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Consider a stipulated ideal cart modeled as a point mass of mass $2\mathrm{kg}$ on a horizontal track in a laboratory inertial frame. Its initial position is $x(0)=0$ and its initial velocity is $v_x(0)=1.5m/s$ to the right. Independent devices exert horizontal forces $+6\mathrm{N}$ and $-6\mathrm{N}$ on this same cart. Its upward support force balances its downward gravitational weight; no other forces contribute. Rotation and deformation are excluded from this model.After $4\mathrm{s}$, the cart is at $6\mathrm{m}$, still moving right at $1.5m/s$. Starting the same model at rest instead would leave it at rest. Force balance determines the zero acceleration, while the initial velocity determines whether the cart moves.

For the plotted model, both horizontal devices are switched off simultaneously at $t=2\mathrm{s}$, with no impulse. The support and gravitational weight remain balanced. The individual horizontal forces change, but their sum stays zero on both sides of the switch, so the velocity stays continuous and unchanged. Switching off just one device would be a different model with a nonzero net external force.

An original reduced model makes the calibration explicit. Let the torsion pendulum have restoring torque $-\kappa \theta$, with torsion constant $\kappa$, and viscous damping torque $-b\dot{\theta }$, with $b>0$. Allow an unwanted axial torque ${\tau }_p(t)$:Here ${\omega }_0$ is the undamped natural angular frequency and $Q$ the quality factor. The forced harmonic oscillator response must be included before a twist can be interpreted as a torque.

Using the complex amplitude convention $m_z(t)=\mathrm{Re}(\tilde{m}{e}^{\mathrm{i}\omega t})$ and similarly for $\theta$ and ${\tau }_p$, differentiation multiplies by $\mathrm{i}\omega$. Substitution givesThis is the torque-to-twist transfer function. With ${\tilde{\tau }}_p=0$ and real positive $\tilde{m}=m_1$, the twist amplitude isFor $\gamma <0$, the twist phase relative to $m_z$ is $\pi /2-atan2(b\omega ,\kappa -I{\omega }^2)$. At $\omega ={\omega }_0$ it is zero, although the reaction torque leads $m_z$ by $\pi /2$. A phase measured at the mirror is not automatically the phase of the applied torque.

A fixed-amplitude reference torque ${\tau }_c\cos \omega t$ measures $|H_{\tau }|$ directly. The Einstein-de Haas effect instead contains an extra factor $\omega$ when $m_1$ stays fixed. Consequently its twist amplitude peaks exactly at ${\omega }_0$ in this model, whereas the fixed-torque peak is at ${\omega }_0\sqrt{1-1/(2Q^2)}$ when $Q>1/\sqrt{2}$. To verify the first statement, divide the squared denominator of $A_{\mathrm{EdH}}$ by ${\omega }^2$: the quantity to minimize is $(\kappa /\omega -I\omega {)}^2+b^2$, minimized at ${\omega }_0$. Both response curves tend to zero at high frequency; only the magnetomechanical curve also tends to zero at zero frequency.

For a fully stipulated example, choose $I=2.00\times 10^{-10}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =8.00\times 10^{-10}\mathrm{N}m/rad$, $b=8.00\times 10^{-11}\mathrm{N}\mathrm{m}s/rad$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and $m_1=1.76\times 10^{-4}\mathrm{A}{\mathrm{m}}^2$. These teaching inputs give ${\omega }_0=2.00rad/s$, $Q=5.00$, reaction torque amplitude $2.00\times 10^{-15}\mathrm{N}\mathrm{m}$ at ${\omega }_0$, andAn optical lever at distance $D=1.00\mathrm{m}$ gives small-displacement spot amplitude $x\simeq 2DA=25.0\mu \mathrm{m}$; its full peak-to-peak width is $4DA=50.0\mu \mathrm{m}$. For a fast reversal from $+m_1$ to $-m_1$, the same inputs give $\Delta L_{\mathrm{body}}=-2.00\times 10^{-15}\mathrm{kg}{\mathrm{m}}^2/s$ and initial $\Delta \dot{\theta }=-1.00\times 10^{-5}rad/s$, provided the external angular impulse is negligible.

