Newton's First Law — Inertia and Equilibrium

Spec mapping: OCR H556 Module 3.2 — Forces in action (Newton's three laws of motion; the concept of inertia; conditions for translational equilibrium). Refer to the official OCR H556 specification document for exact wording.

Newton's laws of motion are the foundation of classical mechanics. They explain why objects move the way they do, turning the descriptive language of kinematics (displacement, velocity, acceleration) into a predictive theory of dynamics. This lesson opens the OCR A-Level treatment of Newton's laws with the First Law — the deceptively simple statement that objects keep doing what they are already doing unless something forces them to change.

The First Law is not really a "law" in the modern sense; it is a definition of an inertial frame of reference and a statement that inertia is a real property of matter. Once you understand it properly, the Second and Third Laws (and the whole of mechanics) snap into place. This treatment goes well beyond the GCSE one-liner, anchoring the law in Galilean relativity, free-body analysis and the geometry of equilibrium polygons.

Statement of Newton's First Law

A body remains at rest, or continues to move with constant velocity in a straight line, unless acted upon by a resultant (net) external force.

Three key ideas are packed into this single sentence:

"At rest or constant velocity" — these are physically the same situation. Zero velocity is just one special case of constant velocity, and there is no mechanical experiment that can distinguish between them (the Galilean principle of relativity).
"Constant velocity" means constant speed and direction. A body moving in a circle at constant speed is not in uniform motion in the Newton-1 sense, because its direction is changing — and so its velocity vector is changing, and so a non-zero resultant force must be acting. Circular motion is not a counterexample to Newton 1; it is a direct application of Newton 2 (covered in lesson 2).
"Resultant force" — forces that cancel out do not count. What matters is the vector sum of all forces acting on the body. A skydiver at terminal velocity has two enormous forces on it — weight $mg$ down and drag $F_D$ up — but the resultant is zero, and Newton 1 says the velocity is constant.

The law was already in essence stated by Galileo in his 1638 Discorsi, where he argued (paraphrased) that a ball rolling on a perfectly horizontal frictionless surface would continue indefinitely; Newton crystallised it and made it the first of his three laws because it defines what an inertial frame is.

Inertia — The Property Behind the Law

Inertia is the tendency of an object to resist changes to its state of motion. More massive objects have more inertia and are harder to start, stop, or turn. Mass itself is often defined as a quantitative measure of inertia — specifically, the inertial mass $m$ that appears in $F = ma$ (lesson 2).

In everyday life inertia is obvious:

A heavy shopping trolley is harder to push from rest than an empty one — and harder to stop once moving.
A juggernaut lorry takes far longer to brake than a small car at the same speed: same braking force from the tyres, but ten times the inertia, so $a = F/m$ is ten times smaller.
A cricket ball flying into a net keeps moving until the net exerts a stopping force on it.
A passenger in a braking bus is thrown forward — not because of a "phantom" forward force, but because the passenger's body tries to continue at constant velocity while the bus decelerates beneath them.

Before Newton, the prevailing Aristotelian view was that bodies naturally came to rest and that a continuous force was needed to keep them moving. The Aristotelian intuition is wrong but not stupid — in the everyday world, friction and air resistance do slow moving objects. It took the conceptual leap of separating the body from the forces on it to see that, with friction removed, motion would persist indefinitely. A coasting spacecraft in deep interplanetary space drifts in a straight line at constant speed for billions of years — Newton 1 in its purest form.

Note the difference between inertial mass (the $m$ in $F = ma$ , which quantifies resistance to acceleration) and gravitational mass (the $m$ in $W = mg$ , which quantifies how strongly gravity pulls). Experimentally these two are equal to better than one part in $10^{13}$ — a fact known as the weak equivalence principle, and the cornerstone of Einstein's General Relativity. A-Level treats them as identical and calls both "mass".

Resultant Force and Equilibrium

A body is in translational equilibrium when the resultant force acting on it is zero:

$\sum \vec{F} = \vec{0}$

By Newton 1, in equilibrium the body is either stationary (static equilibrium) or moving with constant velocity (dynamic equilibrium). No mechanical experiment carried out inside the body can distinguish between the two — this is the Galilean principle of relativity, and it survives intact in Einstein's special relativity (where "inside the body" is replaced by "in the body's instantaneous rest frame").

