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BIOLOGY, M7 2021 HSC 14 MC

The human PRNP gene codes for the prion protein. Misfolding of this protein may be the result of ingesting tissue that contains misfolded prion protein or a mutation. Accumulation of misfolded prion protein causes serious diseases such as Creutzfeldt-Jakob disease (CJD).

Which of the following statements best classifies CJD?

  1. It is both a genetic disease and an infectious disease.
  2. It is a genetic disease only, since it is encoded by a gene.
  3. It is not an infectious disease because the prion is non-cellular.
  4. It is an infectious disease and the normal prion protein is the pathogen.
Show Answers Only

`A`

Show Worked Solution

→ Infectious, as CJD can be caused by an infectious agent, i.e. ingesting tissue that contains the misfolded protein.

→ Genetic, as CJD can be caused by mutation.

`=>A`

♦♦ Mean mark 39%.

BIOLOGY, M8 2021 HSC 13 MC

The photographs show an open and a closed stomate on a leaf surface. When open, stomates allow water vapour to pass out of the leaf.
 

Regulating stomates is a mechanism by which plants maintain water balance.

Which of the following graphs best illustrates this homeostatic mechanism?
 

 

Show Answers Only

`A`

Show Worked Solution

→ Higher availability of water, stomata can open more to release more water through transpiration.

→ Lower availability of water, stomata will partially close to reduce water loss.

`=>A`

Mean mark 53%.

PHYSICS, M6 2020 HSC 34

A charged particle, `q_1`, is fired midway between oppositely charged plates `X` and `Y`, as shown in Figure 1. The voltage between the plates is `V` volts.

The particle strikes plate `Y` at point `P`, a horizontal distance `s` from the edge of the plate. Ignore the effect of gravity.
 

Plate `Y` is then moved to the position shown in Figure 2, with the voltage between the plates remaining the same.

An identical particle, `q_2`, is fired into the electric field at the same velocity, entering the field at the same distance from plate `X` as `q_1`.

  1. Compare the work done on `q_1` and `q_2`.   (3 marks)
  1. Compare the horizontal distances travelled by `q_1` and `q_2` in the electric field.   (3 marks)
Show Answers Only

a.   The work done on `q_2` is `(3)/(2)` times that done on `q_1`.

b.   The horizontal distance travelled by `q_2` is `sqrt(6)` times that of `q_1`.

Show Worked Solution

a.   Work done on `q_1:`

`W=qE((d)/(2))` … (1)

Substitute  `E=(V)/(d)`  into (1)

`W=(qV)/2`

 
Work done on `q_2:`

`W=qE((3d)/(2))` … (1)

Substitute  `E=(V)/(2d)`  into (1)

`W=(3qV)/4`

`(2) -: (1):`

`(W_(2))/(W_(1))=(((3qV)/(4)))/(((qV)/(2)))=(3)/(2)`

→ The work done on `q_2` is `3/2` times that done on `q_1`.
 

♦ Mean mark (a) 47%.

b.   Find the vertical acceleration of `q_1:`

`F` `=qE=(qV)/(d)`
`a` `=(F)/(m)=(qV)/(dm)` … (1)

 
Similarly for `q_2:` 

`F` `=qE=(qV)/(2d)`
`a` `=(F)/(m)=(qV)/(2dm)` … (2)

 
→ The initial horizontal velocity is the same for both `q_1` and `q_2`

→ The horizontal distance travelled by both is proportional to the time taken for them to reach the plate.

→ `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)=(1)/(2)a_(y)t^(2)`  (initial vertical velocity = 0 for both)
 

Time for `q_1` to reach plate `Y:` 

`s_(y)` `=(1)/(2)a_(y)t^(2)`  
`(d)/(2)` `=(1)/(2)((qV)/(dm))t^(2)`    from (1)
`t^(2)` `=(d^(2)m)/(qV)`  
`t` `=sqrt((d^(2)m)/(qV))` …   (3)

 
Time taken for `q_2` to reach plate `Y:` 

`s_(y)` `=(1)/(2)a_(y)t^(2)`  
`(3d)/(2)` `=(1)/(2)((qV)/(2dm))t^(2)`    from (2)
`t^(2)` `=(6d^(2)m)/(qV)`  
`t` `=sqrt((6d^(2)m)/(qV))` …   (4)

 
`(4) -: (3):`

`(t_(2))/(t_(1))=(sqrt((6d^(2)m)/(qV)))/(sqrt((d^(2)m)/(qV)))=sqrt(6)`

→ The time taken for `q_2` to reach plate `Y` is `sqrt(6)` times that taken for `q_1`.

→ The horizontal distance travelled by `q_2` is `sqrt(6)` times that of `q_1`.

♦ Mean mark (b) 43%.

PHYSICS, M6 2020 HSC 33

A strong magnet of mass 0.04 kg falls 0.78 m under the action of gravity from position `X` above a hollow copper cylinder. It then travels a distance of 0.20 m through the cylinder from `Y` to `Z` before falling freely again.
 

The magnet takes 0.5 seconds to pass through the cylinder. The displacement-time graph of the magnet is shown.
 

Analyse the motion of the magnet by applying the law of conservation of energy.

Your analysis should refer to gravity and the copper cylinder, and include both qualitative and quantitative information.   (9 marks)

Show Answers Only

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91`  m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

Show Worked Solution

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91` m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

♦ Mean mark 51%.

PHYSICS M6 2022 HSC 34

Three charged particles, `X, Y` and `Z`, travelling along straight, parallel trajectories at the same speed, enter a region in which there is a uniform magnetic field which causes them to follow the paths shown. Assume that the particles do not exert any significant force on each other.
 


 

Explain the different paths that the particles follow through the magnetic field.   (7 marks)

Show Answers Only

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

`F_(c)` `=F_(b)`  
`(mv^2)/(r)` `=qvB`  
`r` `=(mv)/(qB)`  

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.

Show Worked Solution

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

`F_(c)` `=F_(b)`  
`(mv^2)/(r)` `=qvB`  
`r` `=(mv)/(qB)`  

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.


Mean mark 60%.

PHYSICS M8 2022 HSC 31

Following the Geiger-Marsden experiment, Rutherford proposed a model of the atom.
 


 

Bohr modified this model to explain the spectrum of hydrogen observed in experiments.
 


 

The Bohr-Rutherford model of the atom consists of electrons in energy levels around a positive nucleus.

How do features of this model account for all the experimental evidence above? Support your answer with a sample calculation and a diagram, and refer to energy, forces and photons.  (9 marks)

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The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy  `E=hf`  equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy  `E=hf`  equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

`(1)/(lambda)` `=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`  
  `=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`  
  `=(2 xx1.097 xx10^7)/(9)`  
  `=2.438 xx10^6`  
`lambda` `=410  text{nm}`  

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.

 

Show Worked Solution

The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy  `E=hf`  equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy  `E=hf`  equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

`(1)/(lambda)` `=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`  
  `=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`  
  `=(2 xx1.097 xx10^7)/(9)`  
  `=2.438 xx10^6`  
`lambda` `=410  text{nm}`  

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.


♦ Mean mark 51%.

PHYSICS M7 2022 HSC 30

In a thought experiment, light travels from `X` to a mirror `Y` and back to `X` on a moving train carriage. The path of the light relative to an observer on the train is shown.
 


 

Relative to an observer outside the train, the path of the light is shown below, at three consecutive times as the train carriage moves along the track.
 

  1. Describe qualitatively how the constancy of the speed of light and the thought experiment above led Einstein to predict time dilation.  (3 marks)

  1. The train is travelling with a velocity  `v=0.96 c`. To the observer inside the train, the return journey for the light between `X` and `Y` takes 15 nanoseconds.

  2. How long would this return journey take according to the observer outside the train?  (3 marks)

Show Answers Only

a.  Consider each observer:

→ The observer on the train sees light travel a distance from `X` to `Y` and back to `X` again.

→ The observer outside the train sees light travel a longer path due to the horizontal motion of the train.

→ As the speed of light is constant for both observers, the observer outside the train must observe the light pulse to take a longer time, `t=(text{distance})/(c).`

→ This provided the basis for Einstein’s predictions of time dilation.
 

b.  53.6  ns

Show Worked Solution

a.  Consider each observer:

→ The observer on the train sees light travel a distance from `X` to `Y` and back to `X` again.

→ The observer outside the train sees light travel a longer path due to the horizontal motion of the train.

→ As the speed of light is constant for both observers, the observer outside the train must observe the light pulse to take a longer time, `t=(text{distance})/(c).`

→ This provided the basis for Einstein’s predictions of time dilation.


♦ Mean mark (a) 48%.
b.  `t`

 
 
`=(t_(0))/(sqrt(1-(v^(2))/(c^(2))))`
    `=(15)/(sqrt(1-((0.96c)^(2))/(c^(2))))`
    `=53.6  text{ns}`

PHYSICS M5 2022 HSC 29

An apple was thrown horizontally to the east from the window of a car which was moving with a uniform velocity to the north.

Explain the horizontal and vertical components of the apple's motion during its flight.  (4 marks)

Show Answers Only

Horizontal component:

→ Initially, the horizontal component of the apple’s velocity is the vector sum of the car’s velocity to the north, and the velocity to the east at which the apple was thrown. 

→ There is no horizontal force, and consequently no horizontal acceleration of the apple. So, its horizontal velocity remains constant throughout its flight.
 

Vertical component:

→ The apple has zero initial vertical velocity.

→ There is a downwards force acting on the apple due to gravity. So, the apple accelerates vertically downwards causing its downwards velocity to increase at a constant rate until it hits the floor.

Show Worked Solution

Horizontal component:

→ Initially, the horizontal component of the apple’s velocity is the vector sum of the car’s velocity to the north, and the velocity to the east at which the apple was thrown. 

→ There is no horizontal force, and consequently no horizontal acceleration of the apple. So, its horizontal velocity remains constant throughout its flight.
 

Vertical component:

→ The apple has zero initial vertical velocity.

