Band 5 – Page 3

Functions, 2ADV EQ-Bank 9 MC

The graphs of `y = mx + c` and `y = ax^2` will have no points of intersection for all values of `m, c` and `a` such that

`a > 0 and c > 0`
`m > 0 and c > 0`
`a > 0 and c > -m^2/(4a)`
`a < 0 and c > -m^2/(4a)`

Show Answers Only

`D`

Show Worked Solution

`text(Intersect when:)`

`mx + c`	`= ax^2`
`ax^2-mx-c`	`= 0`

`text(S)text(ince no points of intersection:)`

`Delta`	`< 0`
`m^2-4a(−c)`	`< 0`
`m^2 + 4ac`	`< 0`

`text(Solve for)\ c:`

`:.\ c > (−m^2)/(4a),quada < 0`

`text(or)`

`c < (−m^2)/(4a),quada > 0`

`=> D`

A train departs from Town X at 1:00 pm to travel to Town Y. Its average speed for the journey is 80 km/h, and it arrives at 4:00 pm. A second train departs from Town X at 1:20 pm and arrives at Town Y at 3:30 pm.

What is the average speed of the second train? Give your answer to the nearest kilometre per hour. (2 marks)

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$\text{111 km/h}$

Show Worked Solution

$\text{Distance between towns X and Y using first train}$

$\text{Time taken by first train = 3 hours}$

$\text{Distance}=\text{speed}\times\text{time}=80\times 2=240\ \text{km}$

$\text{Time taken by second train = 2 hours 10 minutes.}$

$\text{2 hours 10 minutes = }\dfrac{130}{60}=\dfrac{13}{6}\ \text{hours.}$

$\text{Find speed of second train using}\ \ s=\dfrac{d}{t}:$

$s=\dfrac{240}{\frac{13}{6}}=240\times \dfrac{6}{13}=110.769…$

$\therefore\ \text{The average speed of the second train is 111 km/h (nearest km/h).}$

Measurement, STD2 EQ-Bank 8 MC

City A is at latitude $ 27^{\circ}\text{S} $ and longitude $ 153^{\circ}\text{E} $. City B is $ 45^{\circ} $ north of City A and $ 38^{\circ} $ east of City A.

What are the latitude and longitude of City B?

$ 18^{\circ}\text{N} $, $ 191^{\circ}\text{E} $
$ 18^{\circ}\text{N} $, $ 115^{\circ}\text{E} $
$ 72^{\circ}\text{S} $, $ 191^{\circ}\text{E} $
$ 18^{\circ}\text{N} $, $ 169^{\circ}\text{W} $

Show Answers Only

$D$

Show Worked Solution

$\text{Latitude of City B}=45^{\circ}-27^{\circ}=18^{\circ}\text{N}$

$\text{Longitude of City B}=153^{\circ}+38^{\circ}=191^{\circ}\text{E}$

$\text{Since longitude}\ >180^{\circ},\ \text{convert to Western hemisphere}:$

$\text{Western longitude}=360^{\circ}-191^{\circ}=169^{\circ}\text{W}$

$\Rightarrow D$

Measurement, STD2 EQ-Bank 7 MC

Island $A$ is located at longitude $104^{\circ}\text{E}$ and Island $B$ is located at longitude $56^{\circ}\text{W}$. Ignoring timezones, estimate the time on Island $B$ when it is 9:20 am on Island $A$?

10:16 pm
8:40 pm
10:40 pm
8:16 am

Show Answers Only

$C$

Show Worked Solution

$\text{Longitudinal difference} = 104^{\circ}+56^{\circ}=160^{\circ}$

$\text{Calculate the time difference (using 15° = 1 hour time difference):}$

$\text{Time Difference} = \dfrac{160}{15} \text{ hours} = 10.666…\ \text{ hours}=10\text{ hours}\ 40\ \text{minutes} $

$\text{Island A is east of Island B}\ \ \Rightarrow\ \ \text{Island A is ahead}$

$\text{Time on Island A}\ =\ 9:20\ \text{am}$

$\text{Time on Island B}$	$=9:20\ \text{am}-10\ \text{hours}\ 40\text{ minutes}$
	$=11:20\ \text{pm}-40\ \text{minutes}$
	$=10:40\ \text{pm (previous day)}$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 9 MC

City $A$ is located at longitude $0^{\circ}$ (on the Prime Meridian) and while City $B$ is located at longitude $118^{\circ}\text{W}$. Ignoring timezones, what is the estimated time in City $B$ when it is 11:00 pm in city $A$?

