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For a Newtonian fluid with velocity $u(z)\hat{\mathbf{x}}$, $\tau =\eta {\partial }_{z}u$ is the $x$ component of traction on a plane whose normal points along $+z$. The opposite face has the opposite traction. The coefficient $\eta$ is dynamic viscosity, with units $\mathrm{Pa}\mathrm{s}$. Faster layers transfer positive $x$ momentum toward slower neighbours; the internal shear stress resists relative sliding.

Take a thin fluid element with horizontal area $A$ and thickness $dz$. The difference between its upper and lower viscous tractions is $A\eta u_{zz}dz$. The pressure gradient supplies $-A{p}_{x}dz$. Incompressibility is satisfied because $u$ has no $x$ dependence, and the convective acceleration vanishes: $(\mathbf{u}\cdot \nabla )\mathbf{u}=u{\partial }_x\mathbf{u}=0$. Equating net force to mass times acceleration therefore gives the relevant component of the Navier-Stokes equation:For steady flow, integrate $\eta u^{\prime \prime }=p^{\prime }$ twice and impose the no-slip boundary condition at each plate, $u(0)=0$ and $u(a)=u_0$. The Couette-Poiseuille flow profile isThe linear term is Couette flow and the parabolic term is plane Poiseuille flow. A negative $p^{\prime }$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,It has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying by the plate width gives the actual volume per time.

With zero pressure gradient, substituting $u=\mathrm{Re}[A{e}^{i(kz-\omega t)}]$ gives $-i\rho \omega =-\eta k^2$. Select the root that decays into fluid occupying $z>0$:The other sign applies to decay in the opposite half-space. With a real wall velocity amplitude $U_0$, the established periodic solution of Stokes' second problem isThe oscillation's amplitude decays exponentially and its phase lags with distance from the plate. This is diffusion of tangential momentum, not a freely propagating undamped elastic wave. Initial transients must have decayed before this purely periodic expression applies.

The fluid-displacement amplitude and the wall-displacement amplitude have the same ratio as their velocity amplitudes, since both are divided by the common $\omega$. Thus for the specified amplitudes,Using the standard approximate water density $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, explicitly an approximation rather than another value printed on the paper, gives $\delta =0.564\mathrm{mm}$ and $z\simeq 2.21\mathrm{mm}$ from either face of a sufficiently large plate. The viscous penetration depth varies as ${\rho }^{-1/2}$ if a more precise water density is used. The ideal parallel-flow solution ignores finite plate edges and tank walls.

An elastic reflection transfers normal momentum $2mv\cos \theta$ to the wall. Multiplying this impulse by the collision flux and integrating givesUsing the radial integral and $\alpha =m/(2k_{B}T)$ yieldsThis agrees with the ideal-gas equation of state. The calculation assumes a stationary wall and elastic normal momentum reversal, appropriate to a gas in equilibrium with it.

A gravitational assist changes a spacecraft's velocity relative to the Sun by scattering it in the gravitational field of a moving planet. In the planet's rest frame, ignore other forces during the encounter and take the spacecraft mass negligible. The incoming and outgoing asymptotic velocities have equal magnitude,because the encounter is conservative, but their directions differ along a hyperbolic orbit. If the planet has approximately constant heliocentric velocity ${\mathbf{V}}_p$, the heliocentric spacecraft velocities are ${\mathbf{v}}_{\mathrm{in},\mathrm{out}}={\mathbf{V}}_p+{\mathbf{u}}_{\mathrm{in},\mathrm{out}}$. Its specific energy change isThus a flyby can increase or decrease heliocentric energy depending on its geometry. The planet exchanges momentum and orbital energy with the spacecraft; its speed change is extremely small because its mass is much greater. There is no energy creation and no increase of asymptotic speed in the planet's own frame.

For a point-mass planet, conservation of energy and angular momentum gives the hyperbolic eccentricity and deflection anglewhere $r_p$ is periapsis distance and $b$ the impact parameter. The velocity-vector change has magnitude $2u_{\infty }\sin (\delta /2)$, so the maximum energy-change magnitude for this fixed deflection is $2V_p{u}_{\infty }\sin (\delta /2)$. A small periapsis increases the turning angle, but collision, atmosphere and tidal forces limit how close a real spacecraft may pass. Multiple encounters allow large changes without expending the propellant that equivalent direct velocity changes would require.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

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.

