Physics HSC Reference Sheet — Explained
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The NESA Physics reference sheet is provided in every HSC Physics exam — but knowing a constant exists is very different from knowing how to apply it under pressure. Below, every constant and formula is explained: what the symbols mean, when to reach for it, a worked example, and practice to try yourself.
In short: NESA provides a data sheet, formula sheet and periodic table in every HSC Physics exam — but knowing a constant exists is very different from applying it under pressure. This guide explains every constant and formula on the sheet, grouped by topic (kinematics, dynamics, circular and projectile motion, gravitation, waves, electricity and magnetism, electromagnetism), with the context for when to use each.
Constants & Data
14The speed at which electromagnetic radiation (light, radio waves, X-rays) travels through a vacuum. It is a universal speed limit and appears in relativity, optics, and electromagnetism.
Used in: hsc-physics-module7-nature-of-light-worked-solutions
The magnitude of the electric charge carried by a single electron (negative) or proton (positive). All electric charges are integer multiples of this elementary charge.
Used in: hsc-physics-module6-electromagnetism-worked-solutions
Rest mass of an electron. Used in quantum mechanics, photoelectric effect calculations, and special relativity problems involving atomic particles.
Used in: hsc-physics-module8-universe-atom-worked-solutions
Rest mass of a proton. Used alongside the electron mass in nuclear physics and mass-energy equivalence calculations.
Used in: hsc-physics-module8-universe-atom-worked-solutions
Rest mass of a neutron. Slightly greater than the proton mass; used in nuclear binding energy and mass defect calculations.
Used in: hsc-physics-module8-universe-atom-worked-solutions
The fundamental quantum of action relating the energy of a photon to its frequency ($E = hf$). It sets the scale of quantum effects.
Used in: hsc-physics-module8-universe-atom-worked-solutions
Proportionality constant in Newton's law of universal gravitation. Its small value reflects that gravity is by far the weakest fundamental force.
Used in: hsc-physics-module5-advanced-mechanics-worked-solutions
Standard gravitational field strength at Earth's surface. Used whenever a problem states 'near Earth's surface' or doesn't give altitude data.
Used in: hsc-physics-module5-advanced-mechanics-worked-solutions
Relates the magnetic field produced by a current to the geometry of the conductor. Appears in Ampere's law and inductance formulae.
Used in: hsc-physics-module6-electromagnetism-worked-solutions
Relates the electric field to the charge distribution producing it. Appears in Coulomb's law and capacitance formulae.
Used in: hsc-physics-module6-electromagnetism-worked-solutions
Total mass of planet Earth. Used in orbital mechanics and gravitational field calculations at various altitudes.
Used in: hsc-physics-module5-advanced-mechanics-worked-solutions
Mean radius of Earth. Used to convert between surface gravity $g$ and the universal gravitational constant $G$, and to find altitudes above the surface.
Used in: hsc-physics-module5-advanced-mechanics-worked-solutions
Speed of sound in air used in HSC calculations (NESA data sheet value).
Proportionality constant in Wien's displacement law, relating the peak wavelength of a black-body spectrum to its temperature. A hotter object peaks at a shorter (bluer) wavelength.
Used in: hsc-physics-module8-universe-atom-worked-solutions
Kinematics
4What each symbol means
$v$ — final velocity (m s⁻¹); $u$ — initial velocity (m s⁻¹); $a$ — acceleration (m s⁻²); $t$ — time (s).
When to use it
Use when you know three of the four variables and need the fourth — and displacement $s$ is not involved.
Units:
m s⁻¹
Worked sample
A car accelerates from rest at $3.0$ m s⁻² for $5.0$ s. Find its final speed.
$v = 0 + (3.0)(5.0) = 15$ m s⁻¹.
Your turn:
A ball is thrown upward at $20$ m s⁻¹. Taking $g = 9.8$ m s⁻², how long before it momentarily stops?
$0 = 20 - 9.8t \Rightarrow t = 2.04$ s.
Assign a positive direction first and be consistent — deceleration means $a$ is negative.
What each symbol means
$s$ — displacement (m); $u$ — initial velocity (m s⁻¹); $a$ — acceleration (m s⁻²); $t$ — time (s).
When to use it
Use when final velocity is unknown and you need displacement (or vice-versa), given time.
Units:
m
Worked sample
A stone is dropped from rest. How far does it fall in $3.0$ s? ($g = 9.8$ m s⁻²)
$s = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = 44.1$ m.
Your turn:
A train at $10$ m s⁻¹ accelerates at $2.0$ m s⁻² for $4.0$ s. Find its displacement.
$s = (10)(4) + \tfrac{1}{2}(2.0)(16) = 40 + 16 = 56$ m.
For free fall from rest, $s = \tfrac{1}{2}gt^2$ is a neat form to remember.
What each symbol means
$v$ — final velocity (m s⁻¹); $u$ — initial velocity (m s⁻¹); $a$ — acceleration (m s⁻²); $s$ — displacement (m).
When to use it
Use when time is not given or not required — links velocity and displacement directly.
Units:
m s⁻¹ (after taking square root)
Worked sample
A cyclist brakes from $15$ m s⁻¹ to rest over $22.5$ m. Find the deceleration.