3:08 pm
6:52 pm
3:52 am
6:52 am

Show Answers Only

$A$

Show Worked Solution

$\text{Longitudinal difference}= 118^{\circ}-0=118^{\circ}$

$\text{Calculate time difference (using 15° = 1 hour):}$

$\text{Time Difference} = \dfrac{118}{15} \text{ hours} = 7.866\text{ hours}=7\text{ hours}\ 52\ \text{minutes} $

$\text{City B is west of City A}\ \ \Rightarrow\ \ \text{City B is behind}$

$\text{Time in City A}\ =\ 11:00\ \text{pm}$

$\therefore\ \text{Time in City B}$	$ = 11:00\ \text{pm}-7\ \text{hours}\ 52\text{ minutes}$
	$=\ 4:00\ \text{pm}-52\ \text{minutes}$
	$=3:08\ \text{pm}$

$\Rightarrow A$

Measurement, STD2 EQ-Bank 26

Use the train timetable below to answer this question.

Emma lives in Berowra and travels to Wyong for an appointment. After her appointment, she needs to attend a meeting in Gosford before returning home to Berowra.

Emma catches the train that departs Wyong at 09:16. How long does this train take to reach Gosford? (1 mark)
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Emma's meeting in Gosford starts at 10:15 am at a venue that is 7 minutes walk from Gosford station. What is the latest train she can catch from Wyong to arrive at her meeting on time? (2 marks)
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After her meeting finishes at 11:20 am, Emma walks back to Gosford station (taking 7 minutes). What is the earliest train she can catch from Gosford to return home to Berowra, and what time will she arrive home? (2 marks)
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a. $17\ \text{minutes}$

b. $09:37\ \text{train}$

c. $\text{11:34 am Gosford train, arrives at Berowra at 12:37 pm}$

Show Worked Solution

a. $\text{Elapsed time:}\ =09:33-09:16 = 17 \text{ minutes} $

b. $\text{Step 1: Meeting starts at 10:15 am}$

$\rightarrow\ 7\text{ minutes to walk from Gosford station to the venue.}$

$\rightarrow\ \text{At Gosford Station by: }\ 10:15-7 \text{ minutes} = 10:08 \text{ am}$

$\text{Step 2: Find latest train arriving at Gosford by 10:08 am}$

$\rightarrow\ \text{Arrival times: } 08:12, 08:33, 08:58, 09:33, 09:58, 10:33$

$\rightarrow\ \text{She must catch the }09:58.$

$\text{Step 3: When does 09:58 depart Wyong?}$

$\rightarrow\ 09:37\ \text{train}$

c. $\text{Step 1: Determine when Emma arrives back at Gosford station}$

$\rightarrow\ \text{Meeting finishes at }11:20 \text{ am}$

$\text{Walking time back to station: 7 minutes}$

$\text{Emma arrives at Gosford station at:}\ 11:20 + 7 \text{ minutes} = 11:27 \text{ am} $

$\text{Step 1: Find the earliest train departing Gosford after 11:27 am}$

$\rightarrow\ \text{Meeting finishes at }11:20 \text{ am}$

$\rightarrow\ \text{Relevant departure times:} …10:34, 10:59, 11:34, 11:59$

$\rightarrow\ \text{Earliest train after 11:27 am is the }11:34 \text{ am}$

$\text{Step 3: Find when this train arrives in Berowra}$

$\rightarrow\ \text{11:34 am Gosford train, arrives at Berowra at }12:37 \text{ pm}$

Statistics, STD2 EQ-Bank 21 MC

Twenty-five people were surveyed about the amount of money they spent on groceries per week, to the nearest dollar.

The results are shown in the frequency table.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Amount Spent} ($) \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \textit{Frequency}\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} 50-99 \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} 100-149 \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\rule{0pt}{2.5ex} 150-199 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 200-249 \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\end{array}

What is the mean amount spent on groceries by these people per week?

$143.50
$148.50
$153.50
$158.50

Show Answers Only

$B$

Show Worked Solution

$\text{Using the class centres}$

$\text{Class centres: } 74.5, 124.5, 174.5, 224.5$

$\text{Total hours}$	$=(74.5 \times 4) + (124.5 \times 9) + (174.5 \times 8) + (224.5 \times 4)$
	$=298 + 1120.5 + 1396 + 898=3712.5$

$\text{Mean amount}=\dfrac{3712.5}{25} =$148.50$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 19 MC

Thirty students were surveyed about the number of hours they spent on homework per week, to the nearest hour.

