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CHEMISTRY, M7 2020 VCE 4 MC

What is the IUPAC name of the molecule shown above?

  1. 3-hydroxy-3-ethyl-propan-1-amine
  2. 3-amino-1-methylpropan-1-ol
  3. 3-hydroxypentan-1-amine
  4. 1-aminopentan-3-ol
Show Answers Only

\(D\)

Show Worked Solution

→ Molecule has a 5-carbon chain with single bonds (“pentan”)

→ Alcohol functional group attached to 3rd carbon (“-3-ol”)

\(\Rightarrow D\)

CHEMISTRY M6 2016 VCE 21*

The ammonium ion \(\ce{NH4+}\) acts as a weak acid according to the equation

\(\ce{NH4+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O+(aq)}\)

Given the \(K_a \ce{(NH4+) = 5.6 \times 10^{-10}}\), determine the [\(\ce{H3O+}\)] of a 0.200 M ammonium chloride solution.   (2 marks)

Show Answers Only

\(1.06 \times 10^{-5}\ \text{mol L}^{-1}\)

Show Worked Solution

\(K_a = \dfrac{\ce{[NH3][H3O+]}}{\ce{[NH4+]}} \)

\(\text{Weak acid assumptions:}\)

\(\ce{[NH4+]_{eq} = 0.200\ M\ \ \text{and}\ \ [NH3] = [H3O+]} \)

\(5.6 \times 10^{-10}\) \(= \dfrac{\ce{[H3O+]^2}}{0.200} \)  
\(\ce{[H3O+]^2}\) \(=0.200 \times 5.6 \times 10^{-10} \)  
\(\ce{[H3O+]}\) \(= \sqrt{1.12 \times 10^{-10}}\)  
  \(=1.06 \times 10^{-5}\ \text{mol L}^{-1} \)  

CHEMISTRY, M7 2016 VCE 7a

Butanoic acid is the simplest carboxylic acid that is also classified as a fatty acid. Butanoic acid may be synthesised as outlined in the following reaction flow chart.
 

  1. Draw the structural formula of but-1-ene in the box provided.  (1 mark)

  2. State the reagent(s) needed to convert but-1-ene to Compound Y in the box provided.  (1 mark)

  3. Write the systematic name of Compound Y in the box provided.  (1 mark)

  4. Write the semi-structural formula of butanoic acid in the box provided.  (1 mark)

  5. Write a balanced half-equation for the conversion of \(\ce{Cr2O7^2– to Cr3+}\).  (2 marks)

Show Answers Only

i.    
       

ii.    \(\ce{H2O}\) and \(\ce{H3PO4}\) (catalyst)

iii.   butan-1-ol or 1-butanol

iv.   \(\ce{CH3CH2CH2COOH}\)

v.   \(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} \)

Show Worked Solution

i.    
       

ii.    \(\ce{H2O}\) and \(\ce{H3PO4}\) (catalyst)

iii.   butan-1-ol or 1-butanol

iv.   \(\ce{CH3CH2CH2COOH}\)

v.   \(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- \rightarrow 2Cr^{3+}(aq) + 7H2O(l)} \)

♦♦ Mean mark (ii) 29%.

CHEMISTRY, M7 2015 VCE 5c

A student mixed salicylic acid with ethanoic anhydride (acetic anhydride) in the presence of concentrated sulfuric acid. The products of this reaction were the painkilling drug aspirin (acetyl salicylic acid) and ethanoic acid.
 

  1. An incomplete structure of the aspirin molecule is shown above.
  2. Complete the structure by filling in the two boxes provided in the diagram.  (2 marks)

  3. Sulfuric acid is used as a catalyst in this reaction.
  4. Explain how a catalyst increases the rate of this reaction.  (2 marks)

Show Answers Only

i.    
       

ii.    Sulphuric acid increases the rate of reaction by:

→ providing an alternative reaction pathway that involves a lower activation energy for the reagents.

→ this increases the likelihood of successful collisions.

Show Worked Solution

i.    
     

♦ Mean mark (i) 48%.

ii.    Sulphuric acid increases the rate of reaction by:

→ providing an alternative reaction pathway that involves a lower activation energy for the reagents.

→ this increases the likelihood of successful collisions.

CHEMISTRY, M7 2016 VCE 2*

What is the correct systematic name for the compound shown above?   (2 marks)

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3,4-dimethylheptane

Show Worked Solution

→ 7 carbon string with single carbon and hydrogen bonds = heptane

→ Two methyl groups attached to carbon 3 and 4 (IUPAC convention)

→ Correct name is: 3,4-dimethylheptane

CHEMISTRY, M7 2015 VCE 5a

A reaction pathway is designed for the synthesis of the compound that has the structural formula shown below.
 

The table below gives a list of available organic reactants and reagents.
 


Complete the reaction pathway design flow chart below. Write the corresponding letter for the structural formula of all organic reactants in each of the boxes provided. The corresponding letter for the formula of other necessary reagents should be shown in the boxes next to the arrows.   (5 marks)

Show Answers Only

Show Worked Solution

CHEMISTRY, M6 2015 VCE 8

Hydrogen sulfide, in solution, is a diprotic acid and ionises in two stages.

\(\ce{H2S(aq) + H2O(l)\rightleftharpoons HS-(aq) + H3O+(aq)}\) \(\quad K_{a1} = 9.6 × 10^{–8} \text{ M}\)

\(\ce{HS–(aq) + H2O(l)\rightleftharpoons S^{2-}(aq) + H3O+(aq)}\) \(\quad K_{a2} = 1.3 × 10^{–14} \text{ M}\)

A student made two assumptions when estimating the pH of a \(0.01 \text{ M}\) solution of \(\ce{H2S}\):

Assumption 1: The pH can be estimated by considering only the first ionisation reaction.

Assumption 2: The concentration of \(\ce{H2S}\) at equilibrium is approximately equal to \(0.01 \text{ M}\).

  1. Explain why these two assumptions are justified.  (2 marks)

  2. Use the two assumptions given above to calculate the pH of a 0.01 M solution of \(\ce{H2S}\).  (3 marks)

  3. Some solid sodium hydrogen sulfide, \(\ce{NaHS}\), is added to a 0.01 M solution of \(\ce{H2S}\).
  4. Predict the effect of this addition on the pH of the hydrogen sulfide solution. Justify your prediction.  (2 marks)

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a.   1st assumption:

→ \(K_{a2}\) is significantly smaller than the first ionisation \(\ce{(K_{a1})}\), making its impact on the \(\ce{[H3O+]}\) / pH level negligible.

2nd assumption:

→ \(K_{a1}\) is very small, making the extent of the ionisation of \(\ce{H2S}\) very small and hence a minimal change in \(\ce{[H2S]}\) results.

b.    \(\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 \)

c.    Adding \(\ce{NaHS}\):

→ Increases the \(\ce{[HS-]}\).

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the \(\ce{[H3O+]}\) and the pH will therefore increase.

Show Worked Solution

a.   1st assumption:

→ \(K_{a2}\) is significantly smaller than the first ionisation \(\ce{(K_{a1})}\), making its impact on the \(\ce{[H3O+]}\) / pH level negligible.

2nd assumption:

→ \(K_{a1}\) is very small, making the extent of the ionisation of \(\ce{H2S}\) very small and hence a minimal change in \(\ce{[H2S]}\) results.
 

♦♦ Mean mark (a) 27%.
b.    \(K_{a1}\) \(=\dfrac{\ce{[HS-][H3O+]}}{\ce{[H2S]}} \)
  \(9.6 \times 10^{-8}\)

\(=\dfrac{\ce{[H3O+]^2}}{0.01}\)
  \(\ce{[H3O+]^2}\) \(=0.01 \times 9.6 \times 10^{-8} \)
  \(\ce{[H3O+]}\) \(=\sqrt{9.6 \times 10^{-10}}\)
    \(=3.1 \times 10^{-5}\ \text{M} \)

 
\(\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 \)
  

c.    Adding \(\ce{NaHS}\):

→ Increases the \(\ce{[HS-]}\).

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the \(\ce{[H3O+]}\) and the pH will therefore increase.

♦♦ Mean mark (c) 35%.

PHYSICS, M1 EQ-Bank 1

A physics student comes across a river which runs north to south and has a current of 3 ms\(^{-1}\) running south.

The student starts on the west side of the river at point A and paddles a kayak at 5 ms\(^{-1}\) directly across the river to finish at point B.

  1. Calculate the angle which he must position the boat to travel in a straight line across the river.   (2 mark)

  2. If the river is 100 metres wide, determine the time it takes for the student to cross the river.   (2 mark)

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a.    \(36.9^{\circ}\)

b.    \(\text{25 seconds}\)

Show Worked Solution

a.   
       

\(\sin \theta\) \(=\dfrac{3}{5}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{3}{5}\Big)=36.9^{\circ}\)  

 
The student must turn 36.9\(^{\circ}\) into the current as shown on the diagram.

 
b.    Using Pythagoras:

\(v=\sqrt{5^2-3^2}=4\ \text{ms}^{-1}\)

\(t=\dfrac{d}{s}=\dfrac{100}{4}=25\ \text{s}\)
 

\(\therefore\) It will take the student 25 seconds to travel from A to B.

