ESAT Physics Section 2027: Six Problem Types Worked

The ESAT Physics module is one of five modules in the Engineering and Science Admissions Test, a computer-based test set by UAT-UK and delivered through Pearson VUE test centres. It contains 27 multiple-choice questions in 40 minutes, with no calculator allowed. This page works through six recognisable Physics question types from the official specification — not topics — each solved by hand end to end, with the reasoning, the units and the specific error that costs marks on that type.

Every problem below either reproduces a calculation style set out in UAT-UK's ESAT Content Specification or is written by Leading Tuition in that style to illustrate it; none is copied from a live UAT-UK paper, and each is marked clearly below as one or the other. Where a number appears without a calculator, that is deliberate: the ESAT Physics module explicitly prohibits calculators, so every official question is built to resolve to a clean value by hand, and the six problems here are built the same way.

Problems Where You Read the Answer Off a Graph, Not a Formula

The specification requires candidates to "interpret distance–time, displacement–time, speed–time and velocity–time graphs" and to "perform calculations using gradients and areas under graphs" (P3.1e–f). That is a distinct skill from applying a named equation: the question gives you a shape, and the physics is in reading it correctly, not in recalling a formula. A large share of marks lost on this type comes from misreading which part of the graph answers the question — gradient for a rate, area for a total.

Our worked example. A cyclist's velocity–time graph shows three phases: uniform acceleration from rest to 8 m/s over the first 4 seconds, a constant 8 m/s for the next 6 seconds, then uniform deceleration back to rest over a final 2 seconds. Find (a) the acceleration during the first phase, (b) the total distance travelled, and (c) the average speed for the whole journey.

Step 1 — acceleration is the gradient. Over the first phase, acceleration = change in velocity ÷ time = (8 − 0) ÷ 4 = 2 m/s².

Step 2 — distance is the area under each phase. Phase 1 is a triangle: ½ × 4 × 8 = 16 m. Phase 2 is a rectangle: 6 × 8 = 48 m. Phase 3 is a triangle: ½ × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m.

Step 3 — average speed uses the totals, not the phase speeds. Average speed = total distance ÷ total time = 72 ÷ (4 + 6 + 2) = 72 ÷ 12 = 6 m/s.

The trap: average speed is not the mean of the speeds reached in each phase — averaging 0, 8 and 8 m/s, or splitting the difference between the maximum and zero, gives a plausible-looking wrong answer. The specification's own definition is explicit: "average speed = total distance ÷ time" (P3.1g), and there is no shortcut that skips finding the total distance first.

Problems With No Numbers Until the Very Last Line

Because no calculator is permitted, a distinct question type avoids awkward arithmetic altogether by giving no absolute numbers at all — only a ratio. The physics is in rearranging the relationship correctly; the answer is always a multiple, never a value in units. Two formulas from the specification produce this type repeatedly: Hooke's law energy (P3.3d) and Boyle's law (P5.2b).

Worked example 1. A spring obeying Hooke's law is stretched to three times its original extension. By what factor does (a) the force needed to hold it there and (b) the elastic potential energy stored change?

Step 1 — force scales directly with extension. Hooke's law is F = kx, so if extension goes from x to 3x, force goes from kx to k(3x) = 3(kx). Force increases by a factor of 3.

Step 2 — energy scales with extension squared. The specification gives elastic potential energy as E = ½kx². Substituting 3x for x: ½k(3x)² = ½k × 9x² = 9 × (½kx²). Energy increases by a factor of 9, not 3.

The trap is applying the force's scaling factor to the energy as well. Because F = kx is linear in x, it is easy to assume every spring quantity scales the same way; E = ½kx² does not, because x is squared.

Worked example 2. A fixed mass of gas at constant temperature is compressed until its volume is one quarter of its original volume. By what factor does its pressure change?

Boyle's law states PV = constant (P5.2b), so P&sub1;V&sub1; = P&sub2;V&sub2;. With V&sub2; = V&sub1; ÷ 4: P&sub1;V&sub1; = P&sub2; × (V&sub1; ÷ 4), which rearranges to P&sub2; = 4P&sub1;. The pressure quadruples. The trap here is direction: a smaller volume must mean a larger pressure, and a candidate working quickly under time pressure sometimes inverts the ratio and divides by 4 instead of multiplying.

