Align the spatial axes and make the origins coincide at $t=t^{\prime }=0$. For an inertial frame moving at velocity $+v$ along $x$, the Lorentz transformation isThe inverse Lorentz transformation follows by interchanging primed and unprimed coordinates and replacing $v$ by $-v$.

By time dilation, the laboratory mean lifetime is $\tau =\gamma {\tau }_0$, where ${\tau }_0=2.6\times 10^{-8}\mathrm{s}$. The exponential decay law evaluated at travel time $d/v$ gives the surviving pion fractionUse the exact product $\gamma v=(400/0.94)c$ from part (a), rather than approximating the speed separately. With $c=2.99792458\times 10^{8}m/s$,Thus about $88.6\%$ emerge without decaying, and about $11.4\%$ decay inside the pipe. This assumes the given common longitudinal speed and neglects losses from other interactions.

The special relativity expression for momentum is $p=\gamma mv$, with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Equal speeds imply equal Lorentz factors, so the pion and proton momenta are in the ratio of their rest masses:The common dimensionless product is $\gamma v/c=p_p/(m_{p}c)=400/0.94\simeq 425.53$, giving $\gamma \simeq 425.53$ and $v/c\simeq 0.99999724$.

A constant displacement proportional to $(1,-3)$ leaves the velocities unchanged and changes $3x+y$ by zero. It is therefore a continuous symmetry of the Lagrangian. The generalized momentum components areNoether's theorem gives the conserved quantityA direct check using the Euler-Lagrange equations isThe normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have length units and $(K_x,K_y)=(1,-3)$ dimensionless, giving $P$ the units of momentum. Alternatively, for the paper's convention with dimensionless parameter and length-valued $K_i$, use $(K_x,K_y)=(\ell ,-3\ell )$; the conserved quantity is then $\ell P$. The conserved content is identical.

Use upward generalized coordinates. Let $h$ be the displacement of the lower pulley and $z$ that of the $4m$ mass. The two inextensible mechanical strings imposeFor ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. HenceUp to a constant, their gravitational potential energy isThe Lagrangian of this compound Atwood machine is therefore

Take $\beta >0$. At a radial turning point, $\dot{r}=0$, and the effective potential givesFor $L\ne 0$, the smaller positive radial turning point is conveniently written without division by $E$:This includes $E=0$, where $r_{\min }=L^2/(2m\beta )$. For $E<0$ there is also an outer radial turning point; the two coincide for the circular orbit at $E=-m{\beta }^2/(2L^2)$. Smaller energies are inaccessible at that $L$. For $E>0$ there is just one positive radial turning point. If $L=0$, there is no centrifugal barrier and an inward trajectory can reach the singular centre; its limiting closest distance is zero.

The requested omission of $\beta$ is not possible. For example, at $E=0$ and fixed nonzero $L,m$, doubling $\beta$ halves $r_{\min }$. Thus $E,L,m$ alone cannot determine closest approach.

For motion in a central potential, conservation of angular momentum gives $L=m{r}^2\dot{\theta }$. In polar coordinates, the conserved mechanical energy isConsequently, for $L\ne 0$,The sign in $\dot{r}$ distinguishes inward and outward motion. Choosing the orientation so $L>0$, it is also the sign of $dr/d\theta$. The expression under the square root must be nonnegative. If $L=0$, the motion is radial and $\theta$ is constant, so $r$ cannot be parametrized by $\theta$ this way.

Let $a$ be the center of mass acceleration downhill and $F$ the uphill static friction. Translation and rotation about the center of mass obeyThe rolling without slipping constraint is $a=r\alpha$. For the coin's moment of inertia $I=m{r}^2/2$, this gives $F=ma/2$, soThe normal force is $N=mg\cos \theta$. Static friction can enforce rolling without slipping only if $F\le \mu N$, givingThe static friction acts uphill to provide the torque; it is not automatically equal to $\mu N$, attaining that limiting value only at the threshold. For finite $\mu \ge 0$ the result lies between $0$ and $\pi /2$.

The free rod has the same mass $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial center of mass at $y=0$ and the impact end at $y=l/2$. At impact the combined center of mass isConservation of momentum gives the combined center of mass velocity $V_C=v_0/2$. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $m{v}_{0}l/4$.

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertiaWith negligible external angular impulse, conservation of angular momentum therefore givesIts sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy isThus $m{v}_0^2/10$, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

For an equal-speed elastic impact on a free rod, let $h$ be the signed distance of the impact point from the rod's center of mass, $I=m{l}^2/12$, and $\mathcal{J}$ the horizontal impulse delivered to the rod. For a smooth impact the impulse is perpendicular to the rod. If the ball's signed final velocity is $u$, the rod's center of mass velocity is $V$ and its angular velocity is $\omega$, momentum balance and angular impulse giveBoth objects have positive final translational velocities for a nonzero impact away from the centre. More explicitly, restitution in an elastic collision gives $V+\omega h-u=v_0$, and henceEqual final speeds thus require $u=V=v_0/2$. One can also see that the alternative $u=-V$ would violate $u+V=v_0$.