One independent mechanical calibration adds known moment of inertia $\Delta I$, keeping $\kappa$ unchanged. If $T_a,T_b$ are the undamped oscillation periods before and after adding it, thenFor $\Delta I=I$ in the stipulated example, $T_a=\pi \mathrm{s}$ and $T_b=\sqrt{2}\pi \mathrm{s}$, recovering the chosen $I$ and $\kappa$. Real damped periods require a damping correction: a free decay has angular envelope $e^{-bt/(2I)}$ and oscillation angular frequency $\sqrt{\kappa /I-(b/(2I){)}^2}$. Measuring that decay determines $b/(2I)$. Calibrating the optical scale, mechanical response and change of magnetic dipole moment separately is what makes the inferred gyromagnetic ratio meaningful.

For a body constrained to rotate about a fixed pivot, a perpendicular impulse $K$ delivered at distance $b$ from the pivot produces $\omega =Kb/I_O$. If the centre of mass is distance $h$ from the pivot, its momentum is $M\omega h$. The pivot impulse is $J_O=M\omega h-K$, and vanishes when $b=I_O/(Mh)$. This impact location is the centre of percussion. The result assumes the stated planar rigid-body geometry and a short impulse; it is not a claim that all forces at the pivot vanish afterward.

In a smooth oblique collision, the impulse is parallel to the contact normal $\mathbf{n}$, while tangential velocity components remain unchanged. Incoming velocities, masses, and a coefficient of restitution do not determine the outgoing velocities until that contact normal, or equivalent scattering data, is specified. For an elastic collision of masses $3m,m$, incoming velocities $(-u,0)$ and $(\sqrt{3}w/2,w/2)$, and outgoing first velocity $(0,h)$, conservation of momentum and conservation of kinetic energy give $w(\sqrt{3}u+h)=2u^2+4h^2$. Thus specifying only the first outgoing $x$ component leaves a family: $h=0$ gives $w=2u/\sqrt{3}$, while $h=u/\sqrt{3}$ gives $w=5\sqrt{3}u/6$.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

For the ideal uniform solid sphere with negligible table impulse during the cue strike, immediate rolling requires $u_0=a{\Omega }_0$, hence the horizontal cue line lies $2a/5$ above the centre. A different inertia distribution changes this height.

A normal impulse $j$ on a free planar rigid body of mass $M$ and moment of inertia $I$, at perpendicular lever arm $b$, changes the contact point's normal velocity by $j(1/M+b^2/I)$. Its effective mass in that direction is therefore $(1/M+b^2/I{)}^{-1}$. This follows by adding the translational change $j/M$ to the rotational contribution $b(jb/I)$.

For an incoming point mass $m$, the inverse reduced contact mass is $1/m+1/M+b^2/I$. A smooth elastic collision with initial relative approach speed $u$ has impulse $2u/(1/m+1/M+b^2/I)$, since the relative normal velocity reverses. This separates effective inertia at a contact from the body's total mass and explains the lever-arm dependence in an equal-speed elastic impact on a free rod.

A short transverse impulse $J$ at lever arm $b$ gives a pivoted rigid body angular velocity $\Omega =Jb/I$ and contact-point velocity $b\Omega =J/m_{\mathrm{eff}}$, where $I$ is its pivot moment of inertia. Thus $m_{\mathrm{eff}}=I/b^2$ plays the role of a translational mass in the impact direction. A point projectile elastically striking an initially stationary body stops immediately after impact precisely when its mass equals this effective value. The pivot can exchange linear momentum while doing no work.

Let a nonrotating disc initially move downwards at speed $u$ and strike a stationary point support offset $d$ from its centre. If the impulse is vertical and the normal coefficient of restitution is one, its upward impulse satisfies $J(1/M+d^2/I)=2u$. The signed upward centre velocity is therefore the title formula, where $I$ is the moment of inertia about the horizontal diameter perpendicular to the impact lever arm. At $d=a/2$ for a thin circular disc, $I=M{a}^2/4=M{d}^2$, so its centre instantaneously stops and all initial kinetic energy becomes rotation. A fixed support can absorb momentum without doing work.

An ideal elastic collision between a sphere and a stationary plane reverses normal approach velocity. On the fully rough, tangentially elastic branch, the tangential contact-point velocity reverses too. If $K_t$ is the signed tangential impulse, translation and spin obey $M\Delta v=K_t$ and $I\Delta \Omega =a{K}_t$, with positive spin anticlockwise. Restoring tangential translational plus rotational kinetic energy givesThe nonzero root is the displayed impulse; the other root is the smooth-contact branch. For $I=2M{a}^2/5$, the map is $v^{\prime }=(3v-4a\Omega )/7$ and $a{\Omega }^{\prime }=-(10v+3a\Omega )/7$. It preserves kinetic energy and the angular momentum about the contact point, $I\Omega -Mav$. An unspecified rough surface need not realize this idealization: finite friction or a dissipative tangential contact law requires additional constitutive data. Setting the outgoing contact slip to zero instead loses $M(v+a\Omega {)}^2/7$ of kinetic energy for this uniform sphere.