$0 = 15^2 + 2a(22.5) \Rightarrow a = -225/45 = -5.0$ m s⁻².
Your turn:
A ball is hit upward at $14$ m s⁻¹. Find the maximum height ($g = 9.8$ m s⁻²).
$0 = 14^2 - 2(9.8)s \Rightarrow s = 196/19.6 = 10$ m.
When the object momentarily stops, $v = 0$, which simplifies the algebra greatly.
What each symbol means
$s$ — displacement (m); $u$ — initial velocity (m s⁻¹); $v$ — final velocity (m s⁻¹); $t$ — time (s).
When to use it
Use when you know both initial and final velocities and the time elapsed but not the acceleration.
Units:
m
Worked sample
An object accelerates from $4$ m s⁻¹ to $10$ m s⁻¹ in $3$ s. Find displacement.
$s = \tfrac{(4+10)}{2} \times 3 = 7 \times 3 = 21$ m.
Your turn:
A runner slows from $8$ m s⁻¹ to $2$ m s⁻¹ in $6$ s. Find the displacement.
$s = \tfrac{(8+2)}{2} \times 6 = 5 \times 6 = 30$ m.
This only works for uniform acceleration. Don't use it for variable acceleration.
Dynamics & Forces
7What each symbol means
$F_{\text{net}}$ — net (resultant) force (N); $m$ — mass (kg); $a$ — acceleration (m s⁻²).
When to use it
Use whenever a net force causes acceleration. Draw a free-body diagram first to identify all forces.
Units:
N (= kg m s⁻²)
Worked sample
A $5.0$ kg box is pushed with $20$ N. Friction is $8$ N. Find the acceleration.
$F_{\text{net}} = 20 - 8 = 12$ N; $a = 12/5.0 = 2.4$ m s⁻².
Your turn:
A $1200$ kg car brakes with a net force of $6000$ N. Find deceleration.
$a = 6000/1200 = 5.0$ m s⁻².
Always substitute the net force, not just one force on the object.
What each symbol means
$p$ — momentum (kg m s⁻¹); $m$ — mass (kg); $v$ — velocity (m s⁻¹).
When to use it
Use in collision and explosion problems. Momentum is always conserved when no external force acts.
Units:
kg m s⁻¹
Worked sample
Find the momentum of a $0.50$ kg ball moving at $12$ m s⁻¹.
$p = (0.50)(12) = 6.0$ kg m s⁻¹.
Your turn:
A $70$ kg person runs at $5.0$ m s⁻¹. Find their momentum.
$p = (70)(5.0) = 350$ kg m s⁻¹.
Momentum is a vector — direction matters. Assign positive to the initial direction of motion.
What each symbol means
$J$ — impulse (N s); $F$ — (average) force (N); $\Delta t$ — time interval (s); $\Delta p = mv - mu$ — change in momentum (kg m s⁻¹).
When to use it
Use when a force acts for a short time (collision, kick, rocket burst) and you want the change in momentum or the average force.
Units:
N s (= kg m s⁻¹)
Worked sample
A $0.20$ kg ball hits a wall at $15$ m s⁻¹ and rebounds at $10$ m s⁻¹. Find the magnitude of impulse.
$J = m(v - u) = 0.20(10 - (-15)) = 0.20 \times 25 = 5.0$ N s.
Your turn:
A $0.5$ kg ball is kicked and its velocity changes from $0$ to $20$ m s⁻¹ in $0.1$ s. Find the average force.
$F = \Delta p / \Delta t = (0.5 \times 20)/0.1 = 100$ N.
When calculating $\Delta p$, use $p_f - p_i$ and watch the sign if the object reverses direction.
What each symbol means
$W$ — work (J); $F$ — force (N); $s$ — displacement (m); $\theta$ — angle between force and displacement.
When to use it
Use when a force is applied at an angle to the direction of motion. If force and displacement are parallel, $\theta = 0$ and $W = Fs$.
Units:
J (= N m)
Worked sample
A $50$ N force at $30°$ to horizontal moves a box $4.0$ m horizontally. Find the work done.
$W = 50 \times 4.0 \times \cos 30° = 200 \times 0.866 = 173$ J.
Your turn:
A $200$ N horizontal force pushes a crate $5.0$ m on a flat floor. Find the work done.
$W = 200 \times 5.0 \times \cos 0° = 1000$ J.
A force perpendicular to displacement ($\theta = 90°$) does zero work — centripetal force is a classic example.
What each symbol means
$P$ — power (W); $W$ — work (J); $t$ — time (s); $F$ — force (N); $v$ — velocity (m s⁻¹).
When to use it
Use $P = W/t$ for total energy delivered over a time. Use $P = Fv$ for instantaneous power when force and velocity are parallel.
Units:
W (= J s⁻¹)
Worked sample
A motor exerts a constant $400$ N force on a vehicle moving at $10$ m s⁻¹. Find the power output.
$P = Fv = 400 \times 10 = 4000$ W $= 4.0$ kW.
Your turn:
A crane lifts $500$ kg through $8.0$ m in $20$ s. Find the power ($g = 9.8$ m s⁻²).
$W = mgh = 500 \times 9.8 \times 8.0 = 39200$ J; $P = 39200/20 = 1960$ W.