→ There is a downwards force acting on the apple due to gravity. So, the apple accelerates vertically downwards causing its downwards velocity to increase at a constant rate until it hits the floor.


Mean mark 51%.

PHYSICS M8 2022 HSC 28

Two steps in the CNO cycle of nuclear fusion are shown.
 

Step `X` releases 1.20 MeV.

The masses in Step `Y` are shown in the table.
 

Propose a reason why Step `Y` releases more energy than Step `X`. Support your answer with calculations.  (3 marks)

Show Answers Only

→ The energy released by a nuclear reaction is proportional to the mass defect.

→ Step `Y` will release more energy if it has a greater mass defect than step `X`.
 

Mass defect (Step `Y`) `=m_(reactants)-m_(pr oducts)`  
  `=1.007 + 13.003-14.003`  
  `=0.007u`  

 

`E` `=0.007 xx931.5`  
  `=6.5  text{MeV}`  

  
Which is greater than the energy released by step `X`.

Show Worked Solution

The energy released by a nuclear reaction is proportional to the mass defect. Step `Y` will release more energy if it has a greater mass defect than step `X`.
 

Mass defect (Step `Y`) `=m_(reactants)-m_(pr oducts)`  
  `=1.007 + 13.003-14.003`  
  `=0.007u`  

 

`E` `=0.007 xx931.5`  
  `=6.5  text{MeV}`  

  
Which is greater than the energy released by step `X`.


Mean mark 51%.

PHYSICS M7 2022 HSC 27

A laser producing red light of wavelength 655 nm is directed onto double slits separated by a distance, `d=5.0 xx 10^{-5} \ text{m}`. A screen is placed behind the double slits.
 


 

  1. Newton proposed a model of light. Use a labelled sketch to show the pattern on the screen that would be expected from Newton's proposed model.   (2 marks)
     

     
  2. When the laser light is turned on, a series of vertical bright lines are seen on the screen.
     
 
  1. Calculate the angle, `\theta`, between the centre line and the bright line at `A`.   (3 marks)
  1. The laser is replaced with one producing green light of wavelength 520 nm.
  2. Explain the difference in the pattern that would be produced.   (2 marks)
Show Answers Only

a.   
       

 
b.   `theta=1.50^@`

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

Show Worked Solution
a.   
     

Mean mark part (a) 53%.

b.   Using  `d  sin  theta=mlambda:`

`5.0 xx10^(-5)  sin  theta` `=2 xx6.55xx10^(-7)`
   `sin  theta` `=(2 xx6.55xx10^(-7))/(5.0 xx10^(-5))`
   `theta` `=1.50^@`

 

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

PHYSICS, M7 2016 HSC 27b

Explain how the result of ONE investigation of the photoelectric effect changed the scientific understanding of the nature of light.   (4 marks)

Show Answers Only

→ One investigation of the photoelectric effect showed that the kinetic energy of emitted photoelectrons increased when the frequency of incident light increased, but did not increase with increased light intensity.

→ This could only be explained by a particle, or photon model of light.

→ The photon model considers light to consist of discrete packets of energy, where the energy of a photon is proportional to its frequency.

→ Thus, higher frequency light → greater photon energy → greater kinetic energy of emitted photoelectrons.

→ The scientific understanding of light was changed from a wave model where light existed only as transverse waves of different wavelengths to a wave-particle duality.

Show Worked Solution

→ One investigation of the photoelectric effect showed that the kinetic energy of emitted photoelectrons increased when the frequency of incident light increased, but did not increase with increased light intensity.

→ This could only be explained by a particle, or photon model of light.

→ The photon model considers light to consist of discrete packets of energy, where the energy of a photon is proportional to its frequency.

→ Thus, higher frequency light → greater photon energy → greater kinetic energy of emitted photoelectrons.

→ The scientific understanding of light was changed from a wave model where light existed only as transverse waves of different wavelengths to a wave-particle duality.


♦ Mean mark 40%.

PHYSICS, M7 2014 HSC 26

  1. Calculate the energy of a photon of wavelength 415 nm.  (2 marks)
  2. An experiment was conducted using a photoelectric cell as shown in the diagram.

The graph plots the maximum kinetic energy of the emitted photoelectrons against radiation frequency for the aluminium surface.

The experiment is planned to be repeated using a voltage of 0.0 V.

Draw a line on the graph to show the predicted results of the planned experiment, and determine the radiation frequency which would produce photoelectrons with a maximum kinetic energy of 1.2 eV using a voltage of 0.0 V.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
a.   `c` `= f lambda`
    `= hf`
    `= (hc)/lambda`
    `= (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1))/(415 xx 10^(-9)m)`
    `= 4.79 xx 10^(-19) J`

 

b.  

`text(To find intercept)`

`4.1 text(V)` `= 4.1 xx 1.602 xx 10^(-19)\ text(J)` `text(of energy required to be)`
`text(supplied by the photon.)`
  `= 6.56 xx 10^(-19)\ text(J)`  
`hf` `= 6.56 xx 10^(-19)\ text(J)`  
`f` `= (6.56 xx 10^(-19))/(6.626 xx 10^(-34))`  
  `= 9.9 xx 10^14`  

`text(Gradient = same as A1)`

`text{From graph maximum KE(eV)}` `= 1.2\ text(eV)`
`text(Frequency)` `= 12.8\ text(Hz)`
  `= 12.8 xx 10^14\ text(Hz)`

PHYSICS M7 2022 HSC 26

Light of frequency  `7.5 xx10^(14) \ text{Hz}`  is incident on a calcium metal sheet which has a work function of  `2.9 \ text{eV}`. Photoelectrons are emitted.

The metal is in a uniform electric field of `5.2 \ text{NC}^{-1}`, perpendicular to the surface of the metal, as shown.
 


  

  1. Show that the maximum kinetic energy of an emitted photoelectron is  `3.2 xx10^(-20) \ text{J}`.  (3 marks)
  1. Calculate the maximum distance, `d`, an emitted photoelectron can travel from the surface of the metal.  (3 marks)
Show Answers Only
  1. Proof (See Worked Solution)
  2. `0.038  text{m.}`
Show Worked Solution
a.    `K_(max)` `=hf-Phi`
    `=6.626 xx10^(-34)xx7.5 xx10^(14)-2.9xx 1.602 xx10^(-19)`
    `=3.2 xx10^(-20)  text{J}`

 

b.   Maximum Distance:

→ The electric field will do `3.2 xx10^(-20)\ text{J}` of work to stop the electron.

`W` `=qEd`     
`3.2 xx10^(-20)` `=1.602 xx10^(-19) xx 5.2 xx d_max`  
`:.d_max` `=(3.2 xx10^(-20))/(1.602 xx10^(-19)xx 5.2)`  
  `=0.038  text{m.}`  

♦ Mean mark (b) 51%.

PHYSICS M5 2022 HSC 25

A rocket is launched vertically from a planet of mass \(M\). After it leaves the atmosphere, the rocket's engine is turned off and it continues to move away from the planet. From this time the rocket's mass is 200 kg. The rocket's speed, \(v\), at two different distances from the planet's centre, \(R\), is shown.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex} \text {Point } \rule[-1ex]{0pt}{0pt}& R\ \text{(m)} & \quad v\left(\text{m s}^{-1}\right) \quad \\
\hline \rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& \quad 4.3 \times 10^6 \quad & 5500 \\
\hline \rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& 2.5 \times 10^7 & 2900 \\
\hline
\end{array}

  1. Show that the magnitude of the change in kinetic energy from point 1 to point 2 is  \(2.2 \times 10^9 \ \text{J}\).  (2 marks)

  1. Determine the mass \(M\) of the planet using the law of conservation of energy.  (3 marks)

Show Answers Only

a.  \(\Delta K=K_f-K_i\)

\(K_f\) \(=\dfrac{1}{2} m v_2^2\)
  \(=\dfrac{1}{2} 200 \times 2900^2\)
  \(=8.41 \times 10^8 \ \text{J}\)
\(K_i\) \(=\dfrac{1}{2} m v_1^2\)
  \(=\dfrac{1}{2} 200 \times 5500^2\)
  \(=3.025 \times 10^9 \ \text{J}\)

 

  \(\therefore \Delta K\) \(=8.41 \times 10^8-3.025 \times 10^9\)
    \(=-2.2 \times 10^9 \ \text{J}\)

This change has a magnitude of \(2.2 \times 10^9 \ \text{J}\).
 

b.   Applying the law of conservation of energy:

→ The magnitude of kinetic energy lost is equal to the magnitude of potential energy gained by the rocket:

\(\Delta U\) \(=U_f-U_i\)  
  \(=-\dfrac{G M m}{r_f}-\left(-\dfrac{G M m}{r_i}\right)\)  

 
\(\left(2.2 \times 10^9\right)=-\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{2.5 \times 10^7}+\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{4.3 \times 10^6}\)
 

\(\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)\)

\(=\left(6.67 \times 10^{-11}\right) 200 M\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^7\right)\right]\)
 

\(\therefore M\) \(=\dfrac{\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)}{200\left(6.67 \times 10^{-11}\right)\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^6\right)\right]}\)  
  \(=8.56 \times 10^{23} \ \text{kg}\)  
Show Worked Solution

a.  \(\Delta K=K_f-K_i\)

\(K_f\) \(=\dfrac{1}{2} m v_2^2\)
  \(=\dfrac{1}{2} 200 \times 2900^2\)
  \(=8.41 \times 10^8 \ \text{J}\)
\(K_i\) \(=\dfrac{1}{2} m v_1^2\)
  \(=\dfrac{1}{2} 200 \times 5500^2\)
  \(=3.025 \times 10^9 \ \text{J}\)

 

  \(\therefore \Delta K\) \(=8.41 \times 10^8-3.025 \times 10^9\)
    \(=-2.2 \times 10^9 \ \text{J}\)

This change has a magnitude of \(2.2 \times 10^9 \ \text{J}\). 

b.   Applying the law of conservation of energy:

→ The magnitude of kinetic energy lost is equal to the magnitude of potential energy gained by the rocket:

\(\Delta U\) \(=U_f-U_i\)  
  \(=-\dfrac{G M m}{r_f}-\left(-\dfrac{G M m}{r_i}\right)\)  

 
\(\left(2.2 \times 10^9\right)=-\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{2.5 \times 10^7}+\dfrac{\left(6.67 \times 10^{-11}\right)(200) M}{4.3 \times 10^6}\)
 

\(\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)\)

\(=\left(6.67 \times 10^{-11}\right) 200 M\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^7\right)\right]\)
 

\(\therefore M\) \(=\dfrac{\left(2.2 \times 10^9\right)\left(2.5 \times 10^7\right)\left(4.3 \times 10^6\right)}{200\left(6.67 \times 10^{-11}\right)\left[\left(2.5 \times 10^7\right)-\left(4.3 \times 10^6\right)\right]}\)  
  \(=8.56 \times 10^{23} \ \text{kg}\)

♦ Mean mark 47%.