Problems That Chain Two Energy Conversions in One Question

A distinct compound type links two of the specification's energy relationships in a single scenario — gravitational potential energy converting to kinetic energy, which is then removed by work done against a resistive force (P3.5, P3.7). The skill is recognising that the two stages share one energy total, so the answer to the second stage depends on correctly finishing the first.

Our worked example. A 2 kg trolley is released from rest at the top of a smooth (frictionless) slope and falls through a vertical height of 5 m to reach the bottom. It then crosses a rough horizontal surface, where a constant friction force of 20 N acts on it, and comes to rest after a distance d. Find d.

Step 1 — gravitational potential energy lost equals kinetic energy gained. The specification fixes gravitational field strength at g = 10 N/kg on Earth (P3.5b), not the 9.8 many candidates carry over from GCSE physics papers. GPE lost = mgh = 2 × 10 × 5 = 100 J. Because the slope is frictionless, all of this becomes kinetic energy at the bottom: KE = 100 J.

Step 2 — work done against friction removes that kinetic energy. Work = force × distance moved in the direction of the force (P3.7a). The trolley comes to rest, so the friction force must do 100 J of work over distance d: 20 × d = 100, giving d = 5 m.

The trap is the value of g. Using 9.8 N/kg instead of the specification's fixed 10 N/kg gives GPE = 98 J and d = 4.9 m — a different final answer, marked wrong against a mark scheme built on the specification's own value. UAT-UK states plainly that content assumes school-taught material, but the constants it fixes for the test are the ones to use in the test, not the more precise value taught elsewhere.

Problems Built on a Circuit With Series and Parallel Sections Together

The specification requires "the current and voltage rules for series and parallel circuits" together, including that "the total resistance of a parallel combination is less than that of any individual resistor" (P1.2i, k). The distinct problem type combines both in one circuit: a parallel section reduced to a single resistance, then treated as one component in a series calculation.

Our worked example. A 4 Ω resistor and a 12 Ω resistor are connected in parallel. This combination is connected in series with a 3 Ω resistor and a 12 V battery of negligible internal resistance. Find the total current drawn from the battery, the current through each resistor in the parallel section, and the power dissipated in the 3 Ω resistor.

Step 1 — reduce the parallel section first. 1/Rᶒ = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3, so Rᶒ = 3 Ω.

Step 2 — add it in series. Total resistance = 3 + 3 = 6 Ω. Using R = V/I (P1.2f) rearranged, total current I = V/R = 12/6 = 2 A.

Step 3 — find the voltage across the parallel section, then split the current. Voltage across the parallel section = I × Rᶒ = 2 × 3 = 6 V. Current through the 4 Ω resistor = 6/4 = 1.5 A; current through the 12 Ω resistor = 6/12 = 0.5 A. These sum to 2 A, matching the total current — a useful check.

Step 4 — power in the series resistor. The 3 Ω resistor carries the full 2 A, so power = I²R = 2² × 3 = 12 W (P1.2m).

The trap: more current flows through the smaller resistor, not the larger one, because current divides inversely with resistance in a parallel branch. Candidates under time pressure sometimes assign the larger share of current to the 12 Ω resistor because it is the "bigger number" on the page.

Problems That Stack Several Stages of Heating Into One Total

The specification separates specific heat capacity, "thermal energy = mass × specific heat capacity × temperature change" (P4.4b), from specific latent heat, which "carries no temperature-change term" during a change of state (P5.3c). A distinct problem type stacks both in sequence: heat a substance to its melting or boiling point, then change its state, and add the two energies.

Our worked example. 1 kg of water at 20°C is heated to its boiling point of 100°C and then completely turned into steam at 100°C. The specific heat capacity of water is 4,200 J/(kg·°C) and the specific latent heat of vaporisation of water is 2,260,000 J/kg. Find the total energy required.

