You know the physics. This guide is about turning it into marks. Each free-response part opens with a task verb, and each verb asks for a different kind of answer.
How to use this guide
This guide is for students in the last months before the exam. Read the verb table, then work through the three worked examples, comparing each weak answer with the strong one. Finish with the practice tasks.
Every question here is original Marlbridge practice, written for this guide, not past exam material. The marks shown are a suggested Marlbridge rubric to help you judge your work, not official scoring.
The free-response section at a glance
Section II has 4 questions in 95 minutes and is worth 50% of the exam. You read the questions on screen and handwrite answers in a paper booklet. A four-function, scientific or graphing calculator is allowed, and you get an equation sheet. The questions always come in this order:
| Question | Type | Points | Suggested time | Main demand |
|---|---|---|---|---|
| 1 | Mathematical Routines | 10 | 20–25 min | Derive and calculate, helped by a diagram or sketch |
| 2 | Translation Between Representations | 12 | 25–30 min | Diagram, equation and graph of one situation, then link them |
| 3 | Experimental Design and Analysis | 10 | 25–30 min | Plan a procedure, then graph given data to find a quantity |
| 4 | Qualitative/Quantitative Translation | 8 | 15–20 min | A justified claim, a derivation, then link the two |
Questions 2 and 4 end with a linking part: say whether two earlier answers agree and why, or use your earlier work to predict a new case. Unless told otherwise, assume an inertial frame, no air resistance, and ideal springs and strings.
The task verbs and what each one wants
The course lists fourteen task verbs that are common in free-response questions. Our own summary:
| Verb | What you produce | How much to write |
|---|---|---|
| Indicate | A bare answer: a direction, a choice, yes or no | A word or phrase |
| Rank | Quantities in order of size, ties shown | One line, such as A > B = C |
| Calculate | A number reached by visible steps | Equation, substitution, answer with units |
| Derive | An expression built from a basic law | Law, then each algebra step, ending in the allowed symbols |
| Determine | A value or a decision | Enough working or reasoning to show how you decided |
| Estimate | A rough value, size or sign from data | The answer; steps not required |
| Draw | A diagram of objects, forces or links | A neat diagram, labelled if asked |
| Sketch | The shape of a relationship | Correct shape, key points and trends |
| Plot | Data points on axes | Accurate points on a sensible scale |
| Label | Names, units, scales or components | Short labels in the right places |
| Justify | Physics reasons supporting a claim | Two to four sentences linking a law to this case |
| Describe | The important features of something | A few precise statements |
| Compare | Likenesses or differences | A statement about both, using “greater”, “less” or “equal” |
| Verify | A check that a law’s conditions are met, or that data support an idea | Name the condition, then show it holds |
The verbs fall into four groups.
Answer-only verbs: indicate, rank
Give the result and stop. A wrong extra reason can make a right answer unclear.
Example. Rank the kinetic energies of carts A (2 kg, 3 m/s), B (4 kg, 2 m/s) and C (1 kg, 4 m/s).
Answer: A > B = C (9 J, 8 J, 8 J). Writing A > B > C loses the mark: ties must be shown.
Working verbs: calculate, derive, determine, estimate
Calculate ends in a number with units. Derive ends in an expression and must begin from a fundamental law (Newton’s second law, the definition of work, conservation of energy or momentum), not from a formula that is already the answer. Determine may end in a number or a decision, with the reasoning shown. Estimate needs no steps.
Example. A 1,200 kg car takes a flat curve of radius 40 m at 15 m/s, with μ_s = 0.50. Determine whether it skids.
Answer: Inward force needed: mv²/r = 1,200 × 15² ÷ 40 = 6,750 N. Most static friction can give: μ_s mg = 0.50 × 1,200 × 9.8 = 5,880 N. Friction cannot supply enough, so the car skids. The comparison is what makes this a determination rather than a guess.
Picture verbs: draw, sketch, plot, label
These score on features, not art. On a free-body diagram, each force is one labelled arrow starting on the object. A sketch is checked for shape, axis crossings, zero slopes and agreement with your other answers. A plot needs a scale that fills most of the grid.
Example. A ball is thrown straight up at 12 m/s and caught at the same height. Sketch v_y against t, up positive.
Answer: A straight line of constant negative slope, from +12 m/s through zero at about 1.2 s to −12 m/s at about 2.4 s. A curve, or a line that “bounces” at zero, loses the shape mark.
Reasoning verbs: justify, describe, compare, verify
These want sentences. An equation shows what happens; a justification says why, naming a law and linking it to this situation.
Example. Two pucks collide on smooth ice. Verify that the system’s horizontal momentum is constant.
Answer: Momentum is constant when the net external force is zero. Each puck’s weight is balanced by the normal force, friction from the ice is negligible, and the pucks’ forces on each other are internal and cancel in pairs. So the net external horizontal force is zero. State the condition, then show it is met.
Reading a multi-part question
Spend a minute on the whole question first.
- Find the system. Underline the objects, the givens and what is negligible.
- Circle every verb. Mark the parts that need words, so you do not answer them with an equation alone.
