Past exam of the physics B course of the University of Cambridge / 2019 / Paper 1 / Section A

For two circular Kepler orbits about the same body, Kepler's third law gives $T^2/r^3=\text{constant}$. The reference radius is measured from Earth's centre:A geostationary orbit is prograde and equatorial, with one sidereal day as its orbital period. Using $T_g\simeq 1436\min$ and the supplied $T_1=109\min$,Using $24\mathrm{h}$ instead changes this estimate by less than one per cent. This is the orbital radius, not the altitude; the corresponding altitude is about $3.00\times 10^4\mathrm{km}$.

There is a numerical inconsistency in the source data. The quoted reference orbit implies $G{M}_E=4{\pi }^2{r}_1^3/T_1^2\simeq 2.56\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, appreciably below the measured terrestrial value, about $3.99\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. Using the latter instead gives the familiar geostationary orbit radius $4.22\times 10^4\mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.

The radial force is $F_r=-dU/dr=-3A{r}^2$. A nontrivial circular orbit therefore requires $A>0$, so that the force is inward. Balancing its magnitude against the required centripetal acceleration givesThusThis is the circular-orbit energy ratio in a power-law potential, $2K=r{U}^{\prime }(r)=sU$ for $U=A{r}^s$, with $s=3$. It uses the zero of potential energy specified by $U=A{r}^3$; adding a constant would change the ratio without changing the motion. For $A<0$, the force is outward and no such free circular orbit exists.

For linear elasticity of an isotropic material, the isotropic elastic modulus relation givesThe printed Young's modulus is $E=10^{-3}\mathrm{GPa}=10^6\mathrm{Pa}$, and the printed Poisson's ratio is $\nu =1.47$. Formal substitution givesWith compressive pressure positive, the linear volume response is $\Delta V/V=-p/B$. Thus a thirty per cent reduction formally requiresHere $1\mathrm{atm}=101325\mathrm{Pa}$. The negative pressure is a formal tensile load, not the requested positive isotropic compression.

The original PDF really prints $1.47$. It is incompatible with a stable ordinary isotropic elastic solid: positive bulk modulus and shear modulus require $-1<\nu <1/2$. Consequently there is no physically admissible positive-pressure answer with the stated constants. If $0.47$ was intended, the corresponding linear estimates would instead beThat is an explicit alternative, not a correction established by the source. A thirty per cent volume strain is also outside a controlled small-strain approximation; accurate rubber behaviour at that deformation needs a finite-strain constitutive law. The initial volume cancels because the specified change is a fractional one.

Closest approach. In the heavy-target approximation, a head-on alpha particle of charge $2e$ converts its initial kinetic energy to repulsive Coulomb potential energy at its turning point. The nuclear radius bound from a Coulomb turning point is thereforeThus the point-projectile interpretation of the radius bound isIt is an upper limit from Coulomb scattering, not a measurement of the actual nuclear radius. If “size” denotes diameter, the associated bound is twice this value. Including nuclear recoil, $E_{\mathrm{cm}}\simeq [207/(207+4)]E_{\alpha }$, increases the closest separation by about two per cent to $4.4\times 10^{-14}\mathrm{m}$.

Count-rate estimate and normalization. The graph has approximate slope $\Delta \ln S/\Delta \ln A\simeq 15/10=3/2$. It therefore supplies the empirical log-log plot relation $S\propto A^{3/2}$. Calibrating its normalization with the lead observation givesThe Rutherford differential cross-section at fixed beam energy and scattering angle instead has a $Z^2$ dependence per target nucleus. More generally, the target normalization in Rutherford scattering rates iswhere $\mathcal{N}$ is the target nuclei per unit area. If the foils have equal $\mathcal{N}$, the theoretical estimate is $19.7$, or about $20$ particles per minute, close to the graphical estimate at its rough level. Equal foil mass per area would instead give $37.8$ per minute because $\mathcal{N}\propto 1/A$.

The source does not specify that target normalization. Thus $23$ per minute is the estimate from its supplied graph; an exact rate prediction from Rutherford theory alone is underdetermined without comparable foil areal densities. The empirical $A^{3/2}$ slope and an exact equal-$\mathcal{N}$ $Z^2$ law should not be presented as identical assumptions.

For a reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.