ions must react with \(\ce{Ag+}\) ion and be precipitated out of the solution as the sodium chloride is Titrant (known volume and known concentration).
  • Only once all of the \(\ce{Cl-}\) ions are used up will silver ions begin to react the chromate ions signally the endpoint of the titration.
  • Hence, \(\ce{Ag2CrO4}\) is the more soluble salt and as the potassium chromate is initially yellow, the colour change must be from yellow to red.

\(\Rightarrow B\)

Filed Under: Inorganic Substances Tagged With: Band 5, smc-3682-70-Precipitation Titrations

CHEMISTRY, M8 2025 HSC 17 MC

The chemical environment of an atom depends on the species surrounding that atom within a molecule.

In which of the following compounds does the number of carbon chemical environments equal the number of proton chemical environments?
 

 

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\(A\)

Show Worked Solution
  • In the diagrams below, each colour represents a different carbon or proton environment (ignoring the oxygen atoms).

  • Only \(A\) has equal number of carbon and proton environments with two of each.

\(\Rightarrow A\)

Filed Under: Organic Substances Tagged With: Band 5, smc-3683-10-C NMR, smc-3683-20-H NMR

CHEMISTRY, M7 2025 HSC 16 MC

A single straight strand of polyester was produced through a condensation reaction of 1000 molecules of 3-hydroxypropanoic acid, \(\ce{HOCH2CH2COOH}\).

What is the approximate molar mass of the strand (in g mol\(^{-1}\) )?

  1. 72 062
  2. 72 080
  3. 90 060
  4. 90 078
Show Answers Only

\(B\)

Show Worked Solution
  • The molar mass of 3-hydroxypropanoic acid \(= 6(1.008) +3(16.00) + 3(12.01) = 90.078\ \text{g mol}^{-1}\).
  • When 1000 molecules of 3-hydroxypropanoic acid undergoes condensation reactions, 999 molecules of water are formed as by-products.
  • Hence the molar mass of the polymer strand can be calculated by:
\(MM_{\text{strand}}\) \(=MM_{\text{monomers}}-MM_{\text{water}}\)  
  \(=90.078 \times 1000 -18.016 \times 999\)  
  \( = 72\ 080\ \text{g mol}^{-1}\)  

\(\Rightarrow B\)

Filed Under: Polymers Tagged With: Band 5, smc-3681-20-Condensation Polymers

CHEMISTRY, M7 2025 HSC 15 MC

Consider the following sequence of reactions.

  • Prop-2-en-1-ol was reacted with hydrogen gas to form liquid \(X\).
  • \(X\) was oxidised, producing liquid \(Y\) that formed bubbles of a gas when reacted with aqueous sodium carbonate.
  • \(Y\) was heated under reflux with methanol and a drop of concentrated sulfuric acid, producing an organic liquid, \(Z\).

This process has been presented in the flow chart below.
 

  

Which option correctly identifies the structures for \(X\), \(Y\) and \(Z\)?
 

Show Answers Only

\(D\)

Show Worked Solution
  • The first reaction that occurs is a hydrogenation addition reaction in which prop-2-en-1-ol reacts under a Pd catalyst to produce propan-1-ol.
  • The primary alcohol propan-1-ol then undergoes oxidation to produce the carboxylic acid propanoic acid. This is confirmed as when it is reacted with sodium carbonate, it undergoes an acid-carbonate reaction to produce carbon dixoide which is observed through the bubbles.
  • The third reaction is an estification reaction in which propanic acid reacts with methanol under relfex to produce methyl-propanoate.
  • All three of the correctly drawn compounds can be observed in \(D\)

\(\Rightarrow D\)

Filed Under: Alcohols, Hydrocarbons and Associated Reactions, Reactions of Organic Acids and Bases Tagged With: Band 5, smc-3677-70-Reactions of Unsaturated Hydrocarbons, smc-3679-30-Oxidation, smc-3680-10-Esters

CHEMISTRY, M5 2025 HSC 14 MC

The equation for the decomposition of hydrogen iodide is shown.

\(\ce{2HI(g)\rightleftharpoons I2(g) + H2(g)} \quad \quad \Delta H=+52 \ \text{kJ mol}^{-1}\)

The equilibrium formed during this reaction was investigated in two experiments carried out at different temperatures. The initial and equilibrium concentrations for both experiments are shown in the table, with only the \(K_{e q}\) for Experiment 1 shown.
 

