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.

The centre-of-momentum frame is an inertial frame in which the system's total momentum is zero. For nonrelativistic particles with total mass $M_T$ and total momentum $\mathbf{P}$, it moves with velocity $\mathbf{V}=\mathbf{P}/M_T$ relative to the original frame.

Under a Galilean boost by a constant velocity $\mathbf{U}$, the total kinetic energy transforms asFor an isolated collision, momentum conservation gives $\Delta \mathbf{P}=0$, and nonrelativistic mass conservation gives $\Delta M_T=0$. ThereforeAn elastic collision with $\Delta K=0$ in one inertial frame has $\Delta K^{\prime }=0$ in every inertial frame. Individual particles' kinetic energies change under the boost, but the total energy difference between the final and initial states does not. This is the Galilean invariance of collision elasticity.

Write $m_e$ for the electron rest mass and $\epsilon =m_e{c}^2$. The incident positron has total energy $E$ and satisfies $p^2{c}^2=E^2-{\epsilon }^2$. Adding the target electron's four-momentum, the squared total centre-of-momentum energy isAt the fixed-target production threshold, all three final particles are at rest relative to one another, so $W=3M_0{c}^2$. ConsequentlyThis is the positron's total energy, not its kinetic energy. It assumes $3M_0\ge 2m_e$ and addresses kinematic allowance; other selection rules could still forbid a proposed reaction. For $M_0{c}^2=135\mathrm{MeV}$ and $m_e{c}^2\simeq 0.510999\mathrm{MeV}$, $E_{\min }\simeq 160.494\mathrm{GeV}$.

At threshold the three products share the centre-of-momentum frame's laboratory velocity. Their common Lorentz factor and speed areFor the subsequent two-photon decay, each photon has rest-frame energy $E_{\gamma }^{\ast }=M_0{c}^2/2$. A Lorentz transformation of a photon emitted at rest-frame angle ${\theta }^{\ast }$ to the boost gives $E_{\gamma }={\gamma }_{\mathrm{CM}}{E}_{\gamma }^{\ast }(1+{\beta }_{\mathrm{CM}}\cos {\theta }^{\ast })$. Therefore the kinematically allowed laboratory range isor numericallyThe two photon energies in a single event are correlated: their sum is the parent's laboratory energy ${\gamma }_{\mathrm{CM}}{M}_0{c}^2$. When one attains the upper endpoint, the other attains the lower endpoint. Angular distributions and detector acceptance can affect which part of this range is observed.

Put $\mu =GM$. Balancing Newton's law of universal gravitation with the required centripetal force gives $m{v}_0^2/r_0=\mu m/r_0^2$, henceThe circular Kepler orbit is stable against small radial perturbations: its effective radial potential is $U_{\mathrm{eff}}=J^2/(2m{r}^2)-\mu m/r$, and at the circular value $J^2=\mu m^2{r}_0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=\mu m/r_0^3>0$. Here $J$ is the conserved angular momentum.

Take horizontal motion towards the second projectile as positive. In projectile motion, both bodies have the same downward gravitational acceleration, so it cancels from their relative motion. Their coordinates areFor a collision at $t>0$, equality of the horizontal coordinates requires $v_2\cos \theta =v_1$. Equality of the heights then requires $v_{2}t\sin \theta =h$. HenceThe upward solution requires $v_2>v_1$. A collision strictly above the ground further requires $t<\sqrt{2h/g}$, or $v_2^2-v_1^2>gh/2$. The horizontal target distance does not independently affect the interception geometry. If the first shot is also required to hit that ground target, its parameters satisfy $l=v_1\sqrt{2h/g}$.

Let $\lambda =1{\mathrm{s}}^{-1}$ be the launch rate and $\mu =\lambda m$. Treat many small impacts as a continuous momentum flux, justified by $m\ll M$. Write $x(t)$ for truck displacement from its initial position and $U=\dot{x}$, with $t=0$ at the first impact. First neglect vertical motion, so every projectile has horizontal velocity $v$.

A fixed launch rate is different from a fixed impact rate on a receding truck. Projectiles reaching the truck at time $t$ were launched at retarded time $s=t-x(t)/v$, after an irrelevant constant initial-flight offset. The moving-target interception rate is thereforeFor an elastic collision with a much heavier truck, each ball transfers momentum approximately $2m(v-U)$. The elastic-ball truck keeps mass $M$, soSolving from rest and integrating once more givesThe rebound reverses the ball's motion relative to the truck; its laboratory velocity $2U-v$ can still be forwards once $U>v/2$. The truck approaches $v$ asymptotically, while its decreasing impact rate prevents exponential approach in laboratory time. The approximation omits the velocity jumps of individual finite-mass impacts.

For sticking snowballs, let $A(t)$ be the instantaneous truck-plus-snow mass, with $A(0)=M$. The variable-mass system obeysIntegrating the second equation gives $AU=v(A-M)$. Therefore $A\dot{A}=\mu M$, and with $z=\sqrt{1+2\mu t/M}$,Retaining the accumulated mass is necessary even though each individual snowball is small.

For the range calculation, take the intended idealization of horizontal launch from height $h$, no air resistance, and interception possible until the projectile reaches the ground. Its flight time is $\tau =\sqrt{2h/g}$ and horizontal range is $D=v\tau$. If the initial truck distance from the boy is $X_0<D$, substitute $x=D-X_0$ in the last expression. The speed at the range limit isFor an initially nearby truck, $X_0=0$ in this point-target model and $a=\sqrt{2\mu \tau /M}$. A different launch angle or finite interception height changes the available range and must be specified separately.

