Solve the differential equation `(dy)/(dx) = -x sqrt(4-y^2)` given that `y(2) = 0`. Give your answer in the form `y = f(x)`. (3 marks)
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Vectors, SPEC2 2023 VCAA 5
The points with coordinates \(A(1,1,2), B(1,2,3)\) and \(C(3,2,4)\) all lie in a plane \(\Pi\). --- 5 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- A second plane, \(\psi\), has the Cartesian equation \(2 x-2 y-z=-18\). --- 4 WORK AREA LINES (style=lined) --- A line \(L\) passes through the origin and is normal to the plane \(\psi\). The line \(L\) intersects \(\psi\) at a point \(D\). --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
PHYSICS, M5 2021 VCE 3
To calculate the mass of distant pulsars, physicists use Newton's law of universal gravitation and the equations of circular motion.
The planet Phobetor orbits pulsar PSR B1257+12 at an orbital radius of 6.9 × 10\(^{10}\) m and with a period of 8.47 × 10\(^6\) s.
Assuming that Phobetor follows a circular orbit, calculate the mass of the pulsar. Show all your working. (3 marks)
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PHYSICS, M6 2021 VCE 2
A schematic side view of one design of an audio loudspeaker is shown in Diagram 1 below. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out. Diagram 2 shows a schematic view of the loudspeaker from the position of the eye shown in Diagram 1. The direction of the current is clockwise, as shown. --- 0 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. b. \(E\) c. \(2.51\ \text{N}\) a. Magnetic fields run from the north pole to the south pole. b. Apply the right-hand rule: \(\Rightarrow E\) c. \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\) \(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)
A.
left
B.
right
C.
up the page
D.
down the page
E.
into the page
F.
out of the page
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♦♦♦ Mean mark 24%.
Statistics, SPEC2 2023 VCAA 6
A forest ranger wishes to investigate the mass of adult male koalas in a Victorian forest. A random sample of 20 such koalas has a sample mean of 11.39 kg. It is known that the mass of adult male koalas in the forest is normally distributed with a standard deviation of 1 kg. --- 3 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- The ranger wants to decrease the width of the 95% confidence interval by 60% to get a better estimate of the population mean. --- 2 WORK AREA LINES (style=lined) --- It is thought that the mean mass of adult male koalas in the forest is 12 kg. The ranger thinks that the true mean mass is less than this and decides to apply a one-tailed statistical test. A random sample of 40 adult male koalas is taken and the sample mean is found to be 11.6 kg. --- 2 WORK AREA LINES (style=lined) --- The ranger decides to apply the one-tailed test at the 1% level of significance and assumes the mass of adult male koalas in the forest is normally distributed with a mean of 12 kg and a standard deviation of 1 kg. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- Suppose that the true mean mass of adult male koalas in the forest is 11.4 kg, and the standard deviation is 1 kg. The level of significance of the test is still 1%. --- 2 WORK AREA LINES (style=lined) ---
CHEMISTRY, M1 EQ-Bank 6
Explain the trend in reactivity with water of the elements in Group 2 as you move down the group. (2 marks)
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PHYSICS, M8 2022 VCE 15
The diagram shows some of the energy levels of excited neon atoms. These energy levels are not drawn to scale. --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- a. \(E=hf=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{640 \times 10^{-9}}=3.106 \times 10^{-19}\ \text{J}\) \(\text{Convert to electron volts} =\dfrac{3.106 \times 10^{-19}}{1.602 \times 10^{-19}}=1.94\ \text{eV}\) b. a. \(E=hf=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{640 \times 10^{-9}}=3.106 \times 10^{-19}\ \text{J}\) \(\text{Convert to electron volts} =\dfrac{3.106 \times 10^{-19}}{1.602 \times 10^{-19}}=1.94\ \text{eV}\) b.
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PHYSICS, M7 2022 VCE 14*
Sam undertakes a photoelectric effect experiment using the apparatus shown in Figure 1. She uses a green filter.
Sam produces a graph of photocurrent, \(I\), in milliamperes, versus voltage, \(V\), in volts, as shown in Figure 2.
- Identify what point \(\text{P}\) represents on the graph in Figure 2. (1 mark)
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- Sam then significantly increases the intensity of the light.
- Sketch the resulting graph on Figure 3. The dashed line in Figure 14 represents the original data. (2 marks)
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- Sam replaces the green filter with a violet filter, keeping the light source at the increased intensity.
- Sketch the resulting graph on Figure 4. The dashed line in Figure 4 represents the original data. (2 marks)
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a. Stopping voltage
b.
c.
Show Worked Solution
a. Stopping voltage
b. Diagram will exhibit the following points:
- Increasing the intensity of light will increase the photocurrent due to more photoelectrons being ejected from the metal.
- However, it will not change the energy of the photoelectrons, hence the stopping voltage remains the same.
c. Diagram will exhibit the following points:
- Changing the colour from green to violet will increase the frequency of the light.