The results are shown in the frequency table.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Hours per week} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \textit{Frequency}\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} 0-4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 5-9 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} 10-14 \rule[-1ex]{0pt}{0pt} & 7 \\
\hline
\rule{0pt}{2.5ex} 15-19 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\end{array}

What is the mean number of hours of homework completed by the students per week?

Show Answers Only

$B$

Show Worked Solution

$\text{Using the class centres}$

$\text{Class centres: } 2, 7, 12, 17$

$\text{Total hours}$	$=(2\times 8) + (7\times 10) + (12\times 7) + (17\times 5)$
	$=16 + 70 + 84 + 85=255$

$\text{Mean hours}=\dfrac{255}{30} =8.5$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 9 MC

A dataset has an interquartile range (IQR) of 18.

The upper quartile (Q₃) is 45.

What is the maximum value that would NOT be classified as an outlier?

Show Answers Only

$D$

Show Worked Solution

$1.5 \times \text{IQR} = 1.5 \times 18 = 27$

$\text{Upper boundary} = Q_3 + 27 = 45 + 27 = 72$

$\text{Values up to and including 72 are not outliers}$

$\text{Values above 72 are outliers}$

$\therefore\ \text{Maximum value that is not an outlier} = 72$

$\Rightarrow D$

BIOLOGY, M6 2025 HSC 30b

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.

'Genetic technologies are beneficial for society.'

Evaluate this statement. (7 marks)

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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 net benefits are substantial.

Health Benefits

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

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

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 net benefits are substantial.

Health Benefits

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

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.

♦ Mean mark 64%.

Financial Maths, STD2 EQ-Bank 18

David and Mary are a couple living together. They each receive the maximum Age Pension payment of $888.50 per fortnight (which includes basic rate, Pension Supplement and Energy Supplement).

Their combined income from part-time work is $520 per fortnight. Their Age Pension is reduced by 25 cents each for every dollar of combined income over $380 per fortnight..

Calculate the reduction in each person's Age Pension payment. (2 marks)
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Calculate their total household fortnightly income including their reduced Age Pension payments and earnings. (2 marks)
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What percentage of their total household income comes from their Age Pension payments? Give your answer to 1 decimal place. (1 mark)
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a. $$35$

b. $$2227$

c. $76.6\%\ \text{(to 1 d.p.)}$

Show Worked Solution

a. $\text{Combined income over free area} = 520-380=$140$

$\text{Reduction for each person} = 0.25 \times 140= $35$

$\therefore\ \text{Each person’s pension is reduced by \$35}$

b. $\text{Reduced pension for each person}=888.50-35= $853.50$

$\text{Combined Age Pension} = 2 \times 853.50=$1707$

$\text{Total household income} = 1707+520=$2227$

c. $\text{Percentage from Age Pension} =\dfrac{1707}{2227}\times 100=76.6\%\ \text{(to 1 d.p.)}$

$\therefore\ 76.6\%\text{ of their household income comes from Age Pension}$

Financial Maths, STD2 EQ-Bank 34

Baron is 19 years old, single with no children, and lives away from his parents' home to study. He receives the maximum Youth Allowance payment of $663.30 per fortnight.

Baron supplements his payments by working part-time and earns $15.50 per hour in a retail position. His Youth Allowance is reduced by 50 cents for each dollar earned over $236 per fortnight.

Baron works 22 hours per fortnight. Calculate his total fortnightly income including his reduced Youth Allowance payment and earnings. (2 marks)
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What is the maximum number of hours Baron can work per fortnight before his Youth Allowance payment is reduced to zero? Give your answer to the nearest whole hour. (2 marks)
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a. $\text{\$951.80 per fortnight}$

b. $\text{101 hours per fortnight}$

Show Worked Solution

a. $\text{Baron’s earnings} = 15.50\times 22=$341$

$\text{Income over free area} = 341-236 = $105$

$\text{Reduction in payment} = 0.50\times 105= $52.50$

$\text{Reduced Youth Allowance} = 663.30-52.50=$610.80$

$\text{Total fortnightly income} = 610.80+341=$951.80$

b. $\text{For payment to reduce to zero, reduction needed} = $663.30$

$\text{Income over free area} = \dfrac{663.30}{0.50}=$1326.60$

$\text{Total earnings needed} = 1326.60+236=$1562.60$

$\text{Hours needed}=\dfrac{1562.60}{15.50}=100.8$

$\therefore\ \text{Baron can work a maximum of 101 hours per fortnight}$

Financial Maths, STD2 EQ-Bank 16

Yuki is single with no children and receives the maximum JobSeeker Payment of $793.60 per fortnight.