Which row in the table correctly compares features of the two experiments?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
 \ \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}\quad \quad \quad K_{e q}\quad \quad\quad& \ \ \text{Temperature of}\ \  \\
\ \rule[-1ex]{0pt}{0pt}& \textit{experiment} \\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}&\text{Lower in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Lower in 2} \\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • The equilibrium constant for experiment two:
  •    \(K_{eq} = \dfrac{\ce{[H2][I2]}}{\ce{[HI]^2}} = \dfrac{0.02 \times 0.02}{0.04^2} = 0.25\)
  • Therefore the \(K_{eq}\) for experiment 1 is lower than the \(K_{eq}\) for experiment 2.
  • For an endothermic reaction (thinking of heat as a reactant), increasing the temperature of the reaction will shift the equilibrium to the products and increase \(K_{eq}\) which is what occured in experiment 2. 
  • Hence, the temperature of the experiment is higher in 2.

\(\Rightarrow B\)

Filed Under: Equilibrium Constant, Factors that Affect Equilibrium Tagged With: Band 5, smc-3670-10-Temperature, smc-3671-10-Find K(eq)

Vectors, EXT2 V1 2025 HSC 16c

Consider the point \(B\) with three-dimensional position vector \(\underset{\sim}{b}\) and the line  \(\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}\), where \(\underset{\sim}{a}\) and \(\underset{\sim}{d}\) are three-dimensional vectors, \(\abs{\underset{\sim}{d}}=1\) and \(\lambda\) is a parameter.

Let \(f(\lambda)\) be the distance between a point on the line \(\ell\) and the point \(B\).

  1. Find \(\lambda_0\), the value of \(\lambda\) that minimises \(f\), in terms of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{d}\).   (2 marks)

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  2. Let \(P\) be the point with position vector  \(\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\).
  3. Show that \(PB\) is perpendicular to the direction of the line \(\ell\).   (1 mark)

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

  4. Hence, or otherwise, find the shortest distance between the line \(\ell\) and the sphere of radius 1 unit, centred at the origin \(O\), in terms of \(\underset{\sim}{d}\) and \(\underset{\sim}{a}\).
  5. You may assume that if \(B\) is the point on the sphere closest to \(\ell\), then \(O B P\) is a straight line.   (3 marks)

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

Show Answers Only

i.    \(\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})\)

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

iii.  \(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Show Worked Solution

i.    \(\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1\)

\(\text{Vector from point \(B\) to a point on \(\ell\)}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

\(f(\lambda)=\text{distance between \(\ell\) and \(B\)}\)

\(f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}\)

\(\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}\)

\(f(\lambda)^2\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot  \underset{\sim}{d}\)
  \(=\lambda^2|\underset{\sim}{d}|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
  \(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

 

\(f(\lambda)^2 \ \ \text{is a concave up quadratic.}\)

\(f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}\)

\(\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})\)
 

ii.    \(P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\)

\(\text{Show} \ \ \overrightarrow{PB} \perp \ell:\)

\(\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\) \(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
  \(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
  \(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
  \(=0\)

  

\(\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. \)
 

iii.   \(\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1\))}\)

\(=\ \text{(shortest distance \(\ell\) to \(O\))}-1\)
 

\(f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to point \(B\)}\)

\(\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to \(0\)}\)

\(\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}\)

\(f\left(\lambda_0\right)\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\) \(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
  \(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\) \(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

 

\(\text {Shortest distance of \(\ell\) to sphere \(\left(d_{\min }\right)\):}\)

\(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Filed Under: Vectors and Vector Equations of Lines Tagged With: Band 5, Band 6, smc-1196-48-Spheres, smc-1196-80-3D vectors

CHEMISTRY, M6 2025 HSC 11 MC

The structures of two substances, \(\text{X}\) and \(\text{Y}\), are shown.
 

Which row of the table correctly classifies these substances as a Brønsted-Lowry acid or a Brønsted-Lowry base?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Brønsted-Lowry} & \textit{Brønsted-Lowry} \\
\textit{acid}\rule[-1ex]{0pt}{0pt}& \textit{base} \\
\hline
\rule{0pt}{2.5ex}\text{-}\rule[-1ex]{0pt}{0pt}&\text{X and Y}\\
\hline
\rule{0pt}{2.5ex}\text{X and Y}\rule[-1ex]{0pt}{0pt}& \text{-}\\
\hline
\rule{0pt}{2.5ex}\text{Y}\rule[-1ex]{0pt}{0pt}& \text{X} \\
\hline
\rule{0pt}{2.5ex}\text{X}\rule[-1ex]{0pt}{0pt}& \text{Y} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.
  • \(X\) is propanoate (the conjugate base of propanoic acid) and is therefore a proton accepter making it a Brønsted-Lowry base.
  • \(Y\) is ethanamine and is considered a weak base where the \(\ce{NH2}\) group can accept a proton to become \(\ce{NH3+}\).

