University Physics Practice Test — 30 Challenging Problems

This practice test covers 30 problems across four major domains of university physics: Classical Mechanics, Thermodynamics, Electromagnetism, and Quantum Mechanics. Each problem requires multi-step reasoning and the application of fundamental principles. Work through the problems with pen and paper before checking the solutions.

• Time limit: 90 minutes (3 minutes per problem)
• Format: Problem-solving — show all working
• Marking: 1 mark per problem, 30 marks total
• Conditions: Attempt without notes. Show all steps in your solutions.
• After the test: Check the solutions at the bottom. Study the derivations for any problems you got wrong.

A particle of mass m slides without friction on the inside surface of a sphere of radius R. Using the angle theta from the vertical as the generalised coordinate, derive the equation of motion and find the condition for the particle to leave the surface.

Solution:

The kinetic energy is $T = \frac{1}{2}m(R\dot{\theta})^2$ and the potential energy (taking the bottom of the sphere as reference) is $U = mgR(1 - \cos\theta)$. The Lagrangian is:

$L = T - U = \frac{1}{2}mR^2\dot{\theta}^2 - mgR(1 - \cos\theta)$

The Euler-Lagrange equation gives:

$mR^2\ddot{\theta} = -mgR\sin\theta$

$\ddot{\theta} = -\frac{g}{R}\sin\theta$

The particle leaves the surface when the normal force becomes zero. From the radial force equation: $mg\cos\theta - N = mR\dot{\theta}^2$. Setting $N = 0$ and using energy conservation: $v^2 = 2gR(\cos\theta_c - \cos\theta)$, the critical angle is $\cos\theta_c = 2/3$, or $\theta_c \approx 48.2°$.

hard — 1 mark

For a one-dimensional harmonic oscillator with mass m and spring constant k, write the Hamiltonian and derive Hamilton’s equations of motion.

Solution:

The Lagrangian is $L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2$.

The generalised momentum is $p = \frac{\partial L}{\partial \dot{x}} = m\dot{x}$, so $\dot{x} = p/m$.

The Hamiltonian is:

$H = p\dot{x} - L = \frac{p^2}{2m} + \frac{1}{2}kx^2$

Hamilton’s equations:

$\dot{x} = \frac{\partial H}{\partial p} = \frac{p}{m}$

$\dot{p} = -\frac{\partial H}{\partial x} = -kx$

Differentiating the first: $\ddot{x} = \dot{p}/m = -kx/m$, giving $\ddot{x} + \omega^2 x = 0$ where $\omega = \sqrt{k/m}$.

hard — 1 mark

A particle moves in a central force field $F(r) = -kr^{-3}$ (attractive). Determine whether closed orbits are possible using the effective potential.

Solution:

The effective potential is:

$V_{eff}(r) = \frac{l^2}{2mr^2} + \frac{k}{2r^2} = \frac{1}{2r^2}\left(\frac{l^2}{m} + k\right)$

where $l$ is the angular momentum. Since $V_{eff} \propto 1/r^2$, the effective potential is purely repulsive (for $k > 0$) or purely attractive (for $k < 0$). There is no potential well, so no stable circular orbits exist. The particle either spirals inward or escapes to infinity — closed orbits are not possible.

hard — 1 mark

A uniform solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle alpha. Find the acceleration of the centre of mass.

Solution:

Forces along the incline: $Mg\sin\alpha - f = Ma$, where $f$ is the static friction force.

Torque about the centre: $fR = I\alpha_{rot} = \frac{1}{2}MR^2 \cdot \frac{a}{R}$, so $f = \frac{1}{2}Ma$.

Substituting: $Mg\sin\alpha - \frac{1}{2}Ma = Ma$, giving:

$a = \frac{2}{3}g\sin\alpha$

The cylinder accelerates at two-thirds the value for a frictionless slide, because some gravitational potential energy is converted to rotational kinetic energy.

medium — 1 mark

Two identical pendulums of mass m and length l are coupled by a spring of constant k attached at a distance h from the pivot. Find the normal mode frequencies.

Solution:

For small oscillations, the equations of motion are:

$ml^2\ddot{\theta}_1 = -mgl\theta_1 - kh^2(\theta_1 - \theta_2)$

$ml^2\ddot{\theta}_2 = -mgl\theta_2 + kh^2(\theta_1 - \theta_2)$

Normal modes: symmetric ($\theta_1 = \theta_2$) and antisymmetric ($\theta_1 = -\theta_2$).

