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Aussie Maths & Science Teachers: Save your time with SmarterEd

ENGINEERING, PPT 2021 HSC 22d

A skateboard rider is at rest at the top of a slope, indicated by point `A` in the diagram.
 

The combined mass of the rider and skateboard is 80 kg.

Calculate the speed of the rider, in metres per second, at point `B` if the average work done by friction against the skateboard is 55 N.    (4 marks)

Show Answers Only

`text{Speed} = 14.57\ text{m/s}`

Show Worked Solution

`m= 80\ text{kg}, \ f= 55\ text{N}`

`PE` `= mgh`  
  `= 80xx10xx27.8`  
  `= 22\ 240\ text{J}`  

  
`text{Losses to friction (W):}`

`W` `= fs`  
  `= 55xx250`  
  `= 13\ 750\ text{J}`  

 

`text{Energy used}` `= PE−text{Frictional losses}`  
  `= 22\ 240−13\ 750`  
  `= 8490\ text{J}`  

 
`text{Calculate speed at point B:}`

`1/2×m×v^2` `= 8490`  
`0.5×80×v^2` `= 8490`  
`v^2` `= 212.25`  
`:.v` `= 14.57\ text{m/s}`  

♦ Mean mark 43%.

ENGINEERING, PPT 2021 HSC 22c

An electric skateboard battery is rated at 0.078 watts at an operating voltage of 48 volts.

Calculate the capacity of the battery in amp hours, if it fully discharges in one hour of use. The power (`P`) is given by the formula  `P=V I`.    (3 marks)

Show Answers Only

`text{Capacity} = 5.85\ text{Ah}`

Show Worked Solution

`text{Power = 0.078 W,  Volts = 48 V}`

`P` `=VI`
`I` `=P÷V`
  `=0.078÷48`

 

`:.\ text{Capacity}` `=(0.078×3600)÷48`
  `=5.85\ text{Ah}`

♦ Mean mark 50%.

ENGINEERING, CS 2021 HSC 20 MC

The galvanic table below shows the relative corrosion potential of a metal when it is in contact with another metal in an aqueous environment.
 

In which of the following situations is the steel component likely to corrode in the shortest amount of time?
 

 

Show Answers Only

`D`

Show Worked Solution

Two key factors affecting the rate of corrosion:

→ First is the difference in reactivity. Since the difference between steel and copper is greater than the difference between steel and aluminium, the answer must be `C` or `D`.

→ Secondly, the reaction is faster if the anode is smaller than the cathode. Therefore the steel must be the rivet, not the sheet.

`=>D`

♦ Mean mark 38%.

ENGINEERING, PPT 2021 HSC 18 MC

The diagram shows a simplified circuit of an electric scooter.
 

What would be the reading of the ammeter in this circuit?

  1. 3.0 A
  2. 6.0 A
  3. 12.0 A
  4. 26.9 A
Show Answers Only

`C`

Show Worked Solution

Find total resistance (`R_t`):

`1/R_t` `=1/R_1+1/R_2`  
`1/R_t` `= 1/(1+2)+1/3 `  
`1/R_t`  `=2/3`  
`R_t` `=3/2 = 1.5\ Ω`  

 
Find current (`I`):

`I` `=V/R_t`  
  `=18/1.5`  
  `=12\ text{A}`  

  
`=>C`

Mean mark 52%.

ENGINEERING, TE 2021 HSC 16 MC

The circuit diagram shown includes two transistors in its configuration.
 

What type of logic gate has the same function as this pair of transistors in the circuit?

  1. OR gate
  2. NOR gate
  3. AND gate
  4. NAND gate
Show Answers Only

`A`

Show Worked Solution

→ If current/signal passes through A, B or both A & B, then the circuit is complete.

→ This is the same requirements as an OR logic gate.

`=>A`

♦ Mean mark 41%.

ENGINEERING, AE 2021 HSC 13 MC

Two identical aircraft, `Q` and `R`, are travelling at 500 km per hour. Aircraft `Q` is at an altitude of 1000 m and aircraft `R` is at an altitude of 10 000 m.

Compared to aircraft `Q`, aircraft `R` generates

  1. less lift and less drag.
  2. less lift and more drag.
  3. more lift and less drag.
  4. more lift and more drag.
Show Answers Only

`A`

Show Worked Solution

→ Flying at higher altitude produces less lift as there is lower air pressure under the wing.

→ The low air pressure at high altitude also decreases drag.

`=>A`

♦ Mean mark 44%.

ENGINEERING, CS 2021 HSC 12 MC

The stress–strain diagram shown is being used to calculate a material property.
 

