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Algebra, STD2 A1 2007 HSC 19 MC v1

Which of the following correctly expresses  \(X\)  as the subject of  \(Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)?

  1. \(X=\dfrac{Y}{\pi}-L\)
  2. \(X=\dfrac{Y}{\pi}-4L\)
  3. \(X=4L-\dfrac{Y}{2\pi}\)
  4. \(X=\dfrac{Y}{8\pi}-\dfrac{L}{4}\)
Show Answers Only

\(B\)

Show Worked Solution
\(Y\) \(=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)
\(\dfrac{Y}{4\pi}\) \(=\dfrac{X}{4}+L\)
\(\dfrac{X}{4}\) \(=\dfrac{Y}{4\pi}-L\)
\(X\) \(=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)\)
\(X\) \(=\dfrac{Y}{\pi}-4L\)

 
\(\Rightarrow B\)

CHEMISTRY, M8 2022 VCE 5*

A chemist uses spectroscopy to identify an unknown organic molecule, Molecule \(\text{J}\), that contains chlorine.

The \({}^{13}\text{C NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

The infra-red (IR) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. Name the functional group that produces the peak at 168 ppm in the \({}^{13}\text{C NMR}\) spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum.   (2 marks)

The mass spectrum of Molecule \(\text{J}\) is shown below
 

  1. The molecular mass of Molecule \(\text{J}\) is 108.5
  2.  Explain the presence of the peak at 110 m/z.  (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. The \({ }^1 \text{H NMR}\) spectrum consists of two singlet peaks.
  2. What information does this give about the molecule?   (2 marks)

  3. Draw a structural formula for Molecule \(\text{J}\) that is consistent with the information provided in parts a–c.   (2 marks)

Show Answers Only

a.   Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

  • Molecule \(\text{J}\) must be an ester. 

b.    The peak at 110 m/z:

  • due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. 

c.    The two singlet peaks indicate:

  • two different hydrogen environments within the molecule.
  • there are no adjacent hydrogen environments.
  • The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. 

d.    Either of the two molecules shown below are correct:

Show Worked Solution

a.    Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

  • Molecule \(\text{J}\) must be an ester. 
♦ Mean mark (a) 41%.

b.    The peak at 110 m/z:

  • due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. 

c.    The two singlet peaks indicate:

  • two different hydrogen environments within the molecule.
  • there are no adjacent hydrogen environments.
  • The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. 
♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d.    Either of the two molecules shown below are correct:
 

♦ Mean mark (d) 40%.

CHEMISTRY, M8 2021 VCE 7*

Two students are given a homework assignment that involves analysing a set of spectra and identifying an unknown compound. The unknown compound is one of the molecules shown below.
 

The \(^{13}\text{C NMR}\) spectrum of the unknown compound is shown below.
 

  1. Based on the number of peaks in the \(^{13}\text{C NMR}\) spectrum above, which compound – \(\text{P, Q, R, S}\) or \(\text{T}\) – could be eliminated as the unknown compound?   (1 mark)

  2. The infra-red (IR) spectrum of the unknown compound is shown below.
     

  1. Identify which of the five compounds (1 or more) can be eliminated on the basis of the IR spectrum. Justify your answer using data from the IR spectrum.   (3 marks)

  2. The mass spectrum of the unknown compound is shown below.
     

  1. Write the chemical formula of the species that produces a peak at m/z = 43.   (1 mark)

  2. Explain why one molecule can produce multiple peaks on a mass spectrum.   (2 mark)

Show Answers Only

a.    The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

  • Compound \(T\) has 5 unique carbon environments and can be eliminated. 

b.    Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

  • Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).
  • Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\). 

c.i.  The unknown compound is compound \(\text{R}\).

  • The chemical species are fragments of the original compound with a positive charge.
  • The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\) 

 ii.   Ions of that molecule must be formed to produce peaks on the mass spectrum.

  • Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

Show Worked Solution

a.    The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

  • Compound \(T\) has 5 unique carbon environments and can be eliminated.

b.    Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

  • Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).
  • Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\). 
♦ Mean mark (b) 43%.
COMMENT: Recognise the difference between OH alcohols and OH acid groups.

c.i.   The unknown compound is compound \(\text{R}\).

  • The chemical species are fragments of the original compound with a positive charge.
  • The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\)
♦ Mean mark (c.i.) 50%.
COMMENT: Ions require a positive charge.

 ii.    Ions of that molecule must be formed to produce peaks on the mass spectrum.

  • Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.
♦ Mean mark (c.ii.) 39%.

CHEMISTRY, M8 2023 VCE 7-1*

Molecule \(\text{V}\) contains only carbon atoms, hydrogen atoms and one oxygen atom.

The mass spectrum of molecule \(\text{V}\) is shown below.
 

  1.  i.  State the molecular formula of molecule \(\text{V}\). Justify your answer by using the information in the mass spectrum.   (2 marks)

  2. ii.  State why there is a small peak at \(\text{m/z} =87\).   (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. State what information the doublet at 1.1 ppm provides about the structure of the molecule.   (1 mark)

The \({ }^{13} \text{C NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. In the space below, draw a structural formula of molecule \(\text{V}\) that is consistent with the information provided in parts a.-c.   (3 marks)

Show Answers Only

a.i.   Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\)

  • Indicated by the parent ion peak at 86 m/z.
  • Molecular formula: \(\ce{C5H10O}\)

a.ii.  Small peak at m/z = 87:

  • A carbon-13 isotope being present in the molecule.

b.    The doublet peak at 1.1 ppm:

  • Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom. 

c.   
       

Show Worked Solution

a.i.   Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\)

  • Indicated by the parent ion peak at 86 m/z.
  • Molecular formula: \(\ce{C5H10O}\) 

a.ii.  Small peak at m/z = 87:

  • A carbon-13 isotope being present in the molecule. 

b.    The doublet peak at 1.1 ppm:

  • Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom. 

c.    From the carbon NMR graph:

  • There are 4 carbon environments, one shifted above 200 ppm indicating a ketone or aldehyde.
  • As there are 5 carbons, 2 of the carbons must have the same environment. 

From the hydrogen NMR graph:

  • There are 3 hydrogen environments.
  • The septet peak indicates there are 6 hydrogens with the same chemical environment on adjacent carbon atoms.

♦ Mean mark (b) 50%.
♦ Mean mark (c) 43%.

CHEMISTRY, M8 2022 VCE 28 MC

The \({ }^{13}\text{C NMR}\) spectrum of an organic compound is shown below.

The organic compound could be

Show Answers Only

\(D\)

Show Worked Solution
  • The \({ }^{13}\text{C NMR}\) has five peaks indicating 5 different carbon environments within the molecule.
  • The peak at 140 indicates the presence of the \(\ce{C=C}\).

\(\Rightarrow D\)

♦ Mean mark 42%.
COMMENT: Solving by elimination is an effective strategy here.

CHEMISTRY, M8 2023 VCE 29 MC

Which one of the following statements about mass spectrometry is always correct?

  1. The relative molecular mass of a molecule is determined from the base peak.
  2. The peaks in a mass spectrum are caused by the presence of different isotopes.
  3. The base peak is formed when an uncharged species is removed from the molecule.
  4. The height of each peak in the mass spectrum is measured relative to the height of the base peak.
Show Answers Only

\(D\)

Show Worked Solution
  • The relative molecular mass of a molecule is determined from the peak with the largest m/z ratio (eliminate \(A\)).
  • The peaks in a mass spectrum is caused by the molecules splitting up into different ions as they are bombarded with electrons. Some peaks that differ by a value of 1 are caused by the presence of different isotopes but this is not for all peaks (eliminate \(B\)).
  • All peaks on the mass spectrometry graph are charged ions (eliminate \(C\)).
  • The base peak is the largest peak on a mass spectrometry graph and therefore the abundances of each peak are relative to the height of the base peak.

\(\Rightarrow D\)

♦ Mean mark 46%.

CHEMISTRY, M8 2023 VCE 16 MC

Consider the following molecule.
 

How many peaks will be observed in a \({ }^{13} \text{C NMR}\) spectrum of this molecule

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

\(C\)

Show Worked Solution
  • There are 7 different carbon environments in the molecule:

\(\Rightarrow C\)

♦ Mean mark 43%.

CHEMISTRY, M8 2014 VCE 16-17 MC

An atomic absorption spectrometer can be used to determine the level of copper in soils. The calibration curve below plots the absorbance of four standard copper solutions against the concentration of copper ions in ppm.

The concentrations of copper ions in the standard solutions were 1.0, 2.0, 3.0 and 4.0 mg L\(^{-1}\). (1 mg L\(^{-1}\) = 1 ppm)
 

Question 16

The concentration of copper in a test solution can be determined most accurately from the calibration curve if it is between

  1. 0.0 ppm and 5.0 ppm.
  2. 0.0 ppm and 4.0 ppm.
  3. 1.0 ppm and 4.0 ppm.
  4. 1.0 ppm and 5.0 ppm.


Question 17

If the test solution gave an absorbance reading of 0.40, what would be the concentration of copper ions in the solution in mol L\(^{-1}\)?

  1. 2.5
  2. 3.9 × 10\(^{-2}\)
  3. 3.9 × 10\(^{-5}\)
  4. 2.5 × 10\(^{-6}\)
Show Answers Only

\(\text{Question 16:}\ C\)

\(\text{Question 17:}\ C\)

Show Worked Solution

Question 16

  • The concentration of copper ions with be most accurate on the calibration curve in between the limits of the concentration of standard solutions used to produce the calibration curve.
  • Thus it will be most accurate between 1.0 ppm and 4.0 ppm.

\(\Rightarrow C\)
 

Question 17

From the graph: Absorbance of 0.4 → 2.5 ppm (2.5 mg L\(^{-1}\))

\(\ce{M(copper ions) = 63.55\ \text{g mol}^{-1}}\)

\(\ce{c(copper ions) = 2.5 \times 10^{-3}\ \text{g L}^{-1} = \dfrac{2.5 \times 10^{-3}}{63.55}=3.9 \times 10^{-5}\ \text{mol L}^{-1}}\)

\(\Rightarrow C\)

♦ Mean mark (Q17) 49%.

CHEMISTRY, M8 2015 VCE 4

UV-visible spectroscopy was used to measure the spectra of two solutions, \(\text{A}\) and \(\text{B}\). Solution \(\text{A}\) was a pink colour, while Solution \(\text{B}\) was a green colour.

The analyst recorded the absorbance of each solution over a range of wavelengths on the same axes. The resultant absorbance spectrum is shown below.
 

  1. If 10.00 mL of Solution \(\text{A}\) was mixed with 10.00 mL of Solution \(\text{B}\), which wavelength should be used to measure the absorbance of Solution \(\text{B}\) in this mixture? Justify your answer.   (2 marks)

The analyst used two sets of standard solutions and blanks to determine the calibration curves for the two solutions. The absorbances were plotted on the same axes. The graph is shown below.
 