- As \(E=hf\), the energy of the photoelectrons will increase, hence there needs to be a greater stopping voltage.
Functions, MET2 2023 VCAA 2
The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.
- Show that \(b=\dfrac{\pi}{15}\) and \(c=75\). (2 marks)
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- Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
- Give your answer in metres, correct to two decimal places. (2 marks)
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- Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\). (1 mark)
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After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.
The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
\(w(t) = \begin {cases}
h(t) &\ \ 0 \leq t < 15 \\
k &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)
where \(k\geq 0, m\geq 0\) and \(n \in R\).
- i.State the values of \(k\) and \(m\). (1 mark)
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ii. Find all possible values of \(n\). (2 marks)--- 4 WORK AREA LINES (style=lined) ---
iii. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints. (3 marks)
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a. \(\text{See worked solution}\)
b. \(\approx 36.80\ \text{m}\)
c. \(8\)
d.i. \(k=135, m=2\)
d.ii. \(n=30p+5,\ p\in Z\)
d.iii.
Show Worked Solution
| a. | \(\text{Period:}\) | \(\dfrac{2\pi}{b}\) | \(=30\) |
| \(\therefore\ b\) | \(=\dfrac{\pi}{15}\) |
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)
| \(-60\ \cos(0)+c\) | \(=15\) |
| \(c\) | \(=75\) |
\(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)
| b. | \(\text{Average height}\) | \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\) |
| \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\) | ||
| \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\) | ||
| \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\) | ||
| \(=36.802\dots\approx 36.80\ \text{m}\) |
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.
c. \(\text{Av rate of change of height}\)
| \(=\dfrac{h(7.5)-h(0)}{7.5}\) | |
| \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\) | |
| \(=\dfrac{75-15}{7.5}=8\) |
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
MARKER’S COMMENT: Common incorrect answer was – 8.
| di. | \(\text{Period is 30 minutes, so after 15 minutes pod}\) \(\text{is at the top of the wheel.}\) |
| \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\) |
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)
| \(\therefore\ \text{Period}\rightarrow\) | \(\dfrac{2\pi}{bm}\) | \(=15\) |
| \(bm\) | \(=\dfrac{2\pi}{15}\) | |
| \(\dfrac{\pi}{15}\cdot m\) | \(=\dfrac{2\pi}{15}\) | |
| \(m\) | \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\) | |
| \(=2\) |
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
| dii. | \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\) \(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\) |
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)
\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)
\(\therefore\ n=30p+5,\ p\in Z\)
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.
diii.
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.
Calculus, MET2 2023 VCAA 1
Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.
- State the coordinates of all axial intercepts of \(f\). (1 mark)
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- Find the coordinates of the stationary points of \(f\). (2 marks)
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-
- Let \(g:R\rightarrow R, g(x)=x-2\).
- Find the values of \(x\) for which \(f(x)=g(x)\). (1 mark)
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-
- Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\). (2 marks)
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- Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places. (1 mark)
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- Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\). (2 marks)
- Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
- Find the possible values of \(a\) and \(b\). (4 marks)
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PHYSICS, M8 2021 VCE 17 MC
Which one of the following is closest to the de Broglie wavelength of a 663 kg motor car moving at 10 m s\(^{-1}\) ?
- \(10^{-37}\) m
- \(10^{-36}\) m
- \(10^{-35}\) m
- \(10^{-34}\) m
PHYSICS, M5 2021 VCE 9-10 MC
Lucy is running horizontally at a speed of 6 m s\(^{-1}\) along a diving platform that is 8.0 m vertically above the water.
Lucy runs off the end of the diving platform and reaches the water below after time \(t\).
She lands feet first at a horizontal distance \(d\) from the end of the diving platform.
Question 9
Which one of the following expressions correctly gives the distance \(d\) ?
- 0.8\(t\)
- 6\(t\)
- 5\(t^2\)
- 6\(t\) + 5\(t^2\)
Question 10
Which one of the following is closest to the time taken, \(t\), for Lucy to reach the water below?
- 0.8 s
- 1.1 s
- 1.3 s
- 1.6 s
PHYSICS, M6 2021 VCE 7*
A mobile phone charger uses a step-down transformer to transform 240 V AC mains voltage to 5.0 V. The mobile phone draws a current of 3.0 A while charging. Assume that the transformer is ideal and that all readings are RMS.
Calculate the current drawn from the mains during charging? (2 marks)
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PHYSICS, M6 2021 VCE 6 MC
A magnet approaches a coil with six turns, as shown in the diagram below. During time interval \(\Delta t\), the magnetic flux changes by 0.05 Wb and the average induced EMF is 1.2 V.
Which one of the following is closest to the time interval \(\Delta t\) ?
