Grade 12 Physics

Resolving a vector is the reverse of building one: the shadow it casts on the x-axis has length |v| cos θ. Simon Stevin demonstrated the underlying parallelogram rule in 1586 with an inclined-plane thought experiment — a closed loop of beads draped over a wedge that would have to move forever if forces did not combine this way — and Newton restated it as Corollary I of the Principia. Every ramp problem since is an application: a 200 N pull on a sled at 25° above the ground drives the sled forward with 200 × cos 25° ≈ 181 N, while the rest of the effort merely lifts.

The classic trap is measuring the angle from the wrong line. The cosine belongs to the component along the axis the angle is measured from; take the angle from the vertical instead and cosine and sine swap places. Check by sanity: at θ = 0° the whole vector lies on x, and cos 0° = 1 delivers exactly that.

The vertical partner of the x-component: |v| sin θ is how much of the arrow points up. A projectile launched at 40 m/s and 30° leaves the muzzle with 40 × sin 30° = 20 m/s of upward speed, and that number alone — not the 40 — decides how high and how long it flies. Together the pair (|v| cos θ, |v| sin θ) is the machinery that lets you replace one awkward diagonal with two independent one-dimensional problems, which is the reason vectors earn their place in physics at all.

Solving for the angle uses arcsin, whose principal branch runs only from −90° to 90°. A vector at 150° has the same y-component as one at 30°, so formula.expert returns the first-quadrant answer; if you know the vector leans left, its true direction is 180° minus what you get. When the whole vector is horizontal, sin θ = 0 and no y-component can tell you how long it is.

SOH — Sine is Opposite over Hypotenuse — is the first line of the most durable mnemonic in mathematics. In a right triangle, the sine of an acute angle is fixed by the triangle's shape alone: every right triangle with a 30° angle has an opposite side exactly half its hypotenuse, no matter the size. That constancy is what makes the ratio a tool. A field example: a 20 m guy wire anchored at 30° to the ground reaches a height of 20 × sin 30° = 10 m up the mast. The idea is ancient — Indian astronomers tabulated the jya (half-chord) around 500 CE, and a translation detour through Arabic and Latin gave us the word sine.

Solving for the angle uses the inverse function: θ = arcsin(o/h). The solver returns the principal branch only, which here is exactly right — the non-right angles of a right triangle are always acute, so the answer between 0° and 90° is the only valid one. The ratio o/h must stay below 1, since a leg can never outgrow the hypotenuse.

CAH — Cosine is Adjacent over Hypotenuse. Like all the trigonometric ratios, it works because the shape of a right triangle is fixed by one acute angle alone: every right triangle with a 60° angle has an adjacent leg exactly half its hypotenuse, whether it is drawn on a napkin or laid out across a field. Similar triangles keep the ratio constant, and that constancy is what turns an angle into a length. The name is a contraction of complementi sinus, the sine of the complement, because $\cos\theta = \sin(90° - \theta)$ — cosine is not a second idea but the same idea viewed from the other acute corner.

A worked instance you can check against a wall. A ladder is meant to stand at about 75° to the ground, the familiar one-out-for-four-up rule. A 6 m ladder therefore has its feet $6\cos 75° = 1.55$ m from the base of the wall, and reaches 5.8 m up it. Run the other way, $\theta = \arccos(a/h)$: a 6 m ladder set 2.0 m out is standing at $\arccos(0.333) = 70.5°$, flatter than it should be.

Three values are worth knowing cold: $\cos 0° = 1$, $\cos 60° = 0.5$, $\cos 90° = 0$. Beyond the triangle, cosine is the universal "how much of this points that way" operator. The component of a force along a direction is $F\cos\theta$; the useful part of an alternating current is the power factor $\cos\varphi$; the projection of any vector onto any axis is a cosine. On the unit circle it is simply the x-coordinate, and the identity $\sin^2\theta + \cos^2\theta = 1$ is the Pythagorean theorem on a triangle of hypotenuse 1.

Where it goes wrong. The most frequent error is picking the wrong leg: the adjacent side is the one touching the angle that is not the hypotenuse — and since the hypotenuse also touches the angle, that phrasing is exactly where people slip. The second is degree-versus-radian mode. This solver handles the conversion, but a phone calculator left in radians returns $\cos(35) = -0.903$, a negative number where a positive one belongs, which is at least loud enough to notice. The third is a domain limit that is really a geometry lesson: $a/h$ can never exceed 1, because a leg cannot outrun the hypotenuse, so an arccos that refuses to evaluate is telling you the two lengths do not form a right triangle. Finally, solving for the hypotenuse divides by $\cos\theta$, which collapses toward zero as the angle nears 90° — near-vertical geometry makes this rearrangement extremely sensitive to a small error in the angle.

Two vectors add head to tail, and the closing side of the parallelogram they form is the resultant. Its length is √(A² + B² + 2AB cos θ), where θ is the angle between the two vectors laid tail to tail. At 90° the cosine term vanishes and the familiar √(A² + B²) returns — 3 and 4 at right angles give exactly 5. At 0° the vectors reinforce completely and R = A + B; at 180° they fight and R = |A − B|. Simon Stevin argued the rule into existence in 1586 with his chain of beads over an inclined plane, and Newton made it Corollary I of the Principia in 1687.

Watch the sign. The law of cosines carries a minus, this formula a plus, and the difference is not a typo: the law of cosines uses the interior angle of the triangle, while θ here is the angle between the vectors, and the two are supplementary. Mixing them up is the single most common error in force-addition problems. Worked backwards: two forces of 5 N and 8 N producing a 7 N resultant must be separated by arccos((49 − 25 − 64)/80) = arccos(−0.5) = 120°.

A vector is an arrow: a direction plus a length. Drop it onto axes and it splits into two ordinary numbers, the components v_x and v_y, which form the legs of a right triangle whose hypotenuse is the arrow itself — so the magnitude is √(v_x² + v_y²). A hiker who walks 3 km east and 4 km north ends up 5 km from camp, not 7 km, because displacement is a vector and its parts add head-to-tail rather than arithmetically. That gap between 5 and 7 is the single most common student error in the whole subject.

Run it backwards and a magnitude plus one component recovers the other: a vector of length 13 with v_y = 5 must have v_x = √(169 − 25) = 12. The solver returns the positive root, since a component's sign is a matter of which way the axis points and the arithmetic cannot know it. Note also that no component may exceed the magnitude — a leg longer than the hypotenuse describes a triangle that does not exist.

Components tell you where a vector ends; the direction angle tells you which way it leans. Since tan θ = v_y/v_x, the naive answer is arctan(v_y/v_x) — and that answer is wrong half the time. The ratio for (−3, −4) is identical to the ratio for (3, 4), so a plain arctan reports a vector pointing southwest as though it pointed northeast. The fix is the two-argument function atan2(v_y, v_x), which keeps both signs and returns the true bearing anywhere in the full −180° to 180° sweep. It entered programming through early Fortran and is now in every standard library precisely because the quadrant bug was so pervasive.

Example: (−1, 1) points up and to the left, and atan2(1, −1) = 135° — correct, where arctan(1/−1) = −45° would have been off by exactly half a turn. Read the other way, a vector known to lie at 30° with v_x = 10 must have v_y = 10 tan 30° ≈ 5.774. Straight up or straight down the tangent blows up, which is the arithmetic's honest way of saying a vertical vector has no run to divide by.

Newton's second law says that the acceleration of an object is proportional to the net force on it and inversely proportional to its mass. Push twice as hard and it speeds up twice as fast; make it twice as heavy and it speeds up half as fast. The equation also defines the unit: one newton is exactly the force that accelerates one kilogram at one metre per second squared, so $1\ \text{N} = 1\ \text{kg·m/s}^2$. Newton himself did not write $F = ma$; he wrote that force is the rate of change of momentum, $F = \mathrm{d}p/\mathrm{d}t$, which is the more general statement and reduces to $ma$ whenever the mass is not changing.

