Band 4 – Page 18

CHEMISTRY, M1 EQ-Bank 2

A student is given a known mixture that contains methanol, water, salt and sand.

Describe a process where the student can separate each component of the mixture. (3 marks)

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→ Each of the components in the mixture have their own unique physical properties which can be exploited to separate them.

A sieve can be used to separate the sand from the mixture
Following this, a distillation set up can be used to remove the methanol from the mixture (as methanol has a lower boiling point than water).
An evaporating basin can then be used to evaporate the water, leaving salt behind.

Show Worked Solution

→ Each of the components in the mixture have their own unique physical properties which can be exploited to separate them.

A sieve can be used to separate the sand from the mixture
Following this, a distillation set up can be used to remove the methanol from the mixture (as methanol has a lower boiling point than water).
An evaporating basin can then be used to evaporate the water, leaving salt behind.

CHEMISTRY, M1 EQ-Bank 1

Complete the following table by providing the physical properties of compounds exploited by various separation methods. (3 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Separation Method} \rule[-1ex]{0pt}{0pt} & \text{Physical Property Exploited} \\
\hline
\rule{0pt}{2.5ex} \text{Filtration:} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \text{Evaporation:} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \text{Distillation:} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

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\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Separation Method} \rule[-1ex]{0pt}{0pt} & \text{Physical Property Exploited} \\
\hline
\rule{0pt}{2.5ex} \text{Filtration:} \rule[-1ex]{0pt}{0pt} & \text{Particle size} \\
\hline
\rule{0pt}{2.5ex} \text{Evaporation:} \rule[-1ex]{0pt}{0pt} & \text{State of matter} \\
\hline
\rule{0pt}{2.5ex} \text{Distillation:} \rule[-1ex]{0pt}{0pt} & \text{Boiling point} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Separation Method} \rule[-1ex]{0pt}{0pt} & \text{Physical Property Exploited} \\
\hline
\rule{0pt}{2.5ex} \text{Filtration:} \rule[-1ex]{0pt}{0pt} & \text{Particle size} \\
\hline
\rule{0pt}{2.5ex} \text{Evaporation:} \rule[-1ex]{0pt}{0pt} & \text{State of matter} \\
\hline
\rule{0pt}{2.5ex} \text{Distillation:} \rule[-1ex]{0pt}{0pt} & \text{Boiling point} \\
\hline
\end{array}

BIOLOGY, M3 2020 VCE 8*

The phylogenetic tree below shows the evolutionary relationship between seven species of cichlid fish.

Fossils of species of fish are more likely to be found than fossils of land-dwelling animals.
Explain why this is the case with reference to two conditions required for the fossilisation of an organism. (2 marks)

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A group of scientists stated that a particular fossilised fish was 5000 years old.
Outline a dating technique that could have been used by the scientists to determine the age of the fossil. (2 marks)

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a. Fossilisation is more likely in water than on land because:

→ Water contains more sediment which allows remains to be covered quicker, reducing disturbance and hiding them from scavengers.

→ Water contains less oxygen and lower temperatures, factors which reduce decomposition of remains.

b. Carbon dating:

→ This technique can be used to validate the age of fossils between 500 and 50,000 years old.

→ It involves measuring the ratio of carbon-14 to carbon-12. Since carbon-14 decays into carbon-12 over time and knowing the half-life of carbon-14, the age of fossils can be accurately calculated.

Show Worked Solution

a. Fossilisation is more likely in water than on land because:

→ Water contains more sediment which allows remains to be covered quicker, reducing disturbance and hiding them from scavengers.

→ Water contains less oxygen and lower temperatures, factors which reduce decomposition of remains.

b. Carbon dating:

→ This technique can be used to validate the age of fossils between 500 and 50,000 years old.

→ It involves measuring the ratio of carbon-14 to carbon-12. Since carbon-14 decays into carbon-12 over time and knowing the half-life of carbon-14, the age of fossils can be accurately calculated.

BIOLOGY, M1 2020 VCE 4c

The diagram below is a simplified example of a signal transduction pathway. Three steps in the pathway are labelled.

Explain whether the pathway shown is initiated at Step 1 by a hydrophobic molecule or by a hydrophilic molecule. (2 marks)

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→ The molecule must be hydrophilic as it does not pass through the plasma membrane but instead attaches to a receptor protein.

Show Worked Solution

→ The molecule does not pass through the plasma membrane but instead attaches to a receptor protein.

→ Therefore, the molecule is hydrophilic.

BIOLOGY, M1 2020 VCE 3c

In plants and algae, photosynthesis is carried out in chloroplasts. It is thought that chloroplasts originated from bacteria.

Describe two features of chloroplasts that support the theory that chloroplasts originated from bacteria. (2 marks)

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Answers can include two of the following:

→ Chloroplasts contain their own DNA, which in itself has many parallels to the DNA found in bacteria, such as it being circular.

→ Chloroplasts contain their own ribosomes.

→ New chloroplasts are made by division of previously existing chloroplasts, similar to the division of bacterial cells.

Show Worked Solution

Answers can include two of the following:

→ Chloroplasts contain their own DNA, which in itself has many parallels to the DNA found in bacteria, such as it being circular.

→ Chloroplasts contain their own ribosomes.

→ New chloroplasts are made by division of previously existing chloroplasts, similar to the division of bacterial cells.

PHYSICS, M3 2017 VCE 14

A light ray from a laser passes from a glucose solution $(n=1.44)$ into the air $(n=1.00)$, as shown in Figure 12.

Calculate the critical angle (total internal reflection) from the glucose solution to the air. (1 mark)

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The light ray strikes the surface at an angle of incidence to the normal of less than the critical angle calculated in part a.
On Figure 12, sketch the ray or rays that should be observed. (2 marks)

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The angle to the normal is increased to a value greater than the critical angle. An observer at point $\text{X}$ in Figure 13 says she cannot see the laser.

Explain why the observer says she cannot see the laser. (2 marks)

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a. $\theta_c=44^{\circ}$

→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.

c. Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

Show Worked Solution

a.	$\sin\theta_c$	$=\dfrac{n_2}{n_1}$
	$\theta_c$	$=\sin^{-1}\Big{(}\dfrac{1.00}{1.44}\Big{)}$
		$=44^{\circ}$

→ As the light strikes the surface at an angle less than that of the critical angle, both reflection and refraction will be observed.

♦♦ Mean mark (b) 37%.

c. Observer cannot see the laser:

→ When the light strikes the surface at an angle greater than the critical angle, Total Internal Reflection will occur.

→ As a result of the total internal reflection, all the light will be reflected off the boundary back into the glucose solution and therefore will not travel to the observer.

PHYSICS, M2 2017 VCE 12

Students are using two trolleys, Trolley $\text{A}$ of mass 4.0 kg and Trolley $\text{B}$ of mass 2.0 kg, to investigate kinetic energy and momentum in collisions.

Before the collision, Trolley $\text{A}$ is moving to the right at 5.0 m s$^{-1}$ and Trolley $\text{B}$ is moving to the right at 2.0 m s$^{-1}$, as shown in Figure 10a. The trolleys collide and lock together, as shown in Figure 10b.

Determine, using calculations, whether the collision is elastic or inelastic. Show your working and justify your answer. (3 marks)

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By the conservation of momentum:

$m_Au_A+m_Bu_B$	$=v(m_A+m_B)$
$4 \times 5 + 2 \times 2$	$=v(4 +2)$
$24$	$=6v$
$v$	$=4\ \text{ms}^{-1}$

For the collision to be elastic, the kinetic energy must be conserved.

$KE_{\text{init}}=\dfrac{1}{2} m_Au_A^2+ \dfrac{1}{2} m_Bu_B^2 =\dfrac{1}{2} \times 4 \times 5^2+\times \dfrac{1}{2} \times 2 \times 2^2=54\ \text{J}$

$KE_{\text{final}}=\dfrac{1}{2} (m_Am_B)v^2=\dfrac{1}{2} \times (4+2) \times 4^2=48\ \text{J}$

→ As the kinetic energy of the system decreases after the collision, the collision is inelastic.

Show Worked Solution

By the conservation of momentum:

$m_Au_A+m_Bu_B$	$=v(m_A+m_B)$
$4 \times 5 + 2 \times 2$	$=v(4 +2)$
$24$	$=6v$
$v$	$=4\ \text{ms}^{-1}$

For the collision to be elastic, the kinetic energy must be conserved.

$KE_{\text{init}}=\dfrac{1}{2} m_Au_A^2+ \dfrac{1}{2} m_Bu_B^2 =\dfrac{1}{2} \times 4 \times 5^2+\times \dfrac{1}{2} \times 2 \times 2^2=54\ \text{J}$

$KE_{\text{final}}=\dfrac{1}{2} (m_A + m_B)v^2=\dfrac{1}{2} \times (4+2) \times 4^2=48\ \text{J}$

→ As the kinetic energy of the system decreases after the collision, the collision is inelastic.

