define and use distance, displacement, speed, velocity and acceleration

1 Physical Quantities, Units & Measurement

1.1 SI Base and Derived Units

1.2 Prefixes

Common SI prefixes are listed below. They are used to express very large or very small quantities.

1.3 Uncertainty & Significant Figures (AO2)

• Read the instrument to the smallest division and add ±1 unit as the absolute uncertainty.
• Relative (fractional) uncertainty = Δx / x.
• When combining quantities:

  • For addition/subtraction, add absolute uncertainties.
  • For multiplication/division, add relative uncertainties.

• Result should be reported to the same number of decimal places (or sig‑figs) as the least precise measurement.

2 Kinematics – Distance, Displacement, Speed, Velocity & Acceleration

2.1 Fundamental Definitions

2.2 Key Relationships

• Speed is the magnitude of velocity: \(v = |\vec{v}|\).
• Acceleration is the time‑derivative of velocity: \(\displaystyle \vec{a}= \frac{d\vec{v}}{dt}\).
• Velocity is the time‑derivative of displacement: \(\displaystyle \vec{v}= \frac{d\vec{s}}{dt}\).
• Speed is the time‑derivative of distance: \(\displaystyle v = \frac{ds}{dt}\).

2.3 Motion Graphs (AO1)

• s–t graph (position vs. time): gradient = instantaneous velocity (sign indicates direction). Area under the curve has no direct physical meaning.
• v–t graph** (velocity vs. time): gradient = acceleration; area under the curve = displacement (signed).
• a–t graph (acceleration vs. time): gradient = rate of change of acceleration (jerk); area under the curve = change in velocity.

Example: Linear v–t graph

From Figure 1:

• Gradient \(= \dfrac{\Delta v}{\Delta t}= \dfrac{20-0}{10-0}=2.0\;\text{m s}^{-2}\) → acceleration.
• Area (right‑angled triangle) \(= \tfrac12\times10\;\text{s}\times20\;\text{m s}^{-1}=100\;\text{m}\) → displacement.

2.4 Constant‑Acceleration Equations (Derivation)

2.5 Summary of Kinematic Equations (AO2)

2.6 Worked Example – Uniform Acceleration

Problem: A car starts from rest and accelerates uniformly at \(2.0\;\text{m s}^{-2}\) for \(10\;\text{s}\). Determine (i) final velocity, (ii) distance travelled, (iii) average speed.

1. \(\displaystyle \vec{v}=0+(2.0)(10)=20.0\;\text{m s}^{-1}\)
2. \(\displaystyle \Delta\vec{s}=0+\tfrac12(2.0)(10)^{2}=100\;\text{m}\)
3. Average speed \(= \dfrac{100\;\text{m}}{10\;\text{s}}=10.0\;\text{m s}^{-1}\).

2.7 Practice Problems (AO1 + AO2)

1. Free‑fall from rest (height = 45 m, \(g=9.81\;\text{m s}^{-2}\)):

   • Time to reach the ground.
   • Velocity just before impact.
   • Average speed during the fall.

   Hint: Use \(\Delta s = \tfrac12 g t^{2}\) with \(\Delta s = 45\;\text{m}\) (downward positive).

2. Horizontal projectile (table height = 1.20 m, horizontal speed = 4.5 m s⁻¹):

   • Time to reach the floor.
   • Horizontal distance travelled.
   • Resultant speed just before impact.

   Hint: Treat vertical and horizontal motions separately.

2.8 Experiment: Determining \(g\) (Free‑Fall)

Objective: Measure the acceleration due to gravity by analysing a v–t graph.

Apparatus

• Digital stop‑watch or motion‑sensor system (e.g. PASCO photogates).
• Metal ball (dense sphere).
• Rigid vertical guide.
• Meter ruler for height measurement.

Method (outline)

1. Measure the vertical distance \(h\) from release point to detector.
2. Release the ball from rest; record the time for several equally spaced distances (e.g. every 0.20 m).
3. Calculate instantaneous velocities \(v = 2h/t\) (or use sensor output) and plot \(v\) against \(t\).
4. The gradient of the straight‑line fit gives the experimental value of \(g\).

Sources of error

• Air resistance – minimise with a dense sphere.
• Reaction time (if manual timing) – use electronic sensors.
• Guide mis‑alignment causing sideways motion.

Typical result: \(g_{\text{exp}} = (9.8 \pm 0.2)\;\text{m s}^{-2}\).

