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The specified rocket fuel fraction gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yieldsThen $h(T)=0$ fixesNumerically,The associated constant mass loss rate is $\alpha =m_0/(3T)\simeq 88.78\mathrm{kg}{\mathrm{s}}^{-1}$. The initial upward thrust acceleration already exceeds $g$ and increases as mass falls, so the downward speed decreases monotonically to zero. This calculation assumes constant lunar gravitational acceleration, constant relative rocket exhaust speed, and negligible drag.

Let $u>0$ be the projectile's final speed, taking its deflection above the $x$-axis. Momentum conservation givesSubstituting these into kinetic energy conservation, $V^2=u^2+2|{\mathbf{v}}_2{|}^2$, gives $3u^2-2Vu\cos 30^{\circ }-V^2=0$. Its positive root is $u=V(\sqrt{3}+\sqrt{15})/6$. HenceThese vectors satisfy ${\mathbf{v}}_1+2{\mathbf{v}}_2=(V,0)$ and $|{\mathbf{v}}_1{|}^2+2|{\mathbf{v}}_2{|}^2=V^2$. For the reflected deflection, reverse both $y$ components.

Let $\theta$ be the displacement from the upward vertical and $v$ the speed of the ball's centre of mass. Its centre follows a circle of radius $R+r$. The rolling without slipping constraint is $v=r\omega$, with $\omega$ the ball's spin angular velocity. The static contact friction does no work on the ball because the instantaneous contact point is stationary. Thus conservation of energy givesProject Newton's second law toward the hemisphere's centre. The inward component of gravity is $Mg\cos \theta$ and the normal force $N$ points outward, soAt contact loss $N=0$. Eliminating $v$ gives the ideal sphere rolling off a hemisphere resultThe radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.

Finite-friction qualification. This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite coefficient of static friction. The tangential force and torque equations giveso $f=(2/7)Mg\sin \theta$ acts uphill, whereasConsequently the required ratio isFor a concrete example, a coefficient of static friction of one allows rolling only up to about $47.54^{\circ }$, where $17\cos \theta -10=2\sin \theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^{\circ }$ answers the source's expressly imposed no-slip idealization.

Choose the initial incident velocity as $u_1\hat{\mathbf{x}}$. The centre-of-mass velocity in the laboratory frame isIn the centre-of-mass frame, the initial velocities are ${\mathbf{u}}_1^{\ast }=m_2{u}_1\hat{\mathbf{x}}/(m_1+m_2)$ and ${\mathbf{u}}_2^{\ast }=-m_1{u}_1\hat{\mathbf{x}}/(m_1+m_2)$. Conservation of momentum and kinetic energy in an elastic collision preserves these two speed magnitudes. Let ${\theta }^{\ast }$ be the outgoing angle of particle 1 in that frame, measured from the initial positive $x$ direction. ThenAdding $\mathbf{V}$ to transform back to the laboratory frame givesThe nonnegative laboratory recoil angle ${\theta }_2$ is measured below the incident axis in this choice of scattering plane. Hence the recoil angle in a two-body elastic collision obeysThe source identity therefore uses ${\theta }^{\ast }$ for the projectile's outgoing direction in the centre-of-mass frame. If instead it denotes particle 2's outgoing direction, the corresponding expression has $1+\cos {\theta }^{\ast }$ in the denominator. A signed laboratory angle for the drawn target trajectory would carry a minus sign. At zero deflection the target remains at rest, so its direction is undefined; ${\theta }_2\to \pi /2$ is a limit.

For a Lorentz transformation with $S^{\prime }$ moving at $+v$ along $x$ relative to $S$,Differentiate along the object's trajectory and divide the spatial differentials by $d{t}^{\prime }$. Using $u_x=dx/dt$ and $u_y=dy/dt$ derives the relativistic velocity-addition formula:In the rest frame of ship A, take $v=c/2$ and the Earth-frame velocity of B as $(0,-c/2)$. Here $\gamma =2/\sqrt{3}$ and the common denominator is one, soThe required relative speed is the magnitude of this two-component velocity, not either component separately:The inverse Lorentz factor check is ${\gamma }_{AB}={\gamma }_A{\gamma }_B(1-{\mathbf{u}}_A\cdot {\mathbf{u}}_B/c^2)=4/3$, which gives the same relative speed.

