Cambridge International AS & A Level Physics (9702) – Core Revision Notes

Learning Objective

Recall and apply fundamental physics principles – especially the principle of conservation of energy – to a wide range of physical situations, and use quantitative reasoning to solve typical exam problems (AO1‑AO3).

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1. Physical Quantities & Units

Base SI Units

Quantity	Symbol	Unit
Length	m	metre (m)
Mass	kg	kilogram (kg)
Time	s	second (s)
Electric current	A	ampere (A)
Thermodynamic temperature	K	kelvin (K)
Amount of substance	mol	mole (mol)
Luminous intensity	cd	candela (cd)

Selected Derived Units

Quantity	Symbol	Expression (SI)
Velocity	v	m s⁻¹
Acceleration	a	m s⁻²
Force	F	N = kg m s⁻²
Energy	E	J = N m = kg m² s⁻²
Power	P	W = J s⁻¹
Pressure	p	Pa = N m⁻²
Electric charge	Q	C = A s
Potential difference	V	V = J C⁻¹
Resistance	R	Ω = V A⁻¹

Prefixes

milli‑ (10⁻³), centi‑ (10⁻²), kilo‑ (10³), mega‑ (10⁶), giga‑ (10⁹) etc.

Scalars vs Vectors

• Scalars: magnitude only (e.g. speed, mass, energy).
• Vectors: magnitude and direction (e.g. displacement \(\vec{s}\), velocity \(\vec{v}\), force \(\vec{F}\)).

Uncertainty & Significant Figures

Propagate uncertainties using the fractional (relative) method; report results with the appropriate number of significant figures (AO2).

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2. Kinematics

Definitions

• Displacement \(\vec{s}\) – vector, SI unit m.
• Velocity \(\vec{v}\) – rate of change of displacement, SI unit m s⁻¹.
• Acceleration \(\vec{a}\) – rate of change of velocity, SI unit m s⁻².

Uniform Acceleration (SUVAT) Equations

\(v = u + at\)

\(s = ut + \tfrac12 at^{2}\)

\(v^{2} = u^{2} + 2as\)

\(s = \tfrac12 (u+v)t\)

Graphical Interpretation

• Area under a velocity–time graph = displacement.
• Gradient of a displacement–time graph = velocity.
• Gradient of a velocity–time graph = acceleration.

Projectile Motion (near‑Earth)

\[ \begin{aligned} x &= vt \quad (\text{horizontal, } a=0)\\ y &= ut + \tfrac12 gt^{2} \quad (\text{vertical, } a=g) \end{aligned} \]

Worked Example

A stone is thrown vertically upward with initial speed \(u = 15\ \text{m s}^{-1}\). Find the maximum height and the total time of flight (ignore air resistance).

\[ \begin{aligned} v &= 0 = u - gt \;\Rightarrow\; t_{\text{up}} = \frac{u}{g}= \frac{15}{9.8}=1.53\ \text{s}\\[4pt] h_{\max} &= ut_{\text{up}}-\tfrac12 g t_{\text{up}}^{2}=15(1.53)-\tfrac12(9.8)(1.53)^{2}=11.5\ \text{m}\\[4pt] t_{\text{total}} &= 2t_{\text{up}} = 3.06\ \text{s} \end{aligned} \]

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3. Dynamics – Forces, Newton’s Laws & Momentum

Newton’s Laws

• 1st law: An object remains at rest or in uniform motion unless acted on by a net external force.
• 2nd law: \(\displaystyle \vec{F}_{\text{net}} = m\vec{a}\).
• 3rd law: For every action there is an equal and opposite reaction.

Forces (common types)

• Weight \(W = mg\).
• Normal reaction \(R\).
• Tension \(T\).
• Friction: \(f = \mu N\) (static \(\mu_{s}\), kinetic \(\mu_{k}\)).
• Spring force \(F = kx\) (Hooke’s law, elastic region).