$P = Fv$ is extremely handy in engine/vehicle problems where you have force and speed.
What each symbol means
$E_k$ — kinetic energy (J); $m$ — mass (kg); $v$ — speed (m s⁻¹).
When to use it
Use to find energy stored in a moving object, or to relate speed before/after a collision or energy conversion.
Units:
J
Worked sample
Find the kinetic energy of a $0.30$ kg cricket ball at $36$ m s⁻¹.
$E_k = \tfrac{1}{2}(0.30)(36)^2 = 0.15 \times 1296 = 194$ J.
Your turn:
A $2.0$ kg rock falls from rest through $10$ m. Find its speed just before impact ($g = 9.8$ m s⁻²).
Using energy: $E_k = mgh = 2.0 \times 9.8 \times 10 = 196$ J; $v = \sqrt{2 \times 196 / 2.0} = 14$ m s⁻¹.
KE depends on $v^2$, so doubling speed quadruples kinetic energy — a common exam twist.
What each symbol means
$E_p$ — gravitational PE (J); $m$ — mass (kg); $g$ — gravitational field strength (m s⁻²); $h$ — height above reference level (m).
When to use it
Use near Earth's surface (constant $g$) when an object changes height. The reference level ($h = 0$) is your choice — pick the most convenient point.
Units:
J
Worked sample
How much gravitational PE is gained when a $70$ kg person climbs $15$ m? ($g = 9.8$ m s⁻²)
$E_p = 70 \times 9.8 \times 15 = 10\,290$ J.
Your turn:
A $5.0$ kg ball is released from $2.0$ m. What is its speed at ground level?
$E_p = mgh = 98$ J $= E_k$; $v = \sqrt{2 \times 98 / 5.0} = \sqrt{39.2} \approx 6.3$ m s⁻¹.
For variable gravity (high altitudes), use $E_p = -GMm/r$ instead.
Circular & Projectile Motion
4What each symbol means
$a_c$ — centripetal acceleration (m s⁻²); $v$ — tangential speed (m s⁻¹); $r$ — radius (m); $T$ — period (s).
When to use it
Use for any object moving in a circle at constant speed. The acceleration points toward the centre.
Units:
m s⁻²
Worked sample
A $0.50$ kg ball on a $0.80$ m string moves at $4.0$ m s⁻¹. Find the centripetal acceleration.
$a_c = v^2/r = 16/0.80 = 20$ m s⁻².
Your turn:
A satellite orbits at radius $7.0 \times 10^6$ m with period $5.8 \times 10^3$ s. Find $a_c$.
$a_c = 4\pi^2 \times 7.0 \times 10^6 / (5.8 \times 10^3)^2 \approx 8.2$ m s⁻².
The centripetal force is not a separate force — it is the net inward force provided by gravity, tension, normal force, etc.
What each symbol means
$F_c$ — centripetal force (N); $m$ — mass (kg); $v$ — tangential speed (m s⁻¹); $r$ — radius (m); $T$ — period (s).
When to use it
Use to find the net inward force required to keep an object moving in a circle — or to identify what provides it (gravity, friction, tension, normal force).
Units:
N
Worked sample
A $1200$ kg car rounds a $50$ m radius curve at $20$ m s⁻¹. Find the required centripetal force.
$F_c = mv^2/r = 1200 \times 400 / 50 = 9600$ N.
Your turn:
A $0.20$ kg ball on a $1.0$ m string completes one circle per $0.5$ s. Find the tension.
$F_c = 4\pi^2 \times 0.20 \times 1.0 / (0.5)^2 = 4\pi^2 \times 0.8 \approx 31.6$ N.
In vertical circles, tension at the bottom must support weight and provide centripetal force; at the top the two subtract.
What each symbol means
$x$ — horizontal displacement (m); $v_x$ — horizontal speed (constant, m s⁻¹); $v_0$ — launch speed (m s⁻¹); $\theta$ — launch angle; $t$ — time (s).
When to use it
Horizontal motion is uniform (constant velocity) — no horizontal acceleration. Apply throughout flight.
Units:
m
Worked sample
A ball launched at $20$ m s⁻¹ horizontally travels for $3.0$ s. Find horizontal range.
$x = 20 \times 3.0 = 60$ m.
Your turn:
A projectile is launched at $30°$ with $v_0 = 40$ m s⁻¹. Find horizontal range if air time is $4.1$ s.
$v_x = 40\cos 30° = 34.6$ m s⁻¹; $x = 34.6 \times 4.1 \approx 142$ m.
Horizontal and vertical motions are completely independent — solve them separately then combine.
What each symbol means
$y$ — vertical displacement (m, upward positive); $v_0$ — launch speed (m s⁻¹); $\theta$ — launch angle; $g$ — gravitational acceleration (m s⁻²); $t$ — time (s).
When to use it
Vertical motion has constant downward acceleration $g$. Use to find height at any time or time of flight.
Units:
m
Worked sample
A ball is launched at $45°$ with $v_0 = 20$ m s⁻¹. Find height at $t = 1.0$ s.
$v_y = 20\sin 45° = 14.14$ m s⁻¹; $y = 14.14(1.0) - \tfrac{1}{2}(9.8)(1.0)^2 = 14.14 - 4.9 = 9.2$ m.
