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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.

Let $L_0=100\mathrm{m}$ be the train's proper length, $D=90\mathrm{m}$ the tunnel's length in the track reference frame, and $v=0.99c$. The Lorentz factor is $\gamma =7.0888$. At one track-frame time, length contraction makes the trainYes, the whole train fits inside at a single track-frame time. Set $t=0$ when the rear enters the near portal. Until the front reaches the far portal, the rear and front are at $vt$ and $vt+L$, respectively. Complete containment lastsa track-frame interval of approximately $256\mathrm{ns}$, counting boundary contact.

In the train's reference frame, its length is $100\mathrm{m}$, while the moving tunnel has simultaneous length $D/\gamma =12.696\mathrm{m}$. No single train-frame time has the whole train between the portals. The two answers refer to different simultaneity slices.

For an explicit check, denote the rear-entry event by $A$ and the front-exit event by $B$. In track coordinates,The Lorentz transformation givesThus the front leaves before the rear enters in train coordinates. These spatially separated events have a frame-dependent order, as allowed by relativity of simultaneity; neither answer asserts a different local event at a portal. No collision with closed doors has been stipulated.

Put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, the common Lorentz factor. The initial total relativistic energy and momentum areAssume that coalescence emits no additional radiation. By four-momentum conservation, the final rest mass $m_f$ is the rest mass of a coalescing relativistic collision:The final velocity is $V=c^2{P}_x/E=v/3$, the centre-of-momentum frame velocity.

For $v\ll c$,The leading term is the classical sum of masses. The first correction is the internal energy retained after sticking, divided by $c^2$: the initial classical kinetic energy is $3m{v}^2/4$, while the common-motion kinetic energy is $\frac{1}{2}(3m/2)(v/3{)}^2=m{v}^2/12$. Their difference is $2m{v}^2/3$, consistent with mass-energy equivalence.

The retained mass at time $t$ is $M(t)=M_0+kt$. Incoming rain has zero horizontal momentum in the ground reference frame. During $dt$, horizontal momentum conservation for the carriage and the rain about to enter givessoThis is a variable-mass system: the entering mass must be accelerated horizontally, even though no external horizontal force acts on the carriage-plus-rain system. Integrating the velocity,For $u_0\ge 0$ this is also the distance travelled. Its $k\to 0$ limit is $u_{0}t$. Vertical impact can dissipate kinetic energy; horizontal momentum conservation does not require kinetic energy conservation.

Use Cartesian coordinates along the edges from the pivot. With surface mass density $M/L^2$, the moment of inertia isEquivalently, the moment of inertia of a uniform square plate about its central perpendicular axis is $M{L}^2/6$, and the parallel axis theorem adds $M(L/\sqrt{2}{)}^2$.

The source figure measures $\theta$ from the upward vertical to $OC$, with $OC=L/\sqrt{2}$. The centre's height above the pivot is $(L/\sqrt{2})\cos \theta$. The smooth fixed pivot does no work. Conservation of energy, with the initially upright configuration as reference, givesFor rotation in the positive $\theta$ direction on the first descent, take the positive root. Its equivalent form is $\dot{\theta }=\sqrt{3\sqrt{2}g/L}\sin (\theta /2)$ for $0<\theta <\pi$.

Strictly, release from rest at a small angle ${\theta }_0$ replaces $1-\cos \theta$ by $\cos {\theta }_0-\cos \theta$. The displayed result is the ${\theta }_0\to 0$ limiting energy relation. Exactly upright and exactly at rest is an unstable equilibrium; without a perturbation it stays there. This distinction prevents the infinitesimal displacement convention from being mistaken for spontaneous motion.

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.

Write the linear drag as ${\mathbf{F}}_d=-b\mathbf{v}$, with $b>0$, and let $m$ be the projectile's mass. Taking upward as positive, Newton's second law givesThe downward terminal velocity is $-v_t$, whereEquivalently, if the drag acceleration is specified as $-k\mathbf{v}$, then $v_t=g/k$. Solving the first-order equation,The highest point occurs at $v_y=0$, soFor an upward launch this root is positive and ${\dot{v}}_y=-g$ there, confirming a maximum height. As $b\to 0$, expanding the logarithm gives the drag-free result $t_{\mathrm{top}}\to u_0\sin \alpha /g$. A horizontal launch has its highest point at launch; a downward launch from the ground does not have a later airborne maximum.

The dimensions of mass density, time and energy are $\mathsf{M}{\mathsf{L}}^{-3}$, $\mathsf{T}$ and $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$. Matching powers in the proposed dimensional analysis givesThusThis is the Sedov-Taylor blast wave scaling for a fixed released energy. Dimensional analysis cannot determine the dimensionless $C$; that requires the fluid equations and thermodynamic assumptions. The assumed list of controlling parameters excludes, for example, an independent ambient sound speed or a continuing energy supply.

A Lorentz invariant has the same value when the same physical system is described in any inertial frame. The components of its energy-momentum four-vector change, but their Minkowski norm does not. In particular the relativistic energy-momentum relation iswhere $M$ is the rest mass. Individual total energy and momentum are frame-dependent; the combination on the left is invariant.

For the fixed-target annihilation, the initial total laboratory energy is $E+M{c}^2$ and its total momentum is the incident particle's $\mathbf{p}$, with $c^2{p}^2=E^2-M^2{c}^4$. The centre-of-mass energy available for new-particle creation is thereforeIn the centre-of-mass frame the total momentum vanishes, so $W$ can supply the new particles' total rest energy at threshold. One does not subtract the annihilated particles' original rest energy. The excess laboratory energy beyond $W$ is associated with motion of the whole final system in the laboratory frame.

For the subsequent elastic collision, the equal and opposite transverse components of final momentum imply equal final momentum magnitudes, and hence equal final energies, for the two identical particles. By conservation of relativistic energy, each hasTheir final momentum magnitude $p_f$ and the incident momentum magnitude $p_i$ obeyLongitudinal momentum conservation gives $p_i=2p_f\cos \theta$. It follows that the equal-angle relativistic elastic scattering relation isThis derivation assumes nonzero incident kinetic energy and the stated symmetric outgoing paths. The angle between those paths is $2\theta$, which is evaluated in the two following parts.

For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.