SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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%.

Measurement, STD2 M2 2022 HSC 29

Sydney is 10 hours ahead of Coordinated Universal Time (UTC +10) and New York is 5 hours behind Coordinated Universal Time (UTC –5).

Tony travels from Sydney to New York. His plane leaves Sydney at 8:20 pm on Wednesday local time and flies non-stop to New York.

The flight takes 20 hours and 24 minutes.

What time and day is it in New York when the plane lands?  (3 marks)

Show Answers Only

`text{1:44 am (Thu)}`

Show Worked Solution

`text{Sydney is 15 hours ahead of New York.}`

`text{Plane’s arrival (Sydney time)}`

`=\ text{8:20 pm + 20h 24 min}`

`=\ text{4:44 pm (Thu)}`
 

`text{Plane’s arrival (New York time)}`

`=\ text{4:44 pm (Thu) – 15 hours}`

`=\ text{1:44 am (Thu)}`

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%.

Probability, STD2 S2 2022 HSC 17

The numbers 1, 2, 3, 4, 5 and 6 are each written on separate cards.

Amy has cards 1, 3 and 5 and Bob has cards 2, 4 and 6.

They play a game in which each person randomly chooses one of their own cards and compares it with the other person's card. The person with the higher card wins.

  1. A partially completed tree diagram is shown.
     
           
     
    Complete the tree diagram and find the probability that Bob wins.  (2 marks)
     
  2. Suppose Amy and Bob play this game 30 times.
  3. How many times would Bob be expected to win?  (1 mark)

Show Answers Only
  1. `6/9`
  2. `20`
Show Worked Solution

a.   
       

`P(text{Bob wins}) = 6/9=2/3`
 

b.    `text{Expected wins}` `=2/3 xx 30`
    `=20`

Algebra, STD2 A4 2022 HSC 22

The formula  `C=100 n+b`  is used to calculate the cost of producing laptops, where `C` is the cost in dollars, `n` is the number of laptops produced and `b` is the fixed cost in dollars.

  1. Find the cost when 1943 laptops are produced and the fixed cost is $20 180.  (1 mark)

  2. Some laptops have some extra features added. The formula to calculate the production cost for these is
  3.      `C=100 n+a n+20\ 180`
  4. where `a` is the additional cost in dollars per laptop produced.
  5. Find the number of laptops produced if the additional cost is $26 per laptop and the total production cost is $97 040.  (2 marks)

Show Answers Only
  1. `$214\ 480`
  2. `610\ text{laptops}`
Show Worked Solution

a.   `text{Find}\ \ C\ \ text{given}\ \ n=1943 and b=20\ 180`

`C` `=100 xx 1943 + 20\ 180`  
  `=$214\ 480`  

 

b.   `text{Find}\ \ n\ \ text{given}\ \ C=97\ 040 and a=26`

`C` `=100 n+a n+20\ 180`  
`97\ 040` `=100n + 26n +20\ 180`  
`126n` `=76\ 860`  
`n` `=(76\ 860)/126`  
  `=610 \ text{laptops}`  

Statistics, STD2 S1 2022 HSC 19

The table shows the types of customer complaints received by an online business in a month.
 

  1. What are the values of `A` and `B`?  (2 marks)

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`


♦ Mean mark part (b) 41%.

Statistics, STD2 S5 2022 HSC 18

The marks in a test were normally distributed. The mean mark was 60 and the standard deviation was 15 .

What was the percentage of marks higher than 90?  (2 marks)

Show Answers Only

`2.5text{%}`

Show Worked Solution
`ztext{-score}` `=(x-mu)/sigma`  
  `=(90-60)/15`  
  `=2`  

 

`text{% marks > 90}\ = 2.5text{%}`


Mean mark 52%.
COMMENT: A poor State result that warrants attention.

Statistics, STD2 S4 2022 HSC 12 MC

For a particular course, the recorded data show a relationship between the number of hours of study per week and the marks achieved out of 100 .

A least-squares regression line is fitted to this dataset. The equation of this line is given by

`M=20+3 H,`

where `M` is the predicted mark and `H` is the number of hours of study per week.

Based on this regression equation, which of the following is correct regarding the predicted mark of a student?

  1. It will be 3 for zero hours of study per week.
  2. It will be 20 for zero hours of study per week.
  3. It will increase by 20 for every additional hour of study per week.
  4. It will increase by 1 for every 3 additional hours of study per week.
Show Answers Only

`B`

Show Worked Solution

`text{Consider Option}\ B:`

`text{If zero hours of study are done per week}\ \ → \ \ H=0`

`:. M=20+(3 xx 0) = 20`

`=>B`

Financial Maths, STD2 F4 2022 HSC 11 MC

In ten years, the future value of an investment will be $150 000. The interest rate is 4% per annum, compounded half-yearly.

