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Put $m_A=M=m$ and $m_B=2m$. At reversal the transferred mass is $q_r=0.45m$, leaving A with $1.55m$ and, after capture of that exhaust, B with $2.45m$. The first burn givesAfter reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this velocity. Applying the rocket equation with reversed thrust to A givesBoth now move toward negative x, but A moves faster in that direction. The asymptotic closing speed isThis conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing speed closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual momentum after collecting that fixed amount.

For the eccentric circular-hole disk inertia, let $D=\pi \rho R^2$ be the mass of the complete disc before removing material, $h=\pi \rho r_o^2$ the missing mass, and $W=D-h$ the wheel's remaining mass. Denote the complete disc's centre by C and the axle/hole centre by H; $CH=r_o$. Assume the hole fits inside the disc, $0<r_o<R/2$. The parallel axis theorem gives the full disc's moment of inertia about H as $\frac{1}{2}D{R}^2+D{r}_o^2$. Subtract the absent little disc about its own centre H:The axle is a support, not part of the rotating wheel's stated mass. Its frictionless bearing does not alter this moment of inertia. For fixed-axis angular velocity $\omega$,The missing material must be subtracted with its actual displacement relative to the axis; treating the wheel as a concentric annulus would give the wrong result.

The supplied PDF labels this equation clarified from original for tutorial purposes. Use the model of an eccentric wheel with missing-mass kinetic energy neglected: neglect the missing disc's contribution to kinetic energy, while retaining its subtraction from gravitational potential energy. Do not silently replace the gravitational mass $W$ by the full-disc mass $D$ as well.

Choose $\theta$ as the clockwise angle of $\overrightarrow{CH}$ from the upward vertical, and set $s=R\theta$. The rolling without slipping constraint makes changes in $s$ equal changes in the distance C travels down the slope. Choose its additive origin so $s=0$ when the hole is vertically above C. Then, up to an irrelevant height constant,The approximate kinetic energy is that of the complete uniform disc rolling at $\dot{s}=R\dot{\theta }$:Subtracting the missing disc's gravitational potential energy givesApply the Euler-Lagrange equation to $\mathcal{L}=T-U$:ThereforeHere $K_1,K_2$ have dimensions of acceleration and $\gamma$ has dimensions of inverse length. The initial conditions are $s(0)=\dot{s}(0)=0$ under the vertical-above interpretation. If an initial orientation is instead specified relative to the slope normal, the same coordinate convention produces a different initial $s$, not different constants. With $r_o\to 0$ the equation becomes $\ddot{s}=(2/3)g\sin \alpha$, the familiar solid-disc result. Retaining the hole's exact kinetic energy would introduce coordinate-dependent inertia and extra ${\dot{s}}^2$ terms; the printed neglect is what permits the requested constant-coefficient form.

Write the lost mass fraction as $f$, with $f=0.02$, and let $\mu =GM$. Initially the circular orbit has $v_0^2=\mu /R$, zero radial velocity, and specific angular momentum $h^2=R^2{v}_0^2=\mu R$. The stipulated rapid symmetric loss leaves the satellite's instantaneous position and velocity unchanged, but reduces the gravitational parameter to ${\mu }^{\prime }=(1-f)\mu$. There is no impulsive kick from contact with ejecta in this model.

The new specific orbital energy isand $h$ remains unchanged because the post-loss force is still central. At a radial turning point,Inserting the known constants and factoring yieldsFor $0<f<1/2$ the two radii are $R$ and $R/(1-2f)$. The initial radius is the pericentre: immediately after the loss, $\ddot{r}=h^2/R^3-{\mu }^{\prime }/R^2=f\mu /R^2>0$, so the satellite begins moving outward. ThereforeEquivalently, the new semi-major axis is $49R/48$ and the orbital eccentricity is $1/49$. These values also satisfy $r_{\min }=a(1-e)$ and $r_{\max }=a(1+e)$. For loss of half the original mass the ideal trajectory becomes parabolic, and larger loss makes it unbound. Slow adiabatic loss would give a different, nearly circular expansion; it is not the instantaneous limit stipulated here.

The figure retains the same angular momentum before and after the change. Once the expelled material is far away, a spherically symmetric outer shell exerts no interior gravitational force by the shell theorem; direct ejecta interaction and the shell's finite transit time are neglected as in the source idealization.

Let the initial velocity be $\mathbf{v}$ and the two final velocities be $\mathbf{u},\mathbf{w}$. In the stated nonrelativistic elastic collision, conservation of momentum and conservation of energy giveSquaring the first equation and comparing it with the second yieldsIf both final velocities are nonzero, their directions are perpendicular. If one is zero, the moving particle retains the initial velocity and a scattering angle for the stationary particle is undefined. This includes the zero-deflection and head-on exchange limits rather than incorrectly assigning a direction to a zero vector. Equal masses and a stationary initial target are essential to the dot-product cancellation.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.