Momentum

• Linear momentum \(\displaystyle \vec{p}=m\vec{v}\).
• Impulse–momentum theorem: \(\displaystyle \vec{J}= \int \vec{F}\,dt = \Delta\vec{p}\).
• Conservation of momentum (isolated system): \(\displaystyle \vec{p}_{i} = \vec{p}_{f}\).

Worked Example – Elastic Collision

A 0.20 kg ball moving at \(8.0\ \text{m s}^{-1}\) collides elastically with a stationary 0.30 kg ball. Find the speeds after collision.

\[ \begin{aligned} &\text{Momentum: }0.20(8)=0.20v_{1}+0.30v_{2}\\ &\text{Kinetic energy: }0.5(0.20)(8^{2})=0.5(0.20)v_{1}^{2}+0.5(0.30)v_{2}^{2}\\ &\Rightarrow v_{1}=2.0\ \text{m s}^{-1},\quad v_{2}=6.0\ \text{m s}^{-1} \end{aligned} \]

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4. Forces, Density & Pressure

Key Concepts

• Weight \(W=mg\).
• Pressure \(p = \dfrac{F}{A}\) (Pa). Hydrostatic pressure \(p = \rho gh\).
• Archimedes’ principle: buoyant force \(F_{B}= \rho_{\text{fluid}} V_{\text{sub}} g\).
• Torque \(\tau = rF\sin\theta\); equilibrium \(\sum \tau =0\), \(\sum \vec{F}=0\).

Worked Example – Floating Block

A rectangular block (density \(800\ \text{kg m}^{-3}\)) of dimensions \(0.10\times0.10\times0.20\ \text{m}\) floats in water (\(\rho_{\text{water}}=1000\ \text{kg m}^{-3}\)). What fraction of its height is submerged?

\[ \frac{\rho_{\text{block}}}{\rho_{\text{water}}}= \frac{h_{\text{sub}}}{h_{\text{total}}}\;\Rightarrow\; h_{\text{sub}} = \frac{800}{1000}\times0.20 =0.16\ \text{m} \]

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5. Work, Energy & Power (Expanded)

Work

• Definition: \(W = \vec{F}\cdot\vec{s}=Fs\cos\theta\) (J).
• Variable force: \(W = \displaystyle\int \vec{F}\cdot d\vec{s}\).
• Positive work increases kinetic energy; negative work decreases it.

Energy Forms (syllabus‑relevant)

Form	Expression	Typical Context
Kinetic energy	\(K=\tfrac12 mv^{2}\)	Moving bodies, projectiles
Gravitational potential energy	\(U_{g}=mgh\) (near Earth)	Raised or falling objects
Elastic potential energy	\(U_{e}=\tfrac12 kx^{2}\)	Springs, stretched strings
Thermal (internal) energy	Related to temperature; \(Q = mc\Delta T\)	Friction, resistive heating
Electrical energy	\(E = qV\) or \(E = Pt\)	Circuit work, batteries

Work–Energy Theorem

\[ W_{\text{net}} = \Delta K = K_{f}-K_{i} \]

Conservation of Mechanical Energy (conservative forces only)

\[ K_{i}+U_{i}=K_{f}+U_{f} \]

Non‑conservative Work

\[ K_{i}+U_{i}+W_{\text{nc}} = K_{f}+U_{f} \] where \(W_{\text{nc}}\) includes work done by friction, air resistance, etc.

Power

• General definition: \(P = \dfrac{dW}{dt}\) (W).
• For a constant force in the direction of motion: \(P = Fv\).
• Electrical: \(P = IV = I^{2}R = \dfrac{V^{2}}{R}\).