PHYSICS, M4 2017 VCE 1

Three charges are arranged in a line, as shown in Figure 1.

Draw an arrow at point $\text{X}$ to show the direction of the resultant electric field at $\text{X}$. If the resultant electric field is zero, write the letter ' $\text{N}$ ' at $\text{X}$. (2 mark)

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(Insert image here)

Show Worked Solution

→ The electric field direction will be from the positive charge to the negative charge.

→The $-Q$ on the right has no effect to the direction of the electric field at $\text{X}$.

(Insert image here)

BIOLOGY, M1 2020 VCE 1

Adrenaline is an amino-acid-based hormone. The image below shows a cell section of an adrenaline-secreting cell examined under a transmission electron microscope.

A secretory granule is a large vesicle formed when several smaller vesicles fuse. Each secretory granule contains a large amount of adrenaline, which is stored until the cell receives a signal to release it.

There are many organelles of the types labelled GA and M visible in the cell section.
Complete the table below by stating the specific function of each organelle with the following labels in the adrenaline-secreting cell. (2 marks)

\begin{array} {|c|l|}
\hline \ \ \ \textbf{Label} & \ \ \ \ \ \ \ \ \ \ \textbf{Specific function in the adrenaline-secreting cell} \ \ \ \ \ \ \ \ \ \ \\
\hline & \\ \text{GA} & \text{......................................................................................................................} \\ & \\ & \text{......................................................................................................................} \\
\hline & \\ \text{M} & \text{......................................................................................................................} \\& \\ & \text{......................................................................................................................} \\
\hline \end{array}

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Adrenaline is secreted from the cell by exocytosis.
Describe this process with reference to the roles of relevant organelles labelled in the cell section above. (2 marks)

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\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \textbf{Label} \rule[-1ex]{0pt}{0pt} & \textbf{Specific function in the adrenaline-secreting cell} \ \ \ \ \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{GA} \rule[-1ex]{0pt}{0pt} & \text{To modify a protein into adrenaline and/or package adrenaline into vesicles to be secreted from cell.} \\
\hline
\rule{0pt}{2.5ex} \text{M} \rule[-1ex]{0pt}{0pt} & \text{To create ATP molecules to be used as energy in the transport or synthesis of adrenaline.} \\
\hline
\end{array}

b. → The mitochondria (M) provides energy for vesicles (V) and secretory granules (SG) to form.

→ They then fuse with the plasma membrane (PM) and release their contents, in this case adrenaline, into the environment.

Show Worked Solution

b. → The mitochondria (M) provides energy for vesicles (V) and secretory granules (SG) to form.

→ They then fuse with the plasma membrane (PM) and release their contents, in this case adrenaline, into the environment.

BIOLOGY, M1 2020 VCE 40 MC

A student investigated the effect of the presence of four different molecules, R, S, T and U, on the rate of cellular respiration in human liver cells. The production of carbon dioxide by the cells was recorded over a five-minute interval. The final concentration of carbon dioxide was recorded. The data collected is shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Molecule present } \rule[-1ex]{0pt}{0pt} & \textbf{Concentration of carbon dioxide (ppm) after five minutes } \\
\hline
\rule{0pt}{2.5ex} \textbf{R} \rule[-1ex]{0pt}{0pt} & 400 \\
\hline
\rule{0pt}{2.5ex} \textbf{S} \rule[-1ex]{0pt}{0pt} & 800 \\
\hline
\rule{0pt}{2.5ex} \textbf{T} \rule[-1ex]{0pt}{0pt} & 600 \\
\hline
\rule{0pt}{2.5ex} \textbf{U} \rule[-1ex]{0pt}{0pt} & 1000 \\
\hline
\end{array}

The student presented the results as a graph.

Which one of the following graphs is the best representation of the results?

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$A$

Show Worked Solution

→ The data is quantitive but not continuous, therefore the column graph is the most appropriate way to represent the results.

$\Rightarrow A$

BIOLOGY, M1 2021 VCE 1

The diagram below shows a small part of a cross-section of the plasma membrane of a cell.

Some substances can move directly through the phospholipid bilayer.

Complete the table below by naming one small hydrophilic substance and one small hydrophobic substance that can move through the phospholipid bilayer. Provide an example of a situation when each small substance would move through the phospholipid bilayer. The first row has been completed for you. (2 marks)

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\begin{array} {|c|l|l|}
\hline \textbf{Name of small} & \textbf{Hydrophilic or} & \ \ \textbf{Situation when substance moves} \ \ \\
\textbf{substance} & \ \ \textbf{hydrophobic} & \ \ \ \ \ \textbf{through phospholipid bilayer} \\
\hline \text{oxygen} & \text{hydrophobic} & \text{Oxygen diffuses out of photosynthetic} \\
& & \text{plant.}\\
\hline & \text{hydrophilic} & \\
& & \\
\hline & \text{hydrophobic} & \\
& & \\
\hline \end{array}

Some very large substances and/or large particles that do not dissolve in the phospholipid bilayer can still move into or out of a cell.

Using one example, explain how the phospholipid bilayer transports these very large substances and/or large particles without the use of channel or carrier proteins either into a cell or out of a cell. (3 marks)

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\begin{array} {|c|l|l|}
\hline \textbf{Name of small} & \textbf{Hydrophilic or} & \ \ \textbf{Situation when substance moves} \ \ \\
\textbf{substance} & \ \ \textbf{hydrophobic} & \ \ \ \ \ \textbf{through phospholipid bilayer} \\
\hline \text{oxygen} & \text{hydrophobic} & \text{Oxygen diffuses out of photosynthetic} \\
& & \text{plant}\\
\hline \text{water} & \text{hydrophilic} & \text{Water diffuses out of a cell during} \\
& & \text{respiration. }\\
\hline \text{carbon dioxide} & \text{hydrophobic} & \text{Carbon dioxide diffuses out of a cell} \\
& & \text{during respiration. } \\
\hline \end{array}

b. Into a cell:

→ Endocytosis is a process which allows large particles such as proteins or antigens to enter a cell.

→ Endocytosis involves the folding over of a membrane to form a vesicle. The contents of the vesicle can then be digested by enzymes within the cell.

Out of a cell:

→ Exocytosis is a process which allows large particles such as proteins, antibodies or neurotransmitters to exit a cell.

→ Exocytosis involves the formation of a vesicle around the materials which need to exit the cell.

→ This vesicle then fuses with the plasma membrane, releasing the contents into the environment.

Show Worked Solution

\begin{array} {|c|l|l|}
\hline \textbf{Name of small} & \textbf{Hydrophilic or} & \ \ \textbf{Situation when substance moves} \ \ \\
\textbf{substance} & \ \ \textbf{hydrophobic} & \ \ \ \ \ \textbf{through phospholipid bilayer} \\
\hline \text{oxygen} & \text{hydrophobic} & \text{Oxygen diffuses out of photosynthetic} \\
& & \text{plant}\\
\hline \text{water} & \text{hydrophilic} & \text{Water diffuses out of a cell during} \\
& & \text{respiration. }\\
\hline \text{carbon dioxide} & \text{hydrophobic} & \text{Carbon dioxide diffuses out of a cell} \\
& & \text{during respiration. } \\
\hline \end{array}

b. Into a cell:

→ Endocytosis is a process which allows large particles such as proteins or antigens to enter a cell.

→ Endocytosis involves the folding over of a membrane to form a vesicle. The contents of the vesicle can then be digested by enzymes within the cell.

Out of a cell:

→ Exocytosis is a process which allows large particles such as proteins, antibodies or neurotransmitters to exit a cell.

→ Exocytosis involves the formation of a vesicle around the materials which need to exit the cell.

→ This vesicle then fuses with the plasma membrane, releasing the contents into the environment.

CHEMISTRY, M4 2022 VCE 2 MC

A fuel undergoes combustion to heat water.

Which of the following descriptions of the energy and enthalpy of combustion, $\Delta H$, of the reaction is correct?

	$\text{Energy}$	$\ \ \Delta H\ \ $
A.	absorbed by the water	negative
B.	released by the water	negative
C.	absorbed by the water	positive
D.	released by the water	positive

Show Answers Only

$A$

Show Worked Solution

→ Combustion is exothermic ($\Delta H$ is negative).

→ Energy released is absorbed by the water.

$\Rightarrow A$

CHEMISTRY, M4 2022 VCE 11*

The graphs shown below are energy profiles for the following reaction.

$\ce{A + B\leftrightharpoons C}$ $\quad \quad \Delta H < 0$

The graphs represent the forward reaction, with and without a catalyst, and the reverse reaction, with and without a catalyst. All graphs are drawn to the same scale.

Which energy profile represents the reverse reaction without a catalyst? Give reasons for your answer. (2 marks)

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→ Since forward reaction is exothermic, reverse reaction is endothermic where energy of reactants < energy of products (eliminate Graphs 3 and 4).

