Interactive resources · new
More physics, one slider at a time
Twenty more live models across the IB Physics syllabus. Predict first, then test it, then take on the challenge.
Exam tips and common errors are original ExaminerPrep material, written from years of teaching physics.
A.2 Inclined plane & friction SL & HL
A block on a rough slope. Change the angle, the mass and the two friction coefficients, push it with a force along the slope or flick it up the slope, then release it. The free-body diagram is drawn to scale, so you can see which force wins.
Will it slip from rest?
Acceleration a
Friction force
Normal force N
Velocity v, time t
Exam tip
Choose axes along and perpendicular to the slope. The weight then splits into mg sin θ down the slope and mg cos θ into the slope, and with no other perpendicular forces N = mg cos θ. Write Newton's second law along the slope only: for a block sliding down, ma = mg sin θ − μdmg cos θ, so a = g(sin θ − μd cos θ). Draw the free-body diagram first and label every force before writing any equation.
Common error
Writing the static friction force as μsN every time. μsN is only the largest value static friction can reach; a block at rest has exactly as much friction as it needs to stay put (here mg sin θ − P, in either direction). Another slip: using N = mg on a slope — the normal force is smaller, mg cos θ.
A.2 Impulse & force–time graphs SL & HL
A ball hits a wall and bounces back. The change in momentum is fixed by the speeds before and after, so the area under the force–time graph is fixed too. Stretch the contact time and watch the peak force fall: the idea behind padding, airbags and crumple zones.
Change in momentum Δp
Impulse = area under F–t
Average force Favg
Peak force Fmax
Kinetic energy lost
Exam tip
Impulse is J = FΔt = Δp for a constant force; when the force varies, the impulse is the area under the force–time graph. Take a positive direction first and give every velocity a sign: a ball arriving at −10 m s⁻¹ and leaving at +6 m s⁻¹ has Δv = +16 m s⁻¹, not 4 m s⁻¹. The average force is the impulse divided by the contact time.
Common error
Saying an airbag or a crumple zone “reduces the impulse” or “absorbs the momentum”. The car's occupant still goes from the same speed to rest, so Δp, and therefore the impulse, is unchanged. What changes is the time over which it happens: a longer Δt gives a smaller average and peak force.
A.3 Work, power & efficiency SL & HL
An electric motor winds a cable and lifts a load at constant speed. Set the electrical power going in and the motor's efficiency: the useful power sets the lift speed through P = Fv, and the energy flow diagram shows where every joule goes.
Useful power
Lift speed v
Time to lift
Energy so far
Wasted so far
Exam tip
Power is the rate of doing work, P = W/t. For a force moving something at constant velocity in the direction of the force, P = Fv; lifting at steady speed needs F = mg, so the useful power is mgv. Efficiency can be found from energies or from powers: η = useful output ÷ total input. Always say which quantity is “useful” and keep η as a decimal (0.60) inside equations.
Common error
Using the motor's input power in P = Fv. Only the useful output power, ηPin, goes into lifting the load. A second slip: thinking the wasted energy disappears. It is still there — mostly as internal energy of the motor and surroundings — so input = useful + wasted at every instant.
A.4 Rolling down a slope HL
Race a hoop, a disc, a solid sphere and a block that slides without friction down the same slope. Each object rolls without slipping, so part of its energy has to go into rotation. Watch the energy bars: the bigger the share that goes into spinning, the slower the object gets down the slope.
Hoop · I = mr²
Disc · I = ½mr²
Solid sphere · I = ⅖mr²
Sliding block (no friction)
Exam tip
For rolling without slipping, link the motions with v = ωr. Energy conservation then gives mgh = ½mv² + ½Iω² = ½mv²(1 + I/mr²), so v = √(2gh / (1 + I/mr²)) at the bottom and the acceleration is a = g sin θ / (1 + I/mr²). The ratio I/mr² depends only on the shape (how the mass is spread out), which is why mass and radius cancel.
Common error
Thinking the friction on a rolling object wastes energy. In rolling without slipping the contact point is momentarily at rest, so static friction does no work: mechanical energy is conserved. Friction's job is to provide the torque that makes the object spin, moving energy into rotation — that is why a rolling object is slower than a block sliding on a frictionless slope, not because energy is lost.
A.5 Spacetime diagrams HL
A spacetime (Minkowski) diagram for frame S, with the axes of frame S′ moving at speed v drawn on top. Place two events, read their coordinates in both frames, and see which quantities every observer agrees on. Drag the events on the diagram or use the sliders.
