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A collision is an interaction between approaching bodies or particles that changes their motion. Momentum conservation and conservation of energy constrain the outcomes; inelastic collisions can transfer mechanical energy into internal degrees of freedom.

In a Keplerian shearing sheet, an epicyclic guiding center $x_0$ carries rotating-frame energy $-3{\Omega }^2{x}_0^2/8$, plus nonnegative radial and vertical oscillation energies. Inelastic collisions dissipate the total Jacobi energy in a shearing sheet, while momentum conservation preserves $\sum x_0$. Thus $\sum x_0^2$ must grow: the ring spreads about its fixed mean radius. The effect is an angular momentum transport process, with some particles moving inward and others outward.

An elastic collision preserves total kinetic energy between the incoming and outgoing states. For isolated colliding bodies, momentum conservation also holds. The kinetic energy condition supplies information beyond momentum conservation alone. ^[ref]

Consider a stipulated one-dimensional example with rightward positive: $m_1=2\mathrm{kg}$, $m_2=1\mathrm{kg}$, incoming velocities $u_1=3m/s$ and $u_2=0$. The total momentum is $P=6\mathrm{kg}m/s$ and total kinetic energy is $K=9\mathrm{J}$. In SI numerical values, outgoing velocities $v_1,v_2$ satisfySubstituting $v_2=6-2v_1$ gives $3(v_1-1)(v_1-3)=0$. The root $v_1=3$, $v_2=0$ repeats the incoming state; the outgoing separating state is $v_1=1m/s$, $v_2=4m/s$. Its momenta add to $2+4=6\mathrm{kg}m/s$ and its kinetic energies add to $1+8=9\mathrm{J}$. The incoming and outgoing relative velocities both have magnitude $3m/s$, consistent with coefficient of restitution $e=1$.

The inelastic collision example below uses exactly these incoming data. Both outcomes preserve total momentum, while only the elastic collision preserves total kinetic energy. These are ideal numerical examples, not experimental measurements.

A useful Franck-Hertz energy-loss model separates three operations: accelerating an electron, losing a discrete energy $\Delta$ in an inelastic collision, and selecting the surviving electrons with a retarding collector. The collector measures an accepted fraction of an outgoing energy distribution. This makes the experiment a small inverse problem: the electric current depends on both the collision physics and the readout.

For a deliberately simple one-channel model, let $u\ge 0$ be the incoming kinetic energy, and let $p(u)$ be the probability that one encounter excites the chemical atom. Set $p(u)=0$ for $u<\Delta$ and $0\le p(u)\le 1$ otherwise. The conditional probability distribution of outgoing kinetic energy $E$ is the discrete loss kernelHere $\delta$ is the Dirac delta function. The kernel has unit integral, no negative outgoing energy, and mean outgoing kinetic energy $u-p(u)\Delta$. It neglects the small recoil loss in an elastic collision, other excitation channels and ionization. The threshold permits excitation; it does not require $p(u)=1$ above threshold. In particular, choosing a single $\Delta =4.9\mathrm{eV}$ here isolates the characteristic channel discussed above without claiming it is mercury's lowest possible excitation.

The Franck-Hertz experiment connects a reproducible loss of electron kinetic energy with excitation of a chemical atom. The original mercury-vapor experiments were performed during 1912–1914 while investigating ionization; they were not designed as a test of the Bohr model. Their later interpretation distinguished internal excitation from removal of an electron. ^[ref].

In a modern mercury tube, a heated cathode emits electrons and a positive perforated grid accelerates them through Hg vapor. A collector slightly below the grid's electric potential rejects slow electrons. The oven controls vapor density and hence the mean free path. Keep heating and retarding settings fixed, sweep the accelerating electric potential difference $V$, and record collector electric current $I(V)$. These measured quantities indirectly probe atomic excitation. ^[ref].

The curve shows repeated current minima, typically about $4.9\mathrm{V}$ apart. In the collision model, an electron gains approximately $eV$ of kinetic energy, with $e$ the positive elementary charge. An inelastic collision transfers about $4.9\mathrm{eV}$ to internal excitation. Near the grid, this can prevent the electron crossing the retarding region, lowering collected electric current. Further acceleration restores collection; repeated excitation losses produce further minima. These represent multiple collisions, rather than equally spaced atomic energy eigenvalues. Contact potentials shift the first minimum, so compare several successive spacings instead of identifying the first dip's voltage with the excitation cost. ^[ref].

