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CHEMISTRY, M4 2014 VCE 3a*

The combustion of ethanol is represented by the following equation.

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)

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Show Worked Solution

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 Figure 18. Note that both patterns shown are to the same scale.
 

The electron diffraction pattern is produced by 3.0 × 10 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)
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a.    → 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.    

Show Worked Solution

a.    → 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 

 
 
   

 

The de Broglie wavelength of the electron is:

 
   
   

 
Frequency of the X-ray:

♦♦♦ 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 15.
 

     

Figure 16 shows the results the students obtained for the maximum kinetic energy 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, . 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 .
    On Figure 17, draw the line that would be obtained if a different metal, with a work function of , were used in the photocell. The original graph is shown as a dashed line.   (2 marks)
     

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a. i.  

    ii.  

   iii.  .

b.  

Show Worked Solution

a.i.  Planck’s constant :

→ Equal to the gradient of the line when is graphed against frequency.


 

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


 

♦ 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 .
 

b.    → The new y-intercept for the graph will be

→ 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 and in Figure 13. 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 , and . 
 

The distance from to is 72.3 cm and the distance from to 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 and .
  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)

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a.    

b.    → Halfway between and   path difference = .

→ Therefore, destructive interference will occur and the signal strength will be a minimum.

c.   → 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 is equal to 3 wavelengths of the microwaves.

 
   
 

 

 

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

b.    → Halfway between and   path difference = .

→ Therefore, destructive interference will occur and the signal strength will be a minimum.
 

♦♦ Mean mark (b) 42%.

c.   → 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)

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→ Einstein’s second postulate states that the speed of light 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 ,

→ For example, if the source and observer were approaching one another then the speed would be greater than and if the source and observer were moving away from one another the speed would be less than .

Show Worked Solution

→ Einstein’s second postulate states that the speed of light 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 ,

→ For example, if the source and observer were approaching one another then the speed would be greater than and if the source and observer were moving away from one another the speed would be less than .

♦ 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 Figure 8.
 

The car starts from rest at point 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 it is travelling at a speed of 3.0 m s.

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

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

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a.    

b.   

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.
 

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. 

 
 
   
   

 
→ As 3 ms is greater than the minimum speed required, a normal force downwards will be present at point .

→ Therefore the car will not fall from the track at point .

Show Worked Solution

a.    Using the law of the conservation of energy:

 
 
 
 
   

 

b.    
 
     
   

♦ 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.
 

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. 

 
 
   
   

 
→ As 3 ms is greater than the minimum speed required, a normal force downwards will be present at point .

→ Therefore the car will not fall from the track at point .

♦♦♦ 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 Figure 5. 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 Figure 5.   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in Figure 5? 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 Figure 5.
  4. What is the average voltage measured across the output terminals for the first quarter turn? Show your working.  (2 marks)

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a.    Alternator.

b.   →

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

c.    

d.    

e.    

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.   →

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

c.    
 

d.    
 

e.    
   
   
♦ 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 series, was launched into a circular orbit at an altitude of 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)

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a.    → Only force acting on a satellite is , the force due to gravity.

→ This force acts on the satellite directly towards the centre of the Earth.v

b.    

Show Worked Solution


a.    → Only force acting on a satellite is , 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.    

PHYSICS, M6 2019 VCE 1

A particle of mass and charge travelling at velocity enters a uniform magnetic field , as shown in Figure 1.
 

  1. Is the charge 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)

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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 .

→ 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 .

→ 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.
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Show Worked Solution

→

→ If the velocity of the electron is doubled, the wavelength of the electron will halve.

→ Using Young’s double slit calculations:

→ Therefore, if lambda decreases so will the fringe spacing .

♦ 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)

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→ 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  ,  where  .

  1. Calculate the average rate of change of between and .  (1 mark)

  2. Calculate the average value of between and .  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions    and the -axis from to .
  4. The heights of the left and right edges of each trapezium are the values of , as shown in the graph below.

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

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a.   

b.   

c.   

Show Worked Solution
a.    
   
   

 

b.   
   
   
   

 

c.    
   
   

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. The mean titration volume was 29.50 mL. The molar mass of citric acid is 192.12 g mol.

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

  1. 1.04 g L
  2. 41.6 g L
  3. 125 g L
  4. 374 g L
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Show Worked Solution

Multiply by molar mass:.

PHYSICS, M8 2020 VCE 17

Figure 17 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)

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a.    

b.    

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he 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.    ,  therefore

→ The photon with the lowest energy will have the highest wavelength.

