Must-know sheet

Physics 1 must-know sheet

The real AP Physics 1 exam gives you an equation sheet with constants and the main equations, and you can use a calculator for the whole exam. This sheet covers what that sheet doesn't tell you: when each relationship applies, units, sign conventions, graph rules, problem-solving checklists and how to analyze lab data.

Units, conversions and significant figures

Units 1, 2, 3, 4, 5, 6, 7, 8

SI units first: meters, kilograms, seconds: Convert everything to m, kg and s before you plug in. Grams to kilograms means dividing by 1000, and centimeters to meters means dividing by 100.
N = kg·m/s², J = N·m, W = J/s, Pa = N/m², Hz = 1/s: The newton, joule, watt, pascal and hertz are built from m, kg and s. Knowing this lets you check that an answer's units make sense, for example that ½mv² comes out in kg·m²/s², which is a joule.
Prefixes: kilo (k) = 10³, centi (c) = 10⁻², milli (m) = 10⁻³, micro (μ) = 10⁻⁶, mega (M) = 10⁶: So 4.0 km = 4.0 × 10³ m, 25 cm = 0.25 m, 30 g = 0.030 kg and 5 mm = 5 × 10⁻³ m.
Speed: divide km/h by 3.6 to get m/s: 1 m/s = 3.6 km/h, and 1 m/s is about 2.24 mi/h. Highway speeds of around 100 km/h are about 28 m/s.
Area and volume: square or cube the length factor: 1 cm² = 10⁻⁴ m² and 1 cm³ = 10⁻⁶ m³, because 1 cm = 10⁻² m. A liter is 1000 cm³ = 10⁻³ m³.
Angles: 1 revolution = 360° = 2π rad: Rotational equations such as s = rθ, v = rω and W = τΔθ only work with angles in radians. To turn revolutions per minute into rad/s, multiply by 2π and divide by 60.
Rotational units: Angle in rad, angular velocity in rad/s, angular acceleration in rad/s², torque in N·m (never written as joules), rotational inertia in kg·m² and angular momentum in kg·m²/s.
Other units to keep straight: Spring constant k is in N/m, momentum and impulse are in kg·m/s = N·s, density is in kg/m³, volume flow rate is in m³/s, and G = 6.67 × 10⁻¹¹ N·m²/kg².
g: use 10 m/s² unless told otherwise: The course says exam questions use g = 10 m/s² whenever a number is needed, and you won't lose credit for using 9.8 or 9.81 m/s² correctly (the exam's sheet lists 9.8). g points down and can also be written as N/kg, the gravitational field strength.
Handy reference values: Water has a density of 1000 kg/m³ (1.0 g/cm³), and atmospheric pressure is about 1.0 × 10⁵ Pa (101 kPa). Problems usually give these, but you should recognize them.
Significant figures: Keep extra digits while you calculate and round only at the end, to about as many significant figures as the given data (usually 2 or 3). Every numerical answer needs a unit.
Check units and extreme cases: Before you trust an answer, check that its units match the quantity you want. Then test an extreme case: if θ = 0 or a mass goes to zero, does the expression behave the way the physics says it should?

Vectors, components and signs

Units 1, 2, 3, 4, 5

Choose a positive direction and write it down: In one dimension, a vector's sign shows its direction. If you call up positive, then the free-fall acceleration is −g (−10 m/s² on the exam) and a downward velocity is negative.
Speeding up or slowing down?: An object speeds up when its velocity and acceleration have the same sign and slows down when they have opposite signs. A negative acceleration does not always mean slowing down.
Components: A_x = A cos θ, A_y = A sin θ: This holds only when θ is measured from the x-axis; if θ is measured from the y-axis, cosine and sine swap. Rebuild a vector with magnitude √(A_x² + A_y²) and angle tan θ = A_y/A_x.
Adding vectors: Add the x-components and the y-components separately. You can add magnitudes directly only when the vectors point the same way; for perpendicular vectors, use the Pythagorean theorem.
Weight on an incline: mg sin θ along, mg cos θ into the surface: Tilt your axes so x runs along the slope. The part of the weight pulling down the slope is mg sin θ and the part pressing into the surface is mg cos θ, which you can check: at θ = 0 nothing pulls along a flat surface.
Sign of work: Work is positive when a force has a component along the displacement, negative when the component points opposite (like kinetic friction on a sliding block), and zero when the force is perpendicular (like the normal force on a level floor or the inward force in uniform circular motion).
Sign of potential energy: ΔU_g = mgΔy is positive when an object rises. U = −Gm₁m₂/r, with zero chosen at infinite separation, is always negative and climbs toward zero as the objects move apart, while spring energy ½kx² is never negative, whether stretched or compressed.
Rotation signs: clockwise or counterclockwise: Pick one sense as positive and give every torque, angular velocity, angular acceleration and angular momentum a sign. AP Physics 1 never asks for the 3D direction of these vectors.
Signs in a bounce: Δp = m(v_f − v_i) with signs included. A ball hitting a wall at 5 m/s and bouncing straight back at 5 m/s has |Δp| = m(10 m/s), twice as much as if it had stopped.

