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).
| 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) |
| 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 |
milli‑ (10⁻³), centi‑ (10⁻²), kilo‑ (10³), mega‑ (10⁶), giga‑ (10⁹) etc.
Propagate uncertainties using the fractional (relative) method; report results with the appropriate number of significant figures (AO2).
| \(v = u + at\) |
| \(s = ut + \tfrac12 at^{2}\) |
| \(v^{2} = u^{2} + 2as\) |
| \(s = \tfrac12 (u+v)t\) |
\[
\begin{aligned}
x &= vt \quad (\text{horizontal, } a=0)\\
y &= ut + \tfrac12 gt^{2} \quad (\text{vertical, } a=g)
\end{aligned}
\]
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).
\[
\begin{aligned}
v &= 0 = u - gt \;\Rightarrow\; t_{\text{up}} = \frac{u}{g}= \frac{15}{9.8}=1.53\ \text{s}\\[4pt]
h{\max} &= ut{\text{up}}-\tfrac12 g t_{\text{up}}^{2}=15(1.53)-\tfrac12(9.8)(1.53)^{2}=11.5\ \text{m}\\[4pt]
t{\text{total}} &= 2t{\text{up}} = 3.06\ \text{s}
\end{aligned}
\]
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.
\[
\begin{aligned}
&\text{Momentum: }0.20(8)=0.20v{1}+0.30v{2}\\
&\text{Kinetic energy: }0.5(0.20)(8^{2})=0.5(0.20)v{1}^{2}+0.5(0.30)v{2}^{2}\\
&\Rightarrow v{1}=2.0\ \text{m s}^{-1},\quad v{2}=6.0\ \text{m s}^{-1}
\end{aligned}
\]
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?
\[
\frac{\rho{\text{block}}}{\rho{\text{water}}}= \frac{h{\text{sub}}}{h{\text{total}}}\;\Rightarrow\;
h_{\text{sub}} = \frac{800}{1000}\times0.20 =0.16\ \text{m}
\]
| 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 |
\[
W{\text{net}} = \Delta K = K{f}-K_{i}
\]
\[
K{i}+U{i}=K{f}+U{f}
\]
\[
K{i}+U{i}+W{\text{nc}} = K{f}+U_{f}
\]
where \(W_{\text{nc}}\) includes work done by friction, air resistance, etc.
\[
\eta = \frac{\text{useful energy output}}{\text{energy input}}\times100\%
\]
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.
\[
\begin{aligned}
h &= L\sin30^{\circ}=1.0\ \text{m}\\
U_{i} &= mgh = 1.2(9.8)(1.0)=11.8\ \text{J}\\
N &= mg\cos30^{\circ}=1.2(9.8)\cos30^{\circ}=10.2\ \text{N}\\
W{f} &= -\mu{k}NL = -0.15(10.2)(2.0) = -3.1\ \text{J}\\
K{f} &= U{i}+W_{f}=8.7\ \text{J}\\
v &= \sqrt{\frac{2K_{f}}{m}} = \sqrt{\frac{2(8.7)}{1.2}}\approx 3.8\ \text{m s}^{-1}
\end{aligned}
\]
\[
U_{e}= \tfrac12 kx^{2}
\]
(derived from \(\displaystyle U{e}= \int{0}^{x} kx'\,dx'\)).
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.
\[
\begin{aligned}
\epsilon &= \frac{1.0\times10^{-3}}{0.50}=2.0\times10^{-3}\\
\sigma &= E\epsilon = (2.0\times10^{11})(2.0\times10^{-3})=4.0\times10^{8}\ \text{Pa}\\
F &= \sigma A = (4.0\times10^{8})(2.0\times10^{-4}) = 8.0\times10^{4}\ \text{N}\\
U_{e} &= \tfrac12 F\Delta L = \tfrac12 (8.0\times10^{4})(1.0\times10^{-3}) = 40\ \text{J}
\end{aligned}
\]
\[
\frac{\partial^{2}y}{\partial x^{2}} = \frac{1}{v^{2}}\frac{\partial^{2}y}{\partial t^{2}}
\]
| 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). |
| 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 |
\[
L = n\frac{\lambda}{2}\quad (n=1,2,3\ldots)
\]
\[
f{n}= n f{1},\qquad f_{1}= \frac{v}{2L}
\]
\[
d\sin\theta = n\lambda\qquad (n=0,\pm1,\pm2\ldots)
\]
where \(d\) is the grating spacing.
Given wave speed \(v = 120\ \text{m s}^{-1}\), find the fundamental frequency.
\[
\lambda{1}=2L = 1.5\ \text{m},\qquad f{1}= \frac{v}{\lambda_{1}} = \frac{120}{1.5}=80\ \text{Hz}
\]
\[
\begin{aligned}
V &= IR \\
P &= IV = I^{2}R = \frac{V^{2}}{R} \\
E &= Pt = VIt
\end{aligned}
\]
Energy supplied by a source: \(E = V I t\). Energy dissipated as heat in a resistor: \(E = I^{2}Rt\).
A 240 V mains supply powers a resistive heater drawing 10 A. Find the power consumed and the energy used in 2 h.
\[
P = VI = 240\times10 = 2400\ \text{W}=2.4\ \text{kW}
\]
\[
E = Pt = 2.4\ \text{kW}\times2\ \text{h}=4.8\ \text{kWh}
\]
\[
V{x}= V{\text{s}}\frac{R{x}}{R{1}+R_{2}}
\]
\[
V = \mathcal{E} - Ir_{\text{int}}
\]
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.
\[
R_{\text{eq}} = \left(\frac{1}{4}+\frac{1}{6}\right)^{-1}= \frac{12}{5}=2.4\ \Omega
\]
\[
R{\text{total}} = R{\text{eq}}+r = 2.4+0.5 = 2.9\ \Omega
\]
\[
I = \frac{\mathcal{E}}{R_{\text{total}}}= \frac{12}{2.9}=4.14\ \text{A}
\]
| 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) |
\[
E{\text{b}} = \left[Zm{p}+(A-Z)m{n}-m{\text{nucleus}}\right]c^{2}
\]
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.)
\[
\Delta m = m{\text{U}}-(m{\text{Th}}+m_{\alpha}) = 238.0508-(234.0436+4.0026)=0.0046\ \text{u}
\]
\[
Q = \Delta m \times 931.5\ \text{MeV}=4.3\ \text{MeV}
\]
Example: A satellite of mass 500 kg orbits Earth at an altitude of 300 km. Find the speed.
\[
r = R_{\oplus}+h = 6.37\times10^{6}+3.0\times10^{5}=6.67\times10^{6}\ \text{m}
\]
\[
v = \sqrt{\frac{GM_{\oplus}}{r}} = \sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6.67\times10^{6}}}=7.73\ \text{km s}^{-1}
\]
Example: 1 mol of an ideal gas expands isothermally from 1.0 L to 3.0 L at 300 K. Find the work done.
\[
W = nRT\ln\frac{V{f}}{V{i}} = (1)(8.314)(300)\ln\frac{3.0}{1.0}= 2.74\ \text{kJ}
\]
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}} = \
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