BIOLOGY, M7 2021 HSC 30

A study compared the incidence of disease and survival of 8134 children who had received the measles vaccine with 8134 children from a neighbouring area who remained unvaccinated against measles. Children in each group were matched for age, sex, size of dwelling, number of siblings and maternal education. The graphs show the number of measles cases among the two groups over three years.


 

The table compares the cause of death and number of deaths of the two groups over the same three years.
 

'A vaccine only protects the community against a specific disease.'

Analyse the data with reference to this statement.   (7 marks)

Show Answers Only

→ These two studies show proof that the vaccine protects against measles.

→ The graphs show the incidence of measles in vaccinated children is extremely low, occasionally zero, compared to unvaccinated children. Since these children are matched for age and other sociocultural/socioeconomic factors, it is highly likely the vaccine is responsible for this difference.

→ The table also supports this conclusion, where 40 children who were unvaccinated died due to measles compared to the 2 who died who were vaccinated.

→ The table also provides evidence that the vaccine may protect children from dying from other diseases, contradicting the statement.

→ The table also uses the same groups as the graph, again, keeping cultural/socioeconomic factors similar. It shows that children vaccinated against measles died at half the rate of unvaccinated children due to diarrhoea and dysentery, and at about a quarter of the rate for oedema.

→ However, the small differences between groups with respect to fever and the small sample size for oedema mean that further research is required.

→ In general, the vaccinated group had less than half the mortality of the unvaccinated group, supporting the conclusion that the vaccine provides a wider protection.

→ However, this statement needs qualification as the data is only with respect to the measles vaccine. 

Show Worked Solution

→ These two studies show proof that the vaccine protects against measles.

→ The graphs show the incidence of measles in vaccinated children is extremely low, occasionally zero, compared to unvaccinated children. Since these children are matched for age and other sociocultural/socioeconomic factors, it is highly likely the vaccine is responsible for this difference.

→ The table also supports this conclusion, where 40 children who were unvaccinated died due to measles compared to the 2 who died who were vaccinated.

→ The table also provides evidence that the vaccine may protect children from dying from other diseases, contradicting the statement.

→ The table also uses the same groups as the graph, again, keeping cultural/socioeconomic factors similar. It shows that children vaccinated against measles died at half the rate of unvaccinated children due to diarrhoea and dysentery, and at about a quarter of the rate for oedema.

→ However, the small differences between groups with respect to fever and the small sample size for oedema mean that further research is required.

→ In general, the vaccinated group had less than half the mortality of the unvaccinated group, supporting the conclusion that the vaccine provides a wider protection.

→ However, this statement needs qualification as the data is only with respect to the measles vaccine. 

BIOLOGY, M7 2021 HSC 28b

An mRNA vaccine has been developed in order to immunise people against a virus. The vaccine contains modified mRNA which codes for the spike protein on the surface of the virus.
 

Explain how this vaccine can lead to active immunity to the virus.   (5 marks)

Show Answers Only

→ The modified mRNA enters the individuals cells and is translated at the ribosomes to form the viral spike protein, which is then released into the body.

→ The protein will be rendered as an antigen, and will trigger a specific immune response by B and T lymphocytes that match the antigen.

→ Once the viral protein has been removed, memory B and T lymphocytes specific to the protein remain, providing active immunity that allows for a rapid future response.

→ If the individual is later exposed to the virus, the same spike proteins present on the virus’ surface will trigger a rapid and large response by the memory cells. This is the basis for providing active immunity.

Show Worked Solution

→ The modified mRNA enters the individuals cells and is translated at the ribosomes to form the viral spike protein, which is then released into the body.

→ The protein will be rendered as an antigen, and will trigger a specific immune response by B and T lymphocytes that match the antigen.

→ Once the viral protein has been removed, memory B and T lymphocytes specific to the protein remain, providing active immunity that allows for a rapid future response.

→ If the individual is later exposed to the virus, the same spike proteins present on the virus’ surface will trigger a rapid and large response by the memory cells. This is the basis for providing active immunity.


♦ Mean mark 43%.

BIOLOGY, M5 2021 HSC 27

Sickle cell anaemia is a genetic disorder. In a family, the parents are both known to be heterozygous for the mutation that causes sickle cell anaemia. The couple has two unaffected children and is now expecting a third child. They have had an allele screening test to determine whether the child will have sickle cell anaemia.

A part of the DNA profile is shown. It shows the alleles present.
 

Use the DNA profile provided to justify whether Child 3 will have sickle cell anaemia.   (3 marks)

Show Answers Only

→ Sickle cell anaemia is a condition which affects homozygous recessive (aa) individuals for a particular allele, which is depicted by the bands in the DNA profile.

→ The mother, father and child 2 are heterozygous (Aa) for sickle cell anaemia, and are unaffected. Child 1 is unaffected and homozygous, proving the bottom band is the dominant allele.

→ Child 3 is homozygous for the other allele, meaning he is homozygous recessive and will  have sickle cell anaemia.

Show Worked Solution

→ Sickle cell anaemia is a condition which affects homozygous recessive (aa) individuals for a particular allele, which is depicted by the bands in the DNA profile.

→ The mother, father and child 2 are heterozygous (Aa) for sickle cell anaemia, and are unaffected. Child 1 is unaffected and homozygous, proving the bottom band is the dominant allele.

→ Child 3 is homozygous for the other allele, meaning he is homozygous recessive and will  have sickle cell anaemia.

Mean mark 51%.

BIOLOGY, M5 2021 HSC 26

Zebra populations are suffering from a reduction in their gene pools due to habitat destruction and increasing isolation. This has led to an increase in the number of offspring born with coat patterns different to that of their parents. An example is shown.

Explain possible reasons for the increase in these offspring.   (4 marks)

Show Answers Only

A reduction in gene flow occurs when populations become smaller and more isolated. This results in the accumulation of mutations and a reduction in the gene pool due to inbreeding.

Many of these mutations are recessive, and the repeat inbreeding expresses these mutations in offspring that receive two copies.

This effect will occur more often as inbreeding increases, with a greater percentage of zebra offspring appearing different from their parents.

Show Worked Solution

→ A reduction in gene flow occurs when populations become smaller and more isolated. This results in the accumulation of mutations and a reduction in the gene pool due to inbreeding.

→ Many of these mutations are recessive, and the repeat inbreeding expresses these mutations in offspring that receive two copies.

→ This effect will occur more often as inbreeding increases, with a greater percentage of zebra offspring appearing different from their parents.

♦ Mean mark 46%.

BIOLOGY, M8 2021 HSC 25

A patient visited an audiologist for a hearing test. The audiologist tested both ears at specific frequencies. The volumes at which each frequency could be heard are shown.
 

  1. Plot the data on the grid provided and include a key.   (3 marks)
     

  1. What conclusions can be drawn about the patient's hearing?   (2 marks)

  2. It is discovered that there is a complete and permanent blockage of the outer ear, but the cochlea is still fully functional.
  3. Justify the use of a suitable technology to assist the patient's hearing.   (3 marks)

Show Answers Only

a.  


 

b.   Right ear is within normal hearing range.

Left ear has a deficit and cannot hear at a normal level.
 

c.    Effective technology: Bone Conduction Implants

Bone conduction implants would prove to be the most effective technology to restore hearing to this patient. 

Bone conduction implants detect sound waves via a microphone, relaying them to a sound processor that converts the waves into vibrations which are then directly transferred to the cochlea. This process bypasses the ear blockage, therefore restoring hearing to the patient.

Show Worked Solution

a.   


 

b.   → Right ear is within normal hearing range.

→ Left ear has a deficit and cannot hear at a normal level.
 

c.    Effective technology: Bone Conduction Implants

→ Bone conduction implants would prove to be the most effective technology to restore hearing to this patient. 

→ Bone conduction implants detect sound waves via a microphone, relaying them to a sound processor that converts the waves into vibrations which are then directly transferred to the cochlea.

→ This process bypasses the ear blockage, therefore restoring hearing to the patient.


♦ Mean mark (c) 43%.

ENGINEERING, CS 2021 HSC 21c

Reinforced concrete is a composite material.

In the box below sketch and label the components of the macrostructure of reinforced concrete.   (3 marks)

 

Show Answers Only

Show Worked Solution

Image should include the following elements:

  • Bulk aggregate
  • Sand
  • Hydrated cement
  • Steel reinforcing bar

 

♦ Mean mark 45%.

PHYSICS M7 2022 HSC 20 MC

In a thought experiment, a car is travelling at a uniform velocity of `0.4 c`. The diagram shows one of the car's wheels as it rolls past a stationary observer at `X`.
 

Consider the instantaneous velocity of different points on the car's wheel relative to the ground. Assume that there is no slippage of the tyre on the road.

At the instant the centre of the wheel, `Q`, passes `X`, how would the observer describe the relativistic length contraction at points `P, Q` and `R` ?