3 Dynamics – Forces, Momentum & Non‑Uniform Motion

3.1 Newton’s Laws of Motion

• First law (inertia): An object remains at rest or in uniform straight‑line motion unless acted on by a net external force.
• Second law: \(\displaystyle \vec{F}_{\text{net}} = m\vec{a}\). The net force is proportional to the mass and the acceleration.
• Third law: For every action there is an equal and opposite reaction.

3.2 Forces and Free‑Body Diagrams (FBDs)

Common forces (vector quantities) to include in an FBD:

• Weight \(\vec{W}=mg\) (downwards).
• Normal reaction \(\vec{N}\) (perpendicular to the surface).
• Applied force \(\vec{F}_{\text{app}}\) (direction as given).
• Friction \(\vec{f}\) – kinetic: \(\displaystyle fk = \muk N\); static: \(\displaystyle fs \le \mus N\).
• Tension \(\vec{T}\) (along a string or rope).
• Air‑drag (often approximated as \(f_d = \tfrac12 C\rho A v^{2}\)).

Example FBD – Block on an inclined plane

3.3 Momentum, Impulse & Collisions

• Linear momentum: \(\displaystyle \vec{p}=m\vec{v}\) (kg m s⁻¹).
• Impulse: \(\displaystyle \vec{J}= \int \vec{F}\,dt = \Delta\vec{p}\).
• Conservation of momentum (isolated system): \(\displaystyle \sum\vec{p}{\text{initial}} = \sum\vec{p}{\text{final}}\).
• Elastic collision – kinetic energy also conserved.
• Inelastic collision – kinetic energy not conserved; momentum still is.

3.4 Non‑Uniform Acceleration

When acceleration varies with time, use calculus or the area‑under‑curve method on v–t graphs:

• Acceleration at any instant: gradient of v–t graph.
• Change in velocity: area under a–t graph.
• Displacement: area under v–t graph (signed).

3.5 Worked Example – Block on a Rough Incline

A 2.0 kg block slides down a 30° incline with coefficient of kinetic friction \(\mu_k = 0.15\). Find the acceleration.

1. Weight components: \(W_{\parallel}=mg\sin30° = 2.0\times9.81\times0.5 = 9.81\;\text{N}\).
2. Normal reaction: \(N = mg\cos30° = 2.0\times9.81\times0.866 = 16.98\;\text{N}\).
3. Friction: \(fk = \muk N = 0.15\times16.98 = 2.55\;\text{N}\) (up the plane).
4. Net force down the plane: \(F{\text{net}} = W{\parallel} - f_k = 9.81 - 2.55 = 7.26\;\text{N}\).
5. Acceleration: \(a = F_{\text{net}}/m = 7.26/2.0 = 3.63\;\text{m s}^{-2}\).

4 Forces, Density & Pressure

4.1 Density

Density \(\rho = \dfrac{m}{V}\) (kg m⁻³). Useful for relating mass and volume in buoyancy problems.

4.2 Pressure

• Definition: \(p = \dfrac{F}{A}\) (Pa = N m⁻²).
• Hydrostatic pressure in a fluid of density \(\rho\): \(\displaystyle p = p_0 + \rho g h\) where \(h\) is depth.
• Pascal’s principle – pressure applied to a confined fluid is transmitted unchanged in all directions.

4.3 Archimedes’ Principle

The upward buoyant force on a body immersed in a fluid equals the weight of the displaced fluid:

\[

F{\text{b}} = \rho{\text{fluid}} g V_{\text{displaced}}

\]

4.4 Worked Example – Floating Block

A wooden block of density \(600\;\text{kg m}^{-3}\) is placed 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}}}\;\Longrightarrow\; \frac{600}{1000}=0.60.

\]

Thus 60 % of the block’s height is below the surface.

5 Work, Energy & Power

5.1 Work

Work done by a constant force \(\vec{F}\) through a displacement \(\vec{s}\) at an angle \(\theta\):

\[

W = F s \cos\theta = \vec{F}\cdot\vec{s}\quad\text{(Joules)}.

\]

For variable forces, \(W = \int \vec{F}\cdot d\vec{s}\).

5.2 Kinetic & Gravitational Potential Energy

• Kinetic energy: \(\displaystyle E_k = \tfrac12 m v^{2}\).
• Gravitational potential energy (near Earth’s surface): \(\displaystyle E_p = m g h\).

