Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / Section B

For the edge-hinged cube with a central spherical cavity, use the parallel-axis theorem: for a parallel axis at perpendicular distance $d$ from the centre-of-mass axis, $I=I_C+M{d}^2$. Let $M=\rho a^3$. The centre lies $d=a/\sqrt{2}$ from the edge hinge, soAt equilibrium the centre is vertically below the hinge. Its gravitational torque at angular displacement $\theta$ is $-Mgd\sin \theta$. Therefore the small-angle equation is $I_h\ddot{\theta }+Mgd\theta =0$, withThe removed sphere has mass $M_s=\pi M/6$ and the same centre. Its hinge inertia isSuperposition leaves the centre at the same location but changes both inertia and weight:The frequency decreases: the removed central material contributes less hinge inertia per unit mass than the original cube average.

Integrating coaxial cylindrical shells givesRolling without slipping has spin speed $u/a$, so $K=m{u}^2/2+I{u}^2/(2a^2)=3m{u}^2/4$.

For the rolling-cylinder loop contact threshold, set $\ell =R-a$, the radius of the centre's circular path, and let $\theta$ run from the loop bottom. The centre starts $h$ above its lowest centre position. Its height gain from the bottom is $\ell (1-\cos \theta )$. Energy conservation givesAt the top, contact requires $N=m(u_t^2/\ell -g)\ge 0$, hence $u_t^2\ge g\ell$. Inserting the energy relation proves the limiting conditionThe same condition keeps $N\ge 0$ elsewhere on the circle, sinceis smallest at the top.

For the centre's acceleration, the inward normal component is $u^2/\ell$ and the tangential component is $-(2g/3)\sin \theta$. If $A=h/\ell -1\ge 7/4$ and $x=\cos \theta$,This convex quadratic has its largest endpoint value at $x=1$. The maximum on the circular loop is at the bottom, directed vertically upward. For the limiting release height,The approach track's curvature is unspecified, so a global bound for an arbitrary approach shape cannot be proved from the sketch alone; the result above concerns the specified circular part and the usual smooth approach interpretation.

The finite-friction qualification of a marginal rolling loop also matters: the contact threshold presumes the stated no-slip idealization. At equality, near the top, required tangential friction vanishes linearly with angular distance from the top but $N$ vanishes quadratically. Thus any finite coefficient of static friction requires a release height above this limiting bound to maintain rolling everywhere; no coefficient is provided. This does not change the requested ideal contact criterion.

For collinear inelastic capture followed by two photons, each particle has $\mathbf{p}=\gamma m\mathbf{u}$ and $E=\gamma m{c}^2$, with $\gamma =(1-u^2/c^2{)}^{-1/2}$. An isolated collision conserves the vector sum of momentum and total energy, including rest energy, internal excitation and radiation. It need not conserve the sum of individual rest masses. At low speed,recovering the classical kinetic energy after subtracting $m{c}^2$.

Here the incoming Lorentz factors are $5/3$ and $5/4$, soThe composite therefore hasIts rest mass exceeds $7m$ because relative kinetic energy has become internal rest energy.

For the subsequent collinear photon pair, $E_{+}+E_{-}=10m{c}^2$ and $(E_{+}-E_{-})/c=mc$. Thus $E_{+}=11m{c}^2/2$, $E_{-}=9m{c}^2/2$, andThe positive-$x$ photon is the higher-frequency one.