A common framework for any core practical: aim, apparatus, method, graph, uncertainties, safety, improvements

CP1 acceleration of free fall

CP2 viscosity by the falling-ball method

CP3 the Young modulus of a wire

CP4 speed of sound with a two-beam oscilloscope

CP5 the vibrating string (length, tension, mass per unit length)

CP6 wavelength of light with a diffraction grating

CP7 resistivity of a wire and CP8 e.m.f. and internal resistance

1. Every AS core practical turns a physics equation into a straight-line graph, and the quantity you want is in the gradient (or the intercept).

2. CP1 h against t² (gradient g/2) · CP3 stress against strain (gradient E) · CP5 f against 1/l · CP6 sin θ against n · CP7 R against l (gradient ρ/A) · CP8 V against I (intercept ε, gradient −r).

3. CP2 uses the force balance at terminal velocity, W = U + F; CP4 finds λ by moving a microphone until two traces are back in phase.

4. The smallest measured quantity (a wire diameter, an extension, a ball radius) usually carries the largest percentage uncertainty.

5. A safety answer names the hazard and the precaution: goggles because a loaded wire may snap; never look along a laser beam.

The practical paper rarely asks you to repeat a method word for word. It changes the object, the liquid, the wire or the source and expects you to adapt. So learn each practical as a pattern under the same seven headings:

Aim: the quantity to be found, in one line.

Apparatus: each instrument with its resolution (metre rule 1 mm, vernier calipers 0.1 mm, micrometer 0.01 mm).

Method: what is varied, what is measured, what is kept constant, how many readings and repeats.

Graph: the axes that give a straight line, and what the gradient or intercept means.

Uncertainties: which measurement limits the result, and whether the error is random or systematic.

Safety: a specific hazard and a specific precaution.

Improvements: a change that reduces the main uncertainty, with the reason.

The AS uncertainty rules (page 3.2) apply throughout: a single reading has an uncertainty of half the resolution, repeats have half the range, and percentage uncertainty = uncertainty ÷ measured value × 100%. At AS you find the percentage uncertainty of a single quantity and say which measurement dominates; combining uncertainties belongs to Unit 6.

Aim: determine g by timing a body falling from rest.

Apparatus: small steel ball, electromagnet, trapdoor switch (or a light gate), electronic timer, metre rule, set square.

Method: switching off the magnet releases the ball and starts the timer; the ball opens the trapdoor and stops it. Measure h from the bottom of the ball to the trapdoor, using the set square to keep the rule vertical. Use five or six heights over the widest range possible (for example 0.20 m to 1.20 m) and repeat each three times.

Graph: from rest, h = ½ g t², so h against t² is a straight line through the origin with gradient g/2. A gradient of 4.87 m s⁻² gives g = 2 × 4.87 = 9.74 m s⁻², 0.71% below 9.81 m s⁻².

Variant: a light gate measures the speed v = (ball diameter) ÷ (time the beam is blocked) after a fall h; then v² = 2gh and the gradient of v² against h is 2g.

Uncertainties: the magnet holds the ball for a moment after switch-off (residual magnetism), making every time slightly too long, a systematic error; reading h to ±0.5 mm and parallax at the scale.

Safety: catch the falling steel ball in a box of sand or cloth; the magnet coil can get warm, so switch it off between drops.

Improvements: use the smallest magnet current that holds the ball; use larger heights so the times are longer and their percentage uncertainty smaller; replace the trapdoor by two light gates.

Aim: find the viscosity η of a liquid such as glycerol from the terminal velocity of a small sphere.

Apparatus: tall, wide tube of the liquid, small steel balls, micrometer, three rubber bands round the tube, metre rule, stopwatch, thermometer, balance and measuring cylinder (for the densities), a magnet on a string to retrieve the balls.

