How to Use the HSC Physics Data and Formulae Sheets
Learn how to find the right equation, check its assumptions, use physical constants, and solve HSC Physics problems. Includes worked calculations, experimental examples, and practice questions with detailed solutions.
You are given the radius and period of a satellite's orbit and asked to calculate its speed. You turn to the HSC Physics formula sheet and find two equations involving orbital motion. Which one should you use?
The challenge isn't finding an equation that contains the right symbols. It's recognising what physical situation the equation describes, what assumptions it makes, and whether the information in the question is sufficient to use it.
That's the skill you need when using the HSC Physics data and formulae sheets. You should be able to locate a relevant relationship, check that it applies, rearrange it if necessary, and use it without blindly trying every equation on the page.
## First, know what NESA gives you
In the HSC Physics examination, you are provided with three reference resources:
- **Data sheet:** physical constants and useful reference values.
- **Formulae sheet:** mathematical relationships used across the Physics course.
- **Periodic table:** element symbols and relative atomic masses.
These are provided in the examination paper. You don't need to bring your own copy.
There is an important syllabus distinction. As at November 2026, the 2017 Physics Stage 6 syllabus still applies to Year 12 students sitting the 2026 HSC. The 2025 Physics 11-12 syllabus begins implementation for Year 11 in 2027, with its first HSC examination in 2028. The 2025 HSC examination paper is therefore a useful official reference for the existing formulae sheet, but students following the new syllabus should use the examination resources issued for their course rather than assume every formula or sheet will remain unchanged.
You are given the radius and period of a satellite's orbit and asked to calculate its speed. You turn to the HSC Physics formula sheet and find two equations involving orbital motion. Which one should you use?
The challenge isn't finding an equation that contains the right symbols. It's recognising what physical situation the equation describes, what assumptions it makes, and whether the information in the question is sufficient to use it.
That's the skill you need when using the HSC Physics data and formulae sheets. You should be able to locate a relevant relationship, check that it applies, rearrange it if necessary, and use it without blindly trying every equation on the page.
## First, know what NESA gives you
In the HSC Physics examination, you are provided with three reference resources:
- **Data sheet:** physical constants and useful reference values.
- **Formulae sheet:** mathematical relationships used across the Physics course.
- **Periodic table:** element symbols and relative atomic masses.
These are provided in the examination paper. You don't need to bring your own copy.
There is an important syllabus distinction. As at November 2026, the 2017 Physics Stage 6 syllabus still applies to Year 12 students sitting the 2026 HSC. The 2025 Physics 11-12 syllabus begins implementation for Year 11 in 2027, with its first HSC examination in 2028. The 2025 HSC examination paper is therefore a useful official reference for the existing formulae sheet, but students following the new syllabus should use the examination resources issued for their course rather than assume every formula or sheet will remain unchanged.
### What does the examination sheet look like?
The 2025 HSC Physics paper provides a useful example. Its reference pages contain:
1. One page of physical constants and values.
2. Two pages of formulae, arranged into four topic areas.
3. One page containing the periodic table.
You can inspect the arrangement in [NESA's 2025 HSC Physics examination paper](https://www.nsw.gov.au/sites/default/files/noindex/2025-11/2025-hsc-physics.pdf).
[IMAGE: An annotated overview of the reference pages at the back of an HSC Physics examination paper. Show the separate data sheet, the two-page formulae sheet with its four clearly labelled topic sections, and the periodic table. Use callouts to distinguish numerical constants from mathematical relationships.]
The most important distinction is this:
**The data sheet tells you the values of useful constants. The formulae sheet tells you how physical quantities are related.**
You often need both to solve one question.
### The four sections of the formulae sheet
The formulae sheet is organised into four sections.
| Section | What you'll find |
|---|---|
| Motion, forces and gravity | Kinematics, Newton's laws, energy, momentum, circular motion, torque, and gravitation |
| Waves and thermodynamics | Wave relationships, interference, refraction, Doppler effect, and heat transfer |
| Electricity and magnetism | Electric fields, circuits, magnetic forces, magnetic flux, induction, and transformers |
| Quantum, special relativity and nuclear | Photon energy, photoelectric effect, de Broglie wavelength, relativity, and radioactive decay |
This organisation is your first clue when searching for an equation.
If a question involves a satellite orbiting Earth, start with motion, forces and gravity. If it involves electrons emitted by light, look in the quantum section.
You don't need to read every equation on the sheet. Narrow down the relevant section first.
## Understanding the data sheet
The data sheet contains constants that you may need during calculations.
You should recognise the most commonly used ones, understand their units, and know which physical situations require them.
