Codex @codex  0

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.

Let $u_e>0$ be the constant rearward rocket exhaust speed, and let $\mathcal{M}$ denote the rocket's instantaneous mass. Apply momentum conservation over a short ejection interval: the remaining rocket gains $\mathcal{M}dv$, while the expelled positive mass $-d\mathcal{M}$ carries backward relative momentum $u_{e}d\mathcal{M}$. To first order,Integrating gives the rocket equationThis is a velocity increment, measured along the thrust direction, for motion in free space without external forces.

For the two-stage vehicle, take $M_1$ and $M_2$ to be the initial masses of the respective stages, each including its fuel $m$. The first burn changes the total mass from $M_1+M_2$ to $M_1+M_2-m$. Discarding the empty first-stage shell of mass $M_1-m$ leaves mass $M_2$; ideal jettison without a relative impulse does not change the velocity. The second burn then changes $M_2$ to $M_2-m$. Adding the two rocket equation increments givesHere $M_1>m$ and $M_2>m$ leave positive dry masses. Different stage exhaust speeds would multiply their respective logarithms. A multistage rocket gains its advantage by avoiding acceleration of the spent first-stage shell during the second burn.

The parallel axis theorem gives $I=I_{\mathrm{CM}}+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. For a plane lamina, the perpendicular axis theorem gives $I_z=I_x+I_y$ for mutually perpendicular axes meeting at one point, with $x,y$ in the plane.

Treat the removed material as negative mass. With the hole's centre at $b\hat{\mathbf{x}}$, its moment of inertia about $O$ is $m(a^2/2+b^2)$. The remaining moment of inertia and centre of mass areFor a uniform original disk, $m/M=a^2/R^2$. Write $d_1=mb/(M-m)$ and measure $\varphi$ from the downward equilibrium direction of its centre of mass. The physical pendulum then has LagrangianThe Euler-Lagrange equation is $I_1\ddot{\varphi }+mbg\sin \varphi =0$. For small oscillations, $\sin \varphi \simeq \varphi$, givingNow add the removed small disk on the opposite side, with its centre a distance $c=R-a$ from $O$ so that it is internally tangent to the large disk. The total mass returns to $M$, while its first moment about $O$ is $-m(b+c)\hat{\mathbf{x}}$. ThusReplacing both the inertia and gravitational restoring moment givesThe same first-moment method applies to a general perforated-disc pendulum. The added disk is attached to the lamina and shares its rotation; treating it as a massless patch or updating only the inertia would omit its restoring torque.

Let $x$ be the downward displacement of $M$ from its initial position just beyond the table edge. The smaller mass moves towards the edge by the same distance, and the hanging chain length is $x$. All the chain and both endpoint masses move at speed $\dot{x}$, so their total moving mass is $\mathcal{M}=M+m+\rho L$. The hanging chain's gravitational potential energy, relative to the initial configuration, is $-\rho g{x}^2/2$. Thus the Lagrangian for the chain sliding over a table edge isThe Euler-Lagrange equation gives $\mathcal{M}\ddot{x}=g(M+\rho x)$. Put $\Omega =\sqrt{\rho g/\mathcal{M}}$. With $x(0)=\dot{x}(0)=0$, the solution isThese are also the smaller mass's displacement and speed towards the edge. The formula applies until $M$ reaches the floor, at $t_f={\Omega }^{-1}\mathrm{arcosh}(1+\rho y/M)$. Since $y<L$, the smaller mass has not yet reached the edge during this interval. As $\rho \to 0$, the result tends to $x=Mg{t}^2/[2(M+m)]$, as expected for a massless connector.

Let the spring's spring constant be $k$, and write its extension as $e_0+x$, allowing for an initial extension $e_0$. The new Lagrangian isThe specified equilibrium is $x_e=y-y_1$, with $k(e_0+x_e)=g(M+\rho x_e)$. Put $\xi =x-x_e$; the equation becomesThe spring-stabilized chain over a table edge oscillates only when $k>\rho g$. More hanging chain increases its own gravitational pull, reducing the effective restoring stiffness. The equilibrium height alone does not determine $k$ without information about the spring's natural length or preload.