A 1400 kg car reaching 100 km/h — that is 27.8 m/s — in 8.5 s has an average acceleration of $a = 27.8/8.5 = 3.27$ m/s², so the net force driving it forward is $F = 1400 \times 3.27 \approx 4580$ N. The engine has to supply more than that, because drag and rolling resistance are pulling the other way and the 4580 N is what is left over after they have taken their share.

Almost every other force relation on this site is this one wearing a hat. Weight is $F = ma$ with $a = g$. Centripetal force is $F = ma$ with $a = v^2/r$. Impulse is this law integrated over time, and the work–energy theorem is it integrated over distance. Its rotational twin, $\tau = I\alpha$, swaps torque for force and moment of inertia for mass and behaves identically.

The word doing the most work in the law is "net", and it is the word most often dropped. A crate being pushed with 200 N across a floor that resists with 150 N of friction accelerates as though 50 N were acting, not 200. If an object moves at constant speed the net force on it is zero, however many forces are actually pushing on it. The second mistake is arithmetically worse: kilograms are not a force. A "70 kg load" is a mass, and the force it exerts on its hanger is $70 \times 9.81 \approx 687$ N. Feeding 70 into this calculator as a force when you meant a mass is wrong by a factor of 9.81. The imperial world hides the same trap behind identical words — the pound-mass and the pound-force are different quantities related by 32.174 ft/s², which is where the notorious $g_c$ conversion factor comes from. Use the unit selectors and let them handle it.

Once a surface is already sliding, friction settles to a nearly constant value proportional to how hard the surfaces are pressed together: f = μₖN. Push a 200 N-loaded crate across a floor with μₖ = 0.3 and it resists with 60 N no matter how fast you shove it. The startling part — that friction depends on load but essentially not on contact area or speed — was established by Guillaume Amontons in 1699 and confirmed in Charles-Augustin de Coulomb's prize-winning 1785 study of rope, axles and rigging for the French navy, work so thorough that dry friction is still called Coulomb friction.

Typical coefficients: rubber on dry asphalt around 0.7, steel on steel about 0.6 lubricated down to 0.05, waxed ski on snow near 0.05, and PTFE on steel about 0.04. The trap is assuming N equals the weight — true only on level ground with no extra push or pull. On a slope N shrinks to mg cos θ, and a downward-angled push or a car's aerodynamic downforce raises it. Note also that kinetic friction always runs slightly below the static maximum, which is why a heavy box lurches forward the instant it breaks free.

Mass and weight are different quantities, and this formula is the exchange rate between them. Mass is how much matter an object contains, measured in kilograms, and it is the same everywhere in the universe. Weight is the gravitational force acting on that mass, measured in newtons, and it changes with where you are standing. The relation $W = mg$ is just Newton's second law with the acceleration set to whatever gravity supplies locally.

A 70 kg person weighs $70 \times 9.80665 \approx 687$ N on Earth. Take the same person to the Moon, where $g = 1.62$ m/s², and they weigh about 113 N — roughly a sixth — while still being made of exactly 70 kg of person. On Mars, $g = 3.72$, the answer is 260 N. Nothing about the body changed; only the field it sits in did.

The $g$ used here, 9.80665 m/s², is standard gravity, a defined constant fixed by the third General Conference on Weights and Measures in 1901 so that engineering calculations would have one agreed number. Real gravity varies: about 9.780 m/s² at the equator and 9.832 at the poles, a 0.5% spread caused by the Earth's rotation and its slightly flattened shape, with smaller local variations from altitude and the density of the rock underfoot. This whole formula is a near-surface shortcut for universal gravitation — put Earth's mass and radius into $g = GM/R^2$ and 9.8 is what comes out.

The everyday confusion is baked into the instruments. A bathroom scale measures force, using a spring or a load cell, then divides by a value of $g$ programmed at the factory and displays the result as kilograms. Take that scale to the Moon and it will confidently report that you have lost five-sixths of your mass. A balance, which compares your weight against known masses, would read correctly anywhere — which is the older and more honest instrument. The same confusion lives in the kilogram-force, a legacy unit equal to 9.80665 N, and in the imperial pound, which does double duty as a mass and a force. One more, because it comes up constantly: astronauts on the space station are not weightless because gravity has run out. At 400 km altitude $g$ is still about 8.7 m/s², nearly 89% of its surface value. They float because they and the station are both in free fall around the Earth together — falling, and missing.

Static friction is the only force in introductory mechanics that is written with an inequality: f ≤ μₛN. It supplies exactly whatever is needed to prevent sliding, up to a ceiling of μₛN, and this formula computes that ceiling. Press a 500 N normal load onto a surface with μₛ = 0.6 and you can push with anything up to 300 N and nothing moves; at 301 N the object breaks free and the weaker kinetic friction takes over.

That inequality is the single biggest trap: plugging μₛN in as "the friction force" on a stationary object is wrong unless the object is on the verge of slipping. The distinction pays real dividends — a car's tyres grip through static friction as long as they roll, which is why ABS pumps the brakes to keep them from locking into a lower-μₖ skid, and why a driven wheel spinning on ice suddenly has far less traction than one that is merely rolling.

Tilt a plank until the block on it just begins to slide: at that angle the down-slope pull mg sin θ exactly equals the friction ceiling μₛ mg cos θ, the mass cancels, and μₛ = tan θ. It is the cheapest friction experiment in existence — no force gauge, no scale, just a protractor. A block that lets go at 31° reports μₛ = tan 31° ≈ 0.60.

The same angle governs bulk solids: pour sand, grain, gravel or cement and the cone stabilises at its angle of repose, roughly 34° for dry sand and 40° for crushed stone. Silo designers, mining engineers and highway embankment crews all size their structures around it, and geologists read it back out of scree slopes. The trap is dimensional intuition — since the mass cancels, a heavy block does not slide at a gentler angle than a light one of the same material, which surprises nearly everyone the first time they see it.

On a slope, gravity still pulls straight down, but the surface can only push back perpendicular to itself — so it carries just the cos θ share of the weight: N = mg cos θ, with g = 9.80665 m/s². A 10 kg block on a 60° ramp presses in with only 10 × 9.80665 × cos 60° ≈ 49 N, half its 98 N weight. Level ground (θ = 0°) recovers N = mg, and a vertical wall (θ = 90°) gives N = 0, which is why nothing rests on a wall.

This is the quiet half of every incline problem, because friction is proportional to N: as the slope steepens, the down-slope pull grows while the friction budget shrinks, and the object eventually lets go. The classic error is using the full weight for N on a ramp, which overestimates friction and predicts that boxes stay put when they will actually slide.

Split the weight along the slope and you get mg sin θ — the force a rope, a brake or friction must hold to keep an object from sliding down. A 20 kg crate on a 30° ramp pulls down-slope with 20 × 9.80665 × sin 30° ≈ 98 N, exactly half its weight, because sin 30° = ½. Galileo built his entire kinematics programme on this: an incline "dilutes" gravity by sin θ, slowing a falling body enough to time it with a water clock in 1604, centuries before anything could time a free fall directly.

Highway grades use the same maths in disguise — a 6% grade means a rise of 6 m per 100 m, θ ≈ 3.43°, so a 40-tonne truck feels about 40,000 × 9.80665 × sin 3.43° ≈ 23 kN pushing it downhill, which is precisely why runaway-truck ramps exist. Pair this with N = mg cos θ and you have the complete free-body diagram for any slope; mixing up which one takes sine and which takes cosine is the standard exam trap, so check the limits: on level ground the down-slope force must vanish, and sin 0° = 0 does exactly that.