→ A catalyst lowers the activation energy required. Graph 1 has a lower activation energy than Graph 2.

→ Therefore, Graph 2 represents the energy profile of the reverse reaction without a catalyst.

Show Worked Solution

→ Since forward reaction is exothermic, reverse reaction is endothermic where energy of reactants < energy of products (eliminate Graphs 3 and 4).

→ A catalyst lowers the activation energy required. Graph 1 has a lower activation energy than Graph 2.

→ Therefore, Graph 2 represents the energy profile of the reverse reaction without a catalyst.

PHYSICS, M3 2017 VCE 14 MC

A teacher stands in the corridor at a short distance from the open door of her classroom, as shown in the diagram below. She can hear her students. but cannot see them.

Which one of the following best explains why the teacher can hear her students?

The speed of sound is much greater than the speed of light.
The speed of sound is comparable with the speed of light.
Sound diffracts because the wavelength of sound is much smaller than the width of the door.
Sound diffracts because the wavelength of sound is comparable with the width of the door.

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$D$

Show Worked Solution

→ Sound waves can range anywhere from 20mm to 17m and thus the some sounds made by the students would have a wavelength of 1 metre.

→ Therefore, sound will diffract through the doorway as it is the same order of magnitude of the size of the door whereas light will not as its wavelengths are much smaller in size.

$\Rightarrow D$

PHYSICS, M4 2017 VCE 1 MC

A group of students is considering how to create a magnetic monopole. Which one of the following is correct?

Break a bar magnet in half.
Pass a current through a long solenoid.
Pass a current through a circular loop of wire.
It is not known how to create a magnetic monopole.

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$D$

Show Worked Solution

→ There is currently no evidence that a magnetic monopole exists.

$\Rightarrow D$

PHYSICS, M3 2018 VCE 12

Optical fibres are constructed using transparent materials with different refractive indices.

Figure 14 shows one type of optical fibre that has a cylindrical core and surrounding cladding. Laser light of wavelength 565 nm is shone from air into the optical fibre ($v=3 \times 10^8$).

Calculate the frequency of the laser light before it enters the optical fibre. (1 mark)

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Calculate the critical angle for the laser light at the cladding-core boundary. Show your working. (2 marks)

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Calculate the speed of the laser light once it enters the core of the optical fibre. Give your answer correct to three significant figures. Show your working. (2 marks)
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a. $f=5.31 \times 10^{14}\ \text{Hz}$

b. $\theta_c=60.3^{\circ}$

c. $v_{\text{x}}=1.80 \times 10^8\ \text{ms}^{-1}$

Show Worked Solution

a.	$f$	$=\dfrac{v}{\lambda}$
		$=\dfrac{3\times 10^8}{565 \times 10^{-9}}$
		$=5.31 \times 10^{14}\ \text{Hz}$

b.	$\theta_c$	$=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)}$
		$=\sin^{-1} \Big{(}\dfrac{1.45}{1.67} \Big{)}$
		$=60.3^{\circ}$

c.	$v_{\text{x}}$	$=\dfrac{c}{n_{\text{x}}}$
		$=\dfrac{3 \times 10^8}{1.67}$
		$=1.80 \times 10^8\ \text{ms}^{-1}$

♦ Mean mark 50%.

PHYSICS, M2 2018 VCE 8

Two blocks, $\text{A}$ of mass 4.0 kg and $\text{B}$ of mass 1.0 kg, are being pushed to the right on a smooth, frictionless surface by a 40 N force, as shown in Figure 10.

Calculate the magnitude of the force on block $\text{B}$ by block $\text{A}$ ($\left.F_{\text {on B by A}}\right)$. Show your working. (2 marks)

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State the magnitude and the direction of the force on block $\text{A}$ by block $\text{B}$ ($\left.F_{\text {on A by B}}\right)$. (2 marks)

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a. $8\ \text{N}$

b. $8\ \text{N}$ to the left.

Show Worked Solution

a. → Using $F=ma$, calculate the acceleration of the entire system:

$a=\dfrac{F}{m}=\dfrac{40}{5}=8\ \text{ms}^{-2}$

→ $F_{\text {on B by A }}$	$=ma$
	$=1 \times 8$
	$=8\ \text{N}$

♦♦ Mean mark (a) 36%.

b. Newton’s third law of motion:

→ $F_{\text {on A by B }}$ will be equal in magnitude and opposite in direction.

→ $F_{\text {on A by B }}= 8\ \text{N}$ to the left.

PHYSICS, M3 2018 VCE 10 MC

A loudspeaker is producing a sound wave of constant frequency. Consider a tiny dust particle 1.0 m in front of the loudspeaker.

Which one of the following diagrams best describes the motion of the dust particle?

Show Answers Only

$D$

Show Worked Solution

→ As a sound wave is a longitudinal wave, the air and dust molecules through which the sound wave passes vibrate parallel to the direction of energy transfer.

→ As the sound is travelling left to right (horizontally), the dust particle will vibrate forwards and backwards in the horizontal plane.

$\Rightarrow D$

PHYSICS, M2 2018 VCE 8-9 MC

A railway truck $\text{X}$ of mass 10 tonnes, moving at 6.0 m s$^{-1}$, collides with a stationary railway truck $\text{Y}$ of mass 5.0 tonnes. After the collision the trucks are joined together and move off as one. The situation is shown below.

$\text{Question 8}$

The final speed of the joined railway trucks after the collision is closest to

$ 2.0 \text{ m s}^{-1}$
$ 3.0 \text{ m s} ^{-1}$
$ 4.0 \text{ m s} ^{-1}$
$ 6.0 \text{ m s} ^{-1}$

$\text{Question 9}$

The collision of the railway trucks is best described as one where

kinetic energy is conserved but momentum is not conserved.
kinetic energy is not conserved but momentum is conserved.
neither kinetic energy nor momentum is conserved.
both kinetic energy and momentum are conserved.

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$\text{Question 8:}\ C$

$\text{Question 9:}\ B$

Show Worked Solution

$\text{Question 8}$

→ By the law of conservation of momentum:

$m_Xu_X+m_Yu_Y$	$=m_Xv_X+m_Yv_Y$
	$=v(m_X + m_Y)\ \ \ (v_X=v_Y) $
$10\ 000 \times 6 +0$	$= v(10\ 000 + 5000)$
$60\ 000$	$=15\ 000v$
$v$	$=4\ \text{ms}^{-1}$

$ \Rightarrow C$

$\text{Question 9}$

→ By the law of conservation of momentum, momentum in the collision is conserved.

→ Kinetic energy conservation:

	$KE_{\text{init}}$	$=\dfrac{1}{2}m_Xu_X^2+\dfrac{1}{2}m_Yu_Y^2$
		$=\dfrac{1}{2} \times 10\ 000 \times 6^2 + \dfrac{1}{2} \times 5000 \times 0^2$
		$=180\ 000\ \text{J}$

$KE_{\text{final}}$	$=\dfrac{1}{2}(m_X+m_Y)v^2$
	$=\dfrac{1}{2} \times 15\ 000 \times 4^2$
	$=120\ 000\ \text{J}$

→ The kinetic energy of the system decreases after the collision and so is not conserved.

$\Rightarrow B$

PHYSICS, M2 2018 VCE 6 MC

Lisa is driving a car of mass 1000 kg at 20 ms$ ^{-1}$ when she sees a dog in the middle of the road ahead of her. She takes 0.50 s to react and then brakes to a stop with a constant braking force. Her speed is shown in the graph below. Lisa stops before she hits the dog.

Which one of the following is closest to the magnitude of the braking force acting on Lisa's car during her braking time?

$6.7 \text{ N}$
$6.7 \text{ kN}$
$8.0 \text{ kN}$
$20.0 \text{ kN}$

Show Answers Only

$C$

Show Worked Solution

→ The deceleration of the car from $0.5\ \text{s}$ to $3.0\ \text{s}$ is:

$a=\dfrac{\Delta v}{\Delta t}= \dfrac{20}{2.5}=8\ \text{ms}^{-2}$

→ Thus the braking force on the car can be calculated:

$F = ma = 1000 \times 8 = 8000\ \text{N} = 8.0\ \text{kN} $

$\Rightarrow C$

PHYSICS, M1 2018 VCE 5*

Four students are pulling on ropes in a four-person tug of war. The relative sizes of the forces acting on the various ropes are $F_{ W }=200 \text{ N} , F_{ X }=240 \text{ N} , F_{ Y }=180 \text{ N}$ and $F_{ Z }=210 \text{ N}$. The situation is shown in the diagram below.

What is the resultant force vector $(F_{\vec{R}})$ acting at the centre of the tug-of-war ropes? (3 marks)

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→ $F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}$ above the horizontal.

Show Worked Solution

→ By resolving the vertical vectors of $F_{ W }=200 \text{ N}$ up and $F_{ Y }=180 \text{ N}$ down, the net force in the vertical direction is $F_v=20\ \text{N}$ up.

