Seventeen lessons on kinematics, forces, energy, thermal physics, circuits, oscillations, waves, fields, radioactive decay and the scientific investigation for the current IB Physics course.
A skills course on calculation and reasoning across the first-assessment-2025 syllabus areas A to E. It does not replace the syllabus. Higher-level-only topics (rigid body mechanics, special relativity, induction, parts of quantum physics) are not covered.
Algebra, basic trigonometry and the idea of a vector.
Course outline
Kinematics: pick the equation from what you know
Choose a constant-acceleration equation by listing the knowns, and read motion graphs correctly.
Projectile motion: two independent motions
Split a projectile into horizontal and vertical parts and recombine them.
Forces, free-body diagrams and Newton's laws
Draw complete free-body diagrams and turn them into equations of motion.
Momentum and impulse: conserve it, then check energy
Use conservation of linear momentum in collisions and impulse in force-time problems.
Work, energy and power: store, transfer, compare
Track energy transfers and compute work, kinetic, potential energy and power.
Circular motion: the force points inward
Link centripetal acceleration to the inward resultant force in circular paths.
Thermal energy: heating, cooling and changing state
Use specific heat capacity and latent heat to compute energy transfers.
Gases: temperature in kelvin and the gas laws
Apply pressure, volume and temperature relationships to ideal gases with kelvin.
Current and circuits: reduce before you calculate
Simplify series and parallel resistor networks and apply Ohm's law with care.
Simple harmonic motion: what sets the period
Link restoring force, period and energy exchange in oscillations.
The wave model: v = f lambda and what each symbol means
Use wavelength, frequency, speed and the transverse and longitudinal distinction.
Count nodes and antinodes in standing waves and relate frequency shifts to relative motion.
Gravitational fields: inverse square and orbits
Apply Newton's law of gravitation and field strength to orbital problems.
Electric and magnetic fields: forces on charges
Compare electric and magnetic forces and find the motion of a charge in each.
Radioactive decay: halve, halve, halve
Use half-life, decay equations and activity with the right units.
Graphs, uncertainty and the scientific investigation
Linearise data, find uncertainties from gradients and structure the investigation report.
Sources and curriculum note
Reviewed October 5, 2026 against the IB subject brief (first assessment 2025). The brief lists Paper 1 at 36 percent, Paper 2 at 44 percent and the internal assessment at 20 percent, with a 3,000-word maximum for the scientific investigation report. Constants are rounded teaching values from the NIST CODATA list (G, R, e); the Coulomb constant, electron mass and Earth GM are rounded supplied values. Use the values in your own current IB data booklet and question. Check the official IB page for changes.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Kinematics: pick the equation from what you know
Learning goal: Choose a constant-acceleration equation by listing the knowns, and read motion graphs correctly.
For motion in a straight line with constant acceleration, five quantities describe the situation: displacement s, initial velocity u, final velocity v, acceleration a and time t. Any three knowns fix the rest. The method is to write the knowns and the wanted quantity, then choose the equation that does not contain the missing fifth quantity. That choice removes most of the guessing.
The four equations are v = u + at, s = ut + at^2/2, v^2 = u^2 + 2as and s = (u + v)t/2. Each leaves out one variable: the first leaves out s, the second leaves out v, the third leaves out t and the fourth leaves out a. If a question gives no time, the third equation is the natural tool. If it gives no final velocity, use the second.
Choose a sign convention before substituting and keep it for the whole problem. If up is positive, a ball thrown upward has positive u and the acceleration of free fall is negative. Quantities with direction (displacement, velocity, acceleration) carry signs. Distance and speed do not. A negative displacement is not an error; it means the object ended below or behind its start.
Graphs carry the same information. On a velocity-time graph the gradient is acceleration and the area between the line and the time axis is displacement, with area below the axis counting as negative. On a displacement-time graph the gradient is velocity. A curved displacement-time graph means velocity is changing. These equations only apply while acceleration is constant, so split the motion into stages when it changes.
Worked example
A trolley starts from rest with a = 3.0 m s-2 over s = 6.0 m. Find v.
u = 0, a = 3.0 m s-2, s = 6.0 m, v unknown, t not given.
The equation without t is v^2 = u^2 + 2as.
v^2 = 0 + 2 x 3.0 x 6.0 = 36 m^2 s-2.
v = 6.0 m s-1 in the direction of the acceleration.
Practice problem and solution
A stone is thrown vertically upward at 19.6 m s-1 from the ground. Use g = 9.8 m s-2. At what time after launch (in seconds) is it 14.7 m above the ground while moving downward?
Up is positive: 14.7 = 19.6t - 4.9t^2, so t^2 - 4t + 3 = 0, giving t = 1 s (rising) or t = 3 s (falling). The stone is falling at t = 3 s.
Mental model: Pick the equation by the missing variable and fix the sign convention first.
Common trap: Treating both quadratic roots as the same event.
2. Projectile motion: two independent motions
Learning goal: Split a projectile into horizontal and vertical parts and recombine them.
A projectile is an object moving only under gravity, ignoring air resistance. The key idea is that horizontal and vertical motion are independent. Horizontally there is no force, so velocity stays constant. Vertically the acceleration is g downward, so the vertical velocity changes by about 9.8 m s-1 each second. Time is the shared quantity that connects the two directions.