### Important constants to recognise
The following values appear on the 2025 HSC Physics data sheet.
| Constant | Symbol | Value |
|---|---|---|
| Earth's gravitational acceleration | \(g\) | \(9.8\ \mathrm{m\,s^{-2}}\) |
| Universal gravitational constant | \(G\) | \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\) |
| Mass of Earth | \(M_E\) | \(6.0\times10^{24}\ \mathrm{kg}\) |
| Radius of Earth | \(r_E\) | \(6.371\times10^6\ \mathrm{m}\) |
| Speed of light | \(c\) | \(3.00\times10^8\ \mathrm{m\,s^{-1}}\) |
| Planck constant | \(h\) | \(6.626\times10^{-34}\ \mathrm{J\,s}\) |
| Charge on an electron | \(q_e\) | \(-1.602\times10^{-19}\ \mathrm{C}\) |
| Mass of an electron | \(m_e\) | \(9.109\times10^{-31}\ \mathrm{kg}\) |
| Mass of a proton | \(m_p\) | \(1.673\times10^{-27}\ \mathrm{kg}\) |
| Mass of a neutron | \(m_n\) | \(1.675\times10^{-27}\ \mathrm{kg}\) |
| Electric permittivity | \(\varepsilon_0\) | \(8.854\times10^{-12}\ \mathrm{A^2\,s^4\,kg^{-1}\,m^{-3}}\) |
| Magnetic permeability | \(\mu_0\) | \(4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}\) |
| Rydberg constant | \(R\) | \(1.097\times10^7\ \mathrm{m^{-1}}\) |
| Atomic mass unit | \(u\) | \(1.661\times10^{-27}\ \mathrm{kg}\) |
| Electron volt | \(1\ \mathrm{eV}\) | \(1.602\times10^{-19}\ \mathrm{J}\) |
The sheet also provides the speed of sound in air, the density and specific heat capacity of water, and Wien's displacement constant.
You don't need to memorise every numerical value. You do need to recognise what each constant represents so that you can find it quickly.
### A common trap: \(g\) versus \(G\)
These two symbols look almost identical, but they represent different quantities.
- \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\) near Earth's surface.
- \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
For an object near Earth's surface, you might use:
\[
F_g=mg
\]
For the gravitational attraction between two masses, use:
\[
F=\frac{GMm}{r^2}
\]
The second equation comes from the formulae sheet. You obtain \(G\) and the relevant masses from the data sheet.
Notice that \(r\) represents the distance between the centres of the two masses. It is not automatically the height above Earth's surface.
That distinction becomes important when working with satellites.
[IMAGE: Earth shown in cross-section with a satellite above its surface. Label Earth's radius from its centre to the surface, satellite altitude from the surface to the satellite, and orbital radius from Earth's centre to the satellite. Show the relationship r = Earth's radius + altitude.]
## How to choose the correct equation
Here is a method you can use for almost any calculation question in the HSC Physics examination.
### Step 1: Identify exactly what you need to calculate
Start with the final sentence of the question.
Are you finding:
- Velocity?
- Acceleration?
- Force?
- Energy?
- Magnetic flux?
- Induced emf?
- Wavelength?
- Electric field strength?
Write down the unknown before looking for an equation.
This stops you from choosing an equation simply because it contains familiar numbers.
### Step 2: List the information you already have
Extract the known quantities and their values.
For example:
A satellite has an orbital radius of \(7.0\times10^6\ \mathrm{m}\) and a period of \(5.8\times10^3\ \mathrm{s}\).
You know:
- \(r=7.0\times10^6\ \mathrm{m}\)
- \(T=5.8\times10^3\ \mathrm{s}\)
You want to find its orbital speed, \(v\).
This immediately narrows down the possible equations.
### Step 3: Identify the physical situation
This is often the most important step.
Think about what is actually happening.
Is the object:
- Accelerating at a constant rate?
- Moving in a circle?
- Falling under gravity?
- Moving through a magnetic field?
- Transferring energy through a circuit?
- Emitting or absorbing photons?
The physical situation determines which equations are valid.
### Step 4: Find an equation that connects your knowns to your unknown
Don't just look for matching symbols. Check that the equation represents the correct physical relationship.
For example, the formulae sheet contains both:
\[
v=\frac{2\pi r}{T}
\]
and
\[
\frac{r^3}{T^2}=\frac{GM}{4\pi^2}
\]
Both describe aspects of orbital motion.
However, the first calculates orbital speed from radius and period. The second relates orbital radius and period to the mass of the central object.
If you are asked for speed and already know \(r\) and \(T\), the first equation is the direct choice.
### Step 5: Check the assumptions
An equation is only useful when its conditions are satisfied.
Some important examples:
| Equation | Important condition |
|---|---|
| \(v=u+at\) | Acceleration is constant |
| \(s=ut+\frac12at^2\) | Acceleration is constant |
| \(F_c=\frac{mv^2}{r}\) | The net inward force produces circular motion |
| \(\Phi=BA\cos\theta\) | The angle is measured between the magnetic field and the area normal |
| \(Q=mc\Delta T\) | No phase change occurs |
| \(K=\frac12mv^2\) | Classical, non-relativistic speeds |
| \(\varepsilon=-N\frac{\Delta\Phi}{\Delta t}\) | Describes average induced emf over the stated time interval |
If the conditions aren't satisfied, the equation may give an incorrect answer even if the substitution looks perfect.
### Step 6: Rearrange before substituting
Once you've selected the equation, rearrange it to make the unknown the subject.
Then substitute your values, keeping units consistent.