PHYSICS, M1 EQ-Bank 2

A boy on a bike is travelling at 8 ms\(^{-1}\) north when he sees a girl on a scooter travelling at 6 ms\(^{-1}\) west, relative to the ground. 

  1. Draw a vector diagram to represent the velocity of the girl relative to the boy.   (2 marks)

  2. Use the vector diagram to calculate the velocity of the girl relative to the boy.   (2 marks)

Show Answers Only

10 ms\(^{-1}\), S36.9\(^{\circ}\)W

Show Worked Solution

a.    \(v_{\text{g}} = 6\ \text{ms}^{-1}\ \text{west} \)

\(-v_{\text{b}} = 8\ \text{ms}^{-1}\ \text{south} \)

\(v_{\text{g rel b}} = v_\text{g} + (-v_\text{b}) \)
 


 

b.    \(v_{\text{g rel b}}=\sqrt{6^2+8^2} =10\ \text{ms}^{-1}\)

  \(\tan \theta\) \(=\dfrac{8}{6}\)
  \(\theta\) \(=\tan^{-1}\Big(\dfrac{8}{6}\Big)=53.1^{\circ}\)

 
\(\therefore\) Velocity of the girl relative to the boy is 10 ms\(^{-1}\), S36.9°W.

PHYSICS, M1 EQ-Bank 11

Car A is travelling north at 40 ms\(^{-1}\) and Car B is travelling at 30 ms\(^{-1}\) east.

Determine the velocity of Car A relative to Car B.   (3 mark)

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50 ms\(^{-1}\), N36.9\(^{\circ}\)W

Show Worked Solution

\(v_{\text{A rel B}}=v_A-v_B =v_A + (-v_B)\) 

\(v_A = 40\ \text{ms}^{-1}\ \text{north}, \ -v_B = 30\ \text{ms}^{-1}\ \text{west} \)

\(v_{\text{A rel B}}=\sqrt{40^2+30^2}=50\ \text{ms}^{-1}\)

\(\tan \theta\) \(=\dfrac{30}{40}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{30}{40}\Big)=36.9^{\circ}\)  

 

\(\terefore\) Velocity of Car A relative to Car B is 50 ms\(^{-1}\), N36.9°W.

PHYSICS, M1 EQ-Bank 10

Determine the magnitude and direction of the resultant velocity vector for an airplane that is simultaneously moving with a velocity of 350\(^{-1}\) directly towards the west and 275 ms\(^{-1}\) towards the northwest.   (3 marks)

Show Answers Only

578 ms\(^{-1}\), N70.3°W

Show Worked Solution

Add vectors to show resultant vector:

Using the cosine rule to find the magnitude of \(v\):

\(v\) \(=\sqrt{350^2+275^2-2 \times 350 \times 275 \times \cos 135°}\)  
  \(=578.1376\dots\ \text{ms}^{-1}\)  

 

Using the sign rule to find \(\theta\):

\(\dfrac{\sin \theta}{275}\) \(=\dfrac{\sin 135°}{578.1376}\)  
\(\sin \theta\) \(=\dfrac{\sin135° \times 275}{578.1376}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{\sin135° \times 275}{578.1376}\Big)\)  
  \(=19.7^{\circ}\)  

 

\(\therefore\) Resultant velocity of the aeroplane is 578 ms\(^{-1}\), N70.3°W.

PHYSICS, M1 EQ-Bank 9

A car is travelling north and approaching an intersection at 50 kmh\(^{-1}\).

While maintaining a constant speed, the car turns left and continues east at 50 kmh\(^{-1}\). 

Using a vector diagram, calculate the change in velocity of the car.   (3 marks)

Show Answers Only

\(\text{70.7 kmh}^{-1}, \text{S45°E} \)

Show Worked Solution

\(v= 50\ \text{kmh}^{-1}\ \text{east},\ \ u= 50\ \text{kmh}^{-1}\ \text{north.}\)

\(\Delta v= v-u = v+(-u),\ \ \text{where}\ -u= 50\ \text{kmh}^{-1}\ \text{south}\)
  

\(\Delta v=\sqrt{50^2 + 50^2}=\sqrt{5000}=70.7\ \text{kmh}^{-1}\)

\(\tan \theta \) \(=\dfrac{50}{50}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{50}{50}\Big)=45^{\circ}\)  

 

Change in velocity of the car = 70.7 \(\text{kmh}^{-1}\), S45\(^{\circ}\)E.

Data Analysis, GEN1 2022 VCAA 9-11 MC

Table 1 summarises the results of a study that compared the effectiveness of individual and group instruction (instructional method) when training future basketball referees.
 

In this table, test grade is the response variable and instructional method is the explanatory variable.
 

Question 9

The variables test grade (A, B, C, D, E) and instructional method (individual, group) are

  1. a numerical and a categorical variable respectively.
  2. both nominal variables.
  3. a nominal and an ordinal variable respectively.
  4. both ordinal variables.
  5. an ordinal and a nominal variable respectively.

 
Question 10

Of the students who received an A grade, the percentage who were instructed individually is closest to

  1. 9%
  2. 22%
  3. 36%
  4. 56%
  5. 64%

 
Question 11

To become a qualified referee, a grade of A or B on the test is required. Those who receive a C, a D or an E will not qualify.

Using column percentages, a new two-way percentage frequency table is constructed from the data in Table 1.

In this new table, qualified to be a referee (yes, no) is the response variable and instructional method (individual, group) is the explanatory variable.

Which one of the following tables correctly displays the data from Table 1?
 

Show Answers Only

\(\text{Question 9:} \ E\)

\(\text{Question 10:} \ C\)

\(\text{Question 11:} \ B\)

Show Worked Solution

\(\text{Question 9} \)

\(\text{Both variables are categorical.}\)

\(\text{Test grade (A, B, C, D, E) → ordinal, as they can be ranked.}\)

\(\text{Instructional method → nominal, has no ranking.}\)

\(\Rightarrow E\)
 

\(\text{Question 10} \)

\(\text{Students who received an A}\ =10+18 = 28 \)

\(\text{% instructed individually}\ = \dfrac{10}{28} \times 100 = 35.714…\% \)

\(\Rightarrow C\)
 

Mean mark (Q10) 56%.

\(\text{Question 11} \)

\(\text{yes (A or B):} \)

\(\text{Individual}\ =\dfrac{10+35}{115} \times 100 \approx 39\% \)

\(\text{Group}\ = \dfrac{18+30}{126} \times 100 \approx 38\% \)
 

\(\text{no (C, D or E):} \)

\(\text{Individual}\ = 100-39 \approx 61\% \)

\(\text{Group}\ = 100-38 \approx 62\% \)

\(\Rightarrow B\)

Data Analysis, GEN1 2022 VCAA 7-8 MC

The association between the weight of a seal's spleen, spleen weight, in grams, and its age, in months, for a sample of seals is non-linear.

This association can be linearised by applying a \(\log _{10}\) transformation to the variable spleen weight.
 

The equation of the least squares line for this scatterplot is

\(\log _{10}\) (spleen weight) = 2.698 + 0.009434 × age

 
Question 7

The equation of the least squares line predicts that, on average, for each one-month increase in the age of the seals, the increase in the value of \(\log _{10}\) (spleen weight) is

  1. 0.009434
  2. 0.01000
  3. 1.020
  4. 2.698
  5. 5.213

 
Question 8

Using the equation of the least squares line, the predicted spleen weight of a 30-month-old seal, in grams, is

  1. 3
  2. 511
  3. 772
  4. 957
  5. 1192
Show Answers Only

\(\text{Question 7:} \ A\)

\(\text{Question 8:} \ D\)

Show Worked Solution

\(\text{Question 7}\)

\(\text{Graph passes through (0, 2.7) and (130, 3.93)} \)

\(\text{Gradient}\ \approx \dfrac{3.93-2.7}{130} \approx 0.0946 \)

\(\Rightarrow A\)
 

\(\text{Question 8}\)

\(\text{Let the predicted spleen weight be}\ w:\)

\(\log_{10} w\) \(=2.698 + 0.009434 \times 30\)  
\(\log_{10} w\) \(=2.98102\)  
\(w\) \(= 10^{2.98102}\)  
  \(=957.2381527\)  

 
\(\Rightarrow D\)

Mean mark (Q8) 56%.

Data Analysis, GEN1 2022 VCAA 6 MC

The histogram below displays the distribution of spleen weight for a sample of 32 seals.

The histogram has a \(\log _{10}\) scale.
 

The number of seals in this sample with a spleen weight of 1000 g or more is

  1. 7
  2. 8
  3. 17
  4. 25
  5. 27
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Given}\ \log_{10} 1000 = 3\ \ (10^3 = 1000) \)

\(\text{Number of seals} = 8 + 9 + 7 + 1 = 25\)

\(\Rightarrow D\)

Data Analysis, GEN1 2022 VCAA 4 MC

The age, in years, of a sample of 14 possums is displayed in the dot plot below.
 

