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  3. Paper 2: short and extended responses

Paper 2: short and extended responses

How Paper 2 questions are structured and marked: setting out calculations, "show that" parts, explanations that reach every mark point, sketches and drawings, and synoptic questions, with fully worked questions and markschemes.

  • IB Physics
  • Exam skills
  • SL and HL

19 min readLast reviewed 14 Oct 2026

Paper 2 is a written paper of structured questions. Each question starts from a concrete situation, such as a satellite, a kettle, a guitar string or a medical tracer, and asks a series of parts that build on one another. Some parts need a calculation, some an explanation, some a sketch or a diagram. At SL the paper covers SL content only; at HL it adds questions on the additional higher level content. The paper usually ends with a longer extended-response question that draws on several themes at once.

Unlike in Paper 1A, marks are given for method as well as for answers. A wrong number early in a question does not doom the rest of it, and a clear explanation in the right physics words earns marks even when it is short. The skill is knowing exactly what each part is asking for and giving that, no more and no less.

Key points

  • The number of marks shows how many separate points or steps the examiner expects.

  • Calculations are set out as equation, substitution, answer with unit, so that method marks can be awarded.

  • A "show that" part needs every step shown, or the answer given to more significant figures than the value in the question.

  • An error made early is penalised once; later parts that use the wrong value correctly still score (error carried forward).

  • Explanations are chains of linked physics statements, one mark per link.

  • Sketches and drawings earn marks for specific features: labelled axes, shape, intercepts, arrow directions and relative lengths.

How a Paper 2 question is built

A question begins with a stem of two to five sentences, often with a diagram or a list of data. The parts that follow usually climb in demand:

  1. State or outline a fact or a definition (1 mark).

  2. Show that a quantity has a given value, which later parts will use (1–2 marks).

  3. Calculate or determine a further quantity, often in several steps (2–3 marks).

  4. Explain, discuss, predict or suggest what happens when something changes, or evaluate a result (2–3 marks).

New information often arrives between parts, as a sentence such as "The heater is now switched off" or "The initial activity of the source is 42 kBq". It applies only to the parts that follow. Reading every lead-in sentence matters, because the later parts cannot be answered without it.

Most parts are worth 1 to 3 marks and almost none more than 4. A typical short question totals 4 to 7 marks; a longer one 8 to 12. The final extended-response question is larger, with many parts spread across several areas of physics.

Exam tip

Use the mark allocation as a guide to length. A 1-mark "state" needs a word or a phrase. A 3-mark "explain" needs three distinct, linked points. Writing a paragraph for one mark wastes time; writing one sentence for three marks gives away marks.

Setting out a calculation

A calculation earns marks in stages, so the working must show each stage clearly. A reliable layout is:

  1. Write the equation you are using, in symbols, as it appears in the data booklet or rearranged for the unknown.

  2. Substitute the values, converted to SI units, so the examiner can see every number you used.

  3. Give the answer with a unit and a sensible number of significant figures, usually 2 or 3, matching the data.

Worked example

A car of mass 1200 kg accelerates uniformly from 8.0 m s⁻¹ to 20 m s⁻¹ over a distance of 140 m. Calculate the resultant force on the car.

Show the solution

Choose the equation. Mass, speeds and distance are known; time is not, so use v2=u2+2asv2=u2+2as for the acceleration, then F=maF=ma.

a=v2−u22s=202−8.022×140=336280=1.2 m s−2F=ma=1200×1.2=1.4×103 NaF​=2sv2−u2​=2×140202−8.02​=280336​=1.2 m s−2=ma=1200×1.2=1.4×103 N​

A markscheme gives one mark for the acceleration and one for the force. If the acceleration were wrong but the force were then found correctly from it, the second mark would still be awarded. A bald "1440 N" gains full marks only if it is correct, so the working is insurance.

Error carried forward

Definition

Error carried forward (ECF) is the marking rule that penalises a wrong value only once: a later part that uses the wrong value with a correct method still receives its marks.