Interpret the quoted muon energy as total relativistic energy, and neglect energy loss in the atmosphere. The Lorentz factor, speed and Earth-frame travel time areThe muon decay law is exponential in proper time. Equivalently, time dilation makes the Earth-frame mean lifetime $\gamma {\tau }_0$. The exponential survival probability over the flight isWith $h=15000\mathrm{m}$ and ${\tau }_0=2.2\times 10^{-6}\mathrm{s}$, the elapsed proper time is $8.841\times 10^{-7}\mathrm{s}$. ThereforeThis uses the stated single incident energy and vertical trajectories. If the number were a kinetic energy, one would instead use $\gamma =1+T/(m{c}^2)$; the distinction is not silently absorbed into the rest energy.

For a boost along $+x$ at speed $v$, the energy-momentum four-vector transforms asThe total four-vector is the sum over the particles. Write $E={\sum }_i{E}_i$ and $\mathbf{P}={\sum }_i{\mathbf{p}}_i$. A centre-of-momentum rest frame exists when this total is timelike, with ${\mathbf{V}}_{\mathrm{CM}}=c^2\mathbf{P}/E$. Boost along $\mathbf{P}$ and use $P=E{V}_{\mathrm{CM}}/c^2$:ThereforeThe timelike qualification is needed: exactly parallel photons alone have null total four-momentum and no such rest frame.

Let $\mathbf{w}={\mathbf{v}}_1-{\mathbf{v}}_2$ be the relative velocity and $\mathbf{V}=(M_1{\mathbf{v}}_1+M_2{\mathbf{v}}_2)/(M_1+M_2)$ the centre-of-mass velocity. Inverting givesSubstitute into the sum of the two kinetic energies. The cross terms cancel, while the coefficient of $|\mathbf{w}{|}^2/2$ is the reduced mass $\mu =M_1{M}_2/(M_1+M_2)$. HenceThe first term is translation of the centre of mass; the second is the kinetic energy measured in the centre-of-momentum frame. For an isolated perfectly inelastic collision, the bodies finish with a common velocity, while their total momentum and thus $\mathbf{V}$ are unchanged. The kinetic energy converted to internal forms is thereforewhere $v$ is their initial relative speed. Total energy remains conserved; it is the translational-plus-relative kinetic energy that decreases.

Write the incoming particle speed as $u>0$, its final signed speed as $v$, and the rod's final centre-of-mass velocity along the impact direction as $V$. Let $b$ be the signed impact distance from the centre, and choose the sign of the angular velocity $\Omega$ so that the contact point's velocity component is $V+b\Omega$. The rod's central moment of inertia is $I=M{L}^2/12$.

The impulse delivered to the rod is $MV$. Conservation of linear and angular momentum givesUsing these in conservation of kinetic energy,This is the equal-mass instance of the effective contact mass calculation. For $u>0$, both final speeds are nonnegative, so equal speeds require $v=V=u/2$. Therefore $M{b}^2/I=2$ andBoth symmetric impact points lie on the rod, since $L/\sqrt{6}<L/2$.

More generally, the relative contact velocity immediately afterwards isThus its magnitude equals the incoming relative speed $u$ and its direction reverses. At the specified impact point, $V+b\Omega =3u/2$ and $v=u/2$, which makes the reversal particularly explicit.

With metric convention $(+---)$, the invariant interval is $c^2\Delta {\tau }^2=c^2\Delta t^2-|\Delta \mathbf{r}{|}^2$ along an inertially moving clock. For constant speed $v$ in the ground frame, $|\Delta \mathbf{r}|=v\Delta t$, soThe aircraft records proper time $\Delta \tau =1000\mathrm{s}$. The ground duration is measured by a network of clocks at rest in the ideal ground frame, with Einstein synchronisation. Its excess isThus the ground-frame duration is longer by about $0.50\mathrm{ns}$. This is the special-relativistic time dilation contribution under the stated constant-speed assumptions; gravitational potential differences and Earth's rotation are outside this model.