The mean and the standard deviation of age for this sample of possums are closest to

  1. mean = 4.25    standard deviation = 2.6
  2. mean = 4.8      standard deviation = 2.4
  3. mean = 4.8      standard deviation = 2.5
  4. mean = 4.9      standard deviation = 2.4
  5. mean = 4.9      standard deviation = 2.5
Show Answers Only

\(E\)

Show Worked Solution

\(\text{By calculator:}\) 

\(\text{Sample standard deviation} = 2.525…\)

\(\text{Mean} = 4.928…\)

\(\Rightarrow E\)

Data Analysis, GEN1 2022 VCAA 1-3 MC

The histogram below displays the distribution of skull width, in millimetres, for 46 female possums.
 

Question 1

The shape of the distribution is best described as

  1. negatively skewed.
  2. approximately symmetric.
  3. negatively skewed with a possible outlier.
  4. positively skewed with a possible outlier.
  5. approximately symmetric with a possible outlier.

 
Question 2

The percentage of the 46 possums with a skull width of less than 55 mm is closest to

  1. 12%
  2. 26%
  3. 39%
  4. 61%
  5. 74%

 
Question 3

The third quartile \((Q_3)\) for this distribution, in millimetres, could be

  1. 55.8
  2. 56.2
  3. 56.9
  4. 57.7
  5. 58.3
Show Answers Only

\(\text{Question 1:} \ C\)

\(\text{Question 2:} \ B\)

\(\text{Question 3:} \ D\)

Show Worked Solution

\(\text{Question 1}\)

The distribution’s centre is in the 56–57 group and if the possible outlier is disregarded, the tail of the distribution is spread more to the left → i.e. negatively skewed with possible outlier.

\(\Rightarrow C\) 


♦♦ Mean mark (Q1) 36%.
COMMENT: 54% of students chose \(E\). The graph distribution however, shows that more than 50% of the data is to the left of the median.

\(\text{Question 2}\)

\(\text{Percentage}\) \(= \dfrac{1+1+2+1+2+5}{46} \times 100\)  
  \(=\dfrac{12}{46} \times 100\)  
  \(=26.086…\)%  

 
\(\Rightarrow B\)
 

\(\text{Question 3}\)

\( Q_3\ =0.75 \times 46 = 34.5 \)

The 34.5th score lies between 57 and 58, therefore, 57.7.

\(\Rightarrow D\)

CHEMISTRY, EQ-Bank 26

Calcium carbonate, when heated, decomposes according to the following chemical equation:

\(\ce{CaCO3 -> CaO(s) + CO2(g) }\)

60.0 grams of calcium carbonate is heated and the calcium oxide residue produced is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaCO3} \rule[-1ex]{0pt}{0pt} & \text{60.0 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaO} \rule[-1ex]{0pt}{0pt} & \text{33.6 grams} \\
\hline
\end{array}

Determine the percentage of carbon dioxide, by weight, released from the calcium carbonate during this process.   (2 marks)

Show Answers Only

\(\%\ce{CO2} = 44.0\%\)

Show Worked Solution

\(\text{Mass of}\ \ce{CaCO3} = 60.0\ \text{grams}\)

\(\text{Mass of}\ \ce{CO2} = 60.0-33.6 = 26.4\ \text{grams}\)

\(\%\ce{CO2}\ \text{released}\ = \dfrac{26.4}{60.0} \times 100\% = 44.0\%\)

CHEMISTRY, M1 EQ-Bank 25

15.00 grams of Silver(\(\text{I}\)) oxide is heated until it decomposes and the silver produced is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{15.00 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{13.96 grams} \\
\hline
\end{array}

Determine the percentage of oxygen, by weight, in the sample of Silver(\(\text{I}\)) oxide.   (2 marks)

Show Answers Only

\(\%\ce{O} = 6.93\%\)

Show Worked Solution

\(\text{Mass of}\ \ce{Ag} = 13.96\ \text{grams}\)

\(\text{Mass of}\ \ce{O} = 15.00-13.96 = 1.04\ \text{grams}\)

\(\%\ce{O} = \dfrac{1.04}{15.00} \times 100\% = 6.93\%\)

CHEMISTRY, M1 EQ-Bank 24

How can the physical properties of components within a mixture be utilised to separate a heterogeneous mixture of sand, salt, and iron filings?   (3 marks)

Show Answers Only

→ Magnetism can be used to separate iron filings from the mixture. Iron filings are magnetic, so you can easily separate them from the non-magnetic sand and salt by moving a magnet over the surface of the mixture.

→ Filtration can be used to separate the sand from the salt. Add water to the remaining mixture of sand and salt, and stir well to dissolve the salt. Since sand is insoluble in water, you can separate it by pouring the mixture through a filter paper placed in a funnel.

→ Evaporation can be used to recover the salt. Finally, to separate the salt from the water, heat the saltwater solution until the water evaporates. The salt will be left behind as a solid residue.

Show Worked Solution

→ Magnetism can be used to separate iron filings from the mixture. Iron filings are magnetic, so you can easily separate them from the non-magnetic sand and salt by moving a magnet over the surface of the mixture.

→ Filtration can be used to separate the sand from the salt. Add water to the remaining mixture of sand and salt, and stir well to dissolve the salt. Since sand is insoluble in water, you can separate it by pouring the mixture through a filter paper placed in a funnel.

→ Evaporation can be used to recover the salt. Finally, to separate the salt from the water, heat the saltwater solution until the water evaporates. The salt will be left behind as a solid residue.

PHYSICS, M1 EQ-Bank 7

 A plane is travelling at 315 ms\(^{-1}\) north when it passes through a dense cloud and slows down to a velocity of 265 ms\(^{-1}\) for safety precautions.

The plane did not change direction and travelled 2.5 km while it was slowing down.

Using north as the positive direction for all calculations, determine:

  1. the change in velocity of the plane.   (1 mark)

  2. the plane's acceleration.   (2 marks)

  3. the time over which the plane slowed down.   (2 marks)

Show Answers Only

a.   \(\text{50 ms}^{-1}\ \text{south}\)

b.   \(\text{5.8 ms}^{-2}\ \text{south}\) 

c.   \(\text{8.62 s}\)

Show Worked Solution
a.     \(\Delta v\) \(=v-u\)
    \(=265-315\)
    \(=-50\ \text{ms}^{-1}\)
    \(=50\ \text{ms}^{-1}\ \text{south}\)

 

b.    Using  \(v^2=u^2 +2as\)  (time is not given):

\(a\) \(=\dfrac{v^2-u^2}{2s}\)  
  \(=\dfrac{(265)^2-(315)^2}{2 \times 2500}\)  
  \(=-5.8\ \text{ms}^{-2}\)  
  \(=5.8\ \text{ms}^{-2}\) to the south.  

 

c.    Using  \(v=u+at\):

\(t\) \(=\dfrac{v-u}{a}\)  
  \(=\dfrac{265-315}{-5.8}\)  
  \(=8.62\ \text{s}\)  

PHYSICS, M1 EQ-Bank 6

A motorbike is travelling to the east at 50 ms\(^{-1}\) when it passes a car travelling west at 60 ms\(^{-1}\).

Calculate the velocity of the motorbike relative to the car.   (2 marks)

Show Answers Only

\(\text{110 ms}^{-1}\ \text{east}\)

Show Worked Solution

Let east be the positive direction and west be the negative direction.

\(v_{\text{m rel c}}\) \(=v_{\text{m}}-v_{\text{c}}\)  
  \(=50-(-60) \)  
  \(=110\ \text{ms}^{-1}\)  

 

Therefore, the velocity of the motorbike relative to the car is 110 ms\(^{-1}\) east.

Mechanics, EXT2 M1 2023 HSC 9 MC

A particle travels along a curve from \(O\) to \(E\) in the \(x y\)-plane, as shown in the diagram.
 

The position vector of the particle is \(r\), its velocity is \( v \), and its acceleration is \( a \).

While travelling from \(O\) to \(E\), the particle is always slowing down.

Which of the following is consistent with the motion of the particle?

  1. \( r \cdot v \leq 0\)  and  \( a \cdot v \geq 0\)
  2. \( r \cdot  v \leq 0\)  and  \( a \cdot v \leq 0\)
  3. \( r \cdot v \geq 0\)  and  \( a \cdot v \geq 0\)
  4. \( r \cdot v \geq 0\)  and  \(a \cdot v \leq 0\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{By elimination:}\)

\(x\ \text{and}\ y\ \text{components of vector}\ r\ \text{are both positive.}\)

\(\text{The velocity of vector}\ v\ \text{is tangential to the curve}\ OE\ \text{and}\)

\(x\ \text{and}\ y\ \text{components of vector}\ v\ \text{are also both positive.}\)

\( r \cdot v \geq 0\ \ \text{(eliminate}\ A\ \text{and}\ B).\)

\(\text{Particle is slowing down so acceleration must be acting, in part,}\)

\(\text{in the opposite direction to velocity.}\)

\( a \cdot v \leq 0\ \ \text{(eliminate}\ C).\)

\(\Rightarrow D\)

Networks, STD2 N3 SM-Bank 24

The network below shows the one-way paths between the entrance, \(A\), and the exit, \(H\), of a children's maze.

The vertices represent the intersections of the one-way paths.

The number on each edge is the maximum number of children who are allowed to travel along that path per minute.

The minimum cut of the network is drawn, showing the maximum flow capacity of the maze is 23 children per minute.
 

One path in the maze is to be changed.