By conservation of kinetic energy in the elastic collision,Thus $I{\omega }^2=m{v}_0^2/2$. Since $I\omega =mVh=m{v}_{0}h/2$, elimination of $\omega$ givesEither side of the center of mass works; the nearer end is a distance $l(1/2-1/\sqrt{6})$ away. The final angular speed is $|\omega |=\sqrt{6}{v}_0/l$. These values also satisfy the restitution condition because the struck point's velocity is $V+\omega h=3v_0/2$, reversing the initial relative contact velocity.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

With the wedge fixed, let $r$ be the signed gap between the cylinder bottom and the surface, positive outwards. The centre height is $s\sin \alpha +(R+r)\cos \alpha$. Retaining all three coordinates before imposing constraints givesUse the holonomic constraints $f_1=r=0$ and $f_2=s+R\theta -s_0=0$. With the convention $d(\partial L/\partial \dot{q})/dt-\partial L/\partial q={\sum }_j{\lambda }_j\partial f_j/\partial q$, the Lagrange multiplier equations areThe constraint gradients identify ${\lambda }_1=N$, the normal reaction, and ${\lambda }_2=f$, the uphill static friction force. The latter supplies the rolling torque. Setting $\ddot{r}=0$ and $\ddot{\theta }=-\ddot{s}/R$ yieldsThe static friction condition $|f|\le {\mu }_{s}N$ therefore becomesfor $0\le \alpha <\pi /2$. Static friction is required to create rotation, but does not dissipate energy here because the contact point is instantaneously at rest on the fixed plane.

Take $s$ positive uphill, $X$ positive to the right, and the cylinder's rotation positive anticlockwise. Contact with the fixed-angle surface determines its normal position; no slipping gives $\dot{\theta }=-\dot{s}/R$. These two constraints leave $X,s$ as the two independent coordinates of the cylinder rolling on a freely translating wedge.

The cylinder centre is at $(X+s\cos \alpha -R\sin \alpha ,s\sin \alpha +R\cos \alpha )$. Since the angle is constant, its squared speed is ${\dot{X}}^2+2\dot{X}\dot{s}\cos \alpha +{\dot{s}}^2$. Including the block's kinetic energy and $I_c=M_c{R}^2/2$, and dropping a constant from the potential, givesThe cyclic coordinate is $X$. Its conserved conjugate momentum is $p_X=(M_b+M_c)\dot{X}+M_c\cos \alpha \dot{s}$, the total horizontal momentum. The two Euler-Lagrange equations areEliminating the accelerations, with $D=3(M_b+M_c)-2M_c{\cos }^2\alpha >0$, givesThus the cylinder rolls downhill and the wedge accelerates rightward. For $M_b\gg M_c$, $\ddot{X}\to 0$ and $\ddot{s}\to -(2/3)g\sin \alpha$, the ordinary fixed-incline result.

For a Lagrangian $L(q_i,{\dot{q}}_i,t)$ expressed in independent generalized coordinates, stationary action gives the Euler-Lagrange equationsIdeal constraints can either be incorporated into the independent coordinates or treated using Lagrange multipliers, as in solution.

The diagram places the force centre on the circumference, not at the centre of the circle. Choose the polar axis through the circle's geometric centre. Its equation is $r=2a\cos \theta$, so $u=\sec \theta /(2a)$ andSubstitute into Binet's equation:Consequently a circular orbit through the force centre is generated byThe result applies to the regular arc where $r>0$ and $J\ne 0$. The force is singular at the point where the geometrical circle meets the origin, so the curve is not a regular periodic orbit continued through that point. Indeed, $U=-C/(4r^4)$ and this trajectory has zero total energy, with kinetic and negative potential terms diverging as the centre is approached.

The effective potential includes the angular part of the kinetic energy. Taking $U(0)=0$ gives $U=k{r}^4/4$, henceFor $J\ne 0$, this diverges at both zero and infinity and has a unique minimum. Finite-energy motion therefore remains between two radial turning points, or at the minimum itself. For $J=0$, the centrifugal barrier is absent, but $U\to \infty$ still bounds every finite-energy trajectory; radial motion can pass through the force centre.

A circular orbit requires $U_{\mathrm{eff}}^{\prime }(r_c)=0$, soIt is stable because $U_{\mathrm{eff}}^{\prime \prime }(r_c)=6k{r}_c^2>0$. These are nonzero-radius orbits for $J\ne 0$; $J=0$ also permits the stationary particle at the origin. Boundedness does not mean that every noncircular orbit is closed.