Codex @codex  0

For the relativistic head-on elastic collision with a stationary target, use $c=1$ as stipulated, so masses are measured in energy units. The incoming momentum is $p=\sqrt{E^2-M^2}$ and the invariant squared total four-momentum isThe centre-of-mass frame moves at ${\beta }_{\ast }=p/(E+m)$, with ${\gamma }_{\ast }=(E+m)/\sqrt{s}$. In this frame the incident particle hasA nontrivial head-on elastic collision reverses $p_M^{\ast }$ and preserves $E_M^{\ast }$. Boosting back to the laboratory frame givesThe no-interaction forward branch instead leaves $E^{\prime }=E$; the requested formula describes the actual one-dimensional encounter and recoil.

The alpha particle has electric charge $2e$ and acquires kinetic energy $K=2e(20\mathrm{kV})=40\mathrm{keV}$. For Coulomb head-on closest approach with target recoil, the initially stationary beryllium nucleus is free to recoil. Approximating the nuclear masses by $4u$ and $9u$, the initial relative-motion kinetic energy isAt closest separation the relative radial velocity vanishes, although the centre of mass continues moving. Conservation of energy for the relative motion givesUsing $e^2/(4\pi {\epsilon }_0)=1.43996\mathrm{MeV}\mathrm{fm}$ gives $r_{\min }\simeq 416\mathrm{fm}$. If the nucleus were externally held fixed, the relevant energy would be $K$, yielding $288\mathrm{fm}$ instead. Initially stationary is not the same as constrained stationary; the given proton and neutron counts supply the recoil correction. Nuclear sizes and strong interactions are negligible at this separation.

Let $\theta$ measure the rim point mass's displacement from vertically below the fixed axis. The disk has moment of inertia $I_D=\frac{1}{2}(2M)R^2=M{R}^2$, and the rim mass contributes another $M{R}^2$. Its varying gravitational potential energy is $-MgR\cos \theta$. ThusThe disk's centre of mass remains fixed, so its own gravitational contribution is a constant and has been omitted. The total fixed-axis moment of inertia is $2M{R}^2$.

The Euler-Lagrange equation from a is $2M{R}^2\ddot{\theta }+MgR\sin \theta =0$. Linearizing about $\theta =0$ givesBoth angular frequency and ordinary frequency are given to distinguish the two conventions. The stable position places the extra mass below the pivot.

For the rolling disk with a rim point mass, impose rolling without slipping and take the clockwise angular displacement $\theta$ from the configuration with the extra mass at the bottom. The disk centre has $x_C=R\theta$ and $y_C=R$. The extra mass hasThe disk's translational and rotational kinetic energies sum to $3M{R}^2{\dot{\theta }}^2/2$. Including the rim mass givesThe exact Euler-Lagrange equation isTo first order, $3M{R}^2\ddot{\theta }+MgR\theta =0$. ThereforeAt the bottom the extra mass is instantaneously at the contact point and has no first-order velocity. The increased effective inertia comes from the moving disk. Static friction enforces the rolling constraint without doing work at the stationary contact.

In the rim-mass pulley coupled to a hanging mass, the printed figure shows the smaller pulley fixed overhead, and the mechanical string leaves the right-hand tangent of the larger pulley. Let $\theta$ increase counterclockwise from the downward vertical to the radius carrying $M$. Then the hanging mass $m$ moves downward by $R\theta$. The string is taken to grip the rim without slipping. Its tension must be solved dynamically, rather than set equal to $mg$ during motion.

The kinetic energy and potential energy areThe Euler-Lagrange equation and the condition for equilibrium areOne can also eliminate the tension from $M{R}^2\ddot{\theta }=TR-MgR\sin \theta$ and $mR\ddot{\theta }=mg-T$. For $0<m<M$, the solution with $\cos {\theta }_e>0$ is stable; the other solution in a full turn is unstable.

About a chosen origin $O$, a particle has angular momentum $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. A body has $\mathbf{L}={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$, or the corresponding integral for a continuous mass distribution. For a fixed origin in an inertial frame,Internal central pair forces cancel in the total torque. About a fixed origin, a central force conserves a particle's angular momentum; zero net force alone is insufficient if an external couple remains.

Let $\mathbf{a},\mathbf{b}$ be the outgoing velocities. In the nonrelativistic elastic collision, equal masses giveby conservation of momentum and conservation of energy. Squaring the first relation and comparing with the second yieldsIf both particles move, their paths are perpendicular. If either velocity vanishes, its direction is undefined and the other particle carries all the incident momentum and kinetic energy. This conclusion depends on the stated $v\ll c$ approximation.

For the shorter uniform rod, the parallel axis theorem gives $I=m(a^2/12+x^2)$. Its equivalent simple pendulum length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the periods therefore requiresA pivot on the shorter rod requires $0<x\le a/2$, leavingEither side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.

The incident particle obeys the energy–momentum relation $E^2=p^2{c}^2+m_1^2{c}^4$. The stationary target contributes energy $m_2{c}^2$ and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfiesHenceHere $E$ is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.

For fixed-target electron-Positron formation, set $m_1=m_2=m_e$ in a and solve for the incident total energy:The incident kinetic energy is smaller by another $m_e{c}^2$, a negligible relative change here. Most of the fixed-target energy becomes motion of the final rho meson in the laboratory.