What material property can be calculated by using the dotted lines shown in this diagram?

  1. Hardness
  2. Toughness
  3. Proof stress
  4. Ultimate tensile stress
Show Answers Only

`C`

Show Worked Solution

→ Proof stress is obtained by drawing a parallel (dotted) line to the start of the graph from a starting point at 0.2% strain and measuring where it subsequently cuts the graph.

`=>C`


♦♦ Mean mark 48%.

BIOLOGY, M7 2020 HSC 15 MC

Four antiviral drugs have been tested in a culture of human cells for their effectiveness in inhibiting infection from a new virus. The toxicity of the antivirals to human cells was also tested.

The ability of the drugs to inhibit viral entry to the cells (% inhibition) and the proportion of the cells killed by the drugs (% toxicity) were recorded at different doses of each drug and shown in the graphs.

From the results shown, which antiviral drug is the safest and most effective at a dose of  1 `mu`mol L¯1?
 


 

Show Answers Only

`C`

Show Worked Solution

→ At 1 `mu`mol L¯1, Drug C inhibited 50% of viral entry, the highest of all antivirals.

→ It was also relatively non-toxic, proving it to be the most effective at that dosage.

`=>C`

Mean mark 52%.

BIOLOGY, M6 2020 HSC 10 MC

A farmer intends to artificially inseminate cows with semen from a bull which has been chosen based on characteristics of colour and muscle mass.

The farmer does not know that the bull is heterozygous for a rare recessive allele not previously present in the farmer's cow population.

The introduction of this recessive allele to the population of cows is an example of

  1. gene flow.
  2. genetic drift.
  3. natural selection.
  4. selective breeding.
Show Answers Only

`A`

Show Worked Solution

→ The introduction of the allele in this fashion was a form of gene immigration (gene flow).

→ It is not the result of a random chance event (genetic drift) nor was it selected by breeders or environmental pressures.

`=>A`

♦ Mean mark 50%.

BIOLOGY, M7 2020 HSC 7 MC

Students designed and conducted an investigation to test for the presence of microbes in THREE different food samples.

They inoculated agar plates with the samples and placed them in an incubator set to 25°C.

Which row of the table represents a valid design for the investigation?
 

Show Answers Only

`A`

Show Worked Solution

→ The different food samples are deliberately selected (independent).

→ The number of microbes is a consequence of the specific food samples (dependent).

→ An agar plate with a food sample would be a base of comparison (experimental control) while temperature would be a controlled variable.

`=>A`

♦ Mean mark 49%.

BIOLOGY, M7 2020 HSC 6 MC

Citrus canker is a bacterial disease that originates in south-east Asia and affects citrus fruit.

What would be the most effective way to prevent the disease from spreading into or across Australia?

  1. Monitor citrus trees and fruit continuously.
  2. Certify orchards before fruit is transported.
  3. Keep citrus trees and fruit entering Australia in quarantine stations until the incubation period has passed.
  4. Inspect citrus trees and fruit entering Australia in quarantine stations before transportation across Australia.
Show Answers Only

`C`

Show Worked Solution

→ Bacterial infections may not be apparent or detected on initial arrival inspection (eliminate D).

→ Options A and B would be difficult to execute and too costly.

→ Quarantining the citrus for the duration of the incubation period should enable the detection of the disease.

`=>C`

♦ Mean mark 46%.

BIOLOGY, M6 2021 HSC 33c

Genetically engineered Atlantic salmon have been produced and approved for aquaculture in the US. These salmon have a transgene that includes a protein-coding sequence from a Chinook salmon's growth hormone gene and the promoter region of an Ocean Pout's antifreeze protein gene. The following diagram provides an overview of the production of the transgenic salmon.
 

Transgenic fish can reproduce and pass on the dominant transgene (T).

Reproduction for aquaculture is strictly controlled using a variety of techniques in order to protect and preserve biodiversity.

Some of these techniques are outlined below. 

1. Homozygous (TT) female (XX) breeding stock are kept in quarantine.

2. The female fish undergo hormone treatment that results in sex reversal and the development of male sex organs and sperm.

3. The sperm produced is collected and used to fertilise eggs obtained from wild-type, non-transgenic salmon.

4. The eggs are treated with pressure shock to prevent the completion of meiosis II. As a result, offspring are triploid (three copies of each chromosome).

All offspring are transgenic female fish and have XXX (XXX fish cannot develop sex organs).

5. Offspring are transported to inland aquaculture tanks to be grown to market size.

Analyse how these techniques protect and preserve biodiversity.   (9 marks)

Show Answers Only

Protecting and Preserving Biodiversity:

→ Biodiversity refers to the variety of genes within the gene pool of a species as well as the variety of species within an ecosystem.