- 0.04 s
- 0.01 s
- 0.25 s
- 0.50 s
PHYSICS, M8 2022 VCE 17
A materials scientist is studying the diffraction of electrons through a thin metal foil. She uses electrons with an energy of 10.0 keV. The resulting diffraction pattern is shown in Figure 19. --- 6 WORK AREA LINES (style=lined) --- The materials scientist then increases the energy of the electrons by a small amount and hence their speed by a small amount. Explain what effect this would have on the de Broglie wavelength of the electrons. Justify your answer. (3 marks) --- 6 WORK AREA LINES (style=lined) ---
Calculus, 2ADV C4 2022 MET1 4
The graph of \(y=x+\dfrac{1}{x}\) is shown over part of its domain.
Use two trapeziums of equal width to approximate the area between the curve, the \(x\)-axis and the lines \(x=1\) and \(x=3\). (2 marks)
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Calculus, 2ADV C2 2023 MET1 1a
Let \(y=\dfrac{x^2-x}{e^x}\).
Find and simplify \(\dfrac{dy}{dx}\). (2 marks)
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L&E, 2ADV E1 2023 MET1 2
Solve \(e^{2x}-12=4e^{x}\) for \(x\ \in\ R\). (3 marks)
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PHYSICS, M6 2022 VCE 6
The diagram shows a simple alternator consisting of a rectangular coil of area 0.060 m\(^{2}\) and 200 turns, rotating in a uniform magnetic field. The magnetic flux through the coil in the vertical position shown in the diagram is 1.2 × 10\(^{-3}\) Wb.
-
Calculate the strength of the magnetic field. Show your working. (2 marks)
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- The rectangular coil rotates at a frequency of 2.5 Hz.
- Calculate the average induced EMF produced in the first quarter of a turn. Begin the quarter with the coil in the vertical position shown in the diagram. (3 marks)
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PHYSICS, M6 2022 VCE 5*
A wind generator provides power to a factory located 2.00 km away, as shown in the diagram. When there is a moderate wind blowing steadily, the generator produces a voltage of 415 V and a current of 100 A. The total resistance of the transmission wires between the wind generator and the factory is 2.00 \(\Omega\). --- 2 WORK AREA LINES (style=lined) --- To operate correctly, the factory's machinery requires a power supply of 40 kW. --- 6 WORK AREA LINES (style=lined) --- The factory's owner decides to limit transmission energy loss by installing two transformers: a step-up transformer with a turns ratio of 1:10 at the wind generator and a step-down transformer with a turns ratio of 10:1 at the factory. Each transformer can be considered ideal. With the installation of the transformers, determine the power, in kilowatts, now supplied to the factory. (3 marks) --- 8 WORK AREA LINES (style=lined) ---
PHYSICS, M6 2022 VCE 3
A schematic diagram of a mass spectrometer that is used to deflect charged particles to determine their mass is shown in the diagram. Positive singly charged ions (with a charge of +1.602 × 10\(^{-19}\) C) are produced at the ion source. These are accelerated between an anode and a cathode. The potential difference between the anode and the cathode is 1500 V. The ions pass into a region of uniform magnetic field, \(B\), and are directed by the field into a semicircular path of diameter \(D\).
- Calculate the increase in the kinetic energy of each ion as it passes between the anode and the cathode. Give your answer in joules. (2 marks)
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Each ion has a mass of 4.80 × 10\(^{-27}\) kg.
- Show that each ion has a speed of 3.16 × 10\(^{5}\) m s\(^{-1}\) when it exits the cathode. Assume that the ion leaves the ion source with negligible speed. Show your working. (2 marks)
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- The region of uniform magnetic field, \(B\), in Figure 3 has a magnitude of 0.10 T.
- Calculate the diameter, \(D\), of the semicircular path followed by the ions within the magnetic field in Figure 3. (3 marks)
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PHYSICS, M5 2022 VCE 2
There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. The diagram shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator. --- 4 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
PHYSICS, M6 2022 VCE 1
The diagram shows four positions (1, 2, 3 and 4) of the coil of a single-turn, simple DC motor. The coil is turning in a uniform magnetic field that is parallel to the plane of the coil when the coil is in Position 1, as shown. When the motor is operating, the coil rotates about the axis through the middle of sides \(L M\) and \(N K\) in the direction indicated. The coil is attached to a commutator. Current for the motor is passed to the commutator by brushes that are not shown in the diagram. --- 3 WORK AREA LINES (style=lined) --- When the coil is in Position 3, in which direction is the current flowing in the side \(KL-\) from \(K\) to \(L\) or --- 2 WORK AREA LINES (style=lined) --- The side \(K L\) of the coil has a length of 0.10 m and experiences a magnetic force of 0.15 N due to the magnetic field, which has a magnitude of 0.5 T. Calculate the magnitude of the current in the coil. (2 marks) --- 5 WORK AREA LINES (style=lined) ---
from \(L\) to \(K\)? (1 marks)
PHYSICS, M8 2022 VCE 17 MC
Gamma radiation is often used to treat cancerous tumours. The energy of a gamma photon emitted by radioactive cobalt-60 is 1.33 MeV.
