How to use this guide
This guide is for students of the algebra-based Physics 1 course who know the physics but lose free-response points. Read the verb tables, then try the examples with the answers covered.
Every example and task here is original Marlbridge practice, not past exam material. The marks shown are a suggested Marlbridge rubric for judging your own work, not official scoring.
The free-response section at a glance
The section has 4 questions in 95 minutes (100 minutes before May 2027) and is 50% of the exam. You read the questions in Bluebook and handwrite answers in a paper booklet. A four-function, scientific or graphing calculator is allowed.
| Question | Type | Points | Suggested time (course description) |
|---|---|---|---|
| 1 | Mathematical Routines | 10 | 20–25 minutes |
| 2 | Translation Between Representations | 12 | 25–30 minutes |
| 3 | Experimental Design and Analysis | 10 | 25–30 minutes |
| 4 | Qualitative/Quantitative Translation | 8 | 15–20 minutes |
Question 1 mixes algebra, numbers and a reasoned claim. Question 2 links a diagram, a derived equation and a graph. Question 3 asks for a procedure, then a graph of given data. Question 4 asks for a reasoned claim and a derivation, then a check that they agree or a justified prediction. You get an equations table, but some results you must derive.
The task verbs: what each one asks for
The course lists fourteen free-response task verbs. They fall into four groups, and the group tells you what kind of thing to put on the page.
Group 1: number and algebra verbs
| Verb | What it asks for | How much to write |
|---|---|---|
| Calculate | A number reached through visible steps: starting equation, substitution, result with unit and sensible significant figures | Two to four lines |
| Derive | An expression built from a basic law through algebra steps, in the symbols allowed | One line per step; box the result |
| Estimate | A rough value, a sign, or bigger/smaller, based on data or evidence | A value; steps optional |
Short example (Unit 2). A 5.0 kg box is pulled across a frictionless floor by a 20 N horizontal force.
- Calculate the acceleration. ΣF = ma, so a = 20 N ÷ 5.0 kg = 4.0 m/s². Law, substitution, unit.
- Derive the acceleration if the floor has a coefficient of kinetic friction μ. Start from ΣF_x = ma. Vertically F_n = mg, so friction is μmg. Then F − μmg = ma, so a = (F − μmg) / m.
- Estimate the time to cross a 10 m room from rest on the frictionless floor. About 2 s (√(2 × 10 ÷ 4.0) ≈ 2.2 s).
Group 2: representation verbs
| Verb | What it asks for | How much to write |
|---|---|---|
| Draw | A diagram: free-body diagram, bar chart, set-up | The picture, labelled where asked |
| Sketch | The shape of a relationship: rises, falls, zero crossings, straight or curved | Axes and key features; no scale |
| Plot | Data points on a scaled graph | Points, scales and units |
| Label | Names, units, scale or components on a picture | Words or symbols only |
Short example (Unit 1). A ball is thrown straight up and caught at the same height. Take +y up; ignore air resistance.
- Sketch velocity against time. A straight line with a negative slope, starting positive, crossing zero at the top and ending at the same size but negative. (Plot would mean placing measured points on a scaled grid.)
- Draw a free-body diagram at the top. One arrow, gravity, pointing down, starting on the object. Velocity and acceleration are not forces, so they do not go on it.
Every representation must agree with your other answers. A sketch that slopes down beside a calculated positive acceleration will lose credit.
Group 3: short-statement verbs
| Verb | What it asks for | How much to write |
|---|---|---|
| Indicate | A choice or fact, with no explanation | A word, letter, tick or arrow |
| Rank | Quantities in order of size, ties shown | A list such as B > A = C |
| Determine | A decision or result from reasoning or calculation | The result plus its working |
Short example (Unit 7). Three block–spring oscillators: A has mass m and spring constant k; B has 2m and k; C has m and 4k.
- Rank the periods. T = 2π√(m/k), so relative periods are A: 1, B: √2 ≈ 1.4, C: ½. T_B > T_A > T_C.