Efficiency

\[ \eta = \frac{\text{useful energy output}}{\text{energy input}}\times100\% \]

Worked Example – Energy Conservation with Friction

A 1.2 kg block slides down a \(30^{\circ}\) rough incline of length \(L=2.0\ \text{m}\). Coefficient of kinetic friction \(\mu_{k}=0.15\). Find the speed at the bottom.

\[ \begin{aligned} h &= L\sin30^{\circ}=1.0\ \text{m}\\ U_{i} &= mgh = 1.2(9.8)(1.0)=11.8\ \text{J}\\ N &= mg\cos30^{\circ}=1.2(9.8)\cos30^{\circ}=10.2\ \text{N}\\ W_{f} &= -\mu_{k}NL = -0.15(10.2)(2.0) = -3.1\ \text{J}\\ K_{f} &= U_{i}+W_{f}=8.7\ \text{J}\\ v &= \sqrt{\frac{2K_{f}}{m}} = \sqrt{\frac{2(8.7)}{1.2}}\approx 3.8\ \text{m s}^{-1} \end{aligned} \]

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6. Deformation of Solids

Stress & Strain

• Stress \(\sigma = \dfrac{F}{A}\) (Pa).
• Strain \(\epsilon = \dfrac{\Delta L}{L}\) (dimensionless).
• Hooke’s law (elastic region): \(\sigma = E\epsilon\), where \(E\) is Young’s modulus.

Energy Stored in a Spring

\[ U_{e}= \tfrac12 kx^{2} \] (derived from \(\displaystyle U_{e}= \int_{0}^{x} kx'\,dx'\)).

Worked Example – Stretching a Steel Rod

A steel rod (Young’s modulus \(E = 2.0\times10^{11}\ \text{Pa}\)) of length \(0.50\ \text{m}\) and cross‑section \(2.0\times10^{-4}\ \text{m}^{2}\) is stretched by \(1.0\ \text{mm}\). Find the applied force and the elastic energy stored.

\[ \begin{aligned} \epsilon &= \frac{1.0\times10^{-3}}{0.50}=2.0\times10^{-3}\\ \sigma &= E\epsilon = (2.0\times10^{11})(2.0\times10^{-3})=4.0\times10^{8}\ \text{Pa}\\ F &= \sigma A = (4.0\times10^{8})(2.0\times10^{-4}) = 8.0\times10^{4}\ \text{N}\\ U_{e} &= \tfrac12 F\Delta L = \tfrac12 (8.0\times10^{4})(1.0\times10^{-3}) = 40\ \text{J} \end{aligned} \]

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7. Waves

Basic Terminology

• Wave – disturbance that transfers energy without permanent displacement of the medium.
• Transverse vs longitudinal.
• Wave speed \(v = f\lambda\).
• Period \(T = 1/f\); angular frequency \(\omega = 2\pi f\).

Wave Equation (1‑D)

\[ \frac{\partial^{2}y}{\partial x^{2}} = \frac{1}{v^{2}}\frac{\partial^{2}y}{\partial t^{2}} \]

Key Phenomena (exam‑relevant)

Phenomenon	Condition / Formula
Reflection	Phase change of \(180^{\circ}\) at a fixed end; none at a free end.
Diffraction	Significant when aperture size \(\sim \lambda\).
Interference	Constructive: \(\Delta r = n\lambda\); Destructive: \(\Delta r = (n+\tfrac12)\lambda\).
Doppler shift	\(f' = f\frac{v\pm v_{o}}{v\pm v_{s}}\) (signs depend on motion direction).

Electromagnetic Spectrum (selected bands)

Region	Wavelength \(\lambda\)	Typical Uses
Radio	\(>10^{-1}\ \text{m}\)	Broadcasting, communications
Microwave	\(10^{-3}–10^{-1}\ \text{m}\)	Radar, cooking
Infrared	\(10^{-6}–10^{-3}\ \text{m}\)	Thermal imaging
Visible	\(4\times10^{-7}–7\times10^{-7}\ \text{m}\)	Human vision
Ultraviolet	\(10^{-8}–4\times10^{-7}\ \text{m}\)	Sterilisation
X‑ray	\(10^{-11}–10^{-8}\ \text{m}\)	Medical imaging
Gamma	\(<10^{-11}\ \text{m}\)	Nuclear processes

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8. Superposition – Standing Waves, Diffraction & Gratings

Standing Waves on a String (both ends fixed)

\[ L = n\frac{\lambda}{2}\quad (n=1,2,3\ldots) \] \[ f_{n}= n f_{1},\qquad f_{1}= \frac{v}{2L} \]

Standing Waves in Air Columns

• Both ends open: same condition as a string.
• Closed at one end: \(L = (2n-1)\frac{\lambda}{4}\) (n = 1,2,3…).