→ Lowest energy spectral line = 668 nm
 

b.    Convert: 588 nm = 588 × 10 m


 

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he 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 m s illuminates a crystal, producing a diffraction pattern as shown in Figure 16. 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)

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a.    

b.    

Show Worked Solution

a.    
   
   
   
   

 

b.    
   
   
   

 

 
   
   
   
   
♦ 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 Figure 14.

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 Figure 15.
  2. On Figure 15, 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 on Figure 15?  (1 mark)

  2. Why does the photocurrent fall to zero at the point labelled on Figure 15 ?  (1 mark)

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a.   
         

b.    The stopping voltage.

c.    → 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.    → 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, and , that are 4.0 × 10 m apart, as shown in Figure 11a.

Rays from the slits meet on a screen 2.00 m from the slits to produce an interference pattern. Point is at the centre of the pattern. Figure 11b shows the pattern obtained on the screen.
 

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

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a.    Bright fringe at point :

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point 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.    

Show Worked Solution

a.    Bright fringe at point :

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point 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    and 

→ is the distance between two bright bands and is the distance from the slits to the screen.

 
 
   
   
♦ Mean mark (b) 40%.

PHYSICS, M7 2020 VCE 11

An astronaut has left Earth and is travelling on a spaceship at 0.800 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)

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a.   

b.   → 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:

.
 

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:

 
 
 

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

b.   → 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

Figure 8 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 to indicate on Figure 8 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.   

Show Worked Solution

a.    Only two forces are acting on the ball: and :
 

   

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

b.   
     

 
 
 
 
   
   
♦ 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  .

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

  1. Find the values of for which .   (2 marks)

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a.
   

b.   

Show Worked Solution

a.
     

b.     

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 , where

is both smooth and continuous at .

The graph of    is shown below, where is the horizontal distance from the start of the ramp and is the height of the ramp. All lengths are in centimetres.

  1. Find for .   (2 marks)

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

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

  4. iii. Find the interval of 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 . The first support has its left edge placed at and the second support has its left edge placed at . 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 and the -axis between and . 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 , where .   (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 , where

  4. is continuous and differentiable at , and is where the car lands on the ground after the jump, such that .
    1. Find the values of and .   (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.   

b.i   

b.ii 

b.iii

c.   

d.   

e.i   

e.ii 

Show Worked Solution

a.   

b.i   


  

b.ii 

b.iii

c.   

 

  

 
 

    
d.   


 

e.i   

  


 

e.ii 

PHYSICS, M6 2020 VCE 3

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

Electrons travelling at the required velocity, , exit the aperture at point , while electrons travelling slower or faster than the required velocity, , 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 is given by  = .   (1 mark)

  2. Calculate the magnitude of the velocity, , of an electron that travels straight through the aperture to point if  = 500 kV m  and  = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – , or – in Figure 2 could electrons travelling faster than arrive?   (1 mark)

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

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a.    Find where forces due to magnetic and electric field are balanced

 
 
 

 
b.    

c.i.    Point

c.ii.  When electrons travel faster than :

→ 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

Show Worked Solution

a.    Find where forces due to magnetic and electric field are balanced

 
 
 
♦ Mean mark (a) 46%.

b.   
 

c.i.    Point
 

c.ii.  When electrons travel faster than :

→ 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

♦ 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 and for this test.   (1 mark)

  2. Find the 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 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)

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a.   

b.   

c.   

d.   

e. 

Show Worked Solution

a.   
 

b.   


 

c.   


 

d.   


 

e.   

♦ 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  seconds it passes through a fixed origin . The student aims to hit the ball into a hole that is 7 m from . When the ball passes through , its path makes an angle of degrees to the forward direction, as shown in the diagram below.
 

The path of the ball seconds after passing through is given by

  for 

where is a unit vector to the right, perpendicular to the forward direction, is a unit vector in the forward direction and displacement components are measured in metres.

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

  2.  i. Find the speed of the ball as it passes through . 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 ? Give your answer in metres, correct to three decimal places.   (2 marks)

Show Answers Only

a.   

b.i. 

b.ii. 

c.   

d.   

Show Worked Solution

a.   

♦ Mean mark (a) 46%.
b.i.    
   
   

 

b.ii. 

 


 

c.    

♦ Mean mark (c) 45%.
d.    
   
♦ Mean mark (d) 48%.

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver nitrate powder into 250.0 mL of 1.00 mol L 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.   

b.   

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.   
 

b.   


 

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 nitrate powder into 200.0 mL of 1.25 mol L 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.   

b.   

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.   
 

b.   


 

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 in oxygen . The balanced chemical equation for this reaction is:

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

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Show Worked Solution

Calculate the number of moles of propane ():

 
 

 Use the stoichiometric ratio to find the moles of produced:

 
 

Calculate the volume of at STP:

 

→ 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 .

