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There are two different issues: whether a law follows logically from the other laws, and whether an experiment can test it without assuming anything else. Newton's laws of motion have empirical content when combined with rules for identifying inertial frames, measuring mass and specifying forces. An experiment tests predictions of that combination. Calling every observed acceleration an otherwise unspecified force would remove the empirical constraint rather than verify Newton's second law. The distinction is central to the empirical testing of Newton's laws of motion.

Newton's first law contributes the claim that a common family of inertial frames exists: bodies with zero net external force have uniform straight motion in those frames. It is not enough to choose coordinates which straighten the trajectory of just one body; an admissible frame must accommodate many independent bodies and interactions. Once that frame and the ordinary constant-positive-mass form of Newton's second law are assumed,This conditional deduction does not establish the existence of the required frame, or independently identify the physical circumstances in which the net force vanishes. Defining an inertial frame and then counting a straight trajectory as proof of the definition would not be a test.

Nevertheless, the law of inertia has testable implications in a network of assumptions. Consider low-friction motion along an air track: determine positions at measured times, reduce the known residual influences, and ask whether the remaining acceleration approaches zero in the same laboratory frame for different bodies. Rotating the frame changes the description through Coriolis forces and centrifugal forces; this dependence must be accounted for. No finite experiment establishes that every unmeasured influence is absent, but varying the apparatus can discriminate an interaction from a frame effect. A failed prediction is evidence against some part of the package, not automatically the isolated verbal statement of Newton's first law.

Newton's second law is more than a definition if the force law and mass assignments constrain predictions before the new acceleration is measured. For instance, independently characterize a spring's reproducible extension response, specify the same interaction in several configurations, and compare the resulting accelerations for different bodies. If its strength and every mass are freely refitted to each outcome, agreement is empty. If they are fixed by an independently checked model and calibration, the law predicts held-out trajectories. Such a test remains conditional on the spring model, initial conditions and frame identification; this is an instance of the Duhem problem, not a reason to deny all empirical content.

Newton's third law supplies a different constraint. For two bodies exchanging forces, Newton's second law giveswhen there is no net external force. Equal-and-opposite interaction forces imply conservation of momentum for the pair. The two second-law equations alone do not require that cancellation: assigning ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$ is algebraically consistent with both second-law equations while violating the third law. This is a logical countermodel, not a proposed isolated physical interaction. Measuring paired impulses or checking the pair's total momentum therefore tests an additional constraint, conditional on the assumed absence of momentum exchange with anything else.

The system boundary matters. In electromagnetism, matter can exchange momentum with the electromagnetic field. Instantaneous mechanical forces between two particles need not satisfy the simple pairwise statement, although total matter-plus-field momentum is conserved under the appropriate closed-system conditions. Likewise, Noether theorem can derive conservation of momentum from a translationally invariant action; that uses an extra symmetry assumption and does not derive the third law from the second alone.

Thus the first law is not an additional differential consequence once the full second-law framework is fixed, and the third law is not a consequence of that framework. All three can contribute to empirical tests, but none is tested in total isolation from frame, interaction and measurement assumptions. This preserves the difference between logical dependence and experimental testability.^[ref]

Idealize the pole and stick as having negligible transverse thickness. Put the pole at the origin, the stick initially along the $y$ axis, and the struck lower end at $y=-l/2$, as in the source diagram. A short horizontal impulse $J$ along $+x$ gives the centre of mass a constant speed $V=J/m$. The angular impulse about the centre is $Jl/2$, giving constant counterclockwise angular velocityThe pole exerts no further force while the stick moves clear of it. Let $\psi =\Omega t$. The trailing upper end has coordinatesThe pole can lie on the translating straight stick again only when the stick becomes horizontal: for other orientations, its intersection with $y=0$ is the moving midpoint $x=Vt>0$. At the first horizontal orientation, $\psi =\pi /2$, the trailing end is atConsequently the grazing clearance of an impulsively struck rod occurs atFor $\beta >1/(2\pi )$ the whole stick has passed the pole by its first horizontal orientation, and the continually advancing centre makes all later passes safe. Equality is the limiting grazing contact of the endpoint; strict avoidance requires the strict inequality. If $\beta <1/(2\pi )$, the pole lies within the stick at that first horizontal pass, so contact occurs before the freely moving trailing end can pass the pole. A uniform thin stick has $\beta =1/12$, below the threshold: sufficient mass near the ends is needed to increase its moment of inertia and reduce the rotation relative to translation.