If instead one idealizes one impact each second, regardless of the truck's changing position, the answers are different: elastic balls give $U=v(1-e^{-2\mu t/M})$ and $x=v[t-M(1-e^{-2\mu t/M})/(2\mu )]$; sticking balls give $U=v\mu t/(M+\mu t)$ and $x=vt-(Mv/\mu )\ln (1+\mu t/M)$. Those expressions assume controlled arrival times, whereas the boxed results respect the stated fixed launch rate.

The initial total momentum is zero, so the outgoing neutron recoils with momentum magnitude $E_{\gamma }/c$. Put $W=(m_p+m_{\pi })c^2$. Conservation of relativistic energy givesFor example the rest energies $938.2721$, $139.5704$ and $939.5654\mathrm{MeV}$ for the proton, charged pion and neutron give this value.^[ref] The result is smaller than the initial-to-final rest-energy difference of about $138.3\mathrm{MeV}$, because some released energy becomes neutron recoil kinetic energy. Atomic binding of a captured pion is neglected in the stated stationary-particle model.

Choose the emitted beam direction as positive $x$. In the original laboratory frame the sensor moves with velocity $-0.2c$; let the laser recede from it with velocity $-\beta c$. The collinear relativistic Doppler shift from source to laboratory and then to sensor givesAt the detection threshold this ratio is $633/1000$. HenceThe minimum collinear speed is therefore about $0.578c$, away from the approaching sensor, while the laser is still aimed towards it. The source must recede fast enough to overcome the sensor's approach-induced blueshift. A larger recession speed gives a longer received wavelength, if the detector has no additional upper cutoff.

As a check, the required relative recession speed is $(1-0.633^2)c/(1+0.633^2)$; relativistic velocity addition gives this same value from $(\beta -0.2)c/(1-0.2\beta )$.

Centre the square on the $xy$ plane, with uniform surface mass density $m/a^2$. For an in-plane axis at angle $\alpha$ to the $x$ direction, the perpendicular distance of $(x,y)$ from the axis is $-x\sin \alpha +y\cos \alpha$. Therefore its moment of inertia isThe mixed integral vanishes by symmetry, while the $x^2$ and $y^2$ integrals are equal. ThusThis is the in-plane isotropy of the square's inertia tensor: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.

Let the motor's constant power be $P>0$ and put $\omega =\dot{\theta }$. The braking coefficient in the proposed formula must describe a retarding torque $k\omega$. This constant-power rotational acceleration obeys the energy balance:Consequently the steady angular velocity is ${\omega }_0=\sqrt{P/k}$, and the ordinary rotational frequency is ${\omega }_0/(2\pi )$. Solving the linear equation for ${\omega }^2$ from rest givesThe printed time formula is missing a minus sign. Its logarithm is negative for $0<\omega <{\omega }_0$, whereas elapsed time must be positive. The terminal speed is approached only as $t\to \infty$.

A constant-power motor starting exactly from rest is an idealization: its driving torque $P/\omega$ would diverge. The energy equation has a regular solution for ${\omega }^2$, although the initial angular acceleration is singular. If $k\omega$ were literally a tangential force, a lever arm would also be needed; the dimensions of the requested expression identify $k$ as a rotational damping coefficient.

In spherical coordinates, the Lagrangian isA central force exerts zero torque, so the angular momentum vector is constant and the orbit lies in a fixed plane. Choose that plane as $\theta =\pi /2$; then $\ell =m{r}^2\dot{\varphi }$ is conserved. Substituting this relation into the energy givesThe effective potential reduces radial motion to a one-dimensional problem. Allowed radii have $E\ge V_{\mathrm{eff}}$, radial turning points satisfy equality, and a circular orbit occurs at $V_{\mathrm{eff}}^{\prime }=0$. A strict local minimum gives stable radial oscillations. Eliminating $\dot{\varphi }$ directly inside the original Lagrangian would require a Routhian; the energy derivation avoids giving the centrifugal term the wrong sign.

The familiar right-angle result requires an elastic collision with conserved horizontal momentum and the appropriate treatment of rotational kinetic energy. In the ideal smooth-sphere model the spins do not change during the collision, so the translational kinetic energy is separately conserved. For equal masses, with initial velocity $\mathbf{u}$ and outgoing velocities ${\mathbf{v}}_1,{\mathbf{v}}_2$,Squaring the first equality and using the second givesThus the two nonzero outgoing directions are perpendicular. In the head-on limiting case one sphere stops, so an angle for its zero velocity is undefined. An ideal elastic model that enforces rolling without slipping at both ends gives the same energy equation: each solid sphere has total kinetic energy $7m{v}^2/10$, and the common factor cancels.

Elasticity is not stated explicitly in the question. It cannot be omitted from the argument: for a smooth equal-mass collision with normal incoming component $u_n$ and coefficient of restitution $e<1$, the outgoing scalar product is $(1-e^2)u_n^2/4>0$. Moreover, after a smooth impact the spheres need not still be rolling; subsequent frictional relaxation can change their directions. The right-angle statement therefore concerns the specified ideal elastic collision, not arbitrary later rolling trajectories.