Her JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight.

Yuki earns $380 per fortnight from casual work. Calculate her reduced JobSeeker Payment. (2 marks)
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How much would Yuki need to earn per fortnight for her JobSeeker Payment to be reduced to $600? (2 marks)
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a. $$678.60\ \text{per fortnight}$

b. $$537.20\ \text{per fortnight}$

Show Worked Solution

a. $\text{Income over free area} = 380-150=$230$

$\text{Reduction in payment} = 0.50\times 230 = $115$

$\text{Reduced JobSeeker Payment} = 793.60-115 = $678.60$

$\therefore\ \text{Yuki receives \$678.60 per fortnight}$

b. $\text{Required reduction} = 793.60-600=$193.60$

$\text{Income over free area} = \dfrac{193.60}{0.50}=$387.20$

$\text{Total income needed} = 387.20+150=$537.20$

$\therefore\ \text{Yuki needs to earn \$537.20 per fortnight}$

Financial Maths, STD1 EQ-Bank 14 MC

Sarah is single with no children and receives the maximum JobSeeker Payment of $793.60 per fortnight.

She earns $350 per fortnight from casual work. The JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight.

What is Sarah's reduced JobSeeker Payment this fortnight?

$593.60
$643.60
$693.60
$743.60

Show Answers Only

$C$

Show Worked Solution

$\text{Maximum JobSeeker Payment (single, no children)} = $793.60$

$\text{Income over free area} = 350-150 = $200$

$\text{Reduction in payment} = 0.50\times 200 = $100$

$\therefore\ \text{Reduced JobSeeker Payment} = 793.60-100 = $693.60$

$\Rightarrow C$

Functions, EXT1′ F1 2007 HSC 3a*

The diagram shows the graph of $y = f(x)$. The line $y = x$ is an asymptote.

Draw separate one-third page sketches of the graphs of the following:

$f(\abs{x})$. (2 marks)

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$f(x)-x$. (2 marks)

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ii.

Show Worked Solution

MARKER’S COMMENT: In part (i), a significant number of students graphed $y=\abs{f(x)}$.

ii.

Functions, EXT1′ F1 2013 HSC 13bii

The diagram shows the graph of a function `f(x).`

Sketch the curve `y = 1/(1-f(x)).` (3 marks)

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Show Worked Solution

`y = 1/(1-f(x))`

MARKER’S COMMENT: Correct working sketches such as `y=-f(x)` and `y=1-f(x)` meant that students could obtain some marks, even if their final sketch was wrong.

`f(x) = 1,\ \ \ y\ text(undefined.)`

`f(x) > 1,\ \ \ y < 0`

`f(x) <= 0, \ \ \ y <= 1`

`\text{Create graph in 3 stages:}`

Financial Maths, STD2 EQ-Bank 32

Ian works in a packaging factory and is paid $0.85 for each box he packs. Last month he worked 160 hours and packed 8960 boxes.

Calculate Ian's total earnings for the month. (1 mark)
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The following month the factory decides to pay Ian a flat hourly rate of $44.50. What percentage increase/decrease is this from his equivalent hourly wage of the previous month. Give your answer correct to 1 decimal place. (2 marks)
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a. $$7616$

b. $\text{6.5% decrease}$

Show Worked Solution

a. $\text{Total Earnings}=0.85\times 8960=$7616$

b. $\text{Hourly rate (last month)}=\dfrac{7616}{160}=$47.60$

$\text{New hourly wage} = $44.50\ \text{(given)}$

$\text{% decrease}=\dfrac{47.60-44.50}{47.60}=0.0651… = 6.5\%\ \text{decrease}$

$\therefore\ \text{Ian’s hourly rate has decreased 6.5%.}$

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$^{-1}$

♦ Mean mark 62%.

$^{13}\text{C NMR}$

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)

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.

Calculate the $K_a$ for the acid used. (3 marks)
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The concentration of the $\ce{NaOH}$ was 0.10 mol L$^{-1}$.
Explain why the pH of the final solution never reached 13. (2 marks)
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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. $\text{Strategy 1:}$

♦♦ Mean mark 40%.

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

$\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.

♦♦♦ Mean mark 19%.

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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Show Answers Only

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

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

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

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

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

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

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

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

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

♦ Mean mark 55%.