Start by resolving the launch velocity v into components: vx = v cos(theta) and vy = v sin(theta), where theta is measured from the horizontal. Then write separate equations. Horizontally, x = vx t. Vertically, use the constant-acceleration equations with a = -g if up is positive. Solve the vertical part for time (for example, the time to return to launch height is 2 vy / g), then put that time into the horizontal equation.
At the top of the path the vertical velocity is zero but the horizontal velocity is not, so the speed there is vx, not zero. The acceleration is still g downward at every point, including the top. For a launch and landing at the same height, the symmetric path means the landing speed equals the launch speed, and the time up equals the time down.
For a launch from a height, the symmetry no longer holds and the landing speed is greater than the launch speed. Use the vertical equation s = ut + at^2/2 with the actual displacement to find the time, which gives a quadratic. Keep the positive root that fits the story. Finally, recombine the components with Pythagoras and an angle if the question asks for the speed or direction of impact. Air resistance reduces the range and makes the path asymmetric; the simple model is an approximation.
Worked example
A ball is kicked at 20 m s-1 at 30 degrees above horizontal from level ground. Use g = 9.8 m s-2. Find the time of flight.
vy = 20 sin 30 = 10 m s-1.
Return to the same height: s = 0 = vy t - 4.9 t^2.
t = 2 vy / g = 20 / 9.8.
t = 2.04 s, about 2.0 s.
Practice problem and solution
A ball is kicked at 20 m s-1 at 30 degrees above horizontal from level ground. Use g = 9.8 m s-2 and ignore air resistance. Give the horizontal range in metres, to three significant figures.
vx = 20 cos 30 = 17.32 m s-1; t = 2.04 s; range = 17.32 x 2.04 = 35.3 m.
Mental model: Split, solve vertically for time, then go horizontally.
Common trap: Using the speed, not its component, in the horizontal equation.
3. Forces, free-body diagrams and Newton's laws
Learning goal: Draw complete free-body diagrams and turn them into equations of motion.
A force is a push or pull on an object, measured in newtons. A free-body diagram shows one object alone, with every force acting on it drawn as an arrow from the centre of the object. Do not draw forces the object exerts on others, and do not draw the object's velocity. Typical forces are weight (mg, downward), the normal force (perpendicular to a surface), tension, friction and an applied push.
Newton's first law says that without a resultant force, velocity stays constant, which includes being at rest. The second law says that the resultant force equals mass times acceleration, F = ma, and that the acceleration is in the direction of the resultant. Apply it along each chosen axis separately. For a block on a slope, choose axes along and perpendicular to the slope, and resolve the weight into mg sin(theta) along the slope and mg cos(theta) perpendicular to it.
Newton's third law is about pairs. If object A exerts a force on object B, then B exerts an equal and opposite force on A. The two forces are the same type, act on different objects and so never cancel in a free-body diagram of a single object. A common mistake is to call weight and the normal force a third-law pair. They act on the same object and often differ, as in a lift that accelerates.
Friction opposes relative motion or its tendency. The maximum static friction is mu_s times the normal force and kinetic friction is mu_k times the normal force. Static friction can take any value up to its maximum, which is why a parked car on a gentle slope has friction equal to the weight component, not to the maximum. For connected objects, write F = ma for each object, and use the fact that a rope of negligible mass has the same tension throughout.
Worked example
A 4.0 kg block is pulled across a table by a 20 N horizontal force. Kinetic friction is 8.0 N. Find the acceleration.
Resultant force = 20 - 8.0 = 12 N.
F = ma, so a = 12 / 4.0.
a = 3.0 m s-2.
The direction is that of the applied force.
Practice problem and solution
A 2.0 kg block slides down a smooth slope inclined at 30 degrees. Use g = 9.8 m s-2. Find its acceleration down the slope in m s-2.
Along the slope the resultant is mg sin 30; a = g sin 30 = 9.8 x 0.5 = 4.9 m s-2.
Mental model: Draw one object, resolve, then apply F = ma.
Common trap: Calling weight and the normal force a third-law pair.
4. Momentum and impulse: conserve it, then check energy
Learning goal: Use conservation of linear momentum in collisions and impulse in force-time problems.
Linear momentum is mass times velocity, p = mv, and is a vector. In a closed system, with no external resultant force, the total momentum before an interaction equals the total after. This holds for collisions and explosions. Choose a positive direction, give each velocity a sign, then write sum of m v before equals sum of m v after. The signs are where most errors occur.
Impulse is the change in momentum, delta p, and equals the resultant force multiplied by the time it acts, F delta t, when the force is constant. In general, impulse is the area under a force-time graph. The same change in momentum can be delivered by a large force over a short time or a smaller force over a longer time. That is why crumple zones and padding reduce the force on a person: they lengthen the time of the collision.
Collisions are classified by what happens to kinetic energy. In an elastic collision the total kinetic energy is unchanged and the relative speed of approach equals the relative speed of separation. In an inelastic collision some kinetic energy becomes other forms such as thermal energy and sound. In a perfectly inelastic collision the objects stick together and move off with one velocity. Momentum is conserved in all of them, as long as the system is closed.
A useful check after finding a final velocity is to calculate kinetic energy before and after. Kinetic energy cannot increase in a collision unless energy is released, as in an explosion. If your answer shows an increase in a simple collision, an error has occurred. Newton's second law can also be written F = delta p / delta t, which connects force to the rate of change of momentum and handles cases where mass changes, such as a jet of water hitting a wall.