  1. The analyst found that, when it was measured at the appropriate wavelength, Solution \(\text{A}\) had an absorbance of 0.2
  2. If Solution \(\text{A}\) was cobalt\(\text{(II)}\) nitrate, \(\ce{Co(NO3)2}\), determine its concentration in mg L\(^{–1}\)   (2 marks)
  3.    \(\ce{ M(Co(NO3)2) = 182.9 g mol^{–1} \quad \quad 1 mM = 10^{–3} M}\)
  4. In another mixture, the pink compound in Solution \(\text{A}\) and the green compound in Solution \(\text{B}\) each have a concentration of approximately 1.5 × 10\(^{-2}\) M.
  5. Could the analyst reliably use both of the calibration curves to determine the concentrations for Solution \(\text{A}\) and Solution \(\text{B}\) by UV-visible spectroscopy? Justify your answer.   (2 marks)

Show Answers Only

a.    Wavelength: 625 nm (600 to 650 was accepted)

  • At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\).

b.    \(1.8 \times 10^3\ \text{mgL}^{-1}\)

c.   \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

  • This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).
  • This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.

Show Worked Solution

a.    Wavelength: 625 nm (600 to 650 was accepted)

  • At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\). 

b.    From the graph:

\(\text{Absorbance of 0.2} \Rightarrow 10 \times 10^{-3}\ \text{mol L}^{-1}\)

\(\ce{n(Co(NO3)2) = 10 \times 10^{-3} \times 182.9 = 1.829\ \text{g L}^{-1} = 1.8 \times 10^3\ \text{mg L}^{-1}}\) 
 

c.   \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

  • This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).
  • This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.
♦♦ Mean mark (c) 30%.

CHEMISTRY, M8 2016 VCE 2*

A common iron ore, fool’s gold, contains the mineral iron pyrite, \(\ce{FeS2}\).

Typically, the percentage by mass of \(\ce{FeS2}\) in a sample of fool’s gold is between 90% and 95%. The actual percentage in a sample can be determined by gravimetric analysis.

The sulfur in \(\ce{FeS2}\) is converted to sulfate ions, \(\ce{SO4^2–}\) as seen below:

\(\ce{4FeS2 + 11O2 \rightarrow 2Fe2O3 + 8SO4^2-}\)

This is then mixed with an excess of barium chloride, \(\ce{BaCl2}\), to form barium sulfate, \(\ce{BaSO4}\), according to the equation

\(\ce{Ba^2+(aq) + SO4^2–(aq)\rightarrow BaSO4(s)}\)

When the reaction has gone to completion, the \(\ce{BaSO4}\) precipitate is collected in a filter paper and carefully washed. The filter paper and its contents are then transferred to a crucible. The crucible and its contents are heated until constant mass is achieved.

The data for an analysis of a mineral sample is as follows.
 

\(\text{initial mass of mineral sample}\) \(\text{14.3 g}\)
\(\text{mass of crucible and filter paper}\) \(\text{123.40 g}\)
\(\text{mass of crucible, filter paper and dry}\ \ce{BaSO4}\) \(\text{174.99 g}\)
\(\ce{M(FeS2)}\) \(\text{120.0 g mol}^{-1}\)
\(\ce{M(BaCl2)}\) \(\text{208.3 g mol}^{-1}\)
\(\ce{M(BaSO4)}\) \(\text{233.4 g mol}^{-1}\)
     
  1. Calculate the percentage by mass of \(\ce{FeS2}\) in this mineral sample.  (4 marks)

  2. State one assumption that was made in completing the calculations for this analysis.  (1 mark)

Show Answers Only

a.  \(\ce{\% FeS2} =92.7\%\ \text{(3 sig.fig.)}\)

b.    Answers could have included one of the following:

  • All the sulfur is converted to \(\ce{SO4^2-}\).
  • Precipitate was pure.
  • No \(\ce{BaSO4}\) was lost in washing the sample.
  • \(\ce{BaSO4}\) was the only precipitate.
  • The sample was fully dehydrated.
Show Worked Solution

a.    \(\ce{m(BaSO4 \text{final})} = 174.99-123.40=51.59\ \text{g}\)

\(\ce{n(BaSO4(s))}=\dfrac{51.59}{233.4}=0.221\ \text{mol}\)

\(\Rightarrow \ce{n(SO4^2-(aq))}=0.221\ \text{mol}\)
 

Molar ratio  \(\ce{FeS2 : SO4^2-} = 1:2\)

\(\Rightarrow \ce{n(FeS2)}=0.221 \times \dfrac{1}{2}=0.1105\ \text{mol}\)

\(\ce{m(FeS2)}=0.1105 \times 120.0 =13.26\ \text{g}\)

\(\ce{\% FeS2} = \dfrac{13.26}{14.3} \times 100=92.7\%\ \text{(3 sig.fig.)}\)
 

b.    Answers could have included one of the following:

  • All the sulfur is converted to \(\ce{SO4^2-}\).
  • Precipitate was pure.
  • No \(\ce{BaSO4}\) was lost in washing the sample.
  • \(\ce{BaSO4}\) was the only precipitate.
  • The sample was fully dehydrated.
♦ Mean mark (b) 39%.

CHEMISTRY, M3 2013 VCE 11*

The following is a student’s summary of catalysts. It contains some correct and incorrect statements.

    1. A catalyst increases the rate of a reaction.
    2. All catalysts are solids.
    3. The mass of a catalyst is the same before and after the reaction.
    4. All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.
    5. A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.
  1. Identify two correct statements.   (1 mark)

  2. Evaluate the student’s summary by identifying two incorrect statements. In each case, explain why it is incorrect.   (4 marks)

Show Answers Only

a.    Statements 1 and 3 are correct.

b.    Answers could include two of the following:

Incorrect statement: All catalysts are solids

  • Catalysts can be solids, liquids, gases or within aqueous solutions.

Incorrect statement: All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.

  • Although solid catalysts orient reactant particles to favor reactions, liquid and gaseous catalysts do not arrange reactants in this manner.

Incorrect statement: A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.

  • A catalyst does not change the relative energy contents of reactants and products.
  • It does, however, lower the activation energy/increases the proportion of successful collisions/provides an alternative reaction pathway with the same overall \(\Delta H.\)
Show Worked Solution

a.    Statements 1 and 3 are correct.

b.    Answers could include two of the following:

Incorrect statement: All catalysts are solids

  • Catalysts can be solids, liquids, gases or within aqueous solutions.

Incorrect statement: All catalysts align the reactant particles in an orientation that is favourable for a reaction to occur.

  • Although solid catalysts orient reactant particles to favor reactions, liquid and gaseous catalysts do not arrange reactants in this manner.

Incorrect statement: A catalyst lowers the enthalpy change of a reaction, enabling more particles to have sufficient energy to successfully react.

  • A catalyst does not change the relative energy contents of reactants and products.
  • It does, however, lower the activation energy/increases the proportion of successful collisions/provides an alternative reaction pathway with the same overall \(\Delta H.\)
♦ Mean mark (b) 51%.

CHEMISTRY, M4 2016 VCE 10a*

A senior Chemistry student created an experiment to calculate the molar heat of combustion of butane.

The experimental steps are as follows:

    1. Measure the initial mass of a butane canister
    2. Measure the mass of a metal can, add 250 mL of water and re-weigh.
    3. Set up the apparatus as in the diagram and measure the initial temperature of the water.
    4. Burn the butane gas for five minutes.
    5. Immediately measure the final temperature of the water.
    6. Measure the final mass of the butane canister when cool.
       

Results 

Quantity Measurement
mass of empty can 52.14 g
mass of can + water before combustion 303.37 g
mass of butane canister before heating 260.15 g
mass of butane canister after heating 259.79 g
initial temperature of water 22.1 °C
final temperature of water 32.7 °C

 

The balanced thermochemical equation for the complete combustion of butane is

\(\ce{2C4H10(g) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)},\ \ \Delta H=-5748\ \text{kJ mol}^{-1}\)

  1. Calculate the amount of heat energy absorbed by the water when it was heated by burning the butane. Give your answer in kilojoules.   (2 marks)

  2. Calculate the experimental value of the molar heat of combustion of butane. Give your answer in kJ mol\(^{–1}\).   (2 marks)
  3. Use the known enthalpy change for butane to calculate the percentage energy loss to the environment.   (2 marks) 
Show Answers Only

i.    \(11.13\ \text{kJ}\)

ii.   \(-1.8 \times 10^{3}\ \text{kJ mol}^{-1} \)

iii.   \(37.4\% \)

Show Worked Solution

i.    \(\ce{m(water) = 303.37-52.14=251.23\ \text{g}}\)

\(\Delta T = 32.7-22.1=10.6^{\circ}\text{C}\)

\(\text{Energy absorbed}\) \(=4.18 \times 251.23 \times 10.6\)  
  \(=1.113 \times 10^{4}\ \text{J}\)  
  \(=11.13\ \text{kJ}\)  

 

ii.   \(\ce{m(C4H10 reacted) = 260.15-259.79=0.36\ \text{g}}\)

\(\ce{n(C4H10 reacted) = \dfrac{0.36}{\text{MM}} = \dfrac{0.36}{58.0} = 0.0062\ \text{mol}}\)

\(\text{Energy released (per mol}\ \ce{C4H10}\text{)} = \dfrac{11.13}{0.0062}=1.8 \times 10^{3}\ \text{mol}\)

\(\text{Experimental molar heat of combustion} = -1.8 \times 10^{3}\ \text{kJ mol}^{-1} \)
 

♦ Mean mark (ii) 42%.

iii.   \(\text{2 moles}\ \ce{C4H10},\ \Delta H = -5748\ \ \Rightarrow\ \ \text{1 mole}\ \ce{C4H10}, \Delta H = -2874\) 

\(\text{% Energy loss}\ = \dfrac{(2874-1.8 \times 10^{3})}{2874} \times 100 = 37.4\% \)

CHEMISTRY, M4 2014 VCE 3a*

The combustion of ethanol is represented by the following equation.

\(\ce{C2H5OH(l) + 3O2(g)\rightarrow 2CO2(g) + 3H2O(l)}\ \ \ \ \ \ \Delta H=-1364\ \text{kJ mol}^{-1}\)

A spirit burner used 1.80 g of ethanol to raise the temperature of 100.0 g of water in a metal can from 25.0 °C to 40.0 °C.
 

Calculate the percentage of heat lost to the environment and to the apparatus.   (5 marks)

Show Answers Only

\(\text{% heat lost}\ = 88.3\% \)

Show Worked Solution

\(\ce{n(CH3CH2OH) = \dfrac{1.80}{46.0} = 0.0391\ \text{mol}}\)

\(\Delta H \ce{(CH3CH2OH) =-1364\ \text{kJ mol}^{-1}}\)

\(\text{Energy released}\ = 0.0391\ \text{mol}\ \times 1364\ \text{kJ mol}^{-1} = 53.4\ \text{kJ}\)

\(\text{Energy absorbed by water}\ = 4.18 \times 100 \times 15.0 = 6.27\ \text{kJ}\)

\(\text{Energy not absorbed by water}\ = 53.4-6.27 = 47.13\ \text{kJ}\)

\(\text{% heat lost}\ = \dfrac{47.13}{53.4} \times 100 = 88.3\% \)

PHYSICS, M8 2019 VCE 17

Students are comparing the diffraction patterns produced by electrons and X-rays, in which the same spacing of bands is observed in the patterns, as shown schematically in the diagram. Note that both patterns shown are to the same scale.
 

