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In a nonrelativistic elastic collision of incident mass $M$ at speed $u$ with initially stationary mass $m$, the centre-of-momentum frame moves at $V=Mu/(M+m)$. The projectile's outgoing velocity in that frame has fixed magnitude $w=mu/(M+m)$. Its laboratory velocity tips lie on a circle of radius $w$ centred at $V\hat{\mathbf{x}}$. For $M>m$, a tangent from the origin gives the largest laboratory scattering angle, $\sin {\theta }_{\max }=w/V=m/M$. This geometrical result follows from Galilean velocity addition and conservation of energy.

Reuse the velocity selector analysis in the Thomson charge-to-mass experiment, but now send positive ions with $q=ze$ into a separate magnetic analyzer. The selector fields $E_s$ and $B_s$ give $v_0=E_s/B_s$. If the analyzer field $B_a$ is perpendicular to the selected velocity, its gyroradius givesfor a semicircular return to the entrance plane. The displayed impact separation $D$ is a diameter. This formula is linear in $r$ at fixed selected speed; the fixed-electric potential difference model above instead gives $m/q\propto r^2$. They impose different experimental conditions.

For a chosen example, take $E_s=2.00\times 10^{4}V/m$, $B_s=0.200\mathrm{T}$, and $B_a=0.500\mathrm{T}$. Thus $v_0=1.00\times 10^{5}m/s$. Approximating an ion mass by $24\mathrm{u}$ and setting $q=e$ gives $r\approx 49.75\mathrm{mm}$ and $D\approx 99.50\mathrm{mm}$. Here $\mathrm{u}$ denotes the unified atomic mass unit. These are stipulated teaching inputs, not a reported Aston measurement or an exact magnesium atomic mass.

A finite accepted speed range broadens the radius distribution. At fixed $m/q$ and $B_a$, $\delta r/r=\delta v/v$. At fixed electric charge and mass, $K=m{v}^2/2$ gives $\delta v/v\simeq \delta K/(2K)$ for a small kinetic energy spread. A one-percent kinetic energy spread therefore produces roughly a half-percent radius spread before other broadening is included. Its width convention must be specified before converting that statement into a mass resolving power.

A velocity selector uses perpendicular electric field and magnetic field to transmit a narrow range of speeds along a chosen beam direction. Assume uniform static electric field $\mathbf{E}=E\hat{\mathbf{y}}$ and magnetic field ${\mathbf{B}}_s=B_s\hat{\mathbf{z}}$, with $E,B_s>0$, and an incident velocity $\mathbf{v}=v\hat{\mathbf{x}}$. The cross product gives $\mathbf{v}\times {\mathbf{B}}_s=-v{B}_s\hat{\mathbf{y}}$, so the transverse Lorentz force isFor an electron, the electric force points downward and the magnetic force upward. A faster electron initially bends upward; a slower one initially bends downward. At the selected speed, both forces cancel along the whole straight trajectory. The cancellation is independent of mass and nonzero electric charge; it selects velocity, not a particular charge-to-mass ratio. A component of velocity parallel to the magnetic field is not constrained by the cancellation, so the incident beam must also have its direction restricted.

A real exit slit accepts a finite range of speeds and incident directions. A zero displacement at one screen alone does not establish zero force throughout the path: a curved trajectory could return to that screen position. The ideal velocity selector interpretation requires a straight aligned beam, known overlapping electric field and magnetic field, and sufficiently narrow entrance and exit slits. Its useful role in the Thomson charge-to-mass experiment teaching model is to supply speed independently of the subsequent magnetic bending.

The vena contracta is the narrowest jet section after an opening. Its area may be smaller than the opening area, so the jet speed and volumetric flow rate require separate comparisons with ideal flow.

Motion can weigh a system whose mass is otherwise inaccessible. For a self-gravitating isolated system in virial equilibrium, let $K$ and $U$ denote equilibrium averages and use its gravitational radius $R_g=-G_N{M}^2/U$ and a mass-weighted mean squared speed relative to its centre of mass. Then $K=M⟨v^2⟩/2$ gives $M=R_g⟨v^2⟩/G_N$. If random velocities are isotropic and mean streaming has been removed, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$, where ${\sigma }_{\mathrm{los}}$ is the one-dimensional line-of-sight velocity dispersion. Observations sample many objects at one epoch; replacing a time average by that sample requires a representative, approximately stationary population, beyond the trajectory identity itself.^[ref]

A practical measurement route uses an optical telescope to locate members and a spectrograph to measure Doppler shifts of identified spectral lines. Their shifts relative to the system's mean give radial velocities; angular separations plus a distance estimate constrain its size. Subtracting instrumental broadening and measurement variance, testing membership, and looking for coherent rotation or subgroups help control the inference. The projected luminous size need not equal $R_g$, and luminous tracers need not have the same distribution as all gravitating mass. Unknown velocity-anisotropy parameters, interlopers, mergers, external tidal forces and appreciable boundary stresses can bias a naive estimate. A large velocity dispersion alone does not establish virial equilibrium.

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted harmonic mean isEqual weights give the harmonic mean, while a zero weight excludes that input. For travel at positive constant speeds $v_i$ over positive leg lengths $d_i$, the weights are fractions of total distance:For $30\mathrm{km}$ at $30km/h$ and $90\mathrm{km}$ at $60km/h$, the weights are $1/4,3/4$, elapsed time is $5/2$ hours, and the average speed is $48km/h$. Weighting the speeds by their time fractions instead gives the same result as a weighted arithmetic mean: $(2/5)30+(3/5)60=48$. The two weight systems describe the same journey in different coordinates and must not be interchanged. ^[ref].

A point mass attached to a light inextensible string winds around a fixed smooth circular pole of radius $R$. Let the initial straight free length be $L$, and assume the string remains taut without slipping over the part already laid on the pole. Since its tension is perpendicular to the point mass's velocity, the speed $v$ stays constant. If the contact point turns through $d\theta$, the free length $s$ obeys $ds=-Rd\theta$, while the involute geometry gives $v=s\dot{\theta }$. Thus $s\dot{s}=-Rv$, or $s^2=L^2-2Rvt$. First contact of the point mass with the pole occurs at $s=0$, after $L^2/(2Rv)$. In the ideal model, the required tension $m{v}^2/s$ diverges at contact; finite string strength or finite particle size limits that idealization.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

For $-\pi \le t\le \pi$, this parametric curve closes with a corner on a unit circular cylinder. Its speed is $\cosh t$, so its arc length is $2\sinh \pi$. The cylindrical surface between the curve and $z=0$ has the same surface area, because unrolling gives the integral ${\int }_{-\pi }^{\pi }\cosh tdt$.

“Arrives at infinity with zero speed” is an asymptotic statement. For outward radial motion at zero specific orbital energy,Every finite distance takes finite time, but $t(r)\to \infty$ as $r\to \infty$. There is no instant of rest at an infinitely distant destination. Differentiating the displayed solution gives $\ddot{r}=-\mu /r^2$, independently checking Newton's second law. In the synthetic example, reaching $r=4\mathrm{m}$ takes exactly $(4-\sqrt{2})/3\mathrm{s}\approx 0.861929\mathrm{s}$. For nonzero specific angular momentum, the Barker equation gives the corresponding infinite-time limit along a parabolic trajectory. Positive-energy escape also takes infinite time to reach infinity; it retains nonzero asymptotic speed.