Each massless rod has endpoint masses $m$ at distance $\ell$ from its centre, hence total mass $2m$ and moment of inertia $2m{\ell }^2$. In a symmetric opposite-end collision with centres approaching at speed $V$, a horizontal impulse $J$ changes each centre speed by $J/(2m)$ and each contact-point speed by $J/m$. The coefficient of restitution gives $2J/m-2V=2eV$. Both rods rotate in the same sense for the staggered vertical configuration; their centre speeds after impact are $V(3-e)/4$.

// Target: electromagnetism.bigb

Consider identical harmonic oscillators with unit mass, natural angular frequency ${\omega }_0>0$, and positive coupling $\kappa \ll {\omega }_0^2$:An initial impulse on the first gives $x_1(0)=x_2(0)={\dot{x}}_2(0)=0$ and ${\dot{x}}_1(0)=v$. The normal mode coordinates $u=x_1+x_2$ and $w=x_1-x_2$ have angular frequencies ${\omega }_0$ and ${\omega }_a=\sqrt{{\omega }_0^2+2\kappa }$. Solving their initial value problems provesLet $\Delta ={\omega }_a-{\omega }_0\sim \kappa /{\omega }_0$. To leading order in $\kappa /{\omega }_0^2$, sum-to-product identities give the displacement amplitude envelopes $(|v|/{\omega }_0)|\cos (\Delta t/2)|$ and $(|v|/{\omega }_0)|\sin (\Delta t/2)|$. Thus the first nearly complete energy transfer occurs at $\pi /\Delta$, and localization returns at $2\pi /\Delta$. More precisely, at $t=\pi /\Delta$, the first oscillator has exactly zero velocity and residual displacement $\frac{v}{2}(1/{\omega }_0-1/{\omega }_a)\sin ({\omega }_{0}t)$, of order $|v|\kappa /{\omega }_0^3$.

Here transferred energy means kinetic energy plus the uncoupled quadratic potential energy; the coupling potential energy is relatively small. The fast phase need not make the receiving oscillator pass through zero displacement at the transfer time. Expanding $\Delta t$ as small is invalid on the transfer scale $t=O(1/\kappa )$. This lemma complements absolute and signed beat envelopes by keeping that slow phase.

Choose right as positive and let two carts have total moving masses $m_A=1\mathrm{kg}$ and $m_B=2\mathrm{kg}$, including any attached apparatus. At an instant when their repulsive interaction has magnitude $4\mathrm{N}$, assume no other horizontal forces. Then Newton's third law gives $F_{B\mathrm{on}A}=-4\mathrm{N}$ and $F_{A\mathrm{on}B}=+4\mathrm{N}$, while Newton's second law givesThe smaller mass has the larger acceleration magnitude. For a nonzero interaction and zero other horizontal forces, $a_A/a_B=-m_B/m_A$. Meanwhile $d(p_A+p_B)/dt=m_A{a}_A+m_B{a}_B=-4+4=0$ in force units. The two carts' velocities change while total momentum stays constant.

To distinguish an instantaneous force from its accumulated impulse, prescribe an original triangular interaction pulse lasting $T=1\mathrm{s}$:This chosen pulse reaches $4\mathrm{N}$ at $t=0.5\mathrm{s}$; it is a constructed model, not an experimental trace or a claim about a particular bumper's material response. Its positive area is $\frac{1}{2}(1\mathrm{s})(4\mathrm{N})=2\mathrm{N}\mathrm{s}$. The impulse-momentum theorem therefore givesOpposite impulses give opposite momentum changes, while the velocity changes differ because of the unequal masses. The cumulative positive impulse is $I(t)=4N/s{t}^2$ during the first half and $I(t)=2\mathrm{N}\mathrm{s}-4N/s(T-t{)}^2$ during the second half. Thus $J_A(t)=-I(t)$ and $J_B(t)=I(t)$ cancel throughout the pulse, not merely at its endpoint.

