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Let $\dot{M}=500\mathrm{kg}{\mathrm{s}}^{-1}$ and $V=10\mathrm{m}{\mathrm{s}}^{-1}$. Incoming material has zero horizontal momentum, while incorporation requires horizontal momentum $VdM$. Thus the engine supplies force $F=V\dot{M}$, and its power isNo duration or loading distance is given, so a single total work cannot be assigned. Over an interval $t$ in seconds,The remaining $25,000t\mathrm{J}$ becomes internal energy in the inelastic collision that brings the falling material to the wagon's horizontal velocity. Vertical falling kinetic energy, if any, is a separate contribution to dissipation; it does not alter this horizontal work balance. The source alternates between stone and coal; the calculation depends only on the given loading rate.

Use the angle $\theta$ from the vertical as the generalized coordinate. For a uniform ladder, its centre of mass is $(l\sin \theta ,l\cos \theta )$, so $v_G^2=l^2{\dot{\theta }}^2$. Its central moment of inertia is $I_G=m(2l{)}^2/12=m{l}^2/3$. With ideal contact constraints that do no work, the Lagrangian, conjugate momentum, and Euler-Lagrange equation areRelease from rest at ${\theta }_0$ also gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$ by conservation of energy. These results hold while both contacts persist. The ideal wall reaction is normal to the wall; an unspecified tangential wall friction would require an additional generalized force and would prevent determination from the supplied data alone.

Treat the carriage as translating along the circular track, neglecting wheel rotational kinetic energy and its finite size. At the top, gravity and the inward normal force supply the centripetal acceleration:For unilateral contact, $N\ge 0$, so the limiting speed is $v_{\mathrm{top}}^2=gr$. Conservation of energy from rest at height $h$ then givesThis top contact condition is the most restrictive: at angle $\varphi$ measured from the bottom, $N/(mg)=2h/r-2+3\cos \varphi$, whose minimum occurs at the top. A mechanically captive carriage able to sustain an outward rail reaction has a different constraint; the result here uses the usual requirement of positive normal contact.

Using rotational kinetic energy $K=I{\omega }^2/2$ at the two steady endpoint configurations,At fixed angular momentum $J$, $K=J^2/(2I)$ increases when the moment of inertia decreases. The skater supplies work by pulling the arms inward; chemical energy in the muscles is converted into additional rotational kinetic energy, with possible extra dissipation in a real skater. This does not require an external axial torque.

For the ideal rotating point masses, an inward radial force supplies the required centripetal acceleration. During contraction both that inward force and the radial displacement point inward, so their scalar product gives positive work. At the initial and final steady configurations radial motion has ceased; any transient radial kinetic energy must also be included in a full intermediate-time balance. The quoted difference compares the rotational endpoint energies only.

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.

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.

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.

A circular orbit of radius $r_0>0$ is a stationary point of the effective potential:For $L\ne 0$, $\beta k>0$ and $k\ne -2$, this givesThe condition $\beta k>0$ makes the radial force attractive. At $k=-2$, the equation instead requires the fine tuning $L^2/m=-2\beta$; then the centrifugal and potential terms cancel at every radius and there is no isolated restoring minimum. If that equality fails, there is no positive-radius circular orbit for this special power.

The two positive powers below require $\beta >0$; the negative power requires $\beta <0$, so that the force is attractive. To leading order, $r\simeq r_0[1+ϵ\cos (\sqrt{k+2}\varphi +\delta )]$. The azimuthal advance from one radial maximum to the next isFor $k=2$, ${\omega }_r/\Omega =2$: there are two radial oscillations per revolution. The isotropic harmonic oscillator has orbits that are exact ellipses centred on the force centre, and the perturbed circle is such an ellipse.

For $k=7$, ${\omega }_r/\Omega =3$: the leading perturbation has three radial lobes per revolution, with consecutive radial maxima separated by $2\pi /3$. Its small-amplitude sketch is a rounded three-lobed orbit.

For $k=-7/4$, ${\omega }_r/\Omega =1/2$: the radius completes one oscillation while the body makes two revolutions. The leading curve closes after $\Delta \varphi =4\pi$, with a double-wound, generally self-crossing shape; the nearest and furthest points occur in the same spatial direction one revolution apart.

The rational ratios for $k=7$ and $k=-7/4$ describe closure of the linearized near-circular curves. They do not imply that every finite-amplitude orbit closes exactly: higher-order changes in the radial-to-orbital frequency ratio generally produce apsidal precession. Bertrand's theorem identifies the harmonic potential $k=2$ and the Kepler potential $k=-1$ as the exceptional attractive central potentials with all nearby bounded orbits closed.

The lines of centres from either lower cylinder to the upper cylinder make $60^{\circ }$ with the horizontal. Let $N_L,N_R$ be the upward normal forces on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal force. While all three contacts remain closed, the triangular configuration translates horizontally with common acceleration $a$; frictionless normals exert no torques about the cylinder centres.

