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

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

The cover dates this resit to 6 September 2021; its archive member is misleadingly labelled 2020/L/a10287w1.pdf.

Interpret $h$ as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Before impact, rolling without slipping gives $\omega =u/r$, so conservation of mechanical energy givesUse the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If $v$ and $V$ are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution giveDuring the subsequent simple pendulum swing, conservation of mechanical energy givesThe cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then $m+I/r^2=3m/2$. Writing $J$ for the impulse delivered to the bob gives $(3m/2)(v-u)=-J$, $2mV=J$ and $V-v=u$. Hence $V=6u/7$ and $\cos {\theta }_{\max }=25/49$, giving $59.32^{\circ }$. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

Choose the heavier particle's outgoing velocity above the initial direction. Let its speed be $V$, and let the lighter particle have components $(v\cos \theta ,-v\sin \theta )$. Thus $\theta$ and $\varphi$ are nonnegative deflections on opposite sides. Conservation of momentum givesThe elastic collision conserves kinetic energy:Taking the squared magnitude of the vector momentum equation yieldsEliminate $v^2$ with the kinetic energy equation. For a nontrivial collision, $V\ne 0$, soSubstituting back into the two momentum components givesThereforeThe ratio alone does not determine the quadrant: for $M>m$, the lighter particle can move backwards. The two component equations select the correct $0\le \theta \le \pi$. At $m/M=\cos 2\varphi$, the outgoing direction is perpendicular to the initial direction; the displayed tangent has a pole there.

Let $t=0$ be the start of deposition and $I_0=M{R}^2/2$ be the original moment of inertia of a uniform disc. Assume that all deposited sand sticks and remains at radius $r$, with negligible thickness and no losses. The sand's incoming angular momentum about the vertical axis is zero. Neither the bearing nor gravity supplies an axial torque, so conservation of angular momentum givesConsequently the angular velocity and angular acceleration areThe rotational kinetic energy isThus, for $\alpha r^2>0$,This is mass-loaded rotation: the sticking process dissipates mechanical energy even though angular momentum is conserved. Indeed $\dot{K}=-\alpha r^2{\omega }^2/2$. If $r=0$ or $\alpha =0$, no additional axial moment of inertia is deposited and the nonzero initial rotational kinetic energy never halves.

In the inertial centre of mass frame, the Lagrangian mechanics description starts from kinetic energy minus potential energy:Define the two-particle relative coordinate $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$. The centre of mass condition givesSubstitution therefore givesThe Euler-Lagrange equation is simply $\mu \ddot{\mathbf{r}}=-\nabla V(r)$. Thus center-of-mass reduction turns a two-body problem into a one-particle central force problem with reduced mass $\mu$, reducing six position components to three relative components. In an arbitrary inertial frame one adds the independent centre of mass term $\frac{1}{2}(m_1+m_2)|\dot{\mathbf{R}}{|}^2$; that free motion does not affect the relative solution. Since $V$ depends only on separation, conservation of angular momentum further confines a nonradial relative trajectory to a plane.

Interpret the percentage relative to the classical kinetic energy, so that $K_{\mathrm{rel}}=1.02K_{\mathrm{cl}}$. Set $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The relativistic kinetic energy and classical kinetic energy giveFor the positive nonzero speed, insert ${\beta }^2=1-{\gamma }^{-2}$ and cancel $\gamma -1$:It follows thatHere $m{c}^2=938\mathrm{MeV}$ is the proton's rest energy; the resulting relativistic kinetic energy is $12.479\mathrm{MeV}$. The zero-speed equality before cancellation has no meaningful percentage excess because both kinetic energies then vanish.

Take positive velocity towards $B$, and interpret $100\mathrm{MHz}$ as the emitted frequency in the probe's rest frame. The relativistic velocity-addition formula gives the probe's station-frame velocity:For reception at the station, photons propagate backwards while the probe recedes. The longitudinal relativistic Doppler effect givesBefore the probe overtakes $B$, their separation decreases. The probe's velocity in $B$'s inertial frame isThe approaching-source relativistic Doppler effect therefore givesAn independent frame calculation gives the same result: forward photons have station-frame frequency $300\mathrm{MHz}$, and a receiver moving at $0.4c$ measures $300\sqrt{0.6/1.4}\mathrm{MHz}$. If reception occurs after the probe passes $B$, backwards photons instead give $f_B=100\sqrt{7/27}\mathrm{MHz}\simeq 50.92\mathrm{MHz}$. The direction of propagation and the reception geometry are essential to choosing the Doppler factor.

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

Invariant and collision. For a particle of rest mass $m$, special relativity gives relativistic energy and momentumTherefore the relativistic energy-momentum relation isIt is the squared Minkowski norm of the four-momentum, multiplied by $c^2$, and hence is preserved by every Lorentz transformation. Explicitly a boost with speed $U$ along $x$ gives $E^{\prime }=\Gamma (E-U{p}_x)$, $p_x^{\prime }=\Gamma (p_x-UE/c^2)$, with $p_y^{\prime }=p_y$, $p_z^{\prime }=p_z$. Expanding $E^{\prime 2}-c^2{p}_x^{\prime 2}$ gives $E^2-c^2{p}_x^2$ because ${\Gamma }^2(1-U^2/c^2)=1$. Spatial rotations preserve $p^2$, completing the invariance argument for all inertial frames.

For the incoming moving particle, $v=3c/5$ and $\gamma =5/4$. The total laboratory energy and momentum before the perfectly inelastic collision areFour-momentum conservation identifies these with the final particle's energy and momentum. Its rest mass is thereforeThe result exceeds the sum $3m_0$ of the incoming rest masses; some initial kinetic energy has become internal rest energy. Since $p{c}^2/E=v$ for a massive particle,

Two-body decay. Work first in the parent's rest frame, denoted by a star. Four-momentum conservation requires equal and opposite daughter momenta. Their equal rest masses give equal energies:Thus their rest-frame velocities are $v_{\ast \pm }=\pm 3c/\sqrt{10}$. Applying the relativistic velocity-addition formula with boost $v^{\prime }=3c/13$ gives the two laboratory velocities:Their laboratory energies follow directly from the Lorentz transformation of four-momentum:The corresponding signed laboratory momenta areTheir sums recover $E_i$ and $p_i$, and each satisfies $E_{\pm }^2-p_{\pm }^2{c}^2=(m_0{c}^2/2{)}^2$. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically, $v_{+}\simeq 0.967615c$, $v_{-}\simeq -0.919138c$, $E_{+}\simeq 1.980756m_0{c}^2$ and $E_{-}\simeq 1.269244m_0{c}^2$.