To test for equilibrium in two dimensions, resolve all forces into perpendicular components — conventionally horizontal ( $x$ ) and vertical ( $y$ ) — and demand that each component sum be zero:

$\sum F_x = 0 \qquad \sum F_y = 0$

If both component sums vanish, the body is in equilibrium. A geometric equivalent: if you draw the forces tip-to-tail as a vector polygon, the polygon closes — the tip of the last arrow lands exactly on the tail of the first. Equilibrium polygons are a powerful sanity check for three-force problems (see worked example 1).

Worked Example 1 — Lamp Hanging from Two Wires

A lamp of weight $W = 25$ N hangs from two wires, each making an angle of $30°$ with the vertical. Find the tension $T$ in each wire.

By the lamp's left-right symmetry, both tensions are equal in magnitude. Resolving vertically (upwards positive):

$2T \cos 30° - W = 0$ $T = \frac{W}{2 \cos 30°} = \frac{25}{2 \times 0.8660} = 14.4 \text{ N}$

Resolving horizontally, the two horizontal components $T \sin 30°$ point in opposite directions and cancel automatically. The lamp is in equilibrium — it hangs at rest — in perfect agreement with Newton 1.

Sanity check by polygon: draw the three forces tip-to-tail. The weight $W$ points down (length 25 mm at 1 mm/N); each tension points along its wire away from the lamp (length $14.4$ mm). The three arrows form a closed isoceles triangle — confirmation that $\sum \vec{F} = 0$ .

What if the wires made angles of $60°$ with the vertical? Then $T = 25/(2 \cos 60°) = 25/1.0 = 25$ N — the tension equals the full weight, because each wire now contributes only half its strength to the vertical component. As the wires approach horizontal ( $90°$ ), $\cos \theta \to 0$ and $T \to \infty$ . This is why a tightrope walker's wire is anchored very securely: small sag means large tension.

Worked Example 2 — Three-Force Equilibrium

A picture of weight $W = 12$ N hangs from a single nail by two strings attached to the top corners. The left string is horizontal (so it must run over a pulley out of frame); the right string makes an angle of $40°$ above the horizontal. Find both tensions.

Let $T_L$ be the left (horizontal) tension and $T_R$ the right tension at $40°$ above horizontal. Resolving:

Vertical: $T_R \sin 40° - W = 0 \Rightarrow T_R = W/\sin 40° = 12/0.6428 = 18.7$ N.
Horizontal: $T_R \cos 40° - T_L = 0 \Rightarrow T_L = T_R \cos 40° = 18.7 \times 0.7660 = 14.3$ N.

Both tensions positive — both strings genuinely pulling the picture, no sign reversal needed. The vector polygon would be a right-angled triangle: weight down ( $12$ ), left tension left ( $14.3$ ), and right tension up-and-to-the-right ( $18.7$ at $40°$ ) closing the figure.

Diagram — The Newton-1 Decision Tree

flowchart TD
  A[Body of interest] --> B{Identify ALL forces on body}
  B --> C[Vector-sum the forces]
  C --> D{Is resultant zero?}
  D -- "Yes (equilibrium)" --> E[At rest OR constant velocity]
  D -- "No" --> F[Velocity changing — apply Newton 2]
  E --> G[Newton 1 satisfied: no acceleration]
  F --> H[a = F_net / m, direction of F_net]

Free-Body Diagrams

To apply Newton 1 you almost always need a free-body diagram (FBD): a sketch showing only the forces acting on the body of interest, not the forces it exerts on anything else. Each force is drawn as an arrow starting at the body, labelled with its origin and (where calculable) its magnitude.

Typical forces in A-Level questions:

Symbol	Name	Origin
$W$	Weight	Earth's gravity; $W = mg$ , always vertically downward
$N$ or $R$	Normal contact force	Perpendicular to the surface of contact
$T$	Tension	Along a rope, cable or wire, away from the body
$F_f$	Friction	Parallel to surface, opposing relative (or tendency to relative) motion
$F_D$	Drag / air resistance	Opposes motion through a fluid
$U$	Upthrust	Buoyancy, upward from a fluid

A clean FBD lists only forces, never accelerations, and never forces the body exerts on something else (those belong on other free-body diagrams). The acid test: every force on an FBD is one for which you could, in principle, name a Newton-3 partner — and that partner is not on the same diagram.