The electron diffraction pattern is produced by 3.0 × 10\(^3\) eV electrons.

  1. Explain why electrons can produce the same spacing of bands in a diffraction pattern as X-rays.   (3 marks)
  1. Calculate the frequency of X-rays that would produce the same spacing of bands in a diffraction pattern as for the electrons. Show your working.   (4 marks)
Show Answers Only

a.    Electron vs X-ray wavelength:

  • Electrons with momentum exhibit wave-like properties.
  • In this way, moving electrons produce a de Broglie wavelength.
  • As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same. 

b.    \(1.34 \times 10^{19}\ \text{Hz}\)

Show Worked Solution

a.    Electron vs X-ray wavelength:

  • Electrons with momentum exhibit wave-like properties.
  • In this way, moving electrons produce a de Broglie wavelength.
  • As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same. 
♦ Mean mark (a) 49%.

b.    Find velocity of the electrons using  \(E=\dfrac{1}{2}mv^2 :\)

\(3.0 \times 10^3 \times 1.602 \times 10^{-19}=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times v^2\)

\(v^2\) \(=\dfrac{4.806 \times 10^{-16}}{4.5545 \times 10^{-31}}\)  
\(v\) \(=\sqrt{1.055 \times 10^{15}}\)  
  \(=3.25 \times 10^7\ \text{ms}^{-1}\)  

 

The de Broglie wavelength of the electron is:

\(\lambda\) \(=\dfrac{h}{mv}\)  
  \(=\dfrac{6.626 \times 10^{-34}}{3.25 \times 10^7 \times 9.109 \times 10^{-31}}\)  
  \(=2.24 \times 10^{-11}\ \text{m}\)  

 
Frequency of the X-ray:

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{2.24 \times 10^{-11}}=1.34 \times 10^{19}\ \text{Hz}\)

♦♦♦ Mean mark (b) 27%.
COMMENT: Multi-step solutions require clear and logical working.

PHYSICS, M7 2019 VCE 16

Students are studying the photoelectric effect using the apparatus shown in Figure 1.
 

     

Figure 2 shows the results the students obtained for the maximum kinetic energy \((E_{\text{k max }})\) of the emitted photoelectrons versus the frequency of the incoming light.
 

  1. Using only data from the graph, determine the values the students would have obtained for
    1. Planck's constant, \(h\). Include a unit in your answer.  (2 marks)
    1. the maximum wavelength of light that would cause the emission of photoelectrons.   (1 mark)
    1. the work function of the metal of the photocell.   (1 mark)
  1. The work function for the original metal used in the photocell is \(\phi\).
    On Figure 3, draw the line that would be obtained if a different metal, with a work function of \(\dfrac{1}{2} \phi\), were used in the photocell. The original graph is shown as a dashed line.   (2 marks)
     

Show Answers Only

a.i.  \(5.4 \times 10^{-15}\ \text{eV s}\)

   ii.  \(811\ \text{nm}\)

  iii.  \(1.9\ \text{eV}\).

b.  

Show Worked Solution

a.i.  Planck’s constant \((h)\):

  • Equal to the gradient of the line when \(E_{\text{k max}}\) is graphed against frequency.

\(\therefore h=\dfrac{\text{rise}}{\text{run}}=\dfrac{1.25-0}{6 \times 10^{14}-3.7\times 10^{-14}}=5.4 \times 10^{-15}\ \text{eV s}\)
 

a.ii.  Max wavelength = minimum frequency of emitted photoelectron.

\(\lambda=\dfrac{c}{f}=\dfrac{3 \times 10^8}{3.7 \times 10^{14}}=811\ \text{nm}\)
 

♦ Mean mark (a)(ii) 44%.

a.iii.  

   

  • The work function is the y-intercept of the graph, so by extending the graph as shown above, the work function is \(1.9\ \text{eV}\).
     

b.   Constructing the new graph:

  • The new \(y\)-intercept for the graph will be \(-0.95\ \text{eV}\)
  • The gradient of the graph will remain the same (Planck’s constant)
     

PHYSICS, M7 2019 VCE 14*

Students have set up a double-slit experiment using microwaves. The beam of microwaves passes through a metal barrier with two slits, shown as \(\text{S}_1\) and \(\text{S}_2\) in the diagram. The students measure the intensity of the resulting beam at points along the line shown. They determine the positions of maximum intensity to be at the points labelled \(\text{P}_0,\) \(\text{P}_1\), \(\text{P}_2\) and \(\text{P}_3\). 
 

The distance from \(\text{S}_1\) to \(\text{P}_3\) is 72.3 cm and the distance from \(\text{S}_2\) to \(\text{P}_3\) is 80.6 cm.

  1. What is the frequency of the microwaves transmitted through the slits? Show your working.   (2 marks)
  2. The signal strength is at a minimum approximately midway between points \(\text{P}_0\) and \(\text{P}_1\).
  3. Explain the reason why the signal strength would be a minimum at this location.   (2 marks)

  4. The microwaves from the source are polarised.
  5. Explain what is meant by the term 'polarised'. You may use a diagram in your answer.   (2 marks)

Show Answers Only

a.    \(1.08 \times 10^{10}\ \text{Hz}\)

b.    Signal strength at midpoints:

  • Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).
  • Therefore, destructive interference will occur and the signal strength will be a minimum.

c.    Polarised:

  • Light is a transverse wave that can oscillate in any direction.
  • Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.
      

Show Worked Solution

a.    The path difference to \(P_3\) is equal to 3 wavelengths of the microwaves.

\(3\lambda\) \(=0.806-0.723\)  
\(\lambda\) \(=\dfrac{0.083}{3}=0.0277\ \text{m}\)  

 
\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{0.0277}=1.08 \times 10^{10}\ \text{Hz}\)
 

♦♦ Mean mark (a) 35%.
COMMENT: Common error was not converting cm to m.

b.    Signal strength at midpoints:

  • Halfway between \(\text{P}_0\) and \(\text{P}_1\ \ \Rightarrow\)  path difference = \(\frac{\lambda}{2}\).
  • Therefore, destructive interference will occur and the signal strength will be a minimum.
♦♦ Mean mark (b) 42%.

c.    Polarised:

  • Light is a transverse wave that can oscillate in any direction.
  • Light becomes polarised when its plane of oscillation is only in a single direction, as seen in the diagram below.

 

♦ Mean mark (c) 50%.
Comment: Students are encouraged to use diagrams when possible.

PHYSICS, M7 2019 VCE 11

What is the second postulate of Einstein's theory of special relativity regarding the speed of light? Explain how the second postulate differs from the concept of the speed of light in classical physics.   (3 marks)

Show Answers Only
  • Einstein’s second postulate states that the speed of light \((c)\) is constant and independent of the motion of the source and motion of the observer.
  • The speed of light in classical physics was dependant on relative motion between the source and observer leading to different measurements for \((c)\),
  • For example, if the source and observer were approaching one another then the speed would be greater than \(3 \times 10^8\ \text{ms}^{-1}\) and if the source and observer were moving away from one another the speed would be less than \(3 \times 10^8\ \text{ms}^{-1}\).
Show Worked Solution
  • Einstein’s second postulate states that the speed of light \((c)\) is constant and independent of the motion of the source and motion of the observer.
  • The speed of light in classical physics was dependant on relative motion between the source and observer leading to different measurements for \((c)\),
  • For example, if the source and observer were approaching one another then the speed would be greater than \(3 \times 10^8\ \text{ms}^{-1}\) and if the source and observer were moving away from one another the speed would be less than \(3 \times 10^8\ \text{ms}^{-1}\).
♦ Mean mark 40%.
COMMENT: Many students failed to properly explain the classical predictions for c.

PHYSICS, M5 2019 VCE 8

A 250 g toy car performs a loop in the apparatus shown in the diagram below.
 

The car starts from rest at point \(\text{A}\) and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point \(\text{B}\) it is travelling at a speed of 3.0 m s\(^{-1}\).

  1. Calculate the value of \(h\). Show your working.   (3 marks)

  2. Calculate the magnitude of the normal reaction force on the car by the track when it is at point \(\text{B}\). Show your working.   (3 marks)

  3. Explain why the car does not fall from the track at point \(\text{B}\), when it is upside down.   (2 marks)

Show Answers Only

a.    \(0.86\ \text{m}\)

b.    \(8.8\ \text{N}\)

c.    Method 1

  • During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.
  • This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).

Method 2

  • The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 
\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}=\sqrt{9.8 \times 0.2}=1.4\ \text{ms}^{-1}\)  

 

  • As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).
  • Therefore the car will not fall from the track at point \(\text{B}\).

Show Worked Solution

a.    Using the law of the conservation of energy:

\(mgh_A\) \(=mgh_B +\dfrac{1}{2}mv^2_b\)  
\(0.25 \times 9.8 \times h_A\) \(=0.25 \times 9.8 \times 0.4 + \dfrac{1}{2} \times 0.25 \times 3^2\)  
\(2.45h_A\) \(=2.105\)  
\(h_A\) \(=\dfrac{2.105}{2.45}=0.86\ \text{m}\)  

 

b.     \(N + mg\) \(=\dfrac{mv^2}{r}\)
  \(N\) \(=\dfrac{mv^2}{r}-mg\)
    \(=\dfrac{0.25 \times 3^2}{0.2}- 0.25 \times 9.8\) 
    \(=8.8\ \text{N}\)

♦ Mean mark (b) 53%.

c.    Method 1

  • During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.
  • This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).

Method 2

  • The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 
\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}=\sqrt{9.8 \times 0.2}=1.4\ \text{ms}^{-1}\)  

  • As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).
  • Therefore the car will not fall from the track at point \(\text{B}\).
♦♦♦ Mean mark (c) 23%.
COMMENT: Normal force is poorly understood here..

PHYSICS, M6 2019 VCE 7*

Students in a Physics practical class investigate the piece of electrical equipment shown in the diagram. It consists of a single rectangular loop of wire that can be rotated within a uniform magnetic field. The loop has dimensions 0.50 m × 0.25 m and is connected to the output terminals with slip rings. The loop is in a uniform magnetic field of strength 0.40 T.
 

  1. What name best describes the piece of electrical equipment shown in the diagram?   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in the diagram? Explain your answer.   (2 marks)

The students connect the output terminals of the piece of electrical equipment to an oscilloscope. One student rotates the loop at a constant rate of 20 revolutions per second.

  1. Calculate the period of the rotation of the loop.   (1 mark)

  2. Calculate the maximum flux through the loop. Show your working.   (1 mark)

  3. The loop starts in the position shown in the diagram.
  4. What is the average voltage measured across the output terminals for the first quarter turn? Show your working.   (2 marks)

Show Answers Only

a.    Alternator.

b.   \(0\ \text{Wb}\)

  • This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(0.05\ \text{s}\)

d.    \(0.05\ \text{Wb}\)

e.    \(4\ \text{V}\)

Show Worked Solution

a.    The device is an alternator.