- Indicate which has the highest frequency. C.
- Determine the ratio T_B / T_C. 1.41 ÷ 0.50 ≈ 2.8.
“Rank … and justify your ranking” turns a one-line answer into a reasoning answer.
Group 4: reasoning verbs
| Verb | What it asks for | How much to write |
|---|---|---|
| Justify | Reasoning in words, beyond algebra, that supports a claim | Two to four sentences: law, how it applies, conclusion |
| Describe | The features that matter for the question | A few phrases; no “because” unless asked |
| Compare | How two things are alike or different, and which way | One sentence naming both (“larger in X than in Y”) |
| Verify | Show a law’s conditions hold here, or confirm a hypothesis with data | Condition, evidence, conclusion |
Short example (Unit 6). A skater spinning on smooth ice pulls her arms in.
- Verify that her angular momentum about the spin axis is constant. The condition is zero net external torque about that axis. Gravity and the normal force act along the axis, and friction from smooth ice is negligible. The condition holds.
- Justify the claim that her angular speed increases. With no external torque, L = Iω stays constant. Pulling her arms in moves mass closer to the axis, reducing I, so ω must increase.
How to read a multi-part question
- Read the stem and every part before writing. Part (d) often shows what part (b) was for.
- Underline the verb in each part. It decides the format of your answer.
- Note the allowed symbols. “In terms of M, L, x and physical constants” means your final expression may contain only those.
- Note sign conventions and the system, such as “take +x to the right” or “the system is the two carts”, and use exactly those.
- Map the chain in the margin: “(a) FBD → (b) uses (a) → (c) checks (b)”. Watch for “your answer to part (a)”: the two parts must be consistent.
- If a part defeats you, move on. Use a placeholder value or symbol, say so in one line, and keep going.
Planning your time
You have 95 minutes for 40 points: about 2.4 minutes per point. This Marlbridge plan sits inside every suggested range:
| Block | Minutes | Running total |
|---|---|---|
| Skim all four questions | 2 | 2 |
| Question 1 (10 points) | 22 | 24 |
| Question 2 (12 points) | 27 | 51 |
| Question 3 (10 points) | 27 | 78 |
| Question 4 (8 points) | 17 | 95 |
Write the running totals at the top of your booklet. If you pass a checkpoint by more than three minutes, finish your current part and move on. Two easy points on a fresh question are worth more than the last hard point of an old one.
Showing work on calculations
For calculate, a bare number risks every point. Use four lines:
- Law or definition: write the starting equation in symbols, e.g. ½kx² = ½mv².
- Rearrange if needed, still in symbols: v = x√(k/m).
- Substitute numbers with units: v = (0.12 m)√(250 N/m ÷ 0.40 kg).
- Answer with unit and 2 or 3 significant figures: v = 3.0 m/s.
For derive, the starting law is the first line and each step changes one thing. Keep symbols to the end, then check units and one limiting case (“if μ = 0, a = F/m”).
For estimate, steps are optional, but a one-line route helps if your value is off.
What earns and what loses credit
| Earns credit | Loses credit |
|---|---|
| A starting law written before any numbers | A number with no visible source |
| Units on every final answer | A missing or wrong unit |
| A claim, then a reason naming a law and this situation | The claim restated as its own reason |
| Final expressions in the allowed symbols only | Leaving an unknown, such as T or F_n, in the final expression |
| One clear answer | Two answers, one right and one wrong, with no choice made |
| Forces drawn as single arrows from the object | Extra arrows for components, “ma” or “centripetal force” on a free-body diagram |
Worked example 1: a spring launcher (Unit 3, calculate)
Question. On a level, frictionless track, a 0.40 kg block is pushed against a spring (k = 250 N/m), compressing it by 0.12 m, and released. (a) Calculate the speed of the block when it leaves the spring. (b) The same spring, compressed by the same amount, launches a 0.80 kg block instead. Determine its speed and justify your answer without repeating the full calculation.