→ Similarly, the net force in the horizontal direction is $F_h=30\ \text{N}$ to the right.

→ Thus the resultant vector $(F_{\vec{R}})$ can be calculated using the vector diagram below.

→ The magnitude of $F_{\vec{R}}=\sqrt{20^2+30^2}$ and $\theta=\tan^{-1}\Big{(}\dfrac{20}{30}\Big{)}$

→ Hence $F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}$ above the horizontal.

PHYSICS, M3 2019 VCE 15

A student sets up an experiment involving a source of white light, a glass prism and a screen. The path of a single ray of white light when it travels through the prism and onto the screen is shown in Figure 14.

A spectrum of colours is observed by the student on the screen, which is positioned to the right of the prism.

Name and explain the effect observed by the student. (3 marks)

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Points $\text{X}$ and $\text{Y}$ on Figure 14 represent either end of the visible spectrum observed by the student.
Identify the two visible colours observed at point $\text{X}$ and at point $\text{Y}$. (1 mark)

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a. → The observed effect is dispersion.

→ As light enters the glass it slows down as it is entering a denser medium.

→ As the refractive index for different wavelengths of light differs, the angle at which individual wavelengths refract differs slightly.

→ This causes the white light to split up as each wavelength refracts differently through the glass resulting in a rainbow spectrum on the screen.

b. Point $X$ is red.

Point $Y$ is blue\purple.

Show Worked Solution

a. → The observed effect is dispersion.

→ As light enters the glass it slows down as it is entering a denser medium.

→ As the refractive index for different wavelengths of light differs, the angle at which individual wavelengths refract differs slightly.

→ This causes the white light to split up as each wavelength refracts at different angles through the glass resulting in a rainbow spectrum on the screen.

♦ Mean mark 53%.

b. Point $X$ is red.

Point $Y$ is blue\purple.

BIOLOGY, M4 2021 VCE 23 MC

Mass extinction events

allow new species to evolve to occupy available niches.
allow all species that survive to recover and diverge.
result from a constant global climate.
affect all species equally.

Show Answers Only

$A$

Show Worked Solution

→ When a mass extinction event occurs, it allows the remaining species to occupy the previously inhabited niches.

$\Rightarrow A$

CHEMISTRY, M4 EQ-Bank 5

The following reaction represents the conversion of diamond to graphite:

$\ce{2C_{diamond} \rightarrow 2C_{graphite}}$

- $\ce{\Delta $H$_{f}\ C_{diamond} = 1.9 kJ mol^{-1}}$
- $\ce{\Delta $S$_{f}\ C_{diamond} = 2.38 J mol^{-1} K^{-1}}$
- $\ce{\Delta $S$_{f}\ C_{graphite} = 5.74 J mol^{-1} K^{-1}}$

Determine $\Delta G$ at 298K and state whether the reaction is spontaneous or not. (3 marks)

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What does $\Delta G$ indicate about the rate of reaction? (1 mark)

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i. $\Delta G = -5.8025\ \text{kJ}$

→ Reaction is spontaneous.

ii. Rate of reaction and $\Delta G$:

→ Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year

Show Worked Solution

→ Standard enthalpy and entropy of elements in their natural state is 0.

$\Delta H$	$= \Sigma H_{\text{products}}-\Sigma H_{\text{reactants}}$
	$= (2 \times 0)-(2 \times 1.9)$
	$=-3.8\ \text{kJ mol}^{-1} $

$\Delta S$	$=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}$
	$= (2 \times 5.74)-(2 \times 2.38)$
	$= 11.48-4.76 $
	$=6.72\ \text{J mol}^{-1}\ \text{K}^{-1}$

$\Delta G$	$= \Delta H-T\Delta S$
	$= -3.8-(298 \times 0.00672)$
	$= -5.8025\ \text{kJ}$

→ The reaction is spontaneous as $\Delta G < 0$.

ii. Rate of reaction and $\Delta G$:

→ Gibbs Free Energy doesn’t indicate anything about the kinetics of the reaction. While the conversion of diamond to graphite is spontaneous, it occurs over millions of year

CHEMISTRY, M4 EQ-Bank 4

Of the following state changes, explain which represents the smallest increase in entropy and which represents the largest. (3 marks)

1. Sublimation of ice to gas
2. Melting of ice to liquid water
3. Freezing of water to ice

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→ State change (3) has the smallest entropy increase as solids have less entropy than liquids and the freezing of water indicates a decrease in entropy as order increases.

→ State change (2) has a larger increase in entropy than state change (3) as water has more entropy than the solid ice it is formed from.

→ State change (1) has the largest entropy increase as gas has the most entropy compared to solids or liquids.

Show Worked Solution

→ State change (3) has the smallest entropy increase as solids have less entropy than liquids and the freezing of water indicates a decrease in entropy as order increases.

→ State change (2) has a larger increase in entropy than state change (3) as water has more entropy than the solid ice it is formed from.

→ State change (1) has the largest entropy increase as gas has the most entropy compared to solids or liquids.

CHEMISTRY, M4 EQ-Bank 3

A 3.1g sample of $\ce{CaCO3_{(s)}}$ decomposes into $\ce{CaO_{(s)}}$ and $\ce{CO2_{(g)}}$. Entropy values for these chemicals are given below and the molar enthalpy for the reaction is 360 kJ/mol.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Substance}\rule[-1ex]{0pt}{0pt} & \text{Standard Entropy}\ (\Delta S) \\
\hline
\rule{0pt}{2.5ex}\ce{CaCO3}\rule[-1ex]{0pt}{0pt} & \text{92.88 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CaO(s)}\rule[-1ex]{0pt}{0pt} & \text{39.75 J/K} \\
\hline
\rule{0pt}{2.5ex}\ce{CO2(g)}\rule[-1ex]{0pt}{0pt} & \text{213.6 J/K} \\
\hline
\end{array}

Write a chemical equation for the above reaction. (1 mark)

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Calculate the entropy change for this reaction. (2 marks)

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Determine whether the reaction is spontaneous or non-spontaneous at room temperature. (2 marks)

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$\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}$
$\ce{4.97 J K^{-1}}$
$\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}$

Show Worked Solution

i. $\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)}$

ii.	$\Delta S$	$=\Sigma S_{\text{products}}-\Sigma S_{\text{reactants}}$
		$= 213.6 + 39.75-92.88$
		$= 160.47\ \text{J mol}^{-1}\ \text{K}^{-1}$

$\ce{n(CaCO3)}= \dfrac{\text{m}}{\text{MM}} = \dfrac{3.1}{100.09} = 0.03097\ \text{mol} $

$\text{Entropy change}\ = 160.47 \times 0.03097 = 4.97\ \text{J K}^{-1}$

iii. $\text{Room Temperature = 298.15 K}$

$\Delta G$	$=\Delta H-T \Delta S$
	$=360-(298.15 \times 0.16047) $
	$= 312.179\ \text{kJ}$

$\text{Since}\ \Delta G > 0,\ \text{the reaction is not spontaneous.}$

CHEMISTRY, M4 EQ-Bank 2

In the following reactions, predict whether entropy will increase or decrease, giving reasons. (3 marks)

$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}$ (1 mark)
$\ce{CaCO3(s) \rightleftharpoons CaO(g) + CO2(g)}$ (1 mark)
$\ce{2HCl(aq) + Zn(s) \rightleftharpoons ZnCl2(aq) + H2(g)}$ (1 mark)

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i. → Decrease

→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.

ii. → Increase

→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.

iii. → Increase

→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.

Show Worked Solution

i. → Decrease

→ 4 moles of gas becomes 2 moles of gas on the product side, causing the reaction to become more ordered and decreasing in entropy.

ii. → Increase

→ A solid decomposes into 2 moles of gas. This reaction becomes more disordered, and the phase change involves an increase in entropy.

iii. → Increase

→ An aqueous and solid reactant becomes an aqueous and gaseous product. The phase changes overall involve an increase in entropy.

CHEMISTRY, M3 2012 VCE 20*

Consider the following reaction.

$\ce{IO3–(aq) + 5I–(aq) + 6H+(aq) \rightarrow 3I2(s) + 3H2O(l)}$

Determine the half equation for the reduction reaction. (3 marks)

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$\ce{2IO3–(aq) + 12H+(aq) + 10e– → I2(s) + 6H2O(l)}$

Show Worked Solution

→ The oxidation number of $\ce{I}$ changes from +5 in $\ce{IO3^{-}}$ to 0 in $\ce{I2}$ (reduced), and from – 1 in $\ce{I^{-}}$ to 0 in $\ce{I2}$ (oxidised).