Which one of the following is closest to the frequency of the gamma radiation?
- \(1.33 \times 10^{6}\ \text{Hz}\)
- \(3.21 \times 10^{20}\ \text{Hz}\)
- \(3.21 \times 10^{21}\ \text{Hz}\)
- \(2.01 \times 10^{39}\ \text{Hz}\)
PHYSICS, M8 2022 VCE 14 MC
Which one of the following best provides evidence of electrons behaving as waves?
- photoelectric effect
- atomic emission spectra
- atomic absorption spectra
- diffraction of electrons through a crystal
PHYSICS, M6 2022 VCE 5 MC
A simple electricity generator is shown in the diagram below. When the coil is rotated, the output voltage across the slip rings is measured. The graph shows how the output voltage varies with time.
The frequency of rotation of the generator is now doubled.
Which one of the following graphs best represents the output voltage measured across the slip rings?
PHYSICS, M8 2023 VCE 16
Fluorescent lights, when operating, contain gaseous mercury atoms, as shown in Figure 1. Analysis of the light produced by fluorescent lights shows a number of emission spectral lines, including a prominent line representing a wavelength of 436.6 nm. --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- a. \(E=2.84\ \text{eV}\) b. a. \(E=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{436.6 \times 10^{-9}}=4.549 \times 10^{-19}\ \text{J}\) \(\text{Convert to eV:}\) \(E= \dfrac{4.549 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.84\ \text{eV}\) b.
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Calculus, SPEC2 2023 VCAA 4
A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, \(P\), grows according to the logistic model, \(\dfrac{d P}{d t}=P\left(1-\dfrac{P}{1000}\right)\) , where \(t\) is the time in years after the release of the 200 fish. --- 2 WORK AREA LINES (style=lined) --- One form of the solution for \(P\) is \(P=\dfrac{1000}{1+D e^{-t}}\ \), where \(D\) is a real constant. --- 2 WORK AREA LINES (style=lined) --- The farmer releases a batch of \(n\) fish into a second pond, pond 2 , which originally contained no fish. The population, \(Q\), of fish in pond 2 can be modelled by \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\), where \(t\) is the time in years after the \(n\) fish are released. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- The farmer wishes to take 5.5% of the fish from pond 2 each year. The modified logistic differential equation that would model the fish population, \(Q\), in pond 2 after \(t\) years in this situation is \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\) --- 4 WORK AREA LINES (style=lined) --- a. \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\Rightarrow D=4\) c. \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(Q^{\prime}=\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)\) e.ii. \(t=2\) f. g. \(Q=950\) a. \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(\text{Expand (by CAS):}\) \(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\text{When}\ \ t=0, P=200 \text{ (given)}\) \(\text{Solve}\ \ 200=\dfrac{1000}{1+D e^0}\ \ \text{for}\ t:\) \(\Rightarrow D=4\ \ \text {(by CAS):}\) c. \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\) \(\text{At}\ \ t=0, Q=n\): \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(\text{Find } Q \text{ when } t=6:\) \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)\) \(\text{Using the product rule:}\) \begin{aligned} e.ii. \(\text{Max } Q^{\prime} \Rightarrow Q^{\prime \prime}=0\) \(\text{Solve } Q^{\prime \prime}=0 \ \ \text {(by CAS):}\) \(\Rightarrow Q=500\) \(\text{Solve } Q=500 \text{ for } t \text{ (by CAS):}\) \(t=1.99 \ldots=2 \ \text{(nearest year)}\) f. g. \(\text{Solve for } Q \ \text{(by CAS):}\) \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\) \(\Rightarrow Q=950\)
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Show Worked Solution
♦ Mean mark (a) 48%.
Q^{\prime \prime} & =\frac{11}{10}\left[Q^{\prime}\left(1-\frac{Q}{1000}\right)+Q\left(-\frac{1}{1000}\, Q^{\prime}\right)\right] \\
& =\frac{11}{10}\left[\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{Q}{1000}\left(\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)\right)\right] \\
& =\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)
\end{aligned}
♦♦♦ Mean mark (e)(i) 21%.
♦ Mean mark (g) 40%.
PHYSICS, M7 2023 VCE 13
A group of physics students undertake a Young's double-slit experiment using the apparatus shown in the diagram. They use a green laser that produces light with a wavelength of 510 nm. The light is incident on two narrow slits, S\(_1\) and S\(_2\). The distance between the two slits is 100 \( \mu \)m.
An interference pattern is observed on a screen with points P\(_{0}\), P\(_{1}\) and P\(_2\) being the locations of adjacent bright bands, as shown. Point P\(_0\) is the central bright band.
- Calculate the path difference between S\(_{1}\)P\(_{2}\) and S\(_{2}\)P\(_{2}\). Give your answer in metres. Show your working. (2 marks)
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- The green laser is replaced by a red laser.