Direct differentiation gives ${\psi }^{\prime \prime }/\psi =x^2/a^4-1/a^2$. Inserting this into the Schrödinger equation and separately matching the $x^2$ and constant terms yieldsFor $V_0=m{\omega }_0^2/2$, $a^2=\hslash /(m{\omega }_0)$ and $E=\hslash {\omega }_0/2$. The normalizable Gaussian has no nodes, as expected for the nondegenerate one-dimensional ground state of this confining potential.

The normalization integral is $|C{|}^{2}a\sqrt{\pi }=1$. Choose its irrelevant global phase so thatFor the oscillator ground-state density in a comparison wavelength, choose a free comparison particle with the same total energy. It has $p=\sqrt{2mE}=\hslash /a$, hence comparison wavelength $\lambda =h/p=2\pi a$. ConsequentlyThe bound density is localized on scale $a=\lambda /(2\pi )$, with standard deviation $\lambda /(2\pi \sqrt{2})$, rather than being spatially uniform like the density of a single free plane wave. The comparison does not assign constant free-particle momentum to the bound state; its total energy also includes potential energy.

The net force equals the time derivative of momentum in an inertial frame:

For $m>0$, $E={\gamma }_{u}m{c}^2$ and $\mathbf{p}={\gamma }_{u}m\mathbf{u}$, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$. ThusA photon obeys $E=pc$; $E=hf$ and $p=h/\lambda$ agree because $f\lambda =c$ in vacuum.

Take the incident photon along $+x$, and write $\beta =v/c$ for the boost. In the boosted frame the initial momenta areTheir sum vanishes when $1-2\beta =0$, giving $v_{\mathrm{ZM}}=c/2$ and ${\gamma }_v=2/\sqrt{3}$. The initial photon energy there is $m{c}^2/\sqrt{3}$, and total energy is $\sqrt{3}m{c}^2$.

For right-angle Compton scattering in the centre-of-momentum frame, choose the outgoing ZM photon along $+y$. The electron has the opposite momentum $-p^{\prime }\hat{\mathbf{y}}$. Conservation of energy determines its magnitude:Thus ${\mathbf{u}}_e^{\prime }=-(c/2)\hat{\mathbf{y}}$. Inverting the velocity transformation givesConsequentlyThe electron recoils below the incident direction for our choice of photon scattering above it; the reflected configuration reverses both transverse signs. Finally,As a check, $E_{e,\mathrm{lab}}=4m{c}^2/3$, so the final energies sum to the initial $2m{c}^2$. The outgoing laboratory photon makes $60^{\circ }$ with $+x$.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

The de Broglie wavelength fixes the same momentum $p=h/\lambda$, but the nonrelativistic electron has $K=p^2/(2m_e)$:Here $K_e/(m_e{c}^2)\simeq 0.00118$, consistent with the stated nonrelativistic approximation. The photon energy and electron kinetic energy are not interchangeable.

Take inertial axes parallel, with frame $S^{\prime }$ moving at $+v\hat{\mathbf{x}}$ relative to $S$ and origins coincident at zero time. The Lorentz factor is ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$; $\Delta x,\Delta y,\Delta t$ are separations of the same pair of events in $S$, and primed quantities are their separations in $S^{\prime }$. The transverse coordinates are unchanged. For two events on an inertial clock's worldline,The proper time $\Delta \tau$ is that clock's elapsed time; uniform speed $w$ gives $\Delta t={\gamma }_w\Delta \tau$. Coordinate intervals between spatially separated events must instead be Lorentz-transformed with their spatial separation included.

For the round trip in the station frame, the rendezvous distance is $u{t}_{\mathrm{out}}$, and the return has the same speed and distance. Thus $t_{\mathrm{round}}=2t_{\mathrm{out}}=10a/(\sqrt{7}c)$. Its departure and arrival both occur at the station, so $\Delta x=0$ for these endpoints:This uses the return speed as measured by the station. The elapsed time aboard A, apart from an ideal instantaneous turn, would be $\Delta t_{S,\mathrm{round}}/{\gamma }_u$.