→ Reduction equation (electrons on LHS) is:

$\ce{2IO3–(aq) + 12H+(aq) + 10e– → I2(s) + 6H2O(l)}$

CHEMISTRY, M3 2013 VCE 14-15 MC

$\ce{Cu(s) + 4HNO3(aq)\rightarrow Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)}$

$\text{Question 14}$

Which one of the following will not increase the rate of the above reaction?

decreasing the size of the solid copper particles
increasing the temperature of $\ce{HNO3}$ by 20 °C
increasing the concentration of $\ce{HNO3}$
allowing $\ce{NO2}$ gas to escape

$\text{Question 15}$

In the above reaction, the number of successful collisions per second is a small fraction of the total number of collisions.

The major reason for this is that

the nitric acid is ionised in solution.
some reactant particles have too much kinetic energy.
the kinetic energy of the particles is reduced when they collide with the container’s walls.
not all reactant particles have the minimum kinetic energy required to initiate the reaction.

Show Answers Only

$\text{Question 14:}\ D$

$\text{Question 15:}\ D$

Show Worked Solution

$\text{Question 14}$

→ Options $A, B$ and $C$ will all increase the rate of the given chemical reaction.

$\Rightarrow D$

$\text{Question 15}$

→ Successful collisions occur only if the particles involved have at least the minimum kinetic energy required and the correct orientation.

$\Rightarrow D$

CHEMISTRY, M3 2014 VCE 10 MC

Which one of the reactions of hydrochloric acid below is a redox reaction?

$\ce{2HCl(aq) + Fe(s)\rightarrow H2(g) + FeCl2(aq)}$
$\ce{2HCl(aq) + Na2S(s)\rightarrow H2S(g) + 2NaCl(aq)}$
$\ce{2HCl(aq) + MgO(s)\rightarrow MgCl2(aq) + H2O(l)}$
$\ce{2HCl(aq) + K2CO3(s)\rightarrow CO2(g) + 2KCl(aq) + H2O(l)}$

Show Answers Only

$A$

Show Worked Solution

→ $\ce{HCl}$ is the common reactant in all options.

→ Therefore, redox reaction will see the oxidation number of $\ce{H}$ or $\ce{Cl}$ change.

→ In option $A$, the oxidation number of $\ce{H}$ decreases from +1 to 0, while the oxidation number of $\ce{Fe}$ increases from 0 to +2.

$\Rightarrow A$

CHEMISTRY, M3 2015 VCE 6 MC

In which one of the following compounds is sulfur in its lowest oxidation state?

$\ce{SO3}$
$\ce{HSO4–}$
$\ce{SO2}$
$\ce{Al2S3}$

Show Answers Only

$D$

Show Worked Solution

Let $x$ equal the oxidation of sulfur in each of the following calculations.

In $\ce{SO3}$:

$x+ 3 \times -2$	$=0$
$x$	$=6$

In $\ce{HSO4–}$:

$1 + x+ 4 \times -2$	$=-1$
$x$	$=6$

In $\ce{SO2}$:

$x+ 2 \times -2$	$=0$
$x$	$=4$

In $\ce{Al2S3}$:

$2 \times 3+ 3 \times x$	$=0$
$3x$	$=-6$
$x$	$=-2$

→ $D$ has the lowest oxidation number showing it is in the lowest oxidation state.

$\Rightarrow D$

PHYSICS, M2 2019 VCE 9

A proton in an accelerator detector collides head-on with a stationary alpha particle, as shown in Figure 9a and Figure 9b. After the collision, the alpha particle travels at a speed of $4.0 \times 10^6 \text{ m s}^{-1}$. The proton rebounds at $6.0 \times 10^6 \text{ m s}^{-1}$.

Find the speed of the proton before the collision, modelling the mass of the alpha particle, $4m$, to be equal to four times the mass of the proton, $m$. Show your working. Ignore relativistic effects. (3 marks)

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$u_p=1 \times 10^7\ \text{ms}^{-1}$

Show Worked Solution

$mu_p+4mu_a$	$=mv_p+4mv_a$
$mu_p +0$	$=m \times -6 \times 10^6 + 4m \times 4 \times 10^6$
$mu_p $	$=16m \times 10^6-6m \times 10^{6}$
$mu_p$	$=10m \times 10^6$
$u_p$	$=1 \times 10^7\ \text{ms}^{-1}$

PHYSICS, M4 2020 VCE 18a

Students are modelling the effect of the resistance of electrical cables, $r$, on the transmission of electrical power. They model the cables using the circuit shown in Figure 18.

The 24 V DC power supply models the mains power.

Describe the effect of increasing the resistance of the electrical cables, $r$, on the brightness of the constant resistance globe, $R$. (2 marks)

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Increasing the resistance of the electrical cables:

→ Increases the power used by the electrical cables $P=IV$. This results in less power in the circuit for the resistance light globe.

→ Therefore, the brightness of the resistance light globe will decrease.

Show Worked Solution

Increasing the resistance of the electrical cables:

→ Increases the power used by the electrical cables $P=IV$. This results in less power in the circuit for the resistance light globe.

→ Therefore, the brightness of the resistance light globe will decrease.

PHYSICS, M2 2019 VCE 11 MC

An ultralight aeroplane of mass 500 kg flies in a horizontal straight line at a constant speed of 100 ms$^{-1}$.

The horizontal resistance force acting on the aeroplane is 1500 N.

Which one of the following best describes the magnitude of the forward horizontal thrust on the aeroplane?

1500 N
slightly less than 1500 N
slightly more than 1500 N
5000 N

Show Answers Only

$A$

Show Worked Solution

→ There is no net force acting on the aeroplane as it is travelling at a constant speed.

→ The forward horizontal thrust on the plane must be opposite in direction but equal in magnitude to the horizontal resistance force acting on the plane.

$\Rightarrow A$

CHEMISTRY, M2 2013 VCE 8a

In an experiment, 5.85 g of ethanol was ignited with 14.2 g of oxygen.

Write an equation for the complete combustion of ethanol. (1 mark)

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Which reagent is in excess? Calculate the amount, in moles, of the reagent identified as being in excess. (3 marks)

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i. $\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }$

Show Worked Solution

i. $\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }$

♦ Mean mark 43%.

ii. $\ce{MM(C2H5OH) = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\ \text{g mol}^{-1}}$

$\ce{n(C2H5OH) = \dfrac{5.85}{46.0} = 0.1272\ \text{mol}} $

$\ce{n(O2) = \dfrac{14.2}{32.0} = 0.444\ \text{mol}}$

$\text{Reaction ratio}\ \ \ce{C2H5OH : O2 = 1:3}$

$\Rightarrow \ce{n(O2)_{\text{required}} = 3 \times 0.1272 = 0.382\ \text{mol}\ \ (\ce{O2}\ \text{excess}) }$

$\ce{O2_{\text{(excess)}} = 0.444-0.382 = 0.062\ \text{mol}}$

CHEMISTRY, M2 2013 VCE 1 MC

Consider the following.

"Calculate the pressure exerted by 6.9 g of argon in a 0.07500 L container at 11.5 °C."

The number of significant figures that should be expressed in the answer is

Show Answers Only

$A$

Show Worked Solution

→ The data input for calculating the pressure with the least number of significant figures is 6.9 grams (to 2 sig fig).

→ The answer should then be expressed to 2 significant figures.

$\Rightarrow A$

CHEMISTRY, M2 2015 VCE 3 MC

In an experiment, 0.051 mol of sodium hydroxide, $\ce{NaOH}$, reacted completely with 0.017 mol of citric acid, $\ce{C6H8O7}$.

Which one of the following equations correctly represents the reaction between citric acid and the sodium hydroxide solution?

$\ce{NaOH(aq) + C6H8O7(aq)\rightarrow NaC6H7O7(aq) + H2O(l)}$
$\ce{2NaOH(aq) + C6H8O7(aq)\rightarrow Na2C6H6O7(aq) + 2H2O(l)}$
$\ce{3NaOH(aq) + C6H8O7(aq)\rightarrow Na3C6H5O7(aq) + 3H2O(l)}$
$\ce{4NaOH(aq) + C6H8O7(aq)\rightarrow Na4C6H4O7(aq) + 4H2O(l)}$

Show Answers Only

$C$

Show Worked Solution

$\text{Ratio of moles reacting}\ = 0.051 : 0.017 = 3:1 $

$\ce{n(NaOH) : n(C6H8O7) = 3:1} $

$\ce{3NaOH(aq) + C6H8O7(aq)\rightarrow Na3C6H5O7(aq) + 3H2O(l)}$

$\Rightarrow C$

CHEMISTRY, M2 2012 VCE 18 MC

2.1 g of an alkene that contains only one double bond per molecule reacted completely with 8.0 g of bromine, $\ce{Br2}$. The molar mass of bromine, $\ce{Br2}$, is 160 g mol$^{–1}$.

Which one of the following is the molecular formula of the alkene?