First work in the receiver's rest frame. If a source recedes at speed $\beta c$, two successive emissions separated by source proper time $T_0$ occur $\gamma T_0$ apart in this frame. During that interval the source retreats by $\beta c\gamma T_0$, so the arrival interval is $T_R=\gamma T_0(1+\beta )$. The longitudinal relativistic Doppler effect is consequentlyA negative $\beta$ describes approach and gives a blueshift.

Interpret $633\mathrm{nm}$ as the laser's rest-frame wavelength and the stated detector velocity as measured in the initial laboratory frame. Choose $+x$ from source to sensor, with detector velocity $-0.2c$. The required relative recession speed follows fromBy relativistic velocity addition, the source velocity ${\beta }_{S}c$ in the laboratory obeys $({\beta }_S+0.2)/(1+0.2{\beta }_S)=-{\beta }_{\mathrm{rec}}$. HenceThe laser must move away from the sensor at about $0.5784c$ in that laboratory frame. Although the sensor moves toward the source's initial location, the retreating source produces the required net redshift. The frame qualification matters: a velocity permanently fixed relative to the moving source would specify the Doppler factor already, rather than leave it adjustable.

The escaping sand initially has the bucket's velocity, so it produces no relative-velocity thrust. This distinction is essential in a leaking Atwood machine. With upward velocity $v$ for the leaky side, the two bucket equations areTheir sum gives $\dot{v}=g(M_0-M)/(M+M_0)$. Since $\dot{M}=-\alpha$, it follows thatIntegrate from $M_0$, where the velocity is zero:At emptying, $M=(1-ϵ)M_0$, givingThis is upward and positive for $0<ϵ<1$. For no initial sand, $ϵ=0$, the emptying time is zero and the speed is zero. As the dry bucket mass tends to zero,This finite limit refers to the idealized model; the endpoint itself would describe a massless empty bucket. For small $ϵ$, $v_{\mathrm{empty}}\sim g{M}_0{ϵ}^2/(4\alpha )$, consistent with initially zero acceleration that grows as mass is lost.

Take $s$ positive uphill, $X$ positive to the right, and the cylinder's rotation positive anticlockwise. Contact with the fixed-angle surface determines its normal position; no slipping gives $\dot{\theta }=-\dot{s}/R$. These two constraints leave $X,s$ as the two independent coordinates of the cylinder rolling on a freely translating wedge.

The cylinder centre is at $(X+s\cos \alpha -R\sin \alpha ,s\sin \alpha +R\cos \alpha )$. Since the angle is constant, its squared speed is ${\dot{X}}^2+2\dot{X}\dot{s}\cos \alpha +{\dot{s}}^2$. Including the block's kinetic energy and $I_c=M_c{R}^2/2$, and dropping a constant from the potential, givesThe cyclic coordinate is $X$. Its conserved conjugate momentum is $p_X=(M_b+M_c)\dot{X}+M_c\cos \alpha \dot{s}$, the total horizontal momentum. The two Euler-Lagrange equations areEliminating the accelerations, with $D=3(M_b+M_c)-2M_c{\cos }^2\alpha >0$, givesThus the cylinder rolls downhill and the wedge accelerates rightward. For $M_b\gg M_c$, $\ddot{X}\to 0$ and $\ddot{s}\to -(2/3)g\sin \alpha$, the ordinary fixed-incline result.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

The initial angular momentum is $J=ml{v}_0/2$, and the initial total energy isWhen the hanging mass reaches the table, $r=l$. Conservation of angular momentum fixes the tabletop tangential speed there to $J/(ml)=v_0/2$. At the limiting case of just reaching, radial velocity is zero. Equating the two energies,yieldsMore generally, the boundary radial kinetic energy isso reaching requires $v_0^2\ge 4Mgl/(3m)$, with equality for just reaching and strict inequality for nonzero outward radial speed at contact. This condition is also sufficient: $U_J$ is strictly convex, and the allowed interval connecting the initial radius to $l$ has no intervening higher effective potential barrier. At the threshold the initial radial acceleration is outward and $r=l$ is the outer turning point.