For example, if:
\[
v=\frac{2\pi r}{T}
\]
you already have the correct subject, so you can substitute directly.
If you need to calculate radius instead:
\[
r=\frac{vT}{2\pi}
\]
Rearranging first makes your working easier to follow and reduces algebra mistakes.
## Worked example 1: Choosing between orbital equations
A satellite travels in a circular orbit around Earth. Its orbital radius, measured from Earth's centre, is \(7.00\times10^6\ \mathrm{m}\). It completes one orbit in \(5.80\times10^3\ \mathrm{s}\).
Calculate its orbital speed.
[IMAGE: Earth with a satellite travelling in a circular orbit. Show the orbital radius r extending from Earth's centre to the satellite, a tangential velocity arrow v, and the orbital period T. Do not show the calculated speed.]
**Step 1**: Identify the unknown and known quantities.
We need orbital speed, \(v\).
We know:
\[
r=7.00\times10^6\ \mathrm{m}
\]
\[
T=5.80\times10^3\ \mathrm{s}
\]
**Step 2**: Choose the appropriate equation.
The satellite moves in a circular orbit, and we know its radius and period.
The formulae sheet gives:
\[
v=\frac{2\pi r}{T}
\]
This directly connects the three quantities we need.
We don't need the gravitational constant or Earth's mass for this calculation.
**Step 3**: Substitute the values.
\[
v=\frac{2\pi(7.00\times10^6)}
{5.80\times10^3}
\]
\[
v=7.59\times10^3\ \mathrm{m\,s^{-1}}
\]
**Step 4**: State the result.
The satellite's orbital speed is:
\[
\boxed{v=7.59\times10^3\ \mathrm{m\,s^{-1}}}
\]
This is approximately \(7.59\ \mathrm{km\,s^{-1}}\).
The important decision was recognising that the question asked for speed, not the mass of Earth or the gravitational force.
If the question instead asked for the mass of the central object, the other orbital equation would become useful.
Also remember that if the question gives altitude above Earth's surface rather than orbital radius, you must first calculate:
\[
r=R_E+h_{\mathrm{altitude}}
\]
where \(R_E\) is Earth's radius and \(h_{\mathrm{altitude}}\) is the satellite's height above the surface.
## Using the formulae sheet in an experimental investigation
The formulae sheet becomes particularly useful when a question gives you experimental measurements rather than simply asking you to calculate a quantity.
Consider an investigation into the photoelectric effect.
The experimental question is straightforward:
How does the frequency of incident light affect the maximum kinetic energy of electrons emitted from a metal?
### Understanding the apparatus
A photoelectric cell contains two electrodes inside an evacuated tube.
- The cathode is made from a photosensitive metal.
- The anode collects electrons emitted from the cathode.
- Light enters through a window and strikes the cathode.
- An ammeter measures the current through the circuit.
- A variable power supply allows a reverse potential difference to be applied.
- A voltmeter measures the potential difference across the electrodes.
When light strikes the cathode, electrons can be emitted from its surface. These electrons travel towards the anode, producing a measurable current.
[IMAGE: A scientifically accurate photoelectric cell apparatus. Show a monochromatic light beam entering an evacuated glass tube and striking a metal cathode on the left. Show emitted electrons travelling towards the anode on the right. Connect the electrodes to a variable reverse-bias power supply, an ammeter in series, and a voltmeter across the electrodes. Label the light path, cathode, anode, electron movement, and reverse potential.]
### What should the investigation show?
According to the photon model of light, the energy of an individual photon is:
\[
E=hf
\]
where:
- \(E\) is photon energy in joules
- \(h\) is Planck's constant
- \(f\) is light frequency in hertz
The energy transferred to an emitted electron depends on the energy of the photon that strikes the metal.
Some of this energy is required to release the electron from the metal. This minimum energy is called the work function, represented by \(\phi\).
The formulae sheet provides:
\[
K_{\max}=hf-\phi
\]
where \(K_{\max}\) is the maximum kinetic energy of an emitted electron.
Predict what happens if the light becomes more intense but its frequency stays constant.
You might initially think the electrons should leave with more energy because more light is hitting the metal.
However, intensity and frequency do different jobs.
- Increasing frequency increases the energy of each photon.
- Increasing intensity increases the number of photons arriving each second, assuming the frequency remains constant.
Therefore, increasing intensity can increase the number of emitted electrons without increasing their maximum kinetic energy.
### How is maximum kinetic energy measured?
Simply measuring the photocurrent isn't enough.
A larger current tells us that more charge is flowing each second. It doesn't directly tell us the maximum kinetic energy of the electrons.
We need a different measurement.
The revised method applies a reverse potential difference between the electrodes. This potential opposes the movement of the emitted electrons towards the anode.
The reverse voltage is gradually increased until the photocurrent falls to zero.
The minimum reverse potential required to stop even the fastest emitted electrons is called the stopping potential, \(V_s\).
At this point, the maximum kinetic energy of the electrons has been converted into electrical potential energy.