Your turn:
A ball is launched horizontally from a cliff at $15$ m s⁻¹ and hits the ground after $2.0$ s. How high is the cliff?
$y = \tfrac{1}{2}(9.8)(2.0)^2 = 19.6$ m.
At maximum height, vertical velocity $v_y = 0$ — use $v_y = v_0\sin\theta - gt$ to find the time to the peak.
Gravitation & Orbits
4What each symbol means
$F_g$ — gravitational force (N); $G = 6.67 \times 10^{-11}$ N m² kg⁻²; $m_1, m_2$ — masses (kg); $r$ — centre-to-centre distance (m).
When to use it
Use to find the gravitational force between any two masses, or the gravitational field strength $g = GM/r^2$ at distance $r$.
Units:
N
Worked sample
Find the gravitational force between Earth ($6.0 \times 10^{24}$ kg) and a $70$ kg person on the surface ($r = 6.37 \times 10^6$ m).
$F = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 70}{(6.37 \times 10^6)^2} \approx 686$ N, consistent with $mg$.
Your turn:
Two $1.0$ kg masses are $1.0$ m apart. Find the gravitational force between them.
$F = 6.67 \times 10^{-11}$ N.
Doubling the distance quarters the force ($F \propto 1/r^2$). This inverse-square relationship is heavily tested.
What each symbol means
$U$ — gravitational PE (J, always negative); $G$ — gravitational constant; $m_1, m_2$ — masses (kg); $r$ — separation (m).
When to use it
Use for objects far from Earth's surface where $g$ is not constant — satellites, planetary motion, escape velocity.
Units:
J
Worked sample
Find the gravitational PE of a $500$ kg satellite at $r = 7.0 \times 10^6$ m from Earth's centre.
$U = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 500}{7.0 \times 10^6} = -2.86 \times 10^{10}$ J.
Your turn:
Why is gravitational PE always negative?
We set $U = 0$ at infinite separation. Bringing masses closer releases energy, so PE decreases below zero.
The total mechanical energy of an orbit is $E = -Gm_1m_2/(2r)$ — half the PE.
What each symbol means
$v$ — orbital speed (m s⁻¹); $G$ — gravitational constant; $M$ — mass of central body (kg); $r$ — orbital radius (m).
When to use it
Use for circular orbits. Derived by equating gravitational force to centripetal force.
Units:
m s⁻¹
Worked sample
Find the orbital speed of the ISS at $r = 6.77 \times 10^6$ m from Earth's centre.
$v = \sqrt{6.67 \times 10^{-11} \times 6.0 \times 10^{24} / 6.77 \times 10^6} \approx 7680$ m s⁻¹.
Your turn:
How does orbital speed change if the radius doubles?
$v \propto 1/\sqrt{r}$, so $v$ decreases by a factor of $\sqrt{2}$.
Higher orbits are slower — counter-intuitive but a common multi-choice trap.
What each symbol means
$r$ — orbital radius (m); $T$ — orbital period (s); $G$ — gravitational constant; $M$ — mass of central body (kg).
When to use it
Use to find period, radius, or central body mass for any circular orbit. Ratios form is useful: $r_1^3/T_1^2 = r_2^3/T_2^2$.
Units:
m³ s⁻²
Worked sample
Earth orbits the Sun at $1.50 \times 10^{11}$ m with period $3.16 \times 10^7$ s. Find the Sun's mass.
$M = \frac{4\pi^2 r^3}{G T^2} = \frac{4\pi^2 (1.50\times10^{11})^3}{6.67\times10^{-11}(3.16\times10^7)^2} \approx 2.0 \times 10^{30}$ kg.
Your turn:
A planet at twice Earth's orbital radius. Find its period in Earth-years.
$T^2 \propto r^3$; $(T/1)^2 = 2^3 = 8$; $T = \sqrt{8} \approx 2.83$ years.
If comparing two orbits around the same body, use the ratio form to avoid substituting constants.
Waves & Sound
3What each symbol means
$v$ — wave speed (m s⁻¹); $f$ — frequency (Hz); $\lambda$ — wavelength (m).
When to use it
Use for any wave — sound, light, water — relating speed, frequency, and wavelength.
Units:
m s⁻¹
Worked sample
A sound wave has $f = 440$ Hz and $\lambda = 0.773$ m. Find the wave speed.
$v = f\lambda = 440 \times 0.773 = 340$ m s⁻¹ (matching NESA's value).
Your turn:
Light has $v = 3.00 \times 10^8$ m s⁻¹ and $\lambda = 500$ nm. Find $f$.
$f = v/\lambda = 3.00 \times 10^8 / 5.00 \times 10^{-7} = 6.00 \times 10^{14}$ Hz.
When light enters a new medium its speed and wavelength change but its frequency stays constant.
What each symbol means
$f'$ — observed frequency (Hz); $f$ — source frequency (Hz); $v_w$ — speed of sound in the medium (m s⁻¹); $v_o$ — speed of observer (m s⁻¹, positive when moving toward source); $v_s$ — speed of source (m s⁻¹, positive when moving toward observer).
When to use it
Use when a sound source or observer is moving relative to the medium. Use $v_w = 340$ m s⁻¹ unless otherwise stated. Sign convention: velocities are positive when directed toward each other.