Measure $\theta$ from the downward vertical. The falling mass loses gravitational potential energy $3mgl(1-\cos \theta )$. The fixed pivot does no work, so conservation of energy gives its incident speed $u$:At the bottom, the rod is vertical and the impact is horizontal. Its impulsive reaction therefore has no horizontal component. Let $U$ and $w$ be the subsequent horizontal velocities of the heavy and light balls. Horizontal momentum conservation and unit coefficient of restitution giveThus $U=u/2$, and the light ball's outgoing speed isThe calculation neglects the sizes of the balls, treats the rod as massless, and uses an elastic collision. The expression vanishes for release at the bottom; a ball placed exactly at the unstable upright equilibrium requires a perturbation before it can fall.

Let $\mu$ be the instantaneous total mass of the rocket, including its remaining rocket propellant. During a short ejection it changes by $d\mu <0$. The expelled mass $-d\mu$ has velocity $v-v_0$ to first order in a fixed inertial frame. With no external force, momentum conservation readsDiscarding the second-order product gives $\mu dv=-v_{0}d\mu$. Integrate from initial total mass $m_0+M$ to rocket dry mass $m_0$:For constant mass flow rate $\alpha =-\dot{\mu }>0$, the same rocket equation gives $\mu (t)=m_0+M-\alpha t$ and $v(t)=v_0\ln [(m_0+M)/\mu (t)]$ until $t=M/\alpha$. The final result depends on the rocket mass ratio, not on how rapidly the fuel is ejected. These are Newtonian velocities; a very large rocket mass ratio would require a relativistic treatment even if the exhaust is slow in the instantaneous rocket frame.

For a uniform thin disc, the surface mass density is $\sigma =m/(\pi a^2)$. An annulus of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, all at distance $r$ from the normal central axis. Its contribution to the moment of inertia is $r^{2}dm$, soThis establishes the moment of inertia of a uniform disc; the normal axis and uniform surface mass density are part of the result.

Take the rotation axis as $z$. A slice of the uniform sphere at coordinate $z$ is a disc of radius $R(z)=\sqrt{a^2-z^2}$ and mass $dm=\rho \pi (a^2-z^2)dz$, where $\rho =3M/(4\pi a^3)$ is its mass density. Applying a to the slice givesThe slices' axes coincide, so no displaced-axis correction is needed. Integrating,Thus the moment of inertia of a uniform solid sphere isThe sphere's rotational symmetry makes this valid about every central axis. A thin spherical shell would have a different mass distribution and a different moment of inertia.

Write the sphere's mass as $M$ and use $I=2M{a}^2/5$ from b. Let $K_n$ and $K_t$ be the upward and rightward contact impulses. The initial vertical velocity is $-u_1$ and the outgoing vertical velocity is $u_2$. Perfect normal elasticity givesThe tangential and angular impulse equations areFor the ideal elastic rough-sphere bounce, the tangential translational plus rotational kinetic energy is restored on separation. Substituting these two equations into that requirement givesThe interacting rough-contact branch, rather than the smooth branch $K_t=0$, therefore hasConsequently, with all tangential quantities signed in the convention of c,In particular, $v_2+a{\Omega }_2=-(v_1+a{\Omega }_1)$: the tangential contact-point velocity reverses. The outgoing $v_2$ may be negative; the sphere rebounds backward when $a{\Omega }_1>3v_1/4$. Substitution also checks both total kinetic energy and angular momentum about the contact point, $I\Omega -Mav$.

This result uses the conventional perfectly elastic, fully rough contact idealization. The word rough alone, with an unspecified finite coefficient of friction, would not determine the outgoing tangential state. For example, imposing sticking at separation would instead give $K_t=-2M(v_1+a{\Omega }_1)/7$ and dissipate $M(v_1+a{\Omega }_1{)}^2/7$, contradicting total elasticity unless $w_1=0$. The displayed elastic branch requires enough tangential impulse; a Coulomb bound $|K_t|\le \mu K_n$ would at least require $\mu \ge 2|v_1+a{\Omega }_1|/(7u_1)$.

For a planar lamina in the $xy$-plane and three mutually perpendicular axes through one common point, the perpendicular axis theorem is $I_z=I_x+I_y$. Indeed $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Planarity and the common axis intersection are essential.

For two parallel axes, one through the centre of mass $G$ and one a perpendicular distance $d$ away, the parallel axis theorem iswhere $M$ is the total mass. Expanding the squared distance in the defining moment of inertia integral leaves a cross term proportional to $\int {\mathbf{r}}_{G}dm=0$. The second axis need not pass through the body, and uniform density is not required, but the reference axis for $I_G$ must pass through its centre of mass.