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

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

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

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

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

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

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

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

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

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

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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Show Answers Only

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 $\ce{MgCO3}$.

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

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

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

♦♦ Mean mark 40%.

$\ce{Mg(OH)2}$ 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.

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.

♦ Mean mark 39%.

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.

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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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:

Mean mark 51%.

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.

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 $Q$.

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

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Consider a modification to this model in which the Earth's orbit is elliptical.
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} $

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

Mean mark (a) 51%.

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.

♦ Mean mark (b) 46%.

Algebra, STD2 EQ-Bank 28

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.

Show that the value of $k$ is 4. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
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. $W=kt$

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

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

b. $162.5\ \text{minutes}$

Show Worked Solution

a. $W=kt$

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

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

b. $W = 4t$

$\text{When } W = 650:$

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

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

Algebra, STD2 EQ-Bank 30

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.

Show that the value of $k$ is 0.35. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
A builder has a budget of $140 for copper wire. Calculate the maximum length of wire that can be purchased. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

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}$

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

Algebra, STD2 EQ-Bank 22 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?

4.2 L
4.75 L
5.25 L
5.75 L

Show Answers Only

$C$

Show Worked Solution

$P \propto A$

$P=kA$

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

$3.5$	$=k \times 28$
$k$	$=\dfrac{3.5}{28}$
$k$	$=0.125$

$\text{When } A = 42:$

$P=0.125\times 42=5.25$

$\Rightarrow C$

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.

$\Rightarrow C$

♦♦ Mean mark 37%.

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 $X$ ?

Boron-10
Boron-11
Lithium-6
Lithium-10

Show Answers Only

$A$

Show Worked Solution

The alpha decay described can be expressed as:
$\ce{^1_0n + ^{10}_5Bu \rightarrow ^7_3 Li + ^4_2 He}$

$\Rightarrow A$

♦ Mean mark 44%.

PHYSICS, M6 2025 HSC 14 MC

A proton having a velocity of $1 \times 10^6 \ \text{m s}^{-1}$ enters a uniform field with a trajectory that is initially perpendicular to the field.

Which row in the table correctly identifies the field, and its effect on the kinetic energy of the proton?

\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{Type of field}\ \ & \textit{Effect on} \\
\textit{}\rule[-1ex]{0pt}{0pt}& \textit{kinetic energy } \\
\hline
\rule{0pt}{2.5ex}\text{Electric}\rule[-1ex]{0pt}{0pt}&\text{Decreases}\\
\hline
\rule{0pt}{2.5ex}\text{Electric}\rule[-1ex]{0pt}{0pt}& \text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Magnetic}\rule[-1ex]{0pt}{0pt}& \text{Decreases} \\
\hline
\rule{0pt}{2.5ex}\text{Magnetic}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

In a magnetic field, the magnetic force is always perpendicular to the velocity, so it changes the direction of motion but does no work (i.e. there is no change in kinetic energy – eliminate $C$ and $D$).
Since the proton enters perpendicular to the electric field, the electric force acts either upward or downward (depending on field direction), giving the proton an acceleration parallel to the field.
This acceleration adds a new velocity component, increasing the proton’s overall speed and kinetic energy.

$\Rightarrow B$

♦ Mean mark 54%.

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).

$\Rightarrow A$

♦♦ Mean mark 34%.

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?

0
Between 0 and 0.5 A
0.5 A
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.

$\Rightarrow D$

♦ Mean mark 45%.

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 $(\lambda)$ ?

$W$
$X$
$Y$
$Z$

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.

$\Rightarrow C$

♦ Mean mark 40%.

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.

Show that the wavelength of the electrons in this experiment is 182 nm. (2 marks)
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An interference fringe occurs on the screen where constructive interference takes place.

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

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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. $\text{Work done by field}=\Delta K=K_f-K_i$

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

♦ Mean mark (c) 45%.

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.

♦♦♦ Mean mark 37%.

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.

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.

♦ Mean mark 53%.

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.

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.

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.

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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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. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

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

$\text{At point} \ A :$

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

♦♦ Mean mark (b) 34%.

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

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

PHYSICS, M7 2025 HSC 25

A student conducts an experiment to determine the work function of potassium.

The following diagram depicts the experimental setup used, where light of varying frequency is incident on a potassium electrode inside an evacuated tube.