ECF works only if the examiner can see what you did. If a later part needs a value you could not find, use the "show that" value given in the question, or make a sensible estimate and say so. A later part answered with a clearly stated wrong value can still earn full marks.

Common mistake

Rounding intermediate values heavily and then using the rounded value. Keep at least one more significant figure than the final answer through the working, and round only at the end. Rounding 1.24 to 1 in the middle of a calculation can make the final answer wrong by 20 %.

Units, significant figures and powers of ten

  • Convert prefixes before substituting: 25 kPa is 2.5×1042.5×104 Pa, 3.0 mm is 3.0×10−33.0×10−3 m, 450 nm is 4.50×10−74.50×10−7 m.

  • Temperatures in the gas laws and in the Stefan–Boltzmann law are in kelvin.

  • A final answer to 2 or 3 significant figures is almost always acceptable. A part may ask for a particular number of significant figures, and then only that number scores.

  • A power-of-ten error usually costs the final mark only; the method marks remain.

"Show that" parts

A "show that" part gives the answer, so the marks are for the route to it. There are two kinds.

Numerical "show that"

The question gives a rounded target: "Show that the speed of the ball is about 4 m s⁻¹." To earn the marks, show the full substitution into the equation, and give the result to at least one more significant figure than the target, for example 4.3 m s⁻¹. Writing only "v=4v=4 m s⁻¹" proves nothing and earns nothing.

Worked example

A satellite orbits the Earth at a radius of 6.78×1066.78×106 m. The mass of the Earth is 5.97×10245.97×1024 kg. Show that the orbital speed of the satellite is about 8 km s⁻¹.

Show the solution

Physics step. The gravitational force provides the centripetal force: GMmr2=mv2rr2GMm​=rmv2​, so v=GMrv=rGM​​.

v=6.67×10−11×5.97×10246.78×106=7.66×103 m s−1v=6.78×1066.67×10−11×5.97×1024​​=7.66×103 m s−1

The answer is given to three significant figures, 7.66 km s⁻¹, which rounds to the target of about 8 km s⁻¹. Both the substitution and the extra significant figures are shown, so either would satisfy the markscheme.

Algebraic "show that"

A derivation, such as "Show that T2=4π2r3GMT2=GM4π2r3​", is marked on the key physics step (here, equating gravitational and centripetal force, or using v=2πrTv=T2πr​) and on the algebra that follows. Start from a named principle or a data-booklet equation, write each step on its own line, and end with the expression asked for.

Tip

A "show that" value is always usable later, even if you could not show it. If the next part needs the speed, use "about 8 km s⁻¹" (or your own more precise value) rather than leaving the part blank.

Explanations that hit the mark points

An explanation gives reasons or causes. A markscheme breaks it into separate points, each worth one mark, and an answer scores for each point it makes clearly. The best answers are short chains of cause and effect, each link written in the precise words of physics.

Worked example

(3 marks) A fixed mass of gas is sealed in a rigid container. Explain, in terms of molecules, why the pressure of the gas increases when it is heated.

Show the solution

A model answer. Heating increases the temperature, so the mean kinetic energy, and therefore the mean speed, of the molecules increases. The molecules collide with the walls more often, because the volume is constant and they move faster. Each collision also involves a larger change of momentum, so the force on the walls per unit area, the pressure, increases.

Markscheme

mean kinetic energy / speed of molecules increases ✓

more frequent collisions with the walls ✓

greater change of momentum per collision «so greater force on the walls» ✓

Do not accept "molecules collide more" without reference to the walls. Do not accept "molecules expand" or "molecules get hotter".

Several habits make explanations score:

  • Count the marks and plan that many points before writing.

  • Name quantities precisely: "the mean kinetic energy of the molecules", not "the energy"; "the resultant force", not "the force".

  • Link each statement to the next with "so", "therefore" or "because", so that the chain is visible.

  • Refer to the situation in the question, not a general rule only.

  • Answer the actual question. If it asks why the pressure increases, the last link of the chain must be about pressure.