A point mass $m$ strikes a free rod of equal mass, initially at rest, perpendicularly at distance $h$ from its center of mass. Assume a smooth elastic collision and rod moment of inertia $I$ about its centre. The final translational speeds are equal precisely when $h^2=2I/m$. Indeed, momentum conservation then fixes both final velocities to $v_0/2$; angular impulse gives $I\omega =m{v}_{0}h/2$, and conservation of kinetic energy gives $I{\omega }^2=m{v}_0^2/2$. Eliminating $\omega$ proves the formula. A uniform rod has $I=m{l}^2/12$, hence $|h|=l/\sqrt{6}$, which lies within the rod.

A smooth central collision leaves a cylinder's angular velocity unchanged if the floor supplies negligible tangential impulse. Imposing rolling without slipping during the impact instead requires a floor impulse and changes the cylinder's effective translational mass to $m+I/r^2$. Consequently a pre-impact rolling condition alone does not determine the outgoing velocities: the impulsive contact law must also be specified. For a uniform solid cylinder striking a stationary point mass $2m$ in an elastic collision, the target acquires $2u/3$ with negligible floor impulse, and $6u/7$ with an enforced rolling constraint, where $u$ is the incoming centre speed.

For a circular rigid body of radius a, mass M and axial moment of inertia I on a fixed horizontal surface, choose positive spin so that rolling means $v=a\omega$. During purely horizontal frictional relaxation, an impulse J changes $Mv$ by J and $I\omega$ by $-aJ$. Hence $I\omega +Mav$ is constant. At final rolling without slipping, $v_f=(Ia{\omega }_0+M{a}^2{v}_0)/(I+M{a}^2)$. Subtracting initial from final translational-plus-rotational kinetic energy gives the title formula. The coefficient of friction controls the relaxation time, but not this loss once the no-slip state is reached. Vertical landing losses and rolling resistance are excluded.

A thin straight rod of length $l$, mass $m$, central moment of inertia $I=\beta m{l}^2$ and midpoint centre of mass starts beside a pointlike pole at that midpoint. An end impulse perpendicular to the rod gives translation $V=J/m$ and rotation $\Omega =J/(2\beta ml)$. At the first quarter-turn the trailing end is at horizontal distance $\pi \beta l-l/2$ from the pole. Strict clearance requires $\beta >1/(2\pi )$; equality is the grazing limit. The criterion is for free motion before contact, a negligibly thin rod and pole, and an impulse directed away from the pole.

For a uniform square plate of side $l$ hinged along an edge, $h=l/2$ and $I_H=M{l}^2/3$. An impulse $K$ normal to the plate, applied a distance $a$ below the centre, gives $v_G=3K(1+2a/l)/(4M)$ and signed hinge impulse $J_H=K(3a/(2l)-1/4)$. Therefore the hinge impulse vanishes at $a=l/6$ below the centre, a distance $2l/3$ from the hinge. The sign of $J_H$ identifies the reaction direction; its magnitude is $|J_H|$.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

The impulse of a force over a time interval is its vector integral:Its International System of Units unit is $\mathrm{N}\mathrm{s}=\mathrm{kg}m/s$. The impulse-momentum theorem equates the net impulse to the change in momentum. For one fixed-mass point mass, integrate its net force; for a closed collection obeying Newton's third law, integrate the sum of external forces. In one dimension, the signed area under the force versus time graph gives the impulse; negative force contributes negative area. ^[ref]

Three stipulated rightward net force pulses illustrate the same $J=6\mathrm{N}\mathrm{s}$: constant $12\mathrm{N}$ for $0.5\mathrm{s}$, a triangle rising from zero to $24\mathrm{N}$ at $0.25\mathrm{s}$ and falling to zero at $0.5\mathrm{s}$, and constant $3\mathrm{N}$ for $2\mathrm{s}$. Their areas areApplied separately to the same fixed mass $m=2\mathrm{kg}$, each gives $\Delta v=J/m=3m/s$. Their peak forces differ: $12$, $24$, and $3\mathrm{N}$. Thus impulse fixes the final momentum change, while the force history determines how abruptly it occurs.

The mean net force over the chosen interval is $\overline{\mathbf{F}}=\mathbf{J}/(t_f-t_i)$. Extending the interval for a fixed impulse reduces this mean force. A lower peak force additionally depends on the pulse shape: the two short pulses above have the same mean $12\mathrm{N}$ but different peaks. This explains why controlling the duration and shape of a stopping impulse matters in cushioning a collision.