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Choose zero potential energy at the equilibrium position. For the simple harmonic oscillator, the kinetic energy and elastic potential energy give the LagrangianIts canonical momentum is $p=\partial L/\partial \dot{x}=m\dot{x}$. The Legendre transform defining the Hamiltonian consequently givesThe Euler-Lagrange equation is $m\ddot{x}+kx=0$. Thus along the motionThis proves conservation of energy. Equivalently, the Lagrangian has no explicit time dependence, so $dH/dt=-\partial L/\partial t=0$. The Lagrangian itself need not be constant: the kinetic energy and potential energy exchange periodically.

Use the inertial frame in which the departure point and destination are stationary, with separation $d$. The astronauts' elapsed proper time is $\tau =t/\gamma$, whereas $d=vt$. HenceThe Earth-frame travel time is $t=\sqrt{41}\mathrm{yr}\simeq 6.40\mathrm{yr}$ and the speed is approximately $0.625c$. The spacecraft's relativistic kinetic energy is thereforeThe five years are not the time to use in $v=d/t$ in the departure frame. This is the cruising kinetic energy, with departure and arrival acceleration intervals omitted, rather than a complete propulsion energy budget.

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.

Write the electron electric charge as $-e$, with $e>0$ the elementary charge, and choose $\mathbf{B}=B\hat{\mathbf{z}}$ and initial velocity $u\hat{\mathbf{x}}$. In the nonrelativistic approximation, Newton's second law with the magnetic Lorentz force givesThe magnetic force is perpendicular to the velocity, so it does no work: $d(m_e|\mathbf{v}{|}^2/2)/dt=0$. Setting ${\omega }_c=eB/m_e$, the solution with $x(0)=y(0)=0$ isConsequently $x^2+(y-r{)}^2=r^2$, where $r=u/{\omega }_c$: this is a cyclotron orbit. Its frequency and cyclotron radius areThe accelerating voltage changes $u$ and $r$ but cancels out of $f$. Using $m_e=9.1094\times 10^{-31}\mathrm{kg}$ and $e=1.6022\times 10^{-19}\mathrm{C}$ givesOnly the magnitude of the magnetic field is determined; its direction is perpendicular to the orbital plane, with the sense of rotation fixed by the negative electric charge. The claimed voltage independence is a nonrelativistic result: exactly, $\gamma =1+eV/(m_e{c}^2)$ and $f=eB/(2\pi \gamma m_e)$. At $1\mathrm{kV}$ the correction is about $0.2\%$. The stated model also neglects the energy lost to emitted electromagnetic radiation.

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.