Strip friction away and the down-slope force mg sin θ divided by the mass m leaves a = g sin θ — the mass cancels completely, so a marble and a bowling ball slide identically. A 30° ramp yields 9.80665 × 0.5 ≈ 4.90 m/s², exactly half of free fall. This is Galileo's "diluted gravity": by 1604 he had shown that distances on an incline still grow as t², and he extrapolated to θ = 90°, where a = g and the ramp becomes free fall.

Watch the assumption — sliding, not rolling. A ball that rolls without slipping must also spin up its own moment of inertia, so a solid sphere manages only (5/7)g sin θ and a hoop just (1/2)g sin θ. That difference is the whole point of the classic race down a ramp, where a solid cylinder always beats a hollow one of identical mass and radius.

Gravity pulls the block down-slope with mg sin θ while friction drags back with μₖ mg cos θ; divide the difference by m and the mass drops out, leaving a = g(sin θ − μₖ cos θ). A 30° ramp with μₖ = 0.2 gives 9.80665 × (0.5 − 0.2 × 0.866) ≈ 3.20 m/s², a third slower than the frictionless 4.90 m/s². If the bracket comes out negative, the block was never sliding in the first place — the slope sits below the angle of repose, and the honest answer is a = 0.

Solving for θ uses the amplitude-phase identity sin θ − μ cos θ = √(1+μ²)·sin(θ − arctan μ), which is why the rearranged angle carries an arctan and an arcsin. The equation is the working model behind ski-slope grooming, luge run design and conveyor chute angles, all of which are chosen to land the acceleration in a narrow, controllable band.

Under constant acceleration, velocity changes at a steady rate, so the final speed is simply the starting speed plus the acceleration multiplied by the elapsed time. Picture a car merging onto a highway: entering the ramp at 15 m/s and holding a steady 2 m/s² for 5 seconds, it reaches v = 15 + (2)(5) = 25 m/s — right at highway pace. Deceleration works the same way with a negative a, which is how stopping times are estimated from braking data.

This is the first of the SUVAT equations, the toolkit of uniformly accelerated motion that traces back to Galileo's inclined-plane experiments in the early 1600s, where he showed that falling bodies gain equal speed in equal times. Because the relationship is linear in every variable, each of the four rearrangements has exactly one answer — no square roots, no ambiguity — making it the friendliest member of the kinematics family.

An object moving in a circle is accelerating even at constant speed, because its direction keeps changing — and sustaining that requires a net inward force of mv²/r. Swing a ball on a string and the string's tension supplies it; drive through a curve and tire friction does. A 1,200 kg car rounding a 50 m curve at 20 m/s needs $F_c$ = (1200)(20²)/50 = 9,600 N of sideways grip — nearly the car's own weight.

The v² is the part engineers respect: doubling speed quadruples the required force, which is why exit ramps post low advisory speeds and why racetracks bank their turns — banking tilts the road's normal force inward so friction isn't doing all the work. Solving for v takes the principal positive square root, since speed is a magnitude. And centripetal force is not a new kind of force; it's simply the name for whatever real force happens to point toward the centre.

An object going round a circle at perfectly constant speed is nevertheless accelerating, and that sentence is the first thing to make peace with. Acceleration is the rate of change of velocity, and velocity is a vector with a direction as well as a size. Something moving in a circle is having its direction changed continuously, so its velocity is changing continuously, so it is accelerating — even though a speedometer strapped to it would never move. The acceleration points at the centre of the circle, and its size is $a_c = v^2/r$.

A 1200 kg car rounding a 50 m curve at 20 m/s — 72 km/h — experiences $a_c = 20^2/50 = 8$ m/s², about 0.82 g. That is close to the limit of what a good tyre on dry pavement can supply, which is why that corner at that speed feels like it is asking a real question. Take the same corner at 30 m/s and the demand rises to 18 m/s², about 1.8 g, and no ordinary road tyre will hold it.

The derivation is short enough to be worth carrying. Over a small time $\Delta t$, the position vector sweeps through an angle $\Delta\theta = v\Delta t/r$. The velocity vector, always at right angles to the position vector, must rotate through exactly the same angle, and rotating a vector of length $v$ through a small angle changes it by $|\Delta v| = v\Delta\theta$. Put the two together: $|\Delta v| = v^2\Delta t/r$, so $a = v^2/r$. Christiaan Huygens published this result in 1673, and it is what let Newton check the inverse-square law against the Moon's orbit.

There is no outward force, and this is the single most persistent misconception in mechanics. In the ground frame nothing pushes you outward in a turning car. What happens is that your body would continue in a straight line, the car turns underneath you, and the door pushes you inward. The sensation of being flung out is your inertia, not a force. "Centrifugal force" is a bookkeeping term that appears only when you insist on doing the physics in the rotating frame, where it is added artificially so Newton's laws balance. It is a real effect and a useful device; it is not a force in an inertial frame, and there is no third-law partner to it. A more mundane error costs just as much: the $r$ is a radius, not a diameter. A component described as "600 mm diameter" has $r = 0.3$ m, and entering 0.6 halves the answer. Last, this is only the component of acceleration perpendicular to the motion. If the object is also speeding up or slowing down, there is a tangential component too, and the total acceleration is the vector sum of the two.

On a flat curve the only thing pointing toward the centre is tyre friction, so μₛmg must cover mv²/r. The mass cancels — a loaded truck and an empty hatchback with the same tyres can theoretically corner at the same speed — leaving $v_{\text{max}} = \sqrt{\mu_s g r}$. Dry asphalt with μₛ ≈ 0.8 on a 50 m radius allows √(0.8 × 9.80665 × 50) ≈ 19.8 m/s, about 71 km/h.

Now halve the grip: wet asphalt at μₛ ≈ 0.4 drops the limit to 14 m/s, and packed snow near 0.15 leaves only 8.6 m/s on that same curve. Because speed enters as a square root, grip losses are less brutal than they feel, but the flip side is that the required friction grows with v² — 20% more speed demands 44% more grip. The formula also assumes friction is doing nothing else, so a driver braking or accelerating mid-corner is spending part of the same friction budget, the "friction circle" that racing drivers train around.

Tilt a roadway and the surface's normal force gains an inward horizontal component; set the tilt so that component alone supplies mv²/r and the turn works even on ice, since mass cancels and tan θ = v²/rg. A 150 m curve designed for 25 m/s needs arctan(625 ⁄ (150 × 9.80665)) ≈ 23°. Railways call the same idea cant, and the maths dates to the 1830s when engineers first had trains fast enough to overturn.

Real tracks push it hard: Daytona banks 31°, Talladega 33°, and Bristol's short oval 28°, letting stock cars corner at speeds friction alone could never hold. The design is speed-specific — go slower than the design speed and the car tends to slide down the bank, go faster and friction has to make up the difference, which is why highway ramps combine a modest 4–8% superelevation with a posted advisory speed rather than relying on banking alone.

Momentum is mass in motion: multiply how much material is moving by how fast it moves. A 40,000 kg truck creeping along at 1 m/s carries 40,000 kg·m/s of momentum, while a 100 kg bicycle-and-rider at 10 m/s carries only 1,000 — which is why the slow truck is far harder to stop. Newton called momentum the "quantity of motion" and framed his second law around its rate of change, not around ma.

Momentum's real power is conservation: in any collision or explosion with no outside force, the total momentum before equals the total after. That single principle lets crash investigators reconstruct vehicle speeds from wreckage and lets rocket engineers plan burns — the exhaust thrown backward and the craft pushed forward always balance. In this one-dimensional form, direction simply rides along as the sign of v.