Worked Example 3 — Book on a Table

A $2.0$ kg textbook sits at rest on a flat horizontal table. Find the normal contact force between the book and the table.

Weight: $W = mg = 2.0 \times 9.81 = 19.6$ N downwards.
The only other force on the book is $N$ from the table, directed upwards (the table cannot pull the book downwards — contact forces are always compressive in this context).
The book is at rest, so $\sum F = 0$ , hence $N - W = 0$ .
$N = 19.6$ N upwards.

This is an application of Newton 1: because the book is in equilibrium, we know immediately that $N = W$ . Students often confuse this with Newton 3 (covered in lesson 3) — but $W$ and $N$ act on the same body (the book) and are therefore Newton-1 balanced forces, not a Newton-3 action-reaction pair. The Newton-3 partner of $W$ is the gravitational pull of the book on the Earth, and the Newton-3 partner of $N$ is the book's downward push on the table. Both partners act on different bodies.

Worked Example 4 — Lift at Constant Velocity

A $70$ kg person stands in a lift. The lift is moving upwards at a constant $2.0$ m s $^{-1}$ . Find the normal contact force from the floor of the lift on the person.

Constant velocity ⇒ equilibrium ⇒ $\sum F = 0$ .

$W = mg = 70 \times 9.81 = 686.7$ N downwards.
$N - W = 0 \Rightarrow N = 687$ N (3 s.f.) upwards.

Many students wrongly guess that the lift's motion makes a difference — perhaps that the normal force is somehow "bigger because the lift is going up". It does not, and it isn't. Constant velocity is mechanically indistinguishable from rest. Only if the lift accelerates does the normal force differ from the weight, and that is a Newton-2 problem (covered in the next lesson).

Worked Example 5 — The Skydiver at Terminal Velocity

A $80$ kg skydiver falls vertically through still air. After some time the speed reaches a constant value $v_T$ (terminal velocity). What can be said about the forces, and what does Newton 1 tell us?

At terminal velocity, the speed is constant and the direction is vertical (constant). So velocity is constant, $\sum F = 0$ , and:

$F_{\text{drag}}(v_T) = W = mg = 80 \times 9.81 = 785 \text{ N}$

i.e. the drag force has grown to exactly cancel the weight. The skydiver is not floating, and not "weightless" — the weight is still $785$ N. But the resultant is zero, so Newton 1 governs: constant velocity (constant in both magnitude and direction). For a spread-eagle freefall position, $v_T \approx 55$ m s $^{-1}$ (about $200$ km/h); for a head-down streamlined position, $v_T$ can exceed $90$ m s $^{-1}$ .

This worked example is critical because it illustrates the most important conceptual point about Newton 1: balanced forces and zero forces look identical to the body's motion. A passenger on the ISS feels weightless not because $g = 0$ at orbital altitude (it's still around $8.7$ m s $^{-2}$ , only ~12% less than at the surface) but because they and the ISS are in mutual free fall — there is no normal force from any surface. The weight is still acting, but unbalanced.

Inertial and Non-Inertial Frames

An inertial frame is one in which Newton 1 holds — i.e. an isolated body (zero resultant force) moves in a straight line at constant speed. A non-inertial frame is one that is accelerating relative to an inertial frame. Examples: a braking bus (linearly accelerating), a rotating roundabout (centripetally accelerating), a free-falling lift (gravitationally accelerating, but in a special way we'll discuss).

Inside a non-inertial frame, objects appear to accelerate even with no real force acting on them. To save Newton's laws, physicists introduce fictitious (or pseudo, or inertial) forces — the centrifugal force on a roundabout rider, the "forward push" felt by a passenger when their bus brakes. These forces are real experientially (you feel them, they make you stumble) but they have no Newton-3 partner — no body exerts them on you. They are bookkeeping artefacts of working in an accelerating frame.

For A-Level OCR you will always work in an inertial frame, typically the ground frame (laboratory frame). Earth's surface is not a perfect inertial frame — it is rotating, so it is accelerating centripetally toward Earth's axis at up to $\omega^2 R \approx 0.034$ m s $^{-2}$ at the equator. This is small compared with $g = 9.81$ m s $^{-2}$ , but it is measurable: it deflects Foucault pendulums, causes hurricanes to rotate (the Coriolis pseudo-force), and means $g_{\text{measured}}$ at the equator is slightly less than at the poles.