Units:
Hz
Worked sample
An ambulance siren emits $800$ Hz and approaches at $30$ m s⁻¹. The observer is stationary. Find the observed frequency ($v_w = 340$ m s⁻¹).
$f' = 800 \times (340 + 0)/(340 - 30) = 800 \times 340/310 \approx 877$ Hz.
Your turn:
The same ambulance recedes at $30$ m s⁻¹. Find the observed frequency.
$f' = 800 \times (340 + 0)/(340 + 30) = 800 \times 340/370 \approx 735$ Hz.
The frequency is higher when source and observer approach, lower when they recede. Swap $+/-$ signs carefully — the most common error in Doppler questions.
What each symbol means
$T$ — period (s); $f$ — frequency (Hz = s⁻¹).
When to use it
Converts between period and frequency — essential first step in most wave problems.
Units:
s
Worked sample
A wave has period $0.025$ s. Find its frequency.
$f = 1/0.025 = 40$ Hz.
Your turn:
Mains electricity in Australia has $f = 50$ Hz. Find the period.
$T = 1/50 = 0.02$ s $= 20$ ms.
Memorise this — it appears in every wave/circuit topic and takes under five seconds to apply.
Electricity & Magnetism
8What each symbol means
$q$ — charge (C); $I$ — current (A); $t$ — time (s).
When to use it
Use to find total charge delivered by a steady current over a time interval, or to find average current when charge and time are known.
Units:
C
Worked sample
A current of $2.0$ A flows through a circuit for $5.0$ s. Find the total charge delivered.
$q = It = 2.0 \times 5.0 = 10$ C.
Your turn:
A charge of $120$ C passes through a component in $1.0$ min. Find the current.
$I = q/t = 120/60 = 2.0$ A.
Current is the rate of flow of charge. $I = \Delta q / \Delta t$ is the general form; $q = It$ is for constant (DC) current.
What each symbol means
$W$ — work done on the charge (J); $q$ — charge (C); $V$ — potential difference (V).
When to use it
Use to find the energy gained or lost by a charge moving through a potential difference — fundamental to all circuit energy calculations.
Units:
J
Worked sample
An electron ($q = 1.602 \times 10^{-19}$ C) is accelerated through a potential difference of $100$ V. Find the kinetic energy gained.
$W = qV = 1.602 \times 10^{-19} \times 100 = 1.60 \times 10^{-17}$ J.
Your turn:
How much work is done moving a $+3.0\,\mu$C charge through $200$ V?
$W = 3.0 \times 10^{-6} \times 200 = 6.0 \times 10^{-4}$ J.
The electron-volt (eV) is defined from this equation: $1$ eV $= 1.602 \times 10^{-19}$ J is the work done accelerating one electron through $1$ V.
What each symbol means
$F_e$ — electrostatic force (N); $k = 8.99 \times 10^9$ N m² C⁻²; $q_1, q_2$ — charges (C); $r$ — separation (m); $\varepsilon_0$ — permittivity of free space.
When to use it
Use to find the force between two point charges. Like charges repel, unlike attract.
Units:
N
Worked sample
Find the force between $q_1 = +2\,\mu$C and $q_2 = -3\,\mu$C separated by $0.10$ m.
$F = \frac{8.99 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.10)^2} = 5.4$ N (attractive).
Your turn:
Two equal charges repel with $0.90$ N at $0.30$ m. Find each charge.
$q = \sqrt{F r^2 / k} = \sqrt{0.90 \times 0.09 / 8.99 \times 10^9} = 3.0\,\mu$C.
Coulomb's law is mathematically identical to Newton's gravity — same inverse-square structure. Compare them!
What each symbol means
$E$ — electric field (N C⁻¹ = V m⁻¹); $F$ — force on test charge (N); $q$ — test charge (C); $k$ — Coulomb constant; $Q$ — source charge (C); $r$ — distance (m).
When to use it
Use to find the force per unit charge at a point, or to find the force on a charge placed in a known field ($F = qE$).
Units:
N C⁻¹
Worked sample
An electron ($e = 1.602 \times 10^{-19}$ C) is in a field of $5.0 \times 10^4$ N C⁻¹. Find the force on it.
$F = qE = 1.602 \times 10^{-19} \times 5.0 \times 10^4 = 8.0 \times 10^{-15}$ N.
Your turn:
Find the electric field $0.20$ m from a $+4.0\,\mu$C charge.
$E = kQ/r^2 = 8.99\times10^9 \times 4.0\times10^{-6} / 0.04 = 9.0 \times 10^5$ N C⁻¹.
Electric field lines run from positive to negative. The field points in the direction a positive test charge would move.
What each symbol means
$V$ — potential difference (V); $I$ — current (A); $R$ — resistance ($\Omega$).
When to use it
Use for any ohmic (constant resistance) conductor. Rearrange for current or resistance as required.
Units:
V
Worked sample
A $12$ V battery drives a current through a $40\,\Omega$ resistor. Find the current.
$I = V/R = 12/40 = 0.30$ A.
Your turn:
A current of $2.0$ A flows through a $15\,\Omega$ resistor. Find the voltage.