  • This is due to the input being mechanical motion and the output being AC current, whereas an AC motor is the opposite.
♦ Mean mark (a) 44%.
COMMENT: Many students incorrectly answered AC motor.

b.   \(0\ \text{Wb}\)

  • This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(T=\dfrac{1}{f}=\dfrac{1}{20}=0.05\ \text{s}\)
 

d.    \(\Phi=BA=0.4 \times (0.5 \times 0.25)=0.05\ \text{Wb}\)
 

e.     \(\varepsilon\) \(=\dfrac{\Delta \Phi}{\Delta t_{\text{1/4 rotation}}}\)
    \(=\dfrac{0.05}{0.0125}\)
    \(=4\ \text{V}\)
♦ Mean mark (e) 51%.

PHYSICS, M5 2019 VCE 5*

Navigation in vehicles or on mobile phones uses a network of global positioning system (GPS) satellites. The GPS consists of 31 satellites that orbit Earth.

In December 2018, one satellite of mass 2270 kg, from the GPS Block \(\text{IIIA}\) series, was launched into a circular orbit at an altitude of \(20\ 000\) km above Earth's surface.

  1. Identify the type(s) of force(s) acting on the satellite and the direction(s) in which the force(s) must act to keep the satellite orbiting Earth.   (2 marks)
  1. Calculate the period of the satellite to three significant figures. You may use data from the table below in your calculations. Show your working.   (3 marks)

\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text{mass of satellite} \rule[-1ex]{0pt}{0pt}& 2.27 \times 10^3 \ \text{kg} \\
\hline \rule{0pt}{2.5ex}\text{altitude of satellite above Earth's surface} \rule[-1ex]{0pt}{0pt}& 2.00 \times 10^7 \ \text{m} \\
\hline
\end{array}

Show Answers Only

a.    Forces acting on satellites:

  • Only force acting on a satellite is \(F_g\), the force due to gravity.
  • This force acts on the satellite directly towards the centre of the Earth.

b.    \(4.25 \times 10^4\ \text{s}\)

Show Worked Solution

a.    Forces acting on satellites:

  • Only force acting on a satellite is \(F_g\), the force due to gravity.
  • This force acts on the satellite directly towards the centre of the Earth.
♦ Mean mark (a) 46%.
COMMENT: Thrust force is not a requirement for orbit. 

b.    \(\dfrac{r^3}{T^2}=\dfrac{GM}{4\pi^2}\ \ \Rightarrow \ \ T= \sqrt{\dfrac{4 \pi^2r^3}{GM}}\)

\(T=\sqrt{\dfrac{4\pi^2 \times (6.371 \times 10^6 + 2 \times 10^7)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}=42\ 533=4.25\times 10^4\ \text{s}\)

PHYSICS, M6 2019 VCE 1

A particle of mass \(m\) and charge \(q\) travelling at velocity \(v\) enters a uniform magnetic field \(\text{B}\), as shown in the diagram.
 

  1. Is the charge \(q\) positive or negative? Give a reason for your answer.   (1 mark)
  1. Explain why the path of the particle is an arc of a circle while the particle is in the magnetic field.   (2 marks)

Show Answers Only

a.    Using the right-hand rule:

  • Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).
  • As the force on the charge is down the page (back of the hand), the charge must be negative.

b.    For circular motion to occur:

  • The force must be at right-angles to the velocity of the particle.
  • The force must be of constant magnitude \((F=qvB)\).
  • In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.
  • The particle will therefore follow the arc of a circle while in the magnetic field.

Show Worked Solution

a.    Using the right-hand rule:

  • Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).
  • As the force on the charge is down the page (back of the hand), the charge must be negative.
♦ Mean mark (a) 44%.

b.    For circular motion to occur:

  • The force must be at right-angles to the velocity of the particle.
  • The force must be of constant magnitude \((F=qvB)\).
  • In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.
  • The particle will therefore follow the arc of a circle while in the magnetic field.
♦♦♦ Mean mark (b) 27%.

PHYSICS, M8 2019 VCE 15 MC

Electrons pass through a fine metal grid, forming a diffraction pattern.

If the speed of the electrons was doubled using the same metal grid, what would be the effect on the fringe spacing?

  1. The fringe spacing would increase.
  2. The fringe spacing would decrease.
  3. The fringe spacing would not change.
  4. The fringe spacing cannot be determined from the information given.
Show Answers Only

\(B\)

Show Worked Solution
  • \(\lambda=\dfrac{h}{mv} \ \Rightarrow \  \lambda \propto \dfrac{1}{v}\)
  • If the velocity of the electron is doubled, the wavelength of the electron will halve.
  • Using Young’s double slit calculations:
  •    \(\dfrac{d\Delta x}{D}=m \lambda\ \Rightarrow\  \Delta x \propto \lambda\)
  • Therefore, if lambda decreases so will the fringe spacing \((\Delta x)\).

\(\Rightarrow B\)

♦ Mean mark 40%.
COMMENT: Formulas from different syllabus areas required here.

PHYSICS, M8 2023 HSC 27b

The diagram represents one hydrogen emission line from the spectrum of a star.

Explain the changes to this spectral line that would be observed as a result of the star’s rotational velocity. Modify the diagram to support your answer.   (4 marks)

Show Answers Only

  • When a star rotates, one side of the star is moving towards the Earth and one side of the star is moving away from the Earth.
  • As a result of the Doppler Effect, the wavelengths of the light moving towards the earth are shortened and so are blueshifted while the wavelengths of the light moving away from the earth and lengthened and so are red shifted. 
  • The light from the centre of the star is not moving towards or away from the Earth so it is not shifted in any direction.
  • As a result of the simultaneous blue and red shifting of the 656 nm absorption line, the spectral line will become broader as seen in the modified diagram below.

Show Worked Solution

  • When a star rotates, one side of the star is moving towards the Earth and one side of the star is moving away from the Earth.
  • As a result of the Doppler Effect, the wavelengths of the light moving towards the earth are shortened and so are blueshifted while the wavelengths of the light moving away from the earth and lengthened and so are red shifted. 
  • The light from the centre of the star is not moving towards or away from the Earth so it is not shifted in any direction.
  • As a result of the simultaneous blue and red shifting of the 656 nm absorption line, the spectral line will become broader as seen in the modified diagram below.

♦ Mean mark 55%.

Calculus, MET1 2023 VCAA SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

CHEMISTRY, M6 2009 HSC 14 MC

Citric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated 25.0 mL samples of lemon juice with 0.550 mol L\(^{-1}\ \ce{NaOH}\). The mean titration volume was 29.50 mL. The molar mass of citric acid is 192.12 g mol\(^{-1}\).

What was the concentration of citric acid in the lemon juice?

  1. 1.04 g L\(^{-1}\)
  2. 41.6 g L\(^{-1}\)
  3. 125 g L\(^{-1}\)
  4. 374 g L\(^{-1}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\ce{H3A(aq) + 3NaOH(aq) → Na3A(aq) + 3H2O(l)}\)

\(\ce{n(NaOH)=c \times v=0.550 \times 0.0295=0.016225\ \text{mol}}\)

\(\ce{n(citric acid)= \dfrac{0.016225}{3}=0.00541\ \text{mol}}\)

\(\ce{[citric acid]=\dfrac{0.00541}{0.025}=0.216\ \text{molL}^{-1}}\)

Multiply by molar mass:.

\(\text{Concentration (citric acid)}\ = 0.216 \times 192.12 = 41.5\ \text{g L}^{-1}\)

\(\Rightarrow B\)

Functions, MET1 2023 VCAA SM-Bank 3

Find the general solution for  \(2 \sin (x)=\tan (x)\) for \(x \in R\).   (3 marks)

Show Answers Only

\(x=n\pi\ ,\ \Big(2n\pm \dfrac{\pi}{3}\Big)\pi\quad n\in \mathbb{Z}\)

Show Worked Solution

\(2\sin(x)\) \(=\tan(x)\)
\(2\sin(x)\) \(=\dfrac{\sin(x)}{\cos(x)}\quad \cos(x)\neq 0\)
\(2\sin(x)\cos(x)-\sin(x)\) \(=0\)
\(\sin(x)\Big(2\cos(x)-1\Big)\) \(=0\)
\(\therefore \sin(x)=0\quad\) \(\text{or}\) \(\quad 2\cos(x)-1=0\)
\(\therefore x=n\pi\quad\) \(\text{or}\) \(\quad \cos(x)=\dfrac{1}{2}\)
    \(\quad x=2n\pi\pm \dfrac{\pi}{3}=\Bigg(2n\pm \dfrac{1}{3}\Bigg)\pi\quad n\in\mathbb{Z}\)

PHYSICS, M8 2020 VCE 17

The diagram shows the emission spectrum for helium gas.
 

  1. Which spectral line indicates the photon with the lowest energy?  (1 mark)
  1. Calculate the frequency of the photon emitted at the 588 nm line. Show your working.  (2 marks)
  1. Explain why only certain wavelengths and, therefore, certain energies are present in the helium spectrum.  (2 marks)

Show Answers Only

a.    \(668\ \text{nm}\)

b.    \(5.1 \times 10^{14}\ \text{Hz}\)

c.   Wavelengths of \(\ce{He}\) spectrum:

  • Electrons in the helium atom exist within electron shells.
  • Electrons in each shell have a fixed amount of energy. 
  • The electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.
  • When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.
  • As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.

Show Worked Solution

a.    \(E=\dfrac{hc}{\lambda}\),  therefore \(E \propto \dfrac{1}{\lambda}\)

  • The photon with the lowest energy will have the highest wavelength.
  • Lowest energy spectral line = 668 nm

b.    Convert: 588 nm = 588 × 10\(^{-9}\) m

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{588 \times 10^{-9}}=5.1 \times 10^{14}\ \text{Hz}\)
 

c.   Wavelengths of \(\ce{He}\) spectrum:

  • Electrons in the helium atom exist within electron shells.
  • Electrons in each shell have a fixed amount of energy. 
  • The electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.
  • When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.
  • As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.
♦♦ Mean mark (c) 34%.

PHYSICS, M8 2020 VCE 16

A beam of electrons travelling at 1.72 × 10\(^5\) m s\(^{-1}\) illuminates a crystal, producing a diffraction pattern as shown below. Ignore relativistic effects.
 

  1. Calculate the kinetic energy of one of the electrons. Show your working.   (2 marks)

  2. The electron beam is now replaced by an X-ray beam. The resulting diffraction pattern has the same spacing as that produced by the electron beam.