- Read “in terms of” lists. Your final expression may use only those symbols, plus constants such as g and π.
- Look for the chain. Part (b) usually uses part (a), and a later part may ask whether two answers agree. Make your sketch and your equation describe the same physics.
- Watch for a changed property. A heavier block or a longer string means your prediction must come from your earlier equation or graph, and you must say how.
If an early part defeats you, write a named symbol or value in its place and carry on. That way a later part can still show a correct method; a blank shows nothing.
Planning your time
The suggested times total 85–105 minutes, but you have 95. A plan that fits: 2 minutes reading all four questions and circling verbs, then about 22 minutes on Question 1, 27 on Question 2, 27 on Question 3 and 17 on Question 4. That is roughly two and a half minutes per point. If a part stalls you for three minutes, leave a gap and return at the end.
Showing work on calculations
- Start with a law in symbols: ΣF = ma, W = ∫F dx, L = Iω, K + U = constant.
- Apply it here, with only the forces or energies present. Write any integral with its limits before evaluating it.
- Solve in symbols first and substitute at the end. The symbolic line often earns its own point.
- Substitute with units and give the answer with units and 2 or 3 significant figures.
- Check units, sign and limiting cases (what happens as a quantity goes to zero?).
Box the final answer. Cross out rejected work fully, so the marker never has to choose.
What earns and loses credit
| Earns credit | Loses credit |
|---|---|
| A law in symbols, then applied | A number with no starting equation |
| One labelled arrow per force | A “force of motion”, or components drawn as extra arrows |
| Units on every final number | Missing or wrong units |
| An expression in only the allowed symbols | A leftover symbol such as v or t |
| Sentences naming a law and linking it to the case | An equation alone for a justify part |
| A best-fit line on labelled axes | Points joined dot to dot |
| Answers that agree, or a stated reason why not | Contradictory answers with no comment |
Worked example 1: Derive and calculate (Unit 3)
A 0.50 kg block on a smooth horizontal surface is attached to a non-linear spring that exerts F = −βx³, where β = 200 N/m³. It is released from rest at x = A = 0.30 m. (a) Derive its speed at x = 0 in terms of m, β and A. (b) Calculate this speed.
Weak answer: “½mv² = ½kA² with k = βA², so v = A√(βA²/m) = 1.8 m/s.”
What it misses. ½kx² holds only when force is proportional to x. The student used a result outside its conditions, with no fundamental starting point and no integral, so the number is wrong too.
Strong answer. (a) Work–energy theorem, W = ΔK. The force varies with position, so
W = ∫ from A to 0 of (−βx³) dx = [−βx⁴/4] from A to 0 = βA⁴/4
From rest, ½mv² = βA⁴/4, so v = A²√(β/(2m)).
(b) v = 0.30² × √(200 ÷ 1.0) = 1.3 m/s (the spring does 0.405 J of work).
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Starts from W = ΔK or energy conservation |
| 1 | Work as an integral of the given force, correct limits |
| 1 | v = A²√(β/(2m)) |
| 1 | 1.3 m/s with units |
Worked example 2: Compare and justify (Unit 6)
A uniform disk (I = 0.040 kg·m²) spins freely at 6.0 rad/s on a smooth vertical axle. A 0.10 kg lump of clay is dropped straight down and sticks 0.20 m from the axle. (a) Calculate the new angular speed. (b) Compare the kinetic energy before and after, and justify your answer.
Weak answer: “(a) Angular momentum is conserved, so ω = 5.5 rad/s. (b) No energy is added, so kinetic energy stays the same.”
What it misses. Part (a) gives no reason for conservation and no working. Part (b) is wrong: sticking is inelastic, so kinetic energy is not constant, and no values are compared.
Strong answer. (a) System: disk and clay. The smooth axle exerts no torque about the axis, and gravity and the normal force are parallel to it. The falling clay has no angular momentum about the axis. So angular momentum about the axis is constant:
I₀ω₀ = (I₀ + mr²)ω_f, with mr² = 0.10 × 0.20² = 0.0040 kg·m², so ω_f = 0.24 ÷ 0.044 = 5.5 rad/s.
(b) K before = ½ × 0.040 × 6.0² = 0.72 J; K after = ½ × 0.044 × 5.45² = 0.65 J. Kinetic energy decreases. Friction between clay and disk speeds up the clay and slows the disk until they move together. This perfectly inelastic interaction turns some kinetic energy into internal energy of the clay. No external torque acts, so angular momentum is conserved, but the internal forces do negative net work.
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Angular momentum constant, because no external torque about the axis |
| 1 | ω_f = 5.5 rad/s with working |
| 1 | Kinetic energy decreases, supported by values |
| 1 | Inelastic sticking converts kinetic to internal energy |
Worked example 3: Describe, plot and determine (Unit 7)
A student finds the spring constant k of a light spring with masses, a clamp stand and a stopwatch. (a) Describe a suitable procedure. (b) Indicate what to plot for a straight line, then determine k from these results.
| Mass m (kg) | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
|---|---|---|---|---|---|
| Time for 10 oscillations (s) | 4.0 | 5.6 | 6.9 | 7.9 | 8.9 |
Weak answer: “(a) Hang a mass and time it bouncing. (b) Plot T against m. Using m = 0.30 kg, T = 0.69 s in T = 2π√(m/k) gives k = 25 N/m.”