Worked example
A 2.0 kg trolley at 3.0 m s-1 hits a stationary 1.0 kg trolley and they stick together. Find the common velocity.
Momentum before = 2.0 x 3.0 + 0 = 6.0 kg m s-1.
Momentum after = (2.0 + 1.0)v.
3.0 v = 6.0.
v = 2.0 m s-1 in the original direction.
Practice problem and solution
A 0.15 kg ball arrives horizontally at 20 m s-1 and rebounds at 15 m s-1 in the opposite direction. The contact lasts 0.010 s. Find the magnitude of the average force in newtons.
Take the arrival direction as positive: delta p = 0.15 x (-15 - 20) = -5.25 kg m s-1. F = delta p / delta t = 5.25 / 0.010 = 525 N, in the rebound direction.
Mental model: Conserve momentum, then check kinetic energy.
Common trap: Forgetting that a rebound changes the sign of the velocity.
5. Work, energy and power: store, transfer, compare
Learning goal: Track energy transfers and compute work, kinetic, potential energy and power.
Work is the energy transferred by a force. For a constant force F moving an object through displacement s at angle theta to the force, W = F s cos(theta). Work is zero when the force is perpendicular to the motion, as with the normal force on a sliding block. Work is negative when the force opposes the motion, as with friction. The unit is the joule, which equals one newton metre.
Kinetic energy is mv^2/2 and depends on the square of speed, so doubling the speed quadruples the kinetic energy. Gravitational potential energy near the surface changes by mg delta h. The work-energy principle says the net work done on an object equals its change in kinetic energy. This is often faster than using kinematics because it ignores the path and needs only the start and end states, plus the work done by non-conservative forces such as friction.
Conservation of energy states that energy is neither created nor destroyed, only transferred or stored in different forms. For a mechanical system where only gravity does work, the sum of kinetic and gravitational potential energy stays constant. If friction or air resistance acts, mechanical energy decreases and the difference appears as thermal energy. A bookkeeping table with columns for before and after, and rows for kinetic, potential and thermal energy, keeps the process organised.
Power is the rate of transferring energy, P = W / t, in watts. For an object moving at constant velocity under a force F, P = F v. Efficiency is useful energy output divided by total energy input, and cannot exceed 100 percent. Be careful to distinguish energy, a quantity, from power, a rate. A device rated at 2.0 kW that runs for 3.0 hours transfers 6.0 kWh, which is 21.6 MJ, because 1 kWh equals 3.6 MJ.
Worked example
A 0.50 kg ball falls from rest through 20 m. Ignore air resistance and use g = 9.8 m s-2. Find its speed at the bottom.
Loss of potential energy = mgh = 0.50 x 9.8 x 20 = 98 J.
This equals the gain in kinetic energy, mv^2/2 = 98 J.
v^2 = 2 x 98 / 0.50 = 392.
v = 19.8 m s-1, about 20 m s-1.
Practice problem and solution
A motor lifts a 40 kg load at a constant speed of 0.50 m s-1. Use g = 9.8 m s-2. Find the useful power delivered to the load in watts.
Force equals weight = 40 x 9.8 = 392 N; P = F v = 392 x 0.50 = 196 W.
Mental model: Work equals force along motion times distance; energy is conserved.
Common trap: Using the total distance instead of the component along the force.
6. Circular motion: the force points inward
Learning goal: Link centripetal acceleration to the inward resultant force in circular paths.
An object moving in a circle at constant speed is accelerating because its velocity direction keeps changing. The acceleration points toward the centre and has magnitude v^2 / r, which is also 4 pi^2 r / T^2 for period T. The resultant force, called the centripetal force, equals mv^2 / r and also points toward the centre. Centripetal force is not a new force. It is the name of the resultant of real forces, such as tension, friction, gravity or the normal force.
A strong habit is to ask which real force or combination supplies the inward resultant. For a car rounding a flat bend, static friction between the tyres and road points inward. For a satellite in orbit, gravity is the inward force. For a ball on a string swung in a horizontal circle, the horizontal component of tension is inward. Never add centripetal force to a free-body diagram as an extra arrow, and do not draw an outward force for the object in an inertial frame.
Angular speed omega equals 2 pi / T and v = omega r. If the speed doubles with the same radius, the inward force needed becomes four times as large. If the radius doubles with the same speed, the force needed halves. For a car on a flat road, the maximum speed before skidding comes from setting the maximum friction, mu m g, equal to mv^2 / r, which gives v = the square root of mu g r. Mass cancels, so a heavier car has the same limit.
In vertical circles the speed is not constant, but at the top and bottom you can still apply F = mv^2 / r. At the bottom of a loop, tension minus weight supplies the inward resultant, so the tension is larger than the weight. At the top, tension plus weight supplies it. A bucket swung in a vertical circle keeps its water in only if the speed at the top is high enough that the required inward force at least equals the weight, so v^2 / r must be at least g.
Worked example
A 1200 kg car rounds a flat bend of radius 50 m at 10 m s-1. Find the inward force needed.
F = mv^2 / r.
F = 1200 x 10^2 / 50.
F = 120000 / 50 = 2400 N.
Friction between tyres and road must supply at least 2400 N.