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
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Show Worked Solution

    

Calculus, MET2 2023 VCE SM-Bank 1

The function is defined as follows.

  1. Sketch the graph of 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 at the point where .   (1 mark)

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

Newton's method is used to find an approximate -intercept of , with an initial estimate of .

  1. Find the value of .   (1 mark)

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

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

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

Show Answers Only
a.    
  
b.i
  
b.ii
  
c.
  
d.
  
e.i
  
e.ii
Show Worked Solution

a.

b.i   
 
 
 

  

b.ii

 
c.   

 
 


  
 

d.    

 

e.i. 

 
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.   

b.   

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  :


 

b.    To find the new volume when pressure doubles:


 

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 36

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.   → 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.    
   
 
   

 


 

b.   → 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 35

Explain why ethanol is a liquid at room temperature whereas ethane is a gas.    (3 marks)

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→ Ethanol is a polar molecule with an 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 , 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 is a polar molecule with an 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 , 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 34

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)

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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 31

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)

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→ 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?
 



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Show Worked Solution

CHEMISTRY, M1 EQ-Bank 28

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

    • atomic radii
    • first ionisation energy
    • electronegativity

Justify 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
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Show Worked Solution


  

 
 
 

 

Calculus, MET2 2023 VCE SM-Bank 2 MC

Newton's method is being used to approximate the non-zero -intercept of the function with the equation  .  An initial estimate of  is used.

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

  5. The intercept cannot be correctly approximated using Newton's method.
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Show Worked Solution

  

  

Calculus, MET2 2022 VCAA 5

Consider the composite function , where the function is an unknown but differentiable function for all values of .

Use the following table of values for and .

 

  1. Find the value of .   (1 mark)

The derivative of with respect to is given by .

  1. Show that .   (1 mark)

  2. Find the equation of the tangent to at .   (2 marks)

  3. Find the average value of the derivative function between and .   (2 marks)

  4. Find four solutions to the equation for the interval .   (3 marks)

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a.   

b.   

c.   

d.   

e.   

Show Worked Solution
 

a. 
 
   
   

 

b.   
 
   

 

c.     and 

 
 
 

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

  
d.   The average value of between and

Average  
   
   
   

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted , not .

e.  

  or 

(1):         
    
 
 
     
(2):  
 
 

  


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find . Some solved or but not both.

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of of landing on heads and of landing on tails.

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

Mika flips the coin five times.

    1. Find .   (1 mark)
    2. Find (1 mark)
    3. Find , correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for .   (2 marks)

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


  

where is the vertical height reached by the coin flip, in metres, between the coin and the floor, and and are real constants.

    1. State the value of the definite integral .   (1 mark)
    2. Given that    and  , find the values of and .   (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, , with probability density function .
    4. The function is a transformation of the function given by , where is the minimum distance between the coin and the ceiling, and and are real constants.
    5. Find the values of and .   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of of landing on heads and of landing on tails, where is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate .
  3. Let be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable discrete or continuous? Justify your answer.   (1 mark)
    2. If , find an approximate 95% confidence interval for , 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 , 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.        ii.        iii.    (3 d.p.)

a. iv   

b. i.      

b. ii.   

b. iii.

c. i.      ii.      iii.   

Show Worked Solution

a.i 


 

a.ii  By CAS:        

  
a.iii 

  
By CAS:
 

(3 decimal places)
  

a.iv

  

b.i  


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

b.ii  By CAS:


 


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

b.iii 

and


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

c.i    is discrete.

The number of coin flips must be zero or a positive integer so    is countable and therefore discrete.
 

c.ii 


 

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

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.

Complex Numbers, SPEC2 2022 VCAA 2

Two complex numbers and are given by   and  , where .

  1.  i. Given that  , show that  .   (2 marks)

  2. ii. One set of possible values for and is    and  .
  3.     Hence, or otherwise, find the other set of possible values.   (1 mark)

  4. Plot and label the points representing    and    on the Argand diagram below.
     

  1. The ray given by    passes through the midpoint of the line interval that joins the points    and  .
  2. Find, in radians, the value of and plot this ray on the Argand diagram in part b.   (2 marks)

  3. The line interval that joins the points    and    cuts the circle    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)

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a.i. 

a.ii. 

b.   
       

c.   

d.   

Show Worked Solution

a.i. 

 

 

 
 
Mean mark (a) 54%.

 
a.ii.


 

b.   
       


c.    
♦ Mean mark (c) 39%.

d.   

