Newton's laws of motion

Newton's laws of motion consist of inertia in an inertial frame, the relation between net force and change of momentum, and equal-and-opposite pair forces. The following three children state the individual laws.

Testing requires independently constrained force laws, mass assignments, initial conditions and a common family of inertial frames. Defining the net force separately for every observation by $\mathbf{F}=m\mathbf{a}$ cannot test Newton's second law. With interaction parameters fixed, the same equation predicts new trajectories. Likewise, zero net force implies uniform motion within the second-law framework, but the existence and empirical identification of its inertial frames are extra content. Pairwise Newton's third law adds a constraint not supplied by the individual second-law equations.

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

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 actual teaching experiment in Arbor Scientific's The Level Air Track uses one glider, an air track, two photogates, and a ruler. Turn on the blower and place the glider near the middle. Adjust the feet until it does not drift when released. Put the photogates at least half a track length apart and align them with the glider's flag. Measure the length $L$ that interrupts the beams. Give a gentle push, let go, and record each beam-blocking time $\Delta t_i$. The readout is the pair of speeds $s_i=L/\Delta t_i$. The worksheet asks how those speeds compare, which forces remain after the push, and which interaction the air track minimizes.^[ref]

The motivating contrast is between a brief push and the subsequent glide. The air cushion reduces contact friction; it does not remove the downward gravitational weight or the upward support force. Similar successive speeds after release are consistent with a small horizontal net external force, rather than with a continuing push. The experiment approximates the zero net external force model over a finite track and observation interval.

To strengthen the documented procedure, use independent measurement calibration: check the flag length and each timer against a known length and time reference; measuring $L$ sets the length scale but does not calibrate the timer. Repeat launches in both directions without contact during the timed passage, and check for drift at several track locations. These are additional controls, not reported steps or measured results in the worksheet. A residual inclined plane produces a downhill component of gravitational weight with the same laboratory direction on both runs; friction and drag force oppose the direction of travel. Reversing the launch can therefore help distinguish these effects.

A photogate directly measures a beam-blocking time; $L/\Delta t_i$ is an average speed during that passage, not an exact instantaneous velocity. Assign the direction separately. Equal readings at two gates do not prove zero acceleration everywhere between them: opposing changes could cancel. Measurement uncertainty from timer resolution, an incorrectly measured flag length, residual slope, airflow disturbances, and remaining friction or drag force limit the inference. No measured glider speeds are assumed here.

Constant velocity depends on the reference frame. For nonrotating axes and a moving origin of laboratory position $\mathbf{R}(t)$, subtracting that origin givesIf $\mathbf{R}(t)={\mathbf{R}}_0+\mathbf{U}t$ with constant velocity $\mathbf{U}$, the Galilean transformation leaves acceleration unchanged. The cart moving at $1.5m/s$ in the laboratory is at rest in a frame moving right at $1.5m/s$. Both descriptions obey Newton's first law.

Instead choose an origin with $R_x(t)=\frac{1}{2}A{t}^2$, where the stipulated constant $A=0.50m/{\mathrm{s}}^2$ points right. The same cart hasAt $t=3\mathrm{s}$, the cart is instantaneously at rest relative to this accelerating origin, but its relative acceleration is nonzero. In this translating accelerated reference frame, Newton's second law must include the effective inertial force $-mA=-1.0\mathrm{N}$. This term accounts for the accelerating coordinates; the actual external forces on the cart still sum to zero. No rotation-dependent terms are present because the axes remain parallel to the laboratory axes.

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

An idealized front at distance $x>0$ down a uniform slope entrains stationary material so that its moving mass is proportional to $x$. Its linear momentum balance is $d(x\dot{x})/dt=Ax$, where $A=g\sin \alpha >0$. Multiplying by $x\dot{x}$ gives $x^2{\dot{x}}^2=2A{x}^3/3+c$. Every forward-moving branch tends to $\dot{x}=\sqrt{2Ax/3}$ as $x\to \infty$, and $\ddot{x}=A/3-c/x^3\to A/3$. Acceleration is smaller than the constant-mass downslope value because some momentum accelerates newly entrained material.

In the pair-interaction model of Newtonian mechanics, two bodies exert equal and opposite forces on each other:Both vectors are expressed in the same inertial frame at the same time. The subscripts name the agent and the recipient: $A$ acts on $B$, and $B$ acts on $A$. These are forces on different bodies, so they cannot cancel in the net force on either body. Newton's second law still uses all the forces on the particular body whose acceleration is sought. Identifying this third-law interaction pair does not require the bodies to be isolated: additional external forces can act at the same time.

The distinction makes Newton's third law useful both for drawing a free-body diagram and for deriving momentum conservation. A body-by-body calculation retains one member of each third-law interaction pair; a calculation for a collection containing both bodies cancels that pair in the total momentum balance. Equal interaction forces consequently coexist with nonzero, generally unequal accelerations.

Isaac Newton stated the equality and opposition of mutual actions in Law III of his 1687 Philosophiae Naturalis Principia Mathematica. The explanation distinguishes equal opposite changes of momentum from changes of velocity, which are inversely proportional to the bodies' masses when other influences are absent. Corollary III then explains why mutual interactions preserve the signed total momentum. The equations here are modern notation for that distinction. ^[ref]

The instantaneous body-pair equation has a scope: use it for the mechanical interactions assumed by this Newtonian mechanics model. In electromagnetism, bodies can exchange momentum with an electromagnetic field; their instantaneous mutual forces need not sum to zero. Momentum conservation then accounts for electromagnetic field momentum as well as matter, and for momentum crossing the chosen boundary. This complication occurs in classical electromagnetism, without invoking quantum mechanics.^[ref]

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

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.

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.

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.