  1. It is the same at `P, Q` and `R`.
  2. It is zero at `P` and greatest at `R`.
  3. It is equal at `P` and `R`, and least at `Q`.
  4. It is zero at `P` and the same value at `Q` and `R`.
Show Answers Only

`B`

Show Worked Solution

Consider Point `P`:

→ Point `P` is moving at 0.4`c` left relative to point `Q`.

→ Point `P` has zero instantaneous velocity relative to the ground.

→ Zero relativistic length contraction will be observed at `P`.
 

Consider Point `R`:

→ Point `R` is moving at 0.4`c` right relative to point `Q`.

→ Point `R` has an instantaneous velocity of 0.8`c` relative to the ground.

→ Maximum relativistic length contraction will be observed at `R`.

`=>B`


♦♦♦ Mean mark 30%.

PHYSICS M7 2022 HSC 17 MC

Unpolarised light of intensity `I_0` is incident upon a vertically polarised filter. The filtered light then passes through a pair of glasses. The glass have polarising filters, with one side polarised vertically and the other horizontally.

The filter undergoes one complete 360° rotation around point `P`, as shown.
 


 

Which of the following correctly compares `I_y` to the intensity at other positions?

  1. `I_y` never equals `I_x`
  2. `I_y` never equals `I_z`
  3. `I_y` sometimes equals `I_z`
  4. `I_y` sometimes equals `I_0`
Show Answers Only

`C`

Show Worked Solution

Once the filter rotates 45° to the vertical:

`I_(y)=I_(x)cos^(2)45^(@)=(1)/(2)I_(x)`

`I_(z)=I_(x)cos^(2)45^(@)=(1)/(2)I_(x)`

→  `I_(y)` sometimes equals `I_(z)`.

`=>C`


Mean mark 55%.

PHYSICS M6 2022 HSC 12 MC

The diagram shows a region in which there are uniform electric and magnetic fields. A positively charged particle moves in the region at constant velocity.
 


 

What is the direction of the particle's velocity?

  1. Up the page
  2. Down the page
  3. To the left
  4. To the right
Show Answers Only

`D`

Show Worked Solution

→ For the particle to have a constant velocity, the net force acting on it must be zero.

→ The electric field exerts a downwards force on the particle.

→ The magnetic field must exert an upwards force on the particle.

→ Using the right hand palm rule, the direction of the particle’s velocity is to the right.

`=>D`


♦ Mean mark 52%.

PHYSICS, M8 2019 HSC 32

Describe how specific experiments have contributed to our understanding of the electron and ONE other fundamental particle.   (5 marks)

Show Answers Only

Millikan’s Oil Drop Experiment:

→ Millikan’s oil drop experiment involved first measuring the terminal velocity of charged oil droplets in a gravitational field and calculating their mass.

→ An electric field was applied to balance the gravitational field, allowing Millikan to find the electric force and hence, charge on an oil droplet.

→ This allowed him to find the charge on an electron as the smallest difference in charges between two oil drops.

Linear accelerator experiment discovering quarks:

→ An experiment involved using a linear accelerator to speed up and fire a beam of electrons at protons. The scattering pattern of the electrons was analysed and was consistent with protons having an internal structure with both positive and negative charges.

→ This contributed to our understanding of the existence of quarks.
 

Other possible answers could include:

  • Thomson’s experiment showing the mass to charge ratio of an electron.
  • Experiments involving synchrotrons discovering particles predicted by the standard model of matter, such as the Higgs-Boson.
Show Worked Solution

Millikan’s Oil Drop Experiment:

→ Millikan’s oil drop experiment involved first measuring the terminal velocity of charged oil droplets in a gravitational field and calculating their mass.

→ An electric field was applied to balance the gravitational field, allowing Millikan to find the electric force and hence, charge on an oil droplet.

→ This allowed him to find the charge on an electron as the smallest difference in charges between two oil drops.

Linear accelerator experiment discovering quarks:

→ An experiment involved using a linear accelerator to speed up and fire a beam of electrons at protons. The scattering pattern of the electrons was analysed and was consistent with protons having an internal structure with both positive and negative charges.

→ This contributed to our understanding of the existence of quarks.

Other possible answers could include:

  • Thomson’s experiment showing the mass to charge ratio of an electron.
  • Experiments involving synchrotrons discovering particles predicted by the standard model of matter, such as the Higgs-Boson.

♦♦ Mean mark 43%.

PHYSICS, M6 2019 HSC 31

A student suspends an electric ceiling fan from a spring balance.

The fan is switched on, reaching a maximum rotational velocity after ten seconds.
 

  1. Explain the changes that would be observed on the spring balance in the first 15 seconds after the fan is switched on.   (4 marks)
  1. The student predicted that the current through the fan's motor would vary as shown on the graph.   
     
     

Assess the accuracy of the student's prediction.   (4 marks)

Show Answers Only

a.    As the fan rotates it exerts a downwards force on air molecules.

By Newton’s Third Law, an equal and opposite force is exerted on the fan by the air. This counteracts the fan’s weight, decreasing the net downwards force acting on the fan.

For the first ten seconds:

→ The fan’s rotation speed increases.

→ The force on the fan due to the air increases.

→ The force observed on the spring balance decreases.

After ten seconds:

→ The force acting on the fan due to the air remains constant.

→ The recorded force on the spring balance remains constant.
 

b.   Between 0-10 seconds:

The student’s prediction of an increasing current is incorrect.

→ Initially, the current through the motor is at a maximum as the fan’s motor is stationary and it experiences no change in magnetic flux (no back EMF induced).

→ As the fan speeds up, the rate of change of flux through the motor increases.

→ An increasing EMF is induced in the motor (Faraday’s Law).

→ This is back EMF which acts to oppose the current.

→ a decrease in current occurs during the first 10 seconds.
 

Between 10-15 seconds:

The student’s prediction of a constant current is correct.

→ The motor is rotating at a constant speed.

→ The magnitude of induced back EMF remains constant.

→ the current through the motor remains constant.

Show Worked Solution

a.    As the fan rotates it exerts a downwards force on air molecules.

By Newton’s Third Law, an equal and opposite force is exerted on the fan by the air. This counteracts the fan’s weight, decreasing the net downwards force acting on the fan.

For the first ten seconds:

→ The fan’s rotation speed increases.

→ The force on the fan due to the air increases.

→ The force observed on the spring balance decreases.

After ten seconds:

→ The force acting on the fan due to the air remains constant.

→ The recorded force on the spring balance remains constant.


♦ Mean mark (a) 48%.

b.   Between 0-10 seconds:

The student’s prediction of an increasing current is incorrect.

→ Initially, the current through the motor is at a maximum as the fan’s motor is stationary and it experiences no change in magnetic flux (no back EMF induced).

→ As the fan speeds up, the rate of change of flux through the motor increases.

→ An increasing EMF is induced in the motor (Faraday’s Law).

→ This is back EMF which acts to oppose the current.

→ a decrease in current occurs during the first 10 seconds.
 

Between 10-15 seconds:

The student’s prediction of a constant current is correct.

→ The motor is rotating at a constant speed.

→ The magnitude of induced back EMF remains constant.

→ the current through the motor remains constant.


♦ Mean mark (b) 39%.

PHYSICS, M5 2019 HSC 30

A ball, initially at rest in position `P`, travels along a frictionless track to point `Q` and then falls to strike the floor below.
 

At the instant the ball leaves the track at `Q` it has a velocity of 1.5 m `text{s}^(-1)` at an angle of 50° to the horizontal.

  1. Calculate the difference in height between `P` and `Q`.   (3 marks)
  1. The ball takes 0.5 s to reach the floor after leaving the track at `Q`.
  2. Calculate the height of `Q` above the floor.   (3 marks)
Show Answers Only

a.   `h=`0.11 m

b.   `h=`1.8 m

Show Worked Solution

a. `Delta U` `=Delta E_(k)`
  `mgDeltah` `=(1)/(2)mv^2-(1)/(2)m u^2`
    `=(1)/(2)mv^2\ \ (u=0)`
  `Delta h` `=(v^2)/(2g)`
    `=(1.5^(2))/(2xx9.8)`
    `=0.1145` m
    `=0.11` m


♦ Mean mark part (a) 41%.

 

b. `u_(y)` `=u sin theta`
    `=1.50  sin 50^(@)`
    `=1.15` m `text{s}^(-1)`
  `s_(y)` `=ut+(1)/(2)at^(2)`
    `=1.15 xx0.5+(1)/(2)(9.80)xx0.5^(2)`
    `=` 1.8 m

 
`:.` Height of Q `=` 1.8 m

PHYSICS, M6 2019 HSC 28

A metal loop, `WXYZ` is connected to a battery and placed in a uniform magnetic field. A current flows through the loop in the direction shown.
 

The loop is then allowed to rotate by 90° about the axis `PQ`.

Compare the forces acting on `WX` and `XY` before and after this rotation.   (3 marks)

Show Answers Only

Originally:

→ The current through  `XY`  is parallel to the magnetic field lines so it experiences no force.

→ `WX` experiences a force given by `F=BIl`  into the page.

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

→ The magnitude and direction of the force on `WX` remains the same.

Show Worked Solution

Originally:

→ The current through  `XY`  is parallel to the magnetic field lines so it experiences no force.

→ `WX` experiences a force given by `F=BIl`  into the page.

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

→ The magnitude and direction of the force on `WX` remains the same.


♦ Mean mark 46%.

Vectors, EXT1 V1 2022 HSC 14c

A video game designer wants to include an obstacle in the game they are developing. The player will reach one side of a pit and must shoot a projectile to hit a target on the other side of the pit in order to be able to cross. However, the instant the player shoots, the target begins to move away from the player at a constant speed that is half the initial speed of the projectile shot by the player, as shown in the diagram below.

The initial distance between the player and the target is `d`, the initial speed of the projectile is `2 u` and it is launched at an angle of `theta` to the horizontal. The acceleration due to gravity is `g`. The launch angle is the ONLY parameter that the player can change.
 

  
 

Taking the position of the player when the projectile is launched as the origin, the positions of the projectile and target at time `t` after the projectile is launched are as follows.