Method: measure each ball's diameter with the micrometer in several directions. Place the first band well below the surface so the ball has reached terminal velocity before timing starts. Release the ball gently on the axis of the tube and time it from A to B and from B to C: equal times confirm a constant (terminal) velocity. Repeat, and record the temperature.

Analysis: at terminal velocity the forces balance, W = U + F, with F = 6πηrv (Stokes' law).

Graph: with balls of several sizes, v against r² is a straight line through the origin of gradient 2g(ρₛ - ρ_f)/9η.

Uncertainties: starting and stopping the stopwatch by eye (reaction time, parallax at the bands); the small radius, which is squared; the temperature, because viscosity falls steeply as the liquid warms.

Safety: wipe up spills at once, as glycerol and oils make the floor slippery; clamp the tall tube so it cannot topple.

Improvements: view the bands at eye level, or film the fall next to a scale and step through the frames; use a longer distance between bands; use a wide tube so the walls do not slow the ball; keep the liquid in a water bath at a known temperature.

Writing the weight and upthrust in terms of the ball's volume and the two densities, then solving for η:

F = 6πηrv is in the formulae list; the expression for η is not, so derive it. Stokes' law holds only for a small sphere moving slowly through the liquid in laminar flow.

WORKED EXAMPLE · QUESTION

A steel ball of diameter 2.00 mm (density 7800 kg m⁻³) falls through glycerol (density 1260 kg m⁻³). Its times to fall the 0.200 m between bands A and B are 19.4 s, 19.9 s and 19.6 s. (a) Calculate the mean time and its uncertainty. (b) Calculate the terminal velocity. (c) Determine the viscosity of the glycerol.

SOLUTION

(a) Mean = 19.63 s; half the range = (19.9 − 19.4) ÷ 2 = 0.25 s. t = 19.6 s ± 0.25 s (1.3%)

(b) Constant speed between the bands: v = (0.200/19.63) = 1.02 ×10⁻² m s⁻¹

(c) With r = 1.00 × 10⁻³ m: η= (2 ×(1.00×10⁻³)² ×9.81 ×(7800 - 1260))/(9 ×0.01019) = 1.40 Pa s

The micrometer reads the diameter to ±0.005 mm, only 0.25% here, so the timing (1.3%) dominates.

Aim: determine the Young modulus E of a metal in the form of a long wire.

Apparatus: a wire 2–3 m long clamped at one end and passing over a pulley at the edge of the bench; mass hanger and slotted masses; a small flag of tape on the wire next to a fixed metre rule (or a vernier scale); micrometer; tape measure; safety goggles.

Method: hang a small load first to straighten the wire and take its reading as zero. Measure the original length l from the clamp to the flag. Measure the diameter at several points and in two directions at right angles; average. Add equal loads, recording the flag position each time; then remove them to check the wire returns (it stayed elastic).

Graph: stress = F/A against strain = Δx/l; the gradient of the straight section is E. Points that curve away at the top show the limit of proportionality has been passed, so leave them out of the gradient.

Uncertainties: the extensions are only a few millimetres, so their percentage uncertainty is large; the diameter of about 0.4 mm read to 0.01 mm is also a few per cent; temperature changes and the wire settling in the clamp.

Safety: goggles, because a loaded wire may snap and whip back; a box of sand or foam under the masses; keep feet clear.

Improvements: a longer wire gives a larger extension for the same strain; a vernier scale reads extensions to 0.1 mm; a second, unloaded reference wire hung alongside (Searle's arrangement) cancels sag and temperature effects.

WORKED EXAMPLE · QUESTION

The wire in the graph above has six diameter readings between 0.37 mm and 0.39 mm, mean 0.38 mm. Its largest load, 40.0 N, gives an extension of 4.2 mm. (a) Calculate the cross-sectional area. (b) Calculate the stress and strain for the 40.0 N load. (c) Use the graph to determine E. (d) Calculate the percentage uncertainty in the diameter.