    Calculate the energy of one X-ray photon. Show your working.   (3 marks)

Show Answers Only

a.    \(0.084\ \text{eV}\)

b.    \(293\ \text{eV}\)

Show Worked Solution

a.     \(KE\) \(=\dfrac{1}{2}mv^2\)
    \(=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times (1.72 \times 10^5)^2\)
    \(=1.3474\times 10^{-20}\ \text{J}\)
    \(=\dfrac{1.3474 \times 10^{-20}}{1.602 \times 10^{-19}}\)
    \(=0.084\ \text{eV}\)

 

b.     \(\lambda_e\) \(=\dfrac{h}{mv}\)
    \(=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1.72 \times 10^5}\)
    \(=4.23 \times 10^{-9}\ \text{m}\)
    \(=\lambda_{\text{x-ray}}\)

 

  \(E_{\text{x-ray}}\) \(=\dfrac{hc}{\lambda}\)
    \(=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.23 \times 10^{-9}}\)
    \(=4.7 \times 10^{-17}\ \text{J}\)
    \(=\dfrac{4.7 \times 10^{-17}}{1.602 \times 10^{-19}}\)
    \(=293\ \text{eV}\)
♦ Mean mark (b) 41%.

PHYSICS, M7 2020 VCE 15

The metal surface in a photoelectric cell is exposed to light of a single frequency and intensity in the apparatus shown in Diagram A.

The voltage of the battery can be varied in value and reversed in direction.
  

  1. A graph of photocurrent versus voltage for one particular experiment is shown in Diagram B.
  2. On Diagram B, draw the trace that would result for another experiment using light of the same frequency but with triple the intensity.  (2 marks)
     

  1. What is a name given to the point labelled \(\text{A}\) on Diagram B?   (1 mark)

  2. Why does the photocurrent fall to zero at the point labelled \(\text{A}\) on Diagram B?   (1 mark)

Show Answers Only

a.  The photocurrent will also be tripled.

b.    The stopping voltage.

c.    Zero photocurrent at \(\text{A}\):

  • The energy of the stopping voltage is equal to the most energetic photoelectrons. 
  • Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

Show Worked Solution

a.    The photocurrent will also be tripled.

b.    The stopping voltage.
 

c.    Zero photocurrent at \(\text{A}\):

  • The energy of the stopping voltage is equal to the most energetic photoelectrons. 
  • Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.
♦♦ Mean mark (c) 32%.
COMMENT: The work function of the metal had no relevance to this question.

PHYSICS, M7 2020 VCE 12

In a Young's double-slit interference experiment, laser light is incident on two slits, \(\text{S}_1\) and \(\text{S}_2\), that are 4.0 × 10\(^{-4}\) m apart, as shown in Figure 1.

Rays from the slits meet on a screen 2.00 m from the slits to produce an interference pattern. Point \(\text{C}\) is at the centre of the pattern. Figure 2 shows the pattern obtained on the screen.
 

  1. There is a bright fringe at point \(\text{P}\) on the screen.
  2. Explain how this bright fringe is formed.   (2 marks)
  1. The distance from the central bright fringe at point \(\text{C}\) to the bright fringe at point P is 1.26 × 10\(^{-2}\) m.
  2. Calculate the wavelength of the laser light. Show your working.   (3 marks

Show Answers Only

a.    Bright fringe at point \(\text{P}\):

  • Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.
  • Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.
  • At this point, the peaks of the twos wave constructively interfere and a bright band is formed.

b.    \(630\ \text{nm}\)

Show Worked Solution

a.    Bright fringe at point \(\text{P}\):

  • Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.
  • Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.
  • At this point, the peaks of the twos wave constructively interfere and a bright band is formed.
♦ Mean mark (a) 45%.

b.    Using  \(d\,\sin \theta=m \lambda\)  and  \(\sin \theta=\dfrac{\Delta x}{D}\)

  • \(\Delta x\) is the distance between two bright bands and \(D\) is the distance from the slits to the screen.
  • Using  \(\dfrac{d\Delta x}{D}=m\lambda\):
  •   \(\lambda=\dfrac{d\Delta x}{Dm}=\dfrac{4 \times 10^{-4} \times 1.26 \times 10^{-2}}{2 \times 4}=630\ \text{nm}\)
♦ Mean mark (b) 40%.

PHYSICS, M7 2020 VCE 11

An astronaut has left Earth and is travelling on a spaceship at 0.800\(c\) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

  1. How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working.   (2 marks)
  1. Is the trip time measured by the astronaut in part (a) a proper time? Explain your reasoning.   (2 marks)

Show Answers Only

a.    \(6.46\ \text{years}\)

b.    Proper time measurement

  • The trip time of 6.46 years on the spaceship is a proper time.
  • This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

Show Worked Solution

a.    From the earth’s perspective:

\(\text{Travel time}\ =\dfrac{8.61}{0.8}=10.76\ \text{years}\).

From the astronaut’s perspective:

  • The Earth’s time is going to be dilated compared to his, so the time the astronauts clock will measure is:

\(t=\dfrac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

\(t_0\) \(=t\sqrt{1-\frac{v^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-\frac{(0.8c)^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-0.8^2}\)
  \(=6.46\ \text{years}\)

♦♦♦ Mean mark (a) 29%.
COMMENT: Light years is a measure of distance, not time!

b.    Proper time measurement

  • The trip time of 6.46 years on the spaceship is a proper time.
  • This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.
♦ Mean mark (b) 44%.

PHYSICS, M5 2020 VCE 8

The diagram below shows a small ball of mass 1.8 kg travelling in a horizontal circular path at a constant speed while suspended from the ceiling by a 0.75 m long string.
 


 

  1. Use labelled arrows on the diagram above to indicate the two physical forces acting on the ball.   (2 marks)
  1. Calculate the speed of the ball. Show your working.   (4 marks)
Show Answers Only

a.  
       

b.    \(1.2\ \text{ms}^{-1}\)

Show Worked Solution

a.    Only two forces are acting on the ball: \(F_w\) and \(F_T\):
 

   

Mean mark 58%.
COMMENT: Many students incorrectly identified centripetal force (this is a result of the tension force).

b.   
     

\(\text{Using}\ \ \tan\theta=\dfrac{F_{\text{net}}}{mg}:\)

\( F_{\text{net}}\) \(=mg\,\tan\theta\)  
\(\dfrac{mv^2}{r}\) \(=mg\,\tan\theta\)  
\(\dfrac{v^2}{r}\) \(=g\,\tan\theta\)  
\(\therefore v\) \(=\sqrt{g\,r\,\tan\theta}\)  
  \(=\sqrt{9.8 \times 0.317 \times \tan 25}\ \ ,\ \ (r = 0.75 \times \sin25=0.317\ \text{m})\)  
  \(=1.2\ \text{ms}^{-1}\)  
♦ Mean mark (b) 50%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
 

  1. Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil?   (1 mark)
  1. Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current.   (3 marks)
  1. The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c. 
     

  1. Describe the direction of any induced currents in the coil during these changes. Give your reasoning.   (2 marks)

Show Answers Only

a.    Decrease 

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
Show Worked Solution

a.    Magnetic flux will decrease.

  • The magnetic flux is proportional to how many field lines are passing through a given area.
  • As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.
♦ Mean mark (b) 42%.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

Graphs, MET1 EQ-Bank 1

Let  \(\displaystyle f:[-3,-2) \cup(-2, \infty) \rightarrow R, f(x)=1+\frac{1}{x+2}\).

  1. On the axes below, sketch the graph of \(f\). Label any asymptotes with their equations, and endpoints and axial intercepts with their coordinates.   (3 marks)

  1. Find the values of \(x\) for which \(f(x) \leq 2\).   (2 marks)

Show Answers Only

a.
   

b.   \(x\in [-3, -2)\cap (-1, \infty)\)

Show Worked Solution

a.
     

b.    \(\text{From graph }f(x)\leq -2\ \text{ for}\ -3\leq x <2\ \text{ and when }\ x\geq -1\) 

\(\rightarrow x\in [-3, -2)\cap (-1, \infty)\)

Calculus, MET2 EQ-Bank 2

Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp.

The cross-section of the ramp is modelled by the function \(f\), where

\(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\)

\(f(x)\) is both smooth and continuous at \(x=5\).

The graph of  \(y=f(x)\)  is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres.

  1. Find \(f^{\prime}(x)\) for \(0<x<55\).   (2 marks)

  2.   i. Find the coordinates of the point of inflection of \(f\).   (1 mark)

  3.  ii. Find the interval of \(x\) for which the gradient function of the ramp is strictly increasing.   (1 mark)

  4. iii. Find the interval of \(x\) for which the gradient function of the ramp is strictly decreasing.  (1 mark)

Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below.

  1. Determine the value of the ratio of the area of the trapezoidal cross-sections to the exact area contained between \(f(x)\) and the \(x\)-axis between \(x=5\) and \(x=25\). Give your answer as a percentage, correct to one decimal place.   (3 marks)

  2. Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area for any interval \(x \in[p, q]\), where \(p \geq 25\).   (1 mark)

  3. Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function \(P\), where

      1. \(P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}\)
  4. \(P\) is continuous and differentiable at \(x=55, g(x)=-\frac{1}{16} x^2+b x+c\), and \(x=a\) is where the car lands on the ground after the jump, such that \(P(a)=0\).
    1. Find the values of \(b\) and \(c\).   (2 marks)

    2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Show Answers Only

a.    \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \((25, 20)\)

b.ii  \([25, 55]\)

b.iii \([5, 25]\)

c.    \(98.1\%\)

d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(b=8.75, c=-283.4375\)

e.ii  \(34.10\ \text{cm}\)

Show Worked Solution

a.    \(\text{Using CAS: Define f(x) then}\)

\(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\)

\(\therefore\ \text{Point of inflection at }(25, 20)\)
  

b.ii  \(\text{Gradient function strictly increasing for }x\in [25, 55]\)

b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\)

c.    \(\text{Using CAS:}\)

\(\text{Area}\) \(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
  \(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)

  

\(\therefore\ \text{Estimate : Exact}\) \(=637.5:650\)
  \(=\dfrac{637.5}{650}\times 100\)
  \(=98.0769\%\approx 98.1\%\)

    
d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)
 

e.i   \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\)

\(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\)

\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\) \(=\dfrac{35}{4}\)
\(c\) \(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\) \(=\dfrac{3165}{16}-55b\quad (1)\)

  
\(g'(x)=-\dfrac{x}{8}+b\)

\(\text{and }g'(55)=f'(55)\)

\(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\)

\(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\)

\(\therefore b=\dfrac{35}{4}\quad (2)\)

\(\text{Sub (2) into (1)}\)

\(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\)

\(c=-\dfrac{4535}{16}\)

\(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\)
 

e.ii  \(\text{Using CAS: Solve }g(x)=0|x>55\)

\(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.

A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.

Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.

  1. Write down suitable hypotheses \(H_0\) and \(H_1\) for this test.   (1 mark)

  2. Find the \(p\) value for the test, correct to three decimal places.   (1 mark)

  3. Does the mean mass of the random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 64 empty cans for \(H_0\) not to be rejected? Give your answer correct to two decimal places.   (1 mark)

The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.