`vecr_(P)` `=((2utcostheta),(2utsintheta-g/2t^2))` `text{Projectile}`
     
`vecr_(T)` `=((d+ut),(0))` `text{Target          (Do NOT prove these)}`

 
Show that, for the player to have a chance of hitting the target, `d` must be less than 37% of the maximum possible range of the projectile (to 2 significant figures).  (4 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Projectile’s motion:}`

`x_P=2utcostheta, \ \ y_P=2utsintheta-g/2t^2`

`text{Time of flight}\ \ =>\ \ text{Find}\ \ t\ \ text{when}\ \ y=0:`

`2utsintheta-g/2t^2` `=0`  
`t(2usintheta-g/2 t)` `=0`  
`g/2 t` `=2u sin theta`  
`t` `=(4usintheta)/g`  

 

`text{Range of projectile}\ (R):`

`R` `=2u((4usintheta)/g) costheta`  
  `=(8u^2sinthetacostheta)/g`  
  `=(4u^2sin(2theta))/g`  

 
`=>R_max=(4u^2)/g \ \ (text{when}\ \ theta=45°):`

`text{Projectile will hit target when}\ \ x_P=x_T\ \ text{at}\ \ t=(4usintheta)/g:`

`(4u^2sin(2theta))/g` `=d+u*(4usintheta)/g`  
`d` `=(4u^2sin(2theta))/g-u*(4usintheta)/g`  
  `=(4u^2)/g\ (sin(2theta)-sintheta)`  
  `=R_max*(sin(2theta)-sintheta)`  

 
`d_max\ \ text{occurs when}\ \ (sin(2theta)-sin theta)\ \ text{is MAX}`

`f(theta)` `=sin(2theta)-sintheta`  
`f^(′)(theta)` `=2cos(2theta)-costheta`  
`f^(′′)(theta)` `=-4sin(2theta)+sin theta`  

 
`text{Find SP’s when}\ \ f^(′)(theta)=0:`

`2cos2theta-costheta` `=0`  
`2(2cos^2theta-1)-costheta` `=0`  
`4cos^2theta-costheta-2` `=0`  

 

`costheta` `=(1+-sqrt(1-4xx4xx-2))/(2xx4)`  
  `=(1+-sqrt33)/8`  
`theta` `=32°32′, \ 126°23′`  

 
`f^(′′)(32°32′)~~-3.1<0`

`=>\ text{MAX at}\ \ theta=32°32′`
 

`:.d_max` `=R_max xx (sin65°04′-sin32°32′)`  
  `=(0.369)R_max<0.37R_max`  

♦♦ Mean mark 33%.

Calculus, EXT1 C3 2022 HSC 14a

Find the particular solution to the differential equation  `(x-2)(dy)/(dx)=xy`  that passes through the point `(0,1)`.  (4 marks)

Show Answers Only

`y=(e^x(x-2)^2)/4`

Show Worked Solution
`(x-2)(dy)/(dx)` `=xy`  
`1/y* dy/dx` `=x/(x-2)`  
`int 1/y\ dy` `=int x/(x-2)\ dx`  
`ln|y|` `=int (x-2)/(x-2)+2/(x-2)\ dx`  
  `=int 1+2/(x-2)\ dx`  
  `=x+2ln|x-2|+c`  

 
`text{Passes through (0,1):`

`ln1` `=0+2ln|-2|+c`  
`c` `=-2ln2`  

 

`ln|y|` `=x+2ln|x-2|-2ln2`  
  `=lne^x+ln(x-2)^2-ln2^2`  
  `=ln(e^x((x-2)^2)/4)`  
`|y|` `=(e^x(x-2)^2)/4`  
`:.y` `=(e^x(x-2)^2)/4\ \ (e^x>0,\ \ (x-2)^2>0)`  

♦ Mean mark 43%.

Statistics, EXT1 S1 2022 HSC 13e

A chocolate factory sells 150-gram chocolate bars. There has been a complaint that the bars actually weigh less than 150 grams, so a team of inspectors was sent to the factory to check. They randomly selected 16 bars, weighed them and noted that 8 bars weighed less than 150 grams.

The factory manager claims 80% of the chocolate bars produced by the factory weigh 150 grams or more.

  1. The inspectors used the normal approximation to the binomial distribution to calculate the probability, `cc "P"`, of having at least 8 bars weighing less than 150 grams in a random sample of 16, assuming the factory manager's claim is correct.
  2. Using the attached probability table, calculate the value of `cc "P"`.   (2 marks)

  3. The factory manager disagrees with the method used by the inspectors as described in part (i).
  4. Explain why the method used by the inspectors might not be valid.   (1 mark)

Show Answers Only
  1. `P=0.0013`
  2. `text{See Worked Solutions}`
Show Worked Solution

i.   `text{Let}\ \ X=\ text{number of chocolate bars < 150 grams}`

`X\ ~\ text{Bin}(16, 0.2)`

`mu=np=16xx0.2=3.2`

`sigma^2=np(1-p)=16xx0.2xx0.8=2.56`

`ztext{-score}\ (X=8)=(8-3.2)/sqrt2.56=3`

 

`cc P(X>=8)` `=P(z>=3.0)`  
  `=1-P(z<=3.0)`  
  `=1-0.9987`  
  `=0.0013`  

 

ii.   `np=3.2,\ \ n(1-p)=16xx0.8=12.8`

`text{S}text{ince}\ \ np<5,\ text{the sample size is too small.}`


♦ Mean mark part (i) 37%.
 
Mean mark part (ii) 54%.

Functions, EXT1 F2 2022 HSC 13d

The monic polynomial, `P`, has degree 3 and roots `alpha, \beta, \gamma`.

It is given that

           `alpha^(2)+beta^(2)+gamma^(2)=85\ \ and`

           `P^(')(alpha)+P^(')(beta)+P^(')(gamma)=87.`

Find  `alpha beta+beta gamma+gamma alpha`.  (3 marks)

Show Answers Only

`-2`

Show Worked Solution

`P(x)=x^3-(alpha+beta+gamma)x^2+(alphabeta+alphagamma+betagamma)x-alphabetagamma`

`P^(′)(x)=3x^2-2(alpha+beta+gamma)x+alphabeta+alphagamma+betagamma`

`P^(′)(alpha)+P^(′)(beta)+P^(′)(gamma)`

`=3alpha^2-2(alpha+beta+gamma)alpha+alphabeta+alphagamma+betagamma`

       `+ 3beta^2-2(alpha+beta+gamma)beta+alphabeta+alphagamma+betagamma`

       `+ 3gamma^2-2(alpha+beta+gamma)gamma+alphabeta+alphagamma+betagamma`

`=3(alpha^(2)+beta^(2)+gamma^(2))-2(alpha^(2)+beta^(2)+gamma^(2))`

           `-2(alphabeta+alphagamma+alphabeta+betagamma+alphagamma+betagamma)+3(alphabeta+alphagamma+betagamma)`

`=(alpha^(2)+beta^(2)+gamma^(2))-(alphabeta+alphagamma+betagamma)`

`text{Substituting in given values:}`

`85-(alphabeta+alphagamma+betagamma)` `=87`  
`:.alphabeta+alphagamma+betagamma` `=-2`  

♦♦ Mean mark 33%.

Trigonometry, EXT1 T1 2022 HSC 13c


The function `f` is defined by  `f(x)=\sin (x)`  for all real numbers `x`. Let `g` be the function defined on  `[-1,1]`  by  `g(x)=\arcsin (x)`.

Is `g` the inverse of `f`? Justify your answer.   (2 marks)

Show Answers Only

`text{The domain of}\ \ f(x) = RR`

`text{The range of}\ \ g(x) = [-pi/2,pi/2]`

`text{S}text{ince the domain}\ \ f(x) !=\ text{range of}\ \ g(x),`

`f^(-1)(x)!=g(x)`

Show Worked Solution

`text{The domain of}\ \ f(x) = RR`

`text{The range of}\ \ g(x) = [-pi/2,pi/2]`

`text{S}text{ince the domain}\ \ f(x) !=\ text{range of}\ \ g(x),`

`f^(-1)(x)!=g(x)`


♦♦ Mean mark 24%.

Statistics, EXT1 S1 2022 HSC 12e

A game consists of randomly selecting 4 balls from a bag. After each ball is selected it is replaced in the bag. The bag contains 3 red balls and 7 green balls. For each red ball selected, 10 points are earned and for each green ball selected, 5 points are deducted. For instance, if a player picks 3 red balls and 1 green ball, the score will be  `3xx10-1xx5=25`  points.

What is the expected score in the game?  (2 marks)

Show Answers Only

`-2`

Show Worked Solution

`text{Let}\ \ X=\ text{number of red balls selected}`

`X\ ~\ text{Bin}(4, 0.3)`

`text{Score}\ (X=4)=(0.3)^4xx40=0.324`

`text{Score}\ (X=3)=((4),(3))(0.3)^3(0.7)xx25=1.89`

`text{Score}\ (X=2)=((4),(2))(0.3)^2(0.7)^2xx10=2.646`

`text{Score}\ (X=1)=((4),(1))(0.3)(0.7)^3xx-5=-2.058`

`text{Score}\ (X=0)=(0.7)^4xx-20=-4.802`
 

`:.E(X)` `=0.324+1.89+2.646-2.058-4.802`  
  `=-2`  

♦♦ Mean mark 35%.

Vectors, EXT1 V1 2022 HSC 8 MC

The angle between two unit vectors `underset~a` and `underset~b` is `theta` and  `|underset~a+ underset~b| < 1`.

Which of the following best describes the possible range of values of  `theta` ?