- Describe the effect of this change on the spacing between adjacent bright bands. (1 mark)
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- Explain how Young's double-slit experiment provides evidence for the wave-like nature of light and not the particle-like nature of light. (3 marks)
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PHYSICS, M7 2023 VCE 10
A proton in an accelerator beamline of proper length 4.80 km has a Lorentz factor, \(\gamma\), of 2.00.
- Calculate the speed of the proton relative to the beamline in terms of \(c\), the speed of light in a vacuum. Give your answer to three significant figures. (3 marks)
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- Calculate the length of the beamline in the reference frame of the proton. (1 mark)
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PHYSICS, M5 2023 VCE 9
Giorgos is practising his tennis serve using a tennis ball of mass 56 g. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
Calculus, SPEC2 2023 VCAA 3
The curve given by \(y^2=x-1\), where \(2 \leq x \leq 5\), is rotated about the \(x\)-axis to form a solid of revolution. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- The total surface area of the solid consists of the curved surface area plus the areas of the two circular discs at each end. The 'efficiency ratio' of a body is defined as its total surface area divided by the enclosed volume. --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
Complex Numbers, SPEC2 2023 VCAA 2
Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\). --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- --- 0 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- use De Moivre's theorem to show that --- 8 WORK AREA LINES (style=lined) --- a. \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\) \(\therefore w\ \text{is a root of}\ \ z^7-1=0 \) b. \(\text{Roots of}\ \ z^7-1=0:\) \(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \) c. d.i. d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \) \(\arg(z-1)=\dfrac{9\pi}{14} \) f.i. \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)} \) f.ii. \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \) \((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \) \(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\) \(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\) \(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\) a. \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\) \(\therefore w\ \text{is a root of}\ \ z^7-1=0 \) b. \(\text{Roots of}\ \ z^7-1=0:\) \(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \) c. d.i. d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \) \(\arg(z-1)=\dfrac{9\pi}{14} \) f.i. \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)} \) f.ii. \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \) \((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \) \(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\) \(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\) \(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)
Show Answers Only
e.
\((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=z^7-1\)
Show Worked Solution
e.
\((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)
\(=z^7-1\)
Algebra, MET2 2023 VCAA 16 MC
Let \(f(x)=e^{x-1}\).
Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is
- \(g(x)=e^{x-1}\)
- \(g(x)=e^{(x-2)(x-1)}\)
- \(g(x)=e^{(x+2)(x-1)}\)
- \(g(x)=e^{x(x-2)}\)
- \(g(x)=e^{x(x-3)}\)
Probability, MET2 2023 VCAA 4
A manufacturer produces tennis balls.
The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.
- Find \(\Pr(D>6.8)\), correct to four decimal places. (1 mark)
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- Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
- Give your answer in centimetres, correct to two decimal places. (1 mark)
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Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.
- Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
- Give your answer correct to four decimal places. (1 mark)
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- In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
- Give your answer correct to four decimal places. (2 marks)
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A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.
- Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
- Give your answer correct to four decimal places. (2 marks)
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- The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
- Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places. (2 marks)
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- An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
- The confidence interval is (0.7382, 0.9493), correct to four decimal places.
- Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer. (2 marks)
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A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg) &\ \ 30 \leq v \leq 3\pi^2+30 \\
0 &\ \ \text{elsewhere}
\end{cases}\)
- Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
- Give your answer correct to four decimal places. (1 mark)
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- Find the exact mean serving speed for grade A balls, in metres per second. (1 mark)
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The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).
A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).
- If the mean serving speed for a grade B ball is \(2\pi^2+8\) metres per second, find the values of \(a\) and \(b\). (2 marks)
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Calculus, SPEC2 2023 VCAA 1
Viewed from above, a scenic walking track from point \(O\) to point \(D\) is shown below. Its shape is given by \(f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}\) The minimum turning point of section \(O A B C\) occurs at point \(A\). Point \(B\) is a point of inflection and the curves meet at point \(C(1,0)\). Distances are measured in kilometres. --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- The return track from point \(D\) to point \(O\) follows an elliptical path given by \(x=2 \cos (t)+2, y=(e-2) \sin (t)\), where \(t \in\left[\dfrac{\pi}{2}, \pi\right]\). --- 3 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. \(\text{At}\ C(1,0): \) \(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \) \(a=-1\) \(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \) b. \(\text{Find derivative functions (by calc)}:\) \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \) \(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \) \(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\) c.i. \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \) c.ii. \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \) d. \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\) e. f.i. \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\) f.ii. \(\text{Length}\ = 2.255\) a. \(\text{At}\ C(1,0): \) \(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \) \(a=-1\) \(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \) b. \(\text{Find derivative functions (by calc)}:\) \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \) \(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \) \(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\) c.i. \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} \) \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \) c.ii. \(\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} \) \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \) d. \(\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) \) \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\) e. f.i. \(\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) \) \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\) f.ii. \(\text{Evaluate the integral in part f.i.}\) \(\Rightarrow \text{Length}\ = 2.255\)
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Show Worked Solution
PHYSICS, M6 2023 VCE 7*
Two high-voltage transmission lines span a distance of 260 km between Power Plant A and Town B, as shown in the diagram. Power Plant A provides 350 MW of power. The potential difference at Power Plant A is 500 kV. The current in the transmission lines has a value of 700 A and the power loss in the transmission lines is 20 MW.