$\ce{C5H10}$
$\ce{C4H8}$
$\ce{C3H6}$
$\ce{C2H4}$

Show Answers Only

$C$

Show Worked Solution

→ Since each molecule only contains a single $\ce{C=C}$ bond:

$\ce{n(alkene) = n(Br2) = \dfrac{8.0}{160.0} = 0.05\ \text{mol}}$

$\ce{m(alkene) = \dfrac{2.1}{0.05} = 42\ \text{g mol}^{-1}} $

$\Rightarrow C$

CHEMISTRY, M2 2012 VCE 15*

A sample of the anticancer drug Taxol$^{\circledR}$, $\ce{ C47H51NO14}$, contains 0.157 g of carbon.

Calculate the mass, in grams, of oxygen in the sample. Give your answer correct to three decimal places. (2 marks)

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$\text{0.062 grams}$

Show Worked Solution

$\ce{n(C) = \dfrac{0.157}{12.0} = 0.0131\ \text{mol}} $

$\ce{n(\text{Taxol}) = \dfrac{0.0131}{47}\ \text{mol}} $

$\ce{n(O) = 14 \times n(\text{Taxol}) = 14 \times \dfrac{0.0131}{47} = 0.00390\ \text{mol}} $

$ \therefore \ce{m(O) = 0.00390 \times 16.0 = 0.062\ \text{g (3 d.p.)}}$

PHYSICS, M3 2019 VCE 10*

The horizontal face of a glass block is covered with a film of liquid, as shown below.

A monochromatic light ray is incident on the glass-liquid boundary with an angle of incidence of 62.0°.

Calculate the minimum value of the liquid's refractive index, so that some light will just cross the interface into the liquid. (2 marks)

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$1.55$

Show Worked Solution

$\sin\theta_c$	$=\dfrac{n_2}{n_1}$
$n_2$	$=\sin\theta_c \times n_1$
	$= \sin62^{\circ} \times 1.75$
	$=1.55$

PHYSICS, M3 2019 VCE 9 MC

A monochromatic light ray passes through three different media, as shown in the diagram below.

Assume that $v_1$ is the speed of light in Medium 1, $v_2$ is the speed of light in Medium 2 and $v_3$ is the speed of light in Medium 3.

Which one of the following would best represent the relative speeds in the media?

$v_1>v_2>v_3$
$v_1>v_3>v_2$
$v_3>v_2>v_1$
$v_3>v_1>v_2$

Show Answers Only

$D$

Show Worked Solution

→ From Medium 1 to Medium 2, the light bends towards the normal, this means the light is entering a denser medium and so the light will slow down ($v_1>v_2$).

→ From Medium 2 to Medium 3, the light bends away from the normal, this means the light is entering a less dense medium and so the light will speed up. Hence ($v_2<v_3$).

→ As the angle of the light as it enters medium 3 is greater than the original angle of incidence. Medium three must be less dense than Medium 1, hence the light would travel faster in medium 3, $v_3>v_1$.

$\Rightarrow D$

PHYSICS, M4 2019 VCE 3*

Three charges $(- \text{Q} ,+2 \text{Q} ,-2 \text{Q})$ are placed at the vertices of an isosceles triangle, as shown below.

Draw a force vector diagram to show the direction of the net force on the charge $- \text{Q}$ ? (2 marks)

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Show Worked Solution

→ $- \text{Q}$ will be attracted to $+2 \text{Q}$ charge and repelled from the $-2 \text{Q}$ charge.

→ This results in a force vector diagram as seen below.

BIOLOGY, M1 2022 VCE 4a

Increased protein consumption is a global trend. Chicken eggs are a good source of protein.

Ovalbumin (egg white protein) is formed from chicken cells during egg production.

Describe the role of organelles in the export of ovalbumin from a chicken's cells into an egg. (3 marks)

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→ The production, transport and exportation of ovalbumin requires energy, which is supplied via mitochondria within the cell.

→ After ovalbumin is made, the rough endoplasmic reticulum transports it to the Golgi, where it is packaged into vesicles.

→ The vesicles then fuse with the plasma membrane of the cell and the ovalbumin can be released from the cell via exocytosis into an egg.

Show Worked Solution

→ The production, transport and exportation of ovalbumin requires energy, which is supplied via mitochondria within the cell.

→ After ovalbumin is made, the rough endoplasmic reticulum transports it to the Golgi, where it is packaged into vesicles.

→ The vesicles then fuse with the plasma membrane of the cell and the ovalbumin can be released from the cell via exocytosis into an egg.

BIOLOGY, M3 2022 VCE 34 MC

Consider a hypothetical animal species, Species X, that can neither fly nor swim. Separate populations of Species X live in the same geographical area and often interbreed. After a period of time, a new species, Species Z, arises from Species X. The following list describes some barriers that may have contributed to the formation of Species Z:

1. A newly formed river separated one population of Species X from the other populations.
2. One population of Species X developed a new mating ritual that was not recognised by members of the other populations of Species X.
3. One population of Species X began to breed several weeks later than the other populations of Species X.
4. A volcanic eruption created a mountain that separated one population of Species X from the other populations of Species X.

The barrier(s) that could have led to the formation of Species Z by sympatric speciation is

either barrier 1 or barrier 4 .
either barrier 2 or barrier 3 .
barrier 1 only.
barrier 2 only.

Show Answers Only

$B$

Show Worked Solution

By Elimination

→ Sympatric speciation is the evolution of a new species (Species Z) from an ancestral species (Species X) while both continue to inhabit the same geographical location.

→ Therefore the physical separation by barrier 1 does not fit the definition (Eliminate A and C).

→ Both barrier 2 and 3 are barriers which could lead to the formation of Species Z, while both Species Z and X can continue to occupy the same area.

$\Rightarrow B$

PHYSICS, M3 2020 VCE 13

A 0.8 m long guitar string is set vibrating at a frequency of 250 Hz. The standing wave envelope created in the guitar string is shown in Figure 12.

Calculate the speed of the wave in the guitar string. (2 marks)

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The frequency of the vibration in the guitar string is tripled to 750 Hz.
On the guitar string below, draw the shape of the standing wave envelope now created. (2 marks)

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a. $v=400\ \text{ms}^{-1}$

Show Worked Solution

a. $v$	$=f\lambda$
	$=250 \times 1.6 $
	$=400\ \text{ms}^{-1}$

→ Tripling the frequency means decreasing the wavelength by a factor of 3, hence three half wavelengths will now fit on the string.

PHYSICS, M2 2020 VCE 10

Jacinda designs a computer simulation program as part of her practical investigation into the physics of vehicle collisions. She simulates colliding a car of mass 1200 kg, moving at 10 ms$^{-1}$, into a stationary van of mass 2200 kg. After the collision, the van moves to the right at 6.5 ms$^{-1}$. This situation is shown in Figure 10.

Calculate the speed of the car after the collision and indicate the direction it would be travelling in. Show your working. (4 marks)

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Explain, using appropriate physics, why this collision represents an example of either an elastic or an inelastic collision. (3 marks)

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The collision between the car and the van takes 40 × 10$^{-3}$ seconds.

1. Calculate the magnitude and indicate the direction of the average force on the van by the car. (3 marks)
  
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2. Calculate the magnitude and indicate the direction of the average force on the car by the van. (2 marks)
  
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a. $v_{\text{car}} =1.92\ \text{ms}^{-1}$ to the left.

b. The collision is an inelastic collision as the kinetic energy decreases after the collision.

c.i The average force of $358\ \text{kN}$ is to right as the change in the momentum is also to the right.

c.ii The average force of $358\ \text{kN}$ is to the left as the change in momentum is also to the left.

Show Worked Solution

a. Using the Conservation of Momentum:

$m_{\text{car}}u_{\text{car}}+m_{\text{van}}u_{\text{van}}$	$=m_{\text{car}}v_{\text{car}}+m_{\text{van}}v_{\text{van}}$
$1200 \times 10 + 2200 \times 0$	$=1200 \times v_{\text{car}} + 2200 \times 6.5$
$1200v_{\text{car}}$	$=12\ 000-14\ 300$
	$=-2300$
$v_{\text{car}}$	$=-1.92\ \text{ms}^{-1}$
	$=1.92\ \text{ms}^{-1}\ \text{to the left}$

b.	$KE_{\text{init}}$	$=\dfrac{1}{2}m_{\text{car}}u_{\text{car}}^2+\dfrac{1}{2}m_{\text{van}}u_{\text{van}}^2$
		$=\dfrac{1}{2} \times 1200 \times 10^2 + \dfrac{1}{2} \times 2200 \times 0^2$
		$=60\ 000\ \text{J}$

$KE_{\text{final}}$	$=\dfrac{1}{2}m_{\text{car}}v_{\text{car}}^2+\dfrac{1}{2}m_{\text{van}}v_{\text{van}}^2$
	$=\dfrac{1}{2} \times 1200 \times 1.92^2 + \dfrac{1}{2} \times 2200 \times 6.5^2$
	$=48\ 687\ \text{J}$

→ As the kinetic energy of the system decreases after the collision, the collision is inelastic.

c.i. $\Delta p$	$=F_{net}\Delta t$
$F_{net}$	$=\dfrac{\Delta p}{ \Delta t}$
	$= \dfrac{m\Delta v}{\Delta t}$
	$= \dfrac{2200 \times (6.5-0)}{40 \times 10^{-3}}$
	$=357\ 500\ \text{N}$
	$=358\ \text{kN to the right}$

→ The average force of $358\ \text{kN}$ is to right as the change in the momentum is also to the right.

c.ii. $F_{net}$	$=\dfrac{\Delta p}{ \Delta t}$
	$= \dfrac{m\Delta v}{\Delta t}$
	$= \dfrac{1200 \times (-1.92-10)}{40 \times 10^{-3}}$
	$=-357\ 600\ \text{N}$
	$=-358\ \text{kN}$

→ The average force of $358\ \text{kN}$ is to the left as the change in momentum is also to the left.