\(\Rightarrow A\)

Filed Under: Properties of Acids and Bases, Reactions of Organic Acids and Bases Tagged With: Band 5, smc-3673-10-Arrhenius and Bronsted-Lowry, smc-3680-60-Reactions of Organic Acids and Bases

Mechanics, EXT2 M1 2025 HSC 16b

A particle of mass 1 kg is projected from the origin with a speed of 50 ms\(^{-1}\), at an angle of \(\theta\) below the horizontal into a resistive medium.
 

The position of the particle \(t\) seconds after projection is \((x, y)\), and the velocity of the particle at that time is  \(\underset{\sim}{v}=\displaystyle \binom{\dot{x}}{\dot{y}}\).

The resistive force, \(\underset{\sim}{R}\), is proportional to the velocity of the particle, so that  \(\underset{\sim}{R}=-k \underset{\sim}{v}\), where \(k\) is a positive constant.

Taking the acceleration due to gravity to be 10 ms\(^{-2}\), and the upwards vertical direction to be positive, the acceleration of the particle at time \(t\) is given by:

\(\underset{\sim}{a}=\displaystyle \binom{-k \dot{x}}{-k \dot{y}-10}\).    (Do NOT prove this.) 

Derive the Cartesian equation of the motion of the particle, given  \(\sin \theta=\dfrac{3}{5}\).   (5 marks)

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

Show Answers Only

 

Show Worked Solution

\(\sin \theta=\dfrac{3}{5} \ \Rightarrow \ \cos \theta=\dfrac{4}{5}\)

\(\text{Components of initial velocity:}\)

\(\dot{x}(0)=50\, \cos \theta=50 \times \dfrac{4}{5}=40 \ \text{ms}^{-1}\)

\(\dot{y}(0)=50\, \sin \theta=50 \times \dfrac{3}{5}=-30\ \text{ms}^{-1}\)

\(\text{Horizontal motion:}\)

  \(\dfrac{d \dot{x}}{dt}\) \(=-k \dot{x} \ \ \text{(given)}\)  
\(\dfrac{dt}{d \dot{x}}\) \(=-\dfrac{1}{k \dot{x}}\)  
\(\displaystyle \int dt\) \(=-\dfrac{1}{k} \int \dfrac{1}{\dot{x}}\, d x\)  
\(t\) \(=-\dfrac{1}{k} \ln \dot{x}+c\)  

 
\(\text{When} \ \ t=0, \ \dot{x}=40 \ \Rightarrow \ c=\dfrac{1}{k} \ln 40\)

\(t\) \(=\dfrac{1}{k} \ln 40-\dfrac{1}{k} \ln \abs{\dot{x}}=\dfrac{1}{k} \ln \abs{\dfrac{40}{\dot{x}}}\)
    \(k t\) \(=\ln \abs{\dfrac{40}{\dot{x}}}\)
  \(e^{k t}\) \(=\dfrac{40}{\dot{x}}\)
\(\dot{x}\) \(=40 e^{-k t}\)
\(x\) \(\displaystyle=\int 40 e^{-k t}\, d t\)
  \(=-\dfrac{40}{k} \times e^{-k t}+c\)

 

\(\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{40}{k}\)

   \(x=\dfrac{40}{k}-\dfrac{40}{k} e^{-k t}=\dfrac{40}{k}\left(1-e^{-k t}\right)\ \ldots\ (1)\)
 

\(\text{Vertical Motion }\)

\(\dfrac{d \dot{y}}{dt}\) \(=-k \dot{y}-10 \quad \text{(given)}\)
\(\dfrac{d t}{d \dot{y}}\) \(=-\dfrac{1}{k} \times \dfrac{1}{\dot{y}+\frac{10}{k}}\)
\(t\) \(=-\dfrac{1}{k} \displaystyle \int \dfrac{1}{\dot{y}+\frac{10}{k}} \, d \dot{y}\)
  \(=-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}+c\)

 

\(\text{When} \ \ t=0, \, \dot{y}=-30 \ \ \Rightarrow\ \ c=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}\)

\(t\) \(=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}\)
  \(=\dfrac{1}{k} \ln \abs{\frac{-30+\frac{10}{k}}{\dot{y}+\frac{10}{k}}}\)
  \(e^{k t}\) \(=\abs{\dfrac{-30+\frac{10}{k}}{y+\frac{10}{k}}}\)
\(\dot{y}\) \(=\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10}{k}\)
\(y\) \(=\displaystyle \left(-30+\dfrac{10}{k}\right) \int e^{-kt}\, d t-\int \dfrac{10}{k}\, dt\)
  \(=-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10 t}{k}+c\)

 \(\text{When} \ \ t=0, y=0 \ \Rightarrow \  c=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)\)

  \(y\) \(=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10t}{k}\)
  \(=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right)\left(1-e^{-k t}\right)-\dfrac{10 t}{k}\ \ldots\ (2)\)

 

\(\text {Cartesian equation (using (1) above):}\)