  1. What is the probability that the masses of two randomly selected cans of soft drink differ by no more than 3 grams? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(H_0: \mu=15, \quad H_1: \mu<15\)

b.   \(p=0.027\)

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

d.   \(a=14.95\)

e.  \(\text{Pr}(-3<D<3)=0.329\)

Show Worked Solution

a.    \(H_0: \mu=15, \quad H_1: \mu<15\)
 

b.    \(\mu=15, \ \ \sigma=0.25\)

\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)

\(\text{By CAS:}\)

\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
 

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

\(\text{Evidence is against \(H_0\)  at the \(5 \%\) level.}\)
 

d.    \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)

\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)

\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
 

e.    \(\text{Let}\ \ M=\ \text{mass of one can}\)

\(M \sim N\left(406,5^2\right)\)

\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)

\(\text {Let}\ \ D=M_1-M_2\)

\(E(D)=406-406=0\)

\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times  \text{Var}\left(M_2\right)=50\)

\(\sigma_D=\sqrt{50}\)

\(D \sim N\left(0,(\sqrt{50})^2\right)\)

\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)

♦ Mean mark (e) 46%.

Vectors, SPEC2 2022 VCAA 4

A student is playing minigolf on a day when there is a very strong wind, which affects the path of the ball. The student hits the ball so that at time  \(t=0\) seconds it passes through a fixed origin \(O\). The student aims to hit the ball into a hole that is 7 m from \(O\). When the ball passes through \(O\), its path makes an angle of \(\theta\) degrees to the forward direction, as shown in the diagram below.
 

The path of the ball \(t\) seconds after passing through \(O\) is given by

\(\underset{\sim}{\text{r}}(t)=\dfrac{1}{2} \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{\text{i}}+2 t \underset{\sim}{\text{j}}\)  for  \(t \in[0,5]\)

where \(\underset{\sim}{i}\) is a unit vector to the right, perpendicular to the forward direction, \(\underset{\sim}{j}\) is a unit vector in the forward direction and displacement components are measured in metres.

  1. Find \(\theta\) correct to one decimal place.   (2 marks)

  2.  i. Find the speed of the ball as it passes through \(O\). Give your answer in metres per second, correct to two decimal places.   (2 marks)

  3. ii. Find the minimum speed of the ball, in metres per second, and the time, in seconds, at which this minimum speed occurs.   (2 marks)

  4. Find the minimum distance from the ball to the hole. Give your answer in metres, correct to three decimal places.   (3 marks)

  5. How far does the ball travel during the first four seconds after passing through \(O\) ? Give your answer in metres, correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(\theta =11.1^{\circ}\)

b.i.  \(2.04\ \text{ms}^{-1}\)

b.ii.  \(\text{Minimum speed} =2 \text{ ms} ^{-1}\)

c.   \(\text {Minimum }\abs{d}=0.188\ \text{m}\)

d.   \(8.077\ \text{m}\)

Show Worked Solution

a.    \(r(t)=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 t \underset{\sim}{j}\)

\(\dot{r}(t)=\dfrac{\pi}{8}\, \cos \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 \underset{\sim}{j}\)

\(\dot{r}(0)=\dfrac{\pi}{8}\,\underset{\sim}{i}+2\underset{\sim}{j}\)

\begin{aligned}
\tan (90-\theta) & =\dfrac{2}{\frac{\pi}{8}} \\
90-\theta & =\tan ^{-1}\left(\dfrac{16}{\pi}\right)=78.9^{\circ} \\
\theta & =11.1^{\circ}
\end{aligned}

♦ Mean mark (a) 46%.
b.i.     \(\text{Speed}\) \(=\abs{\dot{r}(0)}\)
    \(=\sqrt{\left(\dfrac{\pi}{8}\right)^2+2^2}\)
    \(=2.04\ \text{ms}^{-1}\ \text{(2 d.p.)}\)

 

b.ii.  \(\abs{\dot{r}}=\sqrt{\left(\dfrac{\pi}{8}\right)^2 \times \cos ^2\left(\dfrac{\pi t}{4}\right)+4}\)

 \(\abs{\dot{r}}\text { is a minimum when}\ \ \cos ^2\left(\dfrac{\pi t}{4}\right)=0\)

\(\Rightarrow t=2\)

\(\therefore\ \text{Minimum speed} =\sqrt{4}=2 \text{ ms} ^{-1}\)
 

c.    \(\text{Ball position}\ \Rightarrow \ \underset{\sim}{r}=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right)\underset{\sim}{i}+2 \, t\underset{\sim}{j}\)

\(\text{Hole position}\ \Rightarrow \ \underset{\sim}{h}=0 \underset{\sim}{i}+7\underset{\sim}{j}\)

\(\underset{\sim}{d}=\underset{\sim}{r}-\underset{\sim}{h}=\dfrac{1}{2}\, \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+(2 t-7)\underset{\sim}{j}\)

\(\abs{\underset{\sim}{d}}=\sqrt{\left(\dfrac{1}{2}\right)^2 \sin ^2\left(\dfrac{\pi t}{4}\right)+(2 t-7)^2}\)

\(\text {Minimum }\abs{d}=0.188\text{ m (3 d.p.)}\)

♦ Mean mark (c) 45%.
d.     \(\text{Distance}\) \(=\displaystyle \int_0^4\left(\frac{\pi}{8}\, \cos \left(\frac{\pi t}{4}\right)\right)^2+4\, dt\)
    \(=8.077\ \text{m  (3 d.p.)}\)
♦ Mean mark (d) 48%.

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)

b.    \(2.06 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
 

b.   \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)

\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)

\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)

\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)

\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

CHEMISTRY, M3 EQ-Bank 11

A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)

b.    \(1.61 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
 

b.   \(\ce{MM(Cu(NO3)2) = 63.55 + 2 \times (14.01 + 16.00 \times 3) = 187.57}\)

\(\ce{n(Cu(NO3)2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.45}{187.57} = 0.01306 \text{ mol}}\)

\(\ce{n(Na2CO3) = c \times V = 1.25 \times 0.200 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCO3) = n(Cu(NO3)2) = 0.01306 \text{ mol}}\)

\(\ce{m(CuCO3) = n \times MM = 0.01306 \times (63.55 + 12.01 + 16.00 \times 3) = 1.61 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

CHEMISTRY, M2 EQ-Bank 3

Carbon dioxide is produced during the combustion of propane \(\ce{(C3H8)}\) in oxygen \(\ce{(O2)}\). The balanced chemical equation for this reaction is:

\(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}\)

If 44.0 grams of propane are completely combusted, calculate the volume of carbon dioxide produced at STP (100 kPa and 0\(^{\circ}\)C).    (3 marks)

Show Answers Only

\(\ce{V(CO2)}\ =68\ \text{L}\)

Show Worked Solution

Calculate the number of moles of propane (\(\ce{C3H8}\)):

  \(\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{44.0\ \text{g}}{44.094\ \text{g mol}^{-1}} = 0.998\ \text{mol}}\)
 

 Use the stoichiometric ratio to find the moles of \(\ce{CO2}\) produced:

  \(\ce{n(CO2) = 3 \times n(C3H8) = 3 \times 0.998 \, mol = 2.994 \, mol}\)
 

Calculate the volume of \(\ce{CO2}\) at STP:

  \(\ce{V(CO2) = n \times 22.71 \, L \, mol^{-1} = 2.994 \, mol \times 22.71 \, L \, mol^{-1} = 68 \, L}\)
 

  • The volume of carbon dioxide produced is 68 litres.

Calculus, MET2 2023 VCE SM-Bank 6 MC

Consider the algorithm below, which uses the bisection method to estimate the solution to an equation in the form \(f(x)=0\).

The algorithm is implemented as follows.
 

Which value would be returned when the algorithm is implemented as given?

  1. -0.351
  2. -0.108
  3. 3.25
  4. 3.5
  5. 4
Show Answers Only

\(D\)

Show Worked Solution

\(\textbf{Define }\text{bisection}\ (f(x),a,b,\max)\ \rightarrow\ (\sin(x), 3, 5, 2)\)

\(\text{Test for }\ i=0, 1\ \rightarrow i<\max=2\)

\(i=0\quad\) \(\text{mid}\) \(=\dfrac{3+5}{2}\) \(=4\quad [\sin(4)<0\ \text{ and }\ \sin(3)\times\sin(4)<0]\)
\(i=1\quad\) \(\text{mid}\) \(=\dfrac{3+4}{2}\) \(=3.5\quad [\sin(3.5)<0\ \text{ and }\ \sin(3)\times\sin(3.5)<0]\)

    

\(\Rightarrow D\)

Calculus, MET2 2023 SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

  1. Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places.   (3 marks)

     

     

  2.  i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\).   (1 mark)

  3. ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a.   (1 mark)

Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

  1. Find the value of \(x_1\).   (1 mark)

  2. Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places.   (1 mark)

  3.  i. Find the value of \(k\), where \(k>1\), such that an initial estimate of  \(x_0=k\)  gives the same value of  \(x_1\)  as found in part \(c\). Give your answer correct to three decimal places.   (2 marks)

  4. ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where  \(x=k\)  on the axes in part a.   (1 mark)

Show Answers Only
a.    
  
b.i

 \(y=2x-3\)
  
b.ii
  
c. \(\dfrac{3}{2}=1.5\)
  
d. \(0.0036\)
  
e.i \(k=2.397\)
  
e.ii
Show Worked Solution

a.

b.i    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g^{\prime}(x)\) \(=\dfrac{3}{x}-1\)
  \(g^{\prime}(1)\) \(=\dfrac{3}{1}-1=2\)

  
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

 
c.    \(\text{Newton’s Method}\)

\(x_1\) \(=x_0-\dfrac{g(x)}{g'(x)}\)
  \(=1-\left(\dfrac{-1}{2}\right)\)
  \(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)
  
 

d.    \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

 

e.i.  \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) \(=1.5\)
\(k>1\ \therefore\ \ k\) \(=2.397\)

 
e.ii

CHEMISTRY, M2 EQ-Bank 2

During a laboratory experiment, a gas is collected in a sealed syringe. Initially, the gas has a volume of 5.0 litres and a pressure of 1.0 atmosphere. 

  1.  Calculate the new pressure inside the syringe when the volume is decreased to 3.0 litres, assuming no temperature change.   (2 marks)

  2. After reaching the pressure calculated in part a, the volume is further decreased so that the pressure inside the syringe doubles. Calculate the final volume of the gas.   (2 marks)

  3. Discuss two potential experimental errors that could affect the accuracy of the observed results compared to the theoretical predictions of Boyle's Law.   (2 marks)

Show Answers Only

a.    \(1.67\ \text{atm}\)

b.    \(1.5\ \text{L}\)

c.    Potential experimental errors could include:

  • Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
  • Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
  • Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.
Show Worked Solution

a.    Using Boyle’s Law  \( P_1V_1 = P_2V_2 \):

\( P_1 = 1.0 \, \text{atm}, \quad V_1 = 5.0 \, \text{L}, \quad V_2 = 3.0 \, \text{L} \)

\(P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.0 \times 5.0}{3.0} = 1.67 \, \text{atm} \)
 

b.    To find the new volume when pressure doubles:

\( P_3 = 2 \times P_2 = 2 \times 1.67 \, \text{atm} = 3.34 \, \text{atm}\)

\( P_1V_1 = P_3V_3 \ \  \Rightarrow \ \  V_3 = \dfrac{P_1 \times V_1}{P_3} = \dfrac{1.0  \times 5.0 }{3.34} \approx 1.5 \, \text{L}\)
 

c.    Potential experimental errors could include:

  • Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
  • Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
  • Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.