For the top cylinder, Newton's second law givesThus $N_L=m(g/\sqrt{3}+a)$ and $N_R=m(g/\sqrt{3}-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, soA contact may push but cannot pull: $N_L,N_R,Q\ge 0$. The limiting inequalities are thereforeAt the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal forces are strictly positive only inside the open interval. The ground normal forces are positive throughout this range. The applied horizontal force on the left cylinder is $F=3ma$, as follows from the whole system's horizontal momentum balance.

Smooth ground, rough cylinder contacts. Write $f$ for the tangential contact force magnitude at each upper-to-lower contact. On the top cylinder, its tangential components point upward and outward, while their horizontal components and torques cancel by symmetry. For a lower cylinder, the tangential force has an inward horizontal component $\sqrt{3}f/2$ and a downward vertical component $f/2$.

The rolling condition is now at the contact between cylinders, not at the ground. At the initial zero velocity, equality of the tangential contact-point accelerations givesThus $fR=\beta m{R}^2|\alpha |$ gives $f=2\beta mw$. The lower cylinder's outward translation equation and the top cylinder's downward equation areUsing $u=\sqrt{3}w$ gives $N=2\sqrt{3}m(1+\beta )w$ and henceFor solid cylinders this is $g/11$. Here the left lower cylinder rotates clockwise while moving left, and the right rotates counterclockwise while moving right. They may slip freely on the smooth ground, so imposing the ground-rolling relation $R|\alpha |=u$ would be incorrect. No slip at the cylinder contacts requires ${\mu }_c\ge f/N=\beta /[\sqrt{3}(1+\beta )]$ initially.

These are the two cases of the initial acceleration of three stacked cylinders. As a check, as $\beta \to 0$ the rotational contribution disappears in both formulas and they approach the smooth-contact release result $w=g/7$.

Rough ground, smooth cylinder contacts. Let $N$ be either upper-to-lower normal force, and $f_g$ the magnitude of the inward ground friction force on each lower cylinder. Pure rolling on the ground requires $|\alpha |=u/R$. The torque equation about a lower cylinder's centre isThe outward translation equation is $mu=N/2-f_g$, giving $N=2m(1+\beta )u$. The top cylinder's vertical equation is $mw=mg-\sqrt{3}N$. Using $u=\sqrt{3}w$ yieldsFor uniform solid cylinders, $\beta =1/2$ and $a_{\mathrm{top},\mathrm{down}}=g/10$. The left lower cylinder rotates counterclockwise while rolling left; the right rotates clockwise while rolling right.

The assumed no-slip ground contact is possible if its coefficient of static friction satisfies ${\mu }_g\ge \beta \sqrt{3}/(10+9\beta )$ at release. This follows from $f_g=\beta m\sqrt{3}w$ and ground normal force $N_g=mg+\sqrt{3}N/2=mw(10+9\beta )$.

Use spherical coordinates about the fixed pivot, with fixed rod length $l$, polar angle $\theta$ measured from the downward vertical, and azimuth $\varphi$. The bob's position iswhen the Cartesian $z$ axis points upward. Differentiation gives kinetic energy $T=m{l}^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$. Taking zero gravitational potential energy at pivot height gives $V=-mgl\cos \theta$. The spherical pendulum's Lagrangian is thereforeThe rod is light, rigid and of fixed length; pivot friction and air drag are neglected. A rigid rod, unlike a string, may sustain either tensile or compressive constraint force.

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.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.

In the frame following the turn, the outward inertial force $m{v}^2/R$ acts at height $h$ and tends to unload the inner rail. The two upward normal forces satisfyThe second equation is moment balance about the midpoint between the rails. ThusFull contact requires a nonnegative inner normal force. Its limiting value is zero, giving the toppling speed on level railsFor strictly positive contact force on both rails, $v<v_0$. This ideal result assumes level rails, negligible suspension motion and enough lateral contact force to follow the curve; lateral sliding could impose a different limit.

Let $u_e>0$ be the constant rearward rocket exhaust speed, and let $\mathcal{M}$ denote the rocket's instantaneous mass. Apply momentum conservation over a short ejection interval: the remaining rocket gains $\mathcal{M}dv$, while the expelled positive mass $-d\mathcal{M}$ carries backward relative momentum $u_{e}d\mathcal{M}$. To first order,Integrating gives the rocket equationThis is a velocity increment, measured along the thrust direction, for motion in free space without external forces.

For the two-stage vehicle, take $M_1$ and $M_2$ to be the initial masses of the respective stages, each including its fuel $m$. The first burn changes the total mass from $M_1+M_2$ to $M_1+M_2-m$. Discarding the empty first-stage shell of mass $M_1-m$ leaves mass $M_2$; ideal jettison without a relative impulse does not change the velocity. The second burn then changes $M_2$ to $M_2-m$. Adding the two rocket equation increments givesHere $M_1>m$ and $M_2>m$ leave positive dry masses. Different stage exhaust speeds would multiply their respective logarithms. A multistage rocket gains its advantage by avoiding acceleration of the spent first-stage shell during the second burn.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.