Impulse is force multiplied by how long that force acts, $J = F\,\Delta t$, and its whole significance is that it equals the change in momentum produced. That is not a separate law — it is Newton's second law in its original form. Newton wrote that force is the rate of change of momentum, $F = \Delta p/\Delta t$, and multiplying both sides by $\Delta t$ gives this. The units confirm it: a newton-second and a kilogram-metre per second are the same thing.

Here is the calculation that motivates every piece of passive safety equipment in a car. A 70 kg occupant travelling at 15 m/s must lose $p = 70 \times 15 = 1050$ kg·m/s of momentum in a frontal impact. That number is fixed by the crash; nothing can reduce it. If the body stops against a rigid dashboard in 0.1 s, the average force is $F = 1050/0.1 = 10\,500$ N. Stretch the same stop to 0.5 s using a crumple zone, a seatbelt that pays out under load, and an airbag, and the force falls to 2100 N. The impulse is identical in both cases. Only the time changed, and the time is the only variable an engineer gets to design.

The same trade-off is why you bend your knees on landing, why a boxer rolls with a punch, why gymnasts land on foam, and why a fall onto concrete is dangerous and the identical fall onto a mattress is not. In the other direction it is how rockets are specified: a model rocket motor's class is its total impulse in newton-seconds, because that is what determines the velocity change it can give a vehicle, regardless of whether it burns fiercely for a moment or gently for several seconds.

Impulse is a vector, and the sign flip on a bounce is the classic error. A 0.15 kg ball thrown at a wall at 20 m/s and rebounding at 20 m/s has not undergone a momentum change of zero, and not one of 3 kg·m/s either. Its momentum went from $+3$ to $-3$, a change of 6 kg·m/s — twice what stopping it dead would have required. A bouncing collision always demands more impulse than a catching one, which is why a bouncy object hits harder than a soft one of the same mass and speed. The second caution is that $F$ here is the average force over the contact. A real impact pulse is peaked, often two to three times the average at its maximum, so a component designed only against the average will be under-rated for the moment that actually breaks it.

With no external force acting, the total momentum of two colliding bodies before the impact equals the total after — always, whether they bounce, stick, or shatter. Kinetic energy is not so obliging; it survives only in a perfectly elastic collision. The principle was nailed down in 1668 when the Royal Society set the collision problem as a challenge and John Wallis, Christopher Wren and Christiaan Huygens independently sent in the answer.

Signs are everything in one dimension: pick a positive direction and stick with it, so a body moving the other way carries a negative velocity. A 2 kg cart at 5 m/s striking a stationary 3 kg cart and slowing to 1 m/s leaves the second cart at (10 + 0 − 2) ⁄ 3 ≈ 2.67 m/s. Because momentum is conserved in every collision, crash investigators use exactly this equation with skid-mark evidence to back out pre-impact speeds, and it works equally well for recoil: the rifle and the bullet start with zero total momentum and must end with zero.

When two bodies lock together on impact they share one final velocity, and momentum conservation hands it to you directly: the combined momentum divided by the combined mass. A 1000 kg car at 20 m/s rear-ending a stationary 1500 kg van and tangling with it leaves the wreck moving at 20000 ⁄ 2500 = 8 m/s. Kinetic energy, by contrast, is not conserved — here 200 kJ goes in and only 80 kJ comes out, the missing 120 kJ spent deforming metal, which is exactly what crumple zones are designed to do.

The classic application is the ballistic pendulum, devised by Benjamin Robins in 1742: fire a bullet into a hanging block, measure how high the block swings, and work backwards through this equation to get the muzzle velocity — the first accurate method of measuring how fast a bullet flies. Perfectly inelastic collisions dissipate the maximum energy any collision can while still conserving momentum, which is why "sticking together" is the worst case for occupant survival and the best case for a crash-test energy budget.

An elastic collision conserves both momentum and kinetic energy, and solving those two equations together gives this closed form for the first body's rebound. Three cases are worth memorising. Equal masses with the target at rest: v₁ = 0 and the bodies swap velocities exactly — the behaviour of a Newton's cradle and of a well-struck cue ball. A light body striking a much heavier one: v₁ ≈ −u₁, a near-perfect bounce back, as when a ball hits a wall. A heavy body striking a light one: v₁ ≈ u₁, it barely notices.

Worked example: a 1 kg ball at 4 m/s hits a stationary 3 kg ball, giving v₁ = ((1 − 3)(4) + 0) ⁄ 4 = −2 m/s — it rebounds at half speed while the heavier ball moves off at 2 m/s. Truly elastic collisions are an idealisation for everyday objects (steel bearings come close, billiard balls reach about 95%), but they are exact for gas molecules and for neutrons scattering in a reactor moderator — which is precisely why moderators use light nuclei like hydrogen or carbon, where each collision strips the most energy.

Newton's experimental law of restitution says the relative speed after a collision is a fixed fraction of the relative speed before: e = separation ⁄ approach. e = 1 is perfectly elastic, e = 0 is perfectly inelastic (the bodies move off together), and everything real lands in between. Newton reported measurements in the Principia itself, swinging pendulum balls together and recording roughly 5/9 for glass and 15/16 for tightly wound wool.

Sports bodies now legislate the number. A regulation basketball must return 1.2–1.4 m when dropped from 1.8 m onto hardwood, an e of about 0.85; a tennis ball tested per ITF rules comes in near 0.75; golf drivers are capped by a related "COR" limit of 0.83 to keep drives in the stadium. Two traps: e is not a property of one object but of the pair of surfaces, and it drops measurably at high impact speeds, which is why the test conditions are specified so precisely.

Drop height sets impact speed through v = √(2gh), the bounce keeps a fraction e of that speed, and the rebound climbs to a height that goes as the speed squared — so the two g's and the two ½'s cancel and you are left with the beautifully simple h₂ = e²h₁. A ball with e = 0.8 dropped from 2 m returns to 0.64 × 2 = 1.28 m, and the ratio of heights is a laboratory measurement anyone can make with a metre stick and a phone camera.

Because the loss compounds, successive bounces form a geometric sequence: 2 m, 1.28 m, 0.82 m, 0.52 m… and the total distance travelled converges even though the number of bounces is infinite — the classic bouncing-ball paradox, which also finishes in finite time. Watch the units in the ratio (any consistent pair works, since e is dimensionless) and remember that e depends on the surface too: the same ball is livelier on hardwood than on carpet.

In physics, work is force applied through a distance — and only the component of force along the motion counts, which is where the cos θ comes from. Pull a sled with 100 N on a rope angled 30° above the snow for 20 m, and you do W = 100 × 20 × cos 30° ≈ 1732 J, not the full 2000 J. The term itself was coined by the French engineer Gaspard-Gustave de Coriolis in 1826, precisely to compare what steam engines and horses could deliver.

Two useful edge cases: a force perpendicular to the motion (θ = 90°) does no work at all — the Moon's orbit costs gravity nothing — and a force opposing the motion, like friction, does negative work, showing up as cos θ below zero. Solving for θ uses the arccos principal branch, 0° to 180°, which conveniently covers the entire physical range of angles between two directions.

Kinetic energy is the energy an object carries because it is moving, $E_k = \tfrac{1}{2}mv^2$, and equivalently it is the work you would have to do to bring it from rest up to that speed — or the work it can do on something else in coming back to rest. The two odd-looking features, the half and the square, both fall out of that second statement. Push with force $F = ma$ through a distance and integrate: $\int ma\,\mathrm{d}x = \int mv\,\mathrm{d}v = \tfrac{1}{2}mv^2$. The square is not a modelling choice, it is what the integral hands back.