Kinematics: choosing the right equation

Unit 1

Constant-acceleration equations: v = v₀ + at, x = x₀ + v₀t + ½at² and v² = v₀² + 2a(x − x₀) work only when acceleration is constant. List what you know (Δx, v₀, v, a, t) and pick the equation that leaves out the quantity you neither know nor need.
Average velocity: Δx/Δt always, (v₀ + v)/2 only sometimes: Average velocity is always displacement divided by time. Averaging the starting and ending velocities gives the right answer only when the acceleration is constant.
Changing acceleration: use graphs, not the equations: When acceleration changes, the constant-acceleration equations fail. AP Physics 1 handles this case with graphs and words: slopes, areas and shapes.
Distance vs. displacement, speed vs. velocity: Distance is the total path length (a scalar); displacement is final minus initial position (a vector). After a round trip, displacement and average velocity are zero but distance and average speed are not.
Free fall: With only gravity acting, the acceleration is g downward the whole time, including at the top where the velocity is momentarily zero. Without air resistance mass doesn't matter, the trip up takes as long as the trip down to the same height, and the speed is the same at the same height.
Projectile setup: two one-dimensional problems linked by time: Horizontally, a_x = 0, so v_x = v₀ cos θ stays constant and x = v_x t. Vertically, a_y = −g and the starting vertical velocity is v₀ sin θ. At the highest point v_y = 0, but v_x is unchanged.
Horizontal launch from height h: The initial vertical velocity is zero, so the fall time is t = √(2h/g) no matter how fast it was thrown sideways. An object dropped from rest and one launched horizontally from the same height land at the same time.
Level-ground projectile shortcuts: Only if it lands at its launch height: time of flight 2v₀ sin θ/g, maximum height (v₀ sin θ)²/(2g), range v₀² sin 2θ/g. Range is largest at 45°, and complementary angles such as 30° and 60° give equal ranges.
Relative velocity in one dimension: Velocity of A relative to the ground = velocity of A relative to B + velocity of B relative to the ground, using signs. Two cars moving the same way close in at the difference of their speeds; moving toward each other, at the sum.
Inertial reference frames: Observers moving at constant velocity relative to each other measure different velocities but the same acceleration, so Newton's laws work in every inertial frame. Assume frames are inertial unless a problem says otherwise.

Reading graphs: slopes, areas and shapes

Units 1, 2, 3, 4, 5, 6

Position vs. time: Slope = velocity. A straight line means constant velocity, a flat line means at rest, a curve bending upward means positive acceleration, and a peak or valley means the object is momentarily stopped and turning around.
Velocity vs. time: Slope = acceleration, and the area between the graph and the time axis = displacement, with area below the axis counting as negative. Where the graph crosses the axis, the object turns around.
Acceleration vs. time: Area = change in velocity. The graph can't tell you the starting velocity, so you need that from somewhere else.
Matching graph shapes: Constant acceleration shows up as a flat a–t line, a sloped straight v–t line and a parabola on the x–t graph. Constant velocity is a zero a–t line, a flat v–t line and a straight sloped x–t line.
Force vs. position: Area = work done by that force. For a spring, F = kx is a straight line through the origin, so the area is the triangle ½kx², the spring's stored energy.
Force vs. time: Area = impulse = change in momentum. A short, tall spike and a long, low hump with the same area give the same Δp, but the long one has a smaller average force.
Momentum vs. time: Slope = net force. A flat line means the net external force is zero and momentum is conserved.
Rotational graphs: On θ–t the slope is ω; on ω–t the slope is α and the area is Δθ; on torque vs. angle the area is the work done; on torque vs. time the area is the angular impulse, ΔL.
Shapes of proportional relationships: y ∝ x is a straight line through the origin, y ∝ x² is a parabola curving upward, y ∝ √x rises and flattens, and y ∝ 1/x drops steeply and levels off toward the axis. Use these when a question asks you to sketch how one quantity depends on another.