\(x\) \(=\dfrac{40}{k}\left(1-e^{-k t}\right)\)
\(\dfrac{k x}{40}\) \(=1-e^{-k t}\)
\(e^{-k t}\) \(=1-\dfrac{k x}{40}\)
\(-k t\) \(=\ln \abs{1-\dfrac{k x}{40}}\)
\(t\) \(=-\dfrac{1}{k} \ln \abs{1-\dfrac{kx}{40}}\)

 

\(y\) \(=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right) \times \dfrac{k x}{40}+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}\)
  \(=\left(\dfrac{1-3 k}{4 k}\right) x+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}\)

Filed Under: Resisted Motion Tagged With: Band 5, Band 6, smc-1061-07-Resistive medium, smc-1061-10-R ~ v

Complex Numbers, EXT2 N1 2025 HSC 7 MC

The complex number \(z\) lies on the unit circle.
 

What is the range of \(\operatorname{Arg}(z-2 i)\) ?

  1. \(\dfrac{\pi}{6} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{5 \pi}{6}\)
  2. \(\dfrac{\pi}{3} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{2 \pi}{3}\)
  3. \(-\dfrac{5 \pi}{6} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{6}\)
  4. \(-\dfrac{2 \pi}{3} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{3}\)
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\(D\)

Show Worked Solution

\(\text{The limits of Arg}(z-2i)\ \where it intersects the unit circle are:}\)
 

\(\sin \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text {Range Arg}(z-2 i):\)

\(-\dfrac{\pi}{2}-\dfrac{\pi}{6} \leqslant \operatorname{Arg}(z-2 i) \leqslant-\dfrac{\pi}{2}+\dfrac{\pi}{6}\)

\(-\dfrac{2 \pi}{3} \leqslant \operatorname{Arg}(2-2 i) \leqslant-\dfrac{\pi}{3}\)

\(\Rightarrow D\)

Filed Under: Argand Diagrams and Mod/Arg form Tagged With: Band 5, smc-1049-10-Cartesian and Argand diagrams

Complex Numbers, EXT2 N2 2025 HSC 6 MC

The complex numbers \(z\) and \(w\) lie on the unit circle. The modulus of  \(z+w\)  is \(\dfrac{3}{2}\).

What is the modulus of  \(z-w\) ?

  1. \(\dfrac{1}{8}\)
  2. \(\dfrac{\sqrt{7}}{2}\)
  3. \(\dfrac{3}{2}\)
  4. \(\dfrac{7}{4}\)
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\(B\)

Show Worked Solution

\(|z|=|w|=1, \quad \abs{z+w}=\dfrac{3}{2} \ \text{(given)}\)

\(\text{Find}\ \ \abs{z-w}:\)

\(\abs{z+\omega}^2=(z+\omega)(\bar{z}+\bar{\omega})=z \bar{z}+z \bar{\omega}+\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z-\omega}^2=(z-\omega)(\bar{z}-\bar{\omega})=\bar{z} z-z \bar{\omega}-\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z+\omega}^2+|z-\omega|^2\) \(=2 z \bar{z}+2 \omega \bar{\omega}=2\abs{z}^2+2\abs{\omega}^2=4\)
\(\dfrac{9}{4}+\abs{z-\omega}^2\) \(=4\)
\(\abs{z-w}^2\) \(=\dfrac{7}{4}\)
\(\abs{z-\omega}\) \(=\dfrac{\sqrt{7}}{2}\)

 
\(\Rightarrow B\)

Filed Under: Geometrical Implications of Complex Numbers Tagged With: Band 5, smc-1052-60-Other problems

Vectors, EXT2 V1 2025 HSC 10 MC

Which of the following gives the same curve as  \(\left(\begin{array}{c}\cos (t) \\ -t \\ \sin (t)\end{array}\right)\) for  \(t \in \mathbb{R}\) ?

  1. \(\left(\begin{array}{c}\cos (2 t) \\ 2 t \\ \sin (2 t)\end{array}\right)\)
  2. \(\left(\begin{array}{c}\cos \left(t^2+\dfrac{\pi}{2}\right) \\ t^2+\dfrac{\pi}{2} \\ \sin \left(t^2+\dfrac{\pi}{2}\right)\end{array}\right)\)
  3. \(\left(\begin{array}{c}\cos \left(t^2\right) \\ -t^2 \\ \sin \left(t^2\right)\end{array}\right)\)
  4. \(\left(\begin{array}{c}\cos \left(2 t+\dfrac{\pi}{2}\right) \\ 2 t+\dfrac{\pi}{2} \\ -\sin \left(2 t+\dfrac{\pi}{2}\right)\end{array}\right)\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Find which option is a re-parametrisation of the given curve.}\)