A profound consequence of the equivalence principle: a freely falling observer in a gravitational field is also in an inertial frame, locally. This insight became the foundation of General Relativity. For A-Level you can rest in the simpler Newtonian picture, but the equivalence principle is a wonderful Oxbridge interview prompt (covered in "Going further" below).

Centripetal Motion — Is This a Newton 1 Violation?

A common A-Level worry: "If a planet orbits the Sun at constant speed in a circle, isn't it moving at constant velocity? Doesn't Newton 1 say no force is needed?" No. Constant speed in a circle is not constant velocity, because velocity is a vector and its direction is constantly changing. The planet is continuously accelerating toward the Sun (centripetal acceleration $a_c = v^2/r$ ), and Newton 1 demands a resultant force in that direction — supplied by gravity. If gravity vanished, the planet would fly off on a straight tangent line (Newton 1), not continue orbiting.

So Newton 1 is fully consistent with circular motion: it just demands that we recognise the centripetal force (gravity, here) as the resultant force responsible for the change in velocity direction. This is covered in detail in Module 5.4.

Synoptic Links

Newton's second law (Module 3.2, next lesson) — Newton 1 is the special case of Newton 2 with $\sum F = 0$ (giving $a = 0$ , hence constant velocity). The two laws are not independent; the first establishes the frame, the second quantifies the dynamics.
Circular motion (Module 5.4) — orbital and centripetal problems are pure Newton 1+2: the centripetal force is not an extra force, it is the resultant of real forces (gravity, tension, normal contact) pointing inward, which produces the inward acceleration $v^2/r$ .
Equilibrium of rigid bodies (Module 3.2, moments lesson) — Newton 1 ( $\sum \vec{F} = 0$ ) is the translational equilibrium condition; rigid bodies also require rotational equilibrium ( $\sum \tau = 0$ about any point). Both are needed for full static analysis.
Special relativity (post-A-Level) — Newton 1 survives intact as the statement that, in any inertial frame, an object with zero resultant 4-force moves on a straight worldline. Galilean relativity is the low-speed limit of special relativity.

Specimen question modelled on the OCR H556 paper format

Question (9 marks): A car of mass $1200$ kg is travelling along a long, straight, horizontal road. The driver has set the cruise control so that the car travels at a constant velocity of $28$ m s $^{-1}$ . The total resistive force on the car (rolling resistance plus air drag) at this speed is $720$ N.

(a) State Newton's first law of motion, and explain why the car can travel at constant velocity even though resistive forces act on it. [3]

(b) Calculate the magnitude of the forward driving force exerted by the engine on the car at this constant velocity. State Newton's law you have used and give a brief justification. [3]

(c) After 30 s of constant-velocity cruising, the driver lifts their foot off the accelerator. The forward driving force then drops to zero. Without doing a full calculation, describe and explain (i) what happens to the car's velocity in the following few seconds and (ii) what eventually happens to the resistive force as the car slows. [3]

AO breakdown

Mark	AO	Awarded for
1	AO1	Statement of Newton 1 in acceptable form
2	AO1	"Resultant force = 0" reasoning for constant velocity
3	AO2	"Engine force balances resistive force" — equilibrium argument
4	AO1	Identification of Newton 1 as the relevant law (or Newton 2 with $a = 0$ )
5	AO2	Force-balance equation $F_{\text{drive}} = F_{\text{resist}}$
6	AO2	$F_{\text{drive}} = 720$ N stated
7	AO1	Car decelerates (velocity decreases) — Newton 1 / 2 with net backward force
8	AO2	Resistive force depends on speed (drag $\propto v^2$ for air resistance)
9	AO2	As $v$ falls, drag falls; deceleration decreases; car eventually slows to a stop (or asymptotically)

AO split: AO1 = 4, AO2 = 5, AO3 = 0.

Mid-band response (4/9):

(a) Newton's first law says a body keeps moving at constant velocity unless a force acts on it. The car has constant velocity because the forces are balanced.

(b) The engine force is $720$ N because it has to match the friction.