Using the formulae sheet's relationship \(W=qV\), we obtain:
\[
K_{\max}=eV_s
\]
Here, \(e\) is the magnitude of the electron's charge:
\[
e=1.602\times10^{-19}\ \mathrm{C}
\]
We use the magnitude because \(V_s\) is being treated as a positive stopping-voltage magnitude. The actual polarity of the applied voltage is opposite to the electrons' initial motion.
Combining this with the photoelectric equation:
\[
eV_s=hf-\phi
\]
Rearranging:
\[
V_s=\frac{h}{e}f-\frac{\phi}{e}
\]
This equation predicts a straight-line relationship between stopping potential and frequency.
The gradient is \(h/e\), and the frequency-axis intercept represents the threshold frequency.
[IMAGE: A graph of stopping potential Vs against incident light frequency f. Show a straight line with a positive gradient, three illustrative experimental data points, and an extrapolated intersection with the frequency axis labelled threshold frequency. Annotate the gradient as h/e and the vertical intercept as -phi/e. Clearly distinguish the illustrative data points from the theoretical line.]
### What the experiment establishes
The revised experiment measures something the original current-only method could not: the maximum energy of the emitted electrons.
The pattern of results allows us to test the photon model quantitatively.
There is also an important limitation. If the light frequency is below the metal's threshold frequency, electrons are not emitted in the ordinary photoelectric effect, regardless of how much the intensity is increased.
In that situation, there is no photoelectric stopping potential to measure.
This is a useful example of how the formulae sheet supports experimental reasoning. You aren't just substituting numbers. You're connecting a physical model to measurable quantities.
## Worked example 2: Finding Planck's constant from experimental data
A student investigates the photoelectric effect using a metal surface. The following illustrative data are collected.
| Frequency (\(\times10^{14}\ \mathrm{Hz}\)) | Stopping potential (V) |
|---:|---:|
| 6.0 | 0.48 |
| 7.0 | 0.90 |
| 8.0 | 1.31 |
Use the first and last measurements to estimate Planck's constant.
**Step 1**: Identify the relationship between the measured quantities.
From the photoelectric equation:
\[
eV_s=hf-\phi
\]
Rearranging:
\[
V_s=\frac{h}{e}f-\frac{\phi}{e}
\]
This has the form of a straight-line equation:
\[
y=mx+c
\]
The gradient is:
\[
m=\frac{h}{e}
\]
Therefore:
\[
h=em
\]
**Step 2**: Calculate the gradient.
Using the first and last data points:
\[
m=\frac{\Delta V_s}{\Delta f}
\]
\[
m=\frac{1.31-0.48}
{(8.0-6.0)\times10^{14}}
\]
\[
m=\frac{0.83}{2.0\times10^{14}}
\]
\[
m=4.15\times10^{-15}\ \mathrm{V\,s}
\]
**Step 3**: Calculate Planck's constant.
The data sheet gives the magnitude of the electron charge:
\[
e=1.602\times10^{-19}\ \mathrm{C}
\]
Therefore:
\[
h=em
\]
\[
h=(1.602\times10^{-19})
(4.15\times10^{-15})
\]
\[
h=6.65\times10^{-34}\ \mathrm{J\,s}
\]
**Step 4**: Interpret the result.
Our estimate is:
\[
\boxed{h=6.65\times10^{-34}\ \mathrm{J\,s}}
\]
The data sheet gives:
\[
h=6.626\times10^{-34}\ \mathrm{J\,s}
\]
The two values are close.
The small difference is consistent with using rounded illustrative measurements and estimating the gradient from only two points.
In a real investigation, you would use a line of best fit through all the experimental measurements rather than relying on two endpoints.
Notice how several decisions were required:
- We identified a linear relationship.
- We recognised that the gradient represented \(h/e\).
- We rearranged the equation to make \(h\) the subject.
- We obtained the electron charge from the data sheet.
- We checked the result against the accepted value.
That's what effective use of the formulae sheet looks like.
## Common mistakes when using the HSC Physics formulae sheet
Knowing where the equations are is only part of the job. You also need to avoid a few common traps.
### 1. Choosing an equation because the symbols look familiar
Two equations can contain the same variables but describe different situations.
For example:
\[
F=qE
\]
describes the electric force on a charge in an electric field.
Meanwhile:
\[
F=qvB\sin\theta
\]
describes the magnetic force on a moving charge in a magnetic field.
Both involve charge, \(q\), and force, \(F\). But one requires an electric field, while the other requires a moving charge and a magnetic field.
Always identify the physical interaction first.
### 2. Confusing symbols that have multiple meanings
Some symbols are reused across different areas of Physics.
| Symbol | Possible meanings |
|---|---|
| \(E\) | Electric field strength or energy |
| \(I\) | Electric current or light intensity |
| \(\lambda\) | Wavelength or radioactive decay constant |
| \(T\) | Period or absolute temperature |
| \(n\) | Refractive index or number of coil turns |
| \(W\) | Work, not weight |
| \(G\) | Universal gravitational constant |
| \(g\) | Gravitational acceleration |
Read the surrounding equation and check the units before deciding what a symbol represents.
### 3. Forgetting what the angle represents
This is especially important in electricity, magnetism, and mechanics.
