Full Paper Walkthrough

Everything in this course comes together in the exam itself, so the best way to finish is to walk through a representative set of questions exactly as you would meet them on a J249 paper — from the quick multiple-choice openers to the levelled six-marker at the end. As we go, we will model the thought process: reading the command word, choosing the method, showing working, checking units, and building an extended answer. This is not a real past paper (use official OCR papers for those) but a realistic mixed set spanning both tiers and increasing in difficulty.

By the end of this lesson you should be able to approach a mixed paper systematically, applying command-word reading, calculation technique and data analysis to each question type, and you should see a full extended-response question answered at three levels.

Question 1 — Multiple Choice (1 mark, AO1)

Which of these is the correct unit for electrical resistance? A volt B ampere C ohm D watt

Thought process: a straight recall question. Resistance is measured in ohms ( $\Omega$ ). The answer is C. Multiple-choice questions are quick marks — read all four options, eliminate the obviously wrong ones, and do not overthink. If unsure, never leave it blank: even a guess has a 1-in-4 chance.

Exam Tip: On multiple choice, eliminate the options you know are wrong first, then choose between what's left. Here, volt, ampere and watt are units of other quantities, so C stands out. Elimination turns a hard-looking question into an easy one.

Question 2 — Short Calculation (3 marks, AO2)

A resistor has a potential difference of 6.0 V across it and a current of 0.5 A through it. Calculate its resistance.

Thought process: the command word is calculate, so I need a value with working and a unit. The quantities are p.d., current and resistance, so the equation is $V = IR$ .

$R = \dfrac{V}{I} \quad\text{[equation]}$ $R = \dfrac{6.0}{0.5} \quad\text{[substitution]}$ $R = 12\ \Omega \quad\text{[answer with unit]}$

Three lines, three marks. Note the unit ( $\Omega$ ) on the answer and that I rearranged $V = IR$ to $R = \dfrac{V}{I}$ before substituting.

Exam Tip: Even a "simple" calculation should be written as three lines — equation, substitution, answer-with-unit. If the arithmetic slips, the first two lines still bank the method marks. A lone "12" risks losing two of the three marks.

Question 3 — Describe (2 marks, AO1)

Describe the motion of the object shown by a distance–time graph that is a horizontal straight line.

Thought process: command word describe — say what is happening, no why needed. A horizontal line on a distance–time graph means the distance is not changing, so the object is stationary / at rest. Two marks: (1) the object is not moving; (2) because the distance stays constant over time.

Exam Tip: Match your answer to the tariff — 2 marks wants two points. For a "describe the motion" question, name the motion (stationary) and justify it from the graph (distance is constant). One word ("stopped") risks only half the marks.

Question 4 — Explain (3 marks, AO2)

Explain why the resistance of a filament lamp increases as the current through it increases.

Thought process: command word explain — I need reasons, a linked chain. The mark scheme wants:

As current increases, the filament's temperature rises (it heats up).
The metal ions vibrate more with the extra thermal energy.
This makes it harder for electrons to pass, so the resistance increases.

Written as a chain with connectives: "As the current increases, the filament heats up, so the positive metal ions vibrate more, which obstructs the flow of electrons, so the resistance increases."

Exam Tip: An explain answer should read as a chain of "because/so" statements, each cause leading to the next effect. If your answer has no linking words, it is probably a description in disguise — and describe-level answers do not earn explain marks.

Question 5 — Data Analysis (4 marks, AO3)

A student measures the extension of a spring for different forces and plots a graph. The graph is a straight line through the origin up to 5.0 N, then curves. (a) What does the straight portion show? (b) One point at 3.0 N does not fit the line — what should the student do? (c) What happens beyond 5.0 N?

Thought process — command words vary by part:

(a) The straight line through the origin shows extension is directly proportional to force (obeys $F = ke$ , Hooke's law).
(b) The point is an anomaly; the student should ignore it when drawing the line of best fit (and could check/repeat that reading).
(c) Beyond 5.0 N the spring has passed its limit of proportionality — extension is no longer proportional to force (the line curves).

This one question tests reading a graph, handling an anomaly, and interpreting a relationship — classic AO3.

Exam Tip: In multi-part data questions, read the command word of each part separately — "what does it show" (interpret), "what should the student do" (evaluate/method), "what happens beyond" (interpret). Each part wants a different kind of answer, and the marks are split across them.

Question 6 — Standard-Form Calculation (3 marks, AO2, Higher)

Light travels at $3 \times 10^{8}$ m/s. Calculate the distance light travels in $4 \times 10^{-6}$ s.

Thought process: calculate, using $v = \dfrac{s}{t}$ rearranged to $s = v \times t$ .

$s = v \times t = (3 \times 10^{8}) \times (4 \times 10^{-6})$ $s = 12 \times 10^{2} = 1.2 \times 10^{3}\text{ m}$

Multiply the numbers ( $3 \times 4 = 12$ ), add the powers ( $8 + (-6) = 2$ ), then tidy $12 \times 10^{2}$ into proper standard form $1.2 \times 10^{3}$ . The answer is $1200$ m.

Exam Tip: After a standard-form calculation, tidy the answer so the number is between 1 and 10 — $12 \times 10^{2}$ becomes $1.2 \times 10^{3}$ . Leaving it as $12 \times 10^{2}$ can cost the final mark on a "give your answer in standard form" question.

Question 6b — A Two-Step Calculation (4 marks, AO2, Higher)

An immersion heater with a power of 50 W is switched on for 120 s to heat 0.50 kg of water. Calculate the temperature rise of the water. (specific heat capacity of water = 4200 J/kg°C)

Thought process: the givens (power, time, mass, specific heat capacity) do not match a single equation, so this is a two-step problem through the intermediate quantity of energy. First find the energy supplied, then feed it into the heat equation.

Step 1 — energy supplied: using $E = P \times t$ , $E = 50 \times 120 = 6000\text{ J}$

Step 2 — temperature rise: using $E = mc\Delta\theta$ , rearranged to $\Delta\theta = \dfrac{E}{mc}$ , $\Delta\theta = \dfrac{6000}{0.50 \times 4200} = \dfrac{6000}{2100} = 2.9\text{ °C (2 s.f.)}$

The key skill here is recognising that energy links the two equations — you cannot leap straight from power to temperature. Spotting that you must calculate an intermediate quantity first is exactly the reasoning the higher-tariff calculations test, and labelling "energy = 6000 J" clearly lets you carry it cleanly into the second step.

Exam Tip: When a calculation's givens span two different equations, look for the intermediate quantity that connects them — here, energy links $E = Pt$ to $E = mc\Delta\theta$ . Calculate it first, label it, then use it. This two-step recognition is a common source of marks lost by students who search in vain for a single magic equation.

Full Paper Walkthrough

Full Paper Walkthrough

Question 1 — Multiple Choice (1 mark, AO1)

Question 2 — Short Calculation (3 marks, AO2)

Question 3 — Describe (2 marks, AO1)

Question 4 — Explain (3 marks, AO2)

Question 5 — Data Analysis (4 marks, AO3)

Question 6 — Standard-Form Calculation (3 marks, AO2, Higher)

Question 6b — A Two-Step Calculation (4 marks, AO2, Higher)

Question 7 — The Extended Response (6 marks, AO1/AO2/AO3)

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