For each frequency of light tested, the voltage in the circuit is varied, and the minimum voltage (called the stopping voltage) required to bring the current in the circuit to zero is recorded.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Frequency of incident light} \ \ & \quad \quad \quad \textit{Stopping voltage} \quad \quad \quad \\
\left(\times 10^{15} \ \text{Hz}\right) \rule[-1ex]{0pt}{0pt}& \text{(V)}\\
\hline
\rule{0pt}{2.5ex}0.9 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline
\rule{0pt}{2.5ex}1.1 \rule[-1ex]{0pt}{0pt}& 2.0 \\
\hline
\rule{0pt}{2.5ex}1.2 \rule[-1ex]{0pt}{0pt}& 2.5 \\
\hline
\rule{0pt}{2.5ex}1.3 \rule[-1ex]{0pt}{0pt}& 3.0 \\
\hline
\rule{0pt}{2.5ex}1.4 \rule[-1ex]{0pt}{0pt}& 3.5 \\
\hline
\rule{0pt}{2.5ex}1.5 \rule[-1ex]{0pt}{0pt}& 4.0 \\
\hline
\end{array}

Construct an appropriate graph using the data provided, and from this, determine the threshold frequency of potassium. (3 marks)

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Using the particle model of light, explain the features shown in the experimental results. (3 marks)

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b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

Show Worked Solution

b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

♦ Mean mark 47%.

PHYSICS, M6 2025 HSC 23

A wire loop is carrying a current of 2 A from $A$ to $B$ as shown. The length of wire within the magnetic field is 5 cm. The loop is free to pivot around the axis. The magnetic field is of magnitude $3 \times 10^{-2}\ \text{T}$ and at right angles to the wire.

Determine the torque produced on the wire loop due to the motor effect. (3 marks)

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Both the current and the magnetic field were changed, and the torque was observed to be in the same direction but twice the magnitude.
What changes to the magnitude of BOTH the current and the magnetic field are required to produce this result? (2 marks)

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a. $\tau=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

Show Worked Solution

a. $\text{Convert units:}\ \ 5 \ \text{cm} \ \Rightarrow \ \dfrac{5}{100}=0.05 \ \text{m}$

$F=BIL=3 \times 10^{-2} \times 2 \times 0.05=0.003 \ \text{N}$

$\text{Convert units:}\ \ 20\ \text{cm} \ \Rightarrow \ \ 0.2 \ \text{m}$

$\tau$	$=F d$
	$=0.003 \times 0.2$
	$=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

Mean mark (a) 51%.

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

♦ Mean mark (b) 45%.

Networks, GEN2 2025 VCAA 16

The map below shows the passages connecting five areas in the gym: entry $(E)$, recovery $(R)$, weights $(W)$, change room $(C)$ and swimming $(S)$.

A network can be constructed from this map.

An adjacency matrix for this network is shown below.

Some values in the matrix are given as $x, y$ and $z$.

\begin{aligned}
& \quad \ \ \ E \ \ \ R \ \ \ W \ \ \ C \ \ \ S\\
&\begin{array}{l}
E\\
R\\
W\\
C\\
S
\end{array}\begin{bmatrix}
0 & x & 2 & 2 & 0\\
x & 1 & y & 2 & 0\\
2 & y & 1 & z & 0\\
2 & 2 & z & 0 & 1\\
0 & 0 & 0 & 1 & 0
\end{bmatrix}
\end{aligned}

In this matrix, the '2' in row $E$, column $C$ indicates that there are two ways of moving from the entry to the change room without passing through another area or backtracking.

Determine the values of $x, y$ and $z$ in the boxes below. (2 marks)

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$x=2, y=4, z=2$

Show Worked Solution

$x=2, y=4, z=2$

$\text{One strategy}$

$\text{Consider the two paths from \(E$ to $R$ ($x$ in the table):}\)

$\text{Path 1: straight line from \(E$ to $R$.}\)

$\text{Path 2: start towards \(E$ then left, right, right, right.}\)

♦ Mean mark 44%.

BIOLOGY, M5 2025 HSC 33

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

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

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

♦ Mean mark (b) 52%.

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.

♦ Mean mark (c) 52%.

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.

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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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.

♦♦ Mean mark (a) 41%.

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.

♦ Mean mark (b) 50%.

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.

The flow chart shows the process of antibody production following exposure to alpha-gal.
Describe the role of $\text{X}$, $\text{Y}$ and $\text{Z}$ in the process of antibody production. (4 marks)

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

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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. In this way the process provides 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.

♦ Mean mark (a) 45%.

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. In this way, the process provides enhanced immune protection against subsequent infections.

Financial Maths, STD2 EQ-Bank 13 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?