Consider these three equations:
\[
W=Fs\cos\theta
\]
\[
F=qvB\sin\theta
\]
\[
\Phi=BA\cos\theta
\]
The angle means something different in each case.
- For work, it is the angle between force and displacement.
- For magnetic force, it is the angle between velocity and magnetic field.
- For magnetic flux, it is the angle between magnetic field and the normal to the surface.
The magnetic flux angle is a particularly common source of errors.
If the magnetic field is perpendicular to the plane of the coil, it is parallel to the area normal, so \(\theta=0^\circ\) and flux is maximum.
If the field is parallel to the plane of the coil, it is perpendicular to the area normal, so \(\theta=90^\circ\) and flux is zero.
### 4. Forgetting unit conversions
The formulae sheet generally uses SI units.
Convert quantities before substituting.
| Given quantity | Convert to |
|---|---|
| 5 cm | 0.05 m |
| 2 km | 2000 m |
| 300 g | 0.300 kg |
| 5 minutes | 300 s |
| 400 nm | \(4.00\times10^{-7}\) m |
| 2 eV | \(3.204\times10^{-19}\) J |
Pay particular attention to prefixes:
- kilo (k) = \(10^3\)
- centi (c) = \(10^{-2}\)
- milli (m) = \(10^{-3}\)
- micro (\(\mu\)) = \(10^{-6}\)
- nano (n) = \(10^{-9}\)
A missed conversion can change your answer by several orders of magnitude.
### 5. Assuming every useful equation is printed
The formulae sheet is not a complete list of every relationship you need to know.
For example, the 2025 sheet includes:
\[
F=\frac{GMm}{r^2}
\]
but does not separately print the familiar near-surface weight relationship:
\[
F_g=mg
\]
You should understand how the second relationship connects to Newton's second law and gravitational acceleration.
Another example is photon energy.
The sheet provides:
\[
E=hf
\]
and:
\[
c=f\lambda
\]
It does not need to print every possible rearrangement. You can combine these relationships to obtain:
\[
E=\frac{hc}{\lambda}
\]
Learning how to derive relationships from the supplied equations is much more useful than memorising a long list of isolated formulas.
### 6. Treating centripetal force as an extra force
The formula:
\[
F_c=\frac{mv^2}{r}
\]
describes the net inward force required for circular motion.
It is not an additional force that you add to gravity, tension, or the normal force.
You must first identify the actual forces acting on the object, then calculate their resultant towards the centre of the circle.
This distinction is essential for vertical circular motion, orbital motion, and objects travelling over curved surfaces.
### 7. Writing only the substituted equation
A correct numerical answer doesn't automatically demonstrate correct physical reasoning.
NESA's 2025 HSC Physics marking feedback specifically emphasises showing the relevant formula, substitution, correct units, and directions for vector quantities.
A reliable calculation should show:
1. The relevant equation.
2. The rearrangement, if required.
3. Substitution of values.
4. Intermediate working.
5. Final answer with units.
6. Direction, where the quantity is a vector.
You can read the [official 2025 HSC Physics marking feedback](https://www.nsw.gov.au/education-and-training/nesa/curriculum/hsc-exam-papers/physics/2025) for examples of the calculation and explanation skills markers expect.
## Questions and solutions
Try each question before reading its solution. The questions gradually move from straightforward equation selection to situations where you need to combine several relationships and examine their assumptions.
### Question 1
A trolley starts from rest and accelerates along a straight, horizontal track at a constant rate of \(1.8\ \mathrm{m\,s^{-2}}\) for \(4.0\ \mathrm{s}\).
Calculate:
(a) The trolley's displacement.
(b) Its final velocity.
[IMAGE: A trolley travelling along a straight horizontal track. Show its initial position, an arrow indicating forward motion, a constant acceleration arrow of 1.8 m/s squared, and the 4.0 s time interval. Do not show the calculated displacement or final velocity.]
### Solution 1
**The trolley travels 14.4 m and reaches a final velocity of \(7.2\ \mathrm{m\,s^{-1}}\).**
The question describes motion with constant acceleration, so we use the kinematics equations from the motion, forces and gravity section.
**Part (a): Displacement**
The known quantities are:
\[
u=0\ \mathrm{m\,s^{-1}}
\]
\[
a=1.8\ \mathrm{m\,s^{-2}}
\]
\[
t=4.0\ \mathrm{s}
\]
The formulae sheet provides:
\[
s=ut+\frac12at^2
\]
Substitute:
\[
s=(0)(4.0)+\frac12(1.8)(4.0)^2
\]
\[
s=0.9\times16
\]
\[
\boxed{s=14.4\ \mathrm{m}}
\]
**Part (b): Final velocity**
Use:
\[
v=u+at
\]
Substitute:
\[
v=0+(1.8)(4.0)
\]
\[
\boxed{v=7.2\ \mathrm{m\,s^{-1}}}
\]
Both equations are valid because the acceleration is constant.
The important point is that we selected the displacement equation for part (a) and the velocity equation for part (b), rather than trying to force one equation to answer both parts.