Weak answer. “(a) F = kx = 30 N, a = F/m = 75 m/s², v² = 2ax so v = 4.2 m/s. (b) Twice the mass so half the speed, 2.1 m/s.”
What it misses. In (a), the spring force falls as the spring relaxes, so the acceleration is not constant and v² = 2ax does not apply. In (b), the number matches only because two errors cancel: the reasoning “half the speed” is wrong, because the speed falls by a factor of √2, not 2.
Strong answer.
(a) System: block + spring, no friction, so mechanical energy is conserved. ½kx² = ½mv², so v = x√(k/m) = (0.12 m)√(250 N/m ÷ 0.40 kg) = 3.0 m/s. (The stored energy is ½ × 250 × 0.12² = 1.8 J.)
(b) The spring stores the same 1.8 J, all of which becomes kinetic energy. Since ½mv² is fixed, v ∝ 1/√m, so doubling the mass multiplies the speed by 1/√2: v = 3.0 × 0.71 = 2.1 m/s.
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Uses energy conservation, ½kx² = ½mv², not constant-acceleration equations |
| 1 | Correct substitution and 3.0 m/s with unit |
| 1 | States the stored energy is unchanged, so ½mv² is the same |
| 1 | Speed falls by a factor of 1/√2, giving 2.1 m/s |
Worked example 2: a windlass holds a bucket (Unit 5, derive)
Question. A well windlass is a horizontal drum of radius r on a fixed axle. A rope wound on the drum holds a bucket of mass m at rest. A crank handle at distance R from the axle is pushed perpendicular to the crank arm with force F. (a) Derive an expression for F in terms of m, r, R and physical constants. (b) For m = 15 kg, r = 0.10 m and R = 0.40 m, calculate F. (c) A student claims a heavier drum needs a larger holding force. Justify whether the student is right.
Weak answer. “(a) F = mg. (b) F = 15 × 9.8 = 147 N. (c) Yes, because a heavier drum is harder to hold.”
What it misses. (a) has no starting law and ignores the lever arms, so it is right only if r = R. (b) inherits the error. (c) restates the claim with no physics, and the claim is wrong.
Strong answer.
(a) The drum and crank are in rotational equilibrium. Take torques about the axle. The rope pulls down with mg at lever arm r; the hand pushes with F at lever arm R.
Στ = FR − mgr = 0, so F = mgr / R.
(b) F = (15 kg)(9.8 m/s²)(0.10 m) ÷ 0.40 m = 37 N.
(c) The student is wrong. The drum’s weight and the axle’s support force both act through the axis: zero lever arm, so no torque, and F is unchanged.
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Starts from Στ = 0 about the axle |
| 1 | Correct lever arms: r for the rope, R for the hand |
| 1 | F = mgr/R, only allowed symbols |
| 1 | 37 N with unit |
| 1 | Drum weight acts at the axis: zero lever arm, no torque, F unchanged |
Worked example 3: two carts stick together (Unit 4, compare)
Question. On a level track with negligible friction, a 2.0 kg cart moving at 3.0 m/s hits a stationary 1.0 kg cart. They stick together. (a) Calculate their speed after the collision. (b) Compare the kinetic energy of the two-cart system before and after the collision, and justify whether the collision is elastic.
Weak answer. “(a) 2.0 m/s. (b) Momentum is conserved, so energy is conserved too. The collision is elastic.”
What it misses. (a) has no working to credit. (b) confuses momentum with kinetic energy, and never says which energy is larger.
Strong answer.
(a) No net external horizontal force on the two-cart system, so momentum is conserved. (2.0 kg)(3.0 m/s) = (3.0 kg)v_f, so v_f = 2.0 m/s.
(b) Before: K = ½(2.0)(3.0)² = 9.0 J. After: K = ½(3.0)(2.0)² = 6.0 J. The kinetic energy after is smaller, by 3.0 J. An elastic collision keeps the system’s kinetic energy the same, so this one is inelastic. The 3.0 J becomes internal energy and sound.
| Point | Suggested Marlbridge rubric |
|---|---|
| 1 | Momentum conservation stated and applied: 2.0 m/s |
| 1 | Both kinetic energies, 9.0 J and 6.0 J |
| 1 | Direction of the comparison: smaller after |
| 1 | Uses the definition of elastic to conclude inelastic |
Common mistakes
- Giving a bare number for calculate, with no starting equation or substitution.
