How to use this guide
This guide is for students who know the Physics 2 content but lose marks when asked to draw, derive, translate or design. It covers the whole course. Try each worked example before reading the strong answer.
Every example and task here is original Marlbridge practice, not past exam material. The mark points are a suggested Marlbridge rubric, not official scoring. All data are invented.
The free-response section at a glance
Section II has 4 questions in 95 minutes and is worth 50% of the exam. Each question type below appears once. You read the questions on screen and write your answers by hand in a paper booklet. A four-function, scientific or graphing calculator is allowed.
| Question type | Points | Suggested time | Main demands |
|---|---|---|---|
| Mathematical Routines | 10 | 20–25 min | Derive, calculate, use a representation, justify a claim |
| Translation Between Representations | 12 | 25–30 min | Diagram, equations and graph, then show they agree or predict |
| Experimental Design and Analysis | 10 | 25–30 min | Design a procedure, then plot given data and use the graph |
| Qualitative/Quantitative Translation | 8 | 15–20 min | Justify a claim, derive an equation, link the two |
Representations that earn credit
A marker can only credit what is on the page.
| Representation | Features to show | Common way to lose the mark |
|---|---|---|
| Field lines (Unit 10) | Start on + and end on −; arrows on every line; line count proportional to charge; no crossings | Equal line counts for unequal charges |
| Equipotentials (Unit 10) | Perpendicular to field lines wherever they cross; closer together where the field is stronger | Lines meeting field lines at an angle |
| Circuit schematic (Unit 11) | Standard symbols; ammeter in series, voltmeter in parallel; labels that match the question | Voltmeter in series; unlabelled resistors |
| Ray diagram (Unit 13) | Two or more principal rays from the object’s top; arrows; extended (virtual) rays dashed | A virtual image where no rays meet |
| PV diagram (Unit 9) | Labelled axes; each state labelled; an arrow on every process; isobars horizontal, isochores vertical | No arrows, so the sign of W is unclear |
| Graph sketch (any unit) | Labelled axes; correct shape, intercepts and asymptotes | A straight line where the model predicts a curve |
Use a ruler, and cross out any abandoned version clearly.
Deriving and checking expressions
A derivation starts from a fundamental law or definition (Newton’s second law, energy conservation, Kirchhoff’s rules, the first law of thermodynamics, Coulomb’s law) and moves step by step to the requested quantity.
- Name the law and write its general form.
- Apply it to this situation: put in the specific forces, energies, currents or potentials.
- Solve with symbols. Numbers too early hide mistakes.
- Use only the symbols the question gives, plus physical constants.
- Check the result in at least two ways.
| Check | How | What it catches |
|---|---|---|
| Units | Replace each symbol by its unit and simplify | A missing factor of length, time or charge |
| Limiting case | Let one quantity go to 0 or to infinity; does the answer make sense? | Wrong dependence, a term in the wrong place |
| Special case | Set two quantities equal; compare with a known result | Asymmetric errors |
Write the check down: it shows you understand the expression and can support a later justification.
Translating between words, equations, graphs and diagrams
The most common translation is between an equation and the shape of a graph.
| Words | Equation | Graph of y against x |
|---|---|---|
| “y is proportional to x” | y = kx | Straight line through the origin |
| “y increases linearly with x” | y = kx + c | Straight line, intercept c |
| “y is inversely proportional to x” | y = k/x | Falling curve that approaches the x-axis |
| “inverse square law” | y = k/x² | Falls to a quarter when x doubles |
| “y approaches a final value” | — | Changes quickly, then levels off |
Describing a graph in words, give its shape, trend and special points (intercepts, asymptotes). Sketching from words, make those features visible.
Qualitative-quantitative translation connects a claim in words to an equation. Make the claim, derive an expression, then point to the part that supports the claim: “The fringe spacing increases when the slits move closer, because Δy = λL/d and d is in the denominator.” When asked why two answers agree, name the shared physics, not just “they match”.
Designing an experiment and linearising data
A sound procedure states:
- Independent variable: the one quantity you change, over at least five values.