$V = IR = 2.0 \times 15 = 30$ V.
Ohm's law only applies to ohmic materials. Light globes and diodes are non-ohmic — resistance changes with temperature or polarity.
What each symbol means
$P$ — power (W); $V$ — voltage (V); $I$ — current (A); $R$ — resistance ($\Omega$).
When to use it
Use to find energy dissipation rate in a circuit component. Which form to use depends on which two quantities are known.
Units:
W
Worked sample
A $60\,\Omega$ resistor carries $0.5$ A. Find the power dissipated.
$P = I^2 R = (0.5)^2 \times 60 = 15$ W.
Your turn:
A $240$ V appliance has a $40\,\Omega$ element. Find power dissipated.
$P = V^2/R = 57600/40 = 1440$ W.
In transmission line problems, power loss is $P_{\text{loss}} = I^2 R$ — transmitting at high voltage + low current minimises loss.
What each symbol means
$F$ — force (N); $B$ — magnetic flux density (T); $I$ — current (A); $l$ — length of conductor in field (m); $\theta$ — angle between conductor and field direction.
When to use it
Use for the force on a straight wire in a uniform magnetic field. If wire is perpendicular to $B$, $\theta = 90°$ and $F = BIl$.
Units:
N
Worked sample
A $0.20$ m wire carries $3.0$ A perpendicular to a $0.50$ T field. Find the force.
$F = BIl\sin 90° = 0.50 \times 3.0 \times 0.20 = 0.30$ N.
Your turn:
A wire is parallel to a magnetic field. What force acts on it?
$F = BIl\sin 0° = 0$ N — no force when current and field are parallel.
Direction is given by the right-hand slap rule (or cross product $\mathbf{F} = I\mathbf{l} \times \mathbf{B}$).
What each symbol means
$F$ — magnetic force (N); $q$ — charge (C); $v$ — speed (m s⁻¹); $B$ — magnetic flux density (T); $\theta$ — angle between velocity and field.
When to use it
Use for individual charged particles moving through a magnetic field — cathode rays, cyclotrons, mass spectrometers.
Units:
N
Worked sample
An electron ($q = 1.6 \times 10^{-19}$ C) moves at $2.0 \times 10^6$ m s⁻¹ perpendicular to $B = 0.10$ T. Find the force.
$F = 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times 0.10 = 3.2 \times 10^{-14}$ N.
Your turn:
Why does the magnetic force never do work on a charged particle?
The magnetic force is always perpendicular to velocity, so $W = F \cdot d \cos 90° = 0$. Speed (and KE) are unchanged.
When the magnetic force provides centripetal force, $qvB = mv^2/r$, giving radius $r = mv/(qB)$ — used in mass spectrometry.
Electromagnetism & Induction
3What each symbol means
$\Phi$ — magnetic flux (Wb = T m²); $B$ — magnetic flux density (T); $A$ — area of loop (m²); $\theta$ — angle between $B$ and normal to the loop.
When to use it
Use as the first step in any electromagnetic induction problem. Flux is what changes to induce an EMF.
Units:
Wb (= T m²)
Worked sample
A $0.10$ T field passes through a $0.20$ m² coil with $B$ perpendicular to the plane. Find $\Phi$.
$\Phi = 0.10 \times 0.20 \times \cos 0° = 0.020$ Wb.
Your turn:
The same coil is tilted so $B$ makes $60°$ with the normal. Find $\Phi$.
$\Phi = 0.10 \times 0.20 \times \cos 60° = 0.010$ Wb.
$\theta$ is measured from the normal to the plane, not from the plane itself — a frequent source of errors.
What each symbol means
$\mathcal{E}$ — induced EMF (V); $N$ — number of turns; $\Delta\Phi$ — change in flux (Wb); $\Delta t$ — time interval (s).
When to use it
Use whenever flux through a coil changes — moving magnets, changing current in a nearby coil, rotating generators.
Units:
V
Worked sample
Flux through a $200$-turn coil changes from $0.05$ Wb to $0.02$ Wb in $0.10$ s. Find the EMF.
$\mathcal{E} = -200 \times (0.02 - 0.05)/0.10 = -200 \times (-0.30) = 60$ V.
Your turn:
A single loop has flux changing at $0.40$ Wb s⁻¹. Find the induced EMF.
$\mathcal{E} = 1 \times 0.40 = 0.40$ V.
The negative sign is Lenz's law — the induced current opposes the change that caused it. In exam questions, state the direction using Lenz's law separately.
What each symbol means
$V_p$ — primary voltage (V); $V_s$ — secondary voltage (V); $N_p$ — primary turns; $N_s$ — secondary turns.
When to use it
Use for ideal transformers to step voltage up or down. Combined with $V_p I_p = V_s I_s$ for current.
Units:
Dimensionless ratio
Worked sample
A transformer has $N_p = 200$, $N_s = 1000$ and $V_p = 240$ V. Find $V_s$.
$V_s = 240 \times 1000/200 = 1200$ V.
Your turn:
A step-down transformer gives $V_s = 12$ V from $V_p = 240$ V. If $N_p = 2000$, find $N_s$.
$N_s = 2000 \times 12/240 = 100$ turns.
A step-up transformer increases voltage but decreases current. Power is conserved in an ideal transformer.