Lorentz factor γ
Event A in S′
Event B in S′
Interval of A (ct)² − x²
A to B
Exam tip
To read a coordinate in S′, draw a line through the event parallel to the other primed axis: parallel to the x′ axis to read ct′ on the ct′ axis, and parallel to the ct′ axis to read x′ on the x′ axis. Both primed axes make the angle θ = tan⁻¹(v/c) with the S axes, and check every reading with x′ = γ(x − vt) and ct′ = γ(ct − vx/c). The spacetime interval (ct)² − x² is the same in every inertial frame.
Common error
Reading primed coordinates by dropping perpendiculars onto the tilted axes, or using the same tick spacing on both sets of axes. The unit length on the ct′ and x′ axes is longer than on the S axes (by a factor √((1 + v²/c²)/(1 − v²/c²))), so a scale drawn for S cannot be reused for S′. Also, events that are simultaneous in S lie on a horizontal line, but events simultaneous in S′ lie on a line parallel to the x′ axis.
B.1 Thermal conduction SL & HL
Energy flows by conduction through a slab whose faces are held at fixed temperatures. In the steady state the rate of transfer is P = kA ΔT/Δx. Change the material, thickness, area and temperatures. You can also add a second layer in series, as in an insulated wall.
Rate of transfer P
Temperature gradient
Interface temperature
Energy in 1 hour
Exam tip
Use P = kA ΔT/Δx with Δx in metres (convert from mm!) and ΔT in kelvin; a temperature difference is the same in K and °C. For layers in series the same power passes through each layer, so find P first, then use it to find the temperature drop across each layer: ΔT₁ = PΔx₁/(k₁A). The layer with the smaller Δx/k has the smaller temperature drop.
Common error
Adding the thermal conductivities of two layers, or averaging them. In series it is Δx/(kA) for each layer that adds, just like resistors in series. Another slip is thinking a good conductor “holds” the heat: in the steady state the temperature at each point is constant and no energy is being stored anywhere in the slab.
B.4 First law & p–V processes HL
An ideal monatomic gas in a cylinder goes from state 1 to state 2 by one of four processes. Follow the path on the p–V diagram, where the shaded area is the work done by the gas, and check the energy balance Q = ΔU + W.
State 1 → state 2
Work done by gas W
Change in internal energy ΔU
Thermal energy in Q
Check
Exam tip
Write the first law as Q = ΔU + W and state your sign convention: Q is energy transferred to the gas and W is work done by the gas. For an ideal monatomic gas ΔU = (3/2)nRΔT for any process, so find ΔU from the temperatures first. W is the area under the p–V path: pΔV at constant pressure, zero at constant volume. For an adiabatic change Q = 0, so W = −ΔU, and pV5/3 = constant.
Common error
Thinking “isothermal” means no thermal energy is transferred. At constant temperature ΔU = 0, so Q = W: a gas expanding isothermally must absorb energy equal to the work it does. It is the adiabatic process that has Q = 0, and there the temperature does change. Also check units: p in Pa and V in m³ give work in J, so convert kPa and 10⁻³ m³ before multiplying.
B.5 Resistivity & I–V characteristics SL & HL
Vary the potential difference across a component and measure the current. A metal wire at constant temperature is ohmic, with R = ρL/A. A filament lamp heats up, so its resistance rises. A diode barely conducts in one direction. Compare R = V/I with the gradient of the graph.
Current I
Resistance R = V/I
1 ÷ gradient ΔV/ΔI
Power P = VI
Behaviour
Exam tip
Resistance is defined as R = V/I at a point on the I–V graph. Draw the line from the origin to that point: its gradient is I/V = 1/R. For an ohmic conductor this line is the graph, so the gradient gives 1/R everywhere. For a wire, R = ρL/A: convert the area to m² (1 mm² = 1 × 10⁻⁶ m²) and remember that A = πd²/4 if you are given the diameter.
Common error
Taking the resistance of a lamp or diode as 1 ÷ (gradient of the tangent). For a non-ohmic component the tangent gradient and V/I are different, and only V/I is the resistance. Another slip: saying a filament lamp “doesn’t obey Ohm’s law because its resistance changes”. Ohm’s law needs constant temperature; the lamp’s filament gets hotter as the current rises.
B.5 Potential divider (LDR & thermistor) SL & HL
Two resistors in series share the supply p.d. in the ratio of their resistances. Replace one of them with a light-dependent resistor or a thermistor and the output p.d. follows the light level or the temperature — the basis of an automatic night light or a frost-protection heater.
