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PHYSICS, M5 2019 HSC 11 MC

A dwarf planet orbits the sun with a period of 40 000 years.

The average distance from the sun to Earth is one astronomical unit.

What is the average distance between this dwarf planet and the sun in astronomical units?

  1. `34`
  2. `200`
  3. `1170`
  4. `8 × 10^(6)`
Show Answers Only

`C`

Show Worked Solution

Kepler’s Third Law:

`(r^3)/(T^2)`  is constant for objects orbiting the same central body. 

`(r_text{Planet}^3)/(T_text{Planet}^2)` `=(r_text{Earth}^3)/(T_text{Earth}^2)`  
`(r_text{Planet}^3)/(40\ 000^2)` `= 1`  
`r_text{Planet}` `=root(3)(40\ 000^2)`  
  `=1170`  

 
`=>C`

PHYSICS, M7 2019 HSC 10 MC

A beam of light passes through two polarisers. The second polariser has a transmission axis at an angle of 30° to that of the first polariser. The intensity of the light beam before and after the second polariser is \( I_0\) and \(I_B\) respectively.
 

Which row of the table correctly identifies the value of \( \dfrac{I_B}{I_0} \), and the model of light demonstrated by this investigation?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-3.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Value of \(\dfrac{I_{ B }}{I_0}\)}\rule[-1ex]{0pt}{0pt} \quad & \textit{Model of light demonstrated} \\
\hline
\rule{0pt}{2.5ex}0.750\rule[-1ex]{0pt}{0pt}&\text{Wave model}\\
\hline
\rule{0pt}{2.5ex}0.750\rule[-1ex]{0pt}{0pt}& \text{Particle model}\\
\hline
\rule{0pt}{2.5ex}0.866\rule[-1ex]{0pt}{0pt}& \text{Wave model} \\
\hline
\rule{0pt}{2.5ex}0.866\rule[-1ex]{0pt}{0pt}& \text{Particle model} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

Applying Malus’ law:

\(I_{B} \) \( =I_{0} \cos^{2} \theta \)
\( \dfrac{I_B}{I_0} \) \(=\cos^{2} 30^{\circ}=0.750 \)

 
Polarisation is a wave property of light.

\( \Rightarrow A\)

PHYSICS, M7 2019 HSC 6 MC

Which graph correctly shows the relationship between the surface temperature of a black body `(T)` and the wavelength `(λ)` at which the maximum intensity of light is emitted?
 

Show Answers Only

`A`

Show Worked Solution
  • Since  `lambda_(max) = (b)/(T), \ \ lambda prop (1)/(T)`  (Wien’s Displacement Law)

`=>A`

PHYSICS, M6 2019 HSC 5 MC

The diagram shows two coils wound around a solid iron rod. Initially the switch is closed.
 

Opening the switch will cause the galvanometer pointer to

  1. remain at a constant reading.
  2. move from a non-zero reading to a zero reading.
  3. move from a zero reading to a non-zero reading, where it remains.
  4. move from a zero reading to a non-zero reading, then back to zero.
Show Answers Only

`D`

Show Worked Solution
  • Originally, there is a constant magnetic field passing through the coil on the right due to the current through the coil on the left.
  • As there is no change in flux through the coil on the right initially the galvanometer shows a zero reading.
  • When the switch is opened, the decrease in magnetic flux through the coil on the right causes a deflection of the galvanometer to a non-zero reading.
  • The galvanometer will return to a reading of zero when the magnetic flux passing through it drops to zero. 

`=>D`

PHYSICS, M8 2019 HSC 4 MC

Four stars, `P`, `Q`, `R` and `S`, are labelled on the Hertzsprung-Russell diagram.
 

Which statement is correct?

  1. `S` has a greater luminosity than `Q`.
  2. `R` is a blue star whereas `S` is a red star.
  3. `S` has a higher surface temperature than `R`.
  4. `P` is at a more advanced stage of its evolution than `R`.
Show Answers Only

`C`

Show Worked Solution
  • Stars further to the left of the Hertzsprung-Russell diagram have greater surface temperatures.

`=>C`

PHYSICS, M8 2019 HSC 3 MC

Geiger and Marsden carried out an experiment to investigate the structure of the atom.

Which diagram identifies the particles they used and the result that they INITIALLY expected?
 

 

Show Answers Only

`C`

Show Worked Solution
  • They used alpha particles. Based on Thomson’s model it was initially expected that all alpha particles would pass through the atom.

`=>C`

PHYSICS, M6 2020 HSC 32

A rope connects a mass on a horizontal surface to a pulley attached to an electric motor as shown.
 

Explain the factors that limit the speed at which the mass can be pulled along the horizontal surface. Use mathematical models to support your answer.   (7 marks)

Show Answers Only
  • The factors limiting the speed of the mass along horizontal surface can be divided into two main parts; 1-Factors involving friction between the mass and the table and 2-Factors involving the force output of the motor and pulley.

Friction between the mass and table

  • The mass will experience a friction force of  `F_(f)=mu N`. Since the mass has no vertical acceleration,  `N=mg\ \ =>\ \ F_(f)=mu mg.`
  • The speed at which the mass can be pulled is limited by the coefficient of kinetic friction between the mass and the table `mu` and the magnitude of the mass `m`.

 Force output of the motor and pulley

  • The power of the motor depends on its torque and the pulley radius `(tau=rF sin theta).`
  • The torque produced by the motor is given by  `tau=NIAB sin theta`
  • The force of the motor and hence maximum speed at which the mass can be pulled is also limited by the number of turns of the coil in the motor, the current passing through the coils, the area of the coils and the magnetic field strength of the stator magnets.

Answers could also contain:

  • Back emf in the motor limiting its maximum speed.
  • The efficiency of the motor.
Show Worked Solution
  • The factors limiting the speed of the mass along horizontal surface can be divided into two main parts; 1-Factors involving friction between the mass and the table and 2-Factors involving the force output of the motor and pulley.

Friction between the mass and table

  • The mass will experience a friction force of  `F_(f)=mu N`. Since the mass has no vertical acceleration,  `N=mg\ \ =>\ \ F_(f)=mu mg.`
  • The speed at which the mass can be pulled is limited by the coefficient of kinetic friction between the mass and the table `mu` and the magnitude of the mass `m`.

 Force output of the motor and pulley

  • The power of the motor depends on its torque and the pulley radius `(tau=rF sin theta).`
  • The torque produced by the motor is given by  `tau=NIAB sin theta`
  • The force of the motor and hence maximum speed at which the mass can be pulled is also limited by the number of turns of the coil in the motor, the current passing through the coils, the area of the coils and the magnetic field strength of the stator magnets.

Answers could also contain:

  • Back emf in the motor limiting its maximum speed.
  • The efficiency of the motor.
Mean mark 58%.

PHYSICS, M5 2020 HSC 31

  1. The orbit of a comet is shown.
     

  1. Account for the changes in velocity of the comet as it completes one orbit from position `P`.   (3 marks)
  1. Two stars, `A` and `B`, of equal mass `m`, separated by a distance `x`, interact gravitationally such that the speed of `A` is constant.
  2. Derive an expression for the speed of `B`.   (3 marks)
Show Answers Only

a.    Changes in Velocity

  • As the comet moves from position `P` towards the sun, its gravitational potential energy (GPE) is converted into kinetic energy.
  • Increase in the comet’s kinetic energy → increase in the comet’s velocity as it moves towards the sun.
  • As the comet moves away from the sun back to` P`, its kinetic energy is converted into GPE.
  • Decrease in kinetic energy → decrease in comet’s velocity.

b.   `v_(B)=sqrt((Gm)/(2x))`

Show Worked Solution

a.    Changes in Velocity

  • As the comet moves from position `P` towards the sun, its gravitational potential energy (GPE) is converted into kinetic energy.
  • Increase in the comet’s kinetic energy → increase in the comet’s velocity as it moves towards the sun.
  • As the comet moves away from the sun back to` P`, its kinetic energy is converted into GPE.
  • Decrease in kinetic energy → decrease in comet’s velocity.

♦ Mean mark (a) 49%.

b.    As the speed of `A` is constant, it must be travelling in a circular orbit.

  • `A` and `B` are both orbiting in a circle around their centre of mass.
  • Since both comets have the same mass, the radius of their circular motion is half of the distance between them.
  • As the centripetal force keeping `B` in orbit is provided by the gravitational force acting on it due to `A`:
`F_(g)` `=F_(c)`  
`(Gm^(2))/(x^(2))` `=(m(v_(B))^(2))/((x)/(2))`  
`(Gm)/x` `=2(v_B)^2`  
`v_(B)` `=sqrt((Gm)/(2x))`  

PHYSICS, M8 2020 HSC 29

In an experiment, alpha particles were fired into a thin sheet of beryllium. Unknown radiation was detected.
 

Further experiments were conducted in which it was observed that the unknown radiation:

    • was not deflected by an electric field
    • caused protons to be ejected from a block of paraffin
    • could not produce the photoelectric effect.

Scientists debated the nature of this unknown radiation, hypothesising that it was gamma radiation.

  1. Explain why the hypothesis was proposed and then rejected, with reference to the observations.   (3 marks)
  1. How did these experiments change the model of the atom?   (2 marks)
Show Answers Only

a.    Hypothesis proposal and subsequent rejection:

  • The unknown radiation was not deflected by an electric field which is consistent with the properties of gamma radiation which is electrically neutral. This observation supported the hypothesis.
  • However, the hypothesis was rejected as the unknown radiation was able to eject protons from paraffin indicating it carried significant momentum. This was inconsistent with the radiation being gamma radiation.
  • Additionally, the unknown radiation was unable to produce the photoelectric effect. This was also inconsistent with the unknown radiation being gamma radiation as the high photon energy of gamma rays would readily eject photoelectrons from a metal. 

b.    Changes to the model of the atom:

  • These experiments were used to demonstrate the existence of neutrons, a neutral particle with mass similar to protons.
  • This changed the model of the atom as the atomic nucleus was represented as containing protons and neutrons rather than just protons.
Show Worked Solution

a.    Hypothesis proposal and subsequent rejection:

  • The unknown radiation was not deflected by an electric field which is consistent with the properties of gamma radiation which is electrically neutral. This observation supported the hypothesis.
  • However, the hypothesis was rejected as the unknown radiation was able to eject protons from paraffin indicating it carried significant momentum. This was inconsistent with the radiation being gamma radiation.
  • Additionally, the unknown radiation was unable to produce the photoelectric effect. This was also inconsistent with the unknown radiation being gamma radiation as the high photon energy of gamma rays would readily eject photoelectrons from a metal. 

b.    Changes to the model of the atom:

  • These experiments were used to demonstrate the existence of neutrons, a neutral particle with mass similar to protons.
  • This changed the model of the atom as the atomic nucleus was represented as containing protons and neutrons rather than just protons.