→ The nature of the transgenic salmon carries the risk of reducing biodiversity and has the potential to do great harm.

→ If transgenic salmon escape captivity, they may have a survival advantage over natural salmon and other species. In this way, they could outcompete other species and reduce species diversity, as well as outcompete natural salmon.

→ As a result of the transgene being dominant, it will be passed onto offspring, and in time reduce the gene pool.

→ The physical separation techniques 1 and 5 prevent the transgenic salmon from being released into the environment, preserving biodiversity.

→ Reproductive techniques such as the hormone treatment in technique 2 allow sperm to be produced that only carries the X chromosome as the parents are genetically female.

→ This sperm is then used to fertilise from wild-type salmon, and all offspring will be female as sperm does not contain a Y chromosome, and contain the transgene.

→ Technique 4 uses pressure shock to prevent meiosis 2, resulting in infertile, triploid offspring. This produces transgenic salmon incapable of breeding, meaning the transgene cannot be passed on, preserving biodiversity within the gene pool.

→ The use of wild-type salmon eggs to produce transgenic fish is also crucial in preserving biodiversity as it reduces the likelihood of the accumulation of mutations via interbreeding of transgenic salmon.

→ The physical separation techniques of quarantine and inland aquaculture tanks coupled with reproductive techniques are essential biotechnologies to protect and preserve biodiversity from the dangers of transgenic salmon to the ecosystem and gene pool.

Show Worked Solution

Protecting and Preserving Biodiversity:

→ Biodiversity refers to the variety of genes within the gene pool of a species as well as the variety of species within an ecosystem.

→ The nature of the transgenic salmon carries the risk of reducing biodiversity and has the potential to do great harm.

→ If transgenic salmon escape captivity, they may have a survival advantage over natural salmon and other species. In this way, they could outcompete other species and reduce species diversity, as well as outcompete natural salmon.

→ As a result of the transgene being dominant, it will be passed onto offspring, and in time reduce the gene pool.

→ The physical separation techniques 1 and 5 prevent the transgenic salmon from being released into the environment, preserving biodiversity.

→ Reproductive techniques such as the hormone treatment in technique 2 allow sperm to be produced that only carries the X chromosome as the parents are genetically female.

→ This sperm is then used to fertilise from wild-type salmon, and all offspring will be female as sperm does not contain a Y chromosome, and contain the transgene.

→ Technique 4 uses pressure shock to prevent meiosis 2, resulting in infertile, triploid offspring. This produces transgenic salmon incapable of breeding, meaning the transgene cannot be passed on, preserving biodiversity within the gene pool.

→ The use of wild-type salmon eggs to produce transgenic fish is also crucial in preserving biodiversity as it reduces the likelihood of the accumulation of mutations via interbreeding of transgenic salmon.

→ The physical separation techniques of quarantine and inland aquaculture tanks coupled with reproductive techniques are essential biotechnologies to protect and preserve biodiversity from the dangers of transgenic salmon to the ecosystem and gene pool.

♦ Mean mark 44%.

BIOLOGY, M8 2021 HSC 32

The flow chart shows negative feedback by the hormones testosterone and inhibin in a human male.
 


 

Some athletes take anabolic steroids to increase their muscle mass and strength. These steroids may be testosterone or a synthetic modification of testosterone.

Explain the changes that would occur in the testes of a male athlete continuously taking anabolic steroids. Support your answer with reference to the flow chart.   (5 marks)

Show Answers Only

→ Naturally, as shown in the diagram above, males produce Releasing Hormone via the hypothalamus, which stimulates the anterior pituitary gland to produce hormones which further stimulate the testes to produce hormones inhibin and testosterone as well as sperm (via testosterone).

→ A male taking anabolic steroids similar to testosterone will see an initial increase in sperm and inhibin.

→ However, it will also impact the negative feedback on both the hypothalamus and anterior pituitary, reducing their release of respective stimulatory hormones, reducing sperm, testosterone and inhibin.

→ Although less inhibin would mean a smaller negative feedback on the anterior pituitary caused by this hormone, the ongoing dose of testosterone would ensure a large overall negative feedback effect. 

→ Over time, with repetitive and excessive use of anabolic steroids, males will produce less sperm, testosterone and inhibin, and the testes will lose their normal function.

Show Worked Solution

→ Naturally, as shown in the diagram above, males produce Releasing Hormone via the hypothalamus, which stimulates the anterior pituitary gland to produce hormones which further stimulate the testes to produce hormones inhibin and testosterone as well as sperm (via testosterone).