Sensor resistance
Current from supply
Output p.d. Vout
Device
Exam tip
For two resistors in series the current is the same, so the p.d. divides in proportion to resistance: Vout = Vin R₂ / (R₁ + R₂), where R₂ is the resistor the output is taken across. To explain a sensor circuit, go step by step: condition changes → sensor resistance changes → its share of the total resistance changes → Vout rises or falls → the device switches. An LDR’s resistance falls as light intensity rises; an NTC thermistor’s resistance falls as temperature rises.
Common error
Forgetting the load. A device connected across the output is in parallel with R₂, so the combined lower resistance is smaller than R₂ and Vout drops below the unloaded value. The simple ratio formula only holds when the load resistance is much larger than R₂. A second slip: thinking that bigger resistors give a bigger output. Only the ratio R₂ : (R₁ + R₂) matters — doubling both resistances leaves Vout unchanged.
B.5 Kirchhoff’s laws SL & HL
Two cells and three resistors in a two-loop network. Choose a direction for each branch current, apply the junction rule and the loop rule, and solve. A negative answer simply means the current flows the other way. Dots show the actual conventional current.
Current I₁ (in R₁)
Current I₂ (in R₂)
Current I₃ (in R₃)
Junction P: I₁ + I₂ − I₃
Loop 1 (left)
Loop 2 (right)
Outer loop
Exam tip
Mark an assumed direction for every branch current before writing anything. Then: junction rule ΣIin = ΣIout at a node (charge is conserved), and loop rule Σε = ΣIR around any closed loop (energy is conserved). Going round a loop, count an emf as positive if you pass through the cell from − to +, and an IR term as positive if you travel in the same direction as the assumed current. Two loops plus one junction give three equations for three unknowns — solve, and keep the signs.
Common error
Changing the sign of a current halfway through. If I₂ comes out as −0.5 A, keep using −0.5 A in every later equation; only at the end interpret it as 0.5 A flowing against the assumed arrow. Another slip: thinking a cell always delivers energy. If the current is driven backwards through a cell (from + to − inside it), that cell is being charged.
C.1 Simple pendulum SL & HL
A point mass on a light string, released from rest. The full equation of motion is integrated step by step and compared with the small-angle SHM model T = 2π√(L/g). At small amplitudes the two agree; at large amplitudes the real pendulum takes longer.
SHM period 2π√(L/g)
Actual period
Angle θ
Speed v
Total energy
Exam tip
The restoring force on the bob is −mg sin θ. Only when θ is small (in radians, sin θ ≈ θ ≈ x/L) does this become proportional to displacement, giving SHM with ω² = g/L and T = 2π√(L/g). Mass cancels, so it does not affect the period. To measure g, time many oscillations, plot T² against L: a straight line through the origin with gradient 4π²/g.
Common error
Saying the period of a pendulum never depends on amplitude. That is only true for small swings. At 20° the period is less than 1 % longer, but at 90° it is about 18 % longer, because sin θ < θ so the restoring force is weaker than the SHM model assumes. A second slip: using degrees in sin θ ≈ θ — the approximation only works with θ in radians.
C.2 Transverse & longitudinal waves SL & HL
The same travelling wave carried by a row of particles in two ways. In a transverse wave the particles oscillate at right angles to the direction the wave travels; in a longitudinal wave they oscillate along it, forming compressions and rarefactions. In both, each particle only oscillates about its own equilibrium position — the wave carries energy, not the particles.
Wavelength λ = v/f
Period T = 1/f
Highlighted particle
Max particle speed 2πfA
Exam tip
Learn the definitions by the direction of oscillation relative to the direction of energy transfer: perpendicular → transverse (waves on a string, all electromagnetic waves), parallel → longitudinal (sound, P-waves). v = fλ links the three quantities; the speed is set by the medium, so changing the frequency changes the wavelength. On a displacement–position graph of a longitudinal wave, a compression sits where the displacement is zero and the particles on either side are displaced towards it.
Common error
Reading a displacement–position graph of a sound wave as a picture of the wave, and putting the compressions at the peaks. A peak means the particle there is displaced furthest in the positive direction, not that the particles are crowded together. A second slip: thinking the particles travel along with the wave. Follow the highlighted particle — it returns to the same place every period.
C.3 Single-slit diffraction SL & HL
Monochromatic light passes through one narrow slit of width b and spreads out. Waves from different parts of the slit interfere, giving a bright, wide central maximum with much fainter maxima on either side. The narrower the slit, the more the light spreads.