Forces and Newton's laws

Unit 2

Free-body diagram checklist: Draw the object as a dot and draw each force as its own arrow starting at the dot, labeled with its type (F_g, F_N, F_T, f_k, F_s). Show only forces exerted on the object by something else, with forces that point the same way drawn side by side rather than overlapping. Never draw components, ma or a separate 'centripetal force', and make relative lengths sensible.
Newton's second law, one axis at a time: ΣF = ma: Add the forces along each axis with their signs and set the total equal to m times the acceleration along that axis. Along an axis where the motion doesn't change (like perpendicular to a ramp), the forces add to zero.
Newton's first law: zero net force means constant velocity: Constant velocity includes being at rest. Motion itself needs no force; only a change in speed or direction does.
Newton's third law pairs: The two forces in a pair are the same type, equal in size, opposite in direction and act on different objects, so they never appear on the same free-body diagram. Weight and the normal force on a book are not a third-law pair.
The normal force isn't always mg: F_N = mg only on a level surface with no other vertical forces or vertical acceleration. On an incline F_N = mg cos θ (with no other perpendicular forces); pushing down on an object at an angle increases F_N and pulling up on it decreases F_N.
Apparent weight (elevator problems): A scale reads the normal force, F_N = m(g + a) with up as positive. It reads more when accelerating upward, less when accelerating downward and zero in free fall, even though gravity still acts.
Universal gravitation: F = Gm₁m₂/r²: Always attractive, equal on both objects and measured between centers. Doubling the distance makes the force one-fourth as big; doubling one mass doubles it.
Gravitational field: g = GM/r²: g at distance r from a planet's center depends on the planet's mass, not yours. Because inertial mass equals gravitational mass, every object in the same place falls with the same acceleration.
Static friction: f_s ≤ μ_s F_N: Static friction is only as big as needed to stop slipping, up to the maximum μ_s F_N, which it reaches only when slipping is about to start. It points opposite the way the surfaces would slip and can point in the direction of motion, as when you walk or a crate rides an accelerating truck.
Kinetic friction: f_k = μ_k F_N: It points opposite the sliding and, in this model, doesn't depend on speed or contact area. μ has no units, and μ_k is usually smaller than μ_s.
Incline shortcuts with friction: A block sliding down at constant speed has μ_k = tan θ; a block just about to slip has μ_s = tan θ. Both come from setting mg sin θ equal to μ mg cos θ.
Ideal strings and pulleys: A massless string has the same tension all along it and pulls away from each object along the string. An ideal (massless, frictionless) pulley just changes the string's direction. A rope with mass has more tension near the top, but you only need to describe that in words.
Connected objects: treat them as one system: With an ideal string, a = (net external force)/(total mass); then use one object's free-body diagram to find the tension. For two hanging masses over an ideal pulley, a = (m₂ − m₁)g/(m₁ + m₂).
Hooke's law: |F_s| = kΔx: Δx is measured from the spring's relaxed length, not from wherever the object started. The spring force always points back toward the relaxed position, which is why it's written F_s = −kΔx.
Center of mass: x_cm = Σmᵢxᵢ / Σmᵢ: It sits closer to the heavier object and at the geometric center of a uniform symmetric object. Internal forces can't change how the center of mass moves; only the net external force can.

Circular motion and orbits

Units 2, 6

Uniform circular motion is accelerated motion: At constant speed the direction still changes, so a_c = v²/r = ω²r points toward the center. The speed is v = 2πr/T, where T is the time for one lap and f = 1/T.
Inward net force: ΣF_in = mv²/r: Take toward the center as positive and add the real forces (tension, gravity, friction, normal force). The inward net force is not an extra force on the diagram, and if it disappears the object moves off along the tangent, not outward.
Car on a flat curve: Static friction provides the inward force, so the top speed without skidding is v = √(μ_s g r), which doesn't depend on the car's mass.
Banked curve with no friction: The horizontal part of the normal force provides the inward force, giving tan θ = v²/(rg). AP Physics 1 calculates banked curves only in this no-friction case.
Top of a vertical loop (inside the track): Both F_N and mg point down toward the center: F_N + mg = mv²/r. The slowest speed that keeps contact (F_N = 0) is v = √(gr).
Bottom of a loop or swing; top of a hill: At the bottom, F_N − mg = mv²/r (or F_T − mg for a string), so you feel heavier and the string tension is greatest. Going over the top of a hill, mg − F_N = mv²/r, so you feel lighter.
Circular orbit speed: v = √(GM/r): Gravity supplies the inward force: GMm/r² = mv²/r. The satellite's mass cancels, and r is measured from the planet's center (planet radius plus altitude).
Kepler's third law: T² = 4π²r³/(GM): Combine the orbit condition with v = 2πr/T. For orbits around the same central mass, T² ∝ r³, so farther satellites take longer. Kepler's first and second laws aren't required.
Weightless in orbit: Astronauts in orbit still feel strong gravity (g is about 8.7 m/s² at the space station's height). They feel weightless because they and the station are in free fall together, so the normal force on them is zero.