### Question 2
A \(900\ \mathrm{kg}\) car travels over the top of a convex bridge. At the highest point, the bridge has a radius of curvature of \(40\ \mathrm{m}\), and the car's speed is \(16\ \mathrm{m\,s^{-1}}\).
Assume the car maintains contact with the bridge.
(a) Calculate the magnitude of the normal force exerted by the bridge on the car.
(b) Determine the maximum speed at which the car can travel over the top before losing contact with the bridge.
Use \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: A car at the highest point of a convex curved bridge. Show the centre of curvature below the car, the radius of curvature of 40 m extending towards that centre, and a horizontal velocity arrow. Show the car, bridge surface, and gravitational vertical orientation, but do not draw the forces or their components.]
### Solution 2
**The normal force is \(3.06\times10^3\ \mathrm{N}\), and the maximum speed before losing contact is approximately \(19.8\ \mathrm{m\,s^{-1}}\).**
The key is recognising that the centripetal force is the net inward force, not an additional force acting on the car.
**Part (a): Normal force**
At the top of the bridge, the centre of curvature is below the car.
Therefore, the inward direction is vertically downwards.
The actual forces are:
- Gravitational force, \(mg\), acting downwards.
- Normal force, \(F_N\), acting upwards.
The resultant inward force is:
\[
mg-F_N=\frac{mv^2}{r}
\]
Rearrange:
\[
F_N=mg-\frac{mv^2}{r}
\]
Substitute:
\[
F_N=(900)(9.8)
-\frac{(900)(16)^2}{40}
\]
\[
F_N=8820-5760
\]
\[
\boxed{F_N=3060\ \mathrm{N}}
\]
The bridge exerts an upward force of \(3060\ \mathrm{N}\) on the car.
**Part (b): Maximum speed before losing contact**
The car loses contact when the normal force becomes zero.
At this instant, gravity provides the entire centripetal force:
\[
mg=\frac{mv^2}{r}
\]
Cancel the mass:
\[
g=\frac{v^2}{r}
\]
Rearrange:
\[
v=\sqrt{gr}
\]
Substitute:
\[
v=\sqrt{(9.8)(40)}
\]
\[
\boxed{v=19.8\ \mathrm{m\,s^{-1}}}
\]
The car must travel below this limiting speed to maintain contact with the bridge.
The tempting mistake is to write \(F_N=mv^2/r\). That ignores gravity, which is already providing part of the required inward force.
### Question 3
A coil containing 30 turns has an area of \(0.020\ \mathrm{m^2}\). It is placed in a uniform magnetic field of \(0.40\ \mathrm{T}\).
Initially, the normal to the coil's surface is parallel to the magnetic field. The coil is then rotated through \(60^\circ\) in \(0.050\ \mathrm{s}\).
(a) Calculate the magnitude of the average induced emf.
(b) The coil is subsequently rotated back to its original position over the same time interval. Explain how the magnitude and polarity of the induced emf change.
[IMAGE: A rectangular coil rotating between the north and south poles of a magnet. Show the uniform magnetic field across the coil, the coil's initial orientation with its area normal parallel to the field, and its final orientation rotated by 60 degrees. Label the 30 turns, coil area, magnetic field, rotation angle, and time interval. Do not show induced emf or the calculated flux.]
### Solution 3
**The magnitude of the average induced emf is \(2.4\ \mathrm{V}\). Reversing the rotation back to the original position produces an emf of the same magnitude but opposite polarity.**
This question requires two equations from the electricity and magnetism section:
\[
\Phi=BA\cos\theta
\]
and
\[
\varepsilon=-N\frac{\Delta\Phi}{\Delta t}
\]
Here:
- \(\Phi\) is magnetic flux through one turn, in webers.
- \(B\) is magnetic field strength, in teslas.
- \(A\) is coil area, in square metres.
- \(\theta\) is the angle between the magnetic field and the area normal.
- \(N\) is the number of turns.
- \(\varepsilon\) is induced emf, in volts.
**Part (a): Average induced emf**
Initially, the area normal is parallel to the magnetic field.
Therefore:
\[
\theta_i=0^\circ
\]
The initial flux is:
\[
\Phi_i=BA\cos0^\circ
\]
\[
\Phi_i=(0.40)(0.020)(1)
\]
\[
\Phi_i=0.0080\ \mathrm{Wb}
\]
After rotation:
\[
\theta_f=60^\circ
\]
The final flux is:
\[
\Phi_f=BA\cos60^\circ
\]
\[
\Phi_f=(0.40)(0.020)(0.5)
\]
\[
\Phi_f=0.0040\ \mathrm{Wb}
\]
The change in flux per turn is:
\[
\Delta\Phi=0.0040-0.0080
\]
\[
\Delta\Phi=-0.0040\ \mathrm{Wb}
\]
Now apply Faraday's law:
\[
\varepsilon=-N\frac{\Delta\Phi}{\Delta t}
\]
\[
\varepsilon=-30\left(
\frac{-0.0040}{0.050}
\right)
\]
\[
\boxed{\varepsilon=2.4\ \mathrm{V}}
\]
This is the average induced emf magnitude.