$1.15
$1.27
$1.55
$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}=$1.15$

$\Rightarrow A$

Measurement, STD2 EQ-Bank 12 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?

500
2000
2500
5000

Show Answers Only

$D$

Show Worked Solution

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

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

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

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

CHEMISTRY, M7 2025 HSC 28

Kevlar and polystyrene are two common polymers.

A section of their structures is shown.

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

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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

♦ Mean mark (a) 50%.

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.

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?

$8.8 \times 10^{-14}$
$5.3 \times 10^{-12}$
$1.1 \times 10^{-11}$
$2.1 \times 10^{-11}$

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:
$\ce{[Ag+]_{\text{sat}} = 2000 \times [Ag+]_{\text{diluted}}} = 2000 \times 0.11 \times 10^{-6} = 2.2 \times 10^{-4}\ \text{mol L}^{-1}$
The equation for the dissolution of $\ce{Ag2C₂O4}$ is
$\ce{Ag2C₂O4 \leftrightharpoons 2Ag+(aq) + C2O4^{2-}(aq)}$
Hence $\ce{[C2O4^{2-}]_{\text{sat}} = 0.5 \times [Ag+]_{\text{sat}}} = 0.5 \times 2.2 \times 10^{-4} = 1.1 \times 10^{-4}$
$K_{sp} = \ce{[Ag+]^2[C2O4^{2-}]} = (2.2 \times 10^{-4})^2(1.1 \times 10^{-4}) = 5.3 \times 10^{-12}$

$\Rightarrow B$

♦ Mean mark 42%.

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$^{-1}$ )?

72 062
72 080
90 060
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}$

$\Rightarrow B$

♦♦ Mean mark 34%.

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 $B$.

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)

--- 12 WORK AREA LINES (style=lined) ---
Let $P$ be the point with position vector $\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}$.
Show that $PB$ is perpendicular to the direction of the line $\ell$. (1 mark)

--- 7 WORK AREA LINES (style=lined) ---
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 $\underset{\sim}{a}$.
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 \(\ell$ and $B$}\)

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

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

♦♦ Mean mark (i) 33%.

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

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

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

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

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

$\text{Show} \ \ \overrightarrow{PB} \perp \ell:$

♦♦♦ Mean mark (ii) 22%.

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

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

$\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 \(\ell$ to $O$)}-1\)

♦♦♦ Mean mark (iii) 4%.

$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 \(\ell$ to sphere $\left(d_{\min }\right)$:}\)

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

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 $\ce{NH3+}$.

$\Rightarrow A$

♦ Mean mark 45%.

BIOLOGY, M5 2025 HSC 18 MC

The fruit fly, Drosophila melanogaster, normally has large round eyes. A mutation of the eye shape gene, found on the X-chromosome, causes the eye to be a narrow slit (bar-eye), which is a dominant allele for eye shape.

Offspring were produced when a bar-eyed male was crossed with a normal female. Males are XY and females are XX.

Which row in the table shows the correct percentage of male and female offspring produced with bar-eye?

	$\quad\quad\quad\textit{Percentage of offspring with bar-eyes}\quad\quad\quad$
	$\quad\textit{Males}\quad$	$\quad\textit{Females}\quad$
A.	$100$	$100$
B.	$0$	$100$
C.	$100$	$0$
D.	$50$	$0$

Show Answers Only

$B$

Show Worked Solution

B is correct: Males inherit Y from father; all females inherit X with bar-eye.

Other Options:

A is incorrect: Males get Y chromosome from father, not X with mutation.
C is incorrect: Males cannot inherit bar-eye from father; females must inherit it.
D is incorrect: All females receive father’s X chromosome with bar-eye allele.

♦ Mean mark 55%.

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)

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

Show Answers Only

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

Show Worked Solution

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

$\text{Components of initial velocity:}$

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

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

♦♦ Mean mark 35%.

$\text{Horizontal motion:}$

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

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

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

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

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

$\text{Vertical Motion }$

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

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

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

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

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

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

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

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

Complex Numbers, EXT2 N1 2025 HSC 7 MC

The complex number $z$ lies on the unit circle.

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

$\dfrac{\pi}{6} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{5 \pi}{6}$
$\dfrac{\pi}{3} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{2 \pi}{3}$
$-\dfrac{5 \pi}{6} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{6}$
$-\dfrac{2 \pi}{3} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{3}$

Show Answers Only

$D$

Show Worked Solution

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

♦ Mean mark 52%.

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

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

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

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

$\Rightarrow D$

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 $z-w$ ?