Common mistake

Vague wording that sounds right but contains no physics: "the heat is trapped", "the particles get excited", "it gets bigger". Markschemes reject these. Say which quantity changes, in which direction, and why.

Describe, outline, explain, discuss

These four command terms ask for different things. Outline asks for a brief account, one or two points. Describe asks for a detailed account of what happens or what is done, without reasons. Explain asks for the reasons. Discuss asks for a balanced review of more than one factor or point of view, ending with a conclusion where appropriate.

Sketches and drawings

"Sketch" and "draw" ask for different levels of precision. A sketch shows the general shape of a graph with its important features labelled, on axes that need not be scaled. A drawing is accurate: free-body diagrams with arrows of the right relative length, field lines that start and end in the right places, ray diagrams drawn with a ruler.

Drawing forces

A free-body diagram shows the forces acting on one body only. Each force is an arrow starting on the body, at its point of application or at the centre of mass, pointing in the direction of the force, with a length that represents its size, and with a label. The free-body diagrams of a skydiver show how relative lengths carry the physics.

Three free-body diagrams of a skydiver drawn as a dot. Left: only the weight F_g, a long arrow down. Middle: weight down and a shorter fluid resistance F_d up. Right: weight down and fluid resistance up, the two arrows equal in length.Three free-body diagrams of a skydiver drawn as a dot. Left: only the weight F_g, a long arrow down. Middle: weight down and a shorter fluid resistance F_d up. Right: weight down and fluid resistance up, the two arrows equal in length.
A skydiver who steps off a hovering balloon starts from rest, so at first only the weight acts. Each arrow starts on the body and its length shows the size of the force. At terminal speed the two arrows are equal, because the resultant force is zero.

A markscheme for such a drawing typically gives one mark for both forces in the correct directions with labels, and one for their relative lengths, judged by eye. Arrows drawn from different points, or a force from another body included by mistake, lose marks.

Sketching graphs

For a sketch graph, label both axes with the quantities (and units if the axes are scaled), and show every feature the physics predicts: where the line starts, whether it is straight or curved, its initial gradient, any maximum or minimum, and any asymptote. The sketch of speed against time for a falling object shows all of these.

Sketch graph of speed v against time t for a falling object. The curve starts at the origin with gradient g, shown by a dashed tangent, its gradient decreases, and it levels off towards a dashed horizontal asymptote labelled v_t, the terminal speed.Sketch graph of speed v against time t for a falling object. The curve starts at the origin with gradient g, shown by a dashed tangent, its gradient decreases, and it levels off towards a dashed horizontal asymptote labelled v_t, the terminal speed.
A sketch earns its marks through features: labelled axes, the start at the origin with gradient g, a gradient that decreases, and a curve that approaches but never crosses the terminal speed.

Exam tip

On a graph that already has one curve on it, a sketch of a second curve is marked by comparison: higher or lower, steeper or shallower, the same intercept or a different one. Make the difference obvious, and label the new curve.

Parts that use a graph or a diagram

Many parts say "Determine, using the graph, …" or "taking appropriate measurements from the diagram". These words mean the value must come from the figure, and the working must show where it came from: mark the points read on the graph, draw any tangent or construction line on it, and write the readings in the answer. Values read from a graph cannot be exact, so markschemes accept a range, typically one or two small squares either side of the true reading.

Questions on graphs reward the same habits as Paper 1B. Read the axes, including any prefix or power of ten in the label ("tt / ms", "AA / 103103 Bq"). Use the whole line: a half-life or a period read over several cycles or several halvings is more precise than one read over a single step. Draw a tangent long enough to give a gradient from a large triangle.

051015202530354045500100200300400500600700800900t / hA / Bq400 Bqt = 12 h
  • activity of the sample
The variation of the activity A of a radioactive sample with time t. The activity falls from 800 Bq to 400 Bq in 12 h, and halves again in each further 12 h.