PHYSICS, M8 2021 HSC 35

A spacecraft is powered by a radioisotope generator. Pu-238 in the generator undergoes alpha decay, releasing energy. The decay is shown with the mass of each species in atomic mass units, `u`

\begin{array} {ccccc}
\ce{^{238}Pu} & \rightarrow & \ce{^{234}U} & + & \alpha \\
238.0495\ u &  & 234.0409\ u &  & 4.0026\ u \end{array}

  1. Show that the energy released by one decay is  `9.0 × 10^(-13)` J.   (3 marks)
  1. At launch, the generator contains  `9.0 × 10^24`  atoms of Pu-238. The half-life of Pu-238 is 87.7 years.
  2. Calculate the total energy produced by the generator during the first ten years after launch.  (3 marks)
Show Answers Only
  1. `9.0xx10^(-13)\ \text{J}`
  2. `6.2xx10^(11)\ \text{J}`
Show Worked Solution

a.   `text{Find Energy released:}`

`Delta m` `=(234.0409+4.0026)-238.0495=-0.006\ u`  
`Delta m` `=0.006 xx1.661 xx10^(-27)=9.966 xx10^(-30)\ \text{kg}`  

  
`\text{Using}\ \ E=mc^2:`

`E_text{released}=9.966 xx10^(-30)xx(3xx10^(8))^(2)=9.0 xx10^(-13)\ \text{J}`

   

b.   `\text{Find total energy:}`

`lambda=(ln 2)/(t_((1)/(2)))=(ln 2)/(87.7)=0.0079\ text{year}^(-1)`

`N=N_(0)e^(-lambda t)=9xx10^(24)xxe^(-0.0079 xx10)=8.316 xx10^(24)`

`Delta N=9xx10^(24)-8.316 xx10^(24)=6.84 xx10^(23)`

`E=6.84 xx10^(23)xx9.0 xx10^(-13)=6.2 xx10^(11)\ \text{J}`


♦ Mean mark part (b) 51%.

PHYSICS, M5 2021 HSC 34

A 3.0 kg mass is launched from the edge of a cliff.
 

The kinetic energy of the mass is graphed from the moment it is launched until it hits the ground at `X`. The kinetic energy of the mass is provided for times `t_0, t_1` and `t_2`.
 

  1. Account for the relative values of kinetic energy at `t_0, t_1` and `t_2`.    (4 marks)
  1. The horizontal component of the velocity of the mass during its flight is 13.76 `text{m s}^(-1)`.    (3 marks)
  2. Calculate the time of flight of the mass.
Show Answers Only

a.   Relative values of kinetic energy at `t_0, t_1` and `t_2`:

  • After the mass is launched at ` t_0`, downwards gravitational acceleration decreases the vertical velocity of the mass until it is zero at `t_1`.
  • The kinetic energy of the mass is at a minimum here but not zero due to the horizontal velocity of the mass, which remains constant until the projectile hits the ground.
  • After `t_1`, the kinetic energy of the mass increases as its gravitational potential (GPE) energy is converted to kinetic energy. This occurs as gravity increases the vertical velocity of the mass until it strikes the ground at `t_2`.
  • The GPE of the mass at `t_2` is lower than its GPE at`t_0`.
  • The mass has greater kinetic energy at `t_2` than at `t_0.`

b.   4.8 seconds

Show Worked Solution

a.   Relative values of kinetic energy at `t_0, t_1` and `t_2`:

  • After the mass is launched at ` t_0`, downwards gravitational acceleration decreases the vertical velocity of the mass until it is zero at `t_1`.
  • The kinetic energy of the mass is at a minimum here but not zero due to the horizontal velocity of the mass, which remains constant until the projectile hits the ground.
  • After `t_1`, the kinetic energy of the mass increases as its gravitational potential (GPE) energy is converted to kinetic energy. This occurs as gravity increases the vertical velocity of the mass until it strikes the ground at `t_2`.
  • The GPE of the mass at `t_2` is lower than its GPE at`t_0`.
  • The mass has greater kinetic energy at `t_2` than at `t_0.` 

b.   Find  `u_y` when `t=0`:

`KE` `=(1)/(2)mv^(2)`  
`v^2` `=2 xx 864/(3)=576`  
`v` `=24`  

 
Using Pythagoras:

`24^(2)=u_(y)^(2)+13.76^(2)\ \ =>\ \ u_(y)=19.66\ text{m s}^(-1)`
 
Find  `v_y`  at  `t_2`:`

`1393` `=(1)/(2) xx 3v^(2)`  
`v^2` `=2 xx 1393/(3)=928.7`  
`v` `=30.47`  

 
Using Pythagoras:

`30.47^(2)=v_(y)^(2)+13.76^(2)\ \ =>\ \ v_(y)=27.19\ text{m s}^(-1)`

`v_(y)` `=u_(y)+a_(g)t`  
  `t` `=(27.19-(-19.66))/(9.8)=4.8\ \text{sec}`  

CHEMISTRY, M7 2021 HSC 13 MC

A chemist synthesises a substance using the following pathway.

\[\ce{ X ->[{hydration}] {Y} ->[{oxidation}] Z}\]

What are compounds `text{X, Y, Z}` ?

Show Answers Only

`C`

Show Worked Solution

By elimination:

  • Hydration reaction is an addition reaction that can only occur on alkenes, thus  `X` = prop-1-ene  (eliminate A and B)
  • `Y` = propan-2-ol
  • The oxidation of secondary alcohol creates a ketone, thus `Z` = propanone

`=> C`

CHEMISTRY, M7 2021 HSC 10 MC

The structure of part of a polymer chain is shown.
 

 
Which is the monomer of this polymer?
 

 

Show Answers Only

`B`

Show Worked Solution
  • The polymer formed is a condensation polymer, where the by-product is water (`text{H}_2 text{O}`).
  • From the polymer shown, the alcohol functional group must ie on the second carbon atom, and the carboxylic functional group must lie on the first carbon atom.
  • Thus, could only be formed from monomer B.

`=> B`

PHYSICS, M6 2020 HSC 28

A metal rod sits on a pair of parallel metal rails, 20 cm apart, that are connected by a copper wire. The rails are at 30° to the horizontal.

The apparatus is in a uniform magnetic field of 1 T which is upward, perpendicular to the table.
 

A force, `F`, is applied parallel to the rails to move the rod at a constant speed along the rails. The rod is moved a distance of 30 cm in 2.5 s.

  1. Show that the change in magnetic flux through the circuit while the rod is moving is approximately `5.2 × 10^(-2)` Wb.   (2 marks)
  1. Calculate the emf induced between the ends of the rod while it is moving, and state the direction of flow of the current in the circuit.   (2 marks)
  1. The experiment is repeated without the magnetic field.
  2. Explain why the force required to move the rod is different without the magnetic field.   (3 marks)
Show Answers Only

a.   `phi=5.2 xx10^(-2)\ \text{Wb}`

b.   `epsi=2.1 xx10^(-2)  \text{V}`

The direction of the induced current is anticlockwise as viewed from above.

c.    Repeating experiment is without the magnetic field:

  • When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).
  • Additionally, the force required to move the rod must also overcome the downwards gravitational force.
  • Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.
  • Hence, the force required to move the rod is less without the magnetic field.
Show Worked Solution
a.
`phi` `=BA cos theta`
    `=1xx(0.3 xx0.2)xx cos 30^(@)`
    `=0.05196…`
    `~~5.2 xx10^(-2)` Wb

b.
`epsi` `=-N(Delta phi)/(t)`
    `=-1xx(5.2 xx10^(-2))/(2.5)`
    `=0.208…`
    `=2.1 xx10^(-2)\ \text{V  (V>0)`

 

  • The direction of the induced current is anticlockwise as viewed from above (Lenz’s Law).


♦ Mean mark (b) 51%.

c.    Repeating experiment is without the magnetic field:

  • When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).
  • Additionally, the force required to move the rod must also overcome the downwards gravitational force.
  • Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.
  • Hence, the force required to move the rod is less without the magnetic field.

♦ Mean mark (c) 50%.

PHYSICS, M7 2020 HSC 26

  1. Describe the difference between the spectra of the light produced by a gas discharge tube and by an incandescent lamp.   (2 marks)
  1. The graph shows the curves predicted by two different models, `X` and `Y`, for the electromagnetic radiation emitted by an object at a temperature of 5000 K.
     

    Identify an assumption of EACH model which determines the shape of its curve.   (2 marks)

  1. The diagram shows the radiation curve for a black body radiator at a temperature of 5000 K.
     


     
    On the same diagram, sketch a curve for a black body radiator at a temperature of 4000 K and explain the differences between the curves.   (4 marks)

Show Answers Only

a.   The spectra of light produced by a gas discharge tube will consist of lines only at a few discrete wavelengths.

The spectra of light produced by an incandescent lamp will be a continuous spectrum.

b.   Model `X` black bodies absorb and emit energy continuously.

Model `Y` assumes that black bodies absorb and emit energy in discrete quantities.

c.