♦ Mean mark (d) 40%.

Calculus, SPEC2 2022 VCAA 1

Consider the family of functions with rule  , where .

  1. Write down the equations of the two asymptotes of the graph of when .   (2 marks)

  2. Sketch the graph of    for    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 , the equations of the asymptotes of the graph of  .   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  in terms of .   (2 marks)

  3. Now consider the functions and , where    and  .
  4. The region bounded by the curves of and is rotated about the -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. 

b.   
       

c.i.   

c.ii. 

d.i. 

d.ii. 

Show Worked Solution

a.   


 

b.    
       

 

c.i.


 

c.ii. 


 

d.i 


 

d.ii.

♦♦ Mean mark (d)(ii) 37%.

Statistics, SPEC2 2022 VCAA 18 MC

The time taken, minutes, for a student to travel to school is normally distributed with a mean of 30 minutes and a standard deviation of 2.5 minutes.

Assuming that individual travel times are independent of each other, the probability, correct to four decimal places, that two consecutive travel times differ by more than 6 minutes is

  1. 0.0448
  2. 0.0897
  3. 0.1151
  4. 0.2301
  5. 0.9103
Show Answers Only

Show Worked Solution

♦ Mean mark 42%.

PHYSICS, M5 2020 VCE 11 MC

The International Space Station (ISS) is travelling around Earth in a stable circular orbit, as shown in the diagram below.
 

Which one of the following statements concerning the momentum and the kinetic energy of the ISS is correct?

  1. Both the momentum and the kinetic energy vary along the orbital path.
  2. Both the momentum and the kinetic energy are constant along the orbital path.
  3. The momentum is constant, but the kinetic energy changes throughout the orbital path.
  4. The momentum changes, but the kinetic energy remains constant throughout the orbital path.
Show Answers Only

Show Worked Solution

→ The magnitude of both the kinetic energy and momentum of the ISS remains constant.

→ However, momentum is a vector and as the direction of the ISS is continually changing so is the direction of the momentum of the ISS.

♦♦ Mean mark 31%.
COMMENT: Students need to ensure they consider the direction of vector quantities as well as the magnitudes.

PHYSICS, M8 2021 VCE 19

A simplified diagram of some of the energy levels of an atom is shown in Figure 16.
 

  1. Identify the transition on the energy level diagram in Figure 16 that would result in the emission of a 565 nm photon. Show your working.   (2 marks)

  1. A sample of the atoms is excited into the 9.8 eV state and a line spectrum is observed as the states decay. Assume that all possible transitions occur.

    What is the total number of lines in the spectrum? Explain your answer. You may use the diagram below to support your answer.   (2 marks)
     

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a.    The energy transition from 8.9 eV to 6.7 eV.

b.    9 spectral lines.

Show Worked Solution

a.    

→ The energy transition from 8.9 eV to 6.7 eV will result in the emission of a 565 nm photon.

→ Note: This could have also been shown on the diagram as a downwards arrow from 8.9 eV to 6.7 eV.
 

♦ Mean mark (a) 50%.

b.    Each photon emitted that has its own unique energy will result in a spectrum line.

→ There are 4 possible transitions from 9.8 eV, 3 from 8.9 eV, 2 from 6.7 eV and 1 from 4.9 eV, resulting in 10 different transitions.

→ However, the transition from 9.8 eV to 4.9 eV and 4.9 eV to 0 eV will produce a photon of the same energy, hence they will have the same spectral line. 

→ Therefore, there will be 9 lines in the spectrum.

♦♦ Mean mark (b) 31%.

Functions, EXT1 F1 2023 MET1 7

Consider . Part of the graph of    is shown below.
 

  1. State the range of .   (1 mark)
  2. Sketch the graph of the inverse function  on the axes above. Label any endpoints and axial intercepts with their coordinates.   (2 marks)
  3. Determine the equation of the domain for the inverse function  .   (2 marks)
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a.   

b.   

c.   

Show Worked Solution

a.   

b.   

c.   


 

 


♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as .

PHYSICS, M7 2021 VCE 16

Light can be described by a wave model and also by a particle (or photon) model. The rapid emission of photoelectrons at very low light intensities supports one of these models but not the other.

Identify the model that is supported, giving a reason for your answer.

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→ Model supported: Particle (photon) model

Reasons could include one of the following:

→ Each electron ejection corresponds to the absorption of a single photon.

→ The immediate ejection of a photoelectron corresponds to its direct interaction with the initial photon.

→ Contrary to the wave theory, which suggests that energy accumulates gradually, the emission of photoelectrons at very low light intensities demonstrates that energy is delivered in discrete quanta.