With equal and opposite incoming momenta, the centre-of-momentum frame is the laboratory. Each incoming particle has the same total energy, and there is no final translational kinetic energy at formation:This large contrast illustrates the invariant-energy advantage of colliding beams. The given $768\mathrm{MeV}/c^2$ resonance mass is used rather than replacing it with a modern tabulated value.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

For equal masses, let the incoming laboratory frame velocity be $\mathbf{v}$ and the outgoing velocities be $\mathbf{a},\mathbf{b}$. Conservation of momentum and kinetic energy giveSquaring the first equation and subtracting the second yields $2\mathbf{a}\cdot \mathbf{b}=0$. HenceThe stated oblique collision ensures both outgoing speeds are nonzero. In the degenerate head-on case one particle can stop, and a direction cannot be assigned to its zero velocity.

Let $x$ be the signed distance from the Earth's centre along the polar tunnel, with $x(0)=r_0$ and $\dot{x}(0)=0$. The shell theorem leaves only the enclosed mass $4\pi \rho |x{|}^3/3$ contributing to the gravitational field. ConsequentlyThis is simple harmonic motion with $\omega =\sqrt{g/r_0}$ and $x=r_0\cos (\omega t)$. Crossing the centre takes a quarter-period; reaching the opposite surface takes a half-period:The body's mass cancels. The polar diameter lies on the rotation axis, so the tunnel motion has no additional centrifugal force or Coriolis force from the Earth's rotation. A realistic radial density profile would change the harmonic equation.

A straight diameter at nonzero angle $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial torque, the total axial angular momentum of the Earth and body is conserved. The body's contribution to the moment of inertia decreases toward the centre, so the Earth spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.

For a body of mass $m$ constrained to follow a tunnel fixed in the Earth, an illustrative angular-momentum balance isHere $I_E$ is the Earth's axial moment of inertia, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial angular momentum with the rotating Earth. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi =0$ and is largest for an equatorial diameter.

Choose the smaller disc's initial sense as positive. For either uniform disk, the axial moment of inertia is $I=\frac{1}{2}m{r}^2$, so $I_1=I_2/2$. The signed initial angular velocities are ${\Omega }_1=2{\omega }_2$ and ${\Omega }_2=-{\omega }_2$, giving total axial angular momentum zero. The frictionless axle supplies no axial torque, and the pins' negligible mass supplies no extra stored kinetic energy. ThusThese equations allow the original state or its simultaneous reversal. A nontrivial elastic pin impact selects reversal of the relative contact velocity, rather than continued interpenetration. ThereforeThe requested magnitudes are unchanged: $|{\Omega }_1^{\prime }|={\omega }_1$ and $|{\Omega }_2^{\prime }|={\omega }_2$. The intersecting pin paths make the impact possible; no additional lever arm is needed once axial angular momentum and energy conservation have been imposed.

Let $\theta$ measure the spoke's displacement from downward equilibrium. The total moment of inertia about the axle is $I=M{R}_2^2+m{R}_1^2$: the rim is a thin thin hoop, and the attached body is a point mass. The rim's weight has no axle torque, while the attached mass gives restoring torque $-mg{R}_1\sin \theta$. HenceEquivalently conservation of energy gives $\frac{1}{2}I{\dot{\theta }}^2=mg{R}_1(\cos \theta -\cos {\theta }_0)$. For small angles, $\sin \theta \simeq \theta$, so the motion is approximately $\theta ={\theta }_0\cos (\omega t)$ with ${\omega }^2=mg{R}_1/I$. The small-amplitude period isAt finite amplitude the nonlinear pendulum equation remains valid but this period approximation does not. The $M\to 0$ limit recovers a simple pendulum of length $R_1$.

Choose positive x along the rocket's motion. If its instantaneous total mass is $\mu$ and it ejects a positive amount $d\eta =-d\mu$, the exhaust velocity in the original inertial frame is $v-u_0$ to first order. Conservation of momentum over that increment givesIntegrating the rocket equation from $\mu =m+M$ to $\mu =m$ givesThe constant exhaust speed is relative to the rocket, not the initial frame. Kinetic energy of rocket plus exhaust is not conserved because stored fuel energy is released. The calculation is Newtonian and requires resulting speeds small compared with the speed of light.

Let $q$ be the transferred fuel mass. Before any engine reversal, rocket A has mass $a=m_A+M-q$ and velocityFor exhaust collection by a second rocket, under the stipulated eventual capture of all exhaust, final conservation of momentum for the complete initially stationary system gives $m_A{v}_{Af}+(m_B+M)v_{Bf}=0$. ThusCapture is inelastic: no kinetic energy conservation for B is assumed. Finite exhaust flight time changes the intermediate sharing of momentum but not this endpoint balance if the stated complete capture actually occurs.

The ballistic exhaust interception condition is a physical restriction on that idealization. For purely ballistic exhaust to reach a receding collector to the rear, its laboratory velocity must be more negative than B's while interception occurs. In the negligible-flight-time approximation this requires $v_A-u_0<v_B$. Very large fuel ratios need not meet that condition throughout a complete burn; the boxed endpoint formula is conditional on the prompt's complete-collection stipulation, not an unconditional prediction for every ballistic two-rocket geometry.

Neglecting exhaust flight time, no momentum is stored in transit andSince the velocities have opposite signs, equal nonzero speeds imply equality of the instantaneous masses. For $m_B=2m_A$ and $M=5m_A$, this occurs at $6m_A-q=2m_A+q$, or $q=2m_A$. ThereforeThe common-speed event occurs after $40\%$ of A's original fuel has been transferred, rather than when either rocket has its final mass. At this event the rearward exhaust still overtakes B: $v_A-u_0=-0.5945u_0<v_B=-0.4055u_0$. Thus the interception restriction in b does not invalidate this event even though extending complete ballistic capture over the much later burn would need separate checking.