  • Using `lambda_(max)=(b)/(T)\ \ =>\ \ lambda prop (1)/(T)`
  • Therefore, the 4000 K curve will have a peak wavelength greater than the 5000 K curve.
  • The area under the curve and the intensity at all wavelengths will be less for the 4000 K curve, as the total power output of a black body decreases as its temperature decreases.
Show Worked Solution

a.    Differences:

  • The spectra of light produced by a gas discharge tube will consist of lines only at a few discrete wavelengths.
  • The spectra of light produced by an incandescent lamp will be a continuous spectrum.

♦ Mean mark (a) 39%.

b.    Assumptions of EACH model:

  • Model `X` black bodies absorb and emit energy continuously.
  • Model `Y` assumes that black bodies absorb and emit energy in discrete quantities.

♦ Mean mark (b) 44%.

c.

  • Using `lambda_(max)=(b)/(T)\ \ =>\ \ lambda prop (1)/(T)`
  • Therefore, the 4000 K curve will have a peak wavelength greater than the 5000 K curve.
  • The area under the curve and the intensity at all wavelengths will be less for the 4000 K curve, as the total power output of a black body decreases as its temperature decreases.

PHYSICS, M8 2020 HSC 25

Describe the hydrogen atom in terms of the Standard Model of matter.   (4 marks)

Show Answers Only
  • The hydrogen atom is composed of one proton in the nucleus and one orbiting electron.
  • The proton is a hadron comprised of three quarks, two up quarks each with charge  `+(2)/(3)`  and one down quark with charge `-(1)/(3)`. These quarks are fundamental particles and are bound together by the strong nuclear force.
  • The electron is classified as a lepton and is a fundamental particle. It is held in its orbit by the electromagnetic force.
Show Worked Solution
  • The hydrogen atom is composed of one proton in the nucleus and one orbiting electron.
  • The proton is a hadron comprised of three quarks, two up quarks each with charge  `+(2)/(3)`  and one down quark with charge `-(1)/(3)`. These quarks are fundamental particles and are bound together by the strong nuclear force.
  • The electron is classified as a lepton and is a fundamental particle. It is held in its orbit by the electromagnetic force.

♦ Mean mark 55%.

PHYSICS, M5 2020 HSC 24

The graph shows the vertical displacement of a projectile throughout its trajectory. The range of the projectile is 130 m.
 

Calculate the initial velocity of the projectile.   (4 marks)

Show Answers Only

`37\ text{m s}^(-1)`, at 54° above the horizontal.

Show Worked Solution

From the graph, at  `t=3`, the projectile reaches a maximum height of 44 m:

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^(2)`  
`44` `=u_(y)(3)-(1)/(2)(9.8)(3^(2))`  
`u_(y)` `=29.4\ text{m s}^(-1)`  

  
Find `u_x` given time of flight = 6 s:

`u_(x)=(s_(x))/(t)=(130)/(6)=21.7\ text{m s}^(-1)`
 

Using Pythagoras:

`u^(2)` `=u_(x)^(2)+u_(y)^(2)`  
  `=21.7^(2)+29.4^(2)`  
`u` `=37\ text{m s}^(-1)`  

  
Find launch angle (`theta)`:

`tan theta` `=(u_y)/(u_x)`  
`theta` `=54^(@)`  

  
So,  `u=37\ text{m s}^(-1)`, at 54° above the horizontal.

PHYSICS, M7 2020 HSC 22

A capsule travelling at 12 900 m s ¯1 enters Earth's atmosphere, causing it to rapidly slow down to 400 m s ¯1.

  1. During this re-entry, the capsule reaches a temperature of 3200 K.
  2. What is the peak wavelength of the light emitted by the capsule? (2 marks)
  1. Outline TWO limitations of applying special relativity to the analysis of the motion of the capsule.  (3 marks)
Show Answers Only

a.   `lambda_(max)=9xx10^(-7)` m

b.   Limitations:

  • The speed of the capsule is not close to the speed of light and so the effects of special relativity are insignificant.
  • The capsule is decelerating and so it is in a non-inertial frame of reference, therefore special relativity is not applicable.
Show Worked Solution
a.    `lambda_(max)` `=(b)/(T)`
    `=(2.898 xx10^(-3))/(3200)`
    `=9.056xx10^(-7)  text{m}`
    `=9xx10^(-7)  text{m}`

 

b.   Limitations:

  • The speed of the capsule is not close to the speed of light and so the effects of special relativity are insignificant.
  • The capsule is decelerating and so it is in a non-inertial frame of reference, therefore special relativity is not applicable.

♦ Mean mark (b) 42%.

PHYSICS, M8 2020 HSC 21

  1. Calculate the wavelength of light emitted by an electron moving from energy level 3 to 2 in a Bohr model hydrogen atom.   (2 marks)
  1. Describe the behaviour of electrons in the Bohr model of the atom with reference to the law of conservation of energy.   (3 marks)
Show Answers Only

a.   `6.563 xx10^(-7) text{m}`

b.   Behaviour of electrons in the Bohr model:

  • Bohr’s model describes electrons orbiting the atomic nucleus in discrete energy levels.
  • When these electrons absorb a photon they gain energy and move from lower to higher energy levels.
  • When they move from higher to lower energy levels, they emit energy in the form of a photon.
  • This is consistent with the law of conservation of energy as the energy of absorbed or emitted photons is equal to the difference in energy of the discrete levels between which electrons move.
Show Worked Solution
a.
`(1)/(lambda)` `=R((1)/(n_(f^(2)))-(1)/(n_(i^(2))))`
    `=1.097 xx10^(7)((1)/(2^(2))-(1)/(3^(2)))`
    `=1.524 xx10^(6)`
  `lambda` `=6.563 xx10^(-7) text{m}`

 

b.   Behaviour of electrons in the Bohr model:

  • Bohr’s model describes electrons orbiting the atomic nucleus in discrete energy levels.
  • When these electrons absorb a photon they gain energy and move from lower to higher energy levels.
  • When they move from higher to lower energy levels, they emit energy in the form of a photon.
  • This is consistent with the law of conservation of energy as the energy of absorbed or emitted photons is equal to the difference in energy of the discrete levels between which electrons move.

PHYSICS, M7 2020 HSC 17 MC

In a thought experiment, observer `X` is on a train travelling at a constant velocity of 0.95c relative to the ground. Observer `Y` is standing on the ground outside the train. As observer `X` passes observer `Y`, observer `X` sends a short light pulse towards the sensor.
 

Which statement about the light pulse is correct as observed by `X` or `Y` in their respective frames of reference?

  1. Its velocity observed by `Y` is 0.05c.
  2. `X` sees it travel a shorter distance to the sensor than `Y`.
  3. `X` sees it take a longer time to reach the sensor than `Y`.
  4. Both `X` and `Y` see it travel the same distance in the same amount of time.
Show Answers Only

`C`

Show Worked Solution
  • `Y` observes the length of the carriage to be shorter than `X`  does due to length contraction.
  • As the speed of light is equal for both observers, `X`  sees the light pulse take a longer time to reach the sensor.

`=>C` 

PHYSICS, M5 2020 HSC 15 MC

A rocket returns to Earth for reuse after launching satellites, using its engines to make a controlled landing.

The rocket having a mass of 7800 kg is on approach to the ground, travelling horizontally at 20 m s ¯1 as shown in the diagram, when the engine thrust is changed to 90 000 newtons.
 

Which diagram shows the trajectory of the rocket following this change of thrust?
 

Show Answers Only

`D`

Show Worked Solution
  • Initial horizontal direction → net zero force
  • `W=mg=7800 xx 9.8=76\ 440\ text{N}`
  • Since  `90\ 000\ text{N} > 76\ 440` → net upwards force → net upwards acceleration.
  • Net upwards acceleration with constant horizontal velocity results in an upwards parabolic trajectory.

`=>D`

PHYSICS, M8 2020 HSC 11 MC

Consider the following nuclear reaction.

`\ _(3)^(6)text{Li} +\ _(0)^(1)text{n}rarr \ _(2)^(4)text{He} +\ _(1)^(3)text{H}`

The mass of the reactants is 7.023787704 `u` and the mass of the products is 7.018652532 `u`.

What type of reaction is this?

  1. A fusion reaction in which energy is released
  2. A fusion reaction that requires an input of energy
  3. A transmutation reaction in which energy is released
  4. A transmutation reaction that requires an input of energy
Show Answers Only

`C`

Show Worked Solution
  • The products contain multiple nuclei → the reaction is not a fusion reaction.
  • The mass of the products is less than the mass of the reactants → energy is released.

`=>C`

PHYSICS, M8 2020 HSC 9 MC

Bohr improved on Rutherford's model of the atom.

Which observation by Bohr provided evidence supporting the improvement?

  1. Elements produced unique emission spectra consisting of discrete wavelengths.
  2. The collision of an electron and a positron produced two photons that travelled in opposite directions.
  3. A small percentage of alpha particles fired at a gold foil target were deflected by angles of more than 90 degrees.
  4. A beam of electrons reflected from a nickel crystal produced a pattern of intensity at different angles, consistent with their wave properties.
Show Answers Only

`A`

Show Worked Solution
  • Bohr’s model placed electrons in discrete energy levels explaining why emission spectra consisted of discrete wavelengths.

`=>A`

PHYSICS, M5 2020 HSC 5 MC

A student throws a ball that follows a parabolic trajectory.

What change to the initial velocity would make the ball's time of flight shorter?

  1. Increasing only the vertical component
  2. Decreasing only the vertical component
  3. Increasing only the horizontal component
  4. Decreasing only the horizontal component
Show Answers Only

`B`

Show Worked Solution
  • Only the vertical component of the ball’s velocity impacts its time of flight.
  • Decreasing this decreases the time of flight.

`=> B`

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment \(A\), the light is incident on a pair of narrow slits  5.0 × 10\(^{-5}\) m  apart, producing a pattern on a screen located 3.0 m behind the slits.
 

In experiment \(B\), the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.
 

