Physics formulae

Each formula with its symbols in units, the conditions it needs, and the situations where it gives a wrong answer. The formulae sheet gives the equation; the marks are lost on the conditions.

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33 of 33 formulae

Orbital speed from the periodon the NESA formulae sheet
$v = \frac{2 π r}{T}$
Symbols
$v$ speed around the circle, constant in uniform circular motionm s⁻¹
$r$ radius of the circular path, measured from its centrem
$T$ period, the time for one complete revolutions
Valid when
The object moves around a circle at constant speed, so one circumference is covered in exactly one period.
Not valid when
The speed changes around the path, as it does for a mass swung in a vertical circle, or the radius is taken as a diameter or as the length of a string that is not horizontal.
Rearranged
for T: $T = \frac{2 π r}{v}$ for r: $r = \frac{v T}{2 π}$
Where it turns up
Turning a count of revolutions in a measured time into a speed, the first step in almost every circular motion calculation
Finding the speed of a satellite or a planet from its orbital period
Checking the period predicted by a centripetal force calculation against a stopwatch reading in the practical
Where marks go missing
Using the frequency in revolutions per second where the period is needed, which gives a speed that is too small by a factor of the frequency squared
Using the diameter in place of the radius
Timing a single revolution in a practical, where the reaction time of the timer is a large fraction of the period
Centripetal accelerationon the NESA formulae sheet
$a_{c} = \frac{v ^{2}}{r}$
Symbols
$a_{c}$ acceleration directed at the centre of the circlem s⁻²
$v$ speed around the circlem s⁻¹
$r$ radius of the circular pathm
Valid when
The object moves on a circular path at an instant when its speed is v. For uniform circular motion it holds everywhere on the path and the acceleration is entirely towards the centre.
Not valid when
The path is not a circle of the stated radius. When the speed is also changing, this gives only the part of the acceleration pointing at the centre, and a second, tangential part exists alongside it.
Rearranged
for v: $v = a_{c} r$ for r: $r = \frac{v ^{2}}{a _{c}}$
Where it turns up
Finding how hard an object is being pulled towards the centre, as a multiple of g, before any mass is known
Showing that an object moving at constant speed is nonetheless accelerating, which is the conceptual core of the topic
The first half of a centripetal force calculation, since multiplying by the mass gives the force
Where marks go missing
Forgetting to square the speed
Describing the acceleration as outwards, or as along the direction of motion
Concluding that an object at constant speed has zero acceleration
Where this comes from
Centripetal forceon the NESA formulae sheet
$F_{c} = \frac{m v ^{2}}{r}$
Symbols
$F_{c}$ net force towards the centre needed to keep the object on the circleN
$m$ mass of the objectkg
$v$ speed around the circlem s⁻¹
$r$ radius of the circular pathm
Valid when
The object is moving in a circle of radius r at speed v. The result is the size of the net force towards the centre, whatever real forces happen to supply it.
Not valid when
It is treated as an extra force to add to a free body diagram. It is never a separate force: it is the name for the resultant of the real forces, such as tension, friction, gravity or a normal force, along the radius.
Rearranged
for v: $v = \frac{F _{c} r}{m}$ for r: $r = \frac{m v ^{2}}{F _{c}}$ for m: $m = \frac{F _{c} r}{v ^{2}}$
Where it turns up
Finding the friction a car's tyres must supply on a flat bend, and from that the fastest safe speed
Finding the tension in a string holding a mass in a horizontal circle
Predicting how the required force changes when the speed, mass or radius is changed, which is the investigation in the first dot point
Where marks go missing
Drawing a force labelled centripetal force on a free body diagram alongside the tension or friction that is actually providing it
Saying that the force required doubles when the speed doubles, when it quadruples
Using a speed in km per hour without converting to metres per second
Angular velocityon the NESA formulae sheet
$ω = \frac{Δ θ}{t}$
Symbols
$ω$ angular velocity, the rate at which the angle swept out growsrad s⁻¹
$Δ θ$ angle turned through, in radiansrad
$t$ time taken to turn through that angles
Valid when
The angle is measured in radians. For uniform circular motion the angular velocity is constant, one revolution is 2π radians, and so ω = 2π/T and v = ωr.
Not valid when
The angle is left in degrees or in revolutions, in which case the number produced is not in rad s⁻¹ and the relationship v = ωr gives nonsense.