The optical comparison supplies a second observable: mercury radiation at a wavelength near $253.6\mathrm{nm}$ was detected through a window in the tube. Its inferred photon energy $hc/\lambda \approx 4.89\mathrm{eV}$ agrees with the characteristic collision loss, with $c$ the speed of light. This agreement supports the interpretation as a transition between internal atomic energies. It does not make the collector trace a complete spectrum. In particular, the lowest excited mercury level, labeled $6{}^3{P}_0$, is metastable because an electric-dipole selection rule forbids its direct transition to the ground state; the nearby $6{}^3{P}_1$ level emits the quoted ultraviolet line. Collision excitation and optical detection therefore have different sensitivities to the available excited states. ^[ref].

Initial electron kinetic energy spread and varying collision positions broaden real minima. Repeat traces at different retarding settings and oven temperatures to distinguish excitation from collection effects. This probes a discrete energy-transfer channel, rather than every energy eigenspace, its dimension, or an absolute ground-state energy. ^[ref].

James Franck and Gustav Ludwig Hertz presented their paper on 24 April 1914. Its heated platinum wire sat inside a cylindrical platinum mesh, with a collector 1–2 mm beyond it. A heated bath and pumping prepared the vapor; guard rings suppressed leakage. They differentiated retarding-electric current scans to study outgoing energies.^[ref]

Their 1925 Nobel Prize preceded their lectures of 11 December 1926. James Franck described choosing pressure for many collisions before the mesh but few afterward.^[ref] Gustav Ludwig Hertz explained how a field-free collision region narrowed excitation thresholds.^[ref]

An inelastic collision changes total kinetic energy. In the dissipative case it decreases, transferring energy to internal energy, including deformation and microscopic motion. Isolated colliding bodies still obey momentum conservation; conservation of energy requires accounting for the transferred energy. Bodies need not stick in every inelastic collision. The decrease in translational kinetic energy drives dissipative spreading of a planetary ring.

For the incoming masses and velocities in the elastic collision example, stipulate that the bodies stick. This perfectly inelastic collision has a shared outgoing velocityThe outgoing momenta are $4$ and $2\mathrm{kg}m/s$, totaling $6\mathrm{kg}m/s$. The outgoing kinetic energies are $4$ and $2\mathrm{J}$, totaling $6\mathrm{J}$. Thus $9-6=3\mathrm{J}$ of kinetic energy becomes internal energy within the isolated pair. Nothing in momentum conservation requires this $3\mathrm{J}$ to remain as kinetic energy. ^[ref]

The time-independent particle Lagrangian in a shearing sheet has conserved rotating-frame energyThe velocity-linear Coriolis acceleration term cancels from this expression. Up to the reference-orbit constant, it is the second-order expansion of inertial specific orbital energy minus ${\Omega }_0$ times inertial specific angular momentum. Its negative radial tidal term allows inelastic collisions to lower the total energy while increasing the radial extent of a ring.

Initially every particle has $A=B=0$, so its Jacobi energy in a shearing sheet is $-3{\Omega }_0^2{x}_0^2/8$. During an inelastic collision, positions are fixed at the instant of impact, while momentum conservation preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot{y}+2{\Omega }_{0}x$, and therefore the sum of the epicyclic guiding center positions $x_0=2p_y/{\Omega }_0$, is unchanged. Between collisions, these quantities are individually conserved.

The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total Jacobi energy in a shearing sheet decreases. At any later time, writing $x_{0j}$ for the particles' current epicyclic guiding center positions givesIf $D\ge 0$ is the accumulated energy dissipated in the inelastic collisions, comparison with the initial circular orbits givesThus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto (-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: angular momentum transport moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for dissipative spreading of a planetary ring.

Let $\dot{M}=500\mathrm{kg}{\mathrm{s}}^{-1}$ and $V=10\mathrm{m}{\mathrm{s}}^{-1}$. Incoming material has zero horizontal momentum, while incorporation requires horizontal momentum $VdM$. Thus the engine supplies force $F=V\dot{M}$, and its power isNo duration or loading distance is given, so a single total work cannot be assigned. Over an interval $t$ in seconds,The remaining $25,000t\mathrm{J}$ becomes internal energy in the inelastic collision that brings the falling material to the wagon's horizontal velocity. Vertical falling kinetic energy, if any, is a separate contribution to dissipation; it does not alter this horizontal work balance. The source alternates between stone and coal; the calculation depends only on the given loading rate.

