A-Level · 15 September 2026 · 7 min read
From IGCSE to A-Level Physics: a practical readiness check
Updated 25 September 2026
A strong IGCSE grade is encouraging, but it does not show which foundations still need attention before AS Physics. Algebra, units, graphs and careful explanations become more demanding when several steps appear in the same problem.
Use the original questions below to find useful preparation work. This is a learning check, not an admissions test or a grade prediction. The course guidance here is for students moving into Edexcel International AS Physics; a school’s own entry requirements remain a separate matter.
What should you prepare before the course begins?
Prioritise skills that connect many topics. If rearranging an equation takes all your attention, it becomes harder to think about the motion, circuit or wave in the question. If a graph is only a picture, you miss the physical meaning of its gradient and area.
You do not need to teach yourself the whole AS course in advance. Aim to make familiar IGCSE ideas usable without a worked solution beside you. Attempt these checks on paper first, then read the answers and explain each step aloud.
Check 1: rearrange before substituting
A moving object has kinetic energy E = ½mv². Rearrange for speed v, then find the speed when E = 18 J and m = 0.25 kg. Treat these as given values for this practice calculation.
Multiply by 2, divide by m, then take the square root: v = √(2E/m). Substitution gives v = √(36/0.25) = √144 = 12 m/s. For speed, choose the non-negative root. The error to watch for is taking the square root of only the energy while leaving the mass outside it.
Now change the question: with the same mass, what happens to the energy if speed doubles? It becomes four times as large because speed is squared. Being able to reason about the relationship is as useful as obtaining one numerical answer.
Check 2: resolve a vector using the stated angle
A force of 10 N acts at 30° above the horizontal. Find its horizontal and vertical components. Sketch the force as the diagonal of a right-angled triangle and label the angle before using trigonometry.
The horizontal component is adjacent to the stated angle: Fx = 10 cos 30° ≈ 8.66 N. The vertical component is opposite: Fy = 10 sin 30° = 5 N. Check that your calculator is in degree mode for this question.
If the angle had been measured from the vertical, the component expressions would change. ‘Horizontal means cosine’ is an unreliable rule; adjacent and opposite are defined by the angle in your diagram. An answer with both components larger than 10 N would also fail a quick sense check.
Check 3: read both the gradient and the area
A velocity–time graph rises in a straight line from 2 m/s at t = 0 to 8 m/s at t = 3 s. Find the acceleration and displacement during those three seconds.
The gradient is (8 − 2)/(3 − 0) = 2 m/s². The area is a trapezium: displacement = ½ × (2 + 8) × 3 = 15 m. You can also split it into a 6 m rectangle and a 9 m triangle. Both methods describe the same area.
The graph’s height gives velocity; it does not give distance. If velocity became negative, signed area would contribute negative displacement. Total distance would instead add the magnitudes of the areas on either side of the time axis.
Check 4: explain a change without contradicting yourself
A car travels in a straight line at constant speed on a level road. Is the forward driving force necessarily zero? No. It can balance the opposing resistive forces so that the resultant force is zero.
If resistance stays the same and the driving force increases, the resultant force becomes forward and the car initially accelerates. The explanation needs that chain: forces, resultant force, then change of velocity. Saying that a force ‘keeps the acceleration constant’ without checking what the resistance does goes beyond the information given.
Try the same reasoning on an object falling through air. Identify weight and drag, then explain how their changing balance affects acceleration. Avoid equating motion with a resultant force: an object can move with zero resultant force.
Turn the checks into a two-week bridging plan
Choose four manageable sessions each week, leaving room for school work and rest. A session can be 25–40 minutes if it ends with a checkable result. Spend longer on a weak foundation rather than rushing through every heading.
- —Week 1, session 1: rearrange six equations, including a square and a square root; explain each algebraic step.
- —Week 1, session 2: convert quantities to consistent units, including squared and cubed units.
- —Week 1, session 3: draw force diagrams and resolve vectors from angles measured in different directions.
- —Week 1, session 4: calculate gradients and areas, including their units and physical meanings.
- —Week 2, sessions 1–2: revisit two weak IGCSE topics and attempt fresh questions without the examples open.
- —Week 2, sessions 3–4: complete a short mixed set, explain mistakes and repeat similar questions after a gap.
What to do if several checks are difficult
Separate a forgotten method from a concept you never understood. If you can follow a rearrangement but cannot recreate it, practise short retrieval tasks. If you cannot explain why a force component uses a particular angle, ask for a diagram-based explanation before drilling more calculations.
Take your working to a teacher or tutor. ‘I got 12 m/s but do not understand why we take a square root’ is much more useful than ‘I am bad at physics’. For Edexcel AS tuition, share your school’s starting topics and the checks that were difficult so support can begin at the right point.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
Why do sine and cosine seem to swap in different force-component questions?
The component expressions depend on where the angle is measured. The component adjacent to that angle uses cosine; the opposite component uses sine. Draw the right-angled triangle before calculating instead of memorising that horizontal always means cosine.
For a 10 N force at 30° above the horizontal, the horizontal component is 10 cos 30° ≈ 8.66 N and the vertical component is 10 sin 30° = 5.0 N. If the same force is described as 60° from the vertical, the horizontal expression becomes 10 sin 60° and gives the same 8.66 N. The force has not changed; only the reference angle has.
Why is converting cm² to m² different from converting cm to m?
An area involves two lengths, so the length conversion factor must be squared. Since 1 cm = 0.01 m, 1 cm² = 0.0001 m². For volume, cube the factor: 1 cm³ = 0.000001 m³. Convert the unit as well as the number.
A square 3 cm by 3 cm has area 9 cm². In metres, its sides are 0.03 m, so its area is 0.03 × 0.03 = 0.0009 m². Dividing 9 by only 100 would give an area one hundred times too large. This matters when calculating pressure, density or wire resistivity, where an incorrect area or volume conversion can dominate an otherwise correct method.
How can a velocity–time graph give zero displacement when an object has moved?
Displacement uses signed area under a velocity–time graph, so movement in opposite directions can cancel. Distance adds the magnitudes of those areas. An object can return to its starting position with zero displacement while travelling a nonzero total distance.
If an object moves at +3 m/s for 2 s and then at −2 m/s for 3 s, its signed areas are +6 m and −6 m. Total displacement is 0 m, while total distance is 6 + 6 = 12 m. The average velocity over the five seconds is zero; the average speed is 12/5 = 2.4 m/s. This example makes the difference between scalar and vector averages concrete.