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Descent equation. Take downward velocity $v$ and underwater depth $z$ as positive, with $z=0$ at the lake surface. For a sphere of radius $r$, the pebble mass is $m=4\pi r^3{\rho }_p/3$ and the frontal area is $A=\pi r^2$. Newton's second law includes downward gravitational weight, upward buoyancy and upward quadratic drag:ThusHere $b$ has units of inverse length. For arbitrary signed velocity, the last term is $-bv|v|$, so drag always opposes motion. The displayed downward form suffices for this trajectory.

Limiting speed. At terminal velocity, the net force and $\dot{v}$ vanish. For ${\rho }_p>{\rho }_w$,Use ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_w=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $r=0.020\mathrm{m}$, $C=0.7$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. ThenThe ${\rho }_p-{\rho }_w$ factor is essential: omitting buoyancy would overestimate the limiting speed.

For the numerical descent, assume the pebble is released from rest, neglect air drag, take $g$ and the water mass density as constant, and apply the given constant drag coefficient once it is submerged. Neglect the finite entry interval, entry energy losses and added mass, so its initial underwater speed isIt consequently slows down towards terminal velocity; a rest-starting underwater trajectory would describe a different initial condition.

Plot position $x$ horizontally and $ct$ vertically, with equal length scales, so vacuum light rays have slopes $\pm 1$. A possible cause of the spacetime event $A$ must lie on or inside its past light cone, with $\Delta t=t_A-t_{\mathrm{cause}}>0$. Equivalently, causality requires $|\Delta x|\le c\Delta t$.

From $B$ to $A$, $\Delta x=6\mathrm{m}$ and $\Delta t=10^{-8}\mathrm{s}$. Using $c\simeq 3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the available light-travel distance is only $c\Delta t=3.00\mathrm{m}$. In the $(-,+,+,+)$ metric signature, the Minkowski interval isThere is spacelike separation, so a signal would need speed $6\times 10^8\mathrm{m}{\mathrm{s}}^{-1}\simeq 2c$:From $C$ to $A$, $\Delta x=-2\mathrm{m}$ and $\Delta t=1.5\times 10^{-8}\mathrm{s}$. Now $c\Delta t=4.50\mathrm{m}>|\Delta x|$, givingThis is timelike separation, with a possible signal velocity $-2/(1.5\times 10^{-8})\simeq -1.33\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. Therefore

There is no external horizontal force, so momentum conservation keeps the student–punt centre of mass fixed if the system starts at rest. Let the student's walking direction be positive. If the punt's displacement relative to the water is $X$, the student advances $X+L$, where $L=3\mathrm{m}$ is measured relative to the punt. Constancy of the centre of mass givesThe punt moves $1.04\mathrm{m}$ opposite to the student's walking direction; the student's net displacement relative to the water is $3+X\simeq 1.96\mathrm{m}$. These displacements are relative to the initial state, not distances travelled relative to the punt.

Under the same ideal assumptions, merely changing the walking speed or walking back more slowly cannot produce unlimited travel. Returning the student to the original place on the punt restores the original internal mass distribution and hence the punt's original position. A finite punt bounds the displacement available through internal rearrangement. Propulsion over large distances requires an external horizontal impulse, for example a push against the bed or momentum transferred to the water. Such an interaction changes the isolated, frictionless model rather than evading centre of mass conservation.

Take the initial internal pressure as negligible and the external atmospheric pressure as $p_0=1.01325\times 10^5\mathrm{Pa}$. The window has frontal area $A=\pi (0.400\mathrm{m}{)}^2=0.50265{\mathrm{m}}^2$. Before appreciable air has entered, its inward pressure force is approximately $p_{0}A=5.09\times 10^4\mathrm{N}$. Newton's second law therefore gives the initial collective inward accelerationHere $M=110\mathrm{kg}$ and $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$. Neglect gravity in this much larger pressure-driven acceleration, together with supports and elastic stresses once the glass has broken. The given thickness $d=0.080\mathrm{m}$ implies a plausible average glass mass density $M/(Ad)\simeq 2.74\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Equivalently, a fragment retaining the original thickness has acceleration $p_0/({\rho }_{g}d)$, independent of its frontal area while it experiences the full pressure difference.

For a vacuum-window implosion energy estimate, the ideal pressure-work scale for filling vessel volume $V=1.5{\mathrm{m}}^3$ is $W\simeq p_{0}V$. Assigning all that pressure-volume work to the glass's kinetic energy givesThe same result follows from retaining acceleration $a_0$ over an effective swept distance $V/A\simeq 2.98\mathrm{m}$, so that $v_{\mathrm{est}}^2=2a_{0}V/A$. This is an ideal upper energy-budget estimate for the common speed, or mass-weighted root mean square speed, of all the glass. It assumes an atmospheric reservoir maintains the driving pressure and that the whole vessel's pressure-work budget can accelerate the glass. In practice, air motion, fragmentation, heating and collisions share that energy, the pressure difference falls, and fragments may hit the vessel after a much shorter path. The unspecified vessel shape and fragment trajectories prevent a unique actual maximum speed; concentrating energy into one small fragment also prevents interpreting this collective estimate as a strict bound on every individual fragment.