→ Note: no calculation was required for this question as it is an example of Newton’s third law of motion. Simply stating that the force would be equal in magnitude but opposite in direction earned full marks.

PHYSICS, M3 2020 VCE 14 MC

Students are investigating the diffraction of waves using a ripple tank. Water waves are directed towards barriers with gaps of different sizes, as shown below.

In which one of the following would the greatest diffraction effects be observed?

Show Answers Only

$B$

Show Worked Solution

→ The greatest diffraction patterns are observed when the wavelength is the same size as the slit through which the wave passes.

$ \Rightarrow B$

PHYSICS, M4 2020 VCE 1 MC

The diagram below shows the electric field lines between two charges of equal magnitude.

The best description of the two charges is that the

charges are both positive.
charges are both negative.
charges can be either both positive or both negative.
left-hand charge is positive and the right-hand charge is negative.

Show Answers Only

$A$

Show Worked Solution

→ The field lines show the charges are experiencing a repulsion force hence they must be the same charge.

→ The direction of the field lines are away from the charges so they must be positive charges.

$\Rightarrow A$

PHYSICS, M3 2021 VCE 12

A Physics teacher is conducting a demonstration involving the transmission of light within an optical fibre. The optical fibre consists of an inner transparent core with a refractive index of 1.46 and an outer transparent cladding with a refractive index of 1.42. A single monochromatic light ray is incident on the optical fibre, as shown in Figure 12.

Determine the angle of incidence, $\theta$, at the air-core boundary. Show your working. (2 marks)

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Will any of the initial light ray be transmitted into the cladding? Explain your answer and show any supporting working. (3 marks)

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a. $\theta=51^{\circ}$

b. Find the critical angle for Total Internal Reflection (TIR):

$\sin\theta_c$	$=\dfrac{n_2}{n_1}$
$\theta_c$	$=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)} $
	$=\sin^{-1} \Big{(}\dfrac{1.42}{1.46} \Big{)}$
	$=76.6^{\circ}$

→ The angle of incidence is $90^{\circ}-32^{\circ}=58^{\circ}$.

→ As the angle of incidence is less than the critical angle, TIR will not occur and light will be transmitted into the cladding.

Show Worked Solution

a. Using Snell’s Law:

$n_1\sin\theta_1$	$=n_2\sin\theta_2$
$\theta_1$	$=\sin^{-1} \Big{(}\dfrac{n_2\sin\theta_2}{n_1} \Big{)} $
	$=\sin^{-1}\Big{(}\dfrac{1.46 \times \sin32^{\circ}}{1.0}\Big{)} $
	$=51^{\circ}$

b. Find the critical angle for Total Internal Reflection (TIR):

$\sin\theta_c$	$=\dfrac{n_2}{n_1}$
$\theta_c$	$=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)} $
	$=\sin^{-1} \Big{(}\dfrac{1.42}{1.46} \Big{)}$
	$=76.6^{\circ}$

→ The angle of incidence is $90^{\circ}-32^{\circ}=58^{\circ}$.

→ As the angle of incidence is less than the critical angle, TIR will not occur and light will be transmitted into the cladding.

♦ Mean mark (b) 45%.

PHYSICS, M2 2021 VCE 9a

Abbie and Brian are about to go on their first loop-the-loop roller-coaster ride. As competent Physics students, they are working out if they will have enough speed at the top of the loop to remain in contact with the track while they are upside down at point $\text{C}$, shown in Figure 9.

The highest point of the roller-coaster (point $\text{A}$) is 15 m above point $\text{B}$ and the car starts at rest from point $\text{A}$. Assume that there is negligible friction between the car and the track.

What is the speed of the car at point $\text{B}$ at the bottom of the loop? Show your working. (2 marks)

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$17.1\ \text{ms}^{-1}$

Show Worked Solution

Using the Law of Conservation of Energy:

$mgh$	$=\dfrac{1}{2}mv^2$
$v$	$=\sqrt{2gh}$
	$=\sqrt{2 \times 9.8 \times 15}$
	$=17.1\ \text{ms}^{-1}$

CHEMISTRY, M4 EQ-Bank 1

Find $\Delta G $ at 298.15K for the following reaction:

$\ce{2CO + O2 \rightarrow 2CO2}$

Given that $\Delta H =-128.3\ \text{kJ},\ \ \Delta S = -159.5\ \text{J K}^{-1} $ (2 marks)

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$-80.75\ \text{kJ}$

Show Worked Solution

$\Delta S = -159.5\ \text{J K}^{-1} = -0.1595\ \text{kJ K}^{-1}$

$\text{Using the Gibbs free energy equation:}$

$\Delta G$	$= \Delta H-T \Delta S$
	$=-128.3-(298.15 \times -0.1595) $
	$=-128.3 + 47.554$
	$=-80.75\ \text{kJ}$

PHYSICS, M3 2021 VCE 15 MC

A Physics class is investigating the dispersion of white light using a triangular glass prism.

Which one of the following diagrams best shows the principle of dispersion?

Show Answers Only

$A$

Show Worked Solution

→ Dispersion is a result of white light splitting into different colours when it is refracted.

→ Dispersion occurs during both changes in medium (i.e. from air into glass and from glass into air).

→ The shorter violet wavelengths of light experience a greater refraction than the longer wavelength red light.

$\Rightarrow A$

PHYSICS, M3 2021 VCE 13

The diagram below shows part of a travelling wave.

The wave propagates with a speed of 18 ms$^{-1}$.

What is the amplitude and frequency of the wave? (2 marks)

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Amplitude = 8 cm

Frequency = 300 Hz

Show Worked Solution

→ Amplitude: the distance from the centre line to maximum displacement.

→ Amplitude = 8 cm.

→ Frequency: can be determined using $f=\dfrac{v}{\lambda}$, (wavelength must be in metres)

$f=\dfrac{v}{\lambda}=\dfrac{18}{0.06}=300\ \text{Hz} $

PHYSICS, M4 2021 VCE 3 MC

The diagram below shows two parallel metal plates with opposite charges on each plate. $\text{X , Y}$ and $\text{Z}$ represent different distances from the positive plate.

Which one of the following graphs best shows the electric field strength, $E$, versus the position, $x$, between the two parallel plates?

Show Answers Only

$A$

Show Worked Solution

→ The strength of the electric field is uniform between parallel plates. Hence its magnitude remains constant.

$ \Rightarrow A$

PHYSICS, M3 2022 VCE 16

A small sodium lamp, emitting light of wavelength 589 nm, is viewed at night through two windows from across a street. The glass of one window has a fine steel mesh covering it and the other window is open, as shown in Figure 18. Assume that the sodium lamp is a point source at a distance.

A Physics student is surprised to see a pattern formed by the light passing through the steel mesh but no pattern for the light passing through the open window. She takes a photograph of the observed pattern to show her teacher, who assures her that it is a diffraction pattern.

State the condition that the fine steel mesh must satisfy for a diffraction pattern to form. (1 mark)

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Explain why the condition stated in part a. does not apply to the open window. (2 marks)

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a. Condition to satisfy:

→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.

b. Open window differences:

→ The width of the window is significantly greater than wavelength of the light from the sodium lamp and the width of the glass molecules that make up the glass window is significantly smaller than the light from the sodium lamp.

→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.

Show Worked Solution

a. Condition to satisfy:

→ The size of the gaps from the fine steel mesh must be of the same order of magnitude of the wavelengths of the light from the sodium lamp.

b. Open window differences:

→ These widths contrast greatly to the size of the steel mesh and are not of the same order of magnitude of the light from the sodium lamp, hence no diffraction will occur.

♦ Mean mark (b) 40%.