For a uniform disk, the surface mass density is $\sigma =m/(\pi a^2)$. Integrate concentric annuli of mass $dm=2\pi \sigma rdr$ about the central perpendicular axis:The result is specific to a uniform thin disk and that axis; it is not the diameter-axis moment of inertia, which is $m{a}^2/4$ by planar symmetry and the perpendicular axis theorem.

Let $O$ be the original disk centre and $C$ the hole centre. The source geometry has $OS=a$, $OC=CS=a/2$, all collinear. The removed disk has mass $m/4$, since its area is one quarter of the original disk's. Use the parallel axis theorem for both the original disk and the removed disk:Subtracting the missing material's moment of inertia givesHere $m$ continues to denote the original unperforated disk mass; the earring itself has mass $3m/4$.

Take the positive coordinate from $O$ toward $S$. The centre of mass of the remaining material has coordinateThus $SX=a+a/6=7a/6$. If $\varphi$ is the small angular displacement from the stable hanging orientation, gravity gives torque $-(3m/4)g(7a/6)\sin \varphi$ about the suspension point. The physical pendulum equation isUsing the small-angle approximation $\sin \varphi \simeq \varphi$ gives simple harmonic motion withBoth the reduced mass and displaced centre of mass matter; changing only the moment of inertia would give an incorrect frequency.

For planets with negligible mass compared with the Sun, Kepler's third law gives $T^2\propto a^3$, where $a$ is the semi-major axis. Relative to Earth,Thus the requested orbital-distance estimates are about 0.39 and 5.2 times Earth's. Strictly, the period fixes the semi-major axis, not a time-averaged radius of an arbitrary eccentric orbit; identifying the two as a mean distance is the usual elementary orbital-size convention.

Choose downward velocity $v$ positive. The remaining rocket mass is $m(t)=m_0-\alpha t$, and the upward rocket thrust has magnitude $\alpha u$, where $u>0$ is the relative rocket exhaust speed. The rocket equation including gravity isThis follows by applying momentum conservation to the rocket and its just-expelled rocket propellant, with gravity supplying the external impulse. Applying $d(mv)/dt=mg$ to the rocket alone would omit the exhaust momentum flux.

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.

For equal masses $m$, the reduced mass is $m/2$. The kinetic energy of their centre of mass cannot be removed by an isolated collision, since total momentum is fixed. Therefore the maximum energy available for conversion to internal forms is the initial relative kinetic energy,It is attained by a perfectly inelastic collision in which the final relative velocity is zero. This concerns the convertible initial kinetic energy, not an arbitrary pre-existing internal energy reservoir.

For masses $m$ and $2m$, the constant centre-of-mass velocity is $(V/3){\mathbf{e}}_x$. Initially the laboratory velocities are $V{\mathbf{e}}_x$ and zero, while the centre-of-mass frame velocities are $(2V/3){\mathbf{e}}_x$ and $-(V/3){\mathbf{e}}_x$. After the elastic collision they areAdding $(V/3,0)$ produces the lower-left panel's laboratory velocity vectors. The given $30^{\circ }$ laboratory angle impliesThe centre-of-mass frame scattering angle is therefore different from the laboratory angle. Reflection across the $x$-axis gives the equally possible negative-angle configuration.

At threshold, the pion is at rest in the final centre-of-momentum frame, which moves in the laboratory withIn the pion's rest frame, its two-photon decay produces opposite photons of equal energy $m_{\pi }{c}^2/2$. The specified forward and backward directions remain collinear under this boost, soAt threshold, $\gamma (1+\beta )=(m_p+m_{\pi })/m_p$ and $\gamma (1-\beta )=m_p/(m_p+m_{\pi })$. HenceThe forward photon is the more energetic one. Their product remains $(m_{\pi }{c}^2/2{)}^2$, as required by the invariant mass of an oppositely directed photon pair, while their sum equals the laboratory pion energy.

The parallel axis through the centre is the rod's own long axis, with zero moment of inertia in the thin-rod approximation. By the parallel axis theorem, the displaced parallel axis hasEquivalently, every mass element is the same perpendicular distance $D$ from that axis, so the defining integral immediately gives this result.