Choose the $M$ ball's initial velocity as $+v$; the $3M$ ball then has velocity $-v$. Both have Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Their total relativistic energy and signed relativistic momentum areThe laboratory frame is not the centre-of-momentum frame, despite the equal and opposite speeds. Four-momentum conservation and the relativistic energy-momentum relation give the invariant mass $M_f$ of the combined incoming system:The relativistic collision mass threshold for a final object of rest mass $5M$ is $M_f\ge 5M$. At the threshold,At equality, the outgoing object has laboratory velocity $V=c^2{p}_i/E_i=-v/2$, directed towards the initially heavier ball's motion. Its nonzero final kinetic energy is why setting $4\gamma M{c}^2=5M{c}^2$ would give the wrong threshold.

For the literal rest mass of a coalescing relativistic collision, if there is exactly one outgoing object of exactly fixed rest mass $5M$ and no radiation or other products, four-momentum conservation selects the equality above, rather than allowing every larger $v$. Speeds above this threshold require a larger final invariant mass or additional outgoing energy carriers.

Let $\mathbf{r}$ be the orbiting point mass's position relative to the attracting body, and let its velocity be $\mathbf{v}=\dot{\mathbf{r}}$. Its angular momentum isConservation of angular momentum keeps both the magnitude and direction of $\mathbf{L}$ fixed. Since $\mathbf{L}$ is perpendicular to $\mathbf{r}$, a nonradial orbit lies in a fixed plane. During a short interval $dt$, the radius vector sweeps a thin triangle of areaThus the areal velocity is constant:Equal time intervals give equal swept areas: this is Kepler's second law. In plane polar coordinates, the same statement is $L=m{r}^2\dot{\varphi }$ and $\dot{A}=r^2\dot{\varphi }/2$. The attracting body is at the relevant focus of the ellipse, not its geometric centre. The area argument needs conserved angular momentum, not specifically an ellipse; a central force provides this conservation because its torque $\mathbf{r}\times \mathbf{F}$ vanishes.

The disc carries mass $M_d=2M/3$ and the thin rim carries $M_r=M/3$. About the wheel's central normal axis, add the moment of inertia of a uniform disc and the thin rim's moment of inertia:Treat the wheel as a thin planar lamina, as implied by the disc-and-thin-rim model. Interpret the requested perpendicular axis as perpendicular to the wheel's original axis and tangent to the rim: if the wheel lies in the $xy$ plane, choose the tangent line $x=R,z=0$, parallel to the $y$ diameter. By rotational symmetry, the two diametral moments of inertia are equal. The perpendicular axis theorem therefore givesThe parallel axis theorem then gives the tangent-axis inertia of an axisymmetric lamina:The axis direction matters. If “perpendicular” were intended to mean perpendicular to the wheel's plane, an axis through a rim point would instead be parallel to the original wheel axis, giving $I_{\mathrm{rim},\mathrm{normal}}=I_z+M{R}^2=5M{R}^2/3$. The $4M{R}^2/3$ answer uses a tangent axis perpendicular to the original axis; the $5M{R}^2/3$ value belongs to the different, explicitly identified normal axis. Finite wheel thickness would require additional information for the in-plane moment of inertia.

Energy–momentum transformations and invariant. Let the primed inertial frame move at signed velocity $u$ along the unprimed $x$ axis. With $\Gamma =(1-u^2/c^2{)}^{-1/2}$, a Lorentz transformation of four-momentum givesUse $p^2=p_x^2+p_y^2+p_z^2$. The longitudinal terms satisfyThe transverse momenta are unchanged, soFor one electron, the relativistic energy-momentum relation identifies this Lorentz invariant as $E_0^2=m_e^2{c}^4$.

Symmetric electron scattering. Assume an isolated elastic collision of the two electrons, with no emitted radiation. The incident electron has total energy $E$ and momentum magnitude $p$, so $p^2{c}^2=E^2-E_0^2$; the target electron initially contributes rest energy $E_0$ and zero momentum. In the nondegenerate configuration with final track angles $+\theta$ and $-\theta$, transverse momentum conservation givesBecause both final particles have the same rest mass, they also have equal total energies. Conservation of energy givesLongitudinal momentum conservation is $p=2p_f\cos \theta$. Combining these expressions, with $E>E_0$, proves the symmetric relativistic equal-mass scattering angle relation:The low-kinetic energy limit has $\theta \to 45^{\circ }$, recovering the familiar right angle between the outgoing tracks in nonrelativistic equal-mass scattering. Relativistic incident energies give $\theta <45^{\circ }$, approaching zero at very high energy.

Accelerating potential. For $\theta =30^{\circ }$, ${\tan }^2\theta =1/3=2E_0/(E+E_0)$, so $E=5E_0$. The acceleration from rest supplies kinetic energy, not total energy. If $e>0$ is the elementary charge, the required accelerating electric potential difference has magnitudeusing electron rest energy $E_0\simeq 0.511\mathrm{MeV}$. The electric potential difference is oriented to accelerate the negative electron; the result above specifies its magnitude.

For a sphere with uniform mass density, $M=(4\pi /3)\rho a^3$. Its moment of inertia of a uniform solid sphere is $I=2M{a}^2/5$, and rotation about a diameter gives angular momentum magnitude $L=I\omega$. The spheres have the same angular frequency $\omega$, soThe mass grows by a factor of two, and the extra squared lever arm supplies a further factor of four. Equal angular frequencies alone do not mean equal angular momenta.