For the symmetric mode: $\omega_1 = \sqrt{g/l}$ (spring not stretched).

For the antisymmetric mode: $\omega_2 = \sqrt{g/l + 2kh^2/(ml^2)}$.

hard — 1 mark

A particle of mass $m_1$ moving with velocity $v_0$ collides elastically with a stationary particle of mass $m_2$. Derive the final velocity of $m_2$.

Solution:

Conservation of momentum: $m_1 v_0 = m_1 v_1 + m_2 v_2$.

Conservation of kinetic energy: $\frac{1}{2}m_1 v_0^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$.

From these two equations (or using the coefficient of restitution $e = 1$):

$v_2 = \frac{2m_1}{m_1 + m_2}v_0$

Special cases: if $m_1 = m_2$, $v_2 = v_0$ (complete transfer). If $m_1 \ll m_2$, $v_2 \approx 0$ (light particle bounces back).

medium — 1 mark

A free particle of mass m moves in three dimensions with Lagrangian $L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$. Identify the conserved quantities and explain their physical significance.

Solution:

$L$ does not depend explicitly on $x$, $y$, or $z$, so the conjugate momenta are conserved:

$p_x = \frac{\partial L}{\partial \dot{x}} = m\dot{x} = \text{const}$

$p_y = m\dot{y} = \text{const}$

$p_z = m\dot{z} = \text{const}$

These are the components of linear momentum — translational invariance implies conservation of momentum (Noether’s theorem).

$L$ also does not depend on time explicitly, so the Hamiltonian (total energy) $H = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = T$ is conserved.

medium — 1 mark

The Lorenz equations are $\dot{x} = \sigma(y - x)$, $\dot{y} = rx - y - xz$, $\dot{z} = xy - bz$. For what parameter regime do chaotic solutions appear?

Solution:

The Lorenz system exhibits chaos for the classical parameter values $\sigma = 10$, $b = 8/3$, and $r > 24.74$. The onset of chaos occurs at the homoclinic bifurcation.

For $r < 1$: the origin is the only fixed point (globally stable).

For $1 < r < 24.74$: two additional stable fixed points appear (pitchfork bifurcation). Trajectories converge to one of these.

For $r > 24.74$: the fixed points become unstable, and trajectories exhibit sensitive dependence on initial conditions — the hallmark of chaos. The Lorenz attractor is a strange attractor with fractal dimension approximately 2.06.

hard — 1 mark

An ideal gas undergoes a cyclic process consisting of two isothermal and two adiabatic steps (Carnot cycle). Show that the efficiency depends only on the temperatures of the hot and cold reservoirs.

Solution:

In a Carnot cycle:

• Isothermal expansion at $T_H$: absorbs $Q_H$ from the hot reservoir.
• Adiabatic expansion: temperature drops from $T_H$ to $T_C$.
• Isothermal compression at $T_C$: releases $Q_C$ to the cold reservoir.
• Adiabatic compression: temperature rises from $T_C$ to $T_H$.

For an ideal gas, isothermal processes give $Q = nRT\ln(V_f/V_i)$. The adiabatic steps satisfy $TV^{\gamma-1} = \text{const}$.

From the adiabatic conditions: $\frac{V_2}{V_1} = \frac{V_3}{V_4}$, so:

$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$

The efficiency is:

$\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H}$

This depends only on the reservoir temperatures, not on the working substance.

hard — 1 mark

A system of N independent harmonic oscillators has energy levels $E_n = (n + 1/2)\hbar\omega$. Derive the partition function and the mean energy at temperature T.

Solution:

The single-particle partition function is:

$Z_1 = \sum_{n=0}^{\infty} e^{-\beta(n+1/2)\hbar\omega} = e^{-\beta\hbar\omega/2} \sum_{n=0}^{\infty} e^{-n\beta\hbar\omega}$

$Z_1 = \frac{e^{-\beta\hbar\omega/2}}{1 - e^{-\beta\hbar\omega}} = \frac{1}{2\sinh(\beta\hbar\omega/2)}$

For N distinguishable oscillators: $Z = Z_1^N$.

The mean energy:

$\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta} = N\hbar\omega\left(\frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1}\right)$

The first term is the zero-point energy; the second is the thermal excitation.

hard — 1 mark

A mole of an ideal monatomic gas at 300 K and 1 atm is expanded isothermally to twice its volume. Calculate the change in entropy.