Rearranged
for \omega: $ω = \frac{2 π}{T}$ for v: $v = ω r$
Where it turns up
Describing a rotating object where every point shares one angular velocity but has its own speed
Converting a rotation rate in revolutions per minute into rad s⁻¹
Linking the rotation rate to the linear speed at a given radius through v = ωr
Where marks go missing
Leaving the angle in degrees
Using revolutions per second as if it were rad s⁻¹, which is out by a factor of 2π
Assuming points at different radii on a rotating disc share a speed, when they share only an angular velocity
Design speed of a banked tracknot on the formulae sheet, learn it
$tan θ = \frac{v ^{2}}{r g}$
Symbols
$θ$ banking angle of the track, measured from the horizontal°
$v$ the one speed at which no sideways friction is neededm s⁻¹
$r$ radius of the bendm
$g$ gravitational field strength, 9.8 near the Earth's surfacem s⁻²
Valid when
The vehicle moves in a horizontal circle on a surface banked at θ, and the only forces are its weight and the normal force. It gives the single speed for which the horizontal part of the normal force is exactly the centripetal force.
Not valid when
The vehicle travels faster or slower than the design speed, where friction must act along the slope and the two force equations each gain a friction term. It also assumes the circle is horizontal, not the slope.
Rearranged
for v: $v = r g tan θ$ for \theta: $θ = tan^{- 1} (\frac{v ^{2}}{r g})$ for r: $r = \frac{v ^{2}}{g t a n θ}$
Where it turns up
Designing the banking angle of a road or velodrome for a chosen speed
Explaining why a banked bend is safer than a flat one at the speed it was built for
Showing that the design speed does not depend on the vehicle's mass
Where marks go missing
Resolving the weight into components along and across the slope, as for an inclined plane, rather than resolving the normal force horizontally and vertically
Writing N = mg cos θ, which is the inclined plane result and is wrong here because the acceleration is horizontal, not along the slope
Quoting the equation without the derivation when a question asks for it to be shown
Where this comes from
Torqueon the NESA formulae sheet
$τ = r_{⊥} F = r F sin θ$
Symbols
$τ$ torque, the turning effect of the force about the pivotN m
$r$ distance from the pivot to where the force is appliedm
$F$ size of the applied forceN
$θ$ angle between the force and the line from the pivot to its point of application°
$r_{⊥}$ perpendicular distance from the pivot to the line of action of the forcem
Valid when
A single force acts on a body that can rotate about a fixed pivot or axis. The angle is the one between the force and the lever arm, not between the force and the horizontal.
Not valid when
The angle is measured from something other than the lever arm, or several forces act and only one torque is calculated. Where several forces act, their torques are combined, with clockwise and anticlockwise given opposite signs.
Rearranged
for F: $F = \frac{τ}{r s i n θ}$ for r: $r = \frac{τ}{F s i n θ}$
Where it turns up
Explaining why a door handle is placed far from the hinge and why a longer spanner loosens a tight nut
Finding the force needed at a given point to produce a required torque
The torque on a current carrying coil in a magnetic field in Module 6, which is this same relationship applied to the motor
Where marks go missing
Using cos θ where the angle given is between the force and the lever arm
Measuring the distance from the wrong point, such as from the end of the object rather than the pivot
Giving the unit as joules; torque and work share N m dimensionally but torque is written N m
Newton's law of universal gravitationon the NESA formulae sheet
$F = \frac{GM m}{r ^{2}}$
Symbols
$F$ size of the attractive force each mass exerts on the otherN
$G$ universal gravitational constant, 6.67 × 10⁻¹¹N m² kg⁻²
$M$ one of the two masses, usually the largerkg
$m$ the other masskg
$r$ distance between the centres of the two massesm
Valid when
The two bodies are point masses or spheres whose mass is spread symmetrically, and r is measured between their centres. Newton's third law applies: each body feels the same size of force.
Not valid when
r is taken as an altitude above a surface rather than a distance from the centre, or the bodies are irregular and close together compared with their size.
Rearranged
for r: $r = \frac{GM m}{F}$ for M: $M = \frac{F r ^{2}}{G m}$
Where it turns up
Finding the pull between a planet and its moon, or between the Sun and a planet
Predicting how the force changes when a mass or a separation is scaled
Setting gravity equal to the centripetal force to analyse an orbit
Where marks go missing
Using the altitude above the surface in place of the distance from the centre
Forgetting to square the distance
Believing the larger mass pulls harder on the smaller one than the smaller pulls on the larger
Gravitational field strengthon the NESA formulae sheet
$g = \frac{GM}{r ^{2}}$
Symbols
$g$ gravitational field strength, the force per kilogram at that point, equal to the free fall accelerationN kg⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the body producing the fieldkg
$r$ distance from the centre of that body to the pointm
Valid when
The point is outside a spherically symmetric body, or anywhere around a point mass. It follows from dividing the universal gravitation force by the test mass m.
Not valid when
The point is inside the body, where only the mass closer to the centre contributes, or r is taken from the surface instead of the centre.