CHEMISTRY, M1 EQ-Bank 14

A mixture of sand and salt was provided to a group of students for them to determine its percentage composition by mass.

They added water to the sample before using filtration and evaporation to separate the components.

During the evaporation step, the students noticed white powder ‘spitting’ out of the basin onto the bench, so they turned off the Bunsen burner and allowed the remaining water to evaporate overnight.

After filtering, they allowed the filter paper to dry overnight before weighing. An electronic balance was used to measure the mass of each component to two decimal places.

The results were recorded as shown:

    • Mass of the original sand and salt mixture = 15.73 g
    • Mass of the filter paper = 0.80 g
    • Mass of the dried filter paper after filtering = 11.95 g
    • Mass of the empty evaporating basin = 33.50 g
    • Mass of the evaporating basin after evaporation = 36.60 g
  1. Calculate the percentage composition by mass of sand AND salt in the mixture.   (3 marks)

  2. Consider the following definition of validity
  3. Validity is the degree to which tests measure what was intended, or the accuracy of actions, data and inferences produced from tests and other processes.
  4. Use this definition of to assess the validity of the experiment.   (2 marks)

Show Answers Only

a.    % Sand = 70.88%, % Salt = 19.71%

b.   Experiment validity:

  • The calculations show that the percentages do not add up to 100.
  • Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.
  • Hence, experiment is not valid.
Show Worked Solution
a.     \(\text{Sand mass}\ \) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)
    \(= 11.95-0.80 = 11.15\ \text{g}\)
\(\text{Salt mass}\) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)  
  \(= 36.60-33.50 = 3.10\ \text{g}\)  

 

\(\text{% sand} = \left(\dfrac{\text{Mass of sand}}{\text{Mass of original mixture}}\right) \times 100= \left(\dfrac{11.15 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 70.88\%\)

\(\text{% salt} = \left(\dfrac{\text{Mass of salt}}{\text{Mass of original mixture}}\right) \times 100 = \left(\dfrac{3.10 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 19.71\%\)
 

b.   Experiment validity:

  • The calculations show that the percentages do not add up to 100.
  • Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.
  • Hence, experiment is not valid.

CHEMISTRY, M1 EQ-Bank 10

Explain why ethanol \(\ce{(C2H5OH)}\) is a liquid at room temperature whereas ethane \(\ce{(C2H6)}\) is a gas.    (3 marks)

Show Answers Only
  • Ethanol \(\ce{(C2H5OH)}\) is a polar molecule with an \(\ce{(-OH)}\) group that can form hydrogen bonds.
  • Hydrogen bonds are strong intermolecular forces that require more energy to break, resulting in ethanol being a liquid at room temperature.
  • Ethane \(\ce{(C2H6)}\), however, is a nonpolar molecule. The only intermolecular forces present in ethane are dispersion forces (also known as London dispersion forces), which are much weaker than hydrogen bonds.
  • As a result, ethane remains a gas at room temperature because less energy is required to separate its molecules.
  • The significant difference in the strength of intermolecular forces (hydrogen bonding in ethanol vs. dispersion forces in ethane) explains why ethanol is a liquid and ethane is a gas at room temperature.
Show Worked Solution
  • Ethanol \(\ce{(C2H5OH)}\) is a polar molecule with an \(\ce{(-OH)}\) group that can form hydrogen bonds.
  • Hydrogen bonds are strong intermolecular forces that require more energy to break, resulting in ethanol being a liquid at room temperature.
  • Ethane \(\ce{(C2H6)}\), however, is a nonpolar molecule. The only intermolecular forces present in ethane are dispersion forces (also known as London dispersion forces), which are much weaker than hydrogen bonds.
  • As a result, ethane remains a gas at room temperature because less energy is required to separate its molecules.
  • The significant difference in the strength of intermolecular forces (hydrogen bonding in ethanol vs. dispersion forces in ethane) explains why ethanol is a liquid and ethane is a gas at room temperature.

CHEMISTRY, M1 EQ-Bank 9

Carbon exhibits allotropy, meaning it can exist in different forms with distinct physical properties.

Describe two carbon allotropes, graphite and diamond, and explain how their structural differences result in their distinct physical properties.   (4 marks)

Show Answers Only

Graphite:

  • Graphite consists of layers of carbon atoms arranged in a hexagonal lattice.
  • Each carbon atom is bonded to three others in the same plane, with delocalized electrons between layers.
  • The weak intermolecular forces between these layers allow them to slide over each other easily, contributing to graphite’s softness and its use as a lubricant. 

Diamond:

  • In contrast, diamond features a three-dimensional lattice where each carbon atom forms four strong covalent bonds with other carbon atoms, arranged in a tetrahedral structure.
  • This extensive network of strong bonds throughout the lattice makes diamond one of the hardest natural substances, leading to its use in cutting and drilling tools.
Show Worked Solution

Graphite:

  • Graphite consists of layers of carbon atoms arranged in a hexagonal lattice.
  • Each carbon atom is bonded to three others in the same plane, with delocalized electrons between layers.
  • The weak intermolecular forces between these layers allow them to slide over each other easily, contributing to graphite’s softness and its use as a lubricant. 

Diamond:

  • In contrast, diamond features a three-dimensional lattice where each carbon atom forms four strong covalent bonds with other carbon atoms, arranged in a tetrahedral structure.
  • This extensive network of strong bonds throughout the lattice makes diamond one of the hardest natural substances, leading to its use in cutting and drilling tools.

CHEMISTRY, M1 EQ-Bank 10

Using your knowledge of electronic configurations, explain what happens to atomic radii as you go across a period from left to right in the periodic table.

Identify one element from the first period with a larger atomic radius and one with a smaller atomic radius than Boron.   (3 marks)

Show Answers Only
  • As elements progress from left to right across a period, each successive element has one more proton and electron than the previous one.
  • These additional electrons enter the same energy level, increasing the effective nuclear charge experienced by electrons.
  • However, the stronger pull from the nucleus due to the higher positive charge causes the electrons to be drawn closer, resulting in a decrease in atomic radius.
  • Lithium has a larger atomic radius than Boron, while Neon has a smaller atomic radius than Boron.
Show Worked Solution
  • As elements progress from left to right across a period, each successive element has one more proton and electron than the previous one.
  • These additional electrons enter the same energy level, increasing the effective nuclear charge experienced by electrons.
  • However, the stronger pull from the nucleus due to the higher positive charge causes the electrons to be drawn closer, resulting in a decrease in atomic radius.
  • Lithium has a larger atomic radius than Boron, while Neon has a smaller atomic radius than Boron.

Calculus, MET2 2023 VCE SM-Bank 7 MC

One way of implementing Newton's method using pseudocode, with a tolerance level of 0.001 , is shown below.

The pseudocode is incomplete, with two missing lines indicated by an empty box.
 

  

Which one of the following options would be most appropriate to fill the empty box?
 



Show Answers Only

\(E\)

Show Worked Solution

\(\text{The tolerance}=\pm 0.001\)

\(\Big|\text{next_ x}-\text{prev_ x}\Big|<\ \text{tolerance}\)

\(\therefore\ \textbf{If }\ \ -0.001<\ \text{next_ x}-\text{prev_ x}\ <0.001\ \textbf{Then }\)

\(\text{If true }\textbf{Return }\text{next_ x}\)

\(\Rightarrow E\)

CHEMISTRY, M1 EQ-Bank 8

Using your understanding of periodic trends, explain and predict the differences in the properties of the elements lithium \(\ce{(Li)}\) and fluorine \(\ce{(F)}\) regarding their:

    • atomic radii
    • first ionisation energy
    • electronegativity

Give reasons for your predictions based on their positions in the periodic table and electronic configurations.   (4 marks)

Show Answers Only

Atomic Radii:

  • Lithium, being in the same period but a different group than fluorine, has fewer protons and a less effective nuclear charge, resulting in a larger atomic radius.

First Ionisation Energy:

  • Fluorine has a higher first ionisation energy due to its greater nuclear charge and smaller radius, which strongly attracts electrons.

Electronegativity:

  • Fluorine, being one of the most electronegative elements, has a strong ability to attract bonding electrons.
Show Worked Solution

Atomic Radii:

  • Lithium, being in the same period but a different group than fluorine, has fewer protons and a less effective nuclear charge, resulting in a larger atomic radius.

First Ionisation Energy:

  • Fluorine has a higher first ionisation energy due to its greater nuclear charge and smaller radius, which strongly attracts electrons.

Electronegativity:

  • Fluorine, being one of the most electronegative elements, has a strong ability to attract bonding electrons.

Calculus, MET2 2023 VCE SM-Bank 5 MC

The algorithm below, described in pseudocode, estimates the value of a definite integral using the trapezium rule.
 

Consider the algorithm implemented with the following inputs.

The value of the variable sum after one iteration of the while loop would be closest to

  1. 1.281
  2. 1.289
  3. 1.463
  4. 1.617
  5. 2.136
Show Answers Only

\(C\)

Show Worked Solution

\((f(x), a, b, n)\ \rightarrow\ (\log_{e}x, 1, 3, 10) \)

\(h=\dfrac{b-a}{n}=\dfrac{3-1}{10}=\dfrac{1}{5}\)

\(\text{Sum}=f(a)+f(b)=\log_{e}1+\log_{e}3=\log_{e}3\)
  

\(\text{1st iteration of}\ \textbf{while }\text{loop:}\)

\(x\) \(=a+h=1+\dfrac{1}{5}=\dfrac{6}{5}\)
\(\text{Sum}\) \(=\text{Sum}+2\times f(x)\)
  \(=\log_{e}3+2\times f\Bigg(\dfrac{6}{5}\Bigg)\)
  \(=\log_{e}3+2\times \log_{e}{\Bigg(\dfrac{6}{5}\Bigg)}\)
  \(\approx 1.46325\dots\)

 

\(\Rightarrow C\)

Calculus, MET2 2023 VCE SM-Bank 2 MC

Newton's method is being used to approximate the non-zero \(x\)-intercept of the function with the equation  \(f(x)=\dfrac{x^3}{5}-\sqrt{x}\).  An initial estimate of  \(x_0=1\) is used.

Which one of the following gives the first estimate that would correctly approximate the intercept to three decimal places?