Diffraction Grating Equation

\[ d\sin\theta = n\lambda\qquad (n=0,\pm1,\pm2\ldots) \] where \(d\) is the grating spacing.

Worked Example – Fundamental Frequency of a 0.75 m String

Given wave speed \(v = 120\ \text{m s}^{-1}\), find the fundamental frequency.

\[ \lambda_{1}=2L = 1.5\ \text{m},\qquad f_{1}= \frac{v}{\lambda_{1}} = \frac{120}{1.5}=80\ \text{Hz} \]

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9. Electricity

Fundamental Quantities

• Current \(I\) – charge flow per unit time, \(I = \dfrac{Q}{t}\) (A).
• Potential difference \(V\) – energy per unit charge, \(V = \dfrac{W}{Q}\) (V).
• Resistance \(R\) – opposition to current, \(R = \dfrac{V}{I}\) (Ω).
• Resistivity \(\rho\) – material property, \(R = \rho\frac{L}{A}\).

Key Relations

\[ \begin{aligned} V &= IR \\ P &= IV = I^{2}R = \frac{V^{2}}{R} \\ E &= Pt = VIt \end{aligned} \]

Energy in Electrical Circuits

Energy supplied by a source: \(E = V I t\). Energy dissipated as heat in a resistor: \(E = I^{2}Rt\).

Worked Example – Power Rating of a Heater

A 240 V mains supply powers a resistive heater drawing 10 A. Find the power consumed and the energy used in 2 h.

\[ P = VI = 240\times10 = 2400\ \text{W}=2.4\ \text{kW} \] \[ E = Pt = 2.4\ \text{kW}\times2\ \text{h}=4.8\ \text{kWh} \]

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10. D.C. Circuits

Series & Parallel Connections

• Series: \(R_{\text{eq}} = \sum R\), same current, total voltage is sum.
• Parallel: \(\dfrac{1}{R_{\text{eq}}}= \sum \dfrac{1}{R}\), same voltage, total current is sum.

Kirchhoff’s Laws

• Current Law (KCL): \(\displaystyle \sum I_{\text{in}} = \sum I_{\text{out}}\) at any junction.
• Voltage Law (KVL): \(\displaystyle \sum V_{\text{drops}} = \sum V_{\text{rises}}\) around any closed loop.

Potential Divider

\[ V_{x}= V_{\text{s}}\frac{R_{x}}{R_{1}+R_{2}} \]

Internal Resistance of a Cell

\[ V = \mathcal{E} - Ir_{\text{int}} \]

Worked Example – Current Through a Parallel Network

A 12 V battery (emf \(\mathcal{E}=12\ \text{V}\), internal resistance \(r=0.5\ \Omega\)) supplies two parallel branches: \(R_{1}=4\ \Omega\) and \(R_{2}=6\ \Omega\). Find the total current delivered by the battery.

\[ R_{\text{eq}} = \left(\frac{1}{4}+\frac{1}{6}\right)^{-1}= \frac{12}{5}=2.4\ \Omega \] \[ R_{\text{total}} = R_{\text{eq}}+r = 2.4+0.5 = 2.9\ \Omega \] \[ I = \frac{\mathcal{E}}{R_{\text{total}}}= \frac{12}{2.9}=4.14\ \text{A} \]

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11. Particle Physics

Structure of the Atom

• Protons (\(+e\)), neutrons (neutral), electrons (\(-e\)).
• Nuclear notation: \(\displaystyle _{Z}^{A}\text{X}\) where \(A\) = mass number, \(Z\) = atomic number.