Light & Quantum Physics
6What each symbol means
$E$ — photon energy (J); $h = 6.626 \times 10^{-34}$ J s — Planck's constant; $f$ — frequency (Hz); $c$ — speed of light (m s⁻¹); $\lambda$ — wavelength (m).
When to use it
Use to find the energy of a single photon. The $hc/\lambda$ form is useful when wavelength is given directly.
Units:
J (or eV: divide by $1.602 \times 10^{-19}$)
Worked sample
Find the energy of a photon of green light ($\lambda = 540$ nm).
$E = hc/\lambda = (6.626 \times 10^{-34} \times 3.00 \times 10^8) / (540 \times 10^{-9}) = 3.68 \times 10^{-19}$ J.
Your turn:
Find the frequency of an X-ray photon with $E = 1.60 \times 10^{-15}$ J.
$f = E/h = 1.60 \times 10^{-15} / 6.626 \times 10^{-34} = 2.42 \times 10^{18}$ Hz.
Convert energies to eV for intuition (1 eV = 1.602 × 10⁻¹⁹ J). Visible light is roughly 1.8–3.1 eV.
What each symbol means
$E_k^{\max}$ — maximum kinetic energy of ejected photoelectrons (J); $hf$ — photon energy (J); $W = hf_0$ — work function of the metal (J); $f_0$ — threshold frequency (Hz).
When to use it
Use whenever a photon ejects an electron from a metal surface. Below the threshold frequency, no electrons are emitted regardless of intensity.
Units:
J
Worked sample
Light of $f = 1.0 \times 10^{15}$ Hz hits a metal with work function $3.0$ eV. Find $E_k^{\max}$.
$hf = 6.626\times10^{-34} \times 1.0\times10^{15} = 6.63\times10^{-19}$ J $= 4.14$ eV; $E_k^{\max} = 4.14 - 3.0 = 1.14$ eV.
Your turn:
What is the stopping voltage for those ejected electrons?
$eV_s = E_k^{\max}$; $V_s = 1.14$ V.
Intensity affects the number of photons (current), not their energy. Only frequency determines whether electrons are ejected.
What each symbol means
$\lambda$ — de Broglie (matter) wavelength (m); $h$ — Planck's constant (J s); $p = mv$ — momentum (kg m s⁻¹).
When to use it
Use to find the wave-like wavelength of any particle. Larger momentum means shorter wavelength — electrons in an electron microscope have shorter wavelengths than visible light.
Units:
m
Worked sample
Find the de Broglie wavelength of an electron ($m = 9.11 \times 10^{-31}$ kg) moving at $2.0 \times 10^6$ m s⁻¹.
$\lambda = h/(mv) = 6.626\times10^{-34} / (9.11\times10^{-31} \times 2.0\times10^6) = 3.64 \times 10^{-10}$ m $= 0.364$ nm.
Your turn:
How does doubling a particle's speed affect its de Broglie wavelength?
$\lambda \propto 1/v$, so wavelength halves.
The de Broglie wavelength of a macroscopic object (e.g. a cricket ball) is immeasurably small — wave behaviour is only observable for subatomic particles.
What each symbol means
$\lambda_{\text{max}}$ — peak wavelength (m); $b = 2.898 \times 10^{-3}$ m K — Wien's displacement constant; $T$ — absolute temperature of the black body (K).
When to use it
Use to find the wavelength at which a black body emits most intensely, or to infer a star's surface temperature from its colour. Module 8 (The Universe and the Atom).
Units:
m
Worked sample
The Sun's surface temperature is approximately $5778$ K. Find the peak wavelength of solar radiation.
$\lambda_{\text{max}} = b/T = 2.898 \times 10^{-3} / 5778 \approx 5.01 \times 10^{-7}$ m $= 501$ nm (green-yellow visible light).
Your turn:
A red giant has a peak wavelength of $700$ nm. Estimate its surface temperature.
$T = b/\lambda_{\text{max}} = 2.898 \times 10^{-3} / 7.00 \times 10^{-7} \approx 4140$ K.
Wien's law connects colour to temperature — hotter stars appear blue-white, cooler stars red. Remember $\lambda_{\text{max}} \propto 1/T$.
What each symbol means
$t$ — dilated time measured by stationary observer (s); $t_0$ — proper time (in the moving frame, s); $v$ — relative speed (m s⁻¹); $c$ — speed of light (m s⁻¹); $\gamma$ — Lorentz factor.
When to use it
Use when an event is observed from two frames moving relative to each other. The clock in the moving frame runs slower as seen from the stationary frame.
Units:
s
Worked sample
A muon has a proper lifetime of $t_0 = 2.2\,\mu$s and moves at $0.98c$. Find the observed lifetime.
$\gamma = 1/\sqrt{1-0.98^2} = 1/\sqrt{0.0396} \approx 5.03$; $t = 5.03 \times 2.2 = 11.1\,\mu$s.
Your turn:
At what speed is $\gamma = 2$?
$1/\sqrt{1-v^2/c^2} = 2 \Rightarrow v = c\sqrt{3}/2 \approx 0.866c$.
Always identify which time is the proper time $t_0$ (same location in one frame). Proper time is always the shortest measured time.