Step 1 — heat the water to boiling point. Q&sub1; = mcΔT = 1 × 4,200 × (100 − 20) = 1 × 4,200 × 80 = 336,000 J.

Step 2 — boil the water at constant temperature. Q&sub2; = mL = 1 × 2,260,000 = 2,260,000 J. There is no ΔT term in this stage at all — the temperature does not change while the water boils.

Step 3 — add the two stages. Total energy = 336,000 + 2,260,000 = 2,596,000 J = 2,596 kJ.

The trap is treating the latent heat stage as if it were another specific-heat calculation and multiplying it by a temperature change that does not exist — either inventing a ΔT or reusing the 80°C from Step 1. Specific latent heat already accounts for the full energy needed to change state at a fixed temperature; multiplying it again by a temperature difference overstates the answer and misreads what the quantity means.

What Is Left After Several Half-Lives?

The specification requires candidates to "understand and be able to apply half-life calculations" (P7.4c). The distinct problem type asks for a quantity — activity, mass, or fraction remaining — after a whole number of half-lives, and the working is repeated halving, not a single formula substitution.

Our worked example. A radioactive isotope has a half-life of 12 hours and an initial activity of 800 counts per minute. Find its activity after 48 hours, and the fraction of the original undecayed nuclei remaining at that point.

Step 1 — find the number of half-lives elapsed. 48 ÷ 12 = 4 half-lives.

Step 2 — halve the activity once for each half-life. 800 → 400 → 200 → 100 → 50. After 4 half-lives, activity = 50 counts per minute.

Step 3 — express the fraction remaining as a power of one half. Fraction remaining = (½)⁴ = 1/16.

The trap is treating half-lives as additive rather than multiplicative — assuming that after 4 half-lives, 4 × 50% = 200% has decayed (which is impossible), or that exactly half the sample remains after any number of half-lives rather than after just the first one. The correct fraction after n half-lives is always (½)ⁿ, so after 4 half-lives, 1/16 of the original nuclei remain — not 1/8, and certainly not zero.

Leading Tuition provides specialist ESAT Physics tutoring for students preparing across every one of these problem types, matched to their target course's exact module requirements. Rated 4.8/5 on Trustpilot from 57 reviews, read on 19 August 2026, and 91% of our students achieve their desired grades.

Which Courses Actually Require ESAT Physics

Physics is not a free choice on every ESAT course. At Imperial College London, every Physics-linked course we checked in the current UAT-UK course list for 2027 entry — BSc Physics (F300), MSci Physics (F303), Physics with Theoretical Physics (F325/F390), Aeronautical Engineering (H401) and Mechanical Engineering (H300) — fixes the same module set: Mathematics 1, Mathematics 2 and Physics, with no choice involved. Oxford Physics (F303) and Physics and Philosophy (VF53) fix the identical combination. Oxford's Physics department is explicit that every applicant to those two courses "without exception" must sit the ESAT, a "two-hour test that evaluates a student's ability in both physics and maths", and describes the PAT it replaced as "our previous admissions test" — current Oxford Physics applicants sit the ESAT, not the PAT.

Cambridge treats Physics differently by course. BA/MEng Engineering (H100) fixes Mathematics 1, Mathematics 2 and Physics, exactly as Imperial and Oxford do. BA/MSc Natural Sciences and VetMB instead allow "Mathematics 1 + any two modules chosen from: Biology, Chemistry, Physics, Mathematics 2" — Physics is one option among four, and a Natural Sciences applicant with a strong Chemistry and Biology background could sit the ESAT without ever opening the Physics paper. Check your specific course's requirement before assuming Physics is compulsory.

Three of the seven institutions covered by UAT-UK admissions tests do not currently list an ESAT Physics requirement at all. In UAT-UK's own course list, Durham, the London School of Economics and Warwick require the TMUA, not the ESAT, for every course listed against them — the document states plainly for Durham that "the TMUA is recommended" and for Warwick that "the TMUA is compulsory", with no Physics module in sight. A claim that all seven ESAT universities require Physics does not hold up against the administrator's own current list; confirm your specific course on the official UAT-UK site before assuming either way.

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