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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.

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.

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.

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.

For a launch angle $\theta$, the projectile motion equations give a flight time $2u\sin \theta /g$, a projectile range $R=u^2\sin (2\theta )/g$, and an apex height $u^2{\sin }^2\theta /(2g)$. Thus the ceiling requires ${\sin }^2\theta \le 2gH/u^2$.

Set $z={\sin }^2\theta$. Then $R^2=4u^{4}z(1-z)/g^2$, which increases up to $z=1/2$. The optimum is therefore $z=\min (1/2,2gH/u^2)$, giving the projectile range under a ceilingThe printed expression is valid in the first regime. A higher ceiling leaves the usual $45^{\circ }$ launch admissible, so it cannot reduce the maximum range. The angle in the ceiling-limited regime is $\arcsin \sqrt{2gH/u^2}$. Equality at the roof is allowed in this ideal geometrical model; if touching the roof is forbidden, the same value is a supremum approached from below.

For a particle of rest mass $m_0$ and velocity $\mathbf{u}$, the relativistic energy and momentum areFor a photon of wavelength $\lambda$ travelling along $\hat{\mathbf{n}}$,Here $h$ is Planck constant and $c$ is the speed of light.

Let $\epsilon$ and ${\epsilon }^{\prime }$ be the incident and scattered photon energies. The stationary electron initially has energy $m_e{c}^2$. By four-momentum conservation, its final energy and momentum areUse the electron's relativistic energy-momentum relation, rather than assuming its recoil is nonrelativistic:Cancelling common terms givesSince $\lambda =hc/\epsilon$, the Compton wavelength shift isThe word elastic here means that the final electron has the same rest mass as initially and the complete photon-electron system conserves total energy; the photon alone loses energy to recoil.

For backscattering, $1-\cos \theta =2$, and henceChoose the incident direction as $+x$, so that the scattered photon travels in the $-x$ direction. An observer moving with velocity $+\beta c\hat{\mathbf{x}}$ meets that photon head-on. The Lorentz transformation of four-momentum gives the observed energyThis is the approaching-observer relativistic Doppler effect. Set $D=\epsilon /{\epsilon }^{\prime }=2.511/0.511$ and solve:The observer must move opposite to the scattered photon, along the original incident direction, at approximately $0.9205c$. Moving with the scattered photon would redshift it instead.