For the crash, A's station-frame velocity components are $(3c/5,\sqrt{7}c/5)$, while B's are $(3c/5,0)$. Here ${\gamma }_u=5/3$ and ${\gamma }_v=5/4$. Four-momentum conservation, with no separately escaping radiation, gives the wreckage totals in the station frame:In B's original inertial frame the same totals areThe invariant wreckage mass is $m\sqrt{14/3}$, exceeding $2m$ because internal energy remains in the wreckage. Its momentum cannot be specified without naming a frame; both natural frames have been given.

The Lorentz force determines the trajectory:Thus a magnetic field changes momentum direction but does no work; an electric field can change the particle's kinetic energy. In a uniform electric field with negligible magnetic field and nonrelativistic speed, $\mathbf{a}=q\mathbf{E}/m$ is constant. A transverse initial velocity then gives a parabola.

In a uniform magnetic field, the parallel velocity is unchanged and the perpendicular motion is circular. Nonrelativistically,the combined trajectory is a helix, with circulation reversed when the charge sign reverses. Relativistically $r_L=p_{\perp }/(|q|B)$ and the gyrofrequency is $|q|B/(\gamma m)$.

For perpendicular uniform fields in the magnetic-dominated case, the drift velocity is ${\mathbf{v}}_d=\mathbf{E}\times \mathbf{B}/B^2$, which obeys $\mathbf{E}+{\mathbf{v}}_d\times \mathbf{B}=0$. A particle at this velocity is undeflected: the crossed-field velocity selector selects speed $E/B$ in the appropriate direction, independent of mass and charge. Radiation reaction and collisions are neglected in these ideal trajectories.

In standard configuration the frames are inertial, their corresponding axes are parallel, S${}^{\prime }$ moves at $v$ along the positive $x$ axis of S, and their origins coincide at $t=t^{\prime }=0$. Here $(t,x,y,z)$ and $(t^{\prime },x^{\prime },y^{\prime },z^{\prime })$ label the same event, $c$ is the speed of light and $\gamma =(1-v^2/c^2{)}^{-1/2}$.

A clock fixed in S${}^{\prime }$ has $\Delta x^{\prime }=0$. The inverse time transformation givesso the moving clock accumulates less proper time between its two ticks than the elapsed S-coordinate time.

Without time dilation the flight time from the given altitude would be $66.7\mu \mathrm{s}$, about $30.3$ rest-frame mean lifetimes, giving survival fraction only $e^{-30.3}\simeq 6.8\times 10^{-14}$ even at speed $c$. Relativistic muons can instead reach the ground with appreciable survival. Momentum-selected decay measurements test this clock-rate dependence experimentally.^[ref]

A mean-lifetime survival energy estimate requires a mean flight time no greater than the dilated mean lifetime:This corresponds to survival probability at least about $e^{-1}$. There is no sharp minimum energy for an individual random decay. For a required probability $p$, use $\exp [-H/(\gamma v{\tau }_0)]\ge p$. Energy loss in the atmosphere is neglected in this estimate. In the muon's frame the atmospheric path is length-contracted to $H/\gamma \simeq 660\mathrm{m}$; the ground reaches it in about ${\tau }_0$, giving the same survival probability.

For the two-body decay, use the rest frame of A. Momentum conservation makes B and the photon oppositely directed with equal momentum magnitude $p$. Energy conservation and the energy–momentum relation giveSubstitution and cancellation of $E_B^2$ giveA nonzero emitted photon requires $m_A>m_B$.

About a specified axis, the moment of inertia is $I=\int r_{\perp }^{2}dm$, giving rotational kinetic energy $I{\omega }^2/2$ for fixed-axis rotation. With square coordinates centred on the plate,The in-plane inertia isotropy of a square plate follows using the perpendicular axis theorem, which gives $I_z=I_x+I_y=M{l}^2/6$. For any in-plane central axis at angle $\varphi$, the perpendicular in-plane axis has the same inertia by the square's $90^{\circ }$ rotational symmetry. Their sum is $I_z$, so every such in-plane axis has $I=M{l}^2/12$.