Rearranged
for M: $M = \frac{g r ^{2}}{G}$ for r: $r = \frac{GM}{g}$
Where it turns up
Predicting the surface gravity of another planet from its mass and radius
Finding the field at a satellite's altitude
Comparing fields by ratio without calculating G M, since g is proportional to M over r squared
Where marks go missing
Adding the altitude to nothing, using it alone as r
Thinking g is zero in orbit because astronauts float
Using the mass of the object in the field rather than the mass producing it
Where this comes from
Speed of a circular orbitnot on the formulae sheet, learn it
$v = \frac{GM}{r}$
Symbols
$v$ speed of the orbiting bodym s⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central body being orbitedkg
$r$ orbital radius, measured from the centre of the central bodym
Valid when
The orbit is circular and gravity from the central body is the only force, so gravity supplies the whole centripetal force: GMm/r² = mv²/r.
Not valid when
The orbit is elliptical, where the speed changes around the path, or other forces such as atmospheric drag are significant.
Rearranged
for r: $r = \frac{GM}{v ^{2}}$ for M: $M = \frac{v ^{2} r}{G}$
Where it turns up
Finding the speed of a satellite at a given altitude
Showing that orbital speed does not depend on the satellite's own mass
Showing that higher orbits are slower
Where marks go missing
Including the satellite's mass, which cancels
Using the altitude as r
Concluding that a higher orbit is faster because more energy was needed to reach it
Where this comes from
Kepler's third lawon the NESA formulae sheet
$\frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$
Symbols
$r$ orbital radius, or for an ellipse the semi major axism
$T$ orbital periods
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
Valid when
Several bodies orbit the same central mass, which is much heavier than any of them. The ratio r³/T² is then the same for all of them, which is what lets the central mass be measured.
Not valid when
The bodies being compared orbit different central masses, or the orbiting body's mass is comparable with the central mass, as for a binary star.
Rearranged
for T: $T = 2 π \frac{r ^{3}}{GM}$ for r: $r = 3 \frac{GM T ^{2}}{4 π ^{2}}$ for M: $M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$
Where it turns up
Finding the radius of a geostationary orbit from its period of one sidereal day
Measuring the mass of a planet from the orbit of one of its moons
Predicting the period of a planet from its distance to the Sun
Where marks go missing
Leaving the period in hours or days while using G in SI units
Taking a cube root where a square root is needed, or the reverse
Using the altitude of a satellite as r
Where this comes from
Gravitational potential energy in a radial fieldon the NESA formulae sheet
$U = - \frac{GM m}{r}$
Symbols
$U$ gravitational potential energy of the two body system, zero when they are infinitely far apartJ
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
$m$ mass of the smaller bodykg
$r$ distance between their centresm
Valid when
The zero of potential energy is chosen at infinite separation. Every finite separation then has a negative potential energy, rising towards zero as the bodies move apart.
Not valid when
It is mixed with mgh, whose zero is at a chosen surface. The two are consistent only for changes in energy, and mgh itself only holds for heights small compared with the planet's radius.
Rearranged
for r: $r = - \frac{GM m}{U}$
Where it turns up
Finding the energy needed to move a satellite between orbits
Deriving the escape speed by setting the kinetic energy equal to the size of U
Explaining why a planet in an ellipse speeds up as it approaches the Sun
Where marks go missing
Dropping the minus sign, which reverses every conclusion about energy changes
Squaring r, confusing potential energy with force
Thinking a larger negative number means more energy
Total energy of a circular orbitnot on the formulae sheet, learn it
$U + K = - \frac{GM m}{2 r}$
Symbols
$U$ gravitational potential energyJ
$K$ kinetic energy of the orbiting bodyJ
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
$m$ mass of the orbiting bodykg
$r$ orbital radiusm
Valid when
The orbit is circular, so K = GMm/2r from setting gravity equal to the centripetal force. The total is negative, which is the condition for a bound orbit.
Not valid when
The orbit is elliptical, where K and U trade off around the path and r must be replaced by the semi major axis, or the body is not in orbit at all.
Rearranged
for r: $r = - \frac{GM m}{2 E}$
Where it turns up
Finding the energy needed to lift a satellite from a low orbit to a higher one
Showing that a satellite losing energy to drag spirals inwards and speeds up
Checking that K = −U/2 and E = −K for any circular orbit
Where marks go missing
Losing the minus sign and concluding that a higher orbit has less energy
Adding the full orbital speed kinetic energy to mgh
Assuming a satellite in a higher orbit is faster because it has more total energy
Where this comes from
Escape velocityon the NESA formulae sheet
$v_{esc} = \frac{2 GM}{r}$
Symbols
$v_{esc}$ least launch speed that lets an unpowered object reach infinite distancem s⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the body being escapedkg
$r$ distance from its centre at launch, the radius when launched from the surfacem
Valid when
The object is given its speed at the start and then moves under gravity alone. It follows from setting the kinetic energy equal to the size of the potential energy, so that the total energy is zero.