→ A male taking anabolic steroids similar to testosterone will see an initial increase in sperm and inhibin.

→ However, it will also impact the negative feedback on both the hypothalamus and anterior pituitary, reducing their release of respective stimulatory hormones, reducing sperm, testosterone and inhibin.

→ Although less inhibin would mean a smaller negative feedback on the anterior pituitary caused by this hormone, the ongoing dose of testosterone would ensure a large overall negative feedback effect. 

→ Over time, with repetitive and excessive use of anabolic steroids, males will produce less sperm, testosterone and inhibin, and the testes will lose their normal function.

♦ Mean mark 50%.

BIOLOGY, M8 2021 HSC 31

Millions of people around the world take drugs known as statins, which have been shown to reduce the incidence of heart attacks and strokes in vulnerable patients. However, up to 20% of people stop taking statins due to side-effects such as muscle aches, fatigue, feeling sick and joint pain.

A recent study at a public hospital focused on 60 patients who had all stopped taking statins in the past due to severe side-effects. Patients took statin tablets for four months, placebo tablets for four months and no tablets for four months.

Every day for the year the patients scored, from zero to 100, how bad their symptoms were. The results are shown.
 

Evaluate this study and its results.   (6 marks)

Show Answers Only

→ The data shows that when no tablets were taken, patients still displayed symptoms of the side effects (score 8.0). Taking the statin tablets doubled the severity of the side effects (score 16.3), however, placebo tablets with no active ingredients still produced similar results (score 15.4).

→ This seems to indicate that the side effects may arise as a result of other factors.

→ However, this study cannot be regarded as valid. Only 60 patients were involved in comparison to the millions that take statin tablets.

→ The results were also taken in a qualitative way and cannot be regarded as reliable.

→ The study should also have included a control group of people who have never taken statins to be used for a comparison to improve validity.

→ While the raw data provides data which is valuable and should be followed up with a large, randomised control trial, the study itself cannot be regarded as valid.

Show Worked Solution

→ The data shows that when no tablets were taken, patients still displayed symptoms of the side effects (score 8.0). Taking the statin tablets doubled the severity of the side effects (score 16.3), however, placebo tablets with no active ingredients still produced similar results (score 15.4).

→ This seems to indicate that the side effects may arise as a result of other factors.

→ However, this study cannot be regarded as valid. Only 60 patients were involved in comparison to the millions that take statin tablets.

→ The results were also taken in a qualitative way and cannot be regarded as reliable.

→ The study should also have included a control group of people who have never taken statins to be used for a comparison to improve validity.

→ While the raw data provides data which is valuable and should be followed up with a large, randomised control trial, the study itself cannot be regarded as valid.

♦♦ Mean mark 38%.

Proof, EXT2 P1 2022 HSC 7 MC

Consider the statement `P`.

`P` : For all integers `n \geq 1`, if `n` is a prime number then `(n(n+1))/(2)` is a prime number.

Which of the following is true about this statement and its converse?

  1. The statement `P` and its converse are both true.
  2. The statement `P` and its converse are both false.
  3. The statement `P` is true and its converse is false.
  4. The statement `P` is false and its converse is true.
Show Answers Only

`D`

Show Worked Solution

`text{Statement:}\ \ ∀n in ZZ^+, text{if}\ n\ text{is prime}\ \ =>\ \ (n(n+1))/(2)\ text{is prime.}`

`text{Converse:}\ \ ∀n in ZZ^+, text{if}\ (n(n+1))/(2)\ text{is prime}\ \ =>\ \ n\ text{is prime.}`

`text{Consider prime}\ n=3:`

`(n(n+1))/(2)=(3xx4)/2=6\ \ text{(not prime → Statement is false)}`

`text{Consider}\ \ (n(n+1))/(2):`

`(n(n+1))/(2)\ text{is a prime}\ iff\ n=2\ \ text{(prime)}`

`:.\ text{Converse is true}`

`=>D`


♦♦ Mean mark 37%.

Complex Numbers, EXT2 N2 2022 HSC 6 MC

It is known that a particular complex number `z` is NOT a real number.

Which of the following could be true for this number `z` ?

  1. `bar(z)=iz`
  2. `bar(z)=|z^(2)|`
  3. `text{Re}(iz)= text{Im}(z)`
  4. `text{Arg}(z^(3))= text{Arg}(z)`
Show Answers Only

`A`

Show Worked Solution

`z in CC, z !in RR`

`text{Let}\ \ z=a+ib\ \ =>\ \ barz=a-ib`

`iz=i(a+ib)=-b+ia`

`text{If}\ \ barz=iz,`

`a-ib=-b+ia`

`a=-b\ \ text{(satisfies)}`

`:.∃ a,b (b!=0)\ \ text{such that}\ \ barz=iz`

`=>A`


♦♦ Mean mark 31%.