Some results from experiment \(B\) are shown.
 

\begin{array}{|l|l|c|}
\hline
\rule{0pt}{1.5ex}\textit{Metal sample}\rule[-0.5ex]{0pt}{0pt}& \textit{Work function} \ \text{(J)} & \textit{Photoelectrons observed?} \\
\hline
\rule{0pt}{2.5ex}\text{Nickel}\rule[-1ex]{0pt}{0pt}&8.25 \times 10^{-19}&\text{No}\\
\hline
\rule{0pt}{2.5ex}\text{Calcium}\rule[-1ex]{0pt}{0pt}& 4.60 \times 10^{-19}&\text{Yes}\\
\hline
\end{array}

How do the results from Experiment \(A\) and Experiment \(B\) support TWO different models of light? In your answer, include a quantitative analysis of each experiment.   (9 marks)

Show Answers Only
  • Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.
  • When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.
  • Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.
  • The spacing between adjacent bright bands can be calculated using  \(d \sin \theta=m \lambda\):
  •    \(5 \times 10^{-5} \sin \theta=1 \times 400 \times 10^{-9}\ \ \Rightarrow\ \ \theta=0.46^{\circ}\)
  •    \(s=3 \times \tan (0.46^{\circ})=0.024 \ \text{m}\)
  •  Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:
  •    \(f=\dfrac{c}{\lambda}=\dfrac{3.00 \times 10^8}{400 \times 10^{-9}}=7.50 \times 10^{14} Hz\)
  •    \(E=h f=6.626 \times 10^{-34} \times 7.50 \times 10^{14}=4.97 \times 10^{-19} J\)
  • This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.
  • These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:
  •    \(K_{\max }=h f-\phi=4.97 \times 10^{-19}-4.60 \times 10^{-19}=3.70 \times 10^{-20} \ \text{J}\)
Show Worked Solution
  • Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.
  • When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.
  • Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.
  • The spacing between adjacent bright bands can be calculated using  \(d \sin \theta=m \lambda\):
  •    \(5 \times 10^{-5} \sin \theta=1 \times 400 \times 10^{-9}\ \ \Rightarrow\ \ \theta=0.46^{\circ}\)
  •    \(s=3 \times \tan (0.46^{\circ})=0.024 \ \text{m}\)
  •  Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:
  •    \(f=\dfrac{c}{\lambda}=\dfrac{3.00 \times 10^8}{400 \times 10^{-9}}=7.50 \times 10^{14} Hz\)
  •    \(E=h f=6.626 \times 10^{-34} \times 7.50 \times 10^{14}=4.97 \times 10^{-19} J\)
  • This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.
  • These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:
  •    \(K_{\max }=h f-\phi=4.97 \times 10^{-19}-4.60 \times 10^{-19}=3.70 \times 10^{-20} \ \text{J}\)

♦ Mean mark 52%.

PHYSICS, M6 2021 HSC 30

In an experiment, a proton accelerates from rest between parallel charged plates. The spacing of the plates is 12 cm and the proton is initially positioned at an equal distance from both plates, as shown. Ignore the effect of gravity.
 

  1. The electrical potential energy of the proton is recorded in the following graph for the first 5 cm of its motion.
  2. On the graph, sketch the corresponding kinetic energy of the proton over the same distance.   (2 marks)

 
 

  1. The experiment is repeated using an electron in the place of the proton.
  2. Explain how the motion of the electron would differ from that of the proton.    (3 marks)
Show Answers Only

a.   
           
        
b.   Motions differs from proton motion:

  • The electron will move in the opposite direction to the proton as the sign of its charge is opposite.
  • The electron will experience a force equal in magnitude to the proton as its charge has the same magnitude.
  • Since the mass of the electron is significantly smaller, it will experience a much greater acceleration compared to the proton.
Show Worked Solution

a.   
           
  

b.   Motions differs from proton motion:

  • The electron will move in the opposite direction to the proton as the sign of its charge is opposite.
  • The electron will experience a force equal in magnitude to the proton as its charge has the same magnitude.
  • Since the mass of the electron is significantly smaller, it will experience a much greater acceleration compared to the proton.

PHYSICS, M7 2021 HSC 28

A spaceship travels to a distant star at a constant speed, `v`. When it arrives, 15 years have passed on Earth but 9.4 years have passed for an astronaut on the spaceship.

  1. What is the distance to the star as measured by an observer on Earth?   (3 marks)
  1. Outline how special relativity imposes a limitation on the maximum velocity of the spaceship.   (2 marks)
Show Answers Only

a.    12 ly

b.    According to special relativity, as ` v\ →\ c`:

  •   the momentum of the spaceship approaches infinity
  •   the force required to accelerate the spaceship approaches infinity
  •   maximum velocity is limited to the speed of light.
Show Worked Solution

a.    `t_v =t_0/sqrt((1-(v^2)/(c^2)))`

`15=9.4/sqrt((1-(v^2)/(c^2)))`

`1-(v^2)/(c^2)` `=((9.4)/(15))^2`  
`(v^2)/(c^2)` `=1-((9.4)/(15))^2`  
`v^2` `=0.60729c^2`  
`v` `=0.779c`  

  
Distance to star from Earth observer:

`s=ut=0.779 xx15=12\ text{ly}`
 


♦ Mean mark part (a) 51%.

b.  According to special relativity, as ` v\ →\ c`:

  • the momentum of the spaceship approaches infinity
  • the force required to accelerate the spaceship approaches infinity
  • maximum velocity is limited to the speed of light.

PHYSICS, M6 2021 HSC 27

A student is considering how to levitate a thin metal rod in a strong magnetic field of 1.2 T. The current flowing through the rod will be 2.3 A.

  1. Use a labelled diagram to show a suitable orientation of the current and the magnetic field to achieve this outcome. Include relevant forces in your diagram.   (3 marks)
     
     
  1. Explain why the maximum mass per unit length of the rod cannot exceed 0.282 kg m\(^{-1}\). Support your answer with a calculation.   (3 marks)
Show Answers Only

a.


 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

  • Levitation will occur when:
\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}=0.282\ \text{kg m}^{-1}\)  

 

  • If the mass per unit length exceeds this, the rod would not levitate.
Show Worked Solution

a.

Note: a different alignment with the magnetic force coming out of the page and the current moving left is also correct.
 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

  • Levitation will occur when:
\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}=0.282\ \text{kg m}^{-1}\)  

 

  • If the mass per unit length exceeds this, the rod would not levitate.

PHYSICS, M7 2021 HSC 26

A student performs an experiment to measure Planck's constant, \(h\), using a device that emits specific frequencies of light when specific voltages are applied.

The voltage, \(V\), needed to produce each frequency, \(f\), is given by

\(V=\dfrac{h f}{q_e}\)

where `q_e` is the charge on an electron.

Data from the experiment is shown.
 

\begin{array}{|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Data point}\rule[-1ex]{0pt}{0pt}& \textit{Frequency} & \quad \textit{Voltage} \quad  \\
\text{}\rule[-1ex]{0pt}{0pt}& (\times 10^{14}\,\text{Hz})& \text{(V)} \\
\hline
\hline \rule{0pt}{2.5ex}1 \rule[-1ex]{0pt}{0pt}& 3.5 & 1.3 \\
\hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 4.8 & 1.7 \\
\hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 5.3 & 1.9 \\
\hline \rule{0pt}{2.5ex}4 \rule[-1ex]{0pt}{0pt}& 7.0 & 2.6 \\
\hline
\end{array}

  1. Graph this data on the axes provided. Include a line of best fit.   (3 marks)
     
     

       
  2. The student proposes using data point 1 to calculate a value for Planck's constant.
  3. Justify a better method to calculate Planck's constant from the experimental data provided.   (3 marks)

Show Answers Only

a.   
         
 

b.    Determine gradient using the line of best fit of  \(V=\dfrac{h f}{q_e}\)

  • The gradient is equal to \(\dfrac{V}{f}\), and \(h\) can be calculated as \(h=\) gradient  \(\times q_e\).
  • This method increases the accuracy by taking more data points into account.

Show Worked Solution

a.   
         
 

b.    Determine gradient using the line of best fit of  \(V=\dfrac{h f}{q_e}\)

  • The gradient is equal to \(\dfrac{V}{f}\), and \(h\) can be calculated as \(h=\) gradient  \(\times q_e\).
  • This method increases the accuracy by taking more data points into account.

PHYSICS, M5 2021 HSC 25

A satellite is launched from the surface of Mars into an orbit that keeps it directly above a position on the surface of Mars.

Mass of Mars = `6.39 xx 10^(23)` kg

Length of Martian day = 24 hours and 40 minutes

  1. Identify TWO energy changes as the satellite moves from the surface of Mars into orbit.   (2 marks)
  1. Calculate the orbital radius of the satellite.   (3 marks)
Show Answers Only

a.   The kinetic energy and gravitational potential energy of the satellite both increase as it moves into orbit.

b.   `2.0 xx10^(7)\ \text{m}`

Show Worked Solution

a.   The kinetic energy and gravitational potential energy of the satellite both increase as it moves into orbit.

b.   `(r^(3))/(T^(2))` `=(GM)/(4pi^(2))`  
`r^(3)` `=((6.67 xx10^(-11)xx6.39 xx10^(23))/(4pi^(2)))((24+(40)/(60))xx60 xx60)^(2)`  
`:. r` `=2.0 xx10^(7)\ \text{m}`  

PHYSICS, M6 2021 HSC 24

A stationary coil of 35 turns and cross-sectional area of 0.02 m² is placed between two electromagnets, and connected to a voltmeter as shown. The electromagnets produce a uniform magnetic field of 0.15 T through the coil.
 

The magnitude of the magnetic field is then reduced to zero at a constant rate over a period of 0.4 s.