Not valid when
Air resistance matters, or the craft keeps firing its engine, in which case it can leave at any speed. The direction of launch does not matter, provided the path does not hit the planet.
Rearranged
for M: $M = \frac{v _{esc}^{2} r}{2 G}$
Where it turns up
Finding the escape speed from the surface of a planet or moon
Comparing it with the circular orbit speed at the same radius, which is smaller by a factor of the square root of 2
Explaining why the Moon has no atmosphere
Where marks go missing
Including the object's mass, which cancels
Omitting the factor of 2 and finding the circular orbit speed instead
Thinking escape velocity must be directed straight up
Where this comes from
Velocity after uniform accelerationon the NESA formulae sheet
$v = u + a t$
Symbols
$v$ velocity at the end of the intervalm s⁻¹
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$t$ length of the intervals
Valid when
The acceleration is constant and every symbol refers to the same straight line, with one direction chosen as positive and kept that way.
Not valid when
The acceleration changes during the interval, or symbols from two different directions are mixed. In projectile work this equation belongs to the vertical direction only, because the horizontal acceleration is zero.
Rearranged
for t: $t = \frac{v - u}{a}$ for a: $a = \frac{v - u}{t}$ for u: $u = v - a t$
Where it turns up
Finding the time to the top of a projectile's flight, by setting the vertical velocity to zero
Finding the vertical velocity on impact, which then combines with the horizontal component to give the landing speed and angle
Any single stage of a multi-stage motion question where the acceleration is stated as constant
Where marks go missing
Substituting the launch speed for u in the vertical equation instead of the vertical component of the launch speed
Leaving the acceleration positive after choosing up as positive, which makes the projectile speed up on the way to the top
Using it horizontally in projectile work, where the acceleration is zero and the equation collapses to a statement that the velocity never changes
Displacement under uniform accelerationon the NESA formulae sheet
$s = u t + \frac{1}{2} a t^{2}$
Symbols
$s$ displacement over the interval, not distance travelledm
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$t$ length of the intervals
Valid when
The acceleration is constant and the displacement, initial velocity and acceleration are all measured along the same line in the same positive direction.
Not valid when
The acceleration varies, or the answer wanted is total path length rather than displacement. A projectile thrown up and caught again has zero vertical displacement while having travelled a real distance.
Rearranged
for u: $u = \frac{s}{t} - \frac{1}{2} a t$ for a: $a = \frac{2 ( s - u t )}{t ^{2}}$
Where it turns up
The vertical equation of a projectile, giving height against time
The horizontal equation of a projectile, where the acceleration term vanishes and it reduces to distance equals speed times time
Finding the time of flight from a launch height, by setting the vertical displacement equal to the negative of that height
Where marks go missing
Forgetting the one half on the acceleration term, which doubles every height answer
Setting the vertical displacement to the launch height rather than to minus the launch height when down is negative
Squaring only the t and not the whole bracket when the time is itself an expression
Velocity against displacement, with no timeon the NESA formulae sheet
$v^{2} = u^{2} + 2 a s$
Symbols
$v$ velocity at the end of the intervalm s⁻¹
$u$ velocity at the start of the intervalm s⁻¹
$a$ acceleration, constant throughoutm s⁻²
$s$ displacement over the intervalm
Valid when
The acceleration is constant along the line of motion and the question gives or wants a displacement rather than a time.
Not valid when
The acceleration changes, or the question needs a time. It also loses the sign of the answer, since squaring destroys direction, so the direction has to be reasoned back in from the situation.
Rearranged
for s: $s = \frac{v ^{2} - u ^{2}}{2 a}$ for u: $u = v^{2} - 2 a s$
Where it turns up
Greatest height of a projectile, by setting the final vertical velocity to zero
Vertical velocity on landing when the launch height is known but the time is not wanted
Checking an answer obtained another way, since it uses a different pair of quantities
Where marks go missing
Taking the positive square root without asking which direction the object is actually moving in
Using the launch speed rather than its vertical component when finding the greatest height
Applying it across a whole flight in which the object goes up and then down, where the displacement is small and the equation still answers correctly but is misread as a statement about distance
Resolving a launch velocity into componentsnot on the formulae sheet, learn it
$u_{x} = u cos θ u_{y} = u sin θ$
Symbols
$u$ launch speed, the magnitude of the launch velocitym s⁻¹
$θ$ launch angle, measured from the horizontaldegrees or radians
$u_{x}$ horizontal component, unchanged for the whole flightm s⁻¹
$u_{y}$ vertical component at launchm s⁻¹
Valid when
The launch angle is measured from the horizontal, which is the convention every HSC question uses unless it says otherwise.