SOLUTION

(a) A = πd²/4 = (π×(0.38×10⁻³)²)/4 = 1.13 ×10⁻⁷ m²

(b) Stress = 40.0/(1.134×10⁻⁷) = 3.53×10⁸ Pa; strain = (4.2×10⁻³)/2.400 = 1.75×10⁻³. 3.53 × 10⁸ Pa and 1.75 × 10⁻³

(c) Triangle read from the line: E = ((3.42 - 0.39)×10⁸)/((1.70 - 0.20)×10⁻³) = 2.02 ×10¹¹ Pa

(d) Half the range of the diameters is ±0.01 mm: (0.01/0.38)×100% = 2.6%

Compare the extension: ±0.05 mm on 4.2 mm is only 1.2%.

With a vernier scale, the diameter becomes the limiting measurement.

Aim: measure the speed of sound in air using v = fλ.

Apparatus: signal generator, loudspeaker, two microphones, two-beam (dual-trace) oscilloscope, metre rule.

Method: drive the loudspeaker at a few kilohertz. Connect one microphone to channel 1 and the other to channel 2, both in line with the speaker. Start with the microphones side by side so the traces are in phase. Slide microphone 2 away along the rule: the traces drift out of phase and come back into phase each time the separation grows by one wavelength. Record the position after each of n = 1, 2, 3 … wavelengths.

Graph: distance moved against n is a straight line of gradient λ. Read the period from the screen (divisions × timebase) to find f, then v = fλ. Alternatively, change f and plot λ against 1/f: the gradient is v.

Uncertainties: judging exactly where the traces are in phase; sound reflected from the bench and walls distorting the traces; the frequency dial of the generator.

Safety: keep the sound level low and the session short to protect hearing.

Improvements: move through many wavelengths and divide; measure f from the oscilloscope rather than the dial; use the X–Y mode, where in-phase signals give a sloping straight line that is easier to judge; work away from hard surfaces.

WORKED EXAMPLE · QUESTION

The screen above uses a timebase of 0.1 ms per division. Starting from the in-phase position, microphone 2 is moved 1.10 m, during which the traces come back into phase 8 times. (a) Determine the frequency. (b) Calculate the speed of sound. (c) Check your answer using the lag shown on the screen.

SOLUTION

(a) One period covers 4 divisions: T = 4 × 0.1 ms = 0.40 ms, so f = 1/(0.40×10⁻³) = 2500 Hz

(b) 8 wavelengths in 1.10 m: λ = 1.10 ÷ 8 = 0.1375 m, so v = 2500 × 0.1375 = 344 m s⁻¹

(c) A lag of 1 division (0.1 ms) is a quarter of a period, so the extra path is λ/4 = 0.0344 m: v = 0.0344/(1.0×10⁻⁴) = 344 m s⁻¹; the two methods agree

Aim: investigate how the frequency of the first harmonic depends on the length l, tension T and mass per unit length μ of a string.

Apparatus: string or wire passing over a pulley to a mass hanger (T = mg), two bridges, a vibration generator driven by a signal generator, metre rule, balance.

Method: fix the vibrating length between the bridges. Raise the frequency slowly until the string vibrates in one loop with the largest amplitude (the first harmonic) and read f. Change one variable at a time: move a bridge to change l; add masses to change T; change the string to change μ, finding each μ by weighing a long measured length.

Graph: for the first harmonic λ = 2l, so f = 1/2l√(T/μ). Plot f against 1/l (gradient ½√(T/μ)), f² against T (gradient 1/4l²μ), or f against 1/√μ.

Uncertainties: deciding where the amplitude is largest, because resonance covers a range of frequencies; measuring l between the contact points of the bridges; the calibration of the generator.

Safety: the masses fall if the string breaks, so keep feet clear and use a box underneath; wear goggles with a wire under high tension.

Improvements: approach resonance from above and from below and average the two readings; check the frequency on an oscilloscope; use a wider range of lengths.

