Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge

The released examiner commentary is not a complete worked solution set. It records an accepted ambiguity in question 8(d); question 7(e) also requires care about its unstated drag model.^[ref]

By dimensional analysis, power has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-3}$, mass density has dimensions $\mathsf{M}{\mathsf{L}}^{-3}$, and angular velocity has dimensions ${\mathsf{T}}^{-1}$; radians are dimensionless. ThusMatching the three independent dimensions gives $\alpha =1$, $\beta =3$ and $\gamma -3\alpha =2$, henceWith unchanged mass density, angular velocity and the same performance coefficient (a dimensionless quantity), the pump power affinity law givesThe required power therefore increases by $61.1\%$. This uses the stipulated dependence on only these three quantities, with comparable pump geometry and performance; dimensional analysis alone does not guarantee that unrelated pumps have the same performance coefficient.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

Take the thin lens at $x=0$, with the object on the left and light rays travelling to the right. The positive object distance is $u=12\mathrm{cm}$ and the focal length is $f=8\mathrm{cm}$. The thin lens equation givesThe transverse magnification and signed image height areThus the real image is $24\mathrm{cm}$ beyond the lens, inverted and $10\mathrm{cm}$ high. For the ray tracing, a light ray parallel to the optical axis emerges through the far optical focus; a light ray through the optical centre continues straight in the ideal thin lens model. They intersect at $(24,-10)\mathrm{cm}$, independently reading off both answers from the diagram.

The calculation and diagram use the ideal paraxial approximation implicit in the thin lens equation. The comparatively large object height does not supply enough information to calculate optical aberrations of a real lens.

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed Electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching Electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the Electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

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.