Initially $r_i=l/2$, ${\dot{r}}_i=0$ and ${\dot{\theta }}_i=v_0/r_i$, so the conserved angular momentum is $J=ml{v}_0/2$. For $v_0\ne 0$, the effective potential contains the centrifugal barrierThe finite conserved energy therefore forbids the hole from being reached:The exceptional case $v_0=0$ has no centrifugal barrier. Then $J=0$ and $\ddot{r}=-Mg/(M+m)$; starting at rest at $l/2$, the tabletop mass reaches the hole after $t=\sqrt{(M+m)l/(Mg)}$. The model ends at that event. Thus the usual nonzero initial tangential speed assumption matters.

Treat the asteroid as a point mass with negligible influence on the planet. Outside a spherical planet, Newton's law of universal gravitation gives the specific gravitational potential $-GM/r$. The conserved specific energy and specific angular momentum areAt the periapsis $r_p$, the radial velocity vanishes, so conservation of energy givesThe right-hand side increases with $r_p>0$. The grazing threshold $r_p=R$ is therefore the grazing impact parameter in gravitational focusing relation. For $d>R$ its equivalent speed isThe threshold is customarily called the minimum avoiding speed, with tangency counted as its limiting boundary. If any contact with the surface counts as a hit, it is an infimum, not an attained minimum. For $d\le R$, no finite incoming speed avoids contact: gravitational focusing bends the path inward, and even its infinite-speed straight-line limit has closest approach $d\le R$. The finite radius of the asteroid has been neglected.

For an elastic collision with $M_2>M_1$, the positive root in the laboratory scattering circle formula is the only nonnegative one. With $\epsilon =M_1/M_2\ll 1$,ConsequentlyThe heavy target can absorb a finite momentum with negligible velocity and negligible kinetic energy. Thus the projectile may change direction substantially while retaining its speed; a backward reflection also has $v_1\simeq u_1$.

Set $q=1$ in the laboratory scattering circle formula. For $M_1>M_2$, a real outgoing speed requiresAlso $v_1\ge 0$ excludes the backward-angle interval: the square root is smaller than $|\cos {\theta }_1|$ when $M_2/M_1<1$, so both roots would be negative for $\cos {\theta }_1<0$. ThereforeAt this boundary the two possible speeds coincide, with $v_1/u_1=\sqrt{(M_1-M_2)/(M_1+M_2)}$. Geometrically the outgoing ray is tangent to the laboratory scattering circle. As $M_2/M_1\to 0$ the maximum tends to zero, while it approaches $\pi /2$ as the masses approach equality from $M_1>M_2$.

In the laboratory frame, the initial target is at rest and the outgoing directions are directly related to detector positions. The centre-of-momentum frame has zero total momentum, so the two outgoing momenta are equal and opposite; it cleanly separates bulk motion from relative kinetic energy. Transforming back is needed to predict laboratory angles and speeds. Neither reference frame removes the need for an appropriate collision law.

Take the incident direction as $\hat{\mathbf{x}}$ and put $S=M_1+M_2$. Momentum conservation fixes the centre-of-momentum frame velocity, and a Galilean transformation gives the initial velocities in that frame:After the collision, $M_1{\mathbf{w}}_1^{\prime }+M_2{\mathbf{w}}_2^{\prime }=0$. Hence both particles have the same final-to-initial speed ratio,The relative kinetic energy is multiplied by $q^2$. This common ratio exists independently of how the outgoing direction is selected.

The meaning of restitution matters here. The usual coefficient of restitution $e_n$ is a ratio of normal separation and approach speeds, not total speeds. In a smooth oblique collision, the tangential relative velocity is unchanged. If the incoming relative velocity makes angle $\chi$ to the contact normal, thenFor example, $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}\ne 1/2$. Therefore the requested identification is not generally true with the standard normal definition. The printed equation is valid if its $e$ denotes the relative speed retention in an oblique collision, namely $e=q$, or if an additional collision model supplies that equality. The elastic cases have $q=e_n=1$ and are unaffected by this qualification.

To derive the scattering formula under that convention, write ${\mathbf{v}}_1^{\prime }=\mathbf{V}+{\mathbf{w}}_1^{\prime }$. Its components give the laboratory scattering circleSolving the resulting quadratic for the nonnegative speed yieldsWith $q=e$, this is the requested formula. A real square root and $v_1\ge 0$ are both necessary: a negative algebraic root is not a second physical speed, and $v_1=0$ does not define a scattering angle. The subsequent limits are taken using this laboratory scattering circle.