  1. `0 <= theta < (pi)/(3)`
  2. `0 <= theta < (2pi)/(3)`
  3. `(pi)/(3) < theta <= pi`
  4. `(2pi)/(3) < theta <= pi`
Show Answers Only

`D`

Show Worked Solution

`text{By Elimination:}`

`text{Consider}\ \ underset~a=((1),(0)) and underset~b=((-1),(0))`

`|underset~a+ underset~b| =0 < 1\ \ and\ \ theta=pi`

`text{→ Eliminate A and B}`
 

`text{Consider}\ \ theta=(2pi)/3 and underset~a=((1),(0)),\ \ underset~b=((costheta),(sintheta))`

`underset~a+underset~b=((1+cos((2pi)/3)),(sin((2pi)/3)))=((1/2),(sqrt3/2))`

`|underset~a+ underset~b|^2 = (1/2)^2+(sqrt3/2)^2=1`

`:. theta !=(2pi)/3`

`text{→ Eliminate C}`

`=>D`


♦♦ Mean mark 34%.

Combinatorics, EXT1 A1 2022 HSC 7 MC

The diagram shows triangle `A B C` with points chosen on each of the sides. On side `A B`, 3 points are chosen. On side `A C`, 4 points are chosen. On side `B C`, 5 points are chosen.
 


 

How many triangles can be formed using the chosen points as vertices?

  1. 60
  2. 145
  3. 205
  4. 220
Show Answers Only

`C`

Show Worked Solution

`text{1 point taken from each side:}`

`text{Triangles} = 3 xx 4xx5=60`
 

`text{2 points taken from one side:}`

`text{Triangles}` `=((3),(2))((9),(1))+((4),(2))((8),(1))+((5),(2))((7),(1))`  
  `=145`  

 
`:.\ text{Total triangles} =60+145=205`

`=>C`


♦♦ Mean mark 37%.

Vectors, EXT1 V1 2022 HSC 6 MC

The following diagram shows the vector `underset∼u` and the vectors `underset∼i+underset∼j,-underset∼i+ underset∼j,-underset∼i- underset∼j` and `underset∼i-underset∼j`. 
 


 

Which statement regarding this diagram could be true?

  1. The projection of `underset∼u` onto  `underset∼i+ underset∼j`  is the vector  `1.1 underset∼i+ 1.8 underset∼j`.
  2. The projection of `underset∼u` onto  `-underset∼i+ underset∼j`  is the vector  `-0.4 underset∼i+0.4 underset∼j`.
  3. The projection of `underset∼u` onto  `- underset∼i- underset∼j`  is the vector  `3.2 underset∼i+3.2 underset∼j`. 
  4. The projection of `underset∼u` onto  `underset∼i- underset∼j`  is the vector  `0.5 underset∼i-0.5 underset∼j`. 
Show Answers Only

`B`

Show Worked Solution

`text{Consider each option by tracing projections on the graph:}`
 

`overset(->)(OM)= text(proj)_((-underset~i+underset~j)) underset~u`

`text{Option B’s projection is only possible correct option.}`

`=>B`


♦♦ Mean mark 38%.

 

Functions, EXT1 F2 2022 HSC 3 MC

Let `P(x)` be a polynomial of degree 5. When `P(x)` is divided by the polynomial `Q(x)`, the remainder is `2x+5`.

Which of the following is true about the degree of `Q`?

  1. The degree must be 1.
  2. The degree could be 1.
  3. The degree must be 2.
  4. The degree could be 2.
Show Answers Only

`D`

Show Worked Solution

`text{Given}\ \ P(x)\ \ text{has degree 5}`

`P(x) -: Q(x)\ \ text{has remainder}\ \ 2x+5`

`text{Consider examples to resolve possibilities:}`

`text{eg.}\ \ x^5+2x+5 -: x^3 = x^2+\ text{remainder}\ 2x+5`

`:.\ text{Degree must be 2 is incorrect}`

`Q(x)\ \ text{can have a degree of 2, 3 or 4}`

`=>D`


♦ Mean mark 51%.

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount `$P` is borrowed for a period of `n` months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of `$M` is made. After the `n`th repayment has been made, the amount owing, `$A_n`, is given by

`A_(n)=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`

(Do NOT prove this.)

  1. Jane borrows $200 000 in a reducing-balance loan as described.

  2. The loan is to be repaid in 180 monthly repayments.

  3. Show that  `M` = 1381.16, when rounded to the nearest cent.  (2 marks)

  4. After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.

  5. At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)

  6. Jane continues to make the same monthly repayments.

  7. For how many more months will Jane need to make full monthly payments of $1381.16?  (3 marks)

  8. The final payment will be less than $1381.16.
  9. How much will Jane need to pay in the final payment in order to pay off the loan?   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `83`
  3. `$931.54`
Show Worked Solution

a.   `text{Show}\ \ M=$1381.16`

`A_(n)` `=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`  
`0` `=200\ 000(1.0025)^180-M underbrace((1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^179))_(text(GP where)\ a = 1,\ r = 1.0025,\ n = 180)`  
`0` `=200\ 000(1.0025)^180-M((1(1.0025^180- 1))/(1.0025-1))`  

`M((1.0025^180-1)/(1.0025-1))=200\ 000(1.0025)^180`

`:.M` `=200\ 000(1.0025)^180 -: (1.0025^180-1 )/(1.0025-1)`   
  `=1381.163…`  
  `=$1381.16\ \ text{… as required}`  

 

b.   `P=$100\ 032,\ \ r=1.0035  and M=$1381.16`

`text{Find}\ \ n\ \ text{when}\ \ A_n=0:`

`A_(n)` `=P(1.0035)^(n)-1381.16(1+(1.0035)^(1)+(1.0035)^(2)+cdots+(1.0035)^(n-1))`  
`0` `=100\ 032(1.0035)^n-1381.16 ((1.0035^n- 1)/(1.0035-1))`  
`0` `=100\ 032(1.0035)^n-1381.16/0.0035 (1.0035^n- 1)`  
  `=100\ 032(1.0035)^n-394\ 617(1.0035)^n- 394\ 617`  
`294\ 585(1.0035)^n` `=394\ 617`  
`1.0035^n` `=(394\ 617)/(294\ 585)`  
`n` `=ln((394\ 617)/(294\ 585))/ln1.0035`  
  `=83.674…`  
  `=83\ text{more months with full payment}`  

 


♦♦ Mean mark part (b) 38%.

c.   `text{Find}\ \ A_83:`

`A_83` `=100\ 032(1.0035)^83-1381.16 ((1.0035^83- 1)/(1.0035-1))`  
  `=928.291…`  

 
`text{Interest will be added for the last month:}`

`:.\ text{Final payment}` `=928.291… xx 1.0035`  
  `=$931.54`  

♦♦♦ Mean mark part (c) 14%.

Calculus, 2ADV C3 2022 HSC 31

A line passes through the point  `P(1,2)`  and meets the axes at  `X(x, 0)`  and  `Y(0, y)`, where `x>1`.
 

  1. Show that  `y=(2x)/(x-1)`.  (2 marks)

  2. Find the minimum value of the area of triangle `XOY`.  (4 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `4\ text{u}^2`
Show Worked Solution

a.   `text{Show}\ \ y=(2x)/(x-1)`

`text{S}text{ince}\ \ m_(YP)=m_(PX):`

`(y-2)/(0-1)` `=(2-0)/(1-x)`  
`y-2` `=(-2)/(1-x)`  
`y` `=2-2/(1-x)`  
  `=(2(1-x)-2)/(1-x)`  
  `=(-2x)/(1-x)`  
  `=(2x)/(x-1)\ \ text{… as required}`  

 


♦♦♦ Mean mark part (a) 17%.
COMMENT: `y=(2x)/(x-1)` is the expression of a relationship between the intercepts and not the equation of the line.
b.    `A` `=1/2 xx b xxh`
    `=1/2x((2x)/(x-1))`
    `=(x^2)/(x-1)`

 

`(dA)/dx` `=((x-1)*2x-x^2(1))/((x-1)^2)`  
  `=(2x^2-2x-x^2)/((x-1)^2)`  
  `=(x(x-2))/((x-1)^2)`  

 
`text{SP’s occur when}\ \ (dA)/dx=0:`

`x=0\ \ text{or}\ \ 2`
 

`text{Use 1st derivative test to find max/min:}`

`=>\ text{MIN at}\ \ x=2`

`:.A_min` `=1/2 xx 2 xx (2xx2)/(2-1)`  
  `=4\ text{u}^2`  

♦♦ Mean mark part (b) 29%.

Statistics, 2ADV S3 2022 HSC 30

A continuous random variable \(X\) has cumulative distribution function given by

\(F(x)= \begin{cases}
1 & x>e^3 \\
\ \\
\dfrac{1}{k}\, \ln x & 1 \leq x \leq e^3 . \\
\ \\
0 & x<1\end{cases}\)

  1. Show that  \(k = 3\).  (1 mark)

  2. Given that  \(P(X < c)=2P(X > c)\), find the exact value of \(c\).  (2 marks)

Show Answers Only
  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(e^2\)
Show Worked Solution

a.  \(\text{Show} \ \ k=3\)

\begin{aligned}
{\left[\dfrac{1}{k} \, \ln x\right]_1^{e^3} } & =1 \\
\dfrac{1}{k}\, \ln \left(e^3\right)-\dfrac{1}{k}\, \ln 1 & =1 \\
\dfrac{1}{k}(3)-\frac{1}{k}(0) & =1 \\
k & =3 \ldots \text{as required}
\end{aligned}


♦♦ Mean mark (a) 38%.
b.    \(P(X<c)\) \(=\left[\dfrac{1}{3}\, \ln x\right]_0^c\)
    \(=\dfrac{1}{3}\, \ln c\)

\begin{aligned}
2 P(X>c) & =2 P(1-P(X<c)) \\
& =2\left(1-\frac{1}{3} \ln c\right)
\end{aligned}

\(\text { Given } P(X<c)=2 P(X>c)\)

\begin{aligned}
\dfrac{1}{3}\, \ln c & =2-\dfrac{2}{3}\, \ln c \\
\ln c & =2 \\
\therefore c & =e^2
\end{aligned}


♦♦♦ Mean mark (b) 19%.