A 1500 kg car at 50 km/h — 13.9 m/s — carries $\tfrac{1}{2} \times 1500 \times 13.9^2 \approx 145$ kJ. The same car at 100 km/h, 27.8 m/s, carries about 580 kJ. Twice the speed, four times the energy, and since the brakes can only dissipate energy at roughly a fixed rate per metre of road, roughly four times the distance to stop.

Which form of energy is the "real" one was a genuine dispute. Descartes and his followers backed $mv$; Leibniz argued for what he called vis viva, $mv^2$. Willem 's Gravesande settled the experimental half of it in the 1720s by dropping brass balls into soft clay and finding that a ball arriving twice as fast sank about four times as deep, and Émilie du Châtelet made the theoretical case in the 1740s alongside her translation and commentary on the Principia. Both quantities turned out to matter — momentum $mv$ is conserved in every collision, kinetic energy only in elastic ones — which is why this site has pages for each.

Everything that goes wrong here goes wrong at the square. Doubling the speed does not double the energy, and the intuition that it does is what makes highway speeds feel deceptively similar to city ones. The unit trap follows directly: enter a speed in km/h where the formula wants m/s and you are wrong by $3.6^2 = 12.96$, not by 3.6 — the error is an order of magnitude and it looks plausible. Two smaller ones. Kinetic energy is a scalar with no direction, so two cars closing head-on do not have "negative" energy relative to each other, and you cannot cancel them the way you cancel momenta. And it is frame-dependent: a coffee cup on a train table has zero kinetic energy in your frame and a great deal in the frame of the platform. That is not a flaw; it is why the work–energy theorem only ever deals in changes.

Whatever the forces, the net work done on an object shows up entirely as a change in its kinetic energy: W = ½m(v² − v₀²). Accelerate a 2 kg mass from rest to 10 m/s and exactly 100 J went in, no matter whether it took 1 m of huge force or 100 m of gentle push. Gaspard-Gustave de Coriolis formalised both "work" and the ½mv² form of kinetic energy in 1829, precisely so factory owners could compare what different machines actually delivered.

The theorem's power is that it skips time entirely, making it the fastest route to braking distances: a 1360 kg car slowing from 26.8 m/s to 13.4 m/s sheds about 367 kJ, and dividing that by the braking force gives the stopping distance directly. Sign discipline is the trap — friction and braking do negative work, so W comes out negative whenever the object slows, and this calculator will happily return a negative number to tell you so.

Lifting a mass banks energy in the gravitational field, and near Earth's surface the deposit is simply mgh, with g = 9.80665 m/s². A roller coaster earns its entire ride on the first climb: a 500 kg car hauled 30 m up stores 500 × 9.80665 × 30 ≈ 147 kJ, which the drops and loops then spend as speed. Only differences in height matter — you are free to call the ground floor, the table top, or sea level "zero", as long as you stay consistent.

The same idea runs entire power grids: pumped-storage hydro plants push water uphill when electricity is cheap and let it fall through turbines at peak demand, storing gigawatt-hours as nothing more than elevated water. The formula is a near-surface approximation — it treats g as constant, excellent for heights small compared with Earth's radius.

Robert Hooke published this law in 1676 as a Latin anagram — ceiiinosssttuv — unscrambled two years later to "ut tensio, sic vis": as the stretch, so the force. An ideal spring pushes or pulls back in proportion to how far you displace it from rest. The spring constant k is the stiffness: a 200 N/m spring stretched 0.1 m pulls back with 20 N, while a car's suspension spring might run tens of thousands of N/m.

This calculator uses the magnitude form; strictly the restoring force points opposite the displacement, which is written F = −kx and is what makes released springs oscillate. The law holds only up to the elastic limit — stretch a spring too far and it deforms permanently. Within that limit it underpins spring scales, force gauges, vehicle suspensions, and even the atomic bonds that make solids springy.

Compressing or stretching a spring puts energy into it, and $U = \tfrac{1}{2}kx^2$ says how much. The half is not a fudge factor. By Hooke's law the force you must apply grows steadily from zero at the start of the stretch to $kx$ at the end, so the average force over the whole displacement is $\tfrac{1}{2}kx$, and work is average force times distance. Draw the Hooke's law line on a force-versus-displacement graph and the stored energy is the area beneath it — a triangle, and the area of a triangle carries a half.

A 400 N/m spring compressed 50 mm stores $U = 0.5 \times 400 \times 0.05^2 = 0.5$ J. Compress the same spring 100 mm and it stores 2 J, not 1 — the square again. That is the arithmetic behind a mousetrap, a valve spring, and the reason a bow drawn to full draw stores so much more than one drawn halfway.

The shape of this expression is worth recognising because it recurs everywhere. Kinetic energy is $\tfrac{1}{2}mv^2$; rotational kinetic energy is $\tfrac{1}{2}I\omega^2$; the energy in a capacitor is $\tfrac{1}{2}CV^2$ and in an inductor $\tfrac{1}{2}LI^2$. Every one of them is the integral of a quantity that grows linearly, and every one of them therefore comes out as a half times a coefficient times a square. Spot the pattern once and four formulas stop needing to be memorised separately.

The mistake that matters in real machinery is measuring $x$ from the wrong place. The $x$ in this formula is displacement from the spring's free length — its length when nothing is touching it — not from its installed length and not its total length. A spring installed with 20 mm of preload and then compressed a further 10 mm has not stored $\tfrac{1}{2}k(0.010)^2$. It has gone from 20 mm to 30 mm of deflection, so the energy added is $\tfrac{1}{2}k(0.030^2 - 0.020^2)$ — five times as much. Preloaded springs are everywhere in mechanisms, and this catches people every time. Two lesser traps: spring rates are quoted in N/m and in N/mm, and mixing them is a factor of a thousand in $k$; and the formula holds only inside the elastic limit, so a spring stretched until it takes a permanent set has absorbed energy this equation cannot account for, because some of it went into deforming the metal rather than into recoverable storage.

Power is the rate at which work is done, $P = W/t$. It answers a different question from work: not "how much energy did this take?" but "how fast was it delivered?" Two apprentices carrying identical toolboxes up the same stairs do exactly the same work against gravity, and the one who takes the stairs two at a time develops more power. One watt is one joule per second, which makes the watt a small unit — a person working steadily manages perhaps 75 W, and a fit cyclist holds around 250 W for an hour.

Take 20 kg of tools hauled 12 m up a ladder. The work is $mgh = 20 \times 9.80665 \times 12 \approx 2354$ J regardless of how it is done. Take 40 s over it and $P = 2354/40 \approx 59$ W. Rush it in 15 s and the same job demands 157 W. The energy bill is identical; only the rate has changed, and it is the rate that decides whether a motor is big enough.

James Watt coined horsepower in the 1780s as a sales tool. He needed to tell mill owners how many horses one of his engines would replace, measured a horse turning a mill wheel, and settled on 33 000 foot-pounds per minute — about 745.7 W. It was a marketing unit before it was an engineering one, and it has outlived the argument it was built to win. The same rate appears in two other forms on this site: $P = Fv$ when the work is a force moving something along, and $P = \tau\omega$ when it is a torque turning a shaft.

The commonest error is treating power as though it were energy. A kilowatt-hour is not a unit of power — it is a power multiplied by a time, so it is an energy, equal to 3.6 MJ. A 100 W bulb does not consume "100 watts per hour"; it consumes 100 watts, which over an hour amounts to 0.1 kWh. The phrase "watts per hour" is almost always a symptom that the two ideas have been mixed. Second, watch which horsepower a figure is quoted in: mechanical horsepower is 745.7 W, but metric horsepower — PS, cv, ch — is 735.5 W, and European engine ratings are usually the latter, a 1.4% difference that quietly walks into converted specifications. Third, $t$ must be the time over which the work was actually done, not the length of the shift; a hoist that lifts for 20 s and then sits idle for 10 minutes has a duty cycle, and averaging over the whole ten minutes describes the energy consumption honestly but badly understates the motor the job needs.