What each symbol means
$L$ — contracted length (m, measured in the frame in which the object is moving); $L_0$ — proper length (m, measured in the object's rest frame); $\gamma$ — Lorentz factor; $v$ — relative speed (m s⁻¹).
When to use it
Use when measuring the length of an object that is moving relative to you. Moving objects appear shorter along the direction of motion.
Units:
m
Worked sample
A spaceship has proper length $100$ m and moves at $0.60c$. Find its observed length.
$L = 100 \times \sqrt{1 - 0.36} = 100 \times 0.80 = 80$ m.
Your turn:
A rod at rest is $1.0$ m long. At what speed does it appear $0.5$ m long?
$0.5 = 1.0\sqrt{1 - v^2/c^2} \Rightarrow v = c\sqrt{3}/2 \approx 0.866c$.
Length contraction only occurs along the direction of motion. Dimensions perpendicular to motion are unchanged.
Nuclear Physics & the Standard Model
4What each symbol means
$E$ — energy (J); $m$ — mass (kg); $c = 3.00 \times 10^8$ m s⁻¹ — speed of light.
When to use it
Use to convert between mass and energy in nuclear reactions, pair production/annihilation, and binding energy calculations.
Units:
J
Worked sample
A nuclear reaction releases $\Delta m = 3.0 \times 10^{-29}$ kg of mass. Find the energy released.
$E = \Delta m c^2 = 3.0 \times 10^{-29} \times (3.00 \times 10^8)^2 = 2.7 \times 10^{-12}$ J.
Your turn:
Electron–positron annihilation converts $2m_e$ to photons. Find the total energy ($m_e = 9.11 \times 10^{-31}$ kg).
$E = 2 \times 9.11\times10^{-31} \times (3.00\times10^8)^2 = 1.64 \times 10^{-13}$ J.
In nuclear physics it is more convenient to work in atomic mass units (u), where $1$ u $= 931.5$ MeV/$c^2$. You will usually be given the mass defect in kg.
What each symbol means
$N(t)$ — number of undecayed nuclei at time $t$; $N_0$ — initial number; $t_{1/2}$ — half-life (s or years); $t$ — elapsed time.
When to use it
Use to find the remaining activity or number of nuclei after a given number of half-lives. Also works for activity $A$ and mass $m$ in place of $N$.
Units:
Same unit as $N_0$
Worked sample
A sample has $t_{1/2} = 5730$ years (carbon-14). What fraction remains after $11460$ years?
$n = 11460/5730 = 2$ half-lives; fraction $= (1/2)^2 = 1/4 = 25\%$.
Your turn:
After $3$ half-lives, what percentage of a radioactive sample remains?
$(1/2)^3 = 1/8 = 12.5\%$.
You can substitute activity $A$ or mass $m$ directly in place of $N$ — the same exponential decay law holds.
What each symbol means
$p$ — photon momentum (kg m s⁻¹); $E$ — photon energy (J); $c$ — speed of light (m s⁻¹); $h$ — Planck's constant (J s); $\lambda$ — wavelength (m).
When to use it
Use in Compton scattering and radiation pressure problems where photons transfer momentum to matter.
Units:
kg m s⁻¹
Worked sample
Find the momentum of a photon with $\lambda = 400$ nm.
$p = h/\lambda = 6.626\times10^{-34} / 4.00\times10^{-7} = 1.66 \times 10^{-27}$ kg m s⁻¹.
Your turn:
What is the momentum of a photon of energy $E = 2.0 \times 10^{-19}$ J?
$p = E/c = 2.0\times10^{-19} / 3.0\times10^8 = 6.67\times10^{-28}$ kg m s⁻¹.
Photons have momentum despite having zero rest mass — this is a direct consequence of $E = pc$ for massless particles.
What each symbol means
$E$ — total energy (J); $p$ — momentum (kg m s⁻¹); $c$ — speed of light (m s⁻¹); $m_0$ — rest mass (kg).
When to use it
Use for particles moving at relativistic speeds. For photons ($m_0 = 0$) this gives $E = pc$. For a particle at rest ($p = 0$) it gives $E = m_0 c^2$.
Units:
J
Worked sample
An electron has kinetic energy equal to its rest energy ($m_e c^2 = 0.511$ MeV). Find its total energy.
Total energy $= KE + m_0c^2 = 0.511 + 0.511 = 1.022$ MeV.
Your turn:
For a photon with $E = 1.0$ MeV, find its momentum in MeV/$c$.
$p = E/c = 1.0$ MeV/$c$.
This is the most general energy equation in special relativity — it works for everything from rest to relativistic speeds.
Frequently asked questions
It includes physical constants and data plus the key formulas for kinematics, dynamics, circular and projectile motion, gravitation and orbits, waves, electricity and magnetism, and electromagnetic induction. NESA provides it in every Physics exam.
Yes — NESA supplies the data sheet, formula sheet and a periodic table in every HSC Physics exam, so the constants and formulas are there. The skill is choosing and applying the right one.
Some derivations, definitions and relationships aren't on the sheet, and the sheet won't tell you which formula a question needs — so understanding still matters more than the sheet itself.
Practise problems with it open so you learn where each formula lives and the situations it applies to — then you can find and use it fast in the exam.