- Plugging numbers into a derive answer, or leaving a symbol the question did not allow.
- Writing “because” plus the claim again in a justify answer, instead of naming a law.
- Answering compare with one quantity only, or without saying which is larger.
- Writing paragraphs for indicate or rank: it wastes time and risks contradictions.
- Using constant-acceleration equations when a force changes, such as a spring force.
- Two different answers to one part, with neither crossed out.
- Spending 40 minutes on one question and leaving the last question blank.
Practise it
Try each task before opening the model answer.
Task 1 (Unit 1, sketch and describe). A cyclist rides at a steady 6.0 m/s for 4.0 s, then brakes uniformly to rest in 3.0 s. Sketch velocity against time, and describe how to find the total distance from your sketch.
Model answer
A horizontal line at 6.0 m/s from 0 to 4.0 s, then a straight line down to zero at 7.0 s. The distance is the area under the graph: a rectangle (6.0 × 4.0 = 24 m) plus a triangle (½ × 3.0 × 6.0 = 9.0 m), so 33 m.
Task 2 (Unit 6, derive and predict). A turntable with rotational inertia I₀ spins freely at ω₀. A ring, also with rotational inertia I₀ about the same axis, is dropped gently onto it, centred, and the two then turn together. Derive their common angular speed, and predict the factor by which the rotational kinetic energy changes.
Model answer
No external torque about the axis, so I₀ω₀ = (2I₀)ω, giving ω = ω₀/2. K = ½Iω², so K_new / K_old = (½ × 2I₀ × ω₀²/4) ÷ (½ I₀ω₀²) = ½. Friction between ring and turntable turns the rest into internal energy. With I₀ = 3.0 kg·m² and ω₀ = 2.0 rad/s, K goes from 6.0 J to 3.0 J.
Task 3 (Unit 7, estimate and predict). A heavy load hanging from a crane cable swings like a simple pendulum, taking 4.0 s for each full swing. Estimate the cable length. Then predict the period if the load is lowered so the cable is four times longer.
Model answer
From T = 2π√(L/g), L = gT²/(4π²) ≈ 4.0 m (3.97 m with g = 9.8 m/s²). Since T ∝ √L, four times the length doubles the period: 8.0 s.
Task 4 (Unit 8, rank and justify). Water flows steadily through a horizontal pipe with three sections of cross-sectional area 4.0 cm² (section 1), 2.0 cm² (section 2) and 8.0 cm² (section 3). Rank the water speeds and the pressures in the three sections, and justify your rankings.
Model answer
Speeds: v₂ > v₁ > v₃. The volume flow rate Av is the same everywhere, so speed is inversely proportional to area. If v₁ = 1.5 m/s, then v₂ = 3.0 m/s and v₃ = 0.75 m/s.
Pressures: P₃ > P₁ > P₂. The pipe is horizontal, so by Bernoulli’s equation P + ½ρv² is the same in each section. Where the speed is higher, the pressure is lower. (With these speeds, P₁ − P₂ = 3.4 kPa.)
Where to practise next
Quick questions
Do I lose credit if I give more than the question asks for?
Extra correct physics does no harm, but it costs time. Extra wrong physics can cost credit, because a marker cannot tell which of two statements you mean. Answer what the verb asks, then stop.
What if I cannot get an earlier part?
Do not leave the later parts blank. Use a sensible value or symbol for the missing answer, say so in one line, and carry on. In the suggested Marlbridge rubric, a later part that correctly uses an earlier wrong value still earns its points.
Should I write in full sentences?
For justify, describe and compare, yes: short, complete sentences are clearest. For indicate, label and rank, a single word, letter or list is enough.