$\dfrac{1}{8}$
$\dfrac{\sqrt{7}}{2}$
$\dfrac{3}{2}$
$\dfrac{7}{4}$

Show Answers Only

$B$

Show Worked Solution

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

$\text{Find}\ \ \abs{z-w}:$

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

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

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

$\Rightarrow B$

Mean mark 56%.

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 $t \in \mathbb{R}$ ?

$\left(\begin{array}{c}\cos (2 t) \\ 2 t \\ \sin (2 t)\end{array}\right)$
$\left(\begin{array}{c}\cos \left(t^2+\dfrac{\pi}{2}\right) \\ t^2+\dfrac{\pi}{2} \\ \sin \left(t^2+\dfrac{\pi}{2}\right)\end{array}\right)$
$\left(\begin{array}{c}\cos \left(t^2\right) \\ -t^2 \\ \sin \left(t^2\right)\end{array}\right)$
$\left(\begin{array}{c}\cos \left(2 t+\dfrac{\pi}{2}\right) \\ 2 t+\dfrac{\pi}{2} \\ -\sin \left(2 t+\dfrac{\pi}{2}\right)\end{array}\right)$

Show Answers Only

$D$

Show Worked Solution

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

$\text{Consider option D:}$

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

$\text{Since}\ t \in \mathbb{R}\ \ \Rightarrow \ \ u \in \mathbb{R}$

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

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

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

$\Rightarrow D$

♦ Mean mark 52%.

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 $UVWZ$ ?

Parallelogram
Rectangle
Rhombus
Square

Show Answers Only

$C$

Show Worked Solution

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

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

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

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

$\therefore UW\ \text{and}\ VZ\ \text{are perpendicular.}$

$\Rightarrow C$

♦♦ Mean mark 36%.

Mechanics, EXT2 M1 2025 HSC 8 MC

The graph shows the velocity of a particle as a function of its displacement.

Which of the following graphs best shows the acceleration of the particle as a function of its displacement?

Show Answers Only

$A$

Show Worked Solution

$\text{By elimination:}$

$v(x)\ \Rightarrow\ \text{degree 3 (see graph)},\ \ \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 2}$

$a=v \cdot \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 5 (eliminate C and D)}$

$\text{At}\ \ x=0,\ \ v(x)>0\ \ \text{and}\ \ \dfrac{dv}{dx}<0\ \ \Rightarrow\ \ v \cdot \dfrac{dv}{dx} \neq 0\ \text{(eliminate B)}$

$\Rightarrow A$

♦♦ Mean mark 35%.

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)

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

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

Proof, EXT2 P1 2025 HSC 14d

Positive real numbers $a, b, c$ and $d$ are chosen such that $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ and $\dfrac{1}{d}$ are consecutive terms in an arithmetic sequence with common difference $k$, where $k \in \mathbb{R} , k>0$.

Show that $b+c<a+d$. (3 marks)

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

Show Answers Only

$\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)$

$\text{Show} \ \ b+c<a+d$

$\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)$

$\dfrac{1}{a}<\dfrac{1}{b}$

$\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\ \ldots\ \text{such that}$

$\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)$

$\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab$

$\text {Similarly:}$

$\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd$

$\text{Using \(\ a>b>c>d\ $ (see (1) above):}\)

$a-b>c-d$

$b+c<a+d$

Show Worked Solution

$\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)$

$\text{Show} \ \ b+c<a+d$

$\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)$

$\dfrac{1}{a}<\dfrac{1}{b}$

$\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\ \ldots\ \text{such that}$

$\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)$

♦ Mean mark 39%.

$\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab$

$\text {Similarly:}$

$\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd$

$\text{Using \(\ a>b>c>d\ $ (see (1) above):}\)

$a-b>c-d$

$b+c<a+d$

\(\text{Time on Island B}\)	\(=9:20\ \text{am}-10\ \text{hours}\ 40\text{ minutes}\)
	\(=11:20\ \text{pm}-40\ \text{minutes}\)
	\(=10:40\ \text{pm (previous day)}\)

\(\therefore\ \text{Time in City B}\)	\( = 11:00\ \text{pm}-7\ \text{hours}\ 52\text{ minutes}\)
	\(=\ 4:00\ \text{pm}-52\ \text{minutes}\)
	\(=3:08\ \text{pm}\)

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

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

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

\(\text{Total hours}\)	\(=(2\times 8) + (7\times 10) + (12\times 7) + (17\times 5)\)
	\(=16 + 70 + 84 + 85=255\)

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

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

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