Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section B

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

Let $I_{\perp }$ denote the moment of inertia normal to the disc and $I_d$ that about any central diameter. Rotational symmetry makes the two perpendicular diameter values equal. By the perpendicular axis theorem, $I_{\perp }=2I_d$. The intended angular impulse has magnitude $J=I_{\perp }{\omega }_0$; delivering that magnitude about a diameter instead gives $J=I_d\omega$. ThusThese are principal axes, so angular momentum is parallel to angular velocity in both cases. Equal delivered angular impulse means equal magnitude of the change in angular momentum, not equal rotational kinetic energy; the actual rotational kinetic energy is twice the intended value.

Take $+x$ along the upper ship's initial velocity, and take $+y$ from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial velocities $v_1\hat{\mathbf{x}}$ and $-v_2\hat{\mathbf{x}}$. Their equal masses put each at distance $l/2$ from the centre of mass, soBy momentum conservation, the centre of mass keeps moving with constant velocityIn that frame the ships have opposite velocities of magnitude $u=(v_1+v_2)/2$. Their initial angular momentum about the centre of mass is nonzero even though their trajectories were straight:The tether exerts central internal forces, so conservation of angular momentum givesEach ship follows uniform circular motion of radius $l/2$ around the translating centre of mass, clockwise viewed from $+z$. More explicitly, put the initial centre at the origin and defineThe kinetic-energy decomposition about the centre of mass verifies conservation of mechanical energy directly:At closest approach the relative radial velocity is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its tension is perpendicular to each ship's velocity relative to the centre of mass and does no net work on the pair at fixed $l$.

Each ship needs centripetal acceleration ${\omega }^2(l/2)$, supplied entirely by the tension. ThusThe result has units of force, $\mathrm{N}$.

During quasi-static tether contraction, momentum conservation keeps ${\mathbf{V}}_{\mathrm{CM}}$ constant and conservation of angular momentum keeps $L$ constant. To leading order in the slow radial motion, the kinetic energy isIts derivative at fixed $L$, not fixed $\omega$, isTo compute the power supplied by the tension, use the instantaneous unit vector $\mathbf{n}$ from the lower to the upper ship. Their radial velocities relative to the centre of mass are $\pm (\dot{l}/2)\mathbf{n}$, and the tension forces are $\mp T\mathbf{n}$. ConsequentlyThe translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive work, increasing the rotational kinetic energy.

The slow-contraction approximation matters: for a prescribed smooth $l(t)$ with a taut tether, the exact kinetic energy and tension include radial terms,These still satisfy $d{E}_{\mathrm{exact}}/dt=-T_{\mathrm{exact}}\dot{l}$. Thus the paper's displayed formula is the leading result for slow, smooth contraction, or is exact when the radial kinetic energy is constant; it is not an identity for every possible shrinking schedule.

The mass and surface area of the spherical droplet are $m=(4\pi /3)\rho r^3$ and $A=4\pi r^2$. Constant surface mass accretion rate $\beta$ meansTherefore $\dot{r}=\beta /\rho =\mu$, and the initial radius givesHere $[\beta ]=\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$, $[\rho ]=\mathrm{kg}{\mathrm{m}}^{-3}$, and $[\mu ]=\mathrm{m}{\mathrm{s}}^{-1}$. This surface-fed growth of an accreting sphere law assumes constant droplet mass density.

Choose downward velocity $v$ positive, and assume the collected material is initially at rest in the cloud's inertial frame. This assumption is required for the equation in the question. During $dt$, momentum conservation for the droplet plus the material it collects givesup to second-order increments. Thus the variable-mass system satisfiesSince $m\propto r^3$, $\dot{m}/m=3\dot{r}/r=3\mu /(r_0+\mu t)$, givingThe notation $dv/dt$ is appropriate because $v$ here depends only on time. The extra term accounts for the momentum needed to accelerate newly collected mass; using $m\dot{v}=mg$ would omit it. For incoming material with downward velocity $u$, the more general equation is $m\dot{v}=mg+(u-v)\dot{m}$.

Set $s=r_0+\mu t$. The proposed velocity and its derivative areThe remaining term in the linear ordinary differential equation isso the sum is exactly $g$. Also $v(0)=g(r_0-r_0)/(4\mu )=0$, establishing both the equation and its initial condition:The integrating factor $(r_0+\mu t{)}^3$ explains where the expression comes from:The result has units of velocity. Its continuous $\mu \to 0$ limit is $gt$, despite the apparent $1/\mu$ in the written expression.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

An upward drag force $F_D\ge 0$ changes the variable-mass system equation toFor an ordinary drag force increasing with positive speed, the new curve lies below the no-drag curve. It still starts with slope $g$ if $F_D=0$ at rest. The dashed curve illustrates quadratic drag on a sphere with constant radial growth, using $F_D=\frac{1}{2}{C}_D{\rho }_{\mathrm{air}}\pi r^2{v}^2$ with constant drag coefficient $C_D$ and air mass density ${\rho }_{\mathrm{air}}$. The equation for these dimensionless quantities isThe figure uses $\eta =2$ solely to show the shape; it is not an additional datum of the paper.