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

\(\text{Let}\ \ u=- \left(2t + \dfrac{\pi}{2}\right)\)

\(\text{Since}\ t \in \mathbb{R}\ \ \Rightarrow \ \ u \in \mathbb{R}\)

\(\cos \left(2 t+\dfrac{\pi}{2}\right) = \cos\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \cos\,u\)

\(2 t+\dfrac{\pi}{2} = -\left( -\left(2 t+\dfrac{\pi}{2} \right) \right) = -u\)

\(-\sin \left(2 t+\dfrac{\pi}{2}\right) = \sin\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \sin\,u\)

\(\Rightarrow D\)

Filed Under: Vectors and Vector Equations of Lines Tagged With: Band 5, smc-1196-45-Curves

Complex Numbers, EXT2 N2 2025 HSC 9 MC

The points \(U, V, W\) and \(Z\) represent the complex numbers \(u, v, w\) and \(z\) respectively. It is given that  \(v+z=u+w\)  and  \(u+k i z=w+k i v\)  where  \(k \in \mathbb{R} , k>1\).

Which quadrilateral best describes \(UVWZ\) ?

  1. Parallelogram
  2. Rectangle
  3. Rhombus
  4. Square
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\(C\)

Show Worked Solution

\(\text{Quadrilateral}\ UVWZ\ \ \Rightarrow\ \ \text{Diagonals are \(UW\) and \(VZ\)} \).

\(\text{Given}\ \ v+z=u+w\ \ \Rightarrow\ \ \dfrac{v+z}{2}=\dfrac{u+w}{2}\)

\(\text{Mid-points of diagonals are equal (diagonals bisect).}\)

\(u+kiz\) \(=w+kiv\)  
\(u-w\) \(=ki(v-z)\)  

\(\therefore UW\ \text{and}\ VZ\ \text{are perpendicular.}\)

\(\Rightarrow C\)

Filed Under: Geometrical Implications of Complex Numbers Tagged With: Band 5, smc-1052-30-Quadrilaterals, smc-1052-55-Rotations

Proof, EXT2 P1 2025 HSC 16a

Consider the equation

\(z^n \cos\left[n \theta\right]+z^{n-1} \cos \left[(n-1) \theta\right]+z^{n-2} \cos \left[(n-2) \theta\right]+\cdots+z\, \cos\left[\theta\right]=1\)

where  \(z \in \mathbb{C} , \theta \in \mathbb{R} \), and \(n\) is a positive integer.

Using a proof by contradiction and the triangle inequality, or otherwise, prove that all the solutions to the equation lie outside the circle  \(\abs{z}=\dfrac{1}{2}\)  on the complex plane.   (4 marks)

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\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)
 

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

   \(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)
 

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\  a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Show Worked Solution

\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)
 

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

   \(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)
 

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\  a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Filed Under: Converse, Contradiction and Contrapositive Proof, Proof and Inequalities Tagged With: Band 5, smc-1207-10-Contradiction, smc-1208-55-Triangle inequality

Complex Numbers, EXT2 N2 2025 HSC 15c

  1. Show that
  2.     \(\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.\)   (3 marks)

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  3. Use De Moivre's theorem to show that the sixth roots of \(-1\) are given by 
  4.    \(\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)\)  for  \(k=0,1,2,3,4,5\).   (2 marks)

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  5. Hence, or otherwise, show the solutions to  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)  are 
  6. \(z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)\), and \(i \cot \left(\dfrac{11 \pi}{12}\right)\).   (2 marks)

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i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

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

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

 

Show Worked Solution

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

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

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

Filed Under: Solving Equations with Complex Numbers Tagged With: Band 4, Band 5, Band 6, smc-1050-30-Roots > 3, smc-1050-40-De Moivre and trig identities

Mechanics, EXT2 M1 2025 HSC 15b

A particle moves in simple harmonic motion about the origin with amplitude \(A\), and it completes two cycles per second. When it is \(\dfrac{1}{4}\) metres from the origin, its speed is half its maximum speed.

Find the maximum positive acceleration of the particle during its motion.   (4 marks)

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\(a_{\text{max}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Show Worked Solution

\(v^2=-n^2\left(x^2-A^2\right)\)

\(\operatorname{Period}\ (T)=\dfrac{1}{2} \ \Rightarrow \ \dfrac{2 \pi}{n}=\dfrac{1}{2} \ \Rightarrow \ n=4 \pi\)

\(\text{Max velocity occurs at}\ \  x=0:\)

   \(v_{\text{max}}^2=-(4 \pi)^2\left(0-A^2\right)=16 \pi^2 A^2\)

   \(v_{\text{max}}=\sqrt{16 \pi^2 A^2}=4 \pi A\)
 

\(\text{At} \ \ x=\dfrac{1}{4}, \ v=\dfrac{1}{2} \times 4 \pi A=2 \pi A\)