- Show, using calculations, that the total resistance of the two transmission lines is 41 \(\Omega\). (2 marks)
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- Town B needs a minimum of 480 kV.
- Determine whether 480 kV will be available to Town B. Show your working. (3 marks)
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- Explain what would happen if the electricity between Power Plant A and Town B were to be transmitted at 50 kV instead of 500 kV. Assume that the resistance of the transmission lines is still 41 \(\Omega\). (2 marks)
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PHYSICS, M6 2023 VCE 5
Figure 1 shows a single square loop of conducting wire placed just outside a constant uniform magnetic field, \(B\). The length of each side of the loop is 0.040 m. The magnetic field has a magnitude of 0.30 T and is directed out of the page.
Over a time period of 0.50 s, the loop is moved at a constant speed, \(v\), from completely outside the magnetic field, Figure 1, to completely inside the magnetic field, Figure 2.
- Calculate the average EMF produced in the loop as it moves from the position just outside the region of the field, Figure 1, to the position completely within the area of the magnetic field, Figure 2.
- Show your working. (2 marks)
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- On the small square loop in Figure 3, show the direction of the induced current as the loop moves into the area of the magnetic field. (1 mark)
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Show Answers Only
a. \(9.6 \times 10^{-4}\ \text{V}\)
b.
Show Worked Solution
| a. | \(\varepsilon\) | \(=\dfrac{\Delta \phi}{\Delta t}\) |
| \(=\dfrac{\Delta B A}{\Delta t}\) | ||
| \(=\dfrac{0.3 \times 0.04^2}{0.5}\) | ||
| \(=9.6 \times 10^{-4}\ \text{V}\) |
b.
♦ Mean mark (b) 45%.
Calculus, MET2 2023 VCAA 5
Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).
- Complete a possible sequence of transformations to map \(f\) to \(g\). (2 marks)
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Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.
- Give the domain and range for the inverse of \(g_1\). (2 marks)
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Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).
-
- Find the coordinates of \(P\) and \(Q\), correct to two decimal places. (1 mark)
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- Find the coordinates of \(P\) and \(Q\), correct to two decimal places. (1 mark)
-
- Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
- Give your answer correct to two decimal places. (2 marks)
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Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).
- The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
- Find the value of \(n\). (1 mark)
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Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).
The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)
- What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?\
- Give your answer correct to two decimal places. (1 mark)
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It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).
- Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
- Give your answer correct to two decimal places. (2 marks)
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PHYSICS, M6 2023 VCE 4*
A transformer is used to provide a low-voltage supply for six outdoor garden globes. The circuit is shown in the diagram below. Assume there is no power loss in the connecting wires. The input of the transformer is connected to a power supply that provides an AC voltage of 240 V. The globes in the circuit are designed to operate with an AC voltage of 12 V. Each globe is designed to operate with a power of 20 W. --- 2 WORK AREA LINES (style=lined) --- The globes are turned on. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
PHYSICS, M7 2023 VCE 19*
The diagram below shows the spectrum of light emitted by a hydrogen vapour lamp. The spectral line indicated by the arrow on the diagram is in the visible region of the spectrum.
Calculate the frequency of the light corresponding to the spectral line indicated by the arrow. (2 marks)
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PHYSICS, M7 2023 VCE 18 MC
Which one of the following statements best describes the type of light produced from different types of light sources?
- Light from a laser is coherent and has a very narrow range of wavelengths.
- Light from an incandescent lamp is coherent and has a range of wavelengths.
- Light from an incandescent lamp is incoherent and has a very narrow range of wavelengths.
- Light from a single-colour light-emitting diode (LED) is coherent and contains a very wide range of wavelengths.
PHYSICS, M8 2023 VCE 17 MC
Which one of the following statements best explains why it is possible to compare X-ray and electron diffraction patterns?
- X-rays can exhibit particle-like properties.
- Electrons can exhibit wave-like properties.
- Electrons are a form of high-energy X-rays.
- Both electrons and X-rays can ionise matter.
PHYSICS, M5 2023 VCE 2
Phobos is a small moon in a circular orbit around Mars at an altitude of 6000 km above the surface of Mars. The gravitational field strength of Mars at its surface is 3.72 N kg\(^{-1}\). The radius of Mars is 3390 km. --- 6 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
PHYSICS, M5 2023 VCE 9*
An engineer is designing a banked circular curve of radius 25 m in a new bicycle velodrome.
Diagram A shows the bicycle approaching the banked section, and diagram B shows the front view of a bicycle moving out of the page as it rounds the banked bend.