Which equation will give the present value `(PV)` of the investment?

  1. `PV=(150\ 000)/((1+0.04)^(10))`
  2. `PV=(150\ 000)/((1+0.04)^(20))`
  3. `PV=(150\ 000)/((1+0.02)^(10))`
  4. `PV=(150\ 000)/((1+0.02)^(20))`
Show Answers Only

`D`

Show Worked Solution

`text{Compounding periods}\ = 10 xx 2 = 20`

`text{Compounding rate}\ = (4text{%})/2 = 2text{%} = 0.02`

`PV=(150\ 000)/((1+0.02)^(20))`

`=>D`

PHYSICS, M7 2019 HSC 23

A student investigated the photoelectric effect. The frequency of light incident on a metal surface was varied and the corresponding maximum kinetic energy of the photoelectrons was measured.

The following results were obtained.
 

\begin{array}{|l|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Frequency}\ \left(\times 10^{14} Hz\right) \rule[-1ex]{0pt}{0pt}& 11.2 & 13.5 & 15.2 & 18.6 & 20.0 \\
\hline \rule{0pt}{2.5ex}\textit{Maximum kinetic energy}\ (eV) \rule[-1ex]{0pt}{0pt}& 0.6 & 1.3 & 2.3 & 3.3 & 4.2 \\
\hline
\end{array}

Plot the results on the axes below and hence determine the work function of the metal in electron volts.   (3 marks)
 

Show Answers Only

Work function = 4 eV

Show Worked Solution

The metal has a work function of 4 eV (this is the negative of the \(y\)-intercept).

PHYSICS, M7 2019 HSC 22

Spectra can be used to determine the chemical composition and surface temperature of stars.

Describe how spectra provide information about OTHER features of stars.   (3 marks)

Show Answers Only

→ Spectra provide information about the linear velocity of a star relative to Earth. A shift towards higher wavelengths indicates that a star is moving away from Earth. A shift towards lower wavelengths indicates a star is moving towards earth.

→ Darkened, broader spectral lines indicate that a star has a high density due to high pressures in the star’s atmosphere. Narrower spectral lines indicate a lower density star.

→ Blurring and broadening of the edges of spectral lines indicate a star has a high rotational velocity due to simultaneous red and blue doppler shifting. More defined, narrower spectral lines indicate a star with lower rotational velocity.

Show Worked Solution

→ Spectra provide information about the linear velocity of a star relative to Earth. A shift towards higher wavelengths indicates that a star is moving away from Earth. A shift towards lower wavelengths indicates a star is moving towards earth.

→ Darkened, broader spectral lines indicate that a star has a high density due to high pressures in the star’s atmosphere. Narrower spectral lines indicate a lower density star.

→ Blurring and broadening of the edges of spectral lines indicate a star has a high rotational velocity due to simultaneous red and blue doppler shifting. More defined, narrower spectral lines indicate a star with lower rotational velocity.

PHYSICS, M8 2019 HSC 21

Outline de Broglie's contribution to quantum mechanics. Support your answer with a relevant equation.   (2 marks)

Show Answers Only

de Broglie proposed the dual wave-particle model of matter which postulated that particles, such as electrons possessed wave properties.

The equation to predict the wavelength of such particles is:

`lambda=(h)/(mv)`

Show Worked Solution

de Broglie proposed the dual wave-particle model of matter which postulated that particles, such as electrons possessed wave properties.

The equation to predict the wavelength of such particles is:

`lambda=(h)/(mv)`

PHYSICS, M7 2019 HSC 15 MC

Monochromatic light passes through two slits 1 \(\mu \text{m}\) apart. The resulting diffraction pattern is measured at a distance of 0.3 m.
 


 

This diffraction pattern can be analysed using the equation  \(d \sin \theta=\lambda\).