Calculate the magnitude of the emf induced in the coil.   (3 marks)

Show Answers Only

`0.3\ \text{V}`

Show Worked Solution

`Phi=BA=0.15 xx0.02=0.003\ \text{Wb}`

`epsi=-N(Delta Phi)/(Delta t)=-35((0-0.003))/(0.4)=0.3\ \text{V}`

PHYSICS, M8 2021 HSC 23

Describe how Millikan and Thomson each used fields to determine properties of the electron.   (4 marks)

Show Answers Only

Thomson used an electric field to accelerate electrons. He deflected these electrons using a magnetic field then balanced the force on these electrons using a second electric field, allowing him to calculate the charge to mass ratio of an electron.

Millikan used electric fields to suspend oil droplets by balancing their weight due to Earth’s electric field, allowing him to calculate the charge of an electron.

Show Worked Solution

Thomson used an electric field to accelerate electrons. He deflected these electrons using a magnetic field then balanced the force on these electrons using a second electric field, allowing him to calculate the charge to mass ratio of an electron.

Millikan used electric fields to suspend oil droplets by balancing their weight due to Earth’s electric field, allowing him to calculate the charge of an electron.

PHYSICS, M5 2021 HSC 22

A horizontal disc rotates at a constant rate as shown. Two points on the disc, `X` and `Y`, are labelled. `X` is twice as far away from the centre of the disc as `Y`.
 

Compare the angular and instantaneous velocities of `X` with those of `Y`.   (3 marks)

Show Answers Only
  • The angular velocities `(omega)` of  `X` and `Y` are equal. The direction of instantaneous velocities for `X` and `Y` are the same.
  • Since `v=r omega` the magnitude of the instantaneous velocity of `X` is twice that of `Y`.
Show Worked Solution
  • The angular velocities `(omega)` of  `X` and `Y` are equal. The direction of instantaneous velocities for `X` and `Y` are the same.
  • Since `v=r omega` the magnitude of the instantaneous velocity of `X` is twice that of `Y`.

PHYSICS, M7 2021 HSC 11 MC

What is the peak wavelength of electromagnetic radiation emitted by a person with a body temperature of 37°C (310 K)?

  1. `9.3 × 10^(-6)\ `m
  2. `7.8 × 10^(-5)\ `m
  3. `9.3 × 10^(-3)\ `m
  4. `7.8 × 10^(-2)\ `m
Show Answers Only

`A`

Show Worked Solution

`lambda=(b)/(T)=(2.898 xx10^(-3))/(310)=9.3 xx10^(-6)`

 `=>A`

PHYSICS, M7 2021 HSC 8 MC

Light from a point source is incident upon a circular metal disc, forming a shadow on a screen as shown. A bright spot is observed in the centre of the shadow.
 

The bright spot is caused by a combination of

  1. interference and refraction.
  2. refraction and polarisation.
  3. polarisation and diffraction.
  4. diffraction and interference.
Show Answers Only

`D`

Show Worked Solution
  • Light diffracts around metal disc and constructive interference causes the bright spot.

`=>D`

PHYSICS, M7 2021 HSC 5 MC

The spectrum of an object is shown.
 

Which row of the table correctly identifies the most likely source of the spectrum and the features labelled \(Y\)?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex} \ \rule[-0.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{1.5ex}\text{Source of spectrum}\rule[-0.5ex]{0pt}{0pt}& \textit{Features labelled \(Y\)} \\
\hline
\rule{0pt}{2.5ex}\text{Star}\rule[-1ex]{0pt}{0pt}&\text{Absorption lines}\\
\hline
\rule{0pt}{2.5ex}\text{Discharge tube}\rule[-1ex]{0pt}{0pt}& \text{Absorption lines}\\
\hline
\rule{0pt}{2.5ex}\text{Star}\rule[-1ex]{0pt}{0pt}& \text{Emission lines} \\
\hline
\rule{0pt}{2.5ex}\text{Discharge tube}\rule[-1ex]{0pt}{0pt}& \text{Emission lines} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • Overall shape resembles a blackbody curve so the source of the spectrum is a star. The wavelengths of lower intensity are absorption lines.

\(\Rightarrow A\)

PHYSICS, M7 2021 HSC 4 MC

An astronaut is travelling towards Earth in a spaceship at 0.8c. At regular intervals, a radio pulse is sent from the spaceship to an observer on Earth.

Which quantity would the astronaut and the observer measure to be the same?

  1. Length of the spaceship
  2. Speed of the radio pulses
  3. Momentum of the astronaut
  4. Time interval between the radio pulses
Show Answers Only

`B`

Show Worked Solution
  • Speed of radio waves (light) is equal in all frames of reference. All other quantities depend on the observer.

`=>B`

PHYSICS, M6 2021 HSC 21

A DC motor is constructed from a single loop of wire with dimensions 0.10 m × 0.07 m. The magnetic field strength is 0.40 T and a current of 14 A flows through the loop.
 

  1. Calculate the magnitude of the maximum torque produced by the motor.   (2 marks)
  1. Describe how the magnitude of the torque changes as the loop moves through half a rotation from the position shown.   (2 marks)
Show Answers Only

a.   0.04  Nm

b.    Magnitude of torque changes:

  • The torque is initially at a maximum and decreases to zero after a 90º rotation.
  • The torque then increases to a maximum at 180º from the coil’s original position.
Show Worked Solution
a.    `tau_max` `=nIAB sin theta`
    `=1xx14 xx0.1 xx0.07 xx0.40 xx sin(90^(@))`
    `=0.04`  Nm

 

b.    Magnitude of torque changes:

  • The torque is initially at a maximum and decreases to zero after a 90º rotation.
  • The torque then increases to a maximum at 180º from the coil’s original position.

CHEMISTRY, M7 2021 HSC 7 MC

Methanol undergoes a substitution reaction using hydrogen bromide.

Compared to methanol, the product of this reaction has a

  1. lower boiling point.
  2. lower molecular mass.
  3. greater solubility in water.
  4. different molecular geometry at the carbon atom.
Show Answers Only

`A`

Show Worked Solution
  • The product of the substitution reaction between methanol and hydrogen bromide is bromomethane.
  • Methanol contains an OH functional group and thus can form strong hydrogen bonds.
  • Bromomethane can only form dipole-dipole forces which are weaker than hydrogen bonds. As a result, bromomethane requires less energy to break these intermolecular forces, resulting in a lower boiling point than methanol.

`=> A`

CHEMISTRY, M7 2021 HSC 8 MC

Which diagram shows the expected arrangement of soap anions in an emulsion?
 

Show Answers Only

`D`

Show Worked Solution
  • Soap molecules contain a non-polar hydrophobic hydrocarbon tail and a polar hydrophilic head.
  • The non-polar tail forms dispersion forces with the oil molecule while the polar head forms ion-dipole forces with the polar water molecules.
  • The resulting orientation has the tail within the oil and the head group at the surface of the oil and water.

`=>D`

CHEMISTRY, M7 2021 HSC 3 MC

The structure of a compound is shown.
 


 

What is the preferred IUPAC name of this compound?

  1. `N`-methylpropanamide
  2. `N`-methylpropanamine
  3. `N`-propanylamine
  4. `N`-propylmethanamide
Show Answers Only

`A`

Show Worked Solution

Compound is a secondary amide.

  • Pre-fix (longest carbon chain) → -propan
  • Suffix (functional group) → -amide
  • Alkyl chain bound to the amide nitrogen is treated as a substituent and as it is bound to the nitrogen atom → `N`-methyl
  • Therefore, compound name is `N`-methylpropanamide

`=>A`

CHEMISTRY, M8 2021 HSC 2 MC

Which ion can be detected using a precipitation reaction with silver nitrate?

  1. \(\ce{Ag+}\)
  2. \(\ce{Cl-}\)
  3. \(\ce{Mg^2+}\)
  4. \(\ce{NO3-}\)
Show Answers Only

`B`

Show Worked Solution

Silver Nitrate →`\ text{Ag}^(+)\ text{and}\ text{NO}_(3)^(\ \ -)`

Solubility rules → all nitrates are soluble

  • `text{Ag}^(+)` is involved in the precipitation reaction
  • `text{Ag}^(+)` will form a precipitate with `text{Cl}^(-)`

`=>B`

BIOLOGY, M7 2021 HSC 8 MC

Howard Florey conducted a breakthrough experiment in the development and use of antibiotics. He infected eight mice with Streptococcus bacteria. Four mice were given penicillin and survived while the four untreated mice died.

What conclusion could be drawn from the data obtained?

  1. The experiment should be repeated with more mice.
  2. The use of penicillin causes antibiotic resistance in mice.
  3. Penicillin may be used on humans safely to treat bacterial infections.
  4. Penicillin may have played a role in the survival of the four treated mice.
Show Answers Only

`D`

Show Worked Solution
  • Conclusion must be directed to the research problem.
  • Option A is not a conclusion.
  • Options B and C are definitive beyond the supporting data.

`=>D`

Calculus, MET2 2020 VCAA 5

Let  `f: R to R, \ f(x)=x^{3}-x`.

Let  `g_{a}: R to R`  be the function representing the tangent to the graph of `f` at  `x=a`, where  `a in R`.

Let `(b, 0)` be the `x`-intercept of the graph of `g_{a}`.

  1. Show that  `b= {2a^{3}}/{3 a^{2}-1}`.   (3 marks)

  2. State the values of `a` for which `b` does not exist.    (1 mark)

  3. State the nature of the graph of `g_a` when `b` does not exist.   (1 mark)

  4. i.  State all values of `a` for which  `b=1.1`. Give your answer correct to four decimal places.   (1 mark)

  5. ii. The graph of `f` has an `x`-intercept at (1, 0).
  6.      State the values of  `a`  for which  `1 <= b <= 1.1`.
  7.      Give your answers correct to three decimal places.   (1 mark)

The coordinate `(b, 0)` is the horizontal axis intercept of `g_a`.

Let `g_b` be the function representing the tangent to the graph of `f` at  `x=b`, as shown in the graph below.
 
 
     
 

  1. Find the values of `a` for which the graphs of `g_a` and `g_b`, where `b` exists, are parallel and where  `b!=a`.   (3 marks)

Let  `p:R rarr R, \ p(x)=x^(3)+wx`, where  `w in R`.