There is no external horizontal force, so momentum conservation keeps the student–punt centre of mass fixed if the system starts at rest. Let the student's walking direction be positive. If the punt's displacement relative to the water is $X$, the student advances $X+L$, where $L=3\mathrm{m}$ is measured relative to the punt. Constancy of the centre of mass givesThe punt moves $1.04\mathrm{m}$ opposite to the student's walking direction; the student's net displacement relative to the water is $3+X\simeq 1.96\mathrm{m}$. These displacements are relative to the initial state, not distances travelled relative to the punt.

Under the same ideal assumptions, merely changing the walking speed or walking back more slowly cannot produce unlimited travel. Returning the student to the original place on the punt restores the original internal mass distribution and hence the punt's original position. A finite punt bounds the displacement available through internal rearrangement. Propulsion over large distances requires an external horizontal impulse, for example a push against the bed or momentum transferred to the water. Such an interaction changes the isolated, frictionless model rather than evading centre of mass conservation.

The symbols in the question are diameters: write drum radius $a=R/2$, axle radius $b=r/2$. For a thin hollow drum, $I=M{a}^2$. A rightward horizontal impulse $J$ applied at height $h$ above the centre gives $v=J/M$ and clockwise angular velocity $\omega =Jh/I$. Rolling on the axle requires $v=b\omega$, soThe frictionless rails do not create the required rotation; the initial impulse must give exactly this ratio of linear to angular motion. For the usual case $r<R$, the required height is greater than the drum radius. It cannot be reached by striking the bare cylindrical surface horizontally; an extension or equivalent applied couple would be needed. This geometric limitation should accompany the formal result.

Choose positive horizontal velocity to the right and positive spin angular velocity anticlockwise, as in the source drawing. At the bottom of the sphere, the rotational velocity is then to the right: the initial tangential contact-point velocity isFor the illustrated positive $v_1,{\Omega }_1$, the contact slips to the right, so the tangential contact force acts leftward. The upward normal force acts through the sphere centre and produces no spin torque. The tangential contact force acts at lever arm $a$ and produces a clockwise torque, reducing the positive spin. The weight acts vertically downward through the centre of mass.

During the short impact the contact impulses dominate the weight's impulse. In an ideal elastic rough contact, the tangential deformation can store and return energy; its contact force need not be a dissipative sliding-friction law throughout the collision. For other signed incoming states, reverse the initial tangential force if $w_1<0$; if $w_1=0$, no tangential impulse is required.

Write the sphere's mass as $M$ and use $I=2M{a}^2/5$ from b. Let $K_n$ and $K_t$ be the upward and rightward contact impulses. The initial vertical velocity is $-u_1$ and the outgoing vertical velocity is $u_2$. Perfect normal elasticity givesThe tangential and angular impulse equations areFor the ideal elastic rough-sphere bounce, the tangential translational plus rotational kinetic energy is restored on separation. Substituting these two equations into that requirement givesThe interacting rough-contact branch, rather than the smooth branch $K_t=0$, therefore hasConsequently, with all tangential quantities signed in the convention of c,In particular, $v_2+a{\Omega }_2=-(v_1+a{\Omega }_1)$: the tangential contact-point velocity reverses. The outgoing $v_2$ may be negative; the sphere rebounds backward when $a{\Omega }_1>3v_1/4$. Substitution also checks both total kinetic energy and angular momentum about the contact point, $I\Omega -Mav$.

