Each major syllabus block is presented in a definition → formula → example → typical exam question format.
Key equations are collected in summary tables at the end of each block for quick revision.
Worked examples mirror the style of past Cambridge papers (AO1–AO3) and include step‑by‑step reasoning.
Practical‑skills tips (Paper 3 & 5) are provided in a dedicated appendix.
1. Quantities, Scalars & Vectors
1.1 Fundamental Quantities
Base units (SI): metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd).
Derived units are built from base units (e.g., newton = kg·m s⁻², joule = kg·m² s⁻²).
1.2 Scalars vs Vectors
Property
Scalar
Vector
Magnitude only
✓
✗
Direction required
✗
✓
Examples
mass, speed, energy
displacement, velocity, force
Notation
m, s, E
v, F, a (bold or arrow)
1.3 Vector Operations
Resultant of two vectors (graphical tip‑to‑tail method) or analytically using components.
Dot product \(\mathbf{A}\!\cdot\!\mathbf{B}=AB\cos\theta\) (used for work).
Cross product \(\mathbf{A}\!\times\!\mathbf{B}=AB\sin\theta\) (direction given by right‑hand rule, used for torque).
2. Kinematics – Motion in One and Two Dimensions
2.1 Motion Graphs
Displacement‑time, velocity‑time, and acceleration‑time graphs convey the same information; the slope of a \(v\)–\(t\) graph gives acceleration, the area under a \(v\)–\(t\) graph gives displacement, etc.
Range \(R = \dfrac{u^{2}\sin2\theta}{g}\) for launch speed \(u\) and angle \(\theta\).
Worked Example – Throwing a Ball Horizontally
A ball is thrown horizontally from a 1.5 m high table with a speed of 4.0 m s⁻¹. Find the time to reach the ground and the horizontal distance travelled.
\(N\) is the normal reaction; \(\mu{s},\mu{k}\) are coefficients (dimensionless).
Worked Example – Block on a Rough Incline
A 3.0 kg block rests on a 25° incline. \(\mu{s}=0.35\), \(\mu{k}=0.25\). Find the minimum force parallel to the plane needed to start the block moving up.
Component of weight down the plane: \(W_{\parallel}=mg\sin25^{\circ}=3.0\times9.8\times0.423=12.4\;\text{N}\).
Normal reaction: \(N=mg\cos25^{\circ}=3.0\times9.8\times0.906=26.6\;\text{N}\).
Maximum static friction: \(F{s}^{\max}= \mu{s}N=0.35\times26.6=9.3\;\text{N}\).
Point at which the total weight of a body may be considered to act. For uniform bodies CG = geometric centre; for irregular bodies locate experimentally (e.g., plumb‑line method).
Worked Example – Ladder Against a Smooth Wall
Diagram: ladder length \(L\), angle \(\theta\) with ground, weight \(W=mg\) acting at midpoint.
Translational equilibrium:
\(\sum F{x}: R = F{\text{wall}}\) (horizontal reaction at wall).
\(\sum F_{y}: N = W\) (normal reaction at ground).
Rotational equilibrium about the base:
\[
R\,L\cos\theta = W\,\frac{L}{2}\sin\theta
\;\Rightarrow\;
R = \frac{W}{2}\tan\theta
\]
Result: the wall experiences a horizontal force \(R\) and the ground a vertical reaction \(N=W\).
5. Work, Energy and Power
5.1 Work
Definition: Transfer of energy when a force causes a displacement.
Constant force: \(\displaystyle W = \mathbf{F}\!\cdot\!\mathbf{s}=Fs\cos\theta\).
Variable force: \(\displaystyle W = \int{si}^{s_f}\mathbf{F}\!\cdot\!d\mathbf{s}\).
Units: joule (J) = N·m.
5.2 Kinetic Energy
\[
E_{k}= \frac12 mv^{2}
\]
Work‑Energy Theorem: Net work done on a particle equals the change in its kinetic energy, \(\displaystyle W{\text{net}}=\Delta E{k}\).
Constant force at constant speed: \(P = Fv\cos\theta\).
Electrical form: \(P = IV = I^{2}R = \dfrac{V^{2}}{R}\).
Units: watt (W) = J s⁻¹.
5.6 Energy in Circular Motion (A‑Level preview)
Centrepetal force does no work (always perpendicular to displacement); kinetic energy remains constant if speed is constant.
For a mass \(m\) moving in a circle of radius \(r\) at speed \(v\): \(\displaystyle F_{c}= \frac{mv^{2}}{r}\).
Worked Example – Block on an Incline Pulled by a Horizontal Force
Block mass \(m=5.0\;\text{kg}\), smooth incline \(\alpha=30^{\circ}\), horizontal pull \(F=40\;\text{N}\). The block moves \(s=2.0\;\text{m}\) up the plane in \(t=4.0\;\text{s}\).
Resolve the horizontal force:
\[
F_{\parallel}=F\cos\alpha=34.6\;\text{N},\quad
F_{\perp}=F\sin\alpha=20.0\;\text{N}
\]
Since the plane is smooth, the only force parallel to the plane is the component of the pull and the component of weight \(mg\sin\alpha=24.5\;\text{N}\) down the plane.
Young’s Modulus \(\displaystyle Y = \frac{\sigma}{\varepsilon}\) (Pa).
6.2 Hooke’s Law (Elastic region)
\[
F = kx,\qquad k = \frac{Y A}{L}
\]
Worked Example – Stretching a Wire
A steel wire (length \(L=1.2\;\text{m}\), area \(A=2.0\times10^{-6}\;\text{m}^{2}\)) is loaded with a force of 500 N. Find the extension \(\Delta L\). (Young’s modulus for steel \(Y=2.0\times10^{11}\;\text{Pa}\)).
\[
\Delta L = \frac{FL}{YA}= \frac{500\times1.2}{2.0\times10^{11}\times2.0\times10^{-6}}=1.5\times10^{-3}\;\text{m}=1.5\;\text{mm}
\]
7. Waves – General Properties & Superposition
7.1 Wave Basics
Wave speed: \(\displaystyle v = f\lambda\) (where \(f\) is frequency, \(\lambda\) wavelength).
Transverse vs longitudinal waves.
Medium properties: tension \(T\) and linear mass density \(\mu\) for a string give \(v=\sqrt{T/\mu}\).
7.2 Superposition & Interference
When two waves occupy the same region, the resultant displacement is the algebraic sum of the individual displacements.
Loop rule (conservation of energy): \(\displaystyle \sum V = 0\) around any closed loop (taking signs according to direction).
Worked Example – Mixed Circuit
A 12 V battery supplies a series combination of 4 Ω and a parallel branch of 6 Ω and 12 Ω. Find the total current and the voltage across the 6 Ω resistor.
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