\((2 \pi A)^2\) \(=-(4 \pi)^2\left(\dfrac{1}{16}-A^2\right)\)
\(\dfrac{A^2}{4}\) \(=A^2-\dfrac{1}{16}\)
\(\dfrac{3 A^2}{4}\) \(=\dfrac{1}{16}\)
\(A^2\) \(=\dfrac{1}{12}\)
\(A\) \(=\dfrac{1}{\sqrt{12}}\)

 

\(\text{Max positive acceleration occurs at} \ \ x=-\dfrac{1}{\sqrt{12}}:\)

\(a_{\text{max}}=-n^2 x=-(4 \pi)^2 \times-\dfrac{1}{\sqrt{12}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Filed Under: Simple Harmonic Motion Tagged With: Band 4, Band 5, smc-1059-10-Amplitude / Period, smc-1059-31-Max Speed

Vectors, EXT2 V1 2025 HSC 15a

The adjacent sides of a parallelogram are represented by the vectors  \(\underset{\sim}{a}=4 \underset{\sim}{i}+3 \underset{\sim}{j}-\underset{\sim}{k}\)  and  \(\underset{\sim}{b}=2 \underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k}\).

Show that the area of the parallelogram is \(6 \sqrt{10}\) square units.   (4 marks)

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\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Show Worked Solution

\(A=\dfrac{1}{2}\abs{a}\abs{b} \sin \theta\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \  \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)
 

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=\dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Filed Under: Vectors and Geometry Tagged With: Band 5, smc-1210-30-Quadrilateral, smc-1210-60-2D problems

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that  \(1+w+w^2+\cdots+w^6=0\).

  1. Show that \(w\) is a 7th root of unity.   (1 mark)

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The complex number  \(\alpha=w+w^2+w^4\)  is a root of the equation  \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

  1. Find the other root of  \(x^2+b x+c=0\)  in terms of positive powers of \(w\).  (2 marks)

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  2. Find the numerical value of \(c\).  (1 mark)

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i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7=1\) \(=1\)

ii.   \(w^6+w^5+w^3\)

iii.  \(2\)

Show Worked Solution

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7\) \(=1\)

 
ii.
    \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\) \(=\overline{w+w^2+w^4}\)
  \(=\overline{w}+\overline{w^2}+\overline{w^4}\)
  \(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
  \(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
  \(=w^6+w^5+w^3\)

 

iii.    \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\) \(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
  \(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
  \(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
  \(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
  \(=2\)

Filed Under: Solving Equations with Complex Numbers Tagged With: Band 4, Band 5, smc-1050-10-Quadratic roots, smc-1050-30-Roots > 3, smc-1050-35-Conjugate roots

Calculus, EXT2 C1 2025 HSC 14a

Let  \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta\)  for integers  \(n \geq 0\).

  1. Show that  \(I_n=\dfrac{1}{2 n-1}-I_{n-1}\) for \(n>0\),  given that  \(\dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\).   (3 marks)

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  2. Hence, or otherwise, calculate \(I_2\).   (1 mark)

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i.    \(\text{See Worked Solutions}\)

ii.   \(\dfrac{\pi}{4}-\dfrac{2}{3}\)

Show Worked Solution

i.    \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta \ \ \Rightarrow \ \ \)\(I_{n-1}=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)

\(\text{Show} \ \ I_n=\dfrac{1}{2n-1}-I_{n-1} \ \ \text{for} \ \ n>0\)

\(\text{Given} \ \ \dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\ \ldots\ (1)\)

\(I_n\) \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \cot ^2 \theta \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta\left(\operatorname{cosec}^2 \theta-1\right) \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \operatorname{cosec}^2 \theta \, d \theta-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)
  \(=-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \dfrac{d}{d \theta}(\cot \theta) d \theta-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} \theta}{2 n-1}\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} (\frac{\pi}{2})}{2 n-1}-\dfrac{\cot ^{2 n-1} (\frac{\pi}{4})}{2 n-1}\right]-I_{n-1}\)
  \(=-\left[0-\dfrac{1}{2 n-1}\right]-I_{n-1}\)
  \(=\dfrac{1}{2n-1}-I_{n-1}\)

 

ii.     \(I_2\) \(=\dfrac{1}{3}-I_1\)
    \(=\dfrac{1}{3}-\left[ \dfrac{1}{2-1}-\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} 1 \, d\theta \right]\)
    \(=\dfrac{1}{3}-\left[1-\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)\right]\)
    \(=\dfrac{\pi}{4}-\dfrac{2}{3}\)

Filed Under: Recurrence Relations Tagged With: Band 4, Band 5, smc-1054-10-Trig

Calculus, EXT2 C1 2025 HSC 13d

Evaluate  \(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\), by first using the substitution  \(u=\dfrac{\pi}{2}-x\).   (4 marks)