Examiner commentary: The next-band move is greater precision in the wording of Newton 1 (explicitly "resultant force" rather than "a force"), and an explicit equilibrium equation $F_{\text{drive}} - F_{\text{resist}} = 0$ in part (b) rather than just asserting the answer. Mark 1 awarded for an acceptable but loose Newton-1 statement; Mark 2 not awarded because the candidate did not name "resultant force"; Mark 3 awarded for the balance idea. In (b), Marks 5 and 6 awarded for the correct numerical answer but Mark 4 lost because the candidate did not name Newton 1 (or Newton 2 with $a = 0$ ) as the law applied. In (c), Mark 7 awarded for "slow down"; Marks 8 and 9 lost — no discussion of the velocity-dependence of drag, which is a synoptic discriminator at the mid-to-strong band boundary.

Stronger response (7/9):

(a) Newton's first law: a body remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force. The car has zero resultant force on it — the forward driving force from the engine exactly balances the $720$ N resistive force — so the car keeps moving at constant velocity.

(b) By Newton's first law (or equivalently Newton's second law with $a = 0$ ), if velocity is constant the resultant force is zero. So $F_{\text{drive}} - F_{\text{resist}} = 0$ , giving $F_{\text{drive}} = F_{\text{resist}} = 720$ N forwards.

(c) Without the engine force, the only horizontal force is the $720$ N resistive force backwards. The resultant force is now non-zero, pointing backwards, so by Newton 1 the velocity cannot be constant — the car decelerates. As the car slows, the air-resistance contribution decreases (drag falls with speed), so the resistive force itself shrinks; the deceleration therefore reduces with time. The car eventually slows toward a stop, dominated by rolling resistance once aerodynamic drag becomes negligible.

Examiner commentary: To lift this to top-band the candidate would need to be explicit about the functional form of drag — typically $F_D \propto v^2$ for vehicle aerodynamics — and to note that, with engine off, the deceleration is non-uniform (so SUVAT does not apply directly). Marks 1-7 awarded. Mark 8 partial — the candidate notes that drag decreases with speed but does not state the relationship. Mark 9 partial — qualitatively correct but the candidate does not distinguish between the asymptotic behaviour of pure air drag (approaching but never reaching zero) and the finite stopping time once rolling resistance dominates. A confident strong-band answer with clear synoptic awareness.

Top-band response (9/9):

(a) Newton's first law: an object will remain at rest, or continue to move with constant velocity in a straight line, unless acted upon by a resultant (net) external force. For the car: even though resistive forces act, the engine simultaneously provides a forward driving force; the resultant of the two is zero, so Newton 1 permits — indeed requires — that the velocity be constant. Forces "acting on" a body is not the same as "resultant force on" a body; only the resultant matters.

(b) The car is in dynamic equilibrium (constant velocity ⇒ $\sum \vec{F} = 0$ ). Applying Newton 1 to the car along its direction of motion:

$F_{\text{drive}} - F_{\text{resist}} = 0 \quad \Rightarrow \quad F_{\text{drive}} = 720 \text{ N (forward)}$

Newton's first law (equivalently Newton's second law with $a = 0$ , since for constant mass and zero resultant force $F = ma = 0$ ) gives the equilibrium condition.

(c) (i) With $F_{\text{drive}} = 0$ , the only horizontal force on the car is the $720$ N resistance acting backwards. The resultant force is now $-720$ N (taking forward as positive). By Newton 1, the velocity cannot remain constant: the car decelerates. Initial deceleration is $|a| = F/m = 720/1200 = 0.60$ m s $^{-2}$ .

(ii) The resistive force has two components: rolling resistance (roughly constant) and aerodynamic drag (roughly $\propto v^2$ for vehicle aerodynamics at highway speeds). As $v$ decreases, the drag component shrinks rapidly; the total resistive force therefore decreases, so the deceleration decreases with time. The car's velocity decays in a non-uniform way: initially at $0.60$ m s $^{-2}$ , asymptotically tending toward a slower decay dominated by the (constant) rolling resistance. The car eventually comes to a halt — not in a uniform-deceleration SUVAT scenario but along a curve that is concave-up on a $v$ -vs- $t$ graph.

