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A body force acts throughout a body's or fluid's volume, rather than only through its surface. Gravity is the example used in hydrostatic pressure: a gravitational field $\mathbf{g}$ gives force per unit volume $\rho \mathbf{g}$, where $\rho$ is the mass density. More generally, a body force per unit mass $\mathbf{f}$ contributes $\rho \mathbf{f}$ to momentum conservation. Forces from electromagnetism are another example. Surface pressure forces produce buoyancy on an immersed body, while its body force from gravity acts downward.

For a pure rightgoing acoustic wave, retarded time $\theta =t-x$ and slow distance $X=Mx$ separate fast oscillations from nonlinear steepening. Expanding the equation of state gives $p=\rho +M(\gamma -1){\rho }^2/2+O(M^2)$. The leading perturbations satisfy $u_0={\rho }_0=p_0$. Adding the first-order mass conservation and momentum conservation equations cancels the unknown first corrections and yields the displayed viscous Burgers equation. Inviscid characteristics satisfy $\theta =s-(\gamma +1)F(s)X/2$, $p=F(s)$; loss of invertibility marks shock formation and the end of the smooth inviscid approximation.

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 an impermeable cylinder, an axisymmetric turbulent plume with negligible source volume flux requires a downward return flow. With uniform upward plume velocity $w$ and ambient velocity $v$, volume conservation gives $b^{2}w+(R^2-b^2)v=0$. An entrainment coefficient $\alpha$ based on the relative speed gives $Q^{\prime }=2\alpha b(w-v)$. A specified pressure closure is essential for momentum conservation. If the departure from ambient hydrostatic pressure and wall force are neglected in the whole-section integral, the total specific momentum flux $J=b^2{w}^2+(R^2-b^2)v^2$ obeys $J^{\prime }=b^2{g}^{\prime }$. Homogeneous unmodified ambient and no buoyancy loss give $b^{2}w{g}^{\prime }=B_s$. These are leading integral-model closures; restoring the ambient dynamic pressure can change the singularity criterion.

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.

A classical rocket ejecting fuel at rate $\alpha >0$ and relative exhaust speed $u>0$, while collecting stationary dust of mass $\beta dx$ per traveled distance, obeysThese are mass conservation and momentum conservation for the rocket, incoming dust and outgoing gas. The dust-loading term supplies the momentum needed to accelerate incoming stationary material. Put $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$. While the engines operate and $0\le v<V$, an initially stationary rocket satisfiesFor $\lambda >1$, continuing to $V$ would drive total mass to zero. For $0<\lambda <1$, total mass instead grows without bound and $V$ is approached only after infinite time under indefinitely maintained thrust. In both cases a finite fuel supply runs out before that limiting speed. Total stored mass must not be confused with remaining fuel.

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.

For $S[\varphi ]=\frac{1}{2}\int (\partial \varphi {)}^2$ and a source appearing as $-\int J\varphi$ in the exponent, the positive quadratic operator is $A=-\Delta$. Completing the square gives $Z[J]=\mathcal{N}\exp [\frac{1}{2}J{A}^{-1}J]$. If one writes it with a negative quadratic source exponent, its kernel must be $G=-A^{-1}={\Delta }^{-1}$. In the plane $G(z,w)=(2\pi {)}^{-1}\log (|z-w|/L)$, up to harmonic additions and infrared conventions. The physical covariance is its negative. Zero modes are regulated or separated; oscillatory sources in a string path integral produce momentum conservation from the constant mode.

Fluid mechanics studies the motion and forces of liquids and gases. Its central simplification is the continuum approximation: replace individual molecules by bulk properties defined at every position and time. The averaging region must contain many molecules while remaining small compared with the distances over which those properties vary. A fluid element is such a small material parcel, rather than a single molecule. This connects fluid mechanics to continuum mechanics and makes calculus useful for describing flow. ^[ref]

Three basic fields are the velocity field $\mathbf{u}(\mathbf{x},t)$, the mass density $\rho (\mathbf{x},t)$ and the pressure $p(\mathbf{x},t)$. The first gives a parcel's instantaneous velocity; the second gives mass per unit volume. The pressure contributes an inward normal traction $-p\mathbf{n}$ across a surface with outward unit normal $\mathbf{n}$. Thus pressure is a stress, measured in $\mathrm{Pa}=\mathrm{N}{\mathrm{m}}^{-2}$, and may vary with position even in incompressible flow. Additional viscous stresses can act tangentially. ^[ref]

The guiding questions are what must be conserved and which forces matter. Mass conservation constrains the velocity field; momentum conservation determines its response to pressure gradients, viscosity and body forces. An incompressible flow model, a Reynolds number estimate and Bernoulli's principle answer different parts of a problem. Their appeal is that a few carefully stated assumptions can determine useful quantities before the entire velocity field is known.