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Show Worked Solution

\(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\)

\(\text{Let} \ \ u=\dfrac{\pi}{2}-x \ \ \Rightarrow\ \ \dfrac{du}{dx}=-1 \ \ \Rightarrow\ \ du=-dx\)

\(\text{Limits:} \ \ u=\dfrac{\pi}{2}\ \ \Rightarrow\ \ x=0, \ \ u=0\ \ \Rightarrow\ \ x=\dfrac{\pi}{2}\)

\(I\) \(=-\displaystyle \large{\int_{\small{\dfrac{\pi}{2}}}^0}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)}\,dx\)
  \(=\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\cos x+\sin x}\, d x\)

 

\(\text{Add \(I\) (swap variable from \(x\) to \(u\)) to original integral:}\)

\(2I=\dfrac{\pi}{2} \displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{1}{1+\sin u+\cos u}\, d u\)

\(\text{Substitute} \ \ t=\tan \left(\frac{u}{2}\right), \ \sin u=\dfrac{2t}{1+t^2}, \ \cos u=\dfrac{1-t^2}{1+t^2}\)

\(d t=\dfrac{1}{2} \sec ^2\left(\frac{u}{2}\right)\, du \ \ \Rightarrow\ \ du=\dfrac{2}{1+\tan ^2\left(\frac{u}{2}\right)}\, dt=\dfrac{2}{1+t^2}\, d t\)

\(\text{Limits:} \ \ n=\dfrac{\pi}{2}\  \Rightarrow \ t=1, \ n=0 \ \Rightarrow \ t=0\)

\(2I\) \(=\dfrac{\pi}{2} \displaystyle \int_0^1 \dfrac{1}{1+\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} \times \frac{2}{1+t^2}\,d t\)
\(I\) \(=\displaystyle\frac{\pi}{4} \int_0^1 \frac{2}{1+t^2+2 t+1-t^2}\, d t\)
  \(=\displaystyle \frac{\pi}{4} \int_0^1 \frac{1}{1+t}\, d t\)
  \(=\dfrac{\pi}{4}\Bigl[\ln (1+t)\Bigr]_0^1\)
  \(=\dfrac{\pi}{4}(\ln 2-\ln 1)\)
  \(=\dfrac{\pi \, \ln 2}{4}\)

Filed Under: Substitution and Harder Integration Tagged With: Band 5, smc-1057-10-Trig, smc-1057-50-Substitution given

BIOLOGY, M8 2025 HSC 32

A population lives across three regions, \(A,\ B\) and \(C\).
  

People in community \(B\) developed an environmental disease. An epidemiological study was carried out to determine the risk of developing the disease due to age at exposure. The results of this study are shown in the graph.
  

Design an epidemiological study that could be used to produce the results shown in the graph. Justify the features of your design.   (7 marks)

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Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

Show Worked Solution

Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

Filed Under: Causes and Effects, Epidemiology Tagged With: Band 5, Band 6, smc-3660-11-Environmental/Nutritional, smc-3661-20-Evaluating the Method

BIOLOGY, M6 2025 HSC 30

PAI-1 protein is encoded by the SERPINE 1 gene in humans. Anopheles mosquitoes have been genetically modified to express PAI-1, which blocks the entry of the malarial Plasmodium into the mosquito gut. This disrupts the Plasmodium life cycle, resulting in reduced transmission of malaria. 

  1. Describe a process that could be used to produce mosquitoes which express PAI-1.   (4 marks)

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  2. 'Genetic technologies are beneficial for society.'
  3. Evaluate this statement.   (7 marks)

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a.    Mosquito Production

  • The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
  • The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
  • The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
  • The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
  • Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
  • Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b.    Evaluation Statement

Genetic technologies are highly beneficial for society when evaluated against health improvements and food security criteria. Despite some ethical and environmental concerns requiring careful management, the overall benefits are substantial.

Health Benefits
Evidence supporting includes:

  • Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
  • Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
  • Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

  • Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
  • Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

Weighing these factors shows genetic technologies are substantially beneficial for society. The health improvements and food security gains outweigh the manageable ethical concerns. While challenges like biodiversity impacts and equitable access require ongoing attention, the overall societal benefit remains considerable through life-saving medical applications and enhanced food production.