Practice problem and solution
A satellite orbits at a constant speed of 7.5 km s-1 in a circle of radius 6.8 x 10^6 m. Find its centripetal acceleration in m s-2, to two significant figures.
a = v^2 / r = (7500)^2 / (6.8 x 10^6) = 5.625 x 10^7 / 6.8 x 10^6 = 8.27, about 8.3 m s-2.
Mental model: The centripetal force is the resultant that points inward.
Common trap: Drawing a separate centripetal force in addition to the real forces.
7. Thermal energy: heating, cooling and changing state
Learning goal: Use specific heat capacity and latent heat to compute energy transfers.
Temperature measures the average random kinetic energy of the particles in a substance. Thermal energy, or internal energy in a fuller treatment, depends on the particle energies and the number of particles. A large cool object can hold more thermal energy than a small hot one. Thermal energy transfers spontaneously from a higher temperature to a lower one by conduction, convection or radiation until thermal equilibrium is reached.
The specific heat capacity c of a material is the energy needed to raise the temperature of one kilogram by one kelvin. The energy transferred is Q = m c delta T. A temperature change in kelvin has the same size as in degrees Celsius, so you can use either for delta T, but absolute temperature must be in kelvin elsewhere. Water has a large specific heat capacity compared with metals, which is why a given mass of water warms slowly and stores a lot of energy.
During a change of state, energy is transferred without a change in temperature. The specific latent heat L is the energy needed to change the state of one kilogram, Q = m L. The latent heat of fusion applies to melting and freezing, and the latent heat of vaporization applies to boiling and condensing. On a heating curve of temperature against energy added, sloped sections show the temperature of a single state rising, and flat sections show the energy going into breaking or forming bonds between particles.
Calorimetry problems rest on conservation of energy: the energy lost by the hot object equals the energy gained by the cold one, if no energy escapes. Write m c delta T for each object, with delta T as the magnitude of the temperature change, and set them equal. Real experiments lose energy to the surroundings, so a measured specific heat is usually an overestimate for the sample heated by an electric heater. State that source of systematic error in an evaluation, and propose insulation or a lid as an improvement.
Worked example
How much energy is needed to heat 0.50 kg of water from 20 C to 70 C? Use c = 4200 J kg-1 K-1.
Temperature change = 50 K.
Q = m c delta T.
Q = 0.50 x 4200 x 50.
Q = 105000 J = 1.05 x 10^5 J.
Practice problem and solution
An electric heater supplies 2000 J to melt a block of ice at 0 C. Take the latent heat of fusion of ice as 3.3 x 10^5 J kg-1. What mass of ice melts, in grams, to two significant figures?
m = Q / L = 2000 / 330000 = 0.00606 kg = 6.1 g.
Mental model: Use Q = m c delta T when temperature changes and Q = m L when it does not.
Common trap: Using Q = m c delta T during a change of state.
8. Gases: temperature in kelvin and the gas laws
Learning goal: Apply pressure, volume and temperature relationships to ideal gases with kelvin.
An ideal gas is a model where the particles are point-like, do not interact except in elastic collisions and move randomly. Real gases behave like this at low pressure and high temperature. Pressure arises from particles colliding with the container walls, so it depends on how often and how hard the particles hit. Raising the temperature increases the speed and force of collisions. Increasing the volume reduces the collision frequency per unit area.
For a fixed mass of gas, three relationships follow. At constant temperature, pressure multiplied by volume is constant (Boyle's law). At constant volume, pressure is proportional to absolute temperature. At constant pressure, volume is proportional to absolute temperature. The temperature must be in kelvin, found by adding 273 to the Celsius value for most calculations. Using Celsius gives wrong ratios because zero Celsius is not zero thermal energy.
The combined equation of state is pV = nRT, where n is the amount of substance in moles and R is the molar gas constant, about 8.31 J mol-1 K-1. An equivalent form is pV = N k T with the number of particles N and the Boltzmann constant k. Use pascals for pressure and cubic metres for volume. The conversion 1 cm^3 = 10^-6 m^3 is the most common source of mistakes. Check your volume units before substituting.
Questions that give a gas that changes from state one to state two usually use p1 V1 / T1 = p2 V2 / T2. The ratio form lets units cancel, so any pressure or volume unit is acceptable provided it is the same on both sides, but temperature must still be in kelvin. Interpretation tasks often ask you to explain a change in pressure in terms of particle collisions: state that the particle speed rises, collisions are more frequent and each carries a larger momentum change, so pressure rises.
Worked example
A gas at 300 K has pressure 1.0 x 10^5 Pa. It is heated to 450 K at constant volume. Find the new pressure.
Constant volume, so p1 / T1 = p2 / T2.
p2 = p1 x T2 / T1.
p2 = 1.0 x 10^5 x 450 / 300.
p2 = 1.5 x 10^5 Pa.
Practice problem and solution
An ideal gas of 0.020 mol is at 300 K in a 0.50 L container. Use R = 8.31 J mol-1 K-1. Find the pressure in kPa, to three significant figures.
V = 0.50 x 10^-3 m^3. p = nRT / V = 0.020 x 8.31 x 300 / 5.0 x 10^-4 = 9.97 x 10^4 Pa = 99.7 kPa.
Mental model: Always use kelvin and consistent units.
Common trap: Using degrees Celsius in a gas-law ratio.
9. Current and circuits: reduce before you calculate
Learning goal: Simplify series and parallel resistor networks and apply Ohm's law with care.