Integrating the closed-collection balance in momentum conservation and applying the fundamental theorem of calculus givesFor one fixed-mass point mass, this becomes ${\mathbf{J}}_{\mathrm{net}}=\Delta \mathbf{p}=m\Delta \mathbf{v}$. Zero external impulse implies equal initial and final total momenta; momentum conservation throughout the interval requires zero net external force throughout it.

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]

An interaction vertex joins the field insertions from one interaction term in a Feynman diagram. Its number of incident lines is the number of fields in that term, and its momentum-space Feynman rule includes the interaction coefficient and momentum conservation.

For a fixed collection of two constant-mass bodies in an inertial frame, separate their mutual forces from the resultant external force on each:Adding the two Newton's second law equations and using Newton's third law givesThis is the proof of internal-force cancellation in system momentum. For more bodies, each internal force cancels with its partner once both appear in the sum. A boundary that excludes one interacting body leaves its force in the external force sum. A massive connecting spring must also be included, or its own changing momentum accounted for; its two endpoint forces are not automatically one third-law interaction pair between the carts.

Zero resultant external force gives momentum conservation; complete absence of external forces is sufficient but unnecessary. For constant total mass $M$, the centre of mass has velocity $\mathbf{V}=\mathbf{P}/M$ and obeys $Md\mathbf{V}/dt={\mathbf{F}}_{\mathrm{ext}}$. Internal interactions can move the parts apart while leaving the centre of mass moving uniformly.

Additional external forces do not invalidate the identified mutual pair. At the $4\mathrm{N}$ instant above, suppose a hand also exerts $+3\mathrm{N}$ on $A$. The mutual forces remain $-4\mathrm{N}$ and $+4\mathrm{N}$, but the net forces become $-1\mathrm{N}$ on $A$ and $+4\mathrm{N}$ on $B$. Consequently $a_A=-1m/{\mathrm{s}}^2$, $a_B=+2m/{\mathrm{s}}^2$, $dP/dt=+3\mathrm{N}$, and the centre of mass has acceleration $+1m/{\mathrm{s}}^2$. Pair equality by itself does not promise constant total momentum.

For the material-first nominal stress tensor, the displayed local conservation of energy law balances mechanical work, heat supply and outgoing heat flux. Integrating over a fixed reference material volume and using the divergence theorem gives its integral form. The full total-energy balance includes kinetic energy and external mechanical work; momentum conservation reduces it to this internal-energy balance.

For a half-space initially at rest and a wall given a step velocity $U$, momentum conservation and the Maxwell fluid stress law give $\tau u_{tt}+u_t=(G_0\tau /\rho )u_{yy}$. The displayed decaying Laplace transform solution obeys the wall condition. Inverting on a right-half-plane Bromwich contour yields a shear signal of speed $\sqrt{G_0/\rho }$. The limit $\tau \to 0$ at fixed $G_0\tau =\eta$ gives viscous error-function penetration; $\tau \to \infty$ at fixed $G_0$ gives a sharp elastic front.

For a nonrelativistic point mass of fixed mass $m>0$, measured in an inertial frame, its momentum isMomentum is a vector pointing along the velocity, with International System of Units unit $\mathrm{kg}m/s$. Doubling either mass or velocity doubles momentum. In one dimension, a chosen positive direction makes $p=mv$ a signed quantity: a $2\mathrm{kg}$ body moving right at $3m/s$ has $p=+6\mathrm{kg}m/s$; moving left at the same speed gives $p=-6\mathrm{kg}m/s$. ^[ref]

For a collection, total momentum is the vector sum $\mathbf{P}={\sum }_i{m}_i{\mathbf{v}}_i$. Two equal masses moving with opposite velocities can have zero total momentum while both have positive kinetic energy. Direction is therefore essential when using momentum conservation; adding speeds instead of signed velocities gives the wrong balance.

Newton's second law connects the net force on one fixed-mass point mass to its momentum through ${\mathbf{F}}_{\mathrm{net}}=d\mathbf{p}/dt$. Integrating that change gives impulse. Here $\mathbf{p}$ denotes mechanical momentum; when canonical momentum differs from mass times velocity, the distinction in mechanical momentum below applies. The nonrelativistic formula also has a limited regime: special relativity is needed at speeds comparable with the speed of light.

Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.