Show Worked Solution

a.    Mosquito Production

  • The SERPINE 1 gene is isolated from human DNA using restriction enzymes that cut at specific recognition sites.
  • The same restriction enzymes are used to cut mosquito DNA, creating complementary sticky ends.
  • The human PAI-1 gene is inserted into the mosquito DNA using DNA ligase to form recombinant DNA.
  • The recombinant DNA is introduced into mosquito eggs or embryos using a vector or microinjection technique.
  • Modified mosquitoes are screened to confirm PAI-1 gene expression and successful integration into the genome.
  • Transgenic mosquitoes produce PAI-1 protein that blocks Plasmodium entry into the gut, disrupting malaria transmission.

b.    Evaluation Statement

Genetic technologies are highly beneficial for society when evaluated against health improvements and food security criteria. Despite some ethical and environmental concerns requiring careful management, the overall benefits are substantial.

Health Benefits
Evidence supporting includes:

  • Genetically modified mosquitoes expressing PAI-1 significantly reduce malaria transmission, potentially saving millions of lives annually.
  • Recombinant DNA technology produces insulin and vaccines, improving accessibility to life-saving treatments for diabetes and infectious diseases.
  • Gene therapy offers potential cures for inherited genetic disorders, dramatically improving quality of life for affected individuals.

The health criterion strongly meets beneficial status because these technologies address major global health challenges.

Food Security and Agricultural Benefits
Evidence supporting includes:

  • Genetically modified crops like Bt cotton and Golden Rice increase crop yields and nutritional content, addressing food scarcity.
  • Drought-resistant GM crops enable farming in challenging environments, supporting population growth and farmer livelihoods.

However, concerns exist about reduced genetic diversity and corporate control over seeds, creating inequalities in access.

Final Evaluation

Weighing these factors shows genetic technologies are substantially beneficial for society. The health improvements and food security gains outweigh the manageable ethical concerns. While challenges like biodiversity impacts and equitable access require ongoing attention, the overall societal benefit remains considerable through life-saving medical applications and enhanced food production.

Filed Under: Biotechnology, Genetic Technologies Tagged With: Band 5, Band 6, smc-3653-20-Evaluating Genetic Technology, smc-3654-05-Transgenic Organisms

BIOLOGY, M5 2025 HSC 27

The table shows the number of eggs produced by females in TWO different groups of animals.

\(\textit{Group A}\) \(\textit{Group B}\)
\(\quad\textit{Animal}\quad\) \(\quad\textit{Number of Eggs}\quad\) \(\quad\textit{Animal}\quad\) \(\quad\textit{Number of Eggs}\quad\)
  \(\text{Monotremes}\quad\)  \(1-3\)   \(\text{Crabs}\)   \(1000-2000\)
  \(\text{Snakes}\) \(1-100\)   \(\text{Sea Urchins}\quad\)    \(100\, 000\ \text{to 2 million}\) 
  \(\text{Birds}\) \(1-17\)   \(\text{Squid}\)   \(2000-3000\)

  
Compare the types of fertilisation that occur in group A and group B animals with reference to the data provided.   (4 marks) 

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Group A Animals – Internal Fertilisation

  • Group A animals (monotremes, snakes, birds) use internal fertilisation where sperm meets egg inside the female’s body.
  • The data shows Group A produces few eggs (1-100), as the protected internal environment increases fertilisation success and offspring survival.
  • These animals typically provide greater parental care, further improving survival rates.

Group B Animals – External Fertilisation

  • Group B animals (crabs, sea urchins, squid) use external fertilisation where gametes are released into water.
  • The data shows Group B produces vastly more eggs (1,000-2 million) to compensate for low survival rates.
  • External fertilisation exposes gametes to predation, dilution and environmental factors, reducing fertilisation success.
  • These organisms provide minimal parental care, relying on high egg numbers for species continuity.

Comparison

The dramatic difference in egg numbers (1-100 vs 1,000-2,000,000) directly reflects the fertilisation strategy.

Internal fertilisation prioritises quality through protection and parental care, while external fertilisation relies on quantity to overcome environmental challenges.

Show Worked Solution

Group A Animals – Internal Fertilisation

  • Group A animals (monotremes, snakes, birds) use internal fertilisation where sperm meets egg inside the female’s body.
  • The data shows Group A produces few eggs (1-100), as the protected internal environment increases fertilisation success and offspring survival.
  • These animals typically provide greater parental care, further improving survival rates.

Group B Animals – External Fertilisation

  • Group B animals (crabs, sea urchins, squid) use external fertilisation where gametes are released into water.
  • The data shows Group B produces vastly more eggs (1,000-2 million) to compensate for low survival rates.
  • External fertilisation exposes gametes to predation, dilution and environmental factors, reducing fertilisation success.
  • These organisms provide minimal parental care, relying on high egg numbers for species continuity.

Comparison

The dramatic difference in egg numbers (1-100 vs 1,000-2,000,000) directly reflects the fertilisation strategy.

Internal fertilisation prioritises quality through protection and parental care, while external fertilisation relies on quantity to overcome environmental challenges.

Filed Under: Reproduction Tagged With: Band 5, smc-3647-15-Internal vs. External

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