This result uses the conventional perfectly elastic, fully rough contact idealization. The word rough alone, with an unspecified finite coefficient of friction, would not determine the outgoing tangential state. For example, imposing sticking at separation would instead give $K_t=-2M(v_1+a{\Omega }_1)/7$ and dissipate $M(v_1+a{\Omega }_1{)}^2/7$, contradicting total elasticity unless $w_1=0$. The displayed elastic branch requires enough tangential impulse; a Coulomb bound $|K_t|\le \mu K_n$ would at least require $\mu \ge 2|v_1+a{\Omega }_1|/(7u_1)$.

During the short collision, the pivot's impulse has zero torque about the pivot, so angular momentum conservation about that point givesThe bullet stops rather than becoming embedded: immediately after impact only the rod contributes to the kinetic energy. Because this is an elastic collision,Eliminating $\Omega$ yields $m{L}^2=I$: the projectile mass equals the effective impact mass of a pivoted body, $I/L^2$. Hence the bullet has one third of the rod's mass, andOrdinary momentum conservation for the bullet and rod together would be incorrect: the fixed pivot supplies an external impulse. The short-impact approximation neglects the change in gravitational potential energy during impact.

Choose downward velocity $v$ positive. The remaining rocket mass is $m(t)=m_0-\alpha t$, and the upward rocket thrust has magnitude $\alpha u$, where $u>0$ is the relative rocket exhaust speed. The rocket equation including gravity isThis follows by applying momentum conservation to the rocket and its just-expelled rocket propellant, with gravity supplying the external impulse. Applying $d(mv)/dt=mg$ to the rocket alone would omit the exhaust momentum flux.

With $M_a=M_l=M$ and $l=2a$, a gives $X_{\mathrm{CM}}=a$, while the hand point is $P$ at $X=3a$. Thus the hand-to-CM distance is $d=2a$, and b givesModel the short impact with the hand point fixed but free to rotate about the in-plane axis perpendicular to the handle. Let $K$ be the ball's impulse on the racquet and $b$ its distance from $P$ toward the head. Angular impulse about $P$ gives $I_P\Omega =Kb$. The fixed-pivot kinematics require the centre-of-mass velocity magnitude to be $d\Omega$. The hand's signed impact impulse is consequentlyThe centre of percussion is where $K_P=0$, henceSubtract this from the head-centre distance $3a$ to obtain the requested position:It lies within the circular head. By symmetry its transverse displacement is zero, so the impact is on the head's central line along the handle. This cancels the impulsive hand reaction, not every force on the hand at later times; weight and any sustained grip torque lie outside the short-impact approximation.

Write the incoming particle speed as $u>0$, its final signed speed as $v$, and the rod's final centre-of-mass velocity along the impact direction as $V$. Let $b$ be the signed impact distance from the centre, and choose the sign of the angular velocity $\Omega$ so that the contact point's velocity component is $V+b\Omega$. The rod's central moment of inertia is $I=M{L}^2/12$.

The impulse delivered to the rod is $MV$. Conservation of linear and angular momentum givesUsing these in conservation of kinetic energy,This is the equal-mass instance of the effective contact mass calculation. For $u>0$, both final speeds are nonnegative, so equal speeds require $v=V=u/2$. Therefore $M{b}^2/I=2$ andBoth symmetric impact points lie on the rod, since $L/\sqrt{6}<L/2$.

More generally, the relative contact velocity immediately afterwards isThus its magnitude equals the incoming relative speed $u$ and its direction reverses. At the specified impact point, $V+b\Omega =3u/2$ and $v=u/2$, which makes the reversal particularly explicit.