5.3 Work–Energy Theorem

\[

W{\text{net}} = \Delta Ek = \tfrac12 m v^{2} - \tfrac12 m v_0^{2}.

\]

5.4 Power

\[

P = \frac{W}{t} = \vec{F}\cdot\vec{v}\quad\text{(W)}.

\]

5.5 Conservation of Mechanical Energy

If only conservative forces do work, total mechanical energy is constant:

\[

E{\text{total}} = Ek + E_p = \text{constant}.

\]

5.6 Efficiency

\[

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

\]

5.7 Worked Example – Inclined Plane

A 5.0 kg crate is pulled up a 4.0 m long, 30° incline at constant speed. The coefficient of kinetic friction is 0.20. Find the required pulling force and the power if the motion takes 5.0 s.

1. Weight component down the plane: \(W_{\parallel}=mg\sin30° = 5.0\times9.81\times0.5 = 24.5\;\text{N}\).
2. Normal reaction: \(N = mg\cos30° = 5.0\times9.81\times0.866 = 42.5\;\text{N}\).
3. Friction: \(fk = \muk N = 0.20\times42.5 = 8.5\;\text{N}\).
4. Since speed is constant, net force = 0, so pulling force \(F = W{\parallel}+fk = 24.5+8.5 = 33.0\;\text{N}\).
5. Work done: \(W = Fs = 33.0\times4.0 = 132\;\text{J}\).
6. Power: \(P = W/t = 132/5.0 = 26.4\;\text{W}\).

6 Deformation of Solids

6.1 Stress & Strain

• Normal stress: \(\displaystyle \sigma = \frac{F}{A}\) (Pa).
• Strain (dimensionless): \(\displaystyle \varepsilon = \frac{\Delta L}{L_0}\).

6.2 Hooke’s Law

\[

\sigma = E\,\varepsilon \quad\text{or}\quad F = k\,x,

\]

where \(E\) is Young’s modulus (Pa) and \(k\) is the spring constant (N m⁻¹).

6.3 Energy Stored in a Spring

\[

E_{\text{spring}} = \tfrac12 k x^{2}.

\]

6.4 Elastic vs. Plastic Behaviour

Elastic deformation is reversible; plastic deformation is permanent. The proportional limit marks the end of the linear (Hooke’s law) region.

6.5 Worked Example – Spring‑loaded Door

A door is held closed by a spring with \(k = 150\;\text{N m}^{-1}\). The spring is compressed 0.20 m when the door is shut. Find the force exerted on the door and the energy stored.

• Force: \(F = kx = 150\times0.20 = 30\;\text{N}\).
• Energy: \(E = \tfrac12 kx^{2} = 0.5\times150\times0.20^{2}=3.0\;\text{J}\).

7 Waves

7.1 General Wave Properties

• Wave speed: \(\displaystyle v = f\lambda\) where \(f\) is frequency and \(\lambda\) wavelength.
• Period: \(T = 1/f\).
• Amplitude: maximum displacement from equilibrium.
• Transverse vs. longitudinal: particle motion ⟂ or ∥ to direction of propagation.

7.2 Wave Types in the Syllabus

• Mechanical waves (sound, seismic, string).
• Electromagnetic (EM) waves – travel at \(c = 3.00\times10^{8}\;\text{m s}^{-1}\) in vacuum.
• Light as an EM wave – visible spectrum 400–700 nm.

7.3 Reflection, Refraction & Diffraction

• Law of reflection: angle of incidence = angle of reflection.
• Snell’s law: \(\displaystyle n1\sin\theta1 = n2\sin\theta2\).
• Diffraction grating equation: \(\displaystyle d\sin\theta = n\lambda\) (where \(d\) is grating spacing).

7.4 Doppler Effect

\[

f' = f\frac{v\pm vo}{v\pm vs},

\]

where signs depend on whether source/observer move towards or away from each other.

7.5 Superposition (see Section 8)

7.6 Worked Example – Sound Speed in Air

Given temperature \(T = 20^{\circ}\text{C}\), estimate the speed of sound using \(v = 331 + 0.6T\) (m s⁻¹).

\[

v = 331 + 0.6\times20 = 343\;\text{m s}^{-1}.

\]

8 Superposition – Interference & Diffraction

8.1 Principle of Superposition

When two or more waves occupy the same region, the resultant displacement at any point is the algebraic sum of the individual displacements.