Divide both sides of $W = Fd$ by time and the $d/t$ turns into velocity, leaving $P = Fv$. It is the same statement as $P = W/t$, rewritten for the common case where a steady force is pushing something along at a steady speed — a car holding a cruise, a conveyor dragging material, a tug pulling a barge. The virtue of this form is that it needs no clock and no distance, only what is happening right now.

A car on the highway is fighting drag and rolling resistance. If those total 600 N at 30 m/s — about 108 km/h — the engine must deliver $P = 600 \times 30 = 18\ 000$ W, or 18 kW, roughly 24 hp, purely to keep the needle where it is. Nothing is accelerating and no height is being gained; that power is going straight into stirring air and warming tyres.

The relation has an unpleasant surprise buried in it for anyone chasing top speed. Aerodynamic drag rises with the square of speed, so the power needed to overcome it rises with the cube. Doubling highway speed takes roughly eight times the power, which is why an engine of twice the output buys only about a 26% higher top speed, and why fuel consumption climbs so steeply above about 90 km/h. The rotational version, $P = \tau\omega$, is the same equation on a shaft and is what a dyno chart is plotting.

The conceptual trap is expecting power to feel like force. At a fixed power the two trade off exactly: a truck in low gear applies enormous force at a crawl, and the same engine in top gear applies a small force at speed, with the identical power in both cases. That is the whole job of a gearbox. It also means the equation misbehaves at the ends — at $v$ near zero it would demand infinite force for any finite power, and what actually limits you there is traction and the clutch, not the engine. Two mechanical cautions as well. $F$ must be the component of force along the motion; for a force at an angle, take $F\cos\theta$ first, exactly as in the work formula. And this is instantaneous power unless both $F$ and $v$ hold steady — during acceleration both are changing, and the average power over the run is not $F_{\text{avg}}v_{\text{avg}}$.

Published in Newton's Principia in 1687, this law unified the heavens and the Earth: the same attraction that drops an apple holds the Moon in orbit. Every pair of masses pulls on each other with a force proportional to both masses and falling off with the square of the distance — double the separation and the pull drops to a quarter. The constant G = 6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻² is tiny, which is why gravity between everyday objects is unnoticeable; Henry Cavendish first measured it in 1798 with a delicate torsion balance, in effect "weighing the Earth".

A quick check: Earth's mass is 5.97 × 10²⁴ kg and its radius 6.37 × 10⁶ m, so the force on a 70 kg person works out to about 687 N — exactly the person's weight, as it must. Solving for r takes the principal positive square root, the only physically meaningful distance.

Divide Newton's law of gravitation by the test mass and what remains is the field itself: g = GM/r², the acceleration any object feels at distance r, regardless of what it is made of. At Earth's surface, g = 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ / (6.371 × 10⁶)² ≈ 9.82 m/s² — matching the measured free-fall value once Earth's spin is accounted for. Cavendish's 1798 torsion-balance measurement of G turned this equation around and, as headlines put it, "weighed the Earth."

The inverse-square fall-off is steep near a planet but gentle far away: on the Moon (M = 7.35 × 10²² kg, r = 1.737 × 10⁶ m) the same formula gives 1.63 m/s², a sixth of Earth's pull, which is why Apollo astronauts bounded rather than walked. And at the ISS's altitude, g is still about 8.7 m/s² — nearly nine-tenths of surface gravity. Astronauts float not because gravity is absent, but because they and their station are falling together.

A satellite stays in a circular orbit when gravity supplies exactly the centripetal force it needs — setting GMm/r² equal to mv²/r and cancelling the satellite's mass gives v = √(GM/r). The cancellation is the famous part: orbital speed depends only on the central body's mass and the orbit's radius, never on what is orbiting. The International Space Station, circling Earth (M ≈ 5.97 × 10²⁴ kg) at r ≈ 6.79 × 10⁶ m, moves at √(6.674 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.79 × 10⁶) ≈ 7 660 m/s — about 7.7 km/s, one lap of the planet every 92 minutes.

The inverse square root also explains a counter-intuitive fact: higher orbits are slower. GPS satellites at r ≈ 26 600 km amble along at 3.9 km/s, half the ISS's speed. Astronomers run the formula in reverse constantly — measure a moon's or a star's orbital speed and radius, and M = v²r/G weighs the central body, which is how we know the mass of everything from Jupiter to the black hole at the centre of the Milky Way.

Kepler noticed in 1619 that the square of a planet's year grows with the cube of its distance from the Sun; Newton later showed why — gravity's inverse-square pull makes T² = 4π²r³/(GM). A worked example: a geostationary satellite must orbit once per sidereal day, T = 86 164 s, so $r = (GMT^2/4\pi^2)^{1/3}$ = $(6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 86\,164^2 / 39.48)^{1/3}$ ≈ 4.22 × 10⁷ m — the 35 800 km altitude where every TV broadcast satellite parks.

The rearrangement for M is one of astronomy's sharpest tools: watch anything orbit, time it, measure the orbit's size, and the central mass falls out. The Moon's 27.3-day circuit at 3.84 × 10⁸ m weighs the Earth; Jupiter's moons weigh Jupiter; and the 16-year orbit of the star S2 around Sagittarius A* revealed a central mass of four million Suns packed into a region smaller than our solar system — a black hole.

This is nothing more than speed equals distance over time, applied to a circle. One complete trip round covers a circumference of $2\pi r$, and it takes one period $T$, so the speed is $v = 2\pi r/T$. Everything difficult about circular motion lives in the direction of the velocity, which is changing constantly; the magnitude is this piece of grade-school arithmetic and nothing more.

The International Space Station orbits about 410 km above the surface, and Earth's mean radius is 6371 km, so its orbital radius is $r \approx 6781$ km. It completes one orbit in roughly 93 minutes, which is 5580 s. That gives $v = 2\pi \times 6.781\times10^6 / 5580 \approx 7630$ m/s, or 7.6 km/s — about 27 500 km/h, and the reason its crew see sixteen sunrises a day.

Combine it with the two neighbouring pages and a lot falls out. Since $\omega = 2\pi/T$, this equation is exactly $v = \omega r$ with the period substituted in. Put it into the centripetal acceleration $a_c = v^2/r$ and you get $a_c = 4\pi^2 r/T^2$; set that equal to the gravitational acceleration $GM/r^2$ and rearrange, and $T^2 \propto r^3$ drops out — Kepler's third law, derived in three lines from a circumference and Newton's law of gravitation.

The radius is measured from the centre of rotation, and for orbits that means from the centre of the Earth, not from the ground. Using the ISS's 410 km altitude as $r$ instead of its 6781 km orbital radius understates the speed by a factor of about four and is the most common way this calculation goes wrong. The same principle applies on a smaller scale: for a point on a flywheel, $r$ runs from the shaft axis, and for a car on a banked track it runs to the centre of the curve, not to the inside edge of the road. Two more. $T$ is the time for one full revolution — a rotation rate given in rev/min has to be inverted first, $T = 60/N$ seconds, so 1800 rpm is a period of 0.0333 s, not 1800 of anything. And this describes uniform circular motion. A real planetary orbit is an ellipse on which the speed varies continuously, fastest at perihelion and slowest at aphelion; $2\pi r/T$ with a mean radius gives an average, not the speed at any particular moment.

Kepler's "harmonic law" of 1619 says T² ∝ a³ for everything circling the same central body, and the ratio form lets you compare two orbits without knowing G or the central mass at all. Using Earth as the reference body (T₂ = 1 yr, a₂ = 1 AU), Mars at a₁ = 1.524 AU must take T₁ = √(1.524³) ≈ 1.88 years to circle the Sun — precisely its observed year. Run the other way, an asteroid found with a 5.2-year period must orbit at $(5.2)^{2/3}$ ≈ 3.0 AU, in the heart of the asteroid belt.