Worked example

The graph shows the variation with time tt of the activity AA of a radioactive sample used in a laboratory.

(a) Determine, using the graph, the half-life of the sample. [2]

(b) The sample can be used only while its activity is at least 120 Bq. Deduce whether it can still be used 30 h after t=0t=0. [2]

Show the markscheme

(a) reading the time for the activity to fall from 800 Bq to 400 Bq «or from 400 Bq to 200 Bq» ✓

half-life = 12 «h» ✓

Accept 11–13 h. Award [1 max] for a bald answer with no evidence of the reading on the graph or in the working.

(b) 30 h is 2.5 half-lives, so A=800×(12)2.5≈140A=800×(21​)2.5≈140 «Bq» / reading A≈140A≈140 Bq at t=30t=30 h from the graph ✓

140 Bq is more than 120 Bq, so the sample can still be used ✓

Allow ECF from (a). Do not award MP2 for a bald conclusion.

Deduce and predict: linked marks

In a "deduce" or "predict" part the conclusion mark is usually linked to the reasoning mark. The markscheme awards the conclusion only if it follows from correct, or consistently carried-forward, working. A bald "yes, it can still be used" scores nothing in part (b), even though it is the right answer, because nothing shows how it was reached. Write the calculation or the reason first, then state the conclusion in words that answer the question exactly as it was asked.

Common mistake

Reading a graph at a single point near its start to find a half-life or a period, where a small reading error is a large fraction of the answer. Read across as many halvings or cycles as the graph shows and divide by their number.

Synoptic questions

Many Paper 2 questions combine two or more areas of physics, and the final extended-response question always does. A question on an electric scooter climbing a hill uses circuits, energy and forces; one on a radioactive source powering a cell uses decay, energy and current; one on a satellite uses gravitation, circular motion and energy. These are called synoptic questions.

They are not harder in each step than other questions. The difficulty is recognising which area of physics each part belongs to. Before answering a part, ask which quantity it wants and which equation contains it, then look back through the stem for the values. A list of the given data, with their symbols and SI values, written beside the stem at the start, saves time later.

Tip

Energy is the most common link between themes. When a part involves a change from one kind of situation to another, such as electrical to mechanical or nuclear to thermal, think of the energy transferred per second (power) and the efficiency of the transfer.

Worked question 1: a projectile

A stone is launched from ground level with a speed of 18.0 m s⁻¹ at an angle of 35° above the horizontal. Fluid resistance is negligible.

Worked example

(a) (i) Show that the initial vertical component of the stone’s velocity is about 10 m s⁻¹. [1]

(ii) Calculate the maximum height reached by the stone. [2]

(iii) Determine the horizontal distance travelled by the stone when it returns to ground level. [3]

(b) Determine, using energy considerations, the speed of the stone when it is 3.0 m above the ground. [2]

(c) In practice fluid resistance acts on the stone. Explain how this affects the maximum height and the horizontal distance travelled. [2]

Show the markscheme

(a)(i) uv=18.0sin⁡35∘=10.3uv​=18.0sin35∘=10.3 «m s⁻¹» ✓

Must see the substitution or an answer to at least 3 s.f.

(ii) v2=u2+2asv2=u2+2as with v=0v=0 at the top: 0=10.32−2×9.8×h0=10.32−2×9.8×h ✓

h=5.4h=5.4 «m» ✓

Allow 5.1 m from the use of 10 m s⁻¹. Award [2] for a bald correct answer.

(iii) time of flight t=2×10.39.8=2.1t=9.82×10.3​=2.1 «s» ✓

horizontal component =18.0cos⁡35∘=14.7=18.0cos35∘=14.7 «m s⁻¹» ✓

distance =14.7×2.1=31=14.7×2.1=31 «m» ✓

Allow ECF from MP1. Accept 30–31 m.

(b) 12mv2=12mu2−mgh21​mv2=21​mu2−mgh / v2=18.02−2×9.8×3.0v2=18.02−2×9.8×3.0 ✓

v=16v=16 «m s⁻¹» ✓

Accept 16.3 m s⁻¹. Award [1 max] for a method using kinematics with the vertical component only.