  1. Show that  `p(-x)=-p(x)`  for all  `w in R`.   (1 mark)

A property of the graphs of `p` is that two distinct parallel tangents will always occur at `(t, p(t))` and `(-t,p(-t))` for all  `t!=0`.

  1. Find all values of `w` such that a tangent to the graph of `p` at `(t, p(t))`, for some  `t > 0`, will have an `x`-intercept at `(-t, 0)`.   (1 mark)

  2. Let  `T:R^(2)rarrR^(2),T([[x],[y]])=[[m,0],[0,n]][[x],[y]]+[[h],[k]]`, where  `m,n in R text(\{0})`  and  `h,k in R`.
     
    State any restrictions on the values of `m`, `n`, `h`, and `k`, given that the image of `p` under the transformation `T` always has the property that parallel tangents occur at  `x = -t`  and  `x = t`  for all  `t!=0`.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `a=+-sqrt3/3`
  3. `text(Horizontal line.)`
  4.  i. `a=-0.5052, 0.8084, 1.3468`
  5. ii. `a in (-0.505,-0.500]uu(0.808,1.347)`
  6. `a=+- sqrt5/5`
  7. `text(See Worked Solutions.)`
  8. `w=-5t^2`
  9. `h=0`
Show Worked Solution

a.   `f^{prime}(a) = 3a^2-1`

`g_a(x)\ \ text(has gradient)\ \ 3a^2-1\ \ text(and passes through)\ \ (a, a^3-a)`

`g_a(x)-(a^3-a)` `=(3a^2- 1)(x-a)`  
`g_a(x)` `=(3a^2-1)(x-a)+a^3-a`  

  
`x^{primeprime}-text(intercept occurs at)\ (b,0):`

`0=(3a^2-1)(b-a) + a^3-a`

`(3a^2-1)(b-a)` `=a-a^3`  
`3a^2b-3a^3-b+a` `=a-a^3`  
`b(3a^2-1)` `=a-a^3+3a^3-a`  
`:.b` `=(2a^3)/(3a^2-1)`  

 
b.   `b\ text{does not exist when:}`

♦ Mean mark part (b) 46%.

`(3a^2-1)=0`

`a=+-sqrt3/3`

♦♦ Mean mark part (c) 23%.
 

c.   `text{If}\ \ a=+-sqrt3/3,\ \ g_a^{prime}(x) = 0`

`=>\ text{the graph is a horizontal line (does not cross the}\ xtext{-axis).}`
 

d.i.  `text(Solve)\ {2a^{3}}/{3 a^{2}-1}=1.1\ text(for)\ a:`

`a=-0.5052\ text(or )\ =0.8084\ text(or)\ a=1.3468\ \ text{(to 4 d.p.)}`
  

d.ii.  `text(Solve)\ 1 <= (2a^(3))/(3a^(2)-1) < 1.1\ text(for)\ a:`

♦♦♦ Mean mark part (d)(ii) 13%.

`a in (-0.505,-0.500]uu(0.808,1.347)\ \ text{(to 3 d.p.)}`
 

e.   `f^{prime}(b) = 3b^2-1`

`g_b(x)\ \ text(has gradient)\ \ 3b^2-1\ \ text(and passes through)\ \ (b, b^3-b)`

`g_b(x)-(b^3-b)` `=(3b^2-1)(x-b)`  
`g_b(x)` `=(3b^2-1)(x-b)+b^3-b`  

 
`g_a(x)\ text{||}\ g_b(x)\ \ text{when}`

♦♦♦ Mean mark part (e) 13%.
`3a^2-1` `=3b^2-1`  
  `=3 cdot((2a^3)/(3a^2-1))-1`  

 
`=> a=+-1, +- sqrt5/5, 0`

`text(Test each solution so that)\ \ b!=a :`

`text(When)\ \ a=+-1, 0 \ => \ b=a`

`:. a=+- sqrt5/5`
 

f.    `p(-x)` `=(-x)^3-wx`
    `=-x^3-wx`
    `=-(x^3+wx)`
    `=-p(x)`

 
g.  `p^{prime}(t) = 3t^2+w`

♦♦♦ Mean mark part (g) 3%.

`p(t)\ \ text(has gradient)\ \ 3t^2+w\ \ text(and passes through)\ \ (t, t^3+wt)`

`p(t)-(t^3+wt)` `=(3t^2+w)(x-t)`  
`p(t)` `=(3t^2+w)(x-t) + t^3+wt`  

 
`text{If}\ p(t)\ text{passes through}\ \ (-t, 0):`

`0=(3t^2+w)(-2t) + t^3+wt`

`=>w=-5t^2\ \ (t>0)`
 

h.   `text{Property of parallel tangents is retained under transformation}`

♦♦♦ Mean mark part (h) 2%.

`text{if rotational symmetry remains (odd function).}`

`=>h=0`

`text(No further restrictions apply to)\ m, n\ \ text{or}\ \ k.`

Calculus, MET2 2020 VCAA 4

The graph of the function  `f(x)=2xe^((1-x^(2)))`, where  `0 <= x <= 3`, is shown below.
 

  1. Find the slope of the tangent to `f` at  `x=1`.   (1 mark)

  2. Find the obtuse angle that the tangent to `f` at  `x = 1`  makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree.   (1 mark)

  3. Find the slope of the tangent to `f` at a point  `x =p`. Give your answer in terms of  `p`.   (1 mark)

  4.  i. Find the value of `p` for which the tangent to `f` at  `x=1` and the tangent to `f` at  `x=p`  are perpendicular to each other. Give your answer correct to three decimal places.   (2 marks)

  5. ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at  `x=1`  and  `x=p`  intersect when they are perpendicular. Give your answer correct to two decimal places.   (3 marks)

Two line segments connect the points `(0, f(0))`  and  `(3, f(3))`  to a single point  `Q(n, f(n))`, where  `1 < n < 3`, as shown in the graph below.
 
         
 

  1.   i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
  2.      Find the equation of this line segment in terms of  `n`.   (1 mark)

  3.  ii. The second line segment connects the point `Q(n, f(n))` and the point  `(3, f(3))`, where  `1 < n < 3`.
  4.      Find the equation of this line segment in terms of `n`.   (1 mark)

  5. iii. Find the value of `n`, where  `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
  6.      Give your answer correct to three decimal places.   (3 marks)

Show Answers Only
  1. `-2`
  2. `117^@`
  3. `2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
  4.  i. `0.655`
  5. ii. `(0.80, 2.39)`
  6.   i. `y_1=2e^((1-n^2))x`
  7.  ii. `y_2=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`
  8. iii. `n= 1.088`
Show Worked Solution

a.   `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

 

b.   `text{Solve:}\ tan theta =-2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`
 

c.   `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`
 

d.i.   `text{If tangents are perpendicular:}`

`f^{′}(p) xx-2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`
 

d.ii.  `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

 
e.i.  `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.
 

e.ii.  `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2))-6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3)`  
`y_2` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`  
♦♦ Mean mark part (e)(iii) 28%.

 

e.iii.  `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

Statistics, MET2 2020 VCAA 3

A transport company has detailed records of all its deliveries. The number of minutes a delivery is made before or after its schedule delivery time can be modelled as a normally distributed random variable, `T`, with a mean of zero and a standard deviation of four minutes. A graph of the probability distribution of `T` is shown below.
 

  1. If  `"Pr"(T <= a)=0.6`, find `a` to the nearest minute.   (1 mark)

  2. Find the probability, correct to three decimal places, of a delivery being no later than three minutes after its scheduled delivery time, given that it arrives after its scheduled delivery time.   (2 marks)

  3. Using the model described above, the transport company can make 46.48% of its deliveries over the interval  `-3 <= t <= 2`.
  4. It has an improved delivery model with a mean of `k` and a standard deviation of four minutes.
  5. Find the values of `k`, correct to one decimal place, so that 46.48% of the transport company's deliveries can be made over the interval  `-4.5 <= t <= 0.5`   (3 marks)

A rival transport company claims that there is a 0.85 probability that each delivery it makes will arrive on time or earlier.

Assume that whether each delivery is on time or earlier is independent of other deliveries.

  1. Assuming that the rival company's claim is true, find the probability that on a day in which the rival company makes eight deliveries, fewer than half of them arrive on time or earlier. Give your answer correct to three decimal places.   (2 marks)

  2. Assuming that the rival company's claim is true, consider a day in which it makes `n` deliveries.
    1. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier.   (1 mark)

    2. Hence, or otherwise, find the minimum value of `n` such that there is at least a 0.95 probability that one or more deliveries will not arrive on time or earlier.   (1 mark)
  3. An analyst from a government department believes the rival transport company's claim is only true for deliveries made before 4 pm. For deliveries made after 4 pm, the analyst believes the probability of a delivery arriving on time or earlier is `x`, where  `0.3 <=x <= 0.7`
  4. After observing a large number of the rival transport company's deliveries, the analyst believes that the overall probability that a delivery arrives on time or earlier is actually 0.75
  5. Let the probability that a delivery is made after 4 pm be `y`.
  6. Assuming that the analyst's belief are true, find the minimum and maximum values of `y`.   (2 marks)

Show Answers Only

  1. `a= 1\ text(minute)`
  2. `0.547`
  3. `k=-1.5, -2.5`
  4. `0.003`
  5.  i. `1-0.85^n`
  6. ii. `19`
  7. `2/3`

Show Worked Solution

a.   `T\ ~\ N(0, 4^2)`

`text(Solve (by CAS): Pr)(T<=a) = 0.6`

`:. a= 1\ text(minute)`
 

b.    `text{Pr}(T <= 3∣T > 0)` `=(text{Pr}(0 < T <= 3))/(text{Pr}(T > 0))`
    `=(0.27337 dots)/(0.5)`
    `=0.547\ \ text{(to 3 d.p.)}`

