Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / Section A

For the gravitational-electrostatic balance of equal charged spheres, the shell theorem and Coulomb's law give $F_g=G{M}^2/r^2$ and $F_e=q^2/(4\pi {ϵ}_0{r}^2)$ for well-separated spherical particles. Their separation cancels at balance. Since $M=4\pi \rho a^3/3$,The two charges have the same sign, so electrostatic repulsion opposes gravitational attraction. A zero charge gives no nontrivial force balance. Spherical symmetry of the external charge distribution, or the leading far-field monopole approximation, is understood.

By symmetry, the two grounded plates take equal induced surface charge and the electric field has magnitude $\sigma /(2{ϵ}_0)$ between either plate and the sheet. The grounded-plate potential of a central charged sheet is thereforeAn electron released at zero potential gains kinetic energy $e{V}_{\mathrm{sheet}}$. ThusThe planar infinite-sheet approximation neglects edge effects. Here $e>0$ denotes charge magnitude; the electron's negative charge makes it accelerate toward the positive sheet. Relativistic corrections are below0.1%. Numerical constants use rounded CODATA values, not a claim about the historical handbook: ${ϵ}_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_e=9.10938\times 10^{-31}\mathrm{kg}$. ^[ref]

In the initial centre-of-mass frame, equal fragments acquire opposite velocities $\pm w$ by conservation of momentum. The released energy isThe equal-mass fragmentation energy budget needs $w=500\mathrm{m}{\mathrm{s}}^{-1}$ merely to match the other ship, costing $1.25\mathrm{GJ}$. To close an existing gap requires a little more than this. Using the full reserve givesThe reserve is sufficient to overtake. The rear half moves at $1.368\mathrm{km}{\mathrm{s}}^{-1}$, preserving momentum. The kinetic-energy increase is the same in any inertial frame because total momentum is unchanged. Subsequent capture or braking would require additional interaction details; no target mass is needed for the interception test.

The PDF's apex angle is $30^{\circ }$, so each contacted face makes $\alpha =15^{\circ }$ with the vertical. Let N be the inward normal force per face. Its vertical component is downward, $-N\sin \alpha$; limiting upward static friction contributes $\mu N\cos \alpha$. Symmetric contacts cancel horizontal forces and torque. Incipient lift consequently requiresThis is the usual normal gripping force convention, using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Friction can lift the prism only if $\mu >\tan \alpha$, which is satisfied here.

If “force applied” means the magnitude of the entire finger-contact resultant, including its tangential component, the normal versus resultant force in a frictional pinch givesThe optimal resultant lies at the limiting friction angle $\arctan \mu$ from the inward normal. Stating which force is being measured avoids confusing these two legitimate quantities.

For the double-slit experiment with oblique incidence, the optical path difference has an unknown incident contribution plus $a\sin \theta$ in the outgoing direction. It cancels between adjacent angular features. From a minimum to the next maximum, the path difference changes by $\lambda /2$:This incidence-independent adjacent-fringe wavelength uses the statement that intensity rises directly from the stated zero to the following peak. No absolute interference order or incidence angle is needed; nonadjacent extrema would require the number of intervening fringes.

The central estimate is $c=4.15/1.93-1=1.15026$. With independent input standard errors, the ratio-minus-one uncertainty from first-order uncertainty propagation isThe four independent reported results cluster near1.21, within about one quoted standard error of this estimate, so they are consistent at this precision. Their tight clustering is not evidence that this measurement has the same precision. If the measurements of a and b were correlated, a covariance term $-2a\mathrm{Cov}(a,b)/b^3$ would be needed. ^[ref]