8.2 Constructive & Destructive Interference

• Constructive: path difference = integer multiples of \(\lambda\) → bright fringe.
• Destructive: path difference = \((2n+1)\lambda/2\) → dark fringe.

8.3 Young’s Double‑Slit Experiment

Fringe spacing:

\[

y = \frac{\lambda D}{d},

\]

where \(D\) is the distance to the screen and \(d\) the slit separation.

8.4 Diffraction Grating

Condition for principal maxima:

\[

d\sin\theta = n\lambda\qquad (n = 0, \pm1, \pm2,\dots)

\]

8.5 Worked Example – Grating Spectrum

A diffraction grating has 5000 lines mm⁻¹. Find the angle for the first‑order maximum of red light (\(\lambda = 650\;\text{nm}\)).

• Grating spacing: \(d = 1/(5.0\times10^{6})\;\text{m}=2.0\times10^{-7}\;\text{m}\).
• \(\sin\theta = n\lambda/d = (1)(650\times10^{-9})/(2.0\times10^{-7}) = 0.325\).
• \(\theta = \arcsin(0.325) \approx 19^{\circ}\).

9 Electricity – Charge, Current & Potential

9.1 Fundamental Concepts

• Charge (\(Q\)) – unit coulomb (C). Electron charge \(e = 1.60\times10^{-19}\;\text{C}\).
• Current (\(I\)) – rate of flow of charge: \(I = \dfrac{dQ}{dt}\) (A).
• Potential difference (\(V\)) – energy per unit charge: \(V = \dfrac{W}{Q}\) (V).
• Resistance (\(R\)) – opposition to current: \(R = \dfrac{V}{I}\) (Ω).

9.2 Ohm’s Law

\[

V = IR.

\]

9.3 Power in Electrical Circuits

\[

P = IV = I^{2}R = \frac{V^{2}}{R}.

\]

9.4 Series & Parallel Circuits

• Series: \(I\) same, \(V{\text{total}} = \sum Vi\), \(R{\text{total}} = \sum Ri\).
• Parallel: \(V\) same, \(\dfrac{1}{R{\text{total}}} = \sum \dfrac{1}{Ri}\), \(I{\text{total}} = \sum Ii\).

9.5 Kirchhoff’s Laws (AO2)

• Current law (KCL): Sum of currents entering a junction = sum leaving.
• Voltage law (KVL): Sum of potential differences around any closed loop = 0.

9.6 Worked Example – Mixed Circuit

A 12 V battery supplies a series combination of a 4 Ω resistor and a parallel network of 6 Ω and 12 Ω resistors. Find the total current.

1. Parallel equivalent: \(\displaystyle \frac{1}{Rp}= \frac{1}{6}+\frac{1}{12}= \frac{2+1}{12}= \frac{3}{12}\Rightarrow Rp=4\;\Omega.\)
2. Total resistance: \(R_{\text{tot}} = 4\;(\text{series}) + 4 = 8\;\Omega.\)
3. Total current: \(I = V/R_{\text{tot}} = 12/8 = 1.5\;\text{A}.\)

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10 DC Circuits – Energy, Power & Practical Components

10.1 Potential Divider

For two series resistors \(R1\) and \(R2\) across a supply \(V\), the voltage across \(R_2\) is:

\[

V{R2}= V\frac{R2}{R1+R_2}.

\]

10.2 Capacitors (optional for AS)

• Capacitance: \(C = \dfrac{Q}{V}\) (F).
• Energy stored: \(E = \tfrac12 CV^{2}\).
• In series: \(\displaystyle \frac{1}{C{\text{eq}}}= \sum \frac{1}{Ci}\). In parallel: \(C{\text{eq}} = \sum Ci\).

10.3 Wheatstone Bridge (A‑Level)

Balance condition: \(\displaystyle \frac{R1}{R2} = \frac{R3}{R4}\).

10.4 Worked Example – Voltage Divider

A 10 V source feeds a divider made of 2 kΩ and 3 kΩ resistors. Find the voltage across the 3 kΩ resistor.

\[

V_{3k}=10\frac{3}{2+3}=10\times0.6=6.0\;\text{V}.

\]

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11 Particle Physics – Fundamental Particles & Decays

11.1 Standard Model Overview

• Fermions (matter): 12 quarks (u, d, c, s, t, b) and 6 leptons (e, μ, τ and their neutrinos).
• Bosons (force carriers): photon (γ), gluon (g), W⁺/W⁻, Z⁰, Higgs (H).