WORKED EXAMPLE · QUESTION

The tension in the string above is provided by a 0.500 kg mass. Use the graph to determine the speed of the waves on the string and its mass per unit length.

SOLUTION

Gradient from the triangle: (96 - 16)/(3.0 - 0.5) = 32.0 m s⁻¹ (Hz ÷ m⁻¹ = m s⁻¹).

f = v/2l, so gradient = v/2 and v = 2 × 32.0 = 64.0 m s⁻¹

v = √(T/μ) with T = 0.500 × 9.81 = 4.91 N, so μ= T/v² = 4.905/64.0² = 1.20 ×10⁻³ kg m⁻¹

Aim: determine the wavelength of laser light using a grating with a known number of lines per millimetre.

Apparatus: laser, diffraction grating, screen, metre rule and tape measure.

Method: set the grating at right angles to the beam and the screen parallel to the grating, a distance D of 1–2 m away. Measure the distance between the two first-order spots (2x₁), then between the two second-order spots, and so on. Halve each distance; tan θ = x/D; then nλ = d sin θ with d = 1/(lines per metre).

Graph: sin θ against n is a straight line through the origin with gradient λ/d.

Uncertainties: locating the centre of each spot; measuring D; the grating not being exactly perpendicular to the beam.

Safety: never look along the beam or at its reflections; remove watches and rings and keep shiny surfaces out of the beam path; use a low-power laser and a warning sign.

Improvements: measure across opposite orders rather than from the centre; use a larger D so the spots are further apart; use higher orders; darken the room so the spot centres are easier to see.

WORKED EXAMPLE · QUESTION

A grating with 300 lines per mm is 1.80 m from a screen. The distance between the two first-order spots is 0.582 m; between the two second-order spots it is 1.213 m. (a) Calculate the grating spacing. (b) Determine the wavelength using each order and find the mean. (c) Calculate the highest order that can be seen.

SOLUTION

(a) d = 1/(300×10³ m⁻¹) = 3.33 ×10⁻⁶ m

(b) First order: θ= tan⁻¹(0.291/1.80) = 9.18°, λ= 3.333×10⁻⁶ ×sin 9.18° = 532 nm. Second order: θ= tan⁻¹(0.6065/1.80) = 18.6°, λ= 3.333×10⁻⁶ ×sin 18.6° ÷2 = 532 nm. Mean λ = 532 nm

(c) sin θ cannot exceed 1, so n ≤ d/λ = 6.26: highest order 6

Aim: determine the resistivity ρ of the metal of a wire.

Apparatus: about 1 m of thin wire taped along a metre rule, a sliding clip (crocodile clip or jockey), cell, switch, ammeter, voltmeter, micrometer.

Method: measure the diameter at several points and in two directions; average and find A. Connect the wire in series with the cell, switch and ammeter, with the voltmeter across the length l between the fixed end and the clip. For lengths from about 0.20 m to 0.90 m, close the switch briefly, read V and I, open it again; R = V/I.

Graph: R = ρl/A, so R against l is a straight line of gradient ρ/A; ρ = gradient × A.

Uncertainties: the diameter, which is squared in A; contact resistance at the clip and the leads (a systematic error that gives an intercept); the wire warming, which raises its resistance.

Safety: a thin wire carrying a large current gets hot, so keep the current small and switch off between readings.

Improvements: use a small p.d. so the temperature stays constant; clean the wire where the clip touches it; use the gradient, not a single reading, so a constant contact resistance does not matter.

WORKED EXAMPLE · QUESTION

Five micrometer readings of the wire's diameter are 0.31, 0.32, 0.31, 0.32 and 0.31 mm. The best-fit line on the graph above passes through (0.100 m, 0.73 Ω) and (0.900 m, 5.73 Ω). (a) Calculate the cross-sectional area. (b) Determine the resistivity. (c) Explain why the intercept does not affect your answer.