1st minimum θ = λ/b
Central maximum width
Side maximum width
Central intensity
Exam tip
The first minimum is where θ = λ/b, with θ in radians (small angles). The central maximum runs from −θ to +θ, so its width on a screen a distance D away is 2λD/b — twice the width of each side maximum. To make the pattern spread more, use a narrower slit or a longer wavelength. Diffraction is only noticeable when b is not much larger than λ.
Common error
Using θ = λ/b as the angular width of the whole central maximum. It is the angle from the centre to the first minimum — the half-width — so the full width is 2θ. A second slip: putting θ in degrees into θ = λ/b, or mixing nm and μm without converting both to metres.
C.5 Doppler effect for light & redshift SL & HL
Hydrogen absorbs light at the same four visible wavelengths everywhere in the Universe. Compare a laboratory spectrum with the spectrum of a distant galaxy or star: if the source moves away, every line is shifted to a longer wavelength (redshift); if it approaches, to a shorter one (blueshift). For v ≪ c, Δλ/λ ≈ v/c.
Redshift z = Δλ/λ₀
Hα line (656.3 nm)
Frequency change Hα
Speed as a fraction of c
Exam tip
Use Δλ/λ₀ = v/c with λ₀ the laboratory (rest) wavelength and Δλ = λ_observed − λ₀. A positive Δλ means the source is receding. Pick one line, read both wavelengths carefully, and keep at least 3 s.f. in Δλ — it is a small difference of two large numbers. The same ratio applies to frequency: Δf/f₀ ≈ v/c, but f decreases for a receding source.
Common error
Thinking every line moves by the same number of nanometres. The shift is a fixed fraction of each wavelength, so the 656 nm line moves further than the 410 nm line — check the graph. Also, “redshift” does not mean the light turns red: a blue line becomes a slightly less blue line, and the whole pattern slides along together.
D.1 Gravitational field & potential HL
Around a spherical planet the field strength falls as 1/r² and the potential rises towards zero as −1/r. Field lines point radially inwards; equipotential surfaces are spheres, drawn here at equal steps of potential. Move a mass from point 1 to point 2 and read the work done.
Field strength at 1
Potential at 1 and 2
ΔV = V₂ − V₁
Work done W = mΔV
E_p of m at 1
Exam tip
Outside a spherical mass: g = GM/r² and V = −GM/r, with r measured from the centre, not the surface. The work done moving a mass m between two points is W = mΔV and does not depend on the path. The field is the negative potential gradient, g = −ΔV/Δr: where equipotentials are closer together, the field is stronger.
Common error
Using the height above the surface as r. A satellite 400 km up orbits at r = 6371 + 400 = 6771 km. A second slip: dropping the minus sign. V is zero only at infinity and negative everywhere else, so moving away from a planet makes V less negative — ΔV is positive and work must be done on the mass.
D.1 Kepler's third law SL & HL
Planets on circular orbits around one star. Gravity provides the centripetal force, GMm/r² = mv²/r, which gives v = √(GM/r) and T² = (4π²/GM) r³. Every planet in the system lies on the same straight line of T² against r³; a heavier star makes the line less steep. The orbits are animated at their true relative speeds.
Period of P
Orbital speed of P v = √(GM/r)
Gradient 4π²/GM
Animation
Exam tip
Derive it rather than memorise it: set the gravitational force equal to the centripetal force, GMm/r² = mv²/r, and substitute v = 2πr/T. The planet's mass m cancels, so T depends only on r and the central mass M. To find M from orbit data, plot T² against r³ and use M = 4π²/(G × gradient), or work with one planet. Convert AU and years to metres and seconds first.
Common error
Thinking an outer planet moves faster because it covers a bigger circle. Its path is longer, but its speed is lower (v ∝ 1/√r) and its period is much longer (T ∝ r^1.5). Another slip: assuming a heavier planet orbits faster. At a given radius every planet, moon or satellite has the same period, whatever its own mass.
D.3 Charge between parallel plates SL & HL
A charged particle enters the uniform field between two parallel plates, moving parallel to them. The field E = V/d pushes it with a constant force F = qE, so its path inside the plates is a parabola, exactly like a projectile. Outside the plates the field is taken as zero (edge effects ignored) and the path is straight.
Field strength E
Acceleration a
Time in field
Deflection at exit
Exit angle
Kinetic energy gained
Exam tip
Treat it as a projectile problem. Along the plates there is no force, so the time in the field is t = L/v. Across the plates the acceleration is constant: a = qE/m with E = V/d. Then the deflection is y = ½at² and the perpendicular velocity at the exit is v_y = at, giving the exit angle from tan θ = v_y/v. Write each step separately with units; it keeps the powers of ten under control.