The parallel axis theorem gives hinge inertia $I_H=M{l}^2/3$. During the short impulse gravity is negligible, and the hinge impulse has no moment about the hinge. ThereforeThe hinge impulse of an edge-pivoted square plate follows from linear momentum: $J_H=Mv-K$. Taking its positive direction to be that of the applied impulse,Generally $K$ alone does not supply the centre-of-mass momentum compatible with rotation about the fixed hinge, so the hinge supplies the difference. Its impulse vanishes at the centre of percussion,

Newton's first law identifies inertial frames in which a force-free body maintains constant velocity. The second is $\dot{\mathbf{P}}={\mathbf{F}}_{\mathrm{net}}$ for a closed system; for fixed mass it becomes $m\mathbf{a}=\mathbf{F}$. The third gives equal and opposite pair forces. Summing the second law over a closed collection cancels internal force pairs, so total momentum is conserved when the resultant external force vanishes. A rocket alone is an open variable-mass subsystem, so one must include the momentum carried by exhaust before applying this reasoning.

Choose downward velocity as positive and write $M(t)=M_0-\alpha t$. During $dt$, exhaust mass $dm=\alpha dt$ leaves downward with speed $v+v_E$ in the surface frame. To first order, momentum balance for rocket plus this exhaust givesIntegration, with $v(0)=v_0$, yieldsSetting $g=0$ gives the gravity-free result. This assumes a constant downward relative exhaust speed, constant mass-loss rate and $M_0-\alpha t>0$ throughout the burn; the negative logarithm supplies the upward thrust effect.

Height satisfies $\dot{h}=-v$. Put $u(t)=1-\alpha t/M_0$. Integrating the logarithm and choosing $h(0)=h_0$ givesEquivalently, the last term is $v_E[t+(M_0-\alpha t)\log u/\alpha ]$. Differentiating either form recovers $-v$ and checks the sign.

For this constant-rate retro-rocket landing, at touchdown $u_L=3/4$ and $v_L=0$, henceThe fuel condition gives $\alpha =M_0/(4t_L)=200.30$ kg/s. Requiring $h(t_L)=0$ then givesIn this burn the downward speed decreases monotonically to zero: the initial thrust acceleration already exceeds gravity and increases as fuel is consumed. The result is the idealized vertical landing, without atmospheric drag, changing gravity or a separate terminal burn phase.

In the laboratory frame the incoming particle and the composite move along $+x$, while the target is initially at rest. In the equal-energy two-photon decay of a moving particle, zero total transverse momentum makes the two photon directions symmetric about $+x$. Each photon has energy $2m{c}^2$ and momentum magnitude $2mc$, so longitudinal momentum conservation gives $2(2mc)\cos \theta =2.4mc$. ThereforeTheir momenta can be written $(1.2,\pm 1.6)mc$. The opening angle is not $180^{\circ }$ in this frame, because the parent carried nonzero total momentum.

Boost along $+x$ at $0.6c$ into the centre-of-momentum frame. The composite is stationary. The incoming particle has $p_x^{\prime }=1.25(2.4-0.6\times 3)mc=0.75mc$, while the target has $p_x^{\prime }=-0.75mc$. After decay the two photons have energies $M{c}^2/2=1.6m{c}^2$ and opposite momenta.

Equal laboratory photon energies require the rest-frame photons to have zero momentum component along the boost: $E_{\mathrm{lab}}=\gamma (E^{\prime }+V{p}_x^{\prime })$ must be equal for two opposite photons. Their directions are therefore $+y$ and $-y$ in the centre-of-momentum diagram, perpendicular to the incoming collision axis. Boosting those transverse photons gives exactly the laboratory momenta in part (a). A stationary composite has no direction-of-travel arrow in this frame.

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.

The ground-state energies are $1.50\mathrm{eV}$ and $0.376\mathrm{eV}$, respectively. For a particle in a box of width $L$, the wave function must vanish at both infinite walls. The lowest nonzero standing wave therefore has wavelength $2L$, giving the kinetic energy eigenvalueUsing the electron mass and converting joules to electronvolts givesDoubling the confinement length divides this kinetic energy by four. A zero potential energy inside the box does not imply zero total energy: a constant, zero-momentum wave function cannot satisfy both wall conditions unless it vanishes everywhere. This is a concrete manifestation of quantum confinement and the Heisenberg uncertainty principle.