PHYSICS, M8 2019 HSC 36

A radon-198 atom, initially at rest, undergoes alpha decay. The masses of the atoms involved are shown in atomic mass units `(u)`.  
 

The kinetic energy of the polonium atom produced is  `2.55 × 10^(-14)` J.

By considering mass defect, calculate the kinetic energy of the alpha particle, and explain why it is significantly greater than that of the polonium atom.   (7 marks)

Show Answers Only
`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u  
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.

Show Worked Solution
`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u   
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.


♦♦♦ Mean mark 30%.

PHYSICS, M5 2019 HSC 35

The apparatus shown is attached horizontally to the roof inside a stationary car. The plane of the protractor is perpendicular to the sides of the car.
 

The car was then driven at a constant speed `(v)`, on a horizontal surface, causing the string to swing to the right and remain at a constant angle `(theta)` measured with respect to the vertical.

Describe how the apparatus can be used to determine features of the car's motion. In your answer, derive an expression that relates a feature of the car's motion to the angle `theta`.   (4 marks)

Show Answers Only

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`
Show Worked Solution

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`

♦ Mean mark 52%.

PHYSICS, M8 2019 HSC 34

Use the following information to answer this question.
 

Describe both the production and radiation of energy by the sun. In your answer, include a quantitative analysis of both the power output and the surface temperature of the sun.   (9 marks)

Show Answers Only

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W  
Show Worked Solution

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons, so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W    

♦♦♦ Mean mark 38%.

BIOLOGY, M8 2021 HSC 19-20 MC

Use the following information to answer Questions 19–20.

Alcohol that is consumed is broken down by the following pathway.
 


 

ALDH is the product of the ALDH gene. The normal allele is ALDH1 but many people have the ALDH2 allele, which is associated with an increased risk of cancer.

Scientists have used mouse models to study the effects of ALDH2. They used wild-type mice with ALDH1 alleles and genetically-engineered mice with one or two copies of the ALDH2 allele.

The mice were given a dose of alcohol and the response was measured. The results are shown.
 

 
Question 19

The ALDH alleles differ in their

  1. genome location.
  2. location in gametes.
  3. effect on phenotype.
  4. amino acid composition.

 
Question 20

What do the data show about the effect of the ALDH2 allele?

  1. Enzyme activity is highest in homozygous mice.
  2. Enzyme activity decreases if ALDH2 is present.
  3. Enzyme activity increases if ALDH2 is present.
  4. Enzyme activity is lowest in wild-type mice.
Show Answers Only

`text{Question 19:}\ C`

`text{Question 20:}\ B`

Show Worked Solution

Question 19

→ The graph depicts that the two alleles (ALDH1 and ALDH2) can produce three different phenotypes

→ i.e. slow, moderate and fast alcohol metabolisation.

`=>C`

♦ Mean mark (Q19) 36%.

Question 20

→ The graph indicates that the ALDH2 / ALDH2 group had the slowest metabolism of alcohol shown by the consistently high levels of acetaldehyde over time, suggesting decreased enzyme activity.

`=>B`

♦ Mean mark (Q20) 41%.

BIOLOGY, M5 2021 HSC 18 MC

Which human pedigree shows inheritance of a recessive, sex-linked characteristic?
 

Show Answers Only

`D`

Show Worked Solution

→ In order for the recessive sex-linked inheritance to be true, the mother is a carrier (heterozygous).

→ Thus the daughter can be affected (inheriting an affected chromosome from each parent), but the son can remain unaffected (inheriting the unaffected chromosome from the carrier mother).

`=>D`

♦ Mean mark 45%.

BIOLOGY, M7 2021 HSC 14 MC

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

Which of the following statements best classifies CJD?

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

`A`

Show Worked Solution

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

→ Genetic, as CJD can be caused by mutation.

`=>A`

♦♦ Mean mark 39%.

BIOLOGY, M8 2021 HSC 13 MC

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

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

Which of the following graphs best illustrates this homeostatic mechanism?
 

 

Show Answers Only

`A`

Show Worked Solution

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

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

`=>A`

Mean mark 53%.