Current is the rate of flow of charge, I = q / t, in amperes. Potential difference is the energy transferred per unit charge, V = W / q, in volts. Resistance is R = V / I in ohms. For an ohmic conductor at constant temperature the current is proportional to the potential difference, so a graph of V against I is a straight line through the origin. Filament lamps are not ohmic because their resistance rises as they heat.
In a series circuit the same current passes through every component and the potential differences add up to the source value. The total resistance is the sum, R = R1 + R2 + .... In a parallel circuit, the potential difference is the same across each branch and the currents add up. The total resistance satisfies 1/R = 1/R1 + 1/R2 + ..., so the total is always less than the smallest resistance. Combine parallel groups first, then series.
Power in a circuit element is P = V I, which with Ohm's law gives P = I^2 R and P = V^2 / R. Use whichever form contains the two quantities you know. A real cell has an internal resistance r, so the terminal potential difference is the emf minus I r. The emf is the energy per unit charge supplied by the source. The terminal potential difference drops when the current increases, which is why car headlights dim when the starter motor runs.
Voltmeters are connected in parallel with the component and ideally have very high resistance. Ammeters are connected in series and ideally have very low resistance. A potential divider with two resistors in series gives an output across one resistor of V times R2 / (R1 + R2). When asked how a change affects the circuit, reason in order: decide how the total resistance changes, then the current from the source, then the potential differences across fixed resistors and finally the remainder.
Worked example
A 6.0 ohm and a 3.0 ohm resistor are in parallel across a 12 V source. Find the total current.
1/R = 1/6.0 + 1/3.0 = 0.5.
R = 2.0 ohm.
I = V / R = 12 / 2.0.
I = 6.0 A.
Practice problem and solution
A cell of emf 6.0 V and internal resistance 0.50 ohm drives a current of 2.0 A through an external resistor. Find the terminal potential difference in volts.
V = emf - I r = 6.0 - 2.0 x 0.50 = 5.0 V.
Mental model: Reduce the network first, then use Ohm's law.
Common trap: Adding resistances in parallel instead of using reciprocals.
10. Simple harmonic motion: what sets the period
Learning goal: Link restoring force, period and energy exchange in oscillations.
Simple harmonic motion occurs when the resultant force on an object is proportional to its displacement from equilibrium and directed toward it, F = -kx. The acceleration is therefore a = -omega^2 x, where omega is the angular frequency, omega = 2 pi / T. The minus sign shows that acceleration is opposite to displacement. Acceleration is zero at equilibrium, where speed is greatest, and maximum at the extremes, where the object momentarily stops.
For a mass m on a spring of stiffness k, the period is T = 2 pi times the square root of m / k. For a simple pendulum of length L with small angles, T = 2 pi times the square root of L / g. In both, the period is independent of the amplitude within the small range where the model applies. Doubling the mass on a spring multiplies the period by the square root of two, while doubling the pendulum mass changes nothing.
Displacement, velocity and acceleration graphs are sinusoidal and out of step. If displacement starts at maximum, x = x0 cos(omega t), the velocity is a quarter cycle ahead, and the acceleration is opposite to the displacement. The maximum speed is omega x0 and the maximum acceleration is omega^2 x0. These maximum values are used in many questions. Be sure to match the time origin: starting at equilibrium gives a sine for displacement instead.
Energy in a frictionless oscillator moves between kinetic and potential forms while the total stays constant at k x0^2 / 2. At equilibrium all the energy is kinetic, and at the extremes all of it is potential. With damping, the amplitude decreases because mechanical energy is transferred to thermal energy. Resonance occurs when a driving frequency matches the natural frequency, and the amplitude of the oscillation becomes large. Damping reduces the peak amplitude and broadens the resonance curve.
Worked example
A 0.50 kg mass oscillates on a spring of stiffness 50 N m-1. Find the period.
T = 2 pi sqrt(m / k).
m / k = 0.50 / 50 = 0.010 s^2.
sqrt(0.010) = 0.10 s.
T = 2 pi x 0.10 = 0.63 s.
Practice problem and solution
A pendulum has a period of 2.0 s. Use g = 9.8 m s-2. Find its length in metres, to two significant figures.
L = g T^2 / (4 pi^2) = 9.8 x 4.0 / 39.48 = 0.993 m, about 1.0 m.
Mental model: Restoring force proportional to displacement gives SHM.
Common trap: Using a large amplitude where the formula does not apply.
11. The wave model: v = f lambda and what each symbol means
Learning goal: Use wavelength, frequency, speed and the transverse and longitudinal distinction.
A wave transfers energy without transferring matter. In a transverse wave the particles of the medium oscillate perpendicular to the direction of energy transfer, as in light and waves on a string. In a longitudinal wave they oscillate parallel to it, as in sound, which forms compressions and rarefactions. The wavelength is the distance between successive points in phase, and the period is the time for one complete oscillation at a point.
The wave speed equation is v = f lambda. The frequency is set by the source and does not change when the wave crosses into a new medium. The speed depends on the medium, so the wavelength must change. When a wave enters a medium where it travels more slowly, the wavelength shortens in the same proportion. This is the starting point for refraction questions, and it is worth stating explicitly in answers: frequency fixed, speed changed, so wavelength changed.