Determine the changes in the maximum flow capacity of the network in each of the following changes

  1. the capacity of flow along the edge \(GH\) is increased to 16.   (1 mark)

  2. the capacity of flow along the edge \(C E\) is increased to 12.   (2 marks)

  3. the direction of flow along the edge \(G F\) is reversed.   (2 marks)

Show Answers Only

i.    \(GH ↑ 16,\ \text{minimum cut = 27}\)

\(\text{Change: increases by 4}\)

ii.    \(CE ↑ 12,\ \text{minimum cut = 24}\)

\(\text{Change: increases by 1}\)

iii.   \(GF\ \text{is reversed, minimum cut = 30 (close to exit H)}\)

\(\text{Change: increases by 7}\)

Show Worked Solution

i.    \(GH ↑ 16,\ \text{minimum cut = 27}\)
 

\(\text{Change: increases by 4}\)
 

ii.    \(CE ↑ 12,\ \text{minimum cut = 24}\)
 

\(\text{Change: increases by 1}\)
 

iii.   \(GF\ \text{is reversed, minimum cut = 30 (close to exit H)}\)
 

\(\text{Change: increases by 7}\)

Networks, GEN1 2023 VCAA 33 MC

Consider the following graph.
 

How many of the following five statements are true?

  • The graph is a tree.
  • The graph is connected.
  • The graph contains a path.
  • The graph contains a cycle.
  • The sum of the degrees of the vertices is eight.
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Show Answers Only

\(D\)

Show Worked Solution

The graph is a tree (any two vertices are connected by one edge).  \(\checkmark\)

The graph is connected.  \(\checkmark\)

The graph contains a path.  \(\checkmark\)

The graph contains a cycle.  \(\cross\)

The sum of the degrees of the vertices is eight.  \(\checkmark\)

\(\Rightarrow D\)

Matrices, GEN1 2023 VCAA 29 MC

Matrix \(K\) is a \(3 \times 2\) matrix.

The elements of \(K\) are determined by the rule  \(k_{i j}=(i-j)^2\).

Matrix \(K\) is


A.

 \begin{bmatrix} 0 & 1 & -2 \\
1 & 0 & -1
\end{bmatrix}


B.

 \begin{bmatrix}
0 & 1 & 4 \\
1 & 0 & 1
\end{bmatrix}


C.

\begin{bmatrix}
0 & -1\\
1 & 0\\
4 & 1
\end{bmatrix}


D.

 \begin{bmatrix}
0 & 1\\
1 & 0\\
2 & 1
\end{bmatrix}


E.

 \begin{bmatrix}
0 & 1\\
1 & 0\\
4 & 1
\end{bmatrix}		    
Show Answers Only

\(E\)

Show Worked Solution

\(K\ \text{is a 3 × 2 matrix (eliminate A)}\)

\(k_{ij}\ \text{will all be}\ \geq 0\ \text{(eliminate C)}\)

\(k_{31} = (3-1)^2 = 4\ \text{(eliminate B and D)}\)

\(\Rightarrow E\)

Matrices, GEN1 2023 VCAA 26 MC

Matrix \(P\) is a permutation matrix and matrix \(Q\) is a column matrix.
 

\(P=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\quad \quad Q=\begin{bmatrix}
t \\
e \\
a \\
m \\
s
\end{bmatrix}\)
 

When \(Q\) is multiplied by \(P\), which three letters change position?

  1. \(t, e, a\)
  2. \(e, a, m\)
  3. \(a, m, s\)
  4. \(m, s, t\)
  5. \(e, a, s\)
Show Answers Only

\(B\)

Show Worked Solution

\(P \times Q =\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
t \\
e \\
a \\
m \\
s
\end{bmatrix}
= \begin{bmatrix}
t \\
a \\
m \\
e \\
s
\end{bmatrix}\)

\(\Rightarrow B\)

CHEMISTRY, M3 EQ-Bank 5

"The reason alkali metals produce ignition and the release of light and heat when reacting with cold water is due to the fact they have the energy to combust oxygen in the water".

Explain the validity of the statement with reference to metal reactivity.   (3 marks)

Show Answers Only

The statement is partially correct.

→ Alkali metals have low first ionisation energies and are therefore extremely reactive which enables them to even react with even cold water.

→ However, the statement is incorrect about combustion. Metals and water react in the following generalised equation:

\(\text{Metal + Water}\ \rightarrow\ \text{Metal hydroxide/oxide + Hydrogen gas} \)

→ Although oxygen can combust, the combustion occurring in the reaction of alkali metals is that of hydrogen, which is more flammable than oxygen and produced by the reaction.

→ This is consistent across all metals that react with water, although alkali metals are the only ones that contain enough energy to combust the hydrogen produced.

Show Worked Solution

The statement is partially correct.

→ Alkali metals have low first ionisation energies and are therefore extremely reactive which enables them to even react with even cold water.

→ However, the statement is incorrect about combustion. Metals and water react in the following generalised equation:

\(\text{Metal + Water}\ \rightarrow\ \text{Metal hydroxide/oxide + Hydrogen gas} \)

→ Although oxygen can combust, the combustion occurring in the reaction of alkali metals is that of hydrogen, which is more flammable than oxygen and produced by the reaction.

→ This is consistent across all metals that react with water, although alkali metals are the only ones that contain enough energy to combust the hydrogen produced.

CHEMISTRY, M3 EQ-Bank 4

  1. Write the generalised equation for reactions between metals and dilute acids (1 mark)

  2. Provide one example of the generalised equation from part (i).   (1 mark)

Show Answers Only

i.    \(\text{Metal + Dilute acid}\ \rightarrow\ \text{Salt + Hydrogen gas}\)

ii.    One example (of many possible answers):

→ Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas

\(\ce{2Mg(s) + 2HCl(aq) -> 2MgCl + H2(g) } \)

Show Worked Solution

i.    \(\text{Metal + Dilute acid}\ \rightarrow\ \text{Salt + Hydrogen gas}\)
 

ii.    One example (of many possible answers):

→ Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas

\(\ce{2Mg(s) + 2HCl(aq) -> 2MgCl + H2(g) } \)

CHEMISTRY, M3 EQ-Bank 10

A chemical reaction takes place in a sealed vessel containing limewater.

As the reaction progresses the limewater turns from a clear colourless solution, to a milky white one.

Explain, using chemical equations or otherwise, how the reaction taking place could be either a decomposition or a combustion reaction.   (3 marks)

Show Answers Only

→ Limewater turns milky in the presence of \(\ce{CO2}\).

→ The reaction in the above example could be either a decomposition or combustion reaction because both of these reactions can involve \(\ce{CO2}\) as a product.

→ Possible chemical equations include

Decomposition: \(\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}\)

Combustion: \(\ce{CH4(aq) + 2O2(g) \rightarrow 2H2O(l) + CO2(g)}\)

Show Worked Solution

→ Limewater turns milky in the presence of \(\ce{CO2}\).

→ The reaction in the above example could be either a decomposition or combustion reaction because both of these reactions can involve \(\ce{CO2}\) as a product.

→ Possible chemical equations include

Decomposition: \(\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}\)

Combustion: \(\ce{CH4(aq) + 2O2(g) \rightarrow 2H2O(l) + CO2(g)}\)

CHEMISTRY, M3 SM-Bank 12 MC

Electrolysis and photolysis are examples of which type of reaction? 

  1. Combustion
  2. Acid/Base 
  3. Synthesis
  4. Decomposition 
Show Answers Only

\(D\)

Show Worked Solution

→ Both electrolysis and photolysis involve the breakdown of compounds into simpler substances under decomposition reactions.

\(\Rightarrow D\)

CHEMISTRY, M3 EQ-Bank 9

Two moles of butane \(\ce{C3H8(g)}\) were reacted with 224 grams of oxygen \(\ce{O2(g)}\).

  1. Write the balanced equation for this reaction.   (3 marks)

  2. Determine the mass, in grams, of \(\ce{CO2(g)}\) produced by this reaction.   (1 mark)

Show Answers Only

i.    \(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)

ii.    \(\ce{m(CO2) = 88.02\ \text{g}}\)

Show Worked Solution

i.    Complete combustion equation:

   \(\ce{C3H8(g) + 5O2(g)\ \rightarrow 3CO2(g) + 4H2O(l)} \)

→ Two moles of butane require 10 moles of oxygen to fully combust. 

→ \(\ce{n(O2) = \dfrac {m}{MM}= \dfrac {224}{32} = 7}\)

→ Oxygen is limiting reagent and butane will undergo incomplete combustion according to the following balanced equation:

   \(\ce{2C3H8(g) + 7O2(g)\ \rightarrow 2C(s) + 2CO(g) + 2CO2(g) + 8H2O(l)}\)

 

ii.    Using the equation in part (i), 2 moles of \(\ce{CO2}\) will be produced

\(\ce{m(CO2) = n \times MM = 2 \times 44.01 = 88.02\ \text{g}}\)

Matrices, GEN1 2023 VCAA 27 MC

The following transition matrix, \(T\), models the movement of a species of bird around three different locations, \(M, N\) and \(O\) from one day to the next.

\begin{aligned}
& \quad \ \ \textit{this day} \\
& \quad  M \ \  N \ \ \  O\\
T = & \begin{bmatrix}
\frac{1}{3} & 0 & \frac{9}{10} \\
\frac{1}{3} & 1 & \frac{1}{10} \\
\frac{1}{3} & 0 & 0
\end{bmatrix}\begin{array}{l}
M\\
N\\
O
\end{array}\ \ \ next \ day
\end{aligned}

Which one of the following statements best represents what will occur in the long term?