Not valid when
The angle is given from the vertical, from a slope, or from a line of sight. Then the sine and cosine swap, or the angle has to be converted before either is used.
Rearranged
for u: $u = u_{x}^{2} + u_{y}^{2}$ for \theta: $θ = tan^{- 1} (\frac{u _{y}}{u _{x}})$
Where it turns up
The first line of almost every projectile answer, before any equation of motion is written
Recombining a final horizontal and vertical velocity into a landing speed and a landing angle
Checking that a component is smaller than the speed it came from, which catches a calculator left in radians
Where marks go missing
Swapping sine and cosine, which is caught instantly by asking whether a shallow launch should have a large or a small horizontal component
Leaving the calculator in radian mode, which produces components larger than the launch speed
Resolving once and then forgetting to keep the horizontal component constant for the rest of the flight
Where this comes from
Range of a projectile on level groundnot on the formulae sheet, learn it
$R = \frac{u ^{2} sin 2 θ}{g}$
Symbols
$R$ horizontal distance from launch to landingm
$u$ launch speedm s⁻¹
$θ$ launch angle above the horizontaldegrees or radians
$g$ acceleration due to gravity, 9.8 on the NESA data sheetm s⁻²
Valid when
The projectile lands at exactly the height it was launched from, air resistance is ignored, and the angle is measured from the horizontal.
Not valid when
The launch and landing heights differ by any amount. A ball thrown from a cliff, a shot released at shoulder height, or a serve struck above the far court all break it, and in each case the components have to be worked through instead.
Rearranged
for u: $u = \frac{R g}{s i n 2 θ}$ for \theta: $θ = \frac{1}{2} sin^{- 1} (\frac{R g}{u ^{2}})$
Where it turns up
Showing that the greatest range on level ground comes at 45 degrees, where the sine reaches one
Showing that two launch angles adding to 90 degrees give the same range, because they give the same doubled angle
A fast check on an answer that was properly worked through with components
Where marks go missing
Using it when the launch height and the landing height are different, which is the single most common way marks are lost in this topic
Writing sine of theta rather than sine of two theta, which gives the wrong answer at every angle except the one where they happen to agree
Quoting it in an examination that expects the relationships to be derived, where the marks are for the components and not for a remembered result
Where this comes from
Magnetic flux through an areaon the NESA formulae sheet
$Φ = B_{∥} A = B A cos θ$
Symbols
$Φ$ magnetic flux through the areaWb
$B$ magnetic field strengthT
$B_{∥}$ component of the field along the normal to the areaT
$A$ area the field passes throughm²
$θ$ angle between the field and the normal to the area°
Valid when
The field is uniform over a flat area. θ is measured from the normal, so a loop facing the field has θ = 0 and the largest flux.
Not valid when
The field varies across the area, the surface is curved, or θ is measured from the plane of the loop rather than from its normal.
Rearranged
for B: $B = \frac{Φ}{A c o s θ}$ for A: $A = \frac{Φ}{B c o s θ}$ for \theta: $cos θ = \frac{Φ}{B A}$
Where it turns up
Finding the change in flux when a coil turns, moves or its field changes
Setting up a Faraday's law calculation
Explaining why a loop edge on to a field has no flux through it
Where marks go missing
Measuring θ from the plane of the loop, which swaps cosine for sine
Leaving the area in cm² instead of converting to m²
Multiplying by the number of turns, which belongs in Faraday's law, not in the flux
Where this comes from
Faraday's law of induction, with Lenz's lawon the NESA formulae sheet
$ε = - N \frac{ΔΦ}{Δ t}$
Symbols
$ε$ average induced emfV
$N$ number of turns in the coil1
$ΔΦ$ change in flux through one turnWb
$Δ t$ time taken for the changes
Valid when
Any change in flux through a circuit: a changing field, a changing area or a changing angle. It gives the average emf over Δt; the minus sign is Lenz's law, saying the induced current opposes the change that caused it.
Not valid when
Used for the emf at an instant when the flux is changing at a varying rate, where it gives only the average, or used with the total flux through all N turns and then multiplied by N again.
Rearranged
for \Delta t: $Δ t = \frac{N ∣ΔΦ∣}{∣ ε ∣}$ for N: $N = \frac{∣ ε ∣ Δ t}{∣ΔΦ∣}$ for \varepsilon: $ε = B l v (rod moving across a field)$
Where it turns up
Average emf when a magnet is pushed into a coil or a coil is rotated
The emf across a rod moving along rails, ε = Blv
Explaining which way an induced current flows
Where marks go missing
Using the final flux instead of the change in flux
Forgetting the number of turns
Treating the minus sign as a numerical sign to carry through, instead of as the direction rule
Where this comes from
Ideal transformer turns ratioon the NESA formulae sheet
$\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}$
Symbols
$V_{p}$ voltage across the primary coilV
$V_{s}$ voltage across the secondary coilV
$N_{p}$ turns on the primary coil1
$N_{s}$ turns on the secondary coil1
Valid when
An ideal transformer on alternating current, where every field line from the primary passes through the secondary, so each turn of either coil has the same emf.