Amplitude is the maximum displacement from equilibrium. The energy carried by a wave is proportional to the square of its amplitude, and the intensity, the power per unit area, is also proportional to amplitude squared. For a point source radiating equally in all directions, the power spreads over a sphere of area 4 pi r^2, so the intensity falls with the inverse square of the distance. Doubling the distance reduces the intensity to one quarter.
Graphs are a common source of confusion. A displacement-position graph shows the wave at one instant, from which you read the wavelength. A displacement-time graph shows one point over time, from which you read the period. Do not mix the two when finding the speed. A phase difference of one full wavelength equals 2 pi radians, so a path difference of a quarter wavelength corresponds to a phase difference of pi / 2 radians.
Worked example
A sound wave has a frequency of 440 Hz and a speed of 340 m s-1 in air. Find its wavelength.
v = f lambda.
lambda = v / f.
lambda = 340 / 440.
lambda = 0.77 m.
Practice problem and solution
The intensity of a point source is 8.0 W m-2 at a distance of 2.0 m. What is the intensity, in W m-2, at 4.0 m?
Intensity follows an inverse square law. Doubling the distance gives one quarter of the intensity: 8.0 / 4 = 2.0 W m-2.
Mental model: Frequency is fixed, speed and wavelength change.
Common trap: Treating wavelength as fixed when the medium changes.
Learning goal: Predict how waves bend, spread and combine.
When a wave meets a boundary, it can be reflected, refracted and absorbed. Reflection obeys the law that the angle of incidence equals the angle of reflection, measured from the normal. Refraction occurs when the wave speed changes. Snell's law links the angles: n1 sin(theta1) = n2 sin(theta2), where the refractive index n is the speed of light in a vacuum divided by the speed in the medium. The wave bends toward the normal when entering a medium of higher refractive index.
Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. The effect is most noticeable when the gap is about the same size as the wavelength. A gap much larger than the wavelength produces little spreading. This is why you can hear around a corner, since sound wavelengths are comparable to doorway widths, but cannot see around it, since visible light wavelengths are far smaller than any doorway.
Interference happens when two waves meet and the displacements add, which is the principle of superposition. Constructive interference occurs where the waves arrive in phase, which requires a path difference of a whole number of wavelengths. Destructive interference occurs for a path difference of an odd number of half wavelengths. In the double-slit experiment with slit separation d, a fringe spacing s on a screen at distance D satisfies lambda = s d / D for small angles.
Single-slit diffraction produces a broad central maximum with first minima where the path difference across the slit equals one wavelength, so sin(theta) = lambda / b for slit width b. A narrower slit produces a wider central maximum. A diffraction grating with spacing d gives maxima where d sin(theta) = n lambda, and the maxima are sharper and brighter than in the double slit. Always check whether the question asks about the number of the order or the angle.
Worked example
Light of wavelength 600 nm passes through two slits 0.30 mm apart. A screen is 2.0 m away. Find the fringe spacing.
s = lambda D / d.
lambda = 6.0 x 10^-7 m, d = 3.0 x 10^-4 m, D = 2.0 m.
s = 6.0 x 10^-7 x 2.0 / 3.0 x 10^-4.
s = 4.0 x 10^-3 m = 4.0 mm.
Practice problem and solution
Light of wavelength 500 nm strikes a diffraction grating with 400 lines per millimetre. Find the angle of the first-order maximum, in degrees, to three significant figures.
d = 1 / 400 mm = 2.5 x 10^-6 m. sin(theta) = lambda / d = 5.0 x 10^-7 / 2.5 x 10^-6 = 0.20, theta = 11.5 degrees.
Mental model: Compare wavelength to gap size, and use path difference for interference.
Common trap: Using the number of lines per mm as d.
13. Standing waves, resonance and the Doppler effect
Learning goal: Count nodes and antinodes in standing waves and relate frequency shifts to relative motion.
A standing wave forms when two waves of the same frequency travel in opposite directions and superpose, for example a wave and its reflection. Points of permanent zero displacement are nodes, and points of maximum oscillation are antinodes. Adjacent nodes are half a wavelength apart. Unlike a travelling wave, a standing wave does not transfer net energy along the medium, and all points between two adjacent nodes oscillate in phase.
On a string fixed at both ends there must be a node at each end, so the length is a whole number of half wavelengths, L = n lambda / 2. The fundamental or first harmonic has n = 1 and frequency f1 = v / 2L. Higher harmonics are whole-number multiples of f1. In a pipe open at both ends there are antinodes at both ends, and the harmonics are the same whole-number series. In a pipe closed at one end there is a node at the closed end and an antinode at the open end, so only odd harmonics are present and f1 = v / 4L.
Resonance occurs when a system is driven at its natural frequency and the amplitude of oscillation grows large. Standing waves in tubes and strings are an example. The practical lesson for investigations is that measured values of resonant lengths in tubes need an end correction because the antinode lies slightly beyond the open end. Plotting the resonant length against a suitable quantity and using the gradient, rather than a single point, removes the effect of that constant offset.
The Doppler effect is the change in observed frequency due to relative motion between a source and an observer. For a source moving toward a stationary observer, the wavefronts are compressed and the observed frequency is f' = f v / (v - vs). For a source moving away, the denominator is v + vs. The observed frequency is higher when the source and observer approach and lower when they separate. The effect also applies to light, where it appears as a shift in wavelength used in astronomy.