 

c.   `text(Given)\ \ text{Pr}(-3 <= T <= 2) = 0.4648`

`sigma = 4 text{minutes}`

`=> \ text{Pr}(-4.5 <= T – 1.5 <= 0.5) = 0.4648`

`=> k=-1.5`

`text(By symmetry of the normal distribution)`

`text{Pr}(-2 <= T <= 3) = text{Pr}(-3 <= T <= 2) = 0.4648`

`=> \ text{Pr}(-4.5 <= T – 2.5 <= 0.5) = 0.4648`

`=> k=-2.5`

`:. k=-1.5, -2.5`

 

d.   `text{Let}\ \ X\ ~\ text{Bi}(8, 0.85)`

`text(Solve (by CAS):)`

`text{Pr}(X<=3) = 0.003\ \ text{(to 3 d.p.)}`
 

e.i.   `text{Pr(at least 1 delivery is late)}`

`= 1-\ text{Pr(all deliveries are on time)}`

`=1-0.85^n`
 

e.ii.   `text{Solve for}\ n:`

`1-0.85^n` `<0.95`  
`n` `>18.43…`  

 
`:.n_min=19`
 

f.   `text{Pr(delivery made after 4pm)} = y`

`=>\ text{Pr(delivery made before 4pm)} = 1-y`

`0.85(1-y)+xy` `=0.75`  
`y` `=-(0.1)/(x-0.85)`  
  `=(2)/(17-20 x)`  

 
`text(Given ) 0.3<=x<=0.7:`

`y_min = (2)/(17-20 xx 0.3) = 2/11`

`y_max = (2)/(17-20 xx 0.7) = 2/3`

GEOMETRY, FUR2 2020 VCAA 3

Khaleda manufacturers the face cream in Dhaka, Bangladesh.

Dhaka is located at latitude 24° N and longitude 90° E.

Assume that the radius of Earth is 6400 km.

  1. Write a calculation that shows that the radius of the small circle of Earth at latitude 24° N is 5847 km, rounded to the nearest kilometre.   (1 mark)

Khaleda receives an order from Abu Dhabi, United Arab Emirates (24° N, 54° E).

  1. Find the shortest small circle distance between Dhaka and Abu Dhabi.
  2. Round your answer to the nearest kilometre.   (1 mark)

Khaleda sends the order by plane from Dhaka (24° N, 90° E) to Abu Dhabi (24° N, 54° E).

The flight departs Dhaka at 1.00 pm and arrives in Abu Dhabi 11 hours later.

The time difference between Dhaka and Abu Dhabi is two hours.

  1. What time did the flight arrive in Abu Dhabi?   (1 mark)

A helicopter takes the order from the airport to the customer's hotel.

The hotel is 27 km south and 109 km east of the airport.

  1. Show that the bearing of the hotel from the airport is 104°, correct to the nearest degree.   (1 mark)
  2. After the delivery to the hotel, the helicopter returns to its hangar.
  3. The hangar is located due south of the airport.
  4. The helicopter flies directly from the hotel to the hangar on a bearing of 282° .
  5. How far south of the airport is the hangar?
  6. Round your answer to the nearest kilometre.   (1 mark)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `3674 \ text{km}`
  3. `10:00 \ text{pm}`
  4. `text{See Worked Solution}`
  5. `4 \ text{km}`
Show Worked Solution

a.   `text(Let)\ \ r= \ text(radius of small circle)`
 

`cos \ 24^@` `= r/6400`
`r` `=6400 xx cos 24^@`
  `= 5846.69 …`
  `=5847 \ text{(nearest km)}`

 

b.   `text{Small circle through Dhaka and Abu Dhabi has radius of 5847 km.}`

`text{Longitudinal difference}` `= 90 – 54`
  `= 36^@`

 

`:. \ text{Shortest distance}` `= 36/360 xx 2 xx pi xx 5847`
  `= 3673.77 …`
  `= 3674 \ text{km (nearest km)}`

 

c.   `text{Dhaka} \ (24^@ text{N}, 90^@ \ text{E}) \ text{is further east than Abu Dhabi} \  (24^@ text{N}, 54^@ \ text{E})`

`=> \ text{Dhaka is 2 hours ahead.}`

`:. \ text{Flight arrival time (Abu Dhabi time)}`

`= 1:00 \ text{pm} + 11 \ text{hours} – 2 \ text{hours}`

`= 10:00 \ text{pm}`

 

d.

`tan theta` `= 109/27`
`theta` `=tan^(-1) (109/27) = 76.1^@`

  
`:.\ text{Bearing of hotel from airport}`

`=180 – 76.1`

`=104^@\ \ text{(nearest degree)}`

 

e.

`text{Let X = position of hangar}`

`text{Find OX:}`

`tan 12^@` `= text{OX}/109`
`text{OX}` `= 109 xx tan 12^@`
  `= 23.17 \ text{km}`

 

`:. \ text{AX (distance hangar is south of airport)}`

`= 27 – 23.17`

`= 4 \ text{km (nearest km)}`

GEOMETRY, FUR2 2020 VCAA 2

Khaleda has designed a logo for her business.

The logo contains two identical equilateral triangles,

The side length of each triangle is 4.8 cm, shown in the diagram below.
 

  1. Write a calculation to show that the area of one of the triangles, rounded to the nearest centimetre, is 10 cm2.   (1 mark)

In the logo, the two triangles overlap, as shown below. Part of the logo is shaded and part of the logo is not shaded.
 

  1. What is the area of the entire logo?
  2. Round your answer to the nearest square centimetre.   (1 mark)
  3. What is the ratio of the area of the shaded region to the area of the non-shaded region of the logo?   (1 mark)
  4. The logo is enlarged and printed on the boxes for shipping.
  5. The enlarged logo and the original logo are similar shape.
  6. The area of the enlarged logo is four times the area of the original logo.
  7. What is the height, in centimetres, of the enlarged logo?   (1 mark)

Show Answers Only
  1. `10 \ text{cm}^2 \ (text{nearest cm}^2)`
  2. `15 \ text{cm}^2`
  3. `1:2`
  4. `9.6 \ text{cm}`
Show Worked Solution

a.    `text{Triangle is equilateral (all angles = 60}^@)`

`text{Using the sine rule:}`

`A` `= 1/2 a b sin c`
  `= 1/2 xx 4.8 xx 4.8 xx sin 60^@`
  `= 9.976 …`
  `= 10 text{cm}^2 (text{nearest cm}^2)`

♦♦ Mean mark part (b) 32%.

 

b.   `text{L} text{ogo is made up of 2 identical triangles.}`

`text{Divide each triangle into 4 equal smaller triangles.}`

`text{Total shading = 2 small triangles}\ = 1/2 xx \ text{area of 1 triangle}`

`:. \ text{Area of logo}`

`= 2 xx 10 – 1/2 xx 10`

`= 15 \ text{cm}^2`

♦ Mean mark part (c) 48%.

 

c.    `text{Shaded region}` `: \ text{non-shaded}`
  `text{2 triangles}` `: 4 \ text{triangles}`
  `1` `: 2`

 

d.   `text{Area scale factor = 4 (given)}`

♦♦♦ Mean mark part (d) 17%.

`text{Length scale factor} = sqrt4 = 2`
 

`:. \ text{Height of enlarged logo}`

`= 2 xx \ text{height of original logo}`

`= 2 xx 4.8`

`= 9.6 \ text{cm}`

Calculus, MET2 2020 VCAA 2

An area of parkland has a river running through it, as shown below. The river is shown shaded.

The north bank of the river is modelled by the function  `f_(1):[0,200]rarr R, \ f_(1)(x)=20 cos((pi x)/(100))+40`.

The south bank of the river is modelled by the function  `f_(2):[0,200]rarr R, \ f_(2)(x)=20 cos((pi x)/(100))+30`.

The horizontal axis points east and the vertical axis points north.

All distances are measured in metres.
 

A swimmer always starts at point `P`, which has coordinates  (50, 30).

Assume that no movement of water in the river affects the motion or path of the swimmer, which is always a straight line.

  1. The swimmer swims north from point `P`.
  2. Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river.   (1 mark)

  3. The swimmer swims east from point `P`.
  4. Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river.   (2 marks)

  5. On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
  6. Find this minimum distance. Give your answer in metres, correct to one decimal place.   (2 marks)

  7. Calculate the surface area of the section of the river shown on the graph in square metres.   (1 mark)

  8. A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
  9. Find the area of the "no swimming" zone, correct to the nearest square metre.   (3 marks)

  10. Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule  `y=kf_(1)(x)`, where `k >= 1`.
  11. Find the values of  `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m.   (2 marks)

Show Answers Only
  1. `10\ text{m}`
  2. `16 2/3\ text{m}`
  3. `8.5\ text{m}`
  4. `2000\ text{m}^2`
  5. `837\ text{m²}`
  6. `k in [1, 7/6)`
Show Worked Solution

a.   `text{Since swimmer swims due north,}`

`text{Distance}\ = 40-30=10\ text{m}`

 

b.  `text{Solve} \ f_(1)(x)=30 \ text{or} \ x in[50,100]`

`=> x=200/3`

`:.\ text{Distance to swim (east) to reach north bank}`

`=200/3-50`

`=16 2/3\ text{m}`
 

c.   `text{Let swimmer arrive at north bank at the point}\ \ (x,f_(1)(x))`

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

`d(x)=sqrt((x-50)^(2)+(f_(1)(x)-30)^(2))`

`text{Solve} \ d/dx(d(x))=0 \ text{for} \ x:`

`x=54.47…`

`:. d_min=8.5\ text{m (to 1 d.p.)}`

 

d.   `text{Shaded Area}`

`=int_(0)^(200)(f_(1)(x)-f_(2)(x))\ dx`

`=2000\ text{m}^2`

 

e.   `text{Find}\ \ f_(1)(x) = 30 \ text{for} x in [50,150]:`

♦♦ Mean mark part (e) 35%.