Not valid when
The supply is steady DC, which gives no changing flux and no secondary voltage, or flux linkage is poor, which makes the secondary voltage lower than the ratio predicts.
Rearranged
for V_{s}: $V_{s} = V_{p} \frac{N _{s}}{N _{p}}$ for N_{s}: $N_{s} = N_{p} \frac{V _{s}}{V _{p}}$
Where it turns up
Designing a step-down transformer for a phone charger
Finding the turns ratio needed to step up generator voltage for transmission
Where marks go missing
Inverting the ratio, so a step-up transformer is given fewer secondary turns
Applying it to a DC supply
Where this comes from
Ideal transformer power balanceon the NESA formulae sheet
$V_{p} I_{p} = V_{s} I_{s}$
Symbols
$V_{p}, I_{p}$ primary voltage and currentV, A
$V_{s}, I_{s}$ secondary voltage and currentV, A
Valid when
An ideal transformer, which loses no energy, so the power delivered to the primary equals the power taken from the secondary.
Not valid when
A real transformer with losses, where the output power is less than the input and the efficiency V_sI_s / V_pI_p is below 100 per cent.
Rearranged
for I_{s}: $I_{s} = \frac{V _{p} I _{p}}{V _{s}}$ for I_{p}: $I_{p} = I_{s} \frac{N _{s}}{N _{p}}$
Where it turns up
Finding the current drawn from the mains by a step-down transformer
Showing that stepping up the voltage steps down the current in a transmission line
Calculating the efficiency of a real transformer
Where marks go missing
Assuming current scales with turns the same way voltage does
Believing a step-up transformer gives out more energy than it takes in
Where this comes from
Force on a current-carrying conductor in a magnetic fieldon the NESA formulae sheet
$F = l I_{⊥} B = l I B sin θ$
Symbols
$F$ size of the magnetic force on the conductorN
$l$ length of conductor inside the fieldm
$I$ current in the conductorA
$B$ magnetic field strengthT
$θ$ angle between the conductor and the field°
Valid when
The conductor is straight and the field is uniform along the length that sits inside it. The force is perpendicular to both the conductor and the field, with its direction given by the right hand palm rule.
Not valid when
The field varies along the wire, the wire is curved, or θ is measured from the normal to the field rather than from the field itself.
Rearranged
for B: $B = \frac{F}{l I s i n θ}$ for I: $I = \frac{F}{l B s i n θ}$ for \theta: $sin θ = \frac{F}{l I B}$
Where it turns up
Finding the force on one side of a motor coil
Measuring a magnetic field with a current balance
Predicting when a wire in a field feels no force at all
Where marks go missing
Using the whole length of the wire rather than the length inside the field
Using cos θ, or measuring θ from the perpendicular
Giving the force a direction along the field or along the wire
Where this comes from
Force between two parallel current-carrying wireson the NESA formulae sheet
$\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$
Symbols
$F$ force on a length l of either wireN
$l$ length of wire consideredm
$μ_{0}$ permeability of free space, 4π × 10⁻⁷T m A⁻¹
$I_{1}, I_{2}$ currents in the two wiresA
$r$ perpendicular distance between the wiresm
Valid when
The wires are long, straight and parallel, and their separation is small compared with their length. Currents in the same direction attract; opposite directions repel.
Not valid when
The wires are short compared with their separation, not parallel, or r is measured as anything other than the centre to centre distance between them.
Rearranged
for r: $r = \frac{μ _{0} I _{1} I _{2} l}{2 π F}$ for I_{2}: $I_{2} = \frac{2 π r F}{μ _{0} I _{1} l}$
Where it turns up
Forces between conductors in cables and busbars
Stating the historical definition of the ampere
Showing that the forces on the two wires form a Newton's third law pair
Where marks go missing
Squaring r, as if the force were inverse square
Leaving r in centimetres
Giving the wire with the larger current the larger force
Where this comes from
Wien's displacement lawon the NESA formulae sheet
$λ_{m a x} = \frac{b}{T}$
Symbols
$λ_{m a x}$ wavelength at which a black body's emission is most intensem
$b$ Wien's constant, 2.898 × 10⁻³ m Km K
$T$ absolute temperature of the bodyK
Valid when
A body that emits close to a black body spectrum, such as a star's photosphere, a hot filament or a cavity with a small hole. T must be in kelvin.