During $dt$, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the expelled positive mass $-dM$ has velocity $v-u$ to first order. Subtract the initial momentum $Mv$ from their final momentum to obtain $Mdv+udM$. Equating this to the external impulse gives the rocket equationWith $F=-Mg-kMv$ and $\dot{M}/M=-b$, this reduces to $\dot{v}+kv=ub-g$. Therefore, for initial velocity $v_0$ and $k>0$,For a launch from rest,The inequality makes the initial upward acceleration positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

Take $+y$ upwards in the source diagram. Write $w$ for $B$'s incoming speed, so the initial velocities areA contact-direction assumption is needed for a unique answer. The diagram specifies the incoming velocities but does not specify the contact normal or an outgoing angle. First adopt the usual intended model: the impulse is horizontal, so neither particle changes its $y$ component of velocity. This is the smooth oblique collision model with a horizontal line of centres at impact. Bringing $A$ to rest in the $x$ direction then brings it entirely to rest.

Let $b$ be $B$'s outgoing $x$ component of velocity. Conservation of momentum and conservation of kinetic energy givePut $q=(\sqrt{3}/2)w$. The second equation becomes $3u^2+q^2=(q-3u{)}^2$, so $q=u$. ThereforeThe outgoing kinetic energies areThe outgoing speed of $B$ is $u\sqrt{13/3}$, directed above the negative $x$ axis at angle ${\tan }^{-1}[1/(2\sqrt{3})]\simeq 16.1^{\circ }$. The initial kinetic energy is also $(3/2+2/3)m{u}^2=13m{u}^2/6$, as required for an elastic collision.

For the perfectly inelastic collision with these same incoming velocities, both particles share a common outgoing velocity. Conservation of momentum givesIts direction is again $16.1^{\circ }$ above the negative $x$ axis. The separate translational kinetic energies of the constituents and their total areThe loss of translational kinetic energy isThis becomes internal energy, for example deformation and heating; total energy is conserved. In the elastic collision no kinetic energy is lost and the particles separate with different velocities. In the perfectly inelastic collision relative motion disappears and the common velocity is the original centre of mass velocity.

The limitation of the printed data can be made explicit. Without the horizontal-impulse assumption, write the final velocity of $A$ as $(0,h)$. Then conservation of momentum determinesand conservation of kinetic energy gives onlyThere is a family of possibilities. For a concrete second solution, $h=u/\sqrt{3}$ givesThese velocities conserve both components of momentum and conserve kinetic energy, and they satisfy the specified vanishing $x$ component for $A$, while requiring a different incoming speed. They can arise from a smooth contact normal proportional to $(1,1/\sqrt{3})$. Thus the horizontal-impulse result is conditional, rather than uniquely implied by the printed statement. For any member of this family the corresponding sticking result remains well determined once $w$ is chosen:The last expression is the lost relative kinetic energy in a perfectly inelastic collision of these two point masses.

Interpret $h$ as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Before impact, rolling without slipping gives $\omega =u/r$, so conservation of mechanical energy givesUse the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If $v$ and $V$ are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution giveDuring the subsequent simple pendulum swing, conservation of mechanical energy givesThe cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then $m+I/r^2=3m/2$. Writing $J$ for the impulse delivered to the bob gives $(3m/2)(v-u)=-J$, $2mV=J$ and $V-v=u$. Hence $V=6u/7$ and $\cos {\theta }_{\max }=25/49$, giving $59.32^{\circ }$. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

For an equal-speed elastic impact on a free rod, let $h$ be the signed distance of the impact point from the rod's center of mass, $I=m{l}^2/12$, and $\mathcal{J}$ the horizontal impulse delivered to the rod. For a smooth impact the impulse is perpendicular to the rod. If the ball's signed final velocity is $u$, the rod's center of mass velocity is $V$ and its angular velocity is $\omega$, momentum balance and angular impulse giveBoth objects have positive final translational velocities for a nonzero impact away from the centre. More explicitly, restitution in an elastic collision gives $V+\omega h-u=v_0$, and henceEqual final speeds thus require $u=V=v_0/2$. One can also see that the alternative $u=-V$ would violate $u+V=v_0$.