Worked example
A string of length 0.60 m is fixed at both ends and carries waves of speed 240 m s-1. Find the fundamental frequency.
f1 = v / 2L.
f1 = 240 / (2 x 0.60).
f1 = 240 / 1.2.
f1 = 200 Hz.
Practice problem and solution
An ambulance siren has a frequency of 700 Hz. The ambulance travels at 30 m s-1 toward a stationary observer. The speed of sound is 340 m s-1. Find the observed frequency in Hz, to three significant figures.
f' = f v / (v - vs) = 700 x 340 / (340 - 30) = 238000 / 310 = 768 Hz.
Mental model: Count the nodes and use the boundary condition.
Common trap: Assuming the source emits a different frequency when it moves.
14. Gravitational fields: inverse square and orbits
Learning goal: Apply Newton's law of gravitation and field strength to orbital problems.
Newton's law of universal gravitation says that two point masses attract each other with a force F = G M m / r^2. The gravitational constant G is about 6.67 x 10^-11 N m^2 kg-2. The law is an inverse square law, so doubling the distance reduces the force to one quarter. Spherical masses act as if all their mass is at the centre, so r is measured between the centres, not the surfaces. This is a frequent source of error when a satellite height is given above the surface.
The gravitational field strength g at a point is the force per unit mass, g = F / m = G M / r^2. Near the surface of the Earth it is about 9.8 N kg-1, equal numerically to the free-fall acceleration. The field is a vector pointing toward the mass. At twice the Earth's radius from the centre, the field strength is one quarter of its surface value. Gravitational field lines are closer together where the field is stronger.
For a circular orbit, gravity supplies the centripetal force: G M m / r^2 = m v^2 / r. Solving gives v = sqrt(G M / r), which does not depend on the satellite's mass. The period follows T^2 = 4 pi^2 r^3 / (G M), so T^2 is proportional to r^3, which is Kepler's third law. A satellite in a higher orbit moves more slowly and has a longer period. An astronaut in orbit is not weightless because gravity vanishes; she is in free fall along with the spacecraft.
Gravitational potential energy for two masses is -G M m / r, with zero at infinite separation. The negative sign shows that the masses are bound, and the energy needed to escape from a surface is the positive amount GMm / R. The gravitational potential V = -G M / r is the energy per unit mass. For a circular orbit, the kinetic energy is half the magnitude of the potential energy, so the total energy is negative. Compare orbits by quoting r from the centre of the planet.
Worked example
Mars has mass 6.4 x 10^23 kg and radius 3.4 x 10^6 m. Use G = 6.67 x 10^-11. Find g at its surface.
g = G M / r^2.
GM = 6.67 x 10^-11 x 6.4 x 10^23 = 4.27 x 10^13.
r^2 = 1.156 x 10^13.
g = 3.7 N kg-1.
Practice problem and solution
A satellite orbits Earth at a radius of 7.0 x 10^6 m. Use GM = 3.99 x 10^14 m^3 s-2. Find its orbital speed in m s-1, to three significant figures.
v = sqrt(GM / r) = sqrt(3.99 x 10^14 / 7.0 x 10^6) = sqrt(5.7 x 10^7) = 7550 m s-1.
Mental model: The force falls as 1/r^2 and orbits balance gravity with centripetal force.
Common trap: Using the height above the surface instead of the distance from the centre.
15. Electric and magnetic fields: forces on charges
Learning goal: Compare electric and magnetic forces and find the motion of a charge in each.
A charged particle in an electric field experiences a force F = q E, in the direction of the field for a positive charge and opposite for a negative one. A uniform electric field exists between parallel plates, with E = V / d for potential difference V across separation d. A particle entering such a field perpendicular to it follows a parabolic path, similar to a projectile, because the force and therefore the acceleration are constant in one direction while the speed in the other stays unchanged.
Coulomb's law gives the force between point charges, F = k q1 q2 / r^2, with k approximately 8.99 x 10^9 N m^2 C-2. Like charges repel and unlike charges attract. As with gravity, it is an inverse square law. The field strength around a point charge is E = k Q / r^2. The key contrast with gravity is that electric forces can attract or repel, while gravity only attracts, and the electric force between particles is far larger than their gravitational force.
A magnetic field exerts a force on a moving charge, F = q v B sin(theta), where theta is the angle between the velocity and the field. The force is zero if the charge moves parallel to the field and maximal when it moves perpendicular to it. The direction comes from a hand rule and is perpendicular to both the velocity and the field. Because the force is perpendicular to the velocity, a magnetic force does no work and does not change the speed of the particle, only its direction.
A charge entering a uniform magnetic field perpendicular to the field moves in a circle. The magnetic force supplies the centripetal force, q v B = m v^2 / r, so r = m v / (q B). A faster particle moves on a larger circle, and a larger field gives a smaller circle. A particle with both a velocity component along the field and perpendicular to it moves in a helix. Mass spectrometers use this to separate ions by mass. Always check the sign of the charge when applying the direction rule.
Worked example
An electron (m = 9.11 x 10^-31 kg, q = 1.60 x 10^-19 C) travels at 2.0 x 10^6 m s-1 perpendicular to a 1.0 mT field. Find the radius of its path.
r = m v / (q B).
m v = 9.11 x 10^-31 x 2.0 x 10^6 = 1.822 x 10^-24.
q B = 1.60 x 10^-19 x 1.0 x 10^-3 = 1.60 x 10^-22.
r = 0.0114 m = 1.1 cm.