Not valid when
The source emits a line spectrum, such as a gas discharge tube or a laser, or the temperature is left in degrees Celsius.
Rearranged
for T: $T = \frac{b}{λ _{m a x}}$
Where it turns up
Estimating the surface temperature of a star from its peak wavelength
Explaining why hotter bodies glow blue and cooler ones red
Finding the peak wavelength of radiation from the human body or the Earth
Where marks go missing
Using the temperature in degrees Celsius
Leaving the wavelength in metres when nanometres were asked for, or the reverse
Reading the peak as the colour the body appears, when the eye sees a mixture of wavelengths
Where this comes from
Photon energyon the NESA formulae sheet
$E = h f$
Symbols
$E$ energy of one photonJ
$h$ Planck's constant, 6.626 × 10⁻³⁴ J sJ s
$f$ frequency of the lightHz
Valid when
Any electromagnetic radiation, one photon at a time. Combine it with c = fλ when the wavelength is given.
Not valid when
It is applied to a whole beam, whose energy is the photon energy times the number of photons, or the energy is wanted in electronvolts and the conversion is skipped.
Rearranged
for f: $f = \frac{E}{h}$ for E: $E = \frac{h c}{λ}$
Where it turns up
The energy carried by one photon of a given colour
The number of photons per second in a beam of known power
Planck's quantised oscillators in black body radiation
Where marks go missing
Substituting a wavelength for f
Mixing joules and electronvolts in one line
Thinking a brighter beam has more energetic photons
Where this comes from
Photoelectric equationon the NESA formulae sheet
$K_{m a x} = h f - ϕ$
Symbols
$K_{m a x}$ maximum kinetic energy of an emitted electronJ
$h$ Planck's constant, 6.626 × 10⁻³⁴ J sJ s
$f$ frequency of the incident lightHz
$ϕ$ work function of the metal, the least energy that frees an electronJ
Valid when
Light falling on a clean metal surface, one photon to one electron. The result is the energy of the fastest electrons; most leave with less.
Not valid when
hf is less than φ, where no electrons are emitted and a negative answer has no meaning, or the kinetic energy of a typical electron is wanted rather than the maximum.
Rearranged
for \phi: $ϕ = h f - K_{m a x}$ for f_{0}: $f_{0} = \frac{ϕ}{h}$ for V_{s}: $V_{s} = \frac{K _{m a x}}{e}$
Where it turns up
The stopping voltage for a metal lit at a given frequency
The threshold frequency or wavelength of a metal
Reading h and φ from the gradient and intercept of Millikan's graph
Where marks go missing
Adding the work function instead of subtracting it
Leaving φ in electronvolts while hf is in joules
Expecting a brighter beam to raise K_max
Where this comes from
Time dilationon the NESA formulae sheet
$t = \frac{t _{0}}{1 - \frac{v ^{2}}{c ^{2}}}$
Symbols
$t$ time between two events measured by an observer the clock moves pasts
$t_{0}$ proper time, measured by a clock present at both eventss
$v$ relative speed of the two framesm s⁻¹
$c$ speed of light in a vacuumm s⁻¹
Valid when
Two inertial frames in uniform relative motion. t₀ belongs to the frame in which both events happen at the same place, such as the muon's own frame for its lifetime.
Not valid when
The proper time is assigned to the wrong frame, or the observer is accelerating strongly, as in a round trip with a turnaround, where the two frames are no longer symmetric.
Rearranged
for t_{0}: $t_{0} = t 1 - \frac{v ^{2}}{c ^{2}}$ for v: $v = c 1 - \frac{t _{0}^{2}}{t ^{2}}$
Where it turns up
The lifetime of a fast muon measured in the Earth frame
The time elapsed for astronauts on a fast spacecraft
Explaining the Hafele–Keating atomic clock results
Where marks go missing
Swapping t and t₀, which gives a moving clock that runs fast
Using v in m s⁻¹ with c written as 1
Forgetting the square root
Where this comes from
Length contractionon the NESA formulae sheet
$l = l_{0} 1 - \frac{v ^{2}}{c ^{2}}$
Symbols
$l$ length measured by an observer the object moves pastm
$l_{0}$ proper length, measured in the object's rest framem
$v$ relative speed of the two framesm s⁻¹
$c$ speed of light in a vacuumm s⁻¹
Valid when
Lengths measured along the direction of relative motion, between inertial frames. l₀ belongs to the frame in which the object is at rest.
Not valid when
The length is at right angles to the motion, which is unchanged, or the proper length is assigned to the moving observer.