  1. No birds will remain at location \(M\).
  2. No birds will remain at location \(N\).
  3. All of the birds will end up at location \(M\).
  4. All of the birds will end up at location \(O\).
  5. An equal number of birds will be at all three locations.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Steady state matrix:}\)

\(T^{50} =\begin{bmatrix}
\frac{1}{3} & 0 & \frac{9}{10} \\
\frac{1}{3} & 1 & \frac{1}{10} \\
\frac{1}{3} & 0 & 0
\end{bmatrix}
\approx \begin{bmatrix}
0 & 0 & 0 \\
1 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}\)

\(\Rightarrow A\)

Data Analysis, GEN2 2023 VCAA 2a

The following data shows the sizes of a sample of 20 oysters rated as small, medium or large.

\begin{array} {ccccc}
\text{small} & \text{small} & \text{large} & \text{medium} & \text{medium} \\
\text{medium} & \text{large} & \text{small} & \text{medium} & \text{medium}\\
\text{small} & \text{medium} & \text{small} & \text{small} & \text{medium}\\
\text{medium} & \text{medium} & \text{medium} & \text{small} & \text{large}
\end{array}

  1. Use the data above to complete the following frequency table.   (1 mark)

  1. Use the percentages in the table to construct a percentage segmented bar chart below. A key has been provided.   (1 mark)

     

Show Answers Only

i.    

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}

ii.    
     

Show Worked Solution

i.    

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}

 
ii.    
         

Data Analysis, GEN2 2023 VCAA 2b

An oyster farmer has two farms, \(\text{A}\) and \(\text{B}\).

She takes a random sample of oysters from each of the farms and has the oysters classified as small, medium or large.

The number of oysters of each size is displayed in the two-way table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Oyster size} \rule[-1ex]{0pt}{0pt} & \textbf{Farm A} & \textbf{Farm B}  \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 42 & 114 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 124 & 160 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 44 & 46 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 210 & 320 \\
\hline
\end{array}

  1. Calculate the percentage of the total number of oysters graded as 'large' in this investigation. Round the percentage to the nearest whole number.   (1 mark)

  2. The farmer believes that farm \(\text{A}\) has a greater capacity to grow larger oysters than farm \(\text{B}\). Does the information in the table support the farmer's belief? Explain your conclusion by comparing the values of two appropriate percentages.
  3. Round these percentages to the nearest whole number.   (2 marks)

Show Answers Only

i.    \(\text{Large oysters}\ = \dfrac{44}{210} = \dfrac{90}{530} = 0.1698 = 17\% \)

ii.  \(\text{Large oysters (farm A)}\ = \dfrac{44}{210} = 0.2095 = 21\% \)

\(\text{Large oysters (farm B)}\ = \dfrac{46}{320} = 0.1437 = 14\% \)

\(\therefore\ \text{Since farm A produces a much higher percentage of large oysters,}\)

\(\text{the information in the table supports the farmer’s belief.}\)

Show Worked Solution

i.    \(\text{Large oysters}\ = \dfrac{44}{210} = \dfrac{90}{530} = 0.1698 = 17\% \)
 

ii.  \(\text{Large oysters (farm A)}\ = \dfrac{44}{210} = 0.2095 = 21\% \)

\(\text{Large oysters (farm B)}\ = \dfrac{46}{320} = 0.1437 = 14\% \)
 

\(\therefore\ \text{Since farm A produces a much higher percentage of large oysters,}\)

\(\text{the information in the table supports the farmer’s belief.}\)

Data Analysis, GEN2 2023 VCAA 1

Data was collected to investigate the use of electronic images to automate the sizing of oysters for sale. The variables in this study were:

    • ID: identity number of the oyster
    • weight: weight of the oyster in grams (g)
    • volume: volume of the oyster in cubic centimetres (cm³)
    • image size: oyster size determined from its electronic image (in megapixels)
    • size: oyster size when offered for sale: small, medium or large

The data collected for a sample of 15 oysters is displayed in the table.
 

  1. Write down the number of categorical variables in the table.   (1 mark)

  2. Determine, in grams:
    1. the mean weight of all the oysters in this sample.   (1 mark)

    2. the median weight of the large oysters in this sample.   (1 mark)
  3. When a least squares line is used to model the association between oyster weight and volume, the equation is: 

    1. \(\textit{volume} = 0.780 + 0.953 \times \textit{weight} \)
    1. Name the response variable in this equation.   (1 mark)

    2. Complete the following sentence by filling in the blank space provided.   (1 mark)

      This equation predicts that, on average, each 10 g increase in the weight of an oyster is associated with a ________________ cm³ increase in its volume.
  1. A least squares line can also be used to model the association between an oyster's volume, in cm³, and its electronic image size, in megapixels. In this model, image size is the explanatory variable.
  2. Using data from the table, determine the equation of this least squares line. Use the template below to write your answer. Round the values of the intercept and slope to four significant figures.   (2 marks)

  3. The number of megapixels needed to construct an accurate electronic image of an oyster is approximately normally distributed.
  4. Measurements made on recently harvested oysters showed that:
    • 97.5% of the electronic images contain less than 4.6 megapixels
    • 84% of the electronic images contain more than 4.3 megapixels.
  1. Use the 68-95-99.7% rule to determine, in megapixels, the mean and standard deviation of this normal distribution.   (2 marks)

Show Answers Only

a.    \(\text{Categorical variables = 2 (ID and size)}\)

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)

b.ii.  \(\text{Median}\ = 11.4 \)

c.i.   \(\text{Volume}\)

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)

d.    \(\text{Volume}\ = 0.002857 + 2.571 \times \text{image size} \)

e.    \(s_x = 0.1 \)

\(\bar x = 4.4\) 

Show Worked Solution

a.    \(\text{Categorical variables = 2 (ID and size)}\)
 

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)
 

b.ii.  \(\text{15 data points}\ \ \Rightarrow \ \ \text{Median = 8th data point (in order)}\)

 \(\text{Median}\ = 11.4 \)
 

c.i.   \(\text{Volume}\)
 

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)
 

d.    \(\text{Input the image size column values}\ (x)\ \text{and volume} \)

\(\text{values}\ (y)\ \text{into the calculator:}\)

\(\textit{Volume}\ = 0.002857 + 2.571 \times \textit{image size} \)
 

e.    \(z\text{-score (4.6)}\ = 2\ \ \Rightarrow \bar x + 2 \times s_x = 4.6\ …\ (1)\)

\(z\text{-score (4.3)}\ = -1\ \ \Rightarrow \bar x-s_x = 4.3\ …\ (2)\)

\( (1)-(2) \)

\(3 s_x = 0.3 \ \ \Rightarrow\ \ s_x = 0.1 \)

\(\bar x = 4.4\) 

Recursion and Finance, GEN1 2023 VCAA 20-21 MC

For taxation purposes, Audrey depreciates the value of her $3000 computer over a four-year period. At the end of the four years, the value of the computer is $600.
 

Question 20

If Audrey uses flat rate depreciation, the depreciation rate, per annum is

  1. 10%
  2. 15%
  3. 20%
  4. 25%
  5. 33%

 
Question 21

If Audrey uses reducing balance depreciation, the depreciation rate, per annum is closest to

  1. 10%
  2. 15%
  3. 20%
  4. 25%
  5. 33%
Show Answers Only

\(\text{Question 20:}\ C\)

\(\text{Question 21:}\ E\)

Show Worked Solution

\(\text{Question 20}\)

\(\text{Depreciation value}\ = 3000-600=$2400 \)

\(\text{Depreciation value (per year)}\ = \dfrac{2400}{4} =$600 \)

\(\text{Depreciation rate}\ = \dfrac{600}{3000} \times 100 =20\% \)

\(\Rightarrow C\)
 

\(\text{Question 21}\)

\(A = $600, \ P= $3000,\ n=4\)

\(A\) \(=PR^n\)  
\(600\) \(=3000 \times R^4\)  
\(R^4\) \(=\dfrac{600}{3000} \)  
\(R\) \(=\sqrt[4]{0.2} \)  
  \(=0.668…\)  

 
\(\text{Depreciation rate}\ =1-0.668… = 0.331… \approx 33\% \)

\(\Rightarrow E\)

Recursion and Finance, GEN1 2023 VCAA 18-19 MC

Gus purchases a coffee machine for $15 000 and depreciates its value using the unit cost method.

The rate of depreciation is $0.04 per cup of coffee made.