The bicycle is travelling at 11 m s\(^{-1}\) on the banked section. At this speed there are no sideways frictional forces between the wheels and the road surface.
Determine the angle of the banked bend with the road surface, giving your answer to the nearest degree. (3 marks)
Show Worked Solution
| \(\tan \theta\) | \(=\dfrac{F_c}{mg}\) | |
| \(F_c\) | \(=mg \tan \theta\) | |
| \(\dfrac{mv^2}{r}\) | \(=mg \tan \theta\) | |
| \(\dfrac{v^2}{r}\) | \(=g \tan \theta\) | |
| \(\tan \theta\) | \(=\dfrac{v^2}{gr}=\dfrac{11^2}{9.8 \times 25}\) | |
| \(\theta\) | \(=\tan^{-1}\Big(\dfrac{11^2}{9.8 \times 25}\Big)=26^{\circ}\) |
PHYSICS, M6 2023 VCE 5-6 MC
The diagram below shows a stationary circular coil of conducting wire connected to a low-resistance globe in a uniform, constant magnetic field, \(B\).
Question 5
The magnetic field is switched off.
Which one of the following best describes the globe in the circuit \( \textbf{before} \) the magnetic field is switched off, \( \textbf{during} \) the time the magnetic field is being switched off and \( \textbf{after} \) the magnetic field is switched off?
\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad\text{Before}\quad\rule[-1ex]{0pt}{0pt}&\quad \text{During} \quad& \quad\text{After}\quad\\
\hline
\rule{0pt}{2.5ex}\text{Off}\rule[-1ex]{0pt}{0pt}&\text{On}& \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{On}\rule[-1ex]{0pt}{0pt}& \text{On}& \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{On}\rule[-1ex]{0pt}{0pt}& \text{Off} & \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{Off}\rule[-1ex]{0pt}{0pt}& \text{On} & \text{On}\\
\hline
\end{array}
\end{align*}
Question 6
The radius of the coil is 5 cm and the magnetic field strength is 0.2 T. The coil has 100 loops. Assume that the magnetic field is perpendicular to the area of the coil.
Which one of the following is closest to the magnitude of the magnetic flux through the coil of wire when the magnetic field is switched on?
- 0.0016 Wb
- 0.16 Wb
- 16 Wb
- 1600 Wb
PHYSICS, M6 2023 VCE 1 MC
One type of loudspeaker consists of a current-carrying coil within a radial magnetic field, as shown in the diagram below. \(X\) and \(Y\) are magnetic poles, and the direction of the current, \(I\), in the coil is clockwise as shown.
The force, \(F\), acting on the current-carrying coil is directed into the page.
Which one of the following statements correctly identifies the magnetic polarities of \(X\) and \(Y\)?
- \(X\) is a north pole and \(Y\) is a south pole.
- \(X\) is a south pole and \(Y\) is a north pole.
- Both \(X\) and \(Y\) are north poles.
- Both \(X\) and \(Y\) are south poles.
Statistics, SPEC2 2023 VCAA 19 MC
A company accountant knows that the amount owed on any individual unpaid invoice is normally distributed with a mean of $800 and a standard deviation of $200.
What is the probability, correct to three decimal places, that in a random sample of 16 unpaid invoices the total amount owed is more than $13 500?
- 0.087
- 0.191
- 0.413
- 0.587
- 0.809
Probability, MET2 2023 VCAA 15 MC
Let X be a normal random variable with mean of 100 and standard deviation of 20. Let Y be a normal random variable with mean of 80 and standard deviation of 10.
Which of the diagrams below best represents the probability density functions for X and Y, plotted on the same set of axes?
| A. |
 |
B. |  |
|
| C. | |
D. |  |
|
| E. |
 |
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Vectors, SPEC2 2023 VCAA 17 MC
Consider the vectors \(\underset{\sim}{\text{a}}=\alpha \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}-\underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=3 \underset{\sim}{\text{i}}+\beta \underset{\sim}{\text{j}}+4 \underset{\sim}{\text{k}}\) and \(\underset{\sim}{\text{c}}=2 \underset{\sim}{\text{i}}-7 \underset{\sim}{\text{j}}+\gamma \underset{\sim}{\text{k}}\), where \(\alpha, \beta, \gamma \in R\). If \(\underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\underset{\sim}{\text{c}}\), then
- \(\alpha=-2, \ \beta=-1, \ \gamma=-5\)
- \(\alpha=-1, \ \beta=2, \ \gamma=-1\)
- \(\alpha=1, \ \beta=-2, \ \gamma=-5\)
- \(\alpha=-2, \ \beta=-1, \ \gamma=-1\)
- \(\alpha=1, \ \beta=-2, \ \gamma=5\)
Mechanics, EXT2 M1 SM-Bank 5
A student throws a ball for his dog to retrieve. The position vector of the ball, relative to an origin \(O\) at ground level \(t\) seconds after release, is given by \( \underset{\sim}{\text{r}}{}_\text{B} (t)=5 t \underset{\sim}{\text{i}}+7 t \underset{\sim}{\text{j}}+(15 t-4.9 t^2+1.5) \underset{\sim}{\text{k}} \). Displacement components are measured in metres, where \(\underset{\sim}{\text{i}}\) is a unit vector to the east, \(\underset{\sim}{\text{j}}\) is a unit vector to the north and \( \underset{\sim} {\text{k}}\) is a unit vector vertically up.