In the parent's rest frame, the identical daughters have equal and opposite momenta, and each has total energy $110\mathrm{GeV}$. With daughter rest energy $Q=90\mathrm{GeV}$, each speed in this frame isTransform to the rest frame of one daughter using the collinear relativistic velocity addition law. The other daughter then has speedAn independent invariant mass check is useful. In the rest frame of one daughter the other's total energy satisfies $W^2=2Q^2+2Q{E}^{\prime }$, so $E^{\prime }=178.8889\mathrm{GeV}$ and the relative Lorentz factor is ${\gamma }_{\mathrm{rel}}=E^{\prime }/Q=161/81$. Hence $w/c=\sqrt{1-{\gamma }_{\mathrm{rel}}^{-2}}$, the same result.

The answer is unchanged if the parent is produced by a fixed-target collision. Its laboratory velocity then differs, but its rest mass and two-body decay kinematics do not. The scalar product of the daughters' four-momenta, equivalently their relative Lorentz factor, depends only on $W$ and $Q$. A common boost cannot change the speed of one measured in the rest frame of the other.^[ref]

Idealize the pole and stick as having negligible transverse thickness. Put the pole at the origin, the stick initially along the $y$ axis, and the struck lower end at $y=-l/2$, as in the source diagram. A short horizontal impulse $J$ along $+x$ gives the centre of mass a constant speed $V=J/m$. The angular impulse about the centre is $Jl/2$, giving constant counterclockwise angular velocityThe pole exerts no further force while the stick moves clear of it. Let $\psi =\Omega t$. The trailing upper end has coordinatesThe pole can lie on the translating straight stick again only when the stick becomes horizontal: for other orientations, its intersection with $y=0$ is the moving midpoint $x=Vt>0$. At the first horizontal orientation, $\psi =\pi /2$, the trailing end is atConsequently the grazing clearance of an impulsively struck rod occurs atFor $\beta >1/(2\pi )$ the whole stick has passed the pole by its first horizontal orientation, and the continually advancing centre makes all later passes safe. Equality is the limiting grazing contact of the endpoint; strict avoidance requires the strict inequality. If $\beta <1/(2\pi )$, the pole lies within the stick at that first horizontal pass, so contact occurs before the freely moving trailing end can pass the pole. A uniform thin stick has $\beta =1/12$, below the threshold: sufficient mass near the ends is needed to increase its moment of inertia and reduce the rotation relative to translation.

Take $M$ as the constant mass of the sled and rider when snow is immediately removed. Assume the falling snow has zero horizontal velocity in the ice frame, and its arrival rate $\sigma$ is the specified constant mass flux onto the sled. In a snow-collecting sled, horizontal momentum flux, including the departing snow, determines the change in sled momentum.

The snow remains aboard, so the moving mass grows as $M+\sigma t$. Incoming snow contributes no horizontal momentum, and there is no horizontal external force. ThereforeThe combined sled–snow system follows a variable-mass system balance; setting $(M+\sigma t)\dot{v}=0$ would omit the incoming mass term.

Use a circular Earth orbit of radius $a$ about the Sun, neglect Earth's rotation, the other planets and atmospheric drag, and treat the launch as an initial impulse followed by unpowered flight of constant mass. DefineIn the patched conic approximation, a surface launch at speed $u$ leaves Earth's gravitational field with Earth-relative hyperbolic excess speed $w$, whereThus the necessary solar-orbit change must be supplied in addition to escaping Earth's gravity; simply adding the two escape speeds is incorrect. The asymptotic departure direction can be chosen to minimize $w$. Representative values are $v_E\simeq 29.78\mathrm{km}{\mathrm{s}}^{-1}$ and $v_{e,E}\simeq 11.186\mathrm{km}{\mathrm{s}}^{-1}$.^[ref]

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.

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.

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.

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.

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.

In a quasi-circular spiral transfer, a small tangential thrust $F$ supplies power $F{v}_c$ while the orbit stays approximately circular. Differentiating $E=-\mu m/(2r)$ and using $v_c=\sqrt{\mu /r}$ givesThis is the integrated thrust magnitude per unit constant mass, not the net vector change of orbital velocity.

For comparison, write $u=\sqrt{r_1/r_0}>1$. Adding the two Hohmann transfer impulses and subtracting the slow-transfer result yieldsThe bracket is negative because $(u+1{)}^2<2(u^2+1)$ for $u>1$. Thus the Hohmann transfer uses less total impulse than the slow circular spiral for these ideal coplanar transfers. The Hohmann transfer compared with a slow spiral does not establish a global optimum over every possible transfer family or propulsion constraint.

Use the supplied planet mass $5.9\times 10^{24}\mathrm{kg}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and adopt mean Earth radius $R_E=6371\mathrm{km}$. For a geostationary orbit, use one sidereal day, $T\simeq 86164\mathrm{s}$. The adopted radius and rotation period follow NASA's fact sheet.^[ref]The slightly smaller radius than the usual terrestrial value follows from using the question's smaller Earth mass. For coplanar prograde orbits, the preceding formulas giveThe escape transfer has $a\to \infty$, so its first tangential burn reaches the local escape speed $\sqrt{2\mu /r_0}$. No final circularization at infinity is needed:Thus the geostationary transfer requires about $0.687\mathrm{km}{\mathrm{s}}^{-1}$ more total impulse, or about 22 percent more, despite leaving the spacecraft gravitationally bound. Reaching the outer radius and circularizing there are separate requirements. Changing the starting orbit's inclination would add a plane-change cost absent from this calculation.