What values of \(d\) and \(theta\) should be used in the equation?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-2.5ex]{0pt}{0pt}& \\
\rule{0pt}{3ex}\textbf{A.}\rule[-2.5ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{B.}\rule[-3ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{C.}\rule[-3ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{D.}\rule[-3ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad d \quad \rule[-1ex]{0pt}{0pt}& \theta \\
\hline
\rule{0pt}{2.5ex}0.3 \ \text{m}\rule[-1ex]{0pt}{0pt}&\tan ^{-1}\left(\dfrac{0.08}{0.3}\right)\\
\hline
\rule{0pt}{2.5ex}0.3 \ \text{m}\rule[-1ex]{0pt}{0pt}& \sin ^{-1}\left(\dfrac{0.08}{0.3}\right) \\
\hline
\rule{0pt}{2.5ex}1 \ \mu \text{m}\rule[-1ex]{0pt}{0pt}& \tan ^{-1}\left(\dfrac{0.08}{0.3}\right)\\
\hline
\rule{0pt}{2.5ex}1 \ \mu \text{m} \rule[-1ex]{0pt}{0pt}&\sin ^{-1}\left(\dfrac{0.08}{0.3}\right) \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

\(d\) is the distance between the slits, in this case, \(d=1 \mu \text{m}\)

\(theta\)  is the angular deviation of a maximum, measured from the centre of the slits 

\(\theta=\tan ^{-1}\left(\frac{0.08}{0.3}\right)\)

\(\Rightarrow C\)

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: Factors involving friction between the mass and the table and 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.`

→ So, 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 produced by the motor:

→ The torque produced by the motor is given by  `tau=NIAB sin theta`

→ The torque of the motor and hence maximum speed at which the mass can be pulled is limited by the number of turns `N` of the coil in the motor, the current passing through the coils `I`, the area of the coils `A` and the magnetic field strength of the stator magnets `B`.

→  Via the pulley, the torque generated by the motor generates a force which pulls the mass. As `Tau =rF sin theta`, a smaller pulley radius means that a larger force can be used to pull the mass.

→ So, the speed at which the mass can be pulled is limited by pulley radius.

 
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 m`text{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` m`text{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` m`text{s}^(-1)`

`v_(y)` `=u_(y)+a_(g)t`  
  `t` `=(27.19-(-19.66))/(9.8)`  
  `=4.8` seconds  

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)` Wb

b.   `epsi=2.1 xx10^(-2)  V`

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

c.   Explanation:

→ 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)` V  (V`>`0)

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


♦ Mean mark (b) 51%.

c.   Explanation:

→ 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  m `text{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` m `text {s}^(-1)`  

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

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

   
Using Pythagoras:

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

  
Find launch angle (`theta)`:

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

  
So,  `u=37` m `text{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` N

Since  90 000 N > 76 440 → net upwards force → net upwards acceleration.

Net upwards acceleration with constant horizontal velocity results in an upwards parabolic trajectory.

`=>D`

PHYSICS, M5 2020 HSC 12 MC

In which of the following would the satellite have the greatest escape velocity from Earth?
 

 

Show Answers Only

`D`

Show Worked Solution

`v_(esc)=sqrt((2GM)/(r))`

→ escape velocity is independent of the satellite’s mass

`v_(esc) prop (1)/(sqrt(r))`

→ the smaller the radius, the higher the required escape velocity

`=>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  \(5.0 \times 10^{-5}\) 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:

            \(d \sin \theta\) \(=m \lambda\)  
\(5 \times 10^{-5} \sin \theta\) \(=1 \times 400 \times 10^{-9}\)  
\(\theta\) \(=0.46^{\circ}\)  
\(s\) \(3 \times \tan (0.46)=0.024 \ \text{m}\)  

 
→ 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:

            \(d \sin \theta\) \(=m \lambda\)  
\(5 \times 10^{-5} \sin \theta\) \(=1 \times 400 \times 10^{-9}\)  
\(\theta\) \(=0.46^{\circ}\)  
\(s\) \(3 \times \tan (0.46)=0.024 \ \text{m}\)  

 
→ 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
  1.   12 ly
  2.   According to special relativity, as ` v `→ `c`:
  3.   → the momentum of the spaceship approaches infinity
  4.   → the force required to accelerate the spaceship approaches infinity
  5.   → 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

  1.  
       
     
  2. The gradient of the line of best fit should be determined using the graph  \(V=\dfrac{h f}{q_e}\).
  3. The gradient is equal to  \(\dfrac{V}{f}\),  and \(h\) can be calculated as \(h=\) gradient  \(\times q_e\).
  4. This method increases the accuracy by taking more data points into account.

Show Worked Solution

  1.  
       
     
  2. The gradient of the line of best fit should be determined using the graph \(V=\dfrac{h f}{q_e}\).
    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 V

Show Worked Solution
`Phi` `=BA`  
  `=0.15 xx0.02`  
  `=0.003` Wb  

 

`epsi` `=-N(Delta Phi)/(Delta t)`  
  `=-35((0-0.003))/(0.4)`  
  `=0.3` 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.

\(\Rightarrow A\)