Past exam of the physics 1 paper of the University of Oxford / 2021 (A10287W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover identifies the ordinary sitting of 7 June 2021, despite the public bundle's filename. Solutions cover all ten questions, including every alternative in Section B.

Neglecting the planets' masses compared with the Sun's mass, Newton's law of universal gravitation supplies the centripetal acceleration of each circular orbit:Thus Kepler's third law, calibrated by the Earth's orbital period of one year and orbital radius of one astronomical unit, givesBoth orbits are taken to have the same direction. Their relative angular speed is $2\pi (1/T_J-1/T_S)$, so the Sun-centred interval between successive conjunctions is the synodic periodUsing the supplied rounded orbital radii explains the small difference from the commonly quoted approximately twenty-year interval.

From the Earth, the two lines of sight are displaced differently because Jupiter and Saturn are at different distances. The Earth's annual motion along its orbit therefore shifts the dates of apparent conjunctions by amounts of order months. The broad recurrence remains about twenty years, but successive apparent conjunctions need not be equally spaced: apparent retrograde motion can produce a group of three conjunctions separated by months during one encounter. The Earth's daily rotation adds a much smaller change of viewpoint. This is a geometrical comparison in the stipulated coplanar model; daylight and the horizon can prevent a particular conjunction from being visible at a chosen surface location.

Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.

Choose the vertical $z$ axis downward, and measure the polar angle $\theta$ from it. With azimuth $\varphi$ and fixed rod length $r$, the Cartesian coordinates areThe massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yieldsThe gravitational potential energy is $V=-mgz=-mgr\cos \theta$, up to an arbitrary constant. The Lagrangian of this spherical pendulum is thereforeApplying the Euler-Lagrange equation to the two generalized coordinates givesEquivalently, away from the coordinate singularities at the poles,Since $L$ is independent of $\varphi$, the azimuth is a cyclic coordinate. Its conjugate momentumis the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

Orient the $x$ axis along the displacement from the earlier event to the later one, so $\Delta x=d=1000\mathrm{m}$ and $\Delta t=10^{-6}\mathrm{s}$. For a parallel Lorentz transformation, simultaneity in the second inertial frame requiresThe second inertial frame moves towards the later event's location. The positive value $d^2-c^2\Delta t^2$ establishes spacelike separation, which permits this change of simultaneity. For a general velocity $\mathbf{V}$, the condition is $\mathbf{V}\cdot \Delta \mathbf{r}=c^2\Delta t$: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided $|\mathbf{V}|<c$. Thus the parallel velocity has the smallest possible speed.

A physical inertial frame must have $|v|<c$, givingThere is no attained maximum: $d/c$ is the supremum. For the concrete boundary case $\Delta t=1000\mathrm{m}/c$, the events have null separation and the formula would require $v=c$; no inertial frame can make them simultaneous.

Use the Earth's inertial frame, with the Earth at $x=0$ and the star at $x=D=4.2$ light years, neglecting their relative motion. Take the stated relative velocity of $B$ to be forward along the same direction as $A$. The relativistic velocity addition law givesIn units of light years and years, $c=1$ and the spacecraft worldlines are $x_A=t/4$ and $x_B=2t/3$. Their passages at the star occur atThe return radio wave from $B$ follows the leftward light worldlineIntersecting this with $A$'s worldline gives $t/4=10.5-t$, henceThis is the Earth-frame coordinate time of reception on $A$, as requested; it is not the reading of $A$'s own clock. If an observer on the Earth watched that reception through a returning light signal, they would see it later.

Each transmission from the star takes another $D/c=4.2$ years to reach the Earth. ThereforeAll times are measured from the common passage by the Earth; the spacecraft continue at their stated velocities during the calculation.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

Take $+y$ upwards in the source diagram. Write $w$ for $B$'s incoming speed, so the initial velocities areA contact-direction assumption is needed for a unique answer. The diagram specifies the incoming velocities but does not specify the contact normal or an outgoing angle. First adopt the usual intended model: the impulse is horizontal, so neither particle changes its $y$ component of velocity. This is the smooth oblique collision model with a horizontal line of centres at impact. Bringing $A$ to rest in the $x$ direction then brings it entirely to rest.

Let $b$ be $B$'s outgoing $x$ component of velocity. Conservation of momentum and conservation of kinetic energy givePut $q=(\sqrt{3}/2)w$. The second equation becomes $3u^2+q^2=(q-3u{)}^2$, so $q=u$. ThereforeThe outgoing kinetic energies areThe outgoing speed of $B$ is $u\sqrt{13/3}$, directed above the negative $x$ axis at angle ${\tan }^{-1}[1/(2\sqrt{3})]\simeq 16.1^{\circ }$. The initial kinetic energy is also $(3/2+2/3)m{u}^2=13m{u}^2/6$, as required for an elastic collision.

For the perfectly inelastic collision with these same incoming velocities, both particles share a common outgoing velocity. Conservation of momentum givesIts direction is again $16.1^{\circ }$ above the negative $x$ axis. The separate translational kinetic energies of the constituents and their total areThe loss of translational kinetic energy isThis becomes internal energy, for example deformation and heating; total energy is conserved. In the elastic collision no kinetic energy is lost and the particles separate with different velocities. In the perfectly inelastic collision relative motion disappears and the common velocity is the original centre of mass velocity.

The limitation of the printed data can be made explicit. Without the horizontal-impulse assumption, write the final velocity of $A$ as $(0,h)$. Then conservation of momentum determinesand conservation of kinetic energy gives onlyThere is a family of possibilities. For a concrete second solution, $h=u/\sqrt{3}$ givesThese velocities conserve both components of momentum and conserve kinetic energy, and they satisfy the specified vanishing $x$ component for $A$, while requiring a different incoming speed. They can arise from a smooth contact normal proportional to $(1,1/\sqrt{3})$. Thus the horizontal-impulse result is conditional, rather than uniquely implied by the printed statement. For any member of this family the corresponding sticking result remains well determined once $w$ is chosen:The last expression is the lost relative kinetic energy in a perfectly inelastic collision of these two point masses.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.