Lesson 19: Exam Technique: Structured Problems

You already own the physics; this lesson converts it into marks. IB Physics papers have a distinctive grammar — command terms, mark allocations, "show that" questions, and data-booklet fluency — and examiners' reports show the same avoidable losses every session. This lesson works through two complete structured problems, annotating the technique as it goes, then summarises the Paper 1 and Paper 2 playbooks.

Command terms decoded

• Calculate: obtain a numerical answer showing all working; the marks are usually split substitution → answer → unit.
• Show that: reproduce a given value; you must show *every* step because the answer is given — the working IS the answer. Never use the target value as an input.
• Estimate: use rough values and state assumptions; order of magnitude earns the mark.
• Explain: give a mechanism (because… therefore…), not a description.
• State: one or two words or a value; do not waste time elaborating.
• Sketch: labelled axes, correct shape, key values marked.

The mark allocation is a contract: a [3] needs three distinct creditable points. If your answer to a [3] is one line, it is almost certainly incomplete.

Worked problem 1 — structured mechanics (annotated)

*A ball is projected at 18 m s⁻¹ at 40° above the horizontal from a 1.2 m release height. (a) Show that the time of flight is about 2.5 s. [3] (b) Calculate the horizontal range. [2] (c) Calculate the speed at impact. [3]*

(a) Vertical motion: uy = 18 sin 40° = 11.57 m s⁻¹. Take up as positive; displacement = −1.2 m.
    −1.2 = 11.57 t − 4.905 t² → 4.905t² − 11.57t − 1.2 = 0
    t = [11.57 + √(11.57² + 4 × 4.905 × 1.2)]/(2 × 4.905) = [11.57 + √(133.9 + 23.5)]/9.81
    t = (11.57 + 12.54)/9.81 = 2.46 s ≈ 2.5 s ✓   [all three marks: equation, substitution, root selected]
(b) Range = ux × t = 18 cos 40° × 2.46 = 13.79 × 2.46 = 34 m
(c) At impact vy = 11.57 − 9.81 × 2.46 = −12.6 m s⁻¹; speed = √(13.79² + 12.6²) = √(190 + 158) = 18.6 ≈ 19 m s⁻¹
    Faster route (energy): ½mv² = ½m(18²) + mg(1.2) → v² = 324 + 23.5 = 347.5 → v = 18.6 ✓ same answer, one line.

Technique notes: in (a) discard the negative root explicitly (it corresponds to the "virtual launch" before t = 0) — examiners credit the rejection. In (c), showing two routes is a luxury of practice; in the exam, energy is the reliable one.

Worked problem 2 — experimental question walkthrough

Paper 2 always includes data analysis. Standard moves: (1) read values with uncertainties from the graph or table; (2) compute the derived quantity; (3) propagate uncertainties (Lesson 17 rules); (4) compare with a stated theoretical value within the uncertainty range and conclude. Example: a graph of V against I for a filament lamp curves (resistance rises with temperature): a question asking "state whether the lamp obeys Ohm's law" wants "No: V/I is not constant — the gradient increases with I" — observation plus reason, 2 marks, no calculation at all.

Paper 1 playbook

Paper 1 (multiple choice, no calculator in the current format for Paper 1A—check your session's rules; traditionally calculators are not permitted in Paper 1) rewards: unit analysis to eliminate options; proportional reasoning (doubling, inverse squares) rather than full arithmetic; order-of-magnitude estimates (g ≈ 10, c = 3 × 10⁸); and diagram-first thinking. Bank the easy questions first; a hard question is worth the same 1 mark as an easy one. If two options survive, compare their *ratio* to your estimate rather than recomputing.

Practical worked example — the "show that" discipline

*Show that the escape speed from Earth's surface is about 11 km s⁻¹.* The trap is to write v = √(2GM/R) = 11.2 km s⁻¹ and stop — that earns partial credit because substitution is invisible. Full-credit form:

Escape when Ek + Ep = 0 at infinity: ½mv² − GMm/R = 0 → v = √(2GM/R)
v = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.37 × 10⁶) = √(1.25 × 10⁸) = 1.12 × 10⁴ m s⁻¹ ≈ 11 km s⁻¹ ✓

Every symbol defined by substitution of a booklet value, the arithmetic shown, the target reproduced, the unit correct. That template — relation, substitution, result, sanity check — is the universal IB calculation format.

Nature of Science

Exam technique is really scientific communication: claims proportioned to evidence, working transparent enough to be audited, uncertainties stated. The mark scheme's insistence on shown reasoning mirrors peer review — results that cannot be followed cannot be trusted.

Real-world application

Professional engineers and medics work under the same rules in safety-critical documentation: every figure traceable to an input, every assumption stated, units everywhere. The IB's calculation discipline is industry practice, not exam pedantry.