Examiner commentary: Full marks. The discriminator moves: (i) the explicit "forces acting on the body" vs "resultant force on the body" distinction in part (a) — this is the precise wording that mid-band answers miss; (ii) the framing of part (b) as "dynamic equilibrium" with Newton 1 stated as the law applied; (iii) in part (c), the splitting of resistance into constant rolling and $v^2$ drag terms, with explicit non-uniform-deceleration consequence. The candidate also distinguishes the kinematic regimes correctly: not a SUVAT problem, but a graphical/numerical one. Top-band answers don't just compute — they reason about which laws apply and why.

Common errors and mark-loss patterns

Pedagogical observations from teaching A-Level Newton 1, with no fabricated examiner-report percentages.

Stating Newton 1 imprecisely. "An object stays still until a force acts" misses both the constant-velocity case and the "resultant force" qualifier. OCR mark schemes will penalise both omissions; the canonical wording is "at rest or in uniform motion in a straight line, unless acted upon by a resultant external force".
Confusing "no force" with "zero resultant force". A body in equilibrium may have many forces acting on it (the skydiver at terminal velocity has weight and drag); what matters is that they cancel.
Calling balanced forces a Newton-3 pair. $W$ and $N$ on a book on a table are Newton-1 balanced forces on the same body, not a Newton-3 pair (covered in lesson 3).
Assuming "constant velocity" means "rest". A car cruising at $28$ m s $^{-1}$ on cruise control is in equilibrium just as much as a car parked on the drive.
Forgetting that a change of direction is a change of velocity. Circular motion at constant speed is not Newton-1 motion — it requires a centripetal resultant force.
Mixing mass with weight. Mass (kg) is the inertia coefficient; weight (N) is the gravitational force on the mass. "The weight of a 5 kg book is 5 kg" is wrong twice over.
Skipping the free-body diagram. OCR mark schemes routinely award a mark for the diagram itself. Skipping it both costs that mark and increases the risk of sign or omission errors elsewhere.
Resolving in only one direction in 2-D problems. For 2-D equilibrium you must check both $\sum F_x = 0$ and $\sum F_y = 0$ . Many candidates resolve vertically, find the right vertical answer, and forget the horizontal check.

Going further

Newton's Principia Mathematica (1687) introduced his three laws not as physical axioms but as mathematical propositions from which he then derived, with extraordinary geometric ingenuity, the planetary motions that Kepler had described kinematically a century earlier. The mathematical move from "kinematics tells you the orbit" (Kepler) to "dynamics tells you why" (Newton) is one of the great syntheses in the history of science. Newton's first law was already implicit in Galileo's earlier work on inertia, but Newton elevated it to the first of his laws because it defines the frame in which the others apply.

At undergraduate level, Newton 1 receives a sophisticated reformulation in Lagrangian and Hamiltonian mechanics. Lagrangian mechanics derives the equations of motion from an integral principle — the principle of stationary action, $\delta S = 0$ where $S = \int L\,dt$ and $L = T - V$ — and Newton's first law emerges as the free-particle limit ( $V = 0$ ): in flat space the action-minimising path is a straight line at constant speed. Hamiltonian mechanics generalises this to phase space (position and momentum as canonical variables), leading naturally to quantum mechanics.

The deepest consequence of Newton 1 is the equivalence principle. Because gravitational mass equals inertial mass (verified experimentally to $\sim 10^{-13}$ ), a freely falling observer cannot — by any local mechanical experiment — distinguish their situation from that of an observer floating freely in deep space. Einstein recognised in 1907 ("the happiest thought of my life") that this means a uniform gravitational field is locally equivalent to an accelerating reference frame, and that "gravity" can be reinterpreted as the geometry of curved spacetime. This insight became the foundation of General Relativity (1915). For all everyday purposes Newton 1 remains exact; relativistic corrections matter only for very strong fields (black holes, neutron stars) or very precise measurements (GPS satellite clocks need ~ $38\,\mu$ s/day correction or position fixes would drift by kilometres).

