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Physics Revision Guides

Mechanics and Motion

Year 1 / ASYear 2 / A-Level All Boards (AQA, Edexcel, OCR, WJEC, CCEA) AQA

A-Level Physics revision: Mechanics and Motion. Learning objectives, key points, worked examples and practice questions across AQA, Edexcel, OCR, WJEC and CCEA.

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📌 Key Points

Key Fact: SUVAT: v = u + at, s = ut + ½at^2, v^2 = u^2 + 2as, s = ½(u+v)t -- only for constant a
Key Fact: Newton's 1st: equilibrium -> net force zero; 2nd: F = ma (vector); 3rd: action-reaction pairs equal/opposite
Key Fact: Momentum p = mv; Impulse J = FDeltat = Deltap; Conservation: Σp_before = Σp_after (isolated system)
Key Fact: Work W = F·s = Fs cos θ; KE = ½mv^2; GPE = mgh; Elastic PE = ½kx^2; Power = W/t = Fv
Key Fact: Projectiles: horizontal v constant, vertical a = -g; range = u^2sin2θ/g, max height = u^2sin^2θ/(2g)
Key Fact: Circular: a = v^2/r = ω^2r, F = mv^2/r = mω^2r, ω = 2pi/T = 2pif, v = ωr
Key Fact: Moments: M = F x d (perpendicular distance); Couple: two equal/opposite forces separated by d
Key Fact: Equilibrium: ΣF = 0 and ΣM = 0; Centre of mass: weighted mean of positions
Key Fact: Hooke: F = kx (within limit of proportionality); EPE = ½kx^2 = ½Fx

🎯 Learning Objectives

  • Apply SUVAT equations for constant acceleration
  • Use Newton's three laws of motion for dynamics problems
  • Calculate momentum, impulse, and apply conservation of momentum
  • Analyse work, energy and power including conservation of energy
  • Solve projectile motion problems by separating horizontal/vertical components
  • Analyse circular motion: centripetal force, angular velocity, period
  • Understand moments, couples and equilibrium of rigid bodies
  • Apply Hooke's law and elastic potential energy

💡 Worked Example

Exam-Style Question

Question: A 0.5 kg ball is projected at 20 m/s at 30 deg to horizontal. Find maximum height, range, and speed at impact

Model Answer:

uₓ = 20cos30 deg = 10sqrt3 ≈ 17.32 m/s; uᵧ = 20sin30 deg = 10 m/s. Max height: vᵧ^2 = uᵧ^2 - 2gh -> 0 = 100 - 19.6h -> h = 5.10 m. Time of flight: t = 2uᵧ/g = 20/9.8 = 2.04 s. Range = uₓ x t = 17.32 x 2.04 = 35.3 m. At impact: vᵧ = -10 m/s, v = sqrt(17.32^2 + 10^2) = 20 m/s

❓ Practice Questions

Model answers are being added progressively - questions marked ✗ don't have one yet. Cross-check with your teacher or the official mark scheme.

Questions:

  • A car accelerates from rest at 3 m/s^2 for 8 s. Find final speed and distance✗ answer coming soon
  • A 1200 kg car hits a stationary 800 kg car. They couple. Find speed after impact✗ answer coming soon
  • A projectile launched at 45 deg has range 100 m. Find initial speed✗ answer coming soon
  • A 2 kg mass moves in a circle of radius 0.5 m at 4 m/s. Find centripetal force✗ answer coming soon
  • A uniform beam 4 m long, mass 20 kg, rests on supports at 0.5 m and 3.5 m from ends. Find reactions✗ answer coming soon

🎬 Video Resources

📄 Past Papers & Exam Resources

🔗 Further Reading & Resources

🧠 Flashcards (Spaced Repetition)

📝 Exam Questions by Topic

🎯 Target Tests (Auto-Graded)

🎓 Smart Lesson (Guided)