Calculate the total vertical distance, in metres, travelled by the ball before it hits the ground. Give your answer correct to one decimal place. (3 marks)
Vectors, SPEC2 2023 VCAA 15 MC
If the sum of two unit vectors is a unit vector, then the magnitude of the difference of the two vectors is
- \(0\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(\sqrt{2}\)
- \(\sqrt{3}\)
- \(\sqrt{5}\)
Show Worked Solution
\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)
\(\text{Using the cosine rule for the difference between the two vectors:}\)
| \(c^2\) | \(=a^2+b^2-2ab\, \cos C\) | |
| \(=1+1-2\times -\dfrac{1}{2} \) | ||
| \(=3\) | ||
| \(c\) | \(=\sqrt{3}\) |
\(\Rightarrow D\)
Calculus, SPEC2 2023 VCAA 13 MC
A tourist in a hot air balloon, which is rising vertically at 2.5 m s\(^{-1}\), accidentally drops a phone over the side when the phone is 80 metres above the ground.
Assuming air resistance is negligible, how long in seconds, correct to two decimal places, does it take for the phone to hit the ground?
- 2.86
- 2.98
- 3.79
- 4.04
- 4.30
Mechanics, EXT2 M1 EQ-Bank 6
The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.
Determine the velocity of the particle after \( \log _e(e+1) \) seconds. (3 marks)
Probability, MET2 2023 VCAA 10 MC
A continuous random variable \(X\) has the following probability density function
\(g(x) = \begin {cases}
\dfrac{x-1}{20} &\ \ 1 \leq x < 6 \\
\\
\dfrac{9-x}{12} &\ \ 6 \leq x < 9 \\
\\ 0 &\ \ \ \text{elsewhere}
\end{cases}\)
The value of \(k\) such that \(\text{Pr}(X<k)=0.35\) is
- \(\sqrt{14}-1\)
- \(\sqrt{14}+1\)
- \(\sqrt{15}-1\)
- \(\sqrt{15}+1\)
- \(1-\sqrt{15}\)
Calculus, MET2 2023 VCAA 7 MC
Let \(f(x)=\log_{e}x\), where \(x>0\) and \(g(x)=\sqrt{1-x}\), where \(x<1\).
The domain of the derivative of \((f\circ g)(x)\) is
- \(x\in R\)
- \(x\in (-\infty, 1]\)
- \(x\in (-\infty, 1)\)
- \(x\in (0, \infty)\)
- \(x\in (0, 1)\)
Calculus, MET2 2023 VCAA 5 MC
Which one of the following functions has a horizontal tangent at \((0, 0)\)?
- \(y=x^{-\frac{1}{3}}\)
- \(y=x^{\frac{1}{3}}\)
- \(y=x^{\frac{2}{3}}\)
- \(y=x^{\frac{4}{3}}\)
- \(y=x^{\frac{3}{4}}\)
Algebra, MET2 2023 VCAA 4 MC
Consider the system of simultaneous equations below containing the parameter \(k\).
| \(kx+5y\) | \(=k+5\) |
| \(4x+(k+1)y\) | \(=0\) |
The value(s) of \(k\) for which the system of equations has infinite solutions are
- \(k\in \{-5, 4\}\)
- \(k\in \{-5\}\)
- \(k\in \{4\}\)
- \(k\in R\setminus \{-5, 4\}\)
- \(k\in R\setminus \{-5\}\)
Calculus, MET1 2022 VCAA 8
Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
- State the value of `A\left(\frac{\pi}{3}\right)`. (1 mark)
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- Evaluate `f\left(\frac{\pi}{3}\right)`. (2 marks)
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- Consider the average value of the function `f` over the interval `x \in[0, k]`, where `k \in[0,2]`.
- Find the value of `k` that results in the maximum average value. (2 marks)
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Calculus, MET1 2022 VCAA 7
A tilemaker wants to make square tiles of size 20 cm × 20 cm.
The front surface of the tiles is to be painted with two different colours that meet the following conditions:
- Condition 1 - Each colour covers half the front surface of a tile.
- Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.
An example is shown below.
There are two types of tiles: Type A and Type B.
For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.
The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
- i. Find the area of the front surface of each tile. (1 mark)
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ii. Find the value of `a` so that a Type A tile meets Condition 1. (1 mark)
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Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
- Show that a Type B tile meets Condition 1. (3 marks)
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- Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2. (2 marks)
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