Oxbridge interview prompts that probe Newton 1:

"Two astronauts float in deep space, far from any massive bodies. One holds a tennis ball; she releases it gently. What happens to the ball?" The ball remains where it was released, moving at the same constant velocity as the astronaut who released it (their common reference frame). Newton 1 in its purest form. A good follow-up: "now the astronaut accelerates by firing a thruster. What does the ball do (in her frame)?" In her now-accelerating frame, the ball appears to accelerate backwards — but no force acts on it; the apparent acceleration is a pseudo-force.
"Why does the Earth keep orbiting the Sun without any 'forward push'?" Newton 1 (and 2): the Sun's gravity provides a centripetal (sideways-to-motion) force, which changes the direction of Earth's velocity but not its speed. No forward force is needed; inertia carries Earth forward, gravity steers it sideways. This is exactly the conceptual revolution that lay behind Newton's Principia.
"In an isolated space capsule with the windows blacked out, can you tell whether you are at rest, moving at constant velocity, or in free fall in a gravitational field?" Galilean relativity says no to the first two; the equivalence principle says no to the third. All three are locally indistinguishable inertial frames.

Recommended reading: Six Easy Pieces by Richard Feynman (chapters 1-4 for conceptual clarity); Classical Mechanics by John Taylor; for the historical arc, Newton: A Very Short Introduction (Iliffe).

A-Level misconceptions

"No force ⇒ constant velocity" is sloppy. The correct statement is "zero resultant force ⇒ constant velocity". A body can have many forces acting on it and still be in equilibrium if they vector-sum to zero.
"Constant velocity = at rest" is wrong. They are mechanically indistinguishable (Galilean relativity), but constant velocity at $28$ m s $^{-1}$ is not the same physical state as $v = 0$ — it differs by a Galilean boost.
"Newton 1 fails in circular motion at constant speed" is wrong. Newton 1 is fully consistent with circular motion; it requires a centripetal resultant force, supplied by gravity / tension / normal contact / whatever is doing the steering. Constant speed and constant velocity are not the same.
"A moving body must have a forward force on it" is the Aristotelian instinct, and it is wrong. A puck gliding across frictionless ice needs no forward push because nothing is decelerating it.
"Inertia is the same as momentum" is loose. Inertia is the property of a body (its mass / resistance to acceleration); momentum is the quantity $\vec{p} = m\vec{v}$ , which can be zero even for a high-inertia body if it happens to be at rest.
"Apparent weight = weight" can be wrong. Apparent weight is the normal contact force a supporting surface exerts on you; it equals the true weight only in equilibrium. In a free-falling lift it is zero; in an accelerating lift it is $m(g + a)$ .
"Newton 1 is just an obvious special case of Newton 2" is half-right. Mathematically yes (Newton 1 is Newton 2 with $a = 0$ ); but conceptually Newton 1 plays the unique role of defining the inertial frame in which Newton 2 is then applied. Without Newton 1 you don't have a frame to start writing $F = ma$ in.

Summary

Newton 1: an object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force.
The law defines inertial frames (frames in which it holds) and is the foundation of Galilean (and special) relativity.
Inertia is the tendency of mass to resist changes in motion; inertial mass equals gravitational mass to extreme precision (equivalence principle).
Translational equilibrium: $\sum \vec{F} = 0$ , i.e. $\sum F_x = 0$ and $\sum F_y = 0$ . Equivalent geometric test: the force polygon closes.
A free-body diagram is the standard starting point: only forces on the body of interest are drawn.
Circular motion at constant speed is not a violation of Newton 1: changing direction is a change of velocity, and a centripetal force is responsible.
Constant velocity and rest are mechanically indistinguishable; "apparent weight" and weight differ when there is a non-zero acceleration of the supporting surface.

Reference: OCR A-Level Physics A (H556) specification 3.2 — Forces in action (refer to the official OCR H556 specification document for exact wording).

Newton's First Law — Inertia and Equilibrium

Newton's First Law — Inertia and Equilibrium

Statement of Newton's First Law

Inertia — The Property Behind the Law

Resultant Force and Equilibrium

Worked Example 1 — Lamp Hanging from Two Wires

Worked Example 2 — Three-Force Equilibrium

Diagram — The Newton-1 Decision Tree

Free-Body Diagrams

Worked Example 3 — Book on a Table

Worked Example 4 — Lift at Constant Velocity

Worked Example 5 — The Skydiver at Terminal Velocity

Inertial and Non-Inertial Frames

Centripetal Motion — Is This a Newton 1 Violation?

Synoptic Links

Specimen question modelled on the OCR H556 paper format

AO breakdown

Common errors and mark-loss patterns

Going further

A-Level misconceptions

Summary

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