The law works for any shared centre: Jupiter's moons obey it among themselves, as do Earth's satellites. A geostationary satellite (T = 23.93 h) and the Moon (T = 27.32 d ≈ 655.7 h) give a ratio $a_{\text{Moon}}/a_{\text{geo}} = (655.7/23.93)^{2/3} \approx 9.1$ — and indeed the Moon's 384 400 km orbit is about nine times the 42 164 km geostationary radius. When the ratio form fails, something unseen is tugging: discrepancies in Uranus's motion led astronomers straight to Neptune in 1846.

Escape velocity comes from an energy balance: launch with kinetic energy ½mv² at least equal to the gravitational well's depth GMm/r, and the projectile coasts to infinity with nothing to spare. The projectile's own mass cancels, and the direction of launch doesn't matter — only the speed. For Earth, v = √(2 × 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.371 × 10⁶) ≈ 11 190 m/s, the familiar 11.2 km/s that every Moon-bound Apollo mission had to approach.

Note the √2: escape speed is exactly √2 times circular orbital speed at the same radius. The formula also sorts the solar system's atmospheres — the Moon's gentle 2.4 km/s could not hold onto gas molecules, while Jupiter's crushing 59.5 km/s keeps even hydrogen. Push the logic to its limit by asking where escape velocity reaches the speed of light, and you arrive at the Schwarzschild radius of a black hole.

Gravitational potential energy is negative by convention: the zero is set at infinite separation, and every real pair of masses sits below it, in an energy "well" of depth GMm/r. A 1 000 kg satellite at the ISS's orbital radius, r = 6.79 × 10⁶ m, has U = −6.674 × 10⁻¹¹ × 5.97 × 10²⁴ × 1 000 / 6.79 × 10⁶ ≈ −5.87 × 10¹⁰ J — about 59 GJ that a launch vehicle must partly repay to lift it there from the surface, plus the kinetic energy of orbit.

The minus sign carries real physics. A bound orbit has total energy E = U/2 (the virial theorem's ½ shows up here: kinetic energy equals half the well depth), and escaping requires pumping E up to zero — which is exactly where the escape-velocity formula comes from. It also produces gravity's strangest habit: drag a satellite down and it speeds up, because falling deeper into the well converts potential energy into more kinetic energy than the drag removes.

In 1785 Charles-Augustin de Coulomb hung a charged sphere on a fine torsion wire and measured how hard a second charge twisted it. The result is the electric twin of Newtonian gravity: the force between two point charges grows with the product of the charges and falls off with the square of the distance. Electricity is by far the stronger force: the electric repulsion between two protons is about 10³⁶ times their gravitational attraction — which is why a rubbed balloon can lift paper against the pull of the entire Earth.

Worked example: two 1 µC charges held 10 cm apart feel F = kₑ × (10⁻⁶)² / (0.1)² ≈ 0.9 N — roughly the weight of an apple, from specks of charge. Enter charge magnitudes here; the sign of the product only tells you whether the pair attracts (opposite signs) or repels (like signs).

Current is not a thing that flows. It is a rate — the amount of charge passing a chosen cross-section of the conductor each second — and one ampere means one coulomb per second. Once that is clear, $Q = It$ needs no proof, because it is the definition read backwards: if charge crosses at a steady rate, the total that crossed is the rate multiplied by how long it kept up. The only condition the equation imposes is the word steady. A current that varies has to be integrated, $Q = \int I\,dt$, and this page is the special case where the integral collapses to a rectangle.

A 2 A charger running for one hour moves $2 \times 3600 = 7200\ \text{C}$. Battery ratings are the same arithmetic wearing different units: a phone cell marked 3000 mAh holds 3 Ah, and $3 \times 3600 = 10\,800\ \text{C}$ of deliverable charge. Divide by the elementary charge, $1.602 \times 10^{-19}$ C, and that is about $6.7 \times 10^{22}$ electrons — a number that only sounds absurd until you remember a gram of copper contains ten times as many free ones already sitting in the metal, drifting at well under a millimetre per second.

Since the 2019 redefinition of the SI, this relation is closer to the foundation than it used to be. The ampere is now fixed by declaring the elementary charge to be exactly $1.602176634 \times 10^{-19}$ C, which makes the coulomb a count of charges and the ampere a count per second. Michael Faraday got there experimentally in the 1830s: his laws of electrolysis measure the charge needed to plate out a mole of a substance, and that constant — 96 485 C per mole — is nothing but $Q = It$ run on a plating tank. Electroplating, anodising and battery capacity testing all still bill in ampere-hours for exactly this reason.

Two errors are worth naming, and one convention deserves an apology. The first error is treating milliamp-hours as energy. They are charge; a 3000 mAh cell at 3.7 V holds $3 \times 3.7 = 11.1\ \text{Wh}$, and the same 3000 mAh at 1.2 V holds a third of that, so comparing two batteries by mAh alone tells you very little. The second is applying the equation to a current that is not constant — a motor's inrush, a switching supply's chopped input, or anything on AC, where over a full cycle the net charge transferred is zero even though the current is real all along. As for the convention: current is drawn flowing from plus to minus, while in a metal the electrons actually travel the other way. Benjamin Franklin guessed the sign in the 1750s, a century before anyone knew a charge carrier existed, and he guessed wrong. Nothing in the physics breaks — a deficit of negatives moving left is indistinguishable from positives moving right — but it is a historical accident, not a discovery, and it is worth knowing that it is one.

Capacitance is charge-storing capacity: how many coulombs a capacitor soaks up for every volt you apply across it. The unit, the farad, honors Michael Faraday — and it is a giant. A full farad would have been a bench-filling curiosity for most of electronics history, so practical parts are marked in microfarads, nanofarads, and picofarads; only modern supercapacitors reach whole farads. Doubling the applied voltage doubles the stored charge, but C itself stays fixed — it depends only on geometry and the insulating material between the plates.

Worked example: a 100 µF capacitor charged to 12 V holds Q = CV = 100×10⁻⁶ × 12 = 1.2 mC of charge — about 7.5×10¹⁵ electrons parked on one plate and missing from the other. This defining relation applies to every capacitor, from the tuning capacitor in a radio to the DRAM cell storing one bit in your computer.

Why the half? Charging a capacitor is like stretching a spring: the first coulomb slides on easily, but every later coulomb must be pushed against the voltage the earlier ones built up. The voltage ramps linearly from 0 to V as charge accumulates, so the average push is V/2, and the total work is Q×V/2 = ½CV² — the triangular area under the Q–V line. Equivalent forms E = Q²/(2C) and E = QV/2 follow directly from C = Q/V.

Worked example: a camera-flash capacitor of 1000 µF charged to 300 V stores E = ½ × 0.001 × 300² = 45 J, dumped through the xenon tube in about a millisecond — a burst of tens of kilowatts from a pocket battery. The same math sizes defibrillators and grid-scale supercapacitor banks. Solving for V takes the positive square root, since the formula gives the voltage magnitude.

Magnetic fields are choosy: they push only on charges that move, and only on the component of motion that cuts across the field lines. The force is greatest when velocity and field are perpendicular (θ = 90°), and vanishes entirely for a charge coasting along the field. Because the push is always sideways — perpendicular to both v and B — it does no work; it bends paths into circles and spirals instead of speeding particles up. That steering is the working principle of particle accelerators, mass spectrometers, and the aurora, where solar particles spiral down Earth's field lines to the poles.