What it misses. No independent variable, no repeats, no way to cut timing error. T against m is a curve, not a line, and one data point ignores the rest. The number is close, but the method would not earn the analysis points.
Strong answer. (a) Hang 0.10 kg from the spring, pull it down about 2 cm and release. Time 10 oscillations, divide by 10 for T, and repeat twice to average. Do the same for 0.20 to 0.50 kg. Since T² = (4π²/k)m, plot T² against m; k = 4π² ÷ slope.
(b) Plot T² (vertical) against m (horizontal): T² = 0.16, 0.31, 0.48, 0.62, 0.79 s². Draw a best-fit line. Two points read off the line, (0.10, 0.16) and (0.50, 0.79), give slope = 0.63 ÷ 0.40 = 1.58 s²/kg, so k = 4π² ÷ 1.58 = 25 N/m. The line passes close to the origin, as the model predicts.
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Varies m and measures T over several oscillations |
| 1 | Plots T² against m, with the reason |
| 1 | Slope from the best-fit line, not single data points |
| 1 | k ≈ 25 N/m with units |
Common mistakes
- Answering a justify part with an equation and no sentence naming a law.
- Deriving from a result such as ½kx² or v = √(2gh) when its conditions do not hold.
- Leaving a symbol outside the “in terms of” list in a final expression.
- Writing A > B > C when two quantities are equal.
- Sketching a graph whose shape disagrees with your own earlier equation.
- Taking a slope from two data points instead of the best-fit line.
- Using constant-acceleration equations when the force changes.
- Leaving a later part blank because an earlier answer went wrong.
Practise it
Write a full answer before you open each model answer.
Task 1 (Unit 1). A cart moves with v(t) = 3t² − 12t (m/s). (a) Indicate when, between 0 and 5 s, it turns round, and justify. (b) Calculate its displacement and distance travelled from 0 to 5 s.
Model answer
(a) t = 4 s. v = 3t(t − 4) is negative before 4 s and positive after, so the velocity changes sign there.
(b) Displacement = ∫ from 0 to 5 of (3t² − 12t) dt = 125 − 150 = −25 m. From 0 to 4 s the displacement is −32 m and from 4 to 5 s it is +7 m, so distance = 39 m. Splitting the integral at the turning point earns the distance mark.
Task 2 (Unit 2). A drone of mass m rises straight up from rest. Its rotors give an upward thrust F = F₀ − bt, where F₀ and b are positive constants. (a) Draw a free-body diagram. (b) Derive its acceleration in terms of F₀, b, t, m and g. (c) Calculate its speed at t = 2.5 s for m = 2.0 kg, F₀ = 30 N and b = 4.0 N/s.
Model answer
(a) Two arrows from the drone: thrust F up, weight mg down. F is longer, as the drone speeds up.
(b) Up positive. Newton’s second law: F₀ − bt − mg = ma, so a = (F₀ − bt)/m − g.
(c) a = 15 − 2.0t − 9.8 = 5.2 − 2.0t (m/s²), still positive at 2.5 s. The force changes, so integrate: v = ∫₀^2.5 (5.2 − 2.0t) dt = 13 − 6.25 = 6.8 m/s.
Task 3 (Unit 4). A 0.15 kg ball moving at 20 m/s in −x hits a wall. The wall’s force, in +x, rises steadily from 0 to 600 N in 0.006 s, then falls steadily to 0 at 0.012 s. Calculate the ball’s final velocity.
Model answer
Impulse = area = ½ × 600 × 0.012 = 3.6 N·s in +x. Initial momentum = 0.15 × (−20) = −3.0 kg·m/s. Final momentum = −3.0 + 3.6 = 0.6 kg·m/s, so v = +4.0 m/s, away from the wall. Signs on each momentum earn the method mark.
Task 4 (Unit 5). A rod is pivoted at one end. Separately: A is 10 N at 0.80 m, perpendicular to the rod; B is 20 N at 0.50 m, at 30° to the rod; C is 16 N at 0.50 m, perpendicular. Rank the torques about the pivot and justify.
Model answer
τ_A = τ_C > τ_B. Torque = rF sin φ. A: 0.80 × 10 = 8.0 N·m. C: 0.50 × 16 = 8.0 N·m. B: 0.50 × 20 × sin 30° = 5.0 N·m. B has the largest force, but only its perpendicular component turns the rod.
Where to practise next
Quick questions
Do I lose marks for extra, wrong statements?
They can cost you. If you add a second explanation that contradicts the first, a marker cannot tell which one you mean. Give one clear line of reasoning.
Should I write answers in words or symbols?
Follow the verb. Derive and calculate want symbols and numbers. Justify, describe and compare want sentences, though an equation can support the words.
What if I cannot finish part (a)?
Use a clearly labelled symbol or a guessed value for the missing result and carry on. Your later parts can then still show a correct method, while a blank part shows nothing.