Calculus, 2ADV C4 2022 HSC 28

The graph of the circle  `x^2+y^2=2`  is shown.

The interval connecting the origin, `O`, and the point `(1,1)` makes an angle `theta` with the positive `x`-axis.
 

  1. By considering the value of `theta`, find the exact area of the shaded region, as shown on the diagram.  (2 marks)

Part of the hyperbola  `y=(a)/(b-x)-1`  which passes through the points `(0,0)` and `(1,1)` is drawn with the circle  `x^2+y^2=2`  as shown.
 

  1. Show that  `a=b=2`.  (2 marks)

  2. Using parts (a) and (b), find the exact area of the region bounded by the hyperbola, the positive `x`-axis and the circle as shown on the diagram.   (3 marks)

Show Answers Only
  1. `(pi-2)/4\ text{u}^2`
  2. `text{Proof (See Worked Solutions)}`
  3. `(8ln2+pi-6)/4\ text{u}^2`
Show Worked Solution

a.   `tan theta=1\ \ =>\ \ theta = pi/4`

`text{Using Pythagoras,}`

`r=sqrt(1^2+1^2)=sqrt2`

`text{Shaded Area}` `=A_text{sector}-A_Delta`  
  `=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`  
  `=1/8xxpixx(sqrt2)^2-1/2xx1xx1`  
  `=pi/4-1/2`  
  `=(pi-2)/4\ text{u}^2`  

 


Mean mark (a) 54%.

b.   `text{Show}\ \ a=b=2`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (0,0):`

`0` `=a/(b-0)-1`  
`a/b` `=1`  
`a` `=b`  

 
`y=(a)/(b-x)-1\ \ text{passes through}\ \ (1,1):`

`1` `=a/(b-1)-1`  
`a/(b-1)` `=2`  
`a` `=2(b-1)`  
`a` `=2b-2`  
`b` `=2b-2\ \ text{(using}\ a=b)`  
`b` `=2`  

 
`:.a=b=2`
 


♦ Mean mark (b) 47%.
c.    `int_0^1 2/(2-x)-1\ dx` `=int_0^1 -2 xx (-1)/(2-x)-1\ dx`
    `=[-2ln|2-x|-x]_0^1`
    `=[(-2ln1-1)-(-2ln2-0)]`
    `=-1+2ln2`

 

`:.\ text{Total Area}` `=2ln2-1 + (pi-2)/4`  
  `=(8ln2-4+pi-2)/4`  
  `=(8ln2+pi-6)/4\ text{u}^2`  

♦♦ Mean mark (c) 36%.

Trigonometry, 2ADV T3 2022 HSC 23

The depth of water in a bay rises and falls with the tide. On a particular day the depth of the water, `d` metres, can be modelled by the equation

`d=1.3-0.6 cos((4pi)/(25)t)`,

where `t` is the time in hours since low tide.

  1. Find the depth of water at low tide and at high tide.  (2 marks)

  2. What is the time interval, in hours, between two successive low tides?  (1 mark)

  3. For how long between successive low tides will the depth of water be at least 1 metre?  (3 marks)

Show Answers Only
  1. `text{Low: 0.7 m,  High: 1.9 m}`
  2. `25/2\ text{hours}`
  3. `25/3\ text{hours}`
Show Worked Solution

a.   `text{S}text{ince}\ \ –1<=cos((4pi)/(25)t)<=1:`

`text{Low Tide}\ =1.3-0.6(1)=0.7\ text{m}`

`text{High Tide}\ =1.3-0.6(-1)=1.9\ text{m}`
 

b.   `text{Time between two low tides = Period of equation}\ (n)`

`(2pi)/n` `=(4pi)/25`  
`n/(2pi)` `=25/(4pi)`  
`n` `=25/2\ text{hours}`  

 


♦ Mean mark part (b) 47%.

c.   `text{Find}\ \ t\ \ text{when}\ \ d=1:`

`1.3-0.6 cos((4pi)/(25)t)` `=1`  
`-0.6 cos((4pi)/(25)t)` `=-0.3`  
`cos((4pi)/(25)t)` `=1/2`  

 

`(4pi)/(25)t` `=pi/3,\ \ (5pi)/3`  
`t` `=25/12,\ \ 125/12`  

 
`:.\ text{Time between low tides where water depth}\ >= 1\ text{m}`

`=125/12-25/12`

`=100/12`

`=25/3\ text{hours}`


♦ Mean mark part (c) 46%.

Functions, 2ADV F2 2022 HSC 19

The graph of the function  `f(x)=x^2`  is translated `m` units to the right, dilated vertically by a scale factor of `k` and then translated 5 units down. The equation of the transformed function is  `g(x)=3 x^2-12 x+7`.

Find the values of `m` and `k`.  (3 marks)

Show Answers Only

`m=2, \ \ k=3`

Show Worked Solution

`text{Horizontal translation}\ m\ text{units to the right:}`

`x^2\ → \ (x-m)^2`

`text{Dilated vertically by scale factor}\ k:`

`(x-m)^2\ →\ k(x-m)^2`

`text{Vertical translation 5 units down:}`

`k(x-m)^2\ →\ k(x-m)^2-5`

`y` `=k(x-m)^2-5`  
  `=k(x^2-2mx+m^2)-5`  
  `=kx^2-2kmx+(km^2-5)`  

 
`:.k=3`

`-2km` `=-12`  
`:.m` `=2`  

♦ Mean mark 51%.

Probability, 2ADV S1 2022 HSC 15

In a bag there are 3 six-sided dice. Two of the dice have faces marked  1, 2, 3, 4, 5, 6. The other is a special die with faces marked  1, 2, 3, 5, 5, 5.

One die is randomly selected and tossed.

  1. What is the probability that the die shows a 5?  (1 mark)

  2. Given that the die shows a 5, what is the probability that it is the special die?  (1 mark)

Show Answers Only
  1. `5/18`
  2. `3/5`
Show Worked Solution
a.    `P(5)` `=1/3 xx 1/6 + 1/3 xx 1/6 + 1/3 xx 1/2`
    `=1/18+1/18+1/6`
    `=5/18`

 

b.   `Ptext{(special die given a 5 is rolled)}`

`=(Ptext{(special die)} ∩  P(5))/(P(5))`

`=(1/3 xx 1/2)/(5/18)`

`=1/6 xx 18/5`

`=3/5`


Mean mark (a) 53%.
 
♦ Mean mark (b) 40%.

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

Show Answers Only

a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Algebra, STD2 A4 2022 HSC 24

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.   (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}.`   (2 marks)

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ \ 30\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & & & \\
\hline
\end{array}

 
           

Show Answers Only
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`

b.   

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ 30\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}

 

     

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`


♦♦ Mean mark part (a) 29%.

b.   

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ 30\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}

 

     


♦ Mean mark 44%.

Measurement, STD2 M7 2022 HSC 38

A full container has 4.8 L of a mixture of cordial and water in the ratio 1:3.

After removing 1.2 L of the mixture, more water is added to refill the container.

What is the ratio of cordial to water in the final mixture?  (3 marks)

Show Answers Only

`3:13`

Show Worked Solution

`text{Original mixture ratio}`

`text{C : W = 1.2 litres : 3.6 litres}`

`text{Removing 1.2 L of mixture, removes}`

`text{→ 300 mL cordial and 900 mL water}`
 

`text{Cordial in mixture after refill}`

`=1200-300`

`=900\ text{mL}`
 

`text{Water in mixture after refill}`

`=3600-900+1200`

`=3900\ text{mL}`
 

`:.\ text{C : W (final mixture)}`

`=900:3900`

`=3:13`


♦♦ Mean mark 33%.

Statistics, STD2 S5 2022 HSC 37

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definition)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`


♦♦ Mean mark 28%.

Statistics, STD2 S4 2022 HSC 35

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.  (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`

♦♦ Mean mark 30%.

Measurement, STD2 M1 2022 HSC 34

A composite solid is shown. The top section is a cylinder with a height of 3 cm and a diameter of 4 cm. The bottom section is a hemisphere with a diameter of 6 cm. The cylinder is centred on the flat surface of the hemisphere.
 


 

Find the total surface area of the composite solid in cm², correct to 1 decimal place.  (4 marks)

Show Answers Only

`122.5\ text{cm}^2`

Show Worked Solution
`text{S.A. of Cylinder}` `=pir^2+2pirh`  
  `=pi(2^2)+2pi(2)(3)`  
  `=16pi\ text{cm}^2`  

 

`text{S.A. of Hemisphere}` `=1/2 xx 4pir^2`  
  `=2pi(3^2)`  
  `=18pi\ text{cm}^2`  

 

`text{Area of Annulus}` `=piR^2-pir^2`  
  `=pi(3^2)-pi(2^2)`  
  `=5pi\ text{cm}^2`  

 

`text{Total S.A.}` `=16pi+18pi+5pi`  
  `=39pi`  
  `=122.522…`  
  `=122.5\ text{cm}^2\ \ text{(to 1 d.p.)}`  

♦ Mean mark 50%.

Measurement, STD2 M6 2022 HSC 33

The diagram shows an aeroplane that was flying towards an airport at  `A` on a bearing of `135^@ text{T}`. When it was at point `O`, 20 km away from the airport at  `A`, the flight course was changed. The aeroplane landed at an airport at `B` directly south of `O`. The distance from `O` to `B` is 50 km.
 


 

  1. Show that the distance between the airport at  `A` and the airport at `B` is 38.5 km, correct to 1 decimal place.  (2 marks)

  2. Use the sine rule to find the angle `O B A` to the nearest degree.  (2 marks)

  3. What is the bearing of the airport at `B` from the airport at  `A` ?  (1 mark)

Show Answers Only
  1. `text{Proof}`
  2. `22^@`
  3. `202^@text{T}`
Show Worked Solution

a.   `angle AOB=180-135=45^@`
 

`text{Using the cosine rule:}`

`AB^2` `=20^2+50^2-2xx20xx50xxcos45^@`  
  `=1485.786…`  
`:.AB` `=38.54…`  
  `=38.5\ \text{km (to 1 d.p.)}`  

♦ Mean mark part (a) 50%.