- Dependent variable: the one quantity you measure, and the instrument.
- Controls: what stays the same, and how.
- Equipment: ordinary school laboratory items.
- Repeats: repeat and average to reduce random error.
- Analysis: what you will plot, and what the gradient or intercept tells you.
Rearrange the model into y = mx + c and choose what to plot.
| Model | Plot (y against x) | Gradient | Intercept |
|---|---|---|---|
| B = μ₀I/(2πr) (Unit 12) | B against 1/r | μ₀I/(2π) | 0 |
| ΔV = ε − Ir (Unit 11, internal resistance) | ΔV against I | −r | ε |
| f₁ = v/(2L) (Unit 14) | f₁ against 1/L | v/2 | 0 |
| K_max = hf − φ (Unit 15) | K_max against f | h | −φ |
| 1/sᵢ = 1/f − 1/sₒ (Unit 13) | 1/sᵢ against 1/sₒ | −1 | 1/f |
Take the gradient from two far-apart points on your best-fit line, with units.
Worked example 1: PV diagram and derivation (Unit 9)
Question. An ideal monatomic gas starts in state A (P₀, V₀) and reaches state C (2P₀, 2V₀) by one of two paths. Path ABC: pressure rises at constant volume, then the gas expands at constant pressure. Path ADC: the gas expands at constant pressure, then the pressure rises at constant volume. (a) Draw both paths on one PV diagram. (b) Derive the energy transferred to the gas by heating on each path, in terms of P₀ and V₀. (c) Justify which path needs more heating, using your diagram.
Weak answer. “(a) [Two lines from A to C, no arrows, no state labels.] (b) Q = ΔU = (3/2)nRΔT for both. (c) The same, because ΔU is the same.”
What it misses. No arrows or labels, so no direction. Q = ΔU only when W = 0. The claim is wrong and the diagram is unused.
Strong answer.
(b) First law: ΔU = Q + W, with W the work done on the gas. For a monatomic ideal gas U = (3/2)PV, so for both paths ΔU = (3/2)[(2P₀)(2V₀) − P₀V₀] = (9/2)P₀V₀.
On ABC the only work is during the expansion at 2P₀: W = −(2P₀)(V₀) = −2P₀V₀. So Q_ABC = ΔU − W = (13/2)P₀V₀.
On ADC the expansion is at P₀: W = −P₀V₀, so Q_ADC = (11/2)P₀V₀.
Checks: P₀V₀ has units Pa·m³ = J. If the gas did not expand, W = 0 and Q = ΔU, as expected.
(c) Path ABC. Both paths have the same ΔU because they share the same end states. On ABC the gas expands at higher pressure, so the area under the path, and the work done by the gas, is larger. Heating must supply that extra P₀V₀ (the rectangle between the paths). This agrees with (b): Q_ABC − Q_ADC = P₀V₀.
| Suggested Marlbridge rubric | Point |
|---|---|
| Both paths with correct shapes, arrows and labelled states | 1 |
| ΔU = (9/2)P₀V₀, the same for both paths | 1 |
| Correct W, with sign, for each path | 1 |
| Q_ABC = (13/2)P₀V₀ and Q_ADC = (11/2)P₀V₀ | 1 |
| Claim ABC, justified by equal ΔU and larger area under ABC | 1 |
Worked example 2: field map and limiting case (Unit 10)
Question. A point charge +4q is fixed at x = 0 and −q at x = d. (a) Describe the field-line diagram, including how many lines you would draw. (b) Derive the position on the x-axis where the electric field is zero. (c) Check your answer with a limiting case.
Weak answer. “(a) Lines go from + to −, 8 on each. (b) k(4q)/x² = kq/(d − x)², so x = 2d/3.”
What it misses. Equal line counts imply equal charges. Between the charges both fields point towards −q, so they cannot cancel. No check catches this.
Strong answer.
(a) Draw four times as many lines on +4q, for example 16 leaving +4q and 4 ending on −q; the other 12 go out to large distances. Every line has an arrow, and no lines cross.