Solution:

For an isothermal process, the change in internal energy is zero (ideal gas), so:

$\Delta S = \frac{Q_{rev}}{T} = \frac{W_{rev}}{T} = \frac{nRT\ln(V_2/V_1)}{T} = nR\ln(V_2/V_1)$

$\Delta S = (1)(8.314)\ln(2) = 8.314 \times 0.693 = 5.76 \text{ J K}^{-1}$

The entropy increases because the gas occupies a larger volume, increasing the number of accessible microstates.

easy — 1 mark

Derive the ideal gas law from the kinetic theory of gases, starting from the momentum transfer of molecules colliding with a wall.

Solution:

Consider N molecules of mass m in a container of volume V. A molecule moving with velocity $v_x$ in the x-direction collides elastically with a wall perpendicular to x. The momentum transfer per collision is $2mv_x$.

The number of collisions per unit time with a wall of area A is $\frac{1}{2}nAv_x$ (where n is the number density).

The force on the wall:

$F = 2mv_x \times \frac{1}{2}nAv_x = nmAv_x^2$

Averaging over all molecules: $F = nmA\langle v_x^2 \rangle$. Since $\langle v_x^2 \rangle = \frac{1}{3}\langle v^2 \rangle$:

$P = \frac{F}{A} = \frac{1}{3}nm\langle v^2 \rangle = \frac{2}{3}\frac{N}{V}\langle KE \rangle$

Since $\langle KE \rangle = \frac{3}{2}k_BT$:

$PV = Nk_BT = nRT$

hard — 1 mark

Describe the differences between first-order and second-order phase transitions, giving an example of each.

Solution:

First-order phase transitions involve a discontinuity in the first derivative of the free energy (Gibbs free energy G). There is a latent heat, and the two phases coexist at the transition temperature. Examples: boiling of water (liquid to gas), melting of ice (solid to liquid), and the normal-superconducting transition.

At a first-order transition: $\Delta S \neq 0$ (latent heat $L = T\Delta S$), $\Delta V \neq 0$ (volume change), and there is hysteresis and metastability (superheating, supercooling).

Second-order phase transitions (continuous transitions) have no latent heat and no discontinuity in the first derivatives of G. The second derivatives (heat capacity, compressibility, susceptibility) diverge or have discontinuities. Examples: ferromagnetic transition at the Curie point, superfluid transition in liquid helium, and the lambda transition.

At a second-order transition: $\Delta S = 0$, $\Delta V = 0$, but $C_p$, $\kappa$, and $\chi$ may diverge. The transition is characterised by critical exponents and universality.

medium — 1 mark

For a free electron gas at T = 0, derive the Fermi energy in terms of the electron density n.

Solution:

At T = 0, all states up to the Fermi energy $E_F$ are filled, and all states above are empty.

The number of states with energy less than E:

$N = \frac{V}{3\pi^2}\left(\frac{2mE}{\hbar^2}\right)^{3/2}$

This comes from counting states in k-space: $N = \frac{2V}{(2\pi)^3} \times \frac{4}{3}\pi k_F^3$ (factor of 2 for spin), with $E_F = \frac{\hbar^2 k_F^2}{2m}$.

Solving for $E_F$:

$E_F = \frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3}$

where $n = N/V$ is the electron density. For copper ($n \approx 8.5 \times 10^{28}$ m$^{-3}$), $E_F \approx 7$ eV.

hard — 1 mark

Using Planck’s radiation law, derive the Wien displacement law that relates the peak wavelength of blackbody radiation to temperature.

Solution:

Planck’s law gives the spectral radiance:

$B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/(\lambda k_BT)} - 1}$

To find the peak, set $dB/d\lambda = 0$. Let $x = hc/(\lambda k_BT)$. Then:

$\frac{dB}{d\lambda} = 0 \implies 5(e^x - 1) = xe^x$

Solving numerically: $x \approx 4.965$.

Therefore: $\frac{hc}{\lambda_{max} k_BT} = 4.965$, giving:

$\lambda_{max} T = \frac{hc}{4.965 k_B} \approx 2.898 \times 10^{-3} \text{ m K}$

This is Wien’s displacement law: $\lambda_{max} T = b$ where $b \approx 2.9 \times 10^{-3}$ m K.

hard — 1 mark

State Maxwell’s equations in differential form and explain the physical significance of each.