Choose particle 1's incoming velocity positive and particle 2's negative. The final particle is stationary, so four-momentum conservation requires zero total incoming momentum:It follows that ${\gamma }_1{\beta }_1=1/\sqrt{2}$. Squaring gives ${\beta }_1^2/(1-{\beta }_1^2)=1/2$, henceThe initial total energy becomes the final rest energy. ThereforeThe extra invariant mass is the incoming kinetic energy retained as internal energy of the merged particle: $(M-3m)c^2$. It does not violate conservation of energy or four-momentum conservation; the sum of the separate rest masses is not a conserved quantity in an inelastic collision.

A totally inelastic collision in the sticking sense has zero final relative velocity, so $q=0$. Both velocities are the conserved centre-of-momentum frame velocity:Thus the bodies move together along the original incident direction; the moving projectile has ${\theta }_1=0$. The common outgoing state is stationary in the centre-of-momentum frame. The lost translational kinetic energy iswhich can become internal energy. As explained in part 8, setting only the normal coefficient of restitution to zero in a smooth oblique collision does not force sticking: tangential relative motion can remain.

A perfectly inelastic collision is an inelastic collision in which the bodies stick and share one outgoing velocity. For two isolated point masses, momentum conservation gives $v=(m_1{u}_1+m_2{u}_2)/(m_1+m_2)$. At fixed total momentum, common velocity minimizes the outgoing translational kinetic energy, so sticking gives the greatest loss of that kinetic energy for the specified incoming state.

Oberlin College's documented demonstration uses a track, two collision carts, two force probes set to the $50\mathrm{N}$ range, added $500\mathrm{g}$ bars, and a laptop with a Vernier interface. Preparation consists of entering the measurement calibration, securing the force probes to the carts, and zeroing them. Rubber stoppers serve for elastic collisions, clay for inelastic collisions, and hooks with string for pulling. Comparing the two force records tests the mutual interaction; adding mass to one cart tests whether unequal masses change the equality. ^[ref]

Vernier's force probe uses a bending beam with strain gauges; changing electrical resistance produces a voltage readout calibrated as force. Its manual recommends zeroing in the working orientation. If recalibration is needed, use an unloaded reading and a known suspended gravitational weight; for the $50\mathrm{N}$ range its example is $1\mathrm{kg}$ giving $9.8\mathrm{N}$. Re-zero after changing orientation. The manual quotes calibrated accuracy $\pm 0.2\mathrm{N}$, distinct from the $0.05\mathrm{N}$ resolution of that range, and warns that the interface's data-collection rate can limit collision measurements. ^[ref]

The following sampling and control choices are proposed additions to that documented experiment. Level the track and check unloaded motion in both directions for slope and friction. Use a soft bumper and one interface for simultaneous records; start before contact and stop after separation. A trial sampling rate of $1000{\mathrm{s}}^{-1}$, if supported, would put about 100 samples across a $0.10\mathrm{s}$ interaction. Increase the sampling rate and repeat until the peak and integrated impulse comparison are stable; a large sample count alone does not establish adequate response. Measure each total moving mass, including the force probe and added bars.

Convert both channels to the same right-positive force convention. Oppositely facing force probes can have matching raw signs because their local positive directions differ; check each sign by a gentle push of known direction before the run. Overlay $F_{B\mathrm{on}A}(t)$ and $-F_{A\mathrm{on}B}(t)$, and also examine their signed sum. Integrate both over the same contact interval to compare impulses. Repeat with added mass on one cart, exchange the probes between carts, and check unloaded zeros before and after. These controls help distinguish a physical mass change from a channel offset or gain error.

Agreement within measurement error supports the pair-interaction model over the resolved interval; it does not establish exact equality at arbitrarily short time scales. Misalignment, a loose mounting, sensor-beam dynamics, saturation, unequal timing, and zero drift can spoil the comparison. Track friction and cable pulls are additional external forces; a probe measuring bumper contact need not measure the entire net force on its cart. If a massive connector carries appreciable changing momentum, compare the full cart–connector collection rather than assuming its endpoint forces are an instantaneous cart-to-cart pair. The constructed pulse diagram above illustrates what to compare, and supplies no measurements from this apparatus.