Worked example: a proton (q = 1.602×10⁻¹⁹ C) crossing a 0.5 T field at 10⁶ m/s and 90° feels F = 8×10⁻¹⁴ N — tiny, yet enough to whirl it in a tight circle. Note that θ itself is not solvable here: arcsin cannot tell θ from 180° − θ, so the inversion is ambiguous.

This is the force on one moving charge, $F = qvB\sin\theta$, added up over all the charges in a length of wire. A current $I$ means charge crossing at $I$ coulombs per second, so a length $L$ of conductor holds a quantity of moving charge whose product with its drift speed is exactly $IL$ — the individually feeble pushes on perhaps $10^{22}$ slowly drifting electrons, collected by the metal lattice and delivered to the wire as a whole. That is why $F = BIL\sin\theta$ contains no reference to how many carriers there are or how fast they move: those two factors always multiply out to the current. The sine handles orientation, peaking when the wire lies across the field and vanishing when it lies along it, since a charge coasting parallel to a field feels nothing.

A 0.25 m length of wire carrying 8 A across a 0.4 T field at right angles feels $F = 0.4 \times 8 \times 0.25 = 0.8\ \text{N}$ — about the weight of a coffee mug, from a single conductor. Multiply by a few hundred turns in an armature and you have the torque of a real motor. Tilt the same wire to 30° from the field and the force drops to $0.8 \sin 30° = 0.4\ \text{N}$, half. The direction is perpendicular to both the wire and the field, given by the right-hand rule, and this sideways push is what motor designers arrange to be a torque.

Faraday demonstrated the effect in 1821 with a wire free to rotate around a magnet dipped in mercury — the first electric motor, built to settle an argument about whether electromagnetism could produce continuous motion. Every motor and every loudspeaker since is the same experiment industrialised, the cone driven by a coil of wire hanging in a permanent magnet's gap with the audio signal as $I$. Note that this relation and the motional-EMF page are two faces of one thing: push current through a wire in a field and it moves, move a wire in a field and current appears, and a motor and a generator are the same machine run in opposite directions.

Three cautions. The angle $\theta$ is measured between the wire and the field, and it cannot be solved for on this page — arcsine cannot tell $\theta$ from its supplement $180° - \theta$, so the calculator returns $F$, $B$, $I$ or $L$ but never the angle. $L$ is the length of conductor actually inside the field, not the length of the wire; a metre of lead-in outside the magnet gap contributes nothing. And a note on the right-hand rule: it works with conventional current, drawn flowing from positive to negative, while the electrons in the copper are travelling the other way. Benjamin Franklin fixed that sign a century before the electron was found, and he fixed it backwards. Both descriptions give the same force in the same direction — negative charge moving left is the same current as positive charge moving right — but if you switch to reasoning about electrons you must switch hands too, and mixing the two is the surest way to get a motor turning the wrong way on paper.

This equation is two earlier ones stacked. A long straight wire produces a field circling it at $B = \mu_0 I_1/(2\pi d)$ at distance $d$; a second wire sitting in that field feels $F = B I_2 \ell$. Substitute and you get $F = \mu_0 I_1 I_2 \ell/(2\pi d)$. Each wire sits in the other's field, and by Newton's third law they push on each other equally and oppositely. The direction is the part people find surprising: currents flowing the same way attract, currents flowing opposite ways repel — the reverse of the intuition borrowed from electrostatics, where like charges repel. Note the $1/d$ rather than $1/d^2$: a straight wire's field falls off with the first power of distance because the source is a line rather than a point.

Two wires 10 mm apart, each carrying 10 A, over a 1 m parallel run feel $F = (1.257 \times 10^{-6} \times 10 \times 10 \times 1)/(2\pi \times 0.01) = 2.0\ \text{mN}$ — the weight of a grain of rice, which is why nobody notices it in ordinary wiring. Now put a 20 kA fault through the same pair: the currents appear as a product, so the force scales with the square, and 2 mN becomes 8 kN per metre. That is nearly a tonne of force trying to tear a metre of busbar out of its supports, and it is the reason switchgear bracing is engineered rather than assumed.

From 1948 until 2019 this relation did not merely describe the ampere, it defined it: the ampere was the current which, in two infinitely long parallel conductors one metre apart in vacuum, produced a force of exactly $2 \times 10^{-7}$ newtons per metre. That definition is what made $\mu_0$ exactly $4\pi \times 10^{-7}$ — a defined constant rather than a measured one. The 2019 redefinition of the SI moved the anchor to a fixed value of the elementary charge, and one consequence is that $\mu_0$ is now an experimentally determined quantity with an uncertainty, very slightly different from $4\pi \times 10^{-7}$. This page uses the CODATA value.

The traps are dimensional and geometric. The published version of this law is usually the force per unit length; this page multiplies by $\ell$ to give a total force, so do not apply a per-metre figure and then multiply by the length again. $d$ is the centre-to-centre separation, not the gap between insulation surfaces, and on closely spaced busbars the difference is not small. The result assumes long, straight, parallel conductors — near a bend, a termination or a right-angle crossing the geometry changes and the simple form does not hold. And because the currents enter as a product, an alternating current gives a force that is always attractive or always repulsive but pulses at twice the supply frequency, never reversing: that 120 Hz throb on a 60 Hz system is precisely what makes transformers and reactors hum, and what fatigues busbar supports over years rather than breaking them in an instant.

A single loop of wire makes a field that is strong at its centre and sprawls untidily everywhere else. Wind many loops into a tight helix and something better happens: inside the coil every turn's field points the same way and they add, while outside they point in opposing directions and largely cancel. What is left is a nearly uniform field along the axis, and Ampère's law applied to a rectangular path straddling the wall of a long solenoid gives it as $B = \mu_0 N I / L$. The striking thing about that result is what is missing — the diameter of the coil does not appear. Only the current and the turns per unit length matter, so a broomstick-sized coil and a pencil-sized one with the same winding density and current produce the same interior field.

A 200 mm coil wound with 1000 turns and carrying 5 A gives $B = (1.257 \times 10^{-6} \times 1000 \times 5)/0.2 = 31\ \text{mT}$, roughly six hundred times Earth's field of about 50 µT. Halving the current to 2.5 A halves the field; stretching the same 1000 turns over 400 mm also halves it, because the turns are now half as dense. That second sensitivity is the one people forget, and it is why the equation is better read as $B = \mu_0 n I$ with $n$ the turns per metre.

André-Marie Ampère coined the word solénoïde in the 1820s from the Greek for a channel or pipe, and the coil is still the standard way to make a field you can switch. Relays, contactors, solenoid valves, MRI bores and every particle-physics magnet are variations of it. Wrapping the coil around soft iron multiplies the result by the material's relative permeability — a factor of several thousand for good silicon steel — which is how a modest coil can lift a car, and the equation then reads $B = \mu_r \mu_0 n I$. Superconducting magnets take the other route and simply run enormous current, since with zero resistance there is no $I^2R$ heat to remove.

The formula is an idealisation for a long coil, and it fails where the coil ends. Field lines have to turn around and come back, so at the mouth of a solenoid the axial field falls to about half its interior value, and outside it is weaker still and spread out. Treating a short, fat coil — anything much wider than it is long — with this equation will overstate the field substantially; that geometry needs the exact axial expression or a numerical model. The second failure is saturation. The iron-core multiplication is not a licence to keep raising current: ordinary steels saturate somewhere around 1.5 to 2 T, after which $\mu_r$ collapses toward 1 and every further ampere buys only the modest air-core contribution. Third, watch the length variable: $L$ is the length of the winding, not the length of the wire, and confusing them can be a factor of a thousand. And a coil is an inductor: switching that 31 mT off in a millisecond will produce a voltage spike large enough to arc a contact, which is why a solenoid valve gets a flyback diode.