(b) Between the charges both fields point in the +x direction. Left of +4q, the larger, closer charge always wins. So the zero is right of −q (x > d), where the fields point in opposite directions. Coulomb’s law for each charge:
k(4q)/x² = kq/(x − d)²
Both sides are positive, so take square roots: 2(x − d) = x, giving x = 2d.
(c) For charges +nq and −q the same steps give x = √n·d/(√n − 1), which is 2d for n = 4. If n → 1 (equal and opposite charges), x → ∞: a dipole has no zero point at finite distance, as expected. The answer is a multiple of d, so it is a length.
| Suggested Marlbridge rubric | Point |
|---|---|
| Line counts in the ratio 4 : 1, with arrows and no crossings | 1 |
| Region x > d chosen, with a reason about field directions | 1 |
| Correct equation and x = 2d | 1 |
| A valid limiting case or unit check, stated in words | 1 |
Worked example 3: designing and linearising (Unit 12)
Question. A long straight wire carries a steady current I, but no ammeter is available. (a) Describe a procedure, using a magnetic field sensor and a ruler, to find I from a graph. (b) A student records these fictional data. Choose quantities that give a straight line and use the graph to find I.
| r (m) | 0.020 | 0.030 | 0.040 | 0.060 | 0.080 | 0.100 |
|---|---|---|---|---|---|---|
| B (μT) | 61 | 39 | 30.5 | 19.6 | 15.2 | 11.8 |
Weak answer. “(a) Measure B at different distances. (b) Plot B against r. It curves down, so B decreases with r. Using r = 0.040 m: I = 2πrB/μ₀ = 6.1 A.”
What it misses. No range, no control of orientation, no removal of Earth’s field. A curve has no useful gradient, and one point wastes the other five.
Strong answer.
(a) Point the sensor along the tangent to a circle around the wire. With the current off, zero the sensor to remove Earth’s field. With it on, measure r from the wire’s centre with a ruler for at least five values from 2 cm to 10 cm, averaging three readings of B at each. Keep the current, sensor orientation and position along the wire fixed. Plot B against 1/r.
(b) B = (μ₀I/2π)(1/r), so B against 1/r should be a straight line through the origin with gradient μ₀I/(2π). The values of 1/r are 50.0, 33.3, 25.0, 16.7, 12.5 and 10.0 m⁻¹, and the points lie close to a line. Two points on the best-fit line are (10.0 m⁻¹, 11.8 μT) and (50.0 m⁻¹, 60.5 μT).
Gradient = (60.5 − 11.8) μT ÷ 40.0 m⁻¹ = 1.22 μT·m = 1.22 × 10⁻⁶ T·m.
I = 2π × gradient ÷ μ₀ = 2π(1.22 × 10⁻⁶ T·m) ÷ (4π × 10⁻⁷ T·m/A) ≈ 6.1 A.
| Suggested Marlbridge rubric | Point |
|---|---|
| r changed, B measured, instruments named | 1 |
| Controls stated, including removing Earth’s field | 1 |
| Five or more values with repeats; plan to plot B against 1/r | 1 |
| Labelled axes with units, sensible scales, points plotted correctly | 1 |
| Best-fit straight line drawn | 1 |
| Gradient from points on the line; I ≈ 6.1 A with unit | 1 |
Common mistakes
- Drawing equal numbers of field lines for charges of unequal size.
- Drawing equipotentials that are not perpendicular to field lines.
- Leaving arrows off PV paths, rays or field lines.
- Drawing extended (virtual) rays as solid lines, or an image where no rays meet.
- Putting numbers into the first line of a derivation, then being unable to find the error.
- Saying “the answers agree” without naming the physics that makes them agree.
- Plotting raw variables that give a curve, then trying to read a gradient.
- Taking the gradient from two data points instead of two points on the best-fit line.
Practise it
Task 1 (Unit 11). A battery of emf ε (no internal resistance) is in series with a resistor R. After R, the circuit splits into two parallel branches, R and R_x. A student claims the current in R_x is I_x = εR_x/(R + 2R_x). Use units and a limiting case to show this is wrong, then derive the correct expression.