(c) fluid resistance acts opposite to the velocity / reduces both the horizontal and the vertical components of velocity ✓

so the maximum height and the horizontal distance are both smaller ✓

Do not award MP2 without some reasoning.

Part (b) shows why the command "using energy considerations" matters: the energy method needs only the speed and the height, because energy is a scalar. A kinematics method for the vertical motion alone gives only the vertical component, not the speed.

Worked question 2: heating water

An electric kettle has a power rating of 2.0 kW. It contains 0.80 kg of water at 18 °C. The specific heat capacity of water is 4200 J kg−1 K−14200 J kg−1 K−1 and its specific latent heat of vaporisation is 2.26×106 J kg−12.26×106 J kg−1.

Worked example

(a) Calculate the energy needed to raise the temperature of the water to 100 °C. [2]

(b) Show that the shortest time in which the kettle could heat the water to 100 °C is about 140 s. [1]

The kettle actually takes 160 s to bring the water to 100 °C.

(c) Determine the efficiency of the kettle. [2]

(d) The kettle fails to switch off. Estimate the mass of water that boils away in the next 60 s, stating one assumption you make. [2]

(e) Explain, with reference to the molecules, why the temperature of the water stays at 100 °C while it boils. [2]

Show the markscheme

(a) Q=mcΔT=0.80×4200×(100−18)Q=mcΔT=0.80×4200×(100−18) ✓

Q=2.8×105Q=2.8×105 «J» ✓

Award [2] for a bald correct answer. Accept 2.76 × 10⁵ J.

(b) t=QP=2.76×1052000=138t=PQ​=20002.76×105​=138 «s» ✓

Must see the substitution or an answer to at least 3 s.f.

(c) efficiency =useful energyenergy supplied=2.76×1052000×160=energy supplieduseful energy​=2000×1602.76×105​ ✓

= 0.86 / 86 % ✓

Allow 138/160. Allow ECF from (a).

(d) energy to the water =0.86×2000×60=1.0×105=0.86×2000×60=1.0×105 «J» and m=1.0×1052.26×106=0.046m=2.26×1061.0×105​=0.046 «kg» ✓

assumption: the efficiency is the same as during heating / all the energy reaching the water is used to vaporise it ✓

Allow 0.053 kg if 100 % efficiency is assumed and stated.

(e) the energy supplied increases the «intermolecular» potential energy of the molecules / is used to separate the molecules ✓

the mean kinetic energy of the molecules does not change, so the temperature is constant ✓

Part (d) is an estimate with a stated assumption. Either assumption is acceptable if it is stated and used consistently, which is typical of an "estimate" command. Part (e) needs both halves of the argument: what the energy does, and why that leaves the temperature unchanged.

Worked question 3: a synoptic question on an electric scooter

An electric scooter and its rider have a total mass of 95 kg. The scooter’s battery has an emf of 36 V and an internal resistance of 0.20 Ω. The scooter climbs a straight hill inclined at 4.0° to the horizontal at a constant speed of 5.0 m s⁻¹, and the current in the battery is 15 A.

Worked example

(a) Calculate the terminal potential difference of the battery. [2]

(b) Calculate the power delivered by the battery to the motor. [1]

(c) Show that the rate at which the gravitational potential energy of the scooter and rider increases is about 320 W. [2]

(d) The total resistive force on the scooter is 15 N. Determine the efficiency of the motor. [2]

At the top of the hill the scooter moves onto level ground. The rider brakes and stops from 5.0 m s⁻¹ in 2.5 s.

(e) Calculate the magnitude of the average braking force. [2]

(f) Explain why the terminal potential difference of the battery is lower when the current is larger. [2]

(g) Suggest why the scooter travels a shorter total distance on one charge on a hilly route than on a flat one, even though the scooter descends as far as it climbs. [2]

Show the markscheme

(a) V=ε−Ir=36−15×0.20V=ε−Ir=36−15×0.20 ✓

V=33V=33 «V» ✓

(b) P=VI=33×15=5.0×102P=VI=33×15=5.0×102 «W» ✓

Allow ECF from (a). Accept 495 W.