 

b.   `text{Let}\ \ theta=angleOBA`

`text{Using the sine rule:}`

`sintheta/20` `=sin45^@/38.5`  
`sintheta` `=(sin45^@xx20)/38.5`  
`:.theta` `=sin^(-1)((sin45^@xx20)/38.5)`  
  `=21.55…^@`  
  `=22^@\ \ text{(nearest degree)}`  

♦♦ Mean mark part (b) 39%.

 

c. 

`text{Bearing of}\ B\ text{from}\ \ A`

`=180 + 22`

`=202^@text{T}`


♦♦ Mean mark part (c) 25%.

Measurement, STD2 M1 2022 HSC 32

The diagram shows a park consisting of a rectangle and a semicircle. The semicircle has a radius of 100 m. The dimensions of the rectangle are 200 m and 250 m.

A lake occupies a section of the park as shown. The rest of the park is a grassed section. Some measurements from the end of the grassed section to the edge of the lake are also shown.
 

  1. Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section.  (2 marks)

  2. Hence calculate the approximate area of the lake, to the nearest square metre.   (2 marks)

Show Answers Only
  1. `35\ 500\ text{m}^2`
  2. `30\ 208\ text{m}^2`
Show Worked Solution

a.   `h=100\ text{m}`

`A` `~~h/2(x_1+x_2)+h/2(x_2+x_3)`  
  `~~100/2(160+150)+100/2(150+250)`  
  `~~50xx310+50xx400`  
  `~~35\ 500\ text{m}^2`  

 

b.    `text{Total Area}` `=\ text{Area of Rectangle + Area of Semi-Circle}`
    `= (250 xx 200) + 1/2 xx pi xx 100^2`
    `=65\ 707.96\ text{m}^2`

 

`text{Area of Lake}` `=67\ 708-35\ 500`  
  `=30\ 208\ text{m}^2`  

♦ Mean mark (b) 44%.

Networks, STD2 N3 2022 HSC 31

A wildlife park has 5 main attractions `(A, B, C, D, E)` connected by directional paths. A simple network is drawn to represent the flow through the park's paths. The number of visitors who can access each path at any one time is also shown.
 

   

  1. What is the flow capacity of the cut shown?  (1 mark)

  2. By showing a suitable cut on the diagram below, explain why the network's current maximum flow capacity is less than 40 visitors.  (2 marks)
     

  3. One path is to be increased in capacity so that the overall maximum flow will be 40 visitors at any one time.
  4. Which path could be increased and by how much?  (2 marks)

Show Answers Only
  1. `40`
  2.  
     
  3. `text{Max flow = min cut = 35}`
  4. `AE\ text{or}\ DE\ text{could be increased by 5}`
Show Worked Solution

a.   `text{Flow capacity = 10 + 20 + 10 = 40}`

`text{(DE is not counted as it runs from sink → source)}`
 

b.  
       

`text{Min Cut = Max Flow}`

`text{Max Flow}` `=15+10+10`  
  `=35<40`  


♦ Mean mark part (a) 45%.
♦♦ Mean mark part (b) 33%.

 

c.   `text{Two strategies:}`

  • `AE\ text{could be increased by 5}`
  • `DE\ text{could be increased by 5}`

`text{(both strategies would increase the minimum cut to}`

  `text{40 by increasing the flow to vertex}\ E\ text{to 30)}`


Mean mark 52%.

Financial Maths, STD2 F5 2022 HSC 30

Eli is choosing between two investment options.

A table of future value interest factors for an annuity of $1 is shown.

  1. What is the value of Eli's investment after 10 years using Option 1 ?  (2 marks)

  2. What is the difference between the future values after 10 years using Option 1 and Option 2?  (2 marks)

Show Answers Only
  1. `$45\ 097.17`
  2. `$40.87`
Show Worked Solution

a.   `text{Monthly r/i}\ = 1.2/12=0.1text{%}\ \ =>\ \ r= 0.001`

`text{Compounding periods}\ (n)=12xx10=120`

`FV` `=PV(1+r)^n`  
  `=40\ 000(1+0.001)^120`  
  `=$45\ 097.17`  

 


♦ Mean mark 48%.

b.   `text{Quarterly r/i}\ = 2.4/4=0.6text{%}\ \ =>\ \ r= 0.006`

`text{Compounding periods}\ (N) =4xx10=40`

`text{Annuity factor (from table) = 45.05630}`

`FV` `=1000xx45.05630`  
  `=45\ 056.30`  

 

`text{Difference}` `=45\ 097.17-45\ 056.30`  
  `=$40.87`  

♦ Mean mark 43%.

Measurement, STD2 M1 2022 HSC 28

A dam is in the shape of a triangular prism which is 50 m long, as shown.

Both ends of the dam, `A B C` and `D E F`, are isosceles triangles with equal sides of length 25 metres. The included angles `B A C` and `E D F` are each `150^@`.

 
     

Calculate the number of litres of water the dam will hold when full.  (4 marks)

Show Answers Only

`7\ 812\ 500\ text{L}`

Show Worked Solution

`V=Ah`

`text{Use sine rule to find}\ A:`

`A` `=1/2 ab\ sinC`  
  `=1/2 xx 25 xx 25 xx sin150^@`  
  `=156.25\ text{m}^2`  

 

`:.V` `=156.25 xx 50`  
  `=7812.5\ text{m}^3`  

 

`text{S}text{ince 1 m³ = 1000 litres:}`

`text{Dam capacity}` `=7812.5 xx 1000`  
  `=7\ 812\ 500\ text{L}`  

♦♦ Mean mark 33%.

Financial Maths, STD2 F4 2022 HSC 27

A company purchases a machine for $50 000. The two methods of depreciation being considered are the declining-balance method and the straight-line method.

  1. For the declining-balance method, the salvage value of the machine after `n` years is given by the formula
  2.     `S=V_(0)xx(0.80)^(n),`
  3. where `S` is the salvage value and `V_(0)` is the initial value of the asset.
  4.  i. What is the annual rate of depreciation used in this formula?  (1 mark)
  5. ii. Calculate the salvage value of the machine after 3 years, based on the given formula.  (1 mark)
  6. For the straight-line method, the value of the machine is depreciated at a rate of 12.2% of the purchase price each year.
  7. When will the value of the machine, using this method, be equal to the salvage value found in part (a) (ii)?  (2 marks)
Show Answers Only
  1.  i. `20text{%}`
  2. ii. `$25\ 600`
  3. `text{4 years}`
Show Worked Solution

a.i.  `text{Depreciation rate}\ = 1-0.8=0.2=20text{%}`
 

a.ii.  `text{Find}\ \ S\ \ text{when}\ \ n=3:`

`S` `=V_0 xx (0.80)^n`  
  `=50\ 000 xx (0.80)^3`  
  `=$25\ 600`  

 
b.   `text{Using the SL method}`

`S_n` `= 50\ 000-(0.122 xx 50\ 000)xxn`  
  `=50\ 000-6100n`  

 

`text{Find}\ \ n\ \ text{when}\ \ S_n=$25\ 600`

`25\ 600` `=50\ 000-6100n`  
`6100n` `=24\ 400`  
`n` `=(24\ 400)/6100`  
  `=4\ text{years}`  

♦♦ Mean mark (a.i.) 24%.
COMMENT: A poor State result in part (a.i.) that warrants attention.
 
♦ Mean mark part (b) 38%.

Measurement, STD2 M6 2022 HSC 26

The diagram shows two right-angled triangles, `ABC` and `ABD`,

where `AC=35 \ text{cm},BD=93 \ text{cm}, /_ACB=41^(@)` and `/_ADB=theta`.
 
     

Calculate the size of angle `theta`, to the nearest minute.  (4 marks)

Show Answers Only

`19^@6^{′}`

Show Worked Solution

`text{In}\ Delta ABC:`

`cos 41^@` `=35/(BC)`  
`BC` `=35/(cos 41^@)`  
  `=46.375…`  

 
`angle BCD = 180-41=139^@`
 

`text{Using sine rule in}\ Delta BCD:`

`sin theta/(46.375)` `=sin139^@/93`  
`sin theta` `=(sin 139^@ xx 46.375)/93`  
`:.theta` `=sin^(-1)((sin 139^@ xx 46.375)/93)`  
  `=19.09…`  
  `=19^@6^{′}\ \ text{(nearest minute)}`  

♦ Mean mark 50%.

Financial Maths, STD2 F5 2022 HSC 25

The table shows the future value of an annuity of $1.
 
     

Zal is saving for a trip and estimates he will need $15 000. He opens an account earning 3% per annum, compounded annually.

  1. How much does Zal need to deposit every year if he wishes to have enough money for the trip in 4 years time?  (2 marks)

  2. How much interest will Zal earn on his investment over the 4 years? Give your answer to the nearest dollar.  (2 marks)

Show Answers Only
  1. `$3589.09`
  2. `$660`
Show Worked Solution

a.   `text{Using the table:}\ r=3text{%},\ \ n=4`

`text{Annuity factor}\ = 4.184`

`text{Let}\ \ A=\ text{amount invested each year}`

`FV` `=A xx 4.184`  
`15\ 000` `=A xx 4.184`  
`:.A` `=(15\ 000)/4.184`  
  `=$3585.09`  

 

b.   `text{Total payments}\ = 4 xx 3585.09=$14\ 340.36`

`text{Interest earned}` `=FV-\ text{total payments}`  
  `=15\ 000-14\ 340.36`  
  `=659.64`  
  `=$660\ \ text{(nearest $)}`  

♦♦ Mean mark 33%.

ENGINEERING, PPT 2021 HSC 22a

A baseplate used to support the wheels under a trolley is shown in pictorial view.

Construct, within the dashed-line box provided, a top view of the baseplate. Do NOT dimension.   (3 marks)
 

Show Answers Only

Top view

Show Worked Solution

Top View


♦ Mean mark 42%.