A recurrence relation that models the year-to-year value \(G_n\), in dollars, of the machine is
 

Question 18

A rule for \(G_n\), the value of the machine after \(n\) years is

  1. \(G_n=15\ 000-0.04 n\)
  2. \(G_n=15\ 000+0.04 n\)
  3. \(G_n=15\ 000-1314 n\)
  4. \(G_n=1314-0.04 n\)
  5. \(G_n=1314+0.04 n\)

 
Question 19

The number of cups made by the machine per year is

  1. \(1314\)
  2. \(13\ 686\)
  3. \(15\ 000\)
  4. \(31\ 536\)
  5. \(32\ 850\)
Show Answers Only

\(\text{Question 18:}\ C\)

\(\text{Question 19:}\ E\)

Show Worked Solution

\(\text{Question 18}\)

\(G_n =\ \text{cost}-(\text{cups of coffee}\ \times 0.04) \times n \)

\(G_n =\ 15\ 000-(\text{cups of coffee}\ \times 0.04) \times n \)

\(\text{Option C is the only possible correct answer.}\)

\(\Rightarrow C\)
 

\(\text{Question 19}\)

\(\text{Using the}\ G_n\ \text{rule from question 18:}\)

\(\text{cups of coffee}\ \times 0.04\) \(=1314\)  
\(\text{cups of coffee}\) \(=\dfrac{1314}{0.04}\)  
  \(=32\ 850\)  

 
\(\Rightarrow E\)

Data Analysis, GEN1 2023 VCAA 13-14 MC

The following graph shows a selection of winning times, in seconds, for the women's 800 m track event from various athletic events worldwide. The graph shows one winning time for each calendar year from 2000 to 2022.
 

Question 13

The time series is smoothed using seven-median smoothing.

The smoothed value for the winning time in 2006, in seconds, is closest to

  1. 116.0
  2. 116.4
  3. 116.8
  4. 117.2
  5. 117.6

 
Question 14

The median winning time, in seconds, for all the calendar years from 2000 to 2022 is closest to

  1. 116.8
  2. 117.2
  3. 117.6
  4. 118.0
  5. 118.3
Show Answers Only

\(\text{Question 13:}\ C\)

\(\text{Question 14:}\ B\)

Show Worked Solution

\(\text{Question 13}\)

\(\text{Consider the 2006 data point and 3 data points either side.}\)

\(\text{Median value (of 7 data points) = 116.8}\)

\(\Rightarrow C\)
 

\(\text{Question 14}\)

\(\text{23 data points between 2000 – 2022.}\)

\(\text{Median value = 12th data point (in order) = 117.2}\)

\(\Rightarrow B\)

Data Analysis, GEN1 2023 VCAA 10 MC

A study of Year 10 students shows that there is a negative association between the scores of topic tests and the time spent on social media. The coefficient of determination is 0.72

From this information it can be concluded that

  1. a decreased time spent on social media is associated with an increased topic test score.
  2. less time spent on social media causes an increase in topic test performance.
  3. an increased time spent on social media is associated with an increased topic test score.
  4. too much time spent on social media causes a reduction in topic test performance.
  5. a decreased time spent on social media is associated with a decreased topic test score.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Negative association:}\)

  • \(\text{does not mean causality (eliminate B and D).}\)
  • \(\text{Option A: if time spent on social media ↓, topic test scores ↑}\ \checkmark \)
  • \(\text{Option C: if time spent on social media ↑, topic test scores ↓}\ \cross \)
  • \(\text{Option E: if time spent on social media ↓, topic test scores ↓}\ \cross \)

\(\Rightarrow A\)

Data Analysis, GEN1 2023 VCAA 7-8 MC

A teacher analysed the class marks of 15 students who sat two tests.

The test 1 mark and test 2 mark, all whole number values, are shown in the scatterplot below.

A least squares line has been fitted to the scatterplot.
 

Question 7

The equation of the least squares line is closest to

  1. test 2 mark = – 6.83 + 1.55 × test 1 mark
  2. test 2 mark = 15.05 + 0.645 × test 1 mark
  3. test 2 mark = – 6.78 + 0.645 × test 1 mark
  4. test 2 mark = 1.36 + 1.55 × test 1 mark
  5. test 2 mark = 6.83 + 1.55 × test 1 mark

 
Question 8

The least squares line shows the predicted test 2 mark for each student based on their test 1 mark.

The number of students whose actual test 2 mark was within two marks of that predicted by the line is

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7
Show Answers Only

\(\text{Question 7:}\ A\)

\(\text{Question 8:}\ C\)

Show Worked Solution

\(\text{Question 7}\)

\(\text{By inspection, gradient is greater than 1 (eliminate B and C)}\)

\(\text{LSRL passes through (16, 18):}\)

\(\text{Option A:}\ -6.83 + 1.55 \times 16 = 18.0\ \checkmark \)

\(\Rightarrow A\)
 

\(\text{Question 8}\)

\(\text{5 values are within 1 grid height (measured vertically), or 2 marks,}\)

\(\text{from the LRSR.}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2023 VCAA 6 MC

The histogram below displays the distribution of prices, in dollars, of the cars for sale in a used-car yard.

The histogram has a logarithm (base 10) scale.
 

Six of the cars in the yard have the following prices:

\($2450, \ $3175, \ $4999, \ $8925, \ $10\ 250, \ $105\ 600\)

How many of the six car prices listed above are in the modal class interval?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 6
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Modal class interval is 3.5 – 4.0}\)

\(\text{Take the log}_{10}\ \text{of each car price:}\)

\(\log_{10}2450 = 3.389\ \cross, \ \log_{10}3175 = 3.501\ \checkmark\)

\(\log_{10}4999 = 3.69\ \checkmark, \ \log_{10}8925 = 3.95\ \checkmark\)

\(\log_{10}10\ 250 = 4.01\ \cross, \ \log_{10}105\ 600 = 5.02\ \cross\)

\(\Rightarrow C\)

Data Analysis, GEN1 2023 VCAA 4 MC

The time spent by visitors in a museum is approximately normally distributed with a mean of 82 minutes and a standard deviation of 11 minutes.

2380 visitors are expected to visit the museum today.

Using the 68–95–99.7% rule, the number of these visitors who are expected to spend between 60 and 104 minutes in the museum is

  1. 1128
  2. 1618
  3. 2256
  4. 2261
  5. 2373
Show Answers Only

\(D\)

Show Worked Solution

\(\bar x = 82, \ s_x = 11\)

\(z\text{-score (60)} = \dfrac{x-\bar x}{s_x} = \dfrac{60-82}{11} = -2 \)

\(z\text{-score (104)} = \dfrac{104-82}{11} = 2 \)
 

\(\text{95% of data points lie between}\ z=\pm 2 \)

\(\text{Number of visitors}\ = 95\% \times 2380 = 2261\)

\(\Rightarrow D\)

Data Analysis, GEN1 2023 VCAA 3 MC

Gemma’s favourite online word puzzle allows her 12 attempts to guess a mystery word. Her number of attempts for the last five days is displayed in the table below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Day} \rule[-1ex]{0pt}{0pt} & \textbf{Number of attempts} \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} 4 \rule[-1ex]{0pt}{0pt} & 6 \\
\hline
\rule{0pt}{2.5ex} 5 \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\end{array}

On day six, how many attempts can she make so that the mean number of attempts for these six days is exactly eight?

  1. 5
  2. 6
  3. 7
  4. 8
  5. 9
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Let}\ x=\ \text{number of attempts on day 6}\)

\(8\) \(=\dfrac{x+8+11+5+6+9}{6}\)  
\(8 \times 6\) \(=x+39\)  
\(x\) \(=48-39\)  
  \(=9\)  

 
\(\Rightarrow E\)

Data Analysis, GEN1 2023 VCAA 1-2 MC

The dot plot below shows the times, in seconds, of 40 runners in the qualifying heats of their 800 m club championship.
 

Question 1

The median time, in seconds, of these runners is

  1. 135.5
  2. 136
  3. 136.5
  4. 137
  5. 137

 
Question 2

The shape of this distribution is best described as

  1. positively skewed with one or more possible outliers.
  2. positively skewed with no outliers.
  3. approximately symmetric with one or more possible outliers.
  4. approximately symmetric with no outliers.
  5. negatively skewed with one or more possible outliers.
Show Answers Only

\(\text{Question 1:}\ B\)

\(\text{Question 2:}\ A\)

Show Worked Solution

\(\text{Question 1}\)

\(\text{40 data points}\ \Rightarrow \ \text{Median = average of 20th and 21st data points}\)

\(\text{Median}\ = \dfrac{136 + 136}{2} = 136\)

\(\Rightarrow B\)
 

\(\text{Question 2}\)

\(\text{Distribution is positive skewed (tail stretches to the right)} \)

\(\text{Q}_1 = \dfrac{135+135}{2} = 135\)

\(\text{Q}_3 = \dfrac{138+138}{2} = 138\)

\(\text{IQR} = 138-135=3 \)

\(\text{Outlier (upper fence)}\ = 138+ 1.5 \times 3 = 142.5\)

\(\Rightarrow A\)

PHYSICS, M2 EQ-Bank 3

In the system diagram below, a 5-kilogram mass and masses \(A\) and \(B\) are held by high tensile frictionless wire in static equilibrium.
 