Common error
Using the p.d. V in place of the field strength E in F = qE. The force depends on V/d: move the plates apart at the same p.d. and the deflection falls. Another slip is giving the electron the proton’s mass, or forgetting that a negative charge accelerates towards the positive plate, against the field direction.
D.4 Lenz's law HL
The induced current always flows so that its own magnetic field opposes the change that caused it. Push a magnet towards a copper ring or pull it away and see which pole the ring becomes. Or drop a magnet down a copper tube and a plastic tube: the eddy currents in the copper slow it to a terminal speed, and the lost gravitational energy ends up as internal energy in the tube.
Induced pole (near face)
Current seen from magnet
Force on magnet
Peak current
Exam tip
Explain Lenz's law in three linked steps: (1) the flux linkage through the circuit is changing; (2) by Faraday's law an EMF is induced and, in a closed circuit, a current flows; (3) the current's own magnetic field opposes the change in flux, so the force on the magnet opposes its motion. Then finish with energy: work must be done against that force, and it becomes electrical energy and then internal energy in the conductor.
Common error
Saying the induced field opposes the magnet's field. It opposes the change in flux: when the magnet moves away, the ring attracts it, trying to keep the flux. Another slip is thinking copper is attracted to magnets — it is not magnetic. The plastic tube has no effect only because no current can flow in it; a copper tube with a slit along its length would also slow the magnet much less.
E.1 Rutherford scattering SL & HL
Alpha particles are fired at a single gold nucleus. The only force is the electric (Coulomb) repulsion F = kQq/r², and each path is integrated step by step. Change the impact parameter b, the sideways offset of the beam from the nucleus, or fire a whole beam and count how many particles bounce back.
Scattering angle θ
Closest approach
Head-on d = kQq / E_k
Beam: θ > 90°
Exam tip
For a head-on collision, all the kinetic energy becomes electric potential energy at the closest point: E_k = kQq/d, so d = kQq/E_k. Convert MeV to joules (1 MeV = 1.60 × 10⁻¹³ J) and use Q = Ze and q = 2e. This gives an upper limit for the radius of the nucleus, since the alpha never touched it. When describing the experiment, link each observation to a conclusion: most pass straight through, so the atom is mostly empty space; a very few rebound, so the positive charge and almost all the mass are in a tiny nucleus.
Common error
Drawing the paths as if the alpha bounces off a hard surface. The force acts all along the path, so the tracks are smooth curves (hyperbolas), symmetric about the line of closest approach, and they bend away from the nucleus before they reach it. Another slip is using the charge of the whole atom: the alpha is repelled by the nucleus (+79e for gold), because the electrons have far too little mass to deflect it.
E.3 N–Z chart & decay modes SL & HL
Each square is a nuclide: Z protons and N neutrons. The stable nuclides form a narrow band that starts on the line N = Z and bends upwards, because heavy nuclei need extra neutrons. Move the cursor off the band to predict how the nuclide decays, see each decay as an arrow, or step through the uranium-238 decay series (simplified to its main branch).
Nuclide
N / Z
Likely decay
Daughter
Exam tip
Learn the arrows. α decay: Z falls by 2 and N falls by 2 (A falls by 4). β⁻ decay: a neutron becomes a proton, so N − 1 and Z + 1 (A unchanged), with an electron and an antineutrino emitted. β⁺ decay: a proton becomes a neutron, so Z − 1 and N + 1, with a positron and a neutrino. A nuclide above the stable band has too many neutrons for its Z and moves towards the band by β⁻; one below the band moves by β⁺. Always check that A and Z balance on both sides of the equation.
Common error
Thinking N = Z is the stable line for all nuclei. It is only close for light nuclei; lead-206 has N/Z ≈ 1.5. Another slip is drawing a β⁻ arrow straight up or along the Z axis only: in β decay both N and Z change by one, so on an N–Z chart the arrow is diagonal. Gamma emission does not change N or Z at all, so it does not move the nuclide on the chart.
Understanding the concept is the start. Turning it into marks is the work we do together.
Book a free consultationThese simulations are original ExaminerPrep material. They are not IB resources and do not reproduce IB examination content or mark schemes. Topic references follow the published IB subject guides. ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. “International Baccalaureate”, “IB” and “IB Diploma Programme” are registered trademarks of the IBO, used here for descriptive purposes only.