Model answer
Units: εR_x/(R + 2R_x) has units V·Ω/Ω = V, a potential difference, not a current. Limiting case: as R_x → 0, R_x shorts its branch, so it should carry ε/R; the claim gives 0.
Derivation: R_par = RR_x/(R + R_x), the total current is ε/(R + R_par), and the potential difference across the pair is ΔV = εR_x/(R + 2R_x). The student found ΔV. So I_x = ΔV/R_x = ε/(R + 2R_x). Checks: R_x → 0 gives ε/R; R_x → ∞ gives 0; units V/Ω = A.
Task 2 (Unit 13). An object 2.0 cm tall stands 8.0 cm from a converging lens of focal length 12 cm. Describe the ray diagram, then use the thin-lens equation to find the image distance and height. Show that the two methods agree.
Model answer
Ray diagram: a ray parallel to the axis refracts through the far focal point; a ray through the centre goes straight on. They spread out after the lens, so extend them backwards as dashed lines. They meet on the object’s side, further away: a virtual, upright, enlarged image.
Equation: 1/sᵢ = 1/12 − 1/8.0 = −1/24, so sᵢ = −24 cm. |M| = |sᵢ/sₒ| = 3.0, so the image is 6.0 cm tall. Negative sᵢ means a virtual image on the object’s side, matching the dashed extended rays.
Task 3 (Unit 14). A string is kept at constant tension. Its vibrating length L is changed and the fundamental frequency f₁ measured (fictional data): L = 0.40, 0.50, 0.60, 0.80, 1.00 m gives f₁ = 121, 95, 81, 59, 48 Hz. Choose a linear plot and find the wave speed.
Model answer
Since f₁ = v/(2L), plot f₁ against 1/L: a line through the origin with gradient v/2. The values of 1/L are 2.50, 2.00, 1.67, 1.25 and 1.00 m⁻¹. The best-fit gradient is about 48.6 Hz·m, so v = 2 × 48.6 ≈ 97 m/s. The intercept is close to zero, as the model predicts.
Task 4 (Unit 15). Light falls on a metal with threshold frequency 5.5 × 10¹⁴ Hz. (a) Sketch K_max against f and state what the gradient and horizontal intercept represent. (b) Find the work function in eV. (c) Describe the line for a metal with a larger work function.
Model answer
(a) K_max = hf − φ: zero up to f = 5.5 × 10¹⁴ Hz, then a straight line rising with gradient h. The horizontal intercept is the threshold frequency.
(b) φ = hf₀ = (4.14 × 10⁻¹⁵ eV·s)(5.5 × 10¹⁴ Hz) ≈ 2.3 eV.
(c) The gradient is h for every metal, so the line is parallel, shifted right: higher threshold frequency, lower K_max.
Where to practise next
- First law of thermodynamics study guide (9.4) for PV diagrams and the sign of work.
- Electric potential study guide (10.5) for equipotential maps.
- Compound DC circuits study guide (11.5) for derivations with resistor networks.
- Current-carrying wires practice (12.3) for B = μ₀I/(2πr).
- Images formed by lenses practice (13.4) for ray diagrams.
- Photoelectric effect study guide (15.5) for K_max against f graphs.
- Unit 11 review for mixed circuit questions.
Quick questions
Do I lose credit if my derivation is right but I skip the starting law?
Often, yes. A derivation should start from a fundamental law or definition and show each step. An answer that jumps straight to a final expression gives the marker nothing to award if one step goes wrong.
Should I draw the best-fit line through the origin?
Only if the model says the line passes through the origin and the data agree. Draw the line that best fits your plotted points. A small intercept is normal experimental scatter; a large one may point to a systematic error worth mentioning.
How many limiting cases should I check?
One clear case is usually enough to show the habit; two is better for an expression with several variables. Pick a value that makes the physics obvious, such as a resistance going to zero or a distance becoming very large.