Using a vector diagram, calculate the masses of both \(A\) and \(B\).   (4 mark)

Show Answers Only

\(\text{Mass}_A =3.91\ \text{kg}\)

\(\text{Mass}_B =4.55\ \text{kg}\)

Show Worked Solution

Using the sin rule both \(F_B\) and \(F_A\) can be calculated:

\(\dfrac{F_A}{\sin 48^{\circ}}\) \(=\dfrac{5 \times 9.8}{\sin 72^{\circ}}\)  
\(F_A\) \(=\dfrac{49\,\sin 48^{\circ}}{\sin 72^{\circ}}\)  
  \(=38.3\ \text{N}\)  

 
\(\text{Mass}_A =\dfrac{F}{a}=\dfrac{38.3}{9.8}=3.91\ \text{kg}\)

 

\(\dfrac{F_B}{\sin 60^{\circ}}\) \(=\dfrac{49}{\sin 72^{\circ}}\)
\(F_B\) \(=\dfrac{49\, \sin 60^{\circ}}{\sin 72^{\circ}}\)
  \(=44.6\ \text{N}\)

 

\(\text{Mass}_B =\dfrac{F}{a}=\dfrac{44.6}{9.8}=4.55\ \text{kg}\)

PHYSICS, M2 EQ-Bank 1

A concrete block with a weight of 1000 \(\text{N}\) is lifted with a pulley and chain system.

Initially the block is suspended by the chain as seen below. The block is then pulled to the side using a rope and makes an angle of 40° with the vertical as shown below. Ignore the masses of the chain and rope.
 

  1. State the tension in the chain initially.   (1 mark)
  2. Draw a free body diagram representing the forces acting on the concrete block in the final situation.   (2 marks)
  3. Determine the magnitude of the force being applied on the rope and the tension in the chain in the final situation.   (3 marks)
Show Answers Only

a.    \(1000\ \text{N up.}\)

b. 
             

c.    \(F_{\text{rope}}=839\ \text{N}\)

       \(T_{\text{chain}}=1305.4\ \text{N}\)

Show Worked Solution

a.     → Tension is equal and opposite to the weight force of 1000 \(\text{N}\) down.

        → Hence the tension in the chain is 1000 \(\text{N}\) up.

 
b. 
             

 
c. 
       

Using the triangle above:

\(\tan 40^{\circ}\) \(=\dfrac{F_{\text{rope}}}{1000}\)  
\(F_{\text{rope}}\) \(=1000 \times \tan 40^{\circ}\)  
  \(=839\ \text{N}\)  

 

\( \cos 40^{\circ}\) \(=\dfrac{1000}{T_{\text{chain}}}\)  
\(T_{\text{chain}}\) \(=\dfrac{1000}{\cos 40^{\circ}}\)  
  \(=1305.4\ \text{N}\)  

PHYSICS, M4 EQ-Bank 2

The following circuit has 4 resistors and a 24 V DC supply with a switch.
 

  1. Calculate the value of the single resistor that is equivalent to the 5 \(\Omega\), 20 \(\Omega\), and 40 \(\Omega\) resistors.   (2 marks)

  2. Determine the potential difference across the 15 \(\Omega\) resistor when the switch is closed.   (2 mark)

  3. Determine the current through the 5 \(\Omega\) resistor when the switch is closed.   (2 mark)

Show Answers Only

a.    \(4.6\ \Omega\)

b.    \(18.37\ \text{V}\)

c.   \(1.126\ \text{A}\)

Show Worked Solution
a.     \(R_{series}\) \(= 20 + 40 = 60\ \Omega\)
  \(\dfrac{1}{R_T}\) \(=\dfrac{1}{60}+\dfrac{1}{5}\)
    \(=\dfrac{13}{60}\)
  \(R_T\) \(=\dfrac{60}{13}\)
    \(=4.6\ \Omega\)

 

b.   \(R_{circuit} = 4.6 +15 = 19.6\ \Omega\)

Circuit current:

\(I=\dfrac{V}{R}=\dfrac{24}{19.6}=1.2245\ \text{A}\)
 

Potential difference across the 15 \(\Omega\) resistor:

\(V=IR=1.2245 \times 15=18.37\ \text{V}\)
 

c.   Potential difference across the 5 \(\Omega\) resistor: 

\(V_{(5\ \Omega)}=24-18.37= 5.63\ \text{V}\)
 

Current through the \(5\ \Omega\) resistor:

\(I=\dfrac{V}{R}=\dfrac{5.63}{5}=1.126\ \text{A}\)

PHYSICS, M4 EQ-Bank 1

Determine the magnetic field intensity within a solenoid with 20 coils and a length of 15 cm, given that a direct current of 4 amperes flows through it.   (3 marks)

Show Answers Only

\(6.7 \times 10^{-4}\ \text{T}\)

Show Worked Solution
\(B\) \(=\dfrac{\mu_0 N I}{L}\)  
  \(=\dfrac{4\pi \times 10^{-7} \times 20 \times 4}{0.15}\)  
  \(=6.7 \times 10^{-4}\ \text{T}\)  

PHYSICS, M4 2023 VCE 1

Some physics students are conducting an experiment investigating both electrostatic and gravitational forces. They suspend two equally charged balls, each of mass 4.0 g, from light, non-conducting strings suspended from a low ceiling.

The charged balls repel each other with the strings at an angle of 60°, as shown in Figure 1.
 

There are three forces acting on each ball:

    • a tension force, \(T\)
    • a gravitational force, \(F_{g}\)
    • an electrostatic force, \(F_{E}\).
  1. On Figure 1, using the labels \(T, F_{g}\) and \(F_{E}\), draw each of the three forces acting on each of the charged balls.  (3 marks)

  2. Show that the tension force, \(T\), in each string is \(4.5 \times 10^{-2} \text{ N}\). Use  \(g=9.8 \text{ N kg}^{-1}\).
  3. Show your working.  (2 marks)

  4. Calculate the magnitude of the electrostatic force, \(F_{ E }\). Show your working.  (2 marks)

Show Answers Only

a.   
       

 
b.    System is in an equilibrium state:

→ The sum of the forces must add to zero as seen in the triangle below.
  

→ \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{T}\)  
\(T\) \(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)  
  \(=4.5 \times 10^{-2}\ \text{N}\)  

 
c.   Using the same triangle as part (b): 

\(\tan\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{F_E}\)  
\(F_E\) \(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)  
  \(=2.3 \times 10^{-2}\ \text{N}\)  

Show Worked Solution

a.   
       

 
b.    System is in an equilibrium state:

→ The sum of the forces must add to zero as seen in the triangle below.
  

→ \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{T}\)  
\(T\) \(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)  
  \(=4.5 \times 10^{-2}\ \text{N}\)  

 

c.   Using the same triangle as part (b): 

\(\tan\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{F_E}\)  
\(F_E\) \(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)  
  \(=2.3 \times 10^{-2}\ \text{N}\)  
♦ Mean mark 41%.

BIOLOGY, M7 2022 VCE 27 MC

Which of the following cells responds first when a person receives the MMR vaccination?

  1. helper \( \text{T} \) cells
  2. memory \( \text{T} \) cells
  3. cytotoxic \( \text{T} \) cells
  4. antigen-presenting cells
Show Answers Only

\(D\)

Show Worked Solution

→ As the MMR vaccine contains a weakened version of the Measles, Mumps and Rubella viruses, it is the antigen-presenting cells which first respond.

\(\Rightarrow D\)

PHYSICS, M3 2023 VCE 12

A ray of monochromatic light is incident on a triangular glass prism with a refractive index of 1.52 . The ray is perpendicular to the side \(\text{AB}\) of the glass prism, as shown in Figure 14.
  

The ray of light travels through the glass prism before reaching side \(\text{AC}\).

  1. Calculate the critical angle for the glass prism at the glass-air interface.  (2 marks)

  2. Will the ray of light undergo total internal reflection at side \(\text{AC}\) of the glass prism? Justify your answer.  (2 marks)

Show Answers Only

a.    41°

b.    As the angle of incidence 45° is greater than the critical angle 41°, the total internal reflection will occur on side \(\text{AC}\).

Show Worked Solution

a.     \(\sin \theta_c\) \(=\dfrac{n_2}{n_1}\)
  \(\theta_c\) \(=\sin^{-1}(\dfrac{n_2}{n_1})\)
    \(=\sin^{-1}(\dfrac{1.0}{1.52})\)
    \(=41^{\circ}\)

 

b.    Total internal reflection:

→ The angle of incidence 45° is greater than the critical angle 41°.

→ Therefore, total internal reflection will occur on side \(\text{AC}\).

♦ Mean mark (b) 46%.

PHYSICS, M3 2023 VCE 11

A guitar string of length 0.75 m and fixed at both ends is plucked and a standing wave is produced. The envelope of the standing wave is shown in Figure 13.
 

The speed of the wave along the string is 393 m s\( ^{-1}\).

  1. What is the frequency of the wave?  (1 mark)

  2. Describe how the standing wave is produced on the string fixed at both ends.  (2 marks)

Show Answers Only

a.    262 Hz

b.   Standing wave:

→ When waves encounter fixed ends, they reflect.

→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern

Show Worked Solution

a.   \(f=\dfrac{v}{\lambda}=\dfrac{393}{1.5}=262\ \text{Hz}\)
 

b.   Standing wave:

→ When waves encounter fixed ends, they reflect.

→ If the string’s length is a multiple of half the wavelength, the reflected wave combines with the original wave, resulting in interference that forms a standing wave pattern

♦ Mean mark (b) 47%.

PHYSICS, M2 2023 VCE 8

Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in Figure 8. The slope is 5° to the horizontal.

The combined mass of Maia and the skateboard is 65 kg.
  