Radioactivity

Decay type	Equation	Particle emitted	Change in \(A\) and \(Z\)
Alpha (\(\alpha\))	\(_{Z}^{A}\text{X}\rightarrow\;_{Z-2}^{A-4}\text{Y}+\,_{2}^{4}\alpha\)	\(\alpha\) (He nucleus)	\(A-4,\;Z-2\)
Beta minus (\(\beta^{-}\))	\(_{Z}^{A}\text{X}\rightarrow\;_{Z+1}^{A}\text{Y}+\,_{-1}^{0}\beta^{-}\)	electron	\(A,\;Z+1\)
Beta plus (\(\beta^{+}\))	\(_{Z}^{A}\text{X}\rightarrow\;_{Z-1}^{A}\text{Y}+\,_{+1}^{0}\beta^{+}\)	positron	\(A,\;Z-1\)
Gamma (\(\gamma\))	\(_{Z}^{A*}\text{X}\rightarrow\;_{Z}^{A}\text{X}+\,\gamma\)	photon	none (energy only)

Binding Energy

\[ E_{\text{b}} = \left[Zm_{p}+(A-Z)m_{n}-m_{\text{nucleus}}\right]c^{2} \]

Worked Example – Alpha Decay Energy

Calculate the energy released when \(_{92}^{238}\text{U}\) undergoes alpha decay to \(_{90}^{234}\text{Th}\). Masses: \(m_{\text{U}}=238.0508\ \text{u}\), \(m_{\text{Th}}=234.0436\ \text{u}\), \(m_{\alpha}=4.0026\ \text{u}\). (1 u = 931.5 MeV c⁻².)

\[ \Delta m = m_{\text{U}}-(m_{\text{Th}}+m_{\alpha}) = 238.0508-(234.0436+4.0026)=0.0046\ \text{u} \] \[ Q = \Delta m \times 931.5\ \text{MeV}=4.3\ \text{MeV} \]

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A‑Level Add‑on (Topics 12‑25)

12. Circular Motion & Gravitation

• Centripetal force: \(F_{c}= \dfrac{mv^{2}}{r}=m\omega^{2}r\).
• Universal gravitation: \(F_{g}= G\frac{m_{1}m_{2}}{r^{2}}\).
• Orbital energy: \(E = -\dfrac{GMm}{2r}\) (bound circular orbit).

Example: A satellite of mass 500 kg orbits Earth at an altitude of 300 km. Find the speed.

\[ r = R_{\oplus}+h = 6.37\times10^{6}+3.0\times10^{5}=6.67\times10^{6}\ \text{m} \] \[ v = \sqrt{\frac{GM_{\oplus}}{r}} = \sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6.67\times10^{6}}}=7.73\ \text{km s}^{-1} \]

13. Temperature, Ideal Gases & Thermodynamics

• Ideal‑gas equation: \(pV = nRT\).
• Kinetic theory (average kinetic energy): \(\tfrac32 k_{B}T\).
• First law: \(\Delta U = Q - W\).
• Efficiency of a heat engine: \(\displaystyle \eta = 1-\frac{T_{C}}{T_{H}}\) (Kelvin).

Example: 1 mol of an ideal gas expands isothermally from 1.0 L to 3.0 L at 300 K. Find the work done.

\[ W = nRT\ln\frac{V_{f}}{V_{i}} = (1)(8.314)(300)\ln\frac{3.0}{1.0}= 2.74\ \text{kJ} \]

14. Oscillations (Simple Harmonic Motion)

• Displacement: \(x = A\cos(\omega t + \phi)\).
• Angular frequency: \(\omega = \sqrt{\dfrac{k}{m}}\) (mass‑spring) or \(\omega = \sqrt{\dfrac{g}{L}}\) (simple pendulum, small angles).
• Energy: \(K = \tfrac12 m\omega^{2}(A^{2}-x^{2})\), \(U = \tfrac12 kx^{2}\).

Example: A 0.5 kg mass on a spring (k = 200 N m⁻¹) is set into SHM with amplitude 0.10 m. Find the maximum speed.

\[ \omega = \sqrt{\frac{k}{m}} = \