Solution:

1. Gauss’s law for electricity: $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$

   Electric charges are sources and sinks of electric field lines. The divergence of E at any point is proportional to the charge density.

2. Gauss’s law for magnetism: $\nabla \cdot \mathbf{B} = 0$

   There are no magnetic monopoles. Magnetic field lines always form closed loops.

3. Faraday’s law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$

   A time-varying magnetic field induces an electric field (and hence an EMF). This is the principle behind electromagnetic induction.

4. Ampere-Maxwell law: $\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

   Magnetic fields are generated by currents and by time-varying electric fields (displacement current). The displacement current term was Maxwell’s key insight.

medium — 1 mark

A point charge q is placed at a distance d from the centre of a grounded conducting sphere of radius R. Find the charge distribution on the sphere using the method of images.

Solution:

The image charge method replaces the conducting sphere with an image charge $q'$ at a distance $b$ from the centre. For a grounded sphere:

$q' = -\frac{R}{d}q, \quad b = \frac{R^2}{d}$

The potential outside the sphere is:

$\Phi(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{|\mathbf{r} - d\hat{z}|} + \frac{q'}{|\mathbf{r} - b\hat{z}|}\right)$

The surface charge density on the sphere is:

$\sigma(\theta) = -\epsilon_0 \frac{\partial \Phi}{\partial r}\bigg|_{r=R} = \frac{q(d^2 - R^2)}{4\pi R(d^2 + R^2 - 2dR\cos\theta)^{3/2}}$

hard — 1 mark

Derive the magnetic field inside a long solenoid of n turns per unit length carrying current I.

Solution:

By symmetry, the field inside is uniform and parallel to the axis. Outside, the field is approximately zero (for a long solenoid).

Apply Ampere’s law to a rectangular Amperian loop with one side inside the solenoid (length l) and one side outside:

$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}$

$B \cdot l = \mu_0 \cdot nIl$

$B = \mu_0 nI$

The field is uniform inside and directed along the axis. The direction is given by the right-hand rule: curl your fingers in the direction of current, and your thumb points in the direction of B.

medium — 1 mark

A parallel-plate capacitor with plate area A and separation d is being charged at a rate dQ/dt. Derive the displacement current and show it gives the same magnetic field as the conduction current.

Solution:

The electric field between the plates: $E = \sigma/\epsilon_0 = Q/(A\epsilon_0)$.

The displacement current density:

$J_d = \epsilon_0 \frac{\partial E}{\partial t} = \frac{1}{A}\frac{dQ}{dt} = \frac{I}{A}$

Total displacement current: $I_d = J_d \times A = I$, equal to the conduction current.

For the magnetic field at distance r from the axis (r < plate radius), using the Ampere-Maxwell law:

$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt} = \mu_0 \epsilon_0 \pi r^2 \frac{dE}{dt}$

$B(2\pi r) = \mu_0 \epsilon_0 \pi r^2 \frac{I}{A\epsilon_0} = \frac{\mu_0 I r^2}{A}$

$B = \frac{\mu_0 I r}{2A}$

This matches the field from the conduction current, confirming consistency.

hard — 1 mark

Starting from Maxwell’s equations in free space, derive the wave equation for the electric field and find the speed of electromagnetic waves.

Solution:

In free space ($\rho = 0$, $\mathbf{J} = 0$):

$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$

$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$

Take the curl of Faraday’s law:

$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) = -\mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$

Using the vector identity $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$ and Gauss’s law ($\nabla \cdot \mathbf{E} = 0$ in free space):

$-\nabla^2 \mathbf{E} = -\mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$

$\nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$

This is the wave equation with speed $c = 1/\sqrt{\mu_0 \epsilon_0} \approx 3 \times 10^8$ m/s.

hard — 1 mark

Define the scalar potential phi and vector potential A, and explain the concept of gauge invariance.

Solution:

Since $\nabla \cdot \mathbf{B} = 0$, we can write $\mathbf{B} = \nabla \times \mathbf{A}$ for some vector potential A.

From Faraday’s law: $\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t = -\partial(\nabla \times \mathbf{A})/\partial t$, so $\nabla \times (\mathbf{E} + \partial \mathbf{A}/\partial t) = 0$. Since the curl of a gradient is zero: $\mathbf{E} = -\nabla \phi - \partial \mathbf{A}/\partial t$.

The potentials (phi, A) are not unique. The gauge transformation:

$\mathbf{A}' = \mathbf{A} + \nabla \chi$