PHYSICS, M6 2020 HSC 34

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

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

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

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

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

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

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

Show Worked Solution

a.   Work done on `q_1:`

`W=qE((d)/(2))` … (1)

Substitute  `E=(V)/(d)`  into (1)

`W=(qV)/2`

 
Work done on `q_2:`

`W=qE((3d)/(2))` … (1)

Substitute  `E=(V)/(2d)`  into (1)

`W=(3qV)/4`

Divide  (2) `-: ` (1):

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

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

♦ Mean mark part 47%.

b.   Find the vertical acceleration of `q_1:`

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

 

Similarly for `q_2:` 

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

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

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

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

Time for `q_1` to reach plate `Y:` 

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

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

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

 
Divide  (4) ÷ (3):

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

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

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

♦ Mean mark part 43%.

PHYSICS, M6 2020 HSC 33

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

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

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

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

Show Answers Only

During the first 0.4 seconds:

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

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

→  Quantitatively:

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

 

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

 
As magnet reaches the copper cylinder:

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

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

→  Finding the magnets speed before entering the copper cylinder:

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

 
→  Quantifying the kinetic energy loss:

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

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

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

→  Quantifying the decrease in gravitational potential energy:

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

 

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

Show Worked Solution

During the first 0.4 seconds:

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

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

→  Quantitatively:

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

 

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

 
As magnet reaches the copper cylinder:

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

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

→  Finding the magnets speed before entering the copper cylinder:

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

 
→  Quantifying the kinetic energy loss:

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

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

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

→  Quantifying the decrease in gravitational potential energy:

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

 

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

♦ Mean mark 51%.

PHYSICS M6 2022 HSC 34

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


 

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

Show Answers Only

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

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

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

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


Mean mark 60%.

PHYSICS M8 2022 HSC 31

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


 

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


 

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

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

Show Answers Only

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

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

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

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

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


 

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

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

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

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

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

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

 

Show Worked Solution

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

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

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

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

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


 

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

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

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

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

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

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


♦ Mean mark 51%.

PHYSICS M7 2022 HSC 30

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


 

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

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

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

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

Show Answers Only

a.  Consider each observer:

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

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

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

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

b.  53.6  ns

Show Worked Solution

a.  Consider each observer:

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

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

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

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


♦ Mean mark (a) 48%.
b.  `t`

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

PHYSICS M5 2022 HSC 29

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

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

Show Answers Only

Horizontal component:

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

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

Vertical component:

→ The apple has zero initial vertical velocity.

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

Show Worked Solution

Horizontal component:

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

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

Vertical component:

→ The apple has zero initial vertical velocity.

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


Mean mark 51%.

PHYSICS M8 2022 HSC 28

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

Step `X` releases 1.20 MeV.

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

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

Show Answers Only

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

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

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

 

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

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

Show Worked Solution

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

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

 

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

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


Mean mark 51%.

PHYSICS M7 2022 HSC 27

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


 

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

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

a.   
       
 

b.   `theta=1.50^@`

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

→ `sin theta prop lambda`

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

→ The bright lines will appear closer together.

Show Worked Solution
a.   
     

Mean mark part (a) 53%.

 

b.   Using  `d  sin  theta=mlambda:`

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

 

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

→ `sin theta prop lambda`

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

→ The bright lines will appear closer together.

PHYSICS, M7 2016 HSC 27b

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

Show Answers Only

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

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

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

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

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

Show Worked Solution

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

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

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

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

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


♦ Mean mark 40%.

PHYSICS, M7 2014 HSC 26

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

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

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

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

Show Answers Only

`text(See Worked Solutions)`

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

 

b.  

`text(To find intercept)`

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

`text(Gradient = same as A1)`

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

PHYSICS M7 2022 HSC 26

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

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


  

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

 

b.   Maximum Distance:

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

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

♦ Mean mark (b) 51%.

PHYSICS M5 2022 HSC 25

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

  1. Show that the magnitude of the change in kinetic energy from point 1 to point 2 is  `2.2 xx10^(9) \ text{J}`.  (2 marks)

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

Show Answers Only

a.   `Delta K=K_(f)-K_(i)`

 `K_(f)` `=(1)/(2)mv_(2)^2`
  `=(1)/(2)200 xx2900^2`
  `=8.41 xx10^8  text{J}`
   `K_(i)` `=(1)/(2)mv_(1)^2`
  `=(1)/(2)200 xx5500^2`
  `=3.025 xx10^9  text{J}`

 

     `:.Delta K` `=8.41 xx10^8-3.025 xx10^9`
    `=-2.2 xx10^9  text{J}`

 
This change has a magnitude of `2.2 xx10^9\ text{J}`.
 

b.   Applying the law of conservation of energy:

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

`Delta U` `=U_(f)-U_(i)`  
  `=-(GMm)/(r_(f))- (-(GMm)/(r_(i)))`  

 
`(2.2 xx10^(9))=-((6.67 xx10^(-11))(200)M)/(2.5 xx10^7)+((6.67 xx10^(-11))(200)M)/(4.3 xx10^6)`
 

`(2.2 xx10^(9))(2.5 xx10^7)(4.3 xx10^6)`

`=(6.67 xx10^(-11))200M[(2.5 xx10^7)-(4.3 xx10^7)]`
 

`:.M` `=((2.2 xx10^(9))(2.5 xx10^7)(4.3 xx10^6))/(200(6.67 xx10^(-11))[(2.5 xx10^7)-(4.3 xx10^6)])`  
  `= 8.56 xx10^23\ text(kg)`  
Show Worked Solution

a.   `Delta K=K_(f)-K_(i)`

 `K_(f)` `=(1)/(2)mv_(2)^2`
  `=(1)/(2)200 xx2900^2`
  `=8.41 xx10^8  text{J}`
   `K_(i)` `=(1)/(2)mv_(1)^2`
  `=(1)/(2)200 xx5500^2`
  `=3.025 xx10^9  text{J}`

 

     `:.Delta K` `=8.41 xx10^8-3.025 xx10^9`
    `=-2.2 xx10^9  text{J}`

 
This change has a magnitude of `2.2 xx10^9\ text{J}`.
 

b.   Applying the law of conservation of energy:

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

`Delta U` `=U_(f)-U_(i)`  
  `=-(GMm)/(r_(f))- (-(GMm)/(r_(i)))`  

 
`(2.2 xx10^(9))=-((6.67 xx10^(-11))(200)M)/(2.5 xx10^7)+((6.67 xx10^(-11))(200)M)/(4.3 xx10^6)`
 

`(2.2 xx10^(9))(2.5 xx10^7)(4.3 xx10^6)`

`=(6.67 xx10^(-11))200M[(2.5 xx10^7)-(4.3 xx10^7)]`
 

`:.M` `=((2.2 xx10^(9))(2.5 xx10^7)(4.3 xx10^6))/(200(6.67 xx10^(-11))[(2.5 xx10^7)-(4.3 xx10^6)])`  
  `= 8.56 xx10^23\ text(kg)`  

♦ Mean mark 47%.

BIOLOGY, M7 2021 HSC 30

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


 

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

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

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

Show Answers Only

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

BIOLOGY, M7 2021 HSC 28b

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

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

Show Answers Only

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

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

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

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

Show Worked Solution

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

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

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

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


♦ Mean mark 43%.

BIOLOGY, M5 2021 HSC 27

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

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

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

Show Answers Only

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

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

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

Show Worked Solution

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

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

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

Mean mark 51%.

BIOLOGY, M5 2021 HSC 26

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

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

Show Answers Only

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

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

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

Show Worked Solution

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

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

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

♦ Mean mark 46%.

BIOLOGY, M8 2021 HSC 25

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

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

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

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

Show Answers Only

a.  


 

b.   Right ear is within normal hearing range.

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

c.    Effective technology: Bone Conduction Implants

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

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

Show Worked Solution

a.   


 

b.   → Right ear is within normal hearing range.

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

c.    Effective technology: Bone Conduction Implants

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

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

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


♦ Mean mark (c) 43%.

ENGINEERING, CS 2021 HSC 21c

Reinforced concrete is a composite material.

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

 

Show Answers Only

Show Worked Solution

Image should include the following elements:

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

 

♦ Mean mark 45%.

PHYSICS M7 2022 HSC 20 MC

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

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

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

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

`B`

Show Worked Solution

Consider Point `P`:

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

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

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

Consider Point `R`:

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

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

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

`=>B`


♦♦♦ Mean mark 30%.

PHYSICS M7 2022 HSC 17 MC

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

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


 

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

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

`C`

Show Worked Solution

Once the filter rotates 45° to the vertical:

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

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

→  `I_(y)` sometimes equals `I_(z)`.

`=>C`


Mean mark 55%.

PHYSICS M6 2022 HSC 12 MC

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


 

What is the direction of the particle's velocity?

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

`D`

Show Worked Solution

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

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

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

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

`=>D`


♦ Mean mark 52%.

PHYSICS, M8 2019 HSC 32

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

Show Answers Only

Millikan’s Oil Drop Experiment:

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

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

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

Linear accelerator experiment discovering quarks:

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

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

Other possible answers could include:

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

Millikan’s Oil Drop Experiment:

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

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

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

Linear accelerator experiment discovering quarks:

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

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

Other possible answers could include:

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

♦♦ Mean mark 43%.