**Part (b): Reversing the rotation**
When the coil rotates back from \(60^\circ\) to \(0^\circ\), the magnetic flux increases instead of decreasing.
The change in flux therefore has the opposite sign.
According to Faraday's law, the induced emf also changes sign.
Its magnitude remains \(2.4\ \mathrm{V}\), provided the coil completes the same angular displacement in the same time.
The physical reason is Lenz's law: the induced current acts in a direction that opposes the change in magnetic flux.
A common mistake is to use \(BA\sin\theta\) for magnetic flux without checking what the angle is measured from. The formulae sheet uses the angle between the magnetic field and the area normal.
### Question 4
A satellite of mass \(500\ \mathrm{kg}\) orbits Earth at an altitude of \(400\ \mathrm{km}\) above the surface.
Calculate:
(a) The gravitational acceleration at the satellite's location.
(b) The gravitational force acting on the satellite.
Use the following data sheet values:
\[
G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}
\]
\[
M_E=6.0\times10^{24}\ \mathrm{kg}
\]
\[
R_E=6.371\times10^6\ \mathrm{m}
\]
[IMAGE: Earth with a satellite in a circular orbit 400 km above its surface. Show Earth's centre, Earth's radius, the 400 km altitude measured from the surface, and the full orbital radius measured from Earth's centre. Label the satellite mass as 500 kg. Do not include calculated gravitational acceleration or force.]
### Solution 4
**The gravitational acceleration is approximately \(8.73\ \mathrm{m\,s^{-2}}\), and the gravitational force on the satellite is approximately \(4.37\times10^3\ \mathrm{N}\).**
The main decision is recognising that the satellite is not near Earth's surface in the sense required to use \(g=9.8\ \mathrm{m\,s^{-2}}\) unchanged.
We need the universal gravitational equation.
**Part (a): Gravitational acceleration**
The formulae sheet gives:
\[
F=\frac{GMm}{r^2}
\]
Using Newton's second law:
\[
F=ma
\]
For gravitational acceleration at the satellite's location:
\[
mg_r=\frac{GM_Em}{r^2}
\]
Cancel the satellite's mass:
\[
g_r=\frac{GM_E}{r^2}
\]
Before substituting, we must calculate the distance from Earth's centre.
The question gives altitude, not orbital radius.
\[
r=R_E+h
\]
\[
r=6.371\times10^6+400\times10^3
\]
\[
r=6.771\times10^6\ \mathrm{m}
\]
Now substitute:
\[
g_r=
\frac{(6.67\times10^{-11})
(6.0\times10^{24})}
{(6.771\times10^6)^2}
\]
\[
\boxed{g_r=8.73\ \mathrm{m\,s^{-2}}}
\]
**Part (b): Gravitational force**
Use:
\[
F=mg_r
\]
Substitute:
\[
F=(500)(8.73)
\]
\[
\boxed{F=4.37\times10^3\ \mathrm{N}}
\]
The force is directed towards Earth's centre.
The important distinction is that gravitational acceleration depends on distance from Earth's centre, whereas gravitational force also depends on the satellite's mass.
The tempting approach is to use \(g=9.8\ \mathrm{m\,s^{-2}}\) or substitute \(400\ \mathrm{km}\) directly into the gravitational equation. Both approaches ignore the actual distance from Earth's centre.
### Question 5
A metal surface has a work function of \(3.00\times10^{-19}\ \mathrm{J}\). It is illuminated with monochromatic light of frequency \(8.00\times10^{14}\ \mathrm{Hz}\).
The emitted photoelectrons enter a uniform magnetic field of \(1.50\times10^{-3}\ \mathrm{T}\), with their initial velocities perpendicular to the field.
Assume the emitted electrons have a range of kinetic energies up to the maximum predicted by the photoelectric equation.
(a) Calculate the maximum kinetic energy of an emitted electron.
(b) Calculate the maximum speed and largest circular path radius of the electrons in the magnetic field.
(c) Calculate the de Broglie wavelength of the fastest emitted electron.
(d) The intensity of the incident light is doubled while its frequency remains unchanged. Explain what happens to:
- The maximum kinetic energy.
- The largest circular path radius.
- The stopping potential.
- The number of electrons emitted per second.
Use the following data sheet values:
\[
h=6.626\times10^{-34}\ \mathrm{J\,s}
\]
\[
m_e=9.109\times10^{-31}\ \mathrm{kg}
\]
\[
e=1.602\times10^{-19}\ \mathrm{C}
\]
Assume non-relativistic electron speeds, negligible interactions between emitted electrons, and ideal collection conditions.
[IMAGE: An illuminated metal cathode emitting electrons into a uniform magnetic field directed into the page. Show a monochromatic light beam striking the cathode, electrons emerging from its surface, and several curved trajectories with different radii in the magnetic field. Do not label which trajectory represents the fastest electron or provide calculated values. Include a clearly indicated direction for the initial electron velocities and magnetic field.]