  1. \(x_6\)
  2. \(x_7\)
  3. \(x_8\)
  4. \(x_9\)
  5. The intercept cannot be correctly approximated using Newton's method.
Show Answers Only

\(C\)

Show Worked Solution
\(f(x)\) \(=\dfrac{x^3}{5}-\sqrt{x}\)
\(f'(x)\) \(=\dfrac{3x^2}{5}-\dfrac{1}{2\sqrt{x}}\)

\(\text{For Newton’s Method: }\)

\(x_o\) \(=1\)
\(x_{n+1}\) \(=x_n-\dfrac{f(x_n)}{f'(x_n)}\)

  

\(\text{Check using CAS:}\)

\begin{array} {|c|c|}
\hline
\ \ \ n\ \  \ & x_n \\
\hline
\ 0 \ & 1\\
\hline
\ 1 \ & 1-\dfrac{\dfrac{1^3}{5}-\sqrt{1}}{\dfrac{3\times 1^2}{5}-\dfrac{1}{2\sqrt{1}}}=9\\
\hline
\ 2 \ & 9-\dfrac{\dfrac{9^3}{5}-\sqrt{9}}{\dfrac{3\times 9^2}{5}-\dfrac{1}{2\sqrt{9}}}\approx 6.05\\
\hline
\ 3 \ & 6.05-\dfrac{\dfrac{6.05^3}{5}-\sqrt{6.05}}{\dfrac{3\times 6.05^2}{5}-\dfrac{1}{2\sqrt{6.05}}}\approx 4.13\\
\hline
\ 4 \ & 4.13-\dfrac{\dfrac{4.13^3}{5}-\sqrt{4.13}}{\dfrac{3\times 4.13^2}{5}-\dfrac{1}{2\sqrt{4.13}}}\approx 2.92\\
\hline
\ 5 \ & 2.92-\dfrac{\dfrac{2.92^3}{5}-\sqrt{2.92}}{\dfrac{3\times 2.92^2}{5}-\dfrac{1}{2\sqrt{2.92}}}\approx 2.24\\
\hline
\ 6 \ & 2.24-\dfrac{\dfrac{2.24^3}{5}-\sqrt{2.24}}{\dfrac{3\times 2.24^2}{5}-\dfrac{1}{2\sqrt{2.24}}}\approx 1.96\\
\hline
\ 7 \ & 1.96-\dfrac{\dfrac{1.96^3}{5}-\sqrt{1.96}}{\dfrac{3\times 1.96^2}{5}-\dfrac{1}{2\sqrt{1.96}}}\approx 1.906\\
\hline
\ 8 \ & 1.906-\dfrac{\dfrac{1.906^3}{5}-\sqrt{1.906}}{\dfrac{3\times 1.906^2}{5}-\dfrac{1}{2\sqrt{1.906}}}\approx 1.90366\\
\hline
\end{array}

  
\(\Rightarrow C\)

Calculus, MET2 2022 VCAA 5

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.

`\quad x \quad` `\quad\quad 1/2\quad\quad` `\quad\quad(sqrt{2})/2\quad\quad` `\quad\quad(sqrt{3})/2\quad\quad`
`f(x)` `-2` `5` `3`
`\quad\quad f^{prime}(x)\quad\quad` `7` `0` `1/9`

 

  1. Find the value of `g\left(\frac{\pi}{6}\right)`.   (1 mark)

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

  1. Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`.   (1 mark)

  2. Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`.   (2 marks)

  3. Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`.   (2 marks)

  4. Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`.   (3 marks)

Show Answers Only

a.    `3`

b.    `1/9`

c.    `y=1/9x+3-pi/54`

d.    `-48/pi`

e.    ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution
 

a.  `g(pi/6)`
 `= f(sin(pi/3))`  
  `= f(sqrt3/2)`  
  `= 3`  

 

b.  `g\ ^{prime}(x)` `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`  
`g\ ^{prime}(pi/6)` `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`  
  `= 1/9`  

 

c.   `m = 1/9`  and  `g(pi/6) = 3`

`y  –  y_1` `= m(x-x_1)`  
`y  –  3` `= 1/9(x-pi/6)`  
`y` `= 1/9x + 3-pi/54`  

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

  
d.   The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`  
  `=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`  
  `= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))`  
  `= 24/pi (3-5) = -48/pi`  

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e.   `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\   2 \cos (2 x) = 0\ ….(1)`  or  ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1):   ` 2 \cos (2 x)`  `= 0`      `x \in[0, \pi]`
`\cos (2 x)` `= 0`      `2 x \in[0,2 \pi]`
`2x` `= pi/2 , (3pi)/2`  
`x` `= pi/4 , (3pi)/4`  
     
(2): ` f^{\prime}(\sin (2 x)) ` `= sqrt2/2`  
`2x` `= pi/4 , (3pi)/4`  
`x` `= pi/8 , (3pi)/8`  

  
`:. \  x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

L&E, 2ADV E1 SM-Bank 15

Solve the following equation for \(x\):

\(2^{2x}=3(2^{x+1})-8\).   (3 marks)

Show Answers Only

\(x=1\ \ \text{or} \ \ 2\)

Show Worked Solution
\(2^{2x}\) \(=3(2^{x+1})-8\)  
\(2^{2x}\) \(=3(2 \times 2^{x})-8\)  
\(0\) \(=2^{2x}-6\cdot2^{x}+8\)  

 
\(\text{Let}\ \ X=2^{x}\)

\(X^2-6X+8\) \(=0\)  
\((X-4)(X-2)\) \(=0\)  
\(X\) \(=4\ \ \text{or}\ \ 2\)  

 
\(2^{x}=4\ \ \Rightarrow\ \ x=2\)

\(2^{x}=2\ \ \Rightarrow\ \ x=1\)

Complex Numbers, SPEC2 2022 VCAA 2

Two complex numbers \(u\) and \(v\) are given by  \(u=a+i\)  and  \(v=b-\sqrt{2}i\), where \(a, b \in R\).

  1.  i. Given that  \(uv=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6})i\), show that  \(a^2+(1-\sqrt{3}) a-\sqrt{3}=0\).   (2 marks)

  2. ii. One set of possible values for \(a\) and \(b\) is  \(a=\sqrt{3}\)  and  \(b=\sqrt{2}\).
  3.     Hence, or otherwise, find the other set of possible values.   (1 mark)

  4. Plot and label the points representing  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  on the Argand diagram below.
     

  1. The ray given by  \(\text{Arg}(z)=\theta\)  passes through the midpoint of the line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\).
  2. Find, in radians, the value of \(\theta\) and plot this ray on the Argand diagram in part b.   (2 marks)

  3. The line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  cuts the circle  \(|z|=2\)  into a major and a minor segment.
  4. Find the area of the minor segment, giving your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.i.  \(\text{See worked solutions.}\)

a.ii.  \(a=-1, \ b=-\sqrt{6}\)

b.   
       

c.   \(\theta=-\dfrac{\pi}{24}\)

d.    \(0.69\ \text{u}^2\)

Show Worked Solution

a.i.  \(u v=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6}) i\)

\begin{aligned}
(a+i)(b-\sqrt{2} i) & =a b-\sqrt{2} a i+b i+\sqrt{2} \\
& =a b+\sqrt{2}+(b-\sqrt{2}a) i
\end{aligned}

\(\text {Equating co-efficients: }\)

  \(ab=\sqrt{6} \ \Rightarrow \ b=\dfrac{\sqrt{6}}{a}\ \cdots\ \text {(1)}\)

  \(b-\sqrt{2} a=\sqrt{2}-\sqrt{6}\ \cdots\ \text {(2)}\)

\(\text{Substitute (1) into (2):}\)

\(\dfrac{\sqrt{6}}{a}-\sqrt{2}a=\sqrt{2}-\sqrt{6}\)

\(\sqrt{6}-\sqrt{2} a^2=(\sqrt{2}-\sqrt{6}) a\)

\(\sqrt{2} a^2+(\sqrt{2}-\sqrt{6}) a-\sqrt{6}\) \(=0\)  
\(a^2+(1-\sqrt{3}) a-\sqrt{3}\) \(=0\)  
Mean mark (a) 54%.

 
a.ii. \(\text{Solve } a^2+(1-\sqrt{3}) a-\sqrt{3}=0 \ \ \text{for}\  a :\)

\(a=\sqrt{3} \ \text{ or } -1\)

\(\therefore \text{ Other solution: } a=-1, \ b=-\sqrt{6}\)
 

b.   
       


c.    		 \begin{aligned}
\theta & =\frac{\operatorname{Arg}(u)-\operatorname{Arg}(v)}{2} \\
& =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\pi}{4}\right) \\
& =-\frac{\pi}{24}
\end{aligned}
♦ Mean mark (c) 39%.

d.   \(\text {Sector angle (centre) }=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{5 \pi}{12}\)

\(\text {Area of sector }=\dfrac{\frac{5 \pi}{12}}{2 \pi} \times \pi \times 2^2=\dfrac{5 \pi}{6}\)

\begin{aligned}
\text {Area of segment } & =\frac{5 \pi}{6}-\dfrac{1}{2} \times 2 \times 2 \times \sin \left(\dfrac{5 \pi}{6}\right) \\
& \approx 0.69\ \text{u} ^2
\end{aligned}

♦ Mean mark (d) 40%.

Calculus, SPEC2 2022 VCAA 1

Consider the family of functions \(f\) with rule  \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).

  1. Write down the equations of the two asymptotes of the graph of \(f\) when \(k=1\).   (2 marks)

  2. Sketch the graph of  \(y=f(x)\)  for  \(k=1\)  on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations.   (3 marks)
     

  1.  i. Find, in terms of \(k\), the equations of the asymptotes of the graph of  \(f(x)=\dfrac{x^2}{x-k}\).   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  \(f(x)=\dfrac{x^2}{x-k}\) in terms of \(k\).   (2 marks)

  3. Now consider the functions \(h\) and \(g\), where  \(h(x)=x+3\)  and  \(g(x)=\abs{\dfrac{x^2}{x-1}}\).
  4. The region bounded by the curves of \(h\) and \(g\) is rotated about the \(x\)-axis.
    1. Write down the definite integral that can be used to find the volume of the resulting solid.   (2 marks)

    2. Hence, find the volume of this solid. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only

a.  \(\text {Asymptotes: } x=1,\  y=x+1\)

b.   
       

c.i.   \(\text {Asymptotes: } x=k,\  y=x+k\)

c.ii.  \(\text {Distance }=2 \sqrt{5}|k|\)

d.i.  \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)

d.ii.  \(V=51.42\ \text{u}^3 \)

Show Worked Solution

a.    \(\text {When } k=1 :\)

\(f(x)=\dfrac{x^2}{x-1}=\dfrac{(x+1)(x-1)+1}{(x-1)}=x+1+\dfrac{1}{x-1}\)

\(\text {Asymptotes: } x=1,\  y=x+1\)
 

b.    
       

 

c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)

\(\text {Using part a.}\)

\(\text {Asymptotes: } x=k,\  y=x+k\)
 

c.ii.  \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)

\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)

\(\Rightarrow(2 k, 4 k),(0,0)\)

\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
 

d.i  \(\text {Solve for intersection of graphs (by CAS):}\)

\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)

\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)

\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
 

d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)

♦♦ Mean mark (d)(ii) 37%.