SOLUTION

(a) Mean diameter 0.314 mm: A = (π×(0.314×10⁻³)²)/4 = 7.74 ×10⁻⁸ m²

(b) Gradient = (5.73 - 0.73)/(0.900 - 0.100) = 6.25 Ω m⁻¹, so ρ= 6.25 ×7.74×10⁻⁸ = 4.84 ×10⁻⁷ Ω m

(c) Contact resistance adds the same amount to every R. It shifts the line up but leaves the gradient unchanged.

Aim: determine the e.m.f. ε and internal resistance r of a cell.

Apparatus: the cell, variable resistor, ammeter, voltmeter, switch (a fixed protective resistor in series is useful).

Method: connect the cell in series with the variable resistor, ammeter and switch, with the voltmeter directly across the cell terminals. Close the switch, read V and I, open it; repeat for at least six settings covering the widest range of current.

Graph: ε = I(R + r) gives V = ε − Ir, so V against I is a straight line: intercept on the V-axis = ε, gradient = −r.

Uncertainties: the cell warms, so its internal resistance and e.m.f. drift while current flows; the resolution of the meters at small currents.

Safety: never short-circuit the cell; avoid rechargeable cells with very small r, which can deliver large currents and become hot.

Improvements: switch off between readings so the cell does not heat; take readings quickly; repeat with the readings in reverse order and average.

WORKED EXAMPLE · QUESTION

The best-fit line on the graph above meets the V-axis at 1.55 V and passes through (1.40 A, 0.68 V). (a) State the e.m.f. (b) Determine the internal resistance. (c) The variable resistor is set to 2.0 Ω. Calculate the current and the terminal p.d.

SOLUTION

(a) At I = 0 there are no lost volts: ε = 1.55 V

(b) Gradient = (0.68 - 1.55)/1.40 = -0.621 Ω, and r = −gradient: r = 0.62 Ω

(c) ε = I(R + r): I = 1.55/(2.0 + 0.621) = 0.59 A, and V = IR = 0.591 × 2.0: I = 0.59 A, V = 1.18 V

The largest possible current, with R = 0, would be ε/r ≈ 2.5 A: the reason never to short-circuit the cell.

The theory pages in Units 1 and 2

Each practical belongs to a topic page: free fall (1.1), viscosity (1.7), the Young modulus (1.9), progressive and stationary waves (2.1, 2.2), diffraction gratings (2.4), resistivity (2.8) and e.m.f. (2.9).

1 Planning and 3.2 Uncertainties and graphs

Choosing the micrometer for a wire diameter or a set square for a vertical rule is the same skill in every practical here; gradients from large triangles, half-range uncertainties and percentage uncertainty turn the results into numbers.

The A2 core practicals (Unit 6)

The same seven-heading approach carries on to Core Practicals 9–16, where uncertainties are combined and logarithmic graphs are used.

1. Why is a graph of h against t² used in CP1, and how is g found from it?

2. In CP2, how can you show that the ball has reached terminal velocity before it passes the first timing band?

3. Why is the wire in CP3 long and thin?

4. A grating with 600 lines per mm gives a first-order maximum at 22.3°. Calculate the wavelength.

5. A V–I graph for a battery meets the V-axis at 4.5 V and has a gradient of −1.5 V A⁻¹. State the e.m.f. and the internal resistance.

1. h = ½ g t² for a fall from rest, so h is proportional to t²: a straight line through the origin with gradient g/2. g = 2 × gradient.

2. Time it over two equal distances one after the other (bands A–B and B–C). If the times are equal within their uncertainty, the velocity is constant.

3. A long wire gives a larger extension for the same strain, and a thin wire gives a larger stress for the same load. Both make the extension bigger, so its percentage uncertainty is smaller.

4. d = 1/(600×10³) = 1.67×10⁻⁶ m; λ= dsin 22.3° = 632 nm.

5. ε = 4.5 V (the intercept); r = 1.5 Ω (minus the gradient).