(c) vertical component of velocity =5.0sin⁡4.0∘=5.0sin4.0∘ «=0.349=0.349 m s⁻¹» ✓

rate of gain of Ep=mg×5.0sin⁡4.0∘=95×9.8×0.349=325Ep​=mg×5.0sin4.0∘=95×9.8×0.349=325 «W» ✓

Must see the substitution or an answer to at least 3 s.f.

(d) useful power output =325+15×5.0=400=325+15×5.0=400 «W» ✓

efficiency =400495=0.81=495400​=0.81 / 81 % ✓

Allow ECF from (b) and (c). Allow use of 320 W, giving 0.80.

(e) a=ΔvΔt=5.02.5=2.0a=ΔtΔv​=2.55.0​=2.0 «m s⁻²» ✓

F=ma=95×2.0=190F=ma=95×2.0=190 «N» ✓

Award [2] for a bald correct answer.

(f) there is a potential difference across the internal resistance equal to IrIr / energy is dissipated inside the battery ✓

IrIr is larger when II is larger, so V=ε−IrV=ε−Ir is smaller ✓

OWTTE

(g) energy is dissipated by resistive forces and by the brakes on the way down, so the gravitational potential energy gained is not recovered ✓

the current is larger when climbing, so more energy is dissipated in the internal resistance «power I2rI2r» / the motor efficiency is less than 100 % ✓

The scooter question moves between circuits (a, b, f), energy and power (c, d, g) and Newton’s second law (e). Each part on its own is routine; the skill is recognising which part of the course each one draws on. Part (c) shows the usual "show that" pattern: the true value, 325 W, rounds to the "about 320 W" in the question, and part (d) then uses it.

Exam tip

A "suggest" part, like (g), has several acceptable answers, and the markscheme lists them. Give as many distinct, physically reasoned points as there are marks, each tied to the situation described.

Common ways marks are lost

What happens

How to avoid it

A bald answer to a "show that" part

Show the full substitution and give one more significant figure than the target

The answer to a slightly different question

Reread the last line of the part before writing; check the quantity and the units asked for

One point given for a two-mark "discuss" or "compare"

Count the marks; make that many distinct points

Unconverted units (mm, kPa, °C) in the working

Convert every value to SI before substituting

An explanation that stops one step short

Finish the chain at the quantity the question names

A diagram without labels or with arrows of arbitrary length

Label every arrow and line; make relative lengths match the physics

A blank part because an earlier part went wrong

Use the "show that" value or a clearly stated estimate and carry on

Common mistake

Using a data-booklet equation because its symbols match the question, without checking that its physics applies. The kinematic equations need a constant acceleration; Ek=32kBTEk​=23​kB​T gives the mean energy of one molecule, not of the whole gas. Name the principle before choosing the equation.

Self-check

  • To do:

    I can use the mark allocation of a part to decide how much to write.

  • To do:

    I can set out a calculation as equation, substitution and answer with a unit.

  • To do:

    I can explain how error carried forward protects later marks, and use a "show that" value when I need it.

  • To do:

    I can answer a numerical "show that" with full substitution and an extra significant figure, and an algebraic one from a named principle.

  • To do:

    I can write an explanation as a chain of linked physics statements with one point per mark.

  • To do:

    I can distinguish outline, describe, explain and discuss.

  • To do:

    I can draw a free-body diagram with correctly placed, labelled arrows of sensible relative lengths.

  • To do:

    I can sketch a graph showing all the relevant features: labelled axes, intercepts, gradients and asymptotes.

  • To do:

    I can take readings from a graph or diagram, show where they came from, and write a deduce answer with reasoning before the conclusion.

  • To do:

    I can identify which area of physics each part of a synoptic question draws on.

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