`=>x=200/3, 400/3`

`text{Find}\ \ f_(2)(x) = 30 \ text{for} \ x in [50,150]:`

`=>x=50, 150`

`:.\ text{Area}` `=int_(50)^(150)(30-f_(2)(x))\ dx-int_((200)/(3))^((400)/(3))(30-f_(1)(x))\ dx`  
  `=837\ text{m² (to nearest m²)}`  

♦♦♦Mean mark part (f) 15%.

 

f.   `text{Let}\ \ D(x)=\ text{vertical distance between banks}`

`D(x)` `=kf_(1)(x)-f_(2)(x)`  
  `=20k cos((pi x)/(100))+40k-(20 cos((pi x)/(100))+30)`   
  `=(20k-20)cos((pi x)/(100)) +40k-30`  

  
`text{Given}\ \ D(x)<20 \ text{for}\ x in[0,200]`

`text{Maximum} \ cos((pi x)/(100)) = 1\  text{when}\ \ x=0, 200`

`text{Solve} \ 20k-20+40k-30<20\ \ text{for}\ k:`

`=> k<7/6`

`:. k in [1,7/6)`

GEOMETRY, FUR2 2020 VCAA 1

Khaleda manufactures a face cream. The cream comes in a cylindrical container.

The area of the circular base is 43 cm2. The container has a height of 7 cm, as shown in the diagram below.
 

  1. What is the volume of the container, in cubic centimetres?   (1 mark)
  2. Write a calculation that shows that the radius of the cylindrical container, rounded to one decimal place, is 3.7 cm.   (1 mark)
  3. What is the total surface area of the container, in square centimetres, including the base and the lid?
  4. Round your answer to the nearest square centimetre.   (1 mark)

The diagram below shows the dimensions of a shelf that will display the containers.

 
               

  1. What is the perimeter of the shelf, in centimetres?   (1 mark)
  2. The shelf will display the containers in a single layer. Each container will stand upright on the shelf.
  3. What is the maximum number of containers that can fit on the shelf?   (1 mark)
  4. The shelf sits against a wall at a 90° angle.
  5. The shelf is supported by a 26 cm bracket that forms a 30° angle with the wall, as shown in the diagram below.

 
                 
 

  1. Find the value of  `h`, the distance between the edge of the shelf and the bracket, in centimetres.   (1 mark)

Show Answers Only
  1. `301 \ text{cm}^3`
  2. `3.7 \ text{cm (to d.p)}`
  3. `249 \ text{cm}^2`
  4. `296 \ text{cm}^2`
  5. `75`
  6. `24 \ text{cm}`
Show Worked Solution
a.    `V` `= text{Area of base} xx text{height}`
    `= 43 xx 7`
    `= 301 \ text{cm}^3`

 

b.   `text{Area of base} = 43 \ text{cm}^2`

`pi r^2` `= 43`
`r^2` `= 43/pi`
`r` `= sqrt{43/pi} = 3.699 … = 3.7 \ text{cm (to 1 d.p.)}`

 

c.    `text{S.A.}` `= 2 xx text{base + 2 pi r h}`
    `= 2 xx 43 + 2 xx pi xx 3.7 xx 7`
    `= 248.73 …`
    `= 249 \ text{cm}^2 \ (text{nearest cm}^2)`

 

d.    `text{Perimeter}` `= 2 xx 74 + 4 xx 37`
    `=296 \ text{cm}`

 

e.   `text{Diameter of container} = 2 xx 3.7 = 7.4 \ text{cm}`

♦♦ Mean mark part (e) 27%.

`text{Divide area into 2 sections}`
 


 

`text{S} text{ection 1 dimensions (in containers)}`

`text{Width} = 37/7.4 = 5 \ text{containers}`

`text{Height} = 74/7.4 = 10 \ text{containers}`
 

`text{S} text{ection 2 dimensions}`

`text{Width} = 5`

`text{Height} = 5`
 

`:. \ text{Maximum containers}`

`= 5 xx 10 + 5 xx 5`

`= 75`

 

f.    `sin 30^@` `= x/26`
  `x` `= 13 \ text{cm}`

 

`:. h` `= 37 – x`
  `= 37 – 13`
  `= 24 \ text{cm}`

GRAPHS, FUR2 2020 VCAA 4

Another section of Kyla's business services and details cars and trucks.

Every vehicle is both serviced and detailed.

Each car takes two hours to service and one hour to detail.

Each truck takes three hours to service and three hours to detail.

Let `x` represent the number of cars that are serviced and detailed each day.

Let `y` represent the number of trucks that are serviced and detailed each day.

Past records suggest there are constraints on the servicing and detailing of vehicles each day.

These constraints are represented by Inequalities 1 to 4 below.

`text{Inequality 1}`   `x >= 16`    
`text{Inequality 2}`   `y >= 10`    
`text{Inequality 3}`   `2x + 3y <= 96`   `text{(servicing department)}`
`text{Inequality 4}`   `x + 3y <= 72`   `text{(detailing department)}`
     
  1. Explain the meaning of Inequality 1 in the context of this problem.   (1 mark)
  2. Each employee at the business works eight hours per day.
  3. What is the maximum number of employees who can work in the servicing department each day?   (1 mark)

The graph below shows the feasible region (shaded) that satisfies Inequalities 1 to 4 .
 


 

  1. On a day when 20 cars are serviced and detailed, what is the maximum number of trucks that can be serviced and detailed?   (1 mark)
  2. When servicing and detailing, the business makes a profit of $150 per car and $225 per truck.
  3. List the points within the feasible region that will result in a maximum profit for the day.   (2 marks)

Show Answers Only
  1. `text{At least 16 cars are serviced and detailed each day.}`
  2. `12`
  3. `17 \ text{trucks}`
  4. `(24, 16),(27, 14),(30, 12), \ text{and} (33, 10)`
Show Worked Solution

a.   `text{At least 16 cars are serviced and detailed each day.}`

♦♦ Mean mark part (b) 34%.

 

b.   `text{Consider Inequality 3:} \ 2x + 3y ≤ 96`

`text{Total time} \ ≤ 96 \ text{hours and each employee works 8 hours.}`

`:. \ text{Maximum employees in servicing}`

`=96/8`

`=12`
 

c.    `text{Find} \ y_text{max} \ text{when} \ x= 20`

♦ Mean mark part (c) 42%.

`text{With reference to the feasible region,}`

`y_text{max} \ text{lies on the line of Inequality 4}`

`x + 3y` `= 72`
`20 + 3y` `= 72`
`3y`  `= 52`
`y` `= 17.33`

 
`:. y_text{max} = 17 \ text{trucks (highest integer within the feasible region)}`

 

d.    `text{Profits: $150 per car, $225 per truck}`

♦♦♦ Mean mark part (d) 19%.

`P = 150x + 225y`

`=> \ m_P = – 150/225 = – 2/3`

`text{The objective function} \ (P) \ text{is parallel to}\ \ 2x+3y=96\ \ text{(Inequality 3)}`
 


 

`=> text{Maximum profit occurs at integer co-ordinates on line} \ AB.`

`A\ text(occurs at intersection of:)`

`2x+3y=96 and x+3y=72\ \ =>\ \ x=24`

`B\ text(occurs at intersection of:)`

`2x+3y=96 and y=10\ \ =>\ \ x=33`

`text(Max profit requires)\ \ x in [24,33]`

`text{Test each integer}\ xtext{-value for an integer}\ ytext{-value:}`

`:.\ text{Maximum profit occurs at}`

`text{(24, 16), (27, 14), (30, 12) and (33, 10)}`

GRAPHS, FUR2 2020 VCAA 3

Kyla's business also manufactures car seat covers.

The monthly revenue, `R`, in dollars, from selling `n` seat covers is given by

`R = 80 n`

This relationship is shown on the graph below.
 

The monthly cost, `C`, in dollars, of making `n` seat covers is given by

`C = 36n + 5000`

  1. On the graph above, sketch the monthly cost, `C`, of making `n` seat covers.   (1 mark) 
  2. Find the least number of seat covers that need to be sold in order to make a profit.   (1 mark)

Last month, 180 seat covers were sold and the profit was $2920.

Kyla believes that if she lowers the selling price of the seat covers she will sell 60 more seat covers this month.

  1. If Kyla maintains the same profit of $2920 this month, find the new selling price of the seat covers.   (1 mark)
  2. Kyla finds that the monthly cost, `C`, of making `n` seat covers changes when more than 300 seat covers are made.
  3. The monthly cost, `C`, of making `n` seat covers is now given by
     
              `C={[36 n+5000,0 <= n <= 300],[mn+p,n > 300]:}`
     
    This is represented on the graph below.

       
     
    If 350 seat covers cost $17 100 to make, find the value of `m`.   (1 mark)

Show Answers Only

  1.  
  2. `text{114 seat covers}`
  3. `$69`
  4. `26`
Show Worked Solution

a.   `”Draw graph through endpoints (0, 5000) and (250, 14 000).”`
 

 

b.    `C=36 n+5000`

`R=80 n`

`text{Profit occurs when} \ \ R > C`

`text{Solve for} \ n:`

`80 n` `> 36 n+5000`  
`44n` `>5000`  
`n` `> (5000)/(44)`  
`n` `>113.6`  

 
`:.\ text(114 seat covers is the least to make a profit.)`

♦♦ Mean mark part (c) 31%.

 

c.    `n` `= 240\ \ (text{60 more than last month})`
  `C` `= 36 xx 240 + 5000`
    `= $13\ 640`

 
`text(Let)\ \ x=\ text(selling price where profit is $2920)`

`2920` `= 240x – 13\ 640`
`240x` `= 16\ 560`
`:. x` `= (16\ 560)/240`
  `= $69`
♦♦ Mean mark part (d) 28%.

 

d.   `m \ text{is the gradient of}\ \ C = mn + p \ \ text{when}\ \ p>300.`

`C = mn + p \ \ text{passes through} \ (300, 15\ 800) \ text{and} \ (350, 17\ 100)`

`m` `= {17\ 100 – 15\ 800}/{350 – 300}`
  `= 26`