Rearranged
for l_{0}: $l_{0} = \frac{l}{1 - \frac{v ^{2}}{c ^{2}}}$ for v: $v = c 1 - \frac{l ^{2}}{l _{0}^{2}}$
Where it turns up
The thickness of the atmosphere measured in a muon's frame
The length of a fast spacecraft measured from Earth
Where marks go missing
Contracting a length that is perpendicular to the motion
Dividing by the square root instead of multiplying, which makes the moving object longer
Where this comes from
Relativistic momentumon the NESA formulae sheet
$p_{v} = \frac{m _{0} v}{1 - \frac{v ^{2}}{c ^{2}}}$
Symbols
$p_{v}$ momentum of the particlekg m s⁻¹
$m_{0}$ rest mass of the particlekg
$v$ speed of the particlem s⁻¹
$c$ speed of light in a vacuumm s⁻¹
Valid when
Any particle with rest mass at any speed below c. At everyday speeds it reduces to p = m₀v.
Not valid when
Applied to a particle with no rest mass such as a photon, or used to find a speed at or above c, where the square root is zero or undefined.
Rearranged
for v: $v = \frac{p}{m _{0}^{2} + \frac{p ^{2}}{c ^{2}}}$
Where it turns up
The momentum of electrons or protons in an accelerator
Explaining why no finite force can bring a massive particle to c
Where marks go missing
Using p = m₀v for a particle moving at a large fraction of c
Concluding that the particle's rest mass itself increases
Where this comes from
Mass–energy equivalenceon the NESA formulae sheet
$E = m c^{2}$
Symbols
$E$ energy equivalent to the mass, or released when that mass is convertedJ
$m$ mass, or the mass lost by a systemkg
$c$ speed of light in a vacuumm s⁻¹
Valid when
Any process. The energy a system releases equals its loss of rest mass times c², whether the process is nuclear, chemical or an annihilation.
Not valid when
m is taken as the total mass of the reactants rather than the mass that disappears, or masses in atomic mass units are used without converting to kilograms.
Rearranged
for m: $m = \frac{E}{c ^{2}}$
Where it turns up
The rate at which the Sun loses mass
The photon energy from electron–positron annihilation
The mass change in burning a fuel
Where marks go missing
Using the whole mass of a fuel rather than the tiny mass lost
Forgetting to square c
Believing mass is converted only in nuclear reactions
Where this comes from
Mass defect and binding energynot on the formulae sheet, learn it
$E_{B} = (Z m_{p} + N m_{n} - m_{nucleus}) c^{2}$
Symbols
$E_{B}$ binding energy, the energy needed to separate the nucleus into free nucleonsJ
$Z$ number of protons1
$N$ number of neutrons1
$m_{p}$ mass of a protonkg
$m_{n}$ mass of a neutronkg
$m_{nucleus}$ mass of the nucleuskg
$c$ speed of light in a vacuumm s⁻¹
Valid when
Any nucleus. The bracket is the mass defect: the nucleus is lighter than its separated nucleons by exactly the binding energy divided by c². With masses in u, multiply the defect by 931.5 MeV instead of c².
Not valid when
An atomic mass is used for the nucleus while bare proton masses are used for the parts, which leaves the electrons unbalanced, or the defect in u is multiplied by c² without converting to kilograms.
Rearranged
for \Delta m: $Δ m = Z m_{p} + N m_{n} - m_{nucleus}$ for E_{B}: $E_{B} = Δ m \times 931.5 MeV u^{- 1}$
Where it turns up
The binding energy of helium-4 or iron-56
Binding energy per nucleon, to compare the stability of nuclei
Explaining why fission and fusion both release energy
Where marks go missing
Subtracting the wrong way round, giving a negative defect
Rounding the masses before subtracting, which wipes out the small difference
Mixing atomic and nuclear masses in one calculation
Where this comes from
Energy released in a nuclear reactionnot on the formulae sheet, learn it
$E = (m_{reactants} - m_{products}) c^{2}$
Symbols
$E$ energy released, carried off as kinetic energy of the products and as gamma raysJ
$m_{reactants}$ total mass before the reactionkg
$m_{products}$ total mass after the reactionkg
$c$ speed of light in a vacuumm s⁻¹
Valid when
Any decay or transmutation: alpha and beta decay, fission and fusion. Nucleon number and charge must balance first. A positive answer means energy is released.
Not valid when
The equation is unbalanced, or a neutron released on the product side is left out of the product mass.
Rearranged
for E: $E = Δ m \times 931.5 MeV u^{- 1}$
Where it turns up
The energy of an alpha decay
The energy released per fission of uranium-235
The energy of deuterium–tritium fusion
Where marks go missing
Forgetting the extra neutrons released in fission
Counting the incoming neutron on only one side
Converting u to MeV with 931.5 and then multiplying by c² as well
Where this comes from