By conservation of kinetic energy in the elastic collision,Thus $I{\omega }^2=m{v}_0^2/2$. Since $I\omega =mVh=m{v}_{0}h/2$, elimination of $\omega$ givesEither side of the center of mass works; the nearer end is a distance $l(1/2-1/\sqrt{6})$ away. The final angular speed is $|\omega |=\sqrt{6}{v}_0/l$. These values also satisfy the restitution condition because the struck point's velocity is $V+\omega h=3v_0/2$, reversing the initial relative contact velocity.

The free rod has the same mass $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial center of mass at $y=0$ and the impact end at $y=l/2$. At impact the combined center of mass isConservation of momentum gives the combined center of mass velocity $V_C=v_0/2$. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $m{v}_{0}l/4$.

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertiaWith negligible external angular impulse, conservation of angular momentum therefore givesIts sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy isThus $m{v}_0^2/10$, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

Take upward velocity as positive, and let $w>0$ be the rearward rocket exhaust speed. During $dt$, the ejected parcel has mass $-dm>0$ and laboratory velocity $v-w$ to first order. The momentum change of the rocket and this parcel equals the gravitational impulse:Discarding the second-order product $dmdv$ gives the vertical rocket equationWith $\dot{m}=-\alpha$ and $m=m_0-\alpha t$, integration for free flight starting from rest at $t=0$ givesThe logarithmic term is the ideal Tsiolkovsky rocket equation contribution; $gt$ is the gravity loss. Immediate upward departure requires positive initial acceleration, orAt exact equality the initial acceleration is zero, but further burning reduces the weight and permits departure immediately thereafter. If departure is allowed to be delayed while fuel burns on the pad, the condition is instead $\alpha w>m(t)g$ at some time before burnout. With final rocket dry mass $m_f$, any powered departure before burnout requires $w>g{m}_f/\alpha$; the free-flight formula then starts at the actual departure time $t_{\ell }$ with $m(t_{\ell })$, giving $v=w\ln [m(t_{\ell })/m(t)]-g(t-t_{\ell })$.

For the specified data, the initial mass includes the payload. The rocket fuel fraction givesThe $100\mathrm{kg}$ payload is already part of the $2500\mathrm{kg}$ final mass; it is not the entire rocket dry mass and must not be added a second time. Here $m_{0}g/\alpha =196m/s$, well below the stated rocket exhaust speed, so departure is immediate. The burnout speed isThis is the speed at burnout. Subsequent motion under the assumed uniform gravity has $v(t)=v(T)-g(t-T)$.

Convolution describes the response of a linear time-invariant system to an arbitrary input. If $h$ is its impulse response and $f$ its input,This is a weighted sum of responses to shifted impulses. For integrable functions the operation is commutative and associative. A causal system has $h(t)=0$ for $t<0$, restricting the integral to past inputs. With the Fourier convention $F(\omega )=\int f(t)e^{-i\omega t}dt$, the convolution theorem gives $G=HF$.

For example, an RC low-pass frequency response has $h(t)={\tau }_0^{-1}{e}^{-t/{\tau }_0}\Theta (t)$ and $H(\omega )=(1+i\omega {\tau }_0{)}^{-1}$, where $\Theta$ is the Heaviside step function. Fast input variations are smoothed. The same mathematics describes optical blurring by a point spread function and many other linear response measurements.

Deconvolution tries to recover $f$ from measured $g$ and known $h$. In noise-free data one formally sets $F=G/H$ where $H\ne 0$. Zeros of $H$ lose corresponding information; small values amplify measurement noise. If the measured spectrum is $G+N$, the direct estimate has error $N/H$, which can be very large at frequencies strongly suppressed by the instrument. An inverse is therefore not automatically unique or stable.

One regularized spectral deconvolution reconstruction isThis minimizes $|HF-G{|}^2+\lambda |F{|}^2$ at each frequency and limits noise amplification, at the cost of bias and lost resolution. Choosing the regularization requires assumptions about noise and the desired signal; deconvolution cannot restore components that the data do not determine without such additional information.