PHYSICS, M3 2022 VCE 13

A ray of green light from a light-emitting diode (LED) strikes the surface of a tank of water at an angle of 40.00° to the surface of the water, as shown in Figure 11. The ray arrives at the base of the tank at point $\text{X}$. The depth of the water in the tank is 80.00 cm. The refractive index of green LED light in water is 1.335

Calculate the distance $\text{OX}$. Outline your reasoning and show all your working. Give your answer in centimetres. (4 marks)

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The green LED light is replaced with a narrow beam of white sunlight.
Describe the colour of the light that arrives to the left of point $\text{X}$, at point $\text{X}$ and to the right of point $\text{X}$. (3 marks)
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a. $OX=56\ \text{cm}$

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Light to the left of point X}\rule[-1ex]{0pt}{0pt} & \text{Light at point X} & \text{Light to the right of point X} \\
\hline
\rule{0pt}{2.5ex}\text{Blue\Purple}\rule[-1ex]{0pt}{0pt} & \text{Green} & \text{Red} \\ \text{(lower wavelength)} & & \text{(higher wavelength)} \\
\hline
\end{array}

Show Worked Solution

a. $\text{Using Snell’s Law:}$

$n_1 \sin \theta_1$	$=n_2 \sin \theta_2$
$\sin \theta_2$	$=\dfrac{n_1 \sin \theta_1}{n_2}$
$ \theta_2$	$= \sin^{-1}\Big{(}\dfrac{1 \times \sin 50^{\circ}}{1.335} \Big{)} $
	$=35^{\circ}$

$\tan35^{\circ}$	$=\dfrac{OX}{80}$
$OX$	$=80\times \tan35^{\circ}$
	$=56\ \text{cm}$

♦ Mean mark 48%.

PHYSICS, M2 2022 VCE 7

Kym and Kelly are experimenting with trolleys on a ramp inclined at 25°, as shown in Figure 7. They release a trolley with a mass of 2.0 kg from the top of the ramp. The trolley moves down the ramp, through two light gates and onto a horizontal, frictionless surface. Kym and Kelly calculate the acceleration of the trolley to be 3.2 m s$^{-2}$ using the information from the light gates.

i. Show that the component of the gravitational force of the trolley down the slope is $8.3 \text{ N}$. Use $g=9.8 \text{ m s}^{-2}$. (2 marks)

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ii. Assume that on the ramp there is a constant frictional force acting on the trolley and opposing its motion.
Calculate the magnitude of the constant frictional force acting on the trolley. (2 marks)

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When it reaches the bottom of the ramp, the trolley travels along the horizontal, frictionless surface at a speed of 4.0 m s$^{-1}$ until it collides with a stationary identical trolley. The two trolleys stick together and continue in the same direction as the first trolley.

1. Calculate the speed of the two trolleys after the collision. Show your working and clearly state the physics principle that you have used. (3 marks)

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1. Determine, with calculations, whether this collision is an elastic or inelastic collision. Show your working. (3 marks)

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a.i. See Worked Solutions

a.ii. $F_f=1.9\ \text{N}$

b.i. $2.0\ \text{ms}^{-1}$

b.ii. For the collision to be elastic, the kinetic energy must be conserved.

$KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}$

$KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}$

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

Show Worked Solution

a.i. The gravitational force down the slope:

$F$	$=mg\, \sin \theta$
	$=2.0 \times 9.8 \times \sin 25$
	$=8.3\ \text{N}$

a.ii.	$F_{net}$	$=ma$
		$=2.0 \times 3.2$
		$=6.4$

$6.4$	$=F-F_f$
$F_f$	$=8.3-6.4$
	$=1.9\ \text{N}$

♦ Mean mark (a.ii) 53%.

b.i. By the conservation of momentum:

$m_1u_1+m_2u_2$	$=v(m_1+m_2)$
$2 \times 4 + 2 \times 0$	$=v(2 +2)$
$8$	$=4v$
$v$	$=2\ \text{ms}^{-1}$

b.ii. For the collision to be elastic, the kinetic energy must be conserved.

$KE_{\text{init}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 2 \times 4^2=16\ \text{J}$

$KE_{\text{final}}=\dfrac{1}{2} mv^2=\dfrac{1}{2} \times 4 \times 2^2=8\ \text{J}$

→ As the kinetic energy of the system decreases after the collision, it is not an elastic collision.

PHYSICS, M3 2022 VCE 13*

A travelling wave produced at point $\text{A}$ is reflected at point $\text{B}$ to produce a standing wave on a rope, as represented in the diagram below.

The distance between points $\text{A}$ and $\text{B}$ is 2.4 m. The period of vibration of the standing wave is 1.6 s.

Find the speed of the travelling wave along the rope in metres per second. (2 marks)

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$A$

Show Worked Solution

→ $ \lambda = 1.2$ m

→ $f= \dfrac{1}{T} = \dfrac{1}{1.6} = 0.625$

$v$	$=f \lambda$
	$=0.625 \times 1.2$
	$=0.75\ \text{ms}^{-1}$

BIOLOGY, M4 2022 VCE 20 MC

During their lifetime, plants are exposed to both pathogenic and beneficial microorganisms. Some plants provide shelter within their bodies for beneficial microorganisms.

These beneficial microorganisms help plants that have not been genetically modified to resist pathogenic microorganisms.

The beneficial microorganisms may be providing protection by

Synthesising toxins that kill pathogenic microorganisms.
Providing a permeable physical barrier to pathogenic microorganisms.
Stimulating the production of antibodies against pathogenic microorganisms.
Mobilising the cells of the third line of defence against pathogenic microorganisms.

Show Answers Only

$A$

Show Worked Solution

By Elimination

→ Stimulating antibody production and mobilising cells of the third line of defence are impossible in a plant as they do not contain an adaptive immune system (Eliminate C and D).

→ Microorganisms are not able to create a permeable physical barrier which will allow passage of materials but deter pathogenic microorganisms (Eliminate B).

$\Rightarrow A$

CHEMISTRY, M4 2014 VCE 1

The decomposition of ammonia is represented by the following equation.

$\ce{2NH3(g)\rightleftharpoons N2(g) + 3H2(g) \quad \quad \Delta H = 92.4 \text{kJ mol}^{–1}}$

The activation energy for the uncatalysed reaction is 335 kJ mol$^{-1}$.
The activation energy for the reaction when tungsten is used as a catalyst is 163 kJ mol$^{-1}$.
On the grid provided below, draw a labelled energy profile diagram for the uncatalysed and catalysed reactions. (3 marks)

When osmium is used as a catalyst, the activation energy is 197 kJ mol$^{-1}$.
Which catalyst – osmium or tungsten – will cause ammonia to decompose at a faster rate? Justify your answer in terms of the chemical principles you have studied this year. (2 marks)

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b. → Tungsten will cause ammonia to decompose faster.

→ Faster rate due to lower activation energy.

→ Lower activation energy results in a greater proportion of successful collisions.

Show Worked Solution

b. → Tungsten will cause ammonia to decompose faster.

→ Faster rate due to lower activation energy.

→ Lower activation energy results in a greater proportion of successful collisions.

a.	\(\sin\theta_c\)	\(=\dfrac{n_2}{n_1}\)
	\(\theta_c\)	\(=\sin^{-1}\Big{(}\dfrac{1.00}{1.44}\Big{)}\)
		\(=44^{\circ}\)

\(m_Au_A+m_Bu_B\)	\(=v(m_A+m_B)\)
\(4 \times 5 + 2 \times 2\)	\(=v(4 +2)\)
\(24\)	\(=6v\)
\(v\)	\(=4\ \text{ms}^{-1}\)

\(m_Au_A+m_Bu_B\)	\(=v(m_A+m_B)\)
\(4 \times 5 + 2 \times 2\)	\(=v(4 +2)\)
\(24\)	\(=6v\)
\(v\)	\(=4\ \text{ms}^{-1}\)

a.	\(f\)	\(=\dfrac{v}{\lambda}\)
		\(=\dfrac{3\times 10^8}{565 \times 10^{-9}}\)
		\(=5.31 \times 10^{14}\ \text{Hz}\)

b.	\(\theta_c\)	\(=\sin^{-1}\Big{(}\dfrac{n_2}{n_1}\Big{)}\)
		\(=\sin^{-1} \Big{(}\dfrac{1.45}{1.67} \Big{)}\)
		\(=60.3^{\circ}\)

c.	\(v_{\text{x}}\)	\(=\dfrac{c}{n_{\text{x}}}\)
		\(=\dfrac{3 \times 10^8}{1.67}\)
		\(=1.80 \times 10^8\ \text{ms}^{-1}\)

→ \(F_{\text {on B by A }}\)	\(=ma\)
	\(=1 \times 8\)
	\(=8\ \text{N}\)

\(m_Xu_X+m_Yu_Y\)	\(=m_Xv_X+m_Yv_Y\)
	\(=v(m_X + m_Y)\ \ \ (v_X=v_Y) \)
\(10\ 000 \times 6 +0\)	\(= v(10\ 000 + 5000)\)
\(60\ 000\)	\(=15\ 000v\)
\(v\)	\(=4\ \text{ms}^{-1}\)

\(KE_{\text{final}}\)	\(=\dfrac{1}{2}(m_X+m_Y)v^2\)
	\(=\dfrac{1}{2} \times 15\ 000 \times 4^2\)
	\(=120\ 000\ \text{J}\)