Show Worked Solution

→ Model supported: Particle (photon) model

Reasons could include one of the following:

→ Each electron ejection corresponds to the absorption of a single photon.

→ The immediate ejection of a photoelectron corresponds to its direct interaction with the initial photon.

→ Contrary to the wave theory, which suggests that energy accumulates gradually, the emission of photoelectrons at very low light intensities demonstrates that energy is delivered in discrete quanta.

♦ Mean mark 42%.

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all , where time is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule .

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule .

  1. Show that and .   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of under given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of under the transformation . Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where , in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)

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ai.   

aii.  Minimum population of rabbits

Maximum population of rabbits

aiii.   weeks

b.    See worked solution.

c.    (nearest whole number)

d.    Weeks between the periods is 160

e.    (nearest whole number)

f.    Average rate of change rabbits/week (1 d.p.)

g.    weeks (nearest whole number)  

h.   

Show Worked Solution

ai.  Initial population of rabbits

From graph when

Using formula when

 
  rabbits  


aii. From graph,

Minimum population of rabbits

Maximum population of rabbits

OR

Using formula

Minimum is when

Maximum is when

 
aiii. Number of weeks between maximum populations of rabbits   weeks

 
b.  Period of foxes = period of rabbits = 160:

which is the same period as the rabbit population.

Using the point

Amplitude when : 

 
 
 
 
 

 

  

c.    Using CAS find :

  
     
  

  
Maximum combined population (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

    
 

 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

   →  

  →  

  
Average combined population  [Using CAS]   
  

  

(nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted . Others used average rate of change instead of average value.

f.   Using CAS


 

       
 


 

    
 


 

Av rate of change between the points

  and 


 

Average rate of change rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

,

After testing greatest positive value occurs for  

 
   
   

 
weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved .
A common answer was 41.8, as .
Another common incorrect answer was 76 weeks.

h.   As ,

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of , where .
 

  1. State the equation of the axis of symmetry of the graph of .   (1 mark)

  2. State the derivative of with respect to .   (1 mark)

The tangent to at point has gradient .

  1. Find the equation of the tangent to at point .   (2 marks)

The diagram below shows part of the graph of , the tangent to at point and the line perpendicular to the tangent at point .
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point .   (1 mark)

  2. ii. The line perpendicular to the tangent at point also cuts at point , as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve .   (2 marks)

  4. Another parabola is defined by the rule , where .
  5. A tangent to and the line perpendicular to the tangent at , where , are shown below.

  1. Find the value of , in terms of , such that the shaded area is a minimum.   (4 marks)

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a.   

b.   

c.    

di.   

dii.   Area units²

e.   

Show Worked Solution

a.   Axis of symmetry: 
  

b.       
   

  
c.   At gradient

 
 

 
When 
 


 
Equation of tangent at :

 
 
 


d.i   Gradient of tangent

  gradient of normal  

Equation at

 
 
 


d.ii  Points of intersection of and normal are at  and .

So equate and  to find

 
 
 

   
or

Area  
   
   
  units²  

 

e.     

At   

Gradient of tangent

Gradient of normal

Equation of normal at

 
 
 
 

 
Points of intersection of normal and parabola (Using CAS)

solve

  or 

 
Calculate area using CAS


 
Using CAS Solve derivative of   with respect to to find

solve

 
   and 

Given


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found and failed to subtract or had incorrect terminals.

Calculus, MET2 2022 VCAA 19 MC

A box is formed from a rectangular sheet of cardboard, which has a width of units and a length of units, by first cutting out squares of side length units from each corner and then folding upwards to form a container with an open top.

The maximum volume of the box occurs when is equal to

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Show Worked Solution

Maximum occurs when

Solving for using CAS.

  or 

Testing shape of the cubic using and

Maximum occurs at the smaller value of


♦♦ Mean mark 34%.
MARKER’S COMMENT: 34% of students incorrectly chose option B. 

Calculus, MET2 2022 VCAA 17 MC

A function is continuous on the domain and has the following properties:

  • The average rate of change of between and is positive.
  • The instantaneous rate of change of at is negative.

Therefore, on the interval , the function must be

  1. many-to-one.
  2. one-to-many.
  3. one-to-one.
  4. strictly decreasing.
  5. strictly increasing.
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Show Worked Solution

The first property shows that the point ( , __ ) must be below ( , __ ) to give a positive rate of change.

The second property gives the midpoint of and . But this point has a negative instantaneous rate of change.

Therefore, this shows that the function has turning points and is in the shape of a cubic, so many-to-one is the correct description.

 


♦♦ Mean mark 39%.