Practice problem and solution
A proton (q = 1.60 x 10^-19 C) moves at 3.0 x 10^5 m s-1 perpendicular to a magnetic field of 0.50 T. Find the magnitude of the force in newtons, giving the answer in units of 10^-14 N.
F = q v B = 1.60 x 10^-19 x 3.0 x 10^5 x 0.50 = 2.4 x 10^-14 N.
Mental model: Electric force acts on charges, magnetic only on moving ones.
Common trap: Saying a magnetic force can speed up a charge.
16. Radioactive decay: halve, halve, halve
Learning goal: Use half-life, decay equations and activity with the right units.
Radioactive decay is random and spontaneous. For a large number of unstable nuclei it is impossible to predict when one particular nucleus will decay, but the fraction that decays in a given time is predictable. The half-life is the time for half of the nuclei in a sample to decay, or equivalently the time for the activity to fall to half. After n half-lives the fraction remaining is (1/2)^n. Half-life is not affected by temperature, pressure or chemical state.
The common decay modes are alpha, beta minus and gamma. In alpha decay a nucleus emits a helium nucleus, so the mass number falls by 4 and the proton number by 2. In beta minus decay a neutron changes into a proton, an electron and an antineutrino, so the mass number is unchanged and the proton number rises by 1. Gamma emission releases energy from an excited nucleus without changing A or Z. In nuclear equations the top numbers and bottom numbers must each balance.
Activity is the number of decays per second and is measured in becquerels, where one becquerel is one decay per second. The activity is proportional to the number of undecayed nuclei, A = lambda N, where lambda is the decay constant and lambda = ln 2 / T half. Both the number of nuclei and the activity fall exponentially. A graph of the natural logarithm of activity against time is a straight line with a gradient of minus lambda, a technique used to find the half-life from experimental data.
Measuring decay in the laboratory needs a correction for background radiation, which comes from natural and artificial sources and must be subtracted from every reading. Counts are random, so repeated readings vary, and the relative uncertainty reduces with a larger count. The three radiations differ in penetration: alpha is stopped by paper or a few centimetres of air, beta by a few millimetres of aluminium and gamma is reduced but never fully stopped by thick lead. Take care to link the type of radiation to the right shielding in safety answers.
Worked example
A sample has an activity of 800 Bq. Its half-life is 6.0 hours. Find the activity after 18 hours.
18 hours is 3 half-lives.
Activity after 3 half-lives = 800 x (1/2)^3.
800 / 8 = 100.
Activity = 100 Bq.
Practice problem and solution
A radioactive sample has a half-life of 8.0 days. What fraction of the original nuclei remains after 24 days? Give a decimal.
24 days is 3 half-lives, so the fraction is (1/2)^3 = 1/8 = 0.125.
Mental model: Decay is random for a nucleus but regular for a large sample.
Common trap: Subtracting half-lives from the total instead of halving the sample.
17. Graphs, uncertainty and the scientific investigation
Learning goal: Linearise data, find uncertainties from gradients and structure the investigation report.
The scientific investigation is the internal assessment in IB Physics. The IB subject brief describes it as an open-ended task in which the student gathers and analyses data to answer their own research question, written up as a report with a maximum of 3,000 words. The brief's assessment table gives the internal assessment a weighting of 20 percent of the final grade, against 80 percent for the external papers. The best questions have a measurable independent and dependent variable and a clear relationship that can be tested.
Most physics relationships can be turned into straight lines. If theory says y is proportional to x squared, plot y against x squared. If it says y is proportional to the square root of x, plot y against sqrt(x). If the relationship has the form y = a x^n, plot ln y against ln x and read n from the gradient. A straight line through the origin supports proportionality; a straight line with an intercept shows a linear relationship with a constant offset. Choosing the axes before collecting data keeps the investigation focused.
Every measurement has an uncertainty. A reading error is often half the smallest division for an analogue scale, or the smallest division for a digital one, unless the instrument states otherwise. Random uncertainty appears as scatter in repeated readings and can be reduced by repeating and averaging. Systematic errors, such as a zero error or a stopwatch reaction time bias, shift all readings the same way and do not average out. Name the likeliest source of each type in your evaluation.
Add uncertainties properly. For sums and differences add absolute uncertainties. For products and quotients add percentage uncertainties, and for a power add the percentage uncertainty multiplied by the power. Error bars show uncertainty on a graph. The best-fit line and the maximum and minimum gradient lines, drawn through the error bars, give an uncertainty in the gradient. In the conclusion, compare your result with a literature value using the uncertainty, and suggest specific improvements that would reduce the largest uncertainty.
Worked example
A length is measured as 50.0 cm with an uncertainty of 0.2 cm, and a time as 1.00 s with an uncertainty of 0.02 s. The speed is length divided by time. Find the percentage uncertainty in the speed.
Percentage uncertainty in time = 0.02 / 1.00 = 2 percent.
For a quotient add percentages: 0.4 + 2.
Percentage uncertainty = 2.4 percent.
Practice problem and solution
A quantity is calculated as the square of a length, and the length has a percentage uncertainty of 3 percent. Find the percentage uncertainty in the squared quantity.
For a power the percentage uncertainty is multiplied by the power: 3 x 2 = 6 percent.
Mental model: Choose the axes that make the theory a straight line.
Common trap: Averaging repeats and claiming that a systematic error has gone.