There is a limitation to the usual terminal velocity sketch here. The examiner's commentary describes a curve approaching terminal velocity, but the paper maintains unbounded radius growth and specifies no drag law. A horizontal limit is justified if one additionally treats the radius as approximately fixed, or otherwise assumes a drag-to-mass ratio with a fixed limiting balance. With the growing radius and constant-$C_D$ quadratic drag just written, the late balance instead givesThis curve flattens in the sense that its slope tends to zero, but its speed remains unbounded. Thus a lower, more strongly flattening curve is warranted; a finite terminal speed requires an additional assumption. The no-growth approximation gives the familiar finite value $v_T=\sqrt{8\rho gr/(3C_D{\rho }_{\mathrm{air}})}$ for a specified fixed radius.^[ref]

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

The original diagram has a series circuit containing $R$ and $L$, in parallel with $C$; it is not three separate parallel components. The electrical admittance is thereforeFor $R>0$, voltage and current are in phase when the imaginary part of $Y$ vanishes. Excluding the zero-frequency solution, division by $\omega$ givesA nonzero lossy parallel coil resonance exists only if $L>R^{2}C$. With $R=10\Omega$, $L=0.150\mathrm{H}$ and $C=1.00\times 10^{-5}\mathrm{F}$,For zero electrical resistance, the ideal parallel inductor-capacitor cancellation occurs atThe fractional reduction relative to the zero-resistance coil isAt exactly $R=0$ and $\omega ={\omega }_0$, $Y=0$, so finite applied voltage draws zero net current in the ideal model. The quoted frequency is the limiting parallel-resonance value; the phase of an exactly zero current is undefined.

Let $C_T=C+C_1$ be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put ${\omega }_{\ast }=2\pi (50\mathrm{Hz})$ into the lossy parallel coil resonance condition:The additional capacitance is thereforeThis is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.

For a long solenoid, solenoid inductance obeys $L={\mu }_0{N}^{2}A/\ell$, where $N$ is the number of turns, $A$ the cross-sectional area and $\ell$ the axial winding length. The natural reading of half as many turns occupying half the axial length givesWith unchanged wire material and gauge, halving the wire length also halves its electrical resistance: $R^{\prime }=R/2$. With the total capacitance $C_T$ fixed, the new angular frequency satisfiesThusNeglecting the small resistive correction gives the estimate $\sqrt{2}$.

There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L^{\prime }=L/4$ while $R^{\prime }=R/2$, givingThis corresponds to treating “length of the winding” as wire length while preserving axial geometry. The examiner's report explicitly accepted either inductance scaling, provided the resistance change was included. The first interpretation follows the statement that the axial winding length is halved; the second records the accepted ambiguity rather than silently choosing a different geometry.^[ref]

The Time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified Electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

A scanning tunneling microscope brings a tip and surface, both electrical conductors, close enough to leave only a small vacuum gap. An applied electric potential difference makes occupied Electron states on one side able to feed unoccupied states on the other. Although the gap is a classically forbidden potential barrier, quantum tunnelling through its finite width produces an electric current.

For an approximate rectangular potential barrier of gap width $d$, the transmission probability, and thus the electric current at fixed small bias, varies approximately as $e^{-2\kappa d}$. This strong distance dependence makes the device sensitive to changes in separation on the scale of a chemical atom. Feedback control moves the tip vertically to keep the electric current constant during a scan; the required height changes form an image of the surface, with contrast also depending on its electronic states. The finite gap permits transmitted probability current, whereas the semi-infinite step in part (b) does not.

For an infinite straight wire in vacuum, Ampère's circuital law around a circle of radius $R$ gives $B(2\pi R)={\mu }_{0}I$. ThereforeHere ${\mu }_0$ is the vacuum permeability and the azimuthal direction follows the right-hand rule around the electric current. The magnetic field is measured in $\mathrm{T}$.

Choose the electric current counterclockwise as viewed from $+z$; reversing it reverses the magnetic field. Parameterize the circular wire byThe cross product in the Biot-Savart law isConsequently the transverse terms integrate to zero over $0\le \varphi \le 2\pi$, while the axial term is constant:ThusThe direction is $+z$ on both halves of the axis for this electric current orientation, including $z<0$; it is not “away from the plane” on both sides. The right-hand rule gives the same direction.

Read the original diagram as a clockwise arc carrying electric current from its upper endpoint to its lower endpoint, with tangent semi-infinite leads extending to the right. Their crossing is insulated, so there is no extra junction. Choose $+z$ out of the page. At the centre the arc produces a negative axial magnetic field, and each straight lead a positive one, by the right-hand rule.

For the arc, every element is at distance $R$, with $d\mathbf{l}$ perpendicular to its displacement to the centre. The Biot-Savart law therefore givesFor each straight lead, write $s\ge 0$ for distance along the tangent measured from its endpoint. Its perpendicular distance from the centre is $R$, soThe integral equals $1/R$ because an antiderivative is $s/[R\sqrt{R^2+s^2}]$. Adding both leads and the arc,Using the full infinite-wire magnetic field for either lead would incorrectly double its contribution.

To calculate magnetic-field cancellation of an arc and tangent leads away from the plane, put $a^2=R^2+h^2$. Each point on the arc is distance $a$ from $(0,0,h)$, and the axial component of $d\mathbf{l}\times \mathbf{s}$ still has magnitude $R^{2}d\varphi$. Thus its axial magnetic field isFor either tangent lead the axial numerator remains $Rds$, while the cubed distance is $(R^2+h^2+s^2{)}^{3/2}$. HenceAdding both leads givesFor $ϵ=h/R\ll 1$, a binomial series yieldsThe exact condition for cancellation of the axial component isFor example, keeping $\theta =2$ would leave a small positive axial magnetic field proportional to $h^2/R^2$. Both contributions weaken with height, but the arc factor is $(1+{ϵ}^2{)}^{-3/2}$ whereas the lead factor is $(1+{ϵ}^2{)}^{-1}$. The arc therefore loses a greater fraction of its axial magnetic field. Increasing its angle compensates for this loss. Physically, the arc is all at the same finite radius, whereas distant elements of the semi-infinite leads undergo a smaller fractional change in distance on moving a little out of the plane.

The cancellation here concerns $B_z$ only; off the plane it does not assert that every component of the magnetic field vanishes. The small-height result also stays within the tangent-lead geometry shown in the original paper.