### Solution 5
**The maximum kinetic energy is \(2.30\times10^{-19}\ \mathrm{J}\), the maximum speed is \(7.11\times10^5\ \mathrm{m\,s^{-1}}\), the largest orbit radius is \(2.69\ \mathrm{mm}\), and the fastest electrons have a de Broglie wavelength of approximately \(1.02\ \mathrm{nm}\). Doubling the light intensity increases the emission rate but does not change these maximum-energy quantities or the stopping potential.**
This problem requires relationships from three parts of the formulae sheet:
- Quantum physics: photoelectric energy and de Broglie wavelength.
- Motion and forces: circular motion.
- Electricity and magnetism: magnetic force.
The important conceptual point is that the emitted electrons do not all have the same kinetic energy.
**Part (a): Maximum kinetic energy**
Start with the photoelectric equation:
\[
K_{\max}=hf-\phi
\]
Substitute:
\[
K_{\max}=
(6.626\times10^{-34})
(8.00\times10^{14})
-3.00\times10^{-19}
\]
\[
K_{\max}=5.3008\times10^{-19}
-3.00\times10^{-19}
\]
\[
\boxed{K_{\max}=2.30\times10^{-19}\ \mathrm{J}}
\]
This is the maximum kinetic energy available to an emitted electron.
**Part (b): Maximum speed and largest radius**
First, use the kinetic energy equation:
\[
K=\frac12mv^2
\]
Rearrange:
\[
v=\sqrt{\frac{2K}{m}}
\]
For the fastest electron:
\[
v_{\max}=
\sqrt{
\frac{2(2.3008\times10^{-19})}
{9.109\times10^{-31}}
}
\]
\[
\boxed{v_{\max}=7.11\times10^5\ \mathrm{m\,s^{-1}}}
\]
Now consider the magnetic field.
The magnetic force on a moving charge is:
\[
F=qvB\sin\theta
\]
Since the electron enters perpendicular to the magnetic field:
\[
\theta=90^\circ
\]
Therefore, the magnitude of the magnetic force is:
\[
F=e vB
\]
This force provides the centripetal force:
\[
evB=\frac{mv^2}{r}
\]
Rearrange:
\[
r=\frac{mv}{eB}
\]
For the fastest electron:
\[
r_{\max}=
\frac{(9.109\times10^{-31})
(7.11\times10^5)}
{(1.602\times10^{-19})
(1.50\times10^{-3})}
\]
\[
\boxed{r_{\max}=2.69\times10^{-3}\ \mathrm{m}}
\]
Or:
\[
\boxed{r_{\max}=2.69\ \mathrm{mm}}
\]
Why is this the largest radius?
Because:
\[
r=\frac{mv}{eB}
\]
For a fixed electron mass, charge, and magnetic field, the radius increases with speed.
The fastest emitted electrons therefore follow the largest circular paths.
Slower electrons follow smaller paths.
**Part (c): de Broglie wavelength**
The formulae sheet gives:
\[
\lambda=\frac{h}{mv}
\]
Substitute the values for the fastest electron:
\[
\lambda=
\frac{6.626\times10^{-34}}
{(9.109\times10^{-31})
(7.11\times10^5)}
\]
\[
\boxed{\lambda=1.02\times10^{-9}\ \mathrm{m}}
\]
This is approximately \(1.02\ \mathrm{nm}\).
The wavelength is associated with the electron's momentum. Faster electrons have greater momentum and therefore shorter de Broglie wavelengths.
**Part (d): Doubling the light intensity**
The maximum kinetic energy remains unchanged.
The frequency is unchanged, so the energy of each photon remains:
\[
E=hf
\]
The work function also remains constant.
Therefore:
\[
K_{\max}=hf-\phi
\]
does not change.
The largest circular radius also remains unchanged because the maximum electron speed is unchanged.
The stopping potential remains unchanged because:
\[
eV_s=K_{\max}
\]
However, the number of electrons emitted per second increases.
At the same frequency, doubling the light intensity approximately doubles the photon arrival rate. Under the stated ideal conditions, this approximately doubles the photocurrent.
The tempting prediction is that brighter light produces faster electrons and therefore larger circular paths. That confuses photon number with photon energy.
The frequency determines the energy per photon. The intensity determines how many photons arrive.
There is another subtle trap here.
You cannot assume that every emitted electron travels at \(7.11\times10^5\ \mathrm{m\,s^{-1}}\). That is the maximum speed, not the speed of every electron.
The electrons have a range of kinetic energies, so they produce a range of circular path radii. The fastest electrons define the outermost radius.
Doubling the intensity produces more electrons across that distribution, not a larger maximum radius.
This is why understanding the assumptions behind an equation matters just as much as finding it on the formulae sheet.
## What to practise next
Once you can select equations confidently, the next skill is recognising how those equations translate into experimental graphs.
For example, the photoelectric investigation used:
\[
V_s=\frac{h}{e}f-\frac{\phi}{e}
\]
That relationship allowed us to identify the gradient, calculate Planck's constant, and interpret the threshold frequency.
The same reasoning appears throughout HSC Physics. You may be given experimental measurements, asked to construct a graph, interpret its gradient, or evaluate whether the results support a physical model.
Being familiar with the data and formulae sheets makes those questions much easier to approach because you can focus on the physics rather than spending valuable time searching for an equation.