Common mistakes (examiners' reports, paraphrased)

• No working shown in "show that" questions, or circular use of the given answer.
• Degrees/radian mode errors in SHM and wave calculations — check the calculator before the exam, not during.
• Answers without units, or units inconsistent with substituted values (mm appearing inside a metres calculation).
• Rounding intermediate values to 1 significant figure and drifting outside the accepted range; keep 4 guard digits and round only at the end.
• Leaving the final part (usually the explanation or evaluation) blank — it is often the most accessible mark in the question.

IB Exam Experience

Paper 1-style questions. 1. Which is a correct estimate of the kinetic energy of a 70 kg sprinter at 10 m s⁻¹? A. 35 J B. 350 J C. 3.5 kJ D. 35 kJ. Key: C — ½ × 70 × 100 = 3500 J. Estimation by powers of ten eliminates the rest instantly. 2. A question says "Show that the period is about 1.4 s." The best strategy is to: A. substitute 1.4 s into the formula B. work through the calculation and obtain ≈ 1.4 s C. state the value with a unit D. use the answer in the next part without working. Key: B — and (D) is legitimate only *after* (B) is done, using the given value. 3. In a calculation, a candidate obtains a car acceleration of 850 m s⁻². The most probable error is: A. wrong formula B. a unit slip (for example, mass left in grams or distance in km) C. rounding D. sign convention. Key: B — sanity-checking magnitudes against experience (~a few m s⁻² for cars) flags factor-1000 errors.

Paper 2-style structured question (6 marks). A 60 W heater is used to warm 0.50 kg of a liquid in a beaker. The temperature rises 10 K in 240 s. (a) Show that the electrical energy supplied is about 1.4 × 10⁴ J. [2] (b) Calculate the specific heat capacity of the liquid. [2] (c) The accepted value is 2400 J kg⁻¹ K⁻¹, lower than your answer to (b). Explain why, referring to the energy transfers in the apparatus. [2] *Marking notes:* (a) E = Pt = 60 × 240 = 14 400 J ≈ 1.4 × 10⁴ J (both marks for substitution + result matching the target). (b) c = E/(mΔT) = 14 400/(0.50 × 10) = 2880 ≈ 2900 J kg⁻¹ K⁻¹. (c) Not all the electrical energy reaches the liquid: some heats the beaker, the heater itself and the surroundings, and some is lost by evaporation. The calculation in (b) credits the liquid with energy it never received, so the temperature rise is smaller than the full input would produce and c is overestimated — accept any two correctly linked points for the 2 marks.

IA idea: Choose your IA topic by an exam-technique criterion: pick the phenomenon whose analysis forces you to use the most transferable skill (linearisation plus max/min gradients) — that choice pays marks in both the IA and Paper 2.

Time budgeting and triage

Paper 2 allows roughly 1.5 minutes per mark. Practical consequences: a stuck [3] part gets three to four minutes of genuine attempts, then a written best-effort (with the method shown for ECF) and you move on — returning later with fresh eyes is cheaper than staring. Read the *whole* structured question before starting: later parts often reveal the intended values ("Using your answer in (a)…") or the physical story, and part (d)'s explanation may tell you what part (b) was really about. Underline the command term and the value's expected form (show that ≈, calculate =, estimate ≈ order of magnitude) before lifting the pen.

Worked problem 3 — a synoptic thermal-electric problem (annotated)

*An immersion heater rated 12 V, 36 W is placed in 0.25 kg of water. (a) Calculate the current in the heater. [1] (b) The water warms from 18 °C to 30 °C in 300 s. Calculate the efficiency of the heating process. [3]*

(a) I = P/V = 36/12 = 3.0 A    [1: one step, but write the equation — if the arithmetic slips, the method mark pattern in (b) still protects you]
(b) Energy delivered electrically: E_in = Pt = 36 × 300 = 10 800 J
    Energy to water: E = mcΔT = 0.25 × 4180 × 12 = 12 540 J  ← LARGER than the input!

Stop. The result is impossible — efficiency cannot exceed 100%. The numbers here are deliberately typical of a *misread* question: re-reading shows the heater is rated 36 W but the supply in this part is actually 15 V (a line in the stem that was skipped), making the real power P = V²/R with R = V²/P_rated = 144/36 = 4.0 Ω, so P_actual = 15²/4.0 = 56 W and E_in = 16 900 J, giving efficiency 12 540/16 900 = 74%. The annotation is the lesson: impossible results are alarms, not answers — sanity-check every efficiency (≤ 100%), every speed (≤ c for particles, sensible for mechanisms), every efficiency chain, and when the alarm sounds, re-read the stem before re-doing the arithmetic. This single habit rescues more marks than any formula.