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The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis a perpendicular distance $d$ away:To see the cross term vanish, write each perpendicular displacement as its displacement from the centre of mass plus the fixed offset. In the defining mass integral, the first moment about the centre of mass is zero.

For a thin planar rigid body in the $xy$ plane, the perpendicular axis theorem sayswhere all three axes pass through the same point, $x,y$ lie in the plane, and $z$ is perpendicular to it. This follows directly from $x^2+y^2$ being the squared distance from the $z$ axis. The thin-planar condition and common intersection point are essential; the formula is not a general identity for three-dimensional bodies.

Interpret the quoted muon energy as total relativistic energy, and neglect energy loss in the atmosphere. The Lorentz factor, speed and Earth-frame travel time areThe muon decay law is exponential in proper time. Equivalently, time dilation makes the Earth-frame mean lifetime $\gamma {\tau }_0$. The exponential survival probability over the flight isWith $h=15000\mathrm{m}$ and ${\tau }_0=2.2\times 10^{-6}\mathrm{s}$, the elapsed proper time is $8.841\times 10^{-7}\mathrm{s}$. ThereforeThis uses the stated single incident energy and vertical trajectories. If the number were a kinetic energy, one would instead use $\gamma =1+T/(m{c}^2)$; the distinction is not silently absorbed into the rest energy.

For a Lorentz transformation with $S^{\prime }$ moving at $+v$ along $x$ relative to $S$,Differentiate along the object's trajectory and divide the spatial differentials by $d{t}^{\prime }$. Using $u_x=dx/dt$ and $u_y=dy/dt$ derives the relativistic velocity-addition formula:In the rest frame of ship A, take $v=c/2$ and the Earth-frame velocity of B as $(0,-c/2)$. Here $\gamma =2/\sqrt{3}$ and the common denominator is one, soThe required relative speed is the magnitude of this two-component velocity, not either component separately:The inverse Lorentz factor check is ${\gamma }_{AB}={\gamma }_A{\gamma }_B(1-{\mathbf{u}}_A\cdot {\mathbf{u}}_B/c^2)=4/3$, which gives the same relative speed.

Choose the initial incident velocity as $u_1\hat{\mathbf{x}}$. The centre-of-mass velocity in the laboratory frame isIn the centre-of-mass frame, the initial velocities are ${\mathbf{u}}_1^{\ast }=m_2{u}_1\hat{\mathbf{x}}/(m_1+m_2)$ and ${\mathbf{u}}_2^{\ast }=-m_1{u}_1\hat{\mathbf{x}}/(m_1+m_2)$. Conservation of momentum and kinetic energy in an elastic collision preserves these two speed magnitudes. Let ${\theta }^{\ast }$ be the outgoing angle of particle 1 in that frame, measured from the initial positive $x$ direction. ThenAdding $\mathbf{V}$ to transform back to the laboratory frame givesThe nonnegative laboratory recoil angle ${\theta }_2$ is measured below the incident axis in this choice of scattering plane. Hence the recoil angle in a two-body elastic collision obeysThe source identity therefore uses ${\theta }^{\ast }$ for the projectile's outgoing direction in the centre-of-mass frame. If instead it denotes particle 2's outgoing direction, the corresponding expression has $1+\cos {\theta }^{\ast }$ in the denominator. A signed laboratory angle for the drawn target trajectory would carry a minus sign. At zero deflection the target remains at rest, so its direction is undefined; ${\theta }_2\to \pi /2$ is a limit.

Let $\theta$ be the displacement from the upward vertical and $v$ the speed of the ball's centre of mass. Its centre follows a circle of radius $R+r$. The rolling without slipping constraint is $v=r\omega$, with $\omega$ the ball's spin angular velocity. The static contact friction does no work on the ball because the instantaneous contact point is stationary. Thus conservation of energy givesProject Newton's second law toward the hemisphere's centre. The inward component of gravity is $Mg\cos \theta$ and the normal force $N$ points outward, soAt contact loss $N=0$. Eliminating $v$ gives the ideal sphere rolling off a hemisphere resultThe radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.

Finite-friction qualification. This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite coefficient of static friction. The tangential force and torque equations giveso $f=(2/7)Mg\sin \theta$ acts uphill, whereasConsequently the required ratio isFor a concrete example, a coefficient of static friction of one allows rolling only up to about $47.54^{\circ }$, where $17\cos \theta -10=2\sin \theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^{\circ }$ answers the source's expressly imposed no-slip idealization.

Take the rocket exhaust speed $u$ relative to the hovering ship, with exhaust directed vertically downward. If $\alpha =-\dot{M}>0$ is the mass flow rate, the upward rocket thrust is $u\alpha$. For example, over a short interval $dt$, the expelled mass $\alpha dt$ acquires downward momentum $-u\alpha dt$, giving the opposite recoil to the ship. Balancing its weight requiresSolving this differential equation with $M(0)=M_0$ yields the hovering rocket mass flowThe required mass flow rate decreases because the mass to support decreases. These expressions apply while fuel is available and $g,u$ remain constant. If the rocket dry mass is $M_d>0$, hovering ends at $t=(u/g)\log (M_0/M_d)$; the mathematical limit $M\to 0$ is not a model of an indefinitely fuelled physical ship.

For ordinary conservative classical mechanics, the Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. More generally it is the function whose action $\int Ldt$ satisfies the principle of stationary action. The canonical momenta and Hamiltonian areFor a regular Lagrangian, the first equations can be inverted to express the generalized velocities in terms of $q,p,t$; this defines the Legendre transform on phase space.

Along a solution of the Euler-Lagrange equations, ${\dot{p}}_i=\partial L/\partial q_i$. Taking a total time derivative givesThus an autonomous Lagrangian hasFor the usual time-independent mechanical coordinates and a velocity-independent potential energy, the kinetic energy is quadratic in the generalized velocities, so ${\sum }_i{p}_i{\dot{q}}_i=2T$ and $H=T+V$. The conservation proof itself uses the Euler-Lagrange equations, rather than assuming this special expression for energy.

Take $x$ horizontal, $y$ upward and $g>0$. Resolve Newton's second law into components for projectile motion with linear drag:For $\alpha >0$, let ${\tau }_d=m/\alpha$. An integrating factor $e^{t/{\tau }_d}$ givesThe relaxation time is ${\tau }_d$, and the terminal velocity is $(0,-g{\tau }_d)$. For fixed finite $t$, expanding the exponential function as $\alpha \to 0$ gives the continuous limit $v_x=v_0\cos \theta$ and $v_y=v_0\sin \theta -gt$; the two terms proportional to ${\tau }_d$ cancel before taking the limit.