  1. In Figure 9, the combined system of Maia and the skateboard is modelled as a small box with point \(\text{M}\) at the centre of mass.
  2. Draw and label arrows to represent each of the forces acting on the system - that is, Maia and skateboard as they roll down the slope.  (3 marks)
     


  1. Calculate the magnitude of the total frictional forces acting on Maia and the skateboard.  (2 marks)

Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy.

  1. Explain what happened to both the momentum and the kinetic energy of Maia and the skateboard. No calculations are required.  (2 marks)

Show Answers Only

a.
       

b.     55.5 N

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

Show Worked Solution

a.
       

b.     Total frictional forces:

→ Constant speed means that the force down the slope of the incline is equal to the sum of the frictional forces acting on Maia and the skateboard.

\(F_{\text{down slope}}\) \(=F_f\) \(=mg\, \sin\, \theta\)
    \(=65 \times 9.8 \times \sin\,5^{\circ}\)
    \(=55.5\ \text{N}\)

 

♦ Mean mark (b) 50%.

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

♦♦♦ Mean mark (c) 26%.

BIOLOGY, M4 EQ-Bank 40

"Close monitoring is critical for endangered species and at-risk ecosystems."

Provide an example of an endangered species and an at-risk ecosystem and suggest a policy response that could arise from such monitoring.   (4 marks)

Show Answers Only

Endangered species (one of many possibilities):

→ Tasmanian Devils are an endangered species that has been decimated by Devil Facial Tumor Disease (DFTD) and is intensely monitored.

→ Monitoring provides accurate information of their population numbers and allows informed policy to be made, such as the establishment of Devil’s Ark for devil breeding by non-infected individuals.
 

At-risk ecosystem (one of many possibilities):

→ The Tarkine rainforest in Tasmania is a closely monitored ecosystem.

→ This ecosystem is endangered from climate change with the forest floor underbrush becoming an increasing fire hazard as vegetation dries out. Monitoring allows for the development of an effective bushfire strategy in the area.

Show Worked Solution

Endangered species (one of many possibilities):

→ Tasmanian Devils are an endangered species that has been decimated by Devil Facial Tumor Disease (DFTD) and is intensely monitored.

→ Monitoring provides accurate information of their population numbers and allows informed policy to be made, such as the establishment of Devil’s Ark for devil breeding by non-infected individuals.
 

At-risk ecosystem (one of many possibilities):

→ The Tarkine rainforest in Tasmania is a closely monitored ecosystem.

→ This ecosystem is endangered from climate change with the forest floor underbrush becoming an increasing fire hazard as vegetation dries out. Monitoring allows for the development of an effective bushfire strategy in the area.

PHYSICS, M3 2023 VCE 12 MC

A physics class is investigating the dispersion of white light using a lens, as shown in the diagram below.
 

The students observe the rays \(\text{K–P}\) that have been refracted by the lens.

Which one of the following correctly identifies the colour, red \(\text{(R)}\), green \(\text{(G)}\) or violet \(\text{(V)}\), of the rays \(\text{K–P}\)?
 

  \(\textbf{K}\) \(\textbf{L}\) \(\textbf{M}\) \(\textbf{N}\) \(\textbf{O}\) \(\textbf{P}\)
\(\textbf{A.}\) \(\quad \text{R}\quad \) \(\quad \text{G}\quad \) \(\quad \text{V}\quad \) \(\quad \text{V}\quad \) \(\quad \text{G}\quad \) \(\quad \text{R}\quad \)
\(\textbf{B.}\) \(\text{V}\) \(\text{G}\) \(\text{R}\) \(\text{R}\) \(\text{G}\) \(\text{V}\)
\(\textbf{C.}\) \(\text{V}\) \(\text{G}\) \(\text{R}\) \(\text{V}\) \(\text{G}\) \(\text{R}\)
\(\textbf{D.}\) \(\text{V}\) \(\text{R}\) \(\text{G}\) \(\text{G}\) \(\text{R}\) \(\text{V}\)
Show Answers Only

\(B\)

Show Worked Solution

→ The shorter wavelengths of light (violet) experience a greater refraction so the light rays of \(P\) and \(K\) will be violet. 

→ The longer wavelength light (red) experiences the least refraction and so will be rays \(M\) and \(N\).

\(\Rightarrow B\)

BIOLOGY, M4 EQ-Bank 39

"Biodiversity is the cornerstone of a healthy planet."

Discuss this statement, outlining the key advantages of preserving biodiversity for ecosystems and human well-being?   (4 marks)

Show Answers Only

→ Making the effort to preserve biodiversity creates a range of important benefits to both the ecosystem and us.

→ A well preserved biodiversity provides enormous economic value to society through the bio-resources it produces, such as food, timber, fibres and medicines.

→ Rich biodiversity and a healthy ecosystem also indirectly benefit many other natural processes such as pollination, maintaining clean water and productive soil.

→ Maintaining biodiversity is integral to preserving wild life and ecosystems. By disrupting ecosystems, food webs and chains can be affected, causing drastic changes in populations and even extinction. Preserving ecosystems maintains the aesthetic beauty of nature for current and future generations.

Show Worked Solution

→ Making the effort to preserve biodiversity creates a range of important benefits to both the ecosystem and us.

→ A well preserved biodiversity provides enormous economic value to society through the bio-resources it produces, such as food, timber, fibres and medicines.

→ Rich biodiversity and a healthy ecosystem also indirectly benefit many other natural processes such as pollination, maintaining clean water and productive soil.

→ Maintaining biodiversity is integral to preserving wild life and ecosystems. By disrupting ecosystems, food webs and chains can be affected, causing drastic changes in populations and even extinction. Preserving ecosystems maintains the aesthetic beauty of nature for current and future generations.

PHYSICS, M2 2023 VCE 8 MC

At a swimming pool, Sharukh and Sam, shown below, step off the low diving board at the same time. Over the small distance they fall, air resistance may be ignored. Sharukh and Sam have masses of 80 kg and 60 kg respectively.
 

Which one of the following best explains what happens to Sharukh and Sam as they drop straight down into the water?

  1. Sharukh reaches the surface first because she has a larger mass.
  2. The net force on Sharukh is larger than that on Sam, so Sharukh reaches the surface first.
  3. They both reach the surface together because they both experience the same downward force.
  4. They both reach the surface together because they both experience the same downward acceleration.
Show Answers Only

\(D\)

Show Worked Solution

→ Both Sharukh and Sam fall under the acceleration of gravity which is 9.8 ms\(^{-2}\).

→ They will fall at the same speed and therefore hit the surface at the same time.

\(\Rightarrow D\)

PHYSICS, M3 2017 VCE 16

Standing waves are formed on a string of length 4.0 m that is fixed at both ends. The speed of the waves is 240 m s\(^{-1}\).

  1. Calculate the wavelength of the lowest frequency resonance.  (2 marks)

  2. Calculate the frequency of the second-lowest frequency resonance.  (2 marks)

  3. Explain the physics of how standing waves are formed on the string. Include a diagram in your response.  (3 marks)

Show Answers Only

a.    30 Hz

b.    60 Hz

c.    Standing waves:

→ Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave.

→ The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude.


       

Show Worked Solution

a.     Lowest frequency resonance:

→ Occurs at the maximum wavelength. The maximum wavelength is 8 metres since the half wavelength is the length of the string (4m).

\(f=\dfrac{v}{\lambda}=\dfrac{240}{8}=30\ \text{Hz}\)
 

b.     When \(\lambda = 4: \)

\(f=\dfrac{v}{\lambda}=\dfrac{240}{4}=60\ \text{Hz}\)
 

c.    Standing waves:

→ Produced when a wave travels down a string and is reflected back on itself such that the superposition of the two waves produce an interference pattern to form a standing wave.

→ The two waves must be travelling in opposite directions with the same frequency, wavelength and amplitude.


       

♦ Mean mark (c) 43%.

CHEMISTRY, M4 EQ-Bank 36

The enthalpies of formation for a number of chemical reactions are as follows:

\(\ce{C6H12O6(s)}\)                  \(\Delta H^{\circ}_f = -1271\ \text{kJmol}^{-1}\)

\(\ce{C2H5OH(aq)}\)                \(\Delta H^{\circ}_f = -277.7\ \text{kJmol}^{-1}\)

\(\ce{CO2(g)}\)                         \(\Delta H^{\circ}_f = -393.5\ \text{kJmol}^{-1}\)

Calculate the enthalpy change for the fermentation of glucose (reaction below) using the enthalpies of formation above.

\(\ce{C6H12O6(s) \rightarrow 2C2H5OH(aq) + 2CO2(g)}\)   (3 marks)

Show Answers Only

\(\Delta H^{\circ}_{\text{reaction}} = -71.4\ \text{kJmol}^{-1}\)

Show Worked Solution

\(\Sigma\ \text{Enthalpy (reactants)}\ = 1 \times -1271 = -1271\ \text{kJ}\)

\(\Sigma\ \text{Enthalpy (products)}\ = 2 \times -277.7 + 2 \times -393.5 = -1342.4\ \text{kJ}\)

\(\Delta H^{\circ}_{\text{reaction}}\) \(= \Delta H^{\circ} (\text{products})-\Delta H^{\circ} (\text{reactants}) \)  
  \(=-1342.4-(-1271)\)  
  \(= -71.4\ \text{kJ mol}^{-1}\)