PHYSICS, M6 2019 HSC 31

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

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

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

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

Show Answers Only

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

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

For the first ten seconds:

→ The fan’s rotation speed increases.

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

→ The force observed on the spring balance decreases.

After ten seconds:

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

→ The recorded force on the spring balance remains constant.
 

b.   Between 0-10 seconds:

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

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

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

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

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

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

Between 10-15 seconds:

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

→ The motor is rotating at a constant speed.

→ The magnitude of induced back EMF remains constant.

→ the current through the motor remains constant.

Show Worked Solution

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

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

For the first ten seconds:

→ The fan’s rotation speed increases.

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

→ The force observed on the spring balance decreases.

After ten seconds:

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

→ The recorded force on the spring balance remains constant.


♦ Mean mark (a) 48%.

b.   Between 0-10 seconds:

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

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

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

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

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

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

Between 10-15 seconds:

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

→ The motor is rotating at a constant speed.

→ The magnitude of induced back EMF remains constant.

→ the current through the motor remains constant.


♦ Mean mark (b) 39%.

PHYSICS, M5 2019 HSC 30

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

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

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

a.   `h=`0.11 m

b.   `h=`1.8 m

Show Worked Solution

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


♦ Mean mark part (a) 41%.

 

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

 
`:.` Height of Q `=` 1.8 m

PHYSICS, M6 2019 HSC 28

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

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

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

Show Answers Only

Originally:

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

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

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

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

Show Worked Solution

Originally:

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

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

After a `90^(@)` rotation:

→ `XY` experiences a force to the right.

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


♦ Mean mark 46%.

Vectors, EXT1 V1 2022 HSC 14c

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

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

  
 

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

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

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

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Projectile’s motion:}`

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

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

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

 

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

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

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

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

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

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

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

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

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

 

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

 
`f^(′′)(32°32′)~~-3.1<0`

`=>\ text{MAX at}\ \ theta=32°32′`
 

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

♦♦ Mean mark 33%.

Calculus, EXT1 C3 2022 HSC 14a

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

Show Answers Only

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

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

 
`text{Passes through (0,1):`

`ln1` `=0+2ln|-2|+c`  
`c` `=-2ln2`  

 

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

♦ Mean mark 43%.

Statistics, EXT1 S1 2022 HSC 13e

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

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

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

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

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

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

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

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

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

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

 

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

 

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

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


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

Functions, EXT1 F2 2022 HSC 13d

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

It is given that

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

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

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

Show Answers Only

`-2`

Show Worked Solution

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

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

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

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

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

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

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

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

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

`text{Substituting in given values:}`

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

♦♦ Mean mark 33%.

Trigonometry, EXT1 T1 2022 HSC 13c


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

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

Show Answers Only

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

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

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

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

Show Worked Solution

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

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

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

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


♦♦ Mean mark 24%.

Statistics, EXT1 S1 2022 HSC 12e

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

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

Show Answers Only

`-2`

Show Worked Solution

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

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

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

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

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

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

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

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

♦♦ Mean mark 35%.

Vectors, EXT1 V1 2022 HSC 8 MC

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

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

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

`D`

Show Worked Solution

`text{By Elimination:}`

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

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

`text{→ Eliminate A and B}`
 

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

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

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

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

`text{→ Eliminate C}`

`=>D`


♦♦ Mean mark 34%.

Combinatorics, EXT1 A1 2022 HSC 7 MC

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


 

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

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

`C`

Show Worked Solution

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

`text{Triangles} = 3 xx 4xx5=60`
 

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

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

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

`=>C`


♦♦ Mean mark 37%.

Vectors, EXT1 V1 2022 HSC 6 MC

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


 

Which statement regarding this diagram could be true?

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

`B`

Show Worked Solution

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

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

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

`=>B`


♦♦ Mean mark 38%.

 

Functions, EXT1 F2 2022 HSC 3 MC

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

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

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

`D`

Show Worked Solution

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

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

`text{Consider examples to resolve possibilities:}`

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

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

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

`=>D`


♦ Mean mark 51%.

Financial Maths, 2ADV M1 2022 HSC 32

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

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

(Do NOT prove this.)

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

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

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

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

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

  6. Jane continues to make the same monthly repayments.

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

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

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

a.   `text{Show}\ \ M=$1381.16`

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

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

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

 

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

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

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

 


♦♦ Mean mark part (b) 38%.

c.   `text{Find}\ \ A_83:`

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

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

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

♦♦♦ Mean mark part (c) 14%.