Work, energy and power

Unit 3

Work by a constant force: W = Fd cos θ: θ is the angle between the force and the displacement, and only the force's component along the displacement does work. Work is a scalar measured in joules.
Work–energy theorem: W_net = ΔK: The total work done by all forces on an object equals its change in kinetic energy. Use it when you know the forces and the distance but not the time.
Kinetic energy: K = ½mv²: A scalar that is never negative. Doubling the speed makes it four times as large, and K = p²/(2m) links it to momentum.
Gravitational potential energy near Earth: ΔU_g = mgΔy: Only changes matter, so put U_g = 0 wherever is convenient, often the lowest point. It works only near the surface, where g is constant.
Gravitational potential energy in space: U_g = −Gm₁m₂/r: Use this for orbits and anything far from a planet's surface. It is zero when the objects are infinitely far apart and negative everywhere else.
Spring potential energy: U_s = ½kx²: x is the stretch or compression from the relaxed length. Doubling x makes the stored energy four times as large.
Choose the system first: If Earth is inside your system, use U_g and don't count work by gravity. If Earth is outside, gravity does work W = −mgΔy on the object and there's no U_g. Never count both.
Energy conservation: K_i + U_i + W_ext = K_f + U_f + ΔE_th: The system's energy changes only by the work outside forces do on it. Mechanical energy (K + U) stays constant only when no outside work is done and no friction-like forces act inside the system.
Friction turns mechanical energy into thermal energy: For a block sliding a distance d, the thermal energy produced is f_k d. Air resistance and kinetic friction are nonconservative; gravity and spring forces are conservative, so their work doesn't depend on the path.
Speed after a frictionless drop: v = √(2gh): Any object sliding without friction down a height h from rest reaches this speed whatever the shape of the path, and the mass cancels. It doesn't apply to a rolling object, which also stores rotational energy.
Spring launch: ½kx² = ½mv² gives v = x√(k/m) for a horizontal launch with no friction. For a vertical launch, set ½kx² equal to the gain in K + U_g.
Energy bar charts: Draw bars for K, U_g, U_s and thermal energy at the start and the end, plus a bar for work done by outside forces. The totals on the two sides must balance.
Power: P = ΔE/Δt = W/Δt and P = Fv: Power is the rate of energy transfer, in watts (1 W = 1 J/s). The instantaneous power a force delivers is its component along the velocity times the speed.

Momentum and impulse

Unit 4

Momentum: p = mv: A vector in the direction of the velocity, measured in kg·m/s. A system's total momentum is the vector sum of its parts' momenta, so opposite directions partly cancel.
Impulse–momentum theorem: J = F_avg Δt = Δp: Impulse is measured in N·s, the same as kg·m/s, and points in the direction of Δp, which is the direction of the net force.
Stretching out the stopping time: For the same change in momentum, more time means a smaller average force. That's how airbags, padding and bending your knees protect you.
When momentum is conserved: A system's total momentum stays constant when the net external force on it is zero, or is too small to matter during a brief collision or explosion. If outside forces act only vertically, horizontal momentum is still conserved.
Collision types: In an elastic collision total kinetic energy is the same before and after; in an inelastic one some becomes thermal energy, sound or deformation; in a perfectly inelastic one the objects stick together and the most kinetic energy is lost. Momentum is conserved in all three if the system is isolated.
Objects stick together: v_f = (m₁v₁ + m₂v₂)/(m₁ + m₂): Use signs for direction. If object 2 starts at rest, only the fraction m₁/(m₁ + m₂) of the original kinetic energy remains.
Equal masses, elastic, head-on, one at rest: The velocities swap: the moving object stops and the other moves off with the first object's original velocity.
Explosions and push-offs from rest: Total momentum stays zero, so m₁v₁ = −m₂v₂ and the lighter piece moves faster. Kinetic energy increases, coming from stored energy inside the system.
Center-of-mass velocity: v_cm = Σp / Σm: Total momentum divided by total mass. With no net external force, v_cm stays the same before, during and after a collision or explosion.
Collisions in two dimensions: Momentum is conserved separately in x and in y. On AP Physics 1 you set up these equations and reason with them, but you won't solve them simultaneously.
Split two-step problems (ballistic pendulum): Use momentum conservation for the brief collision and energy conservation before or after it, never energy across a sticking collision. For a bullet that embeds in a hanging block that then rises h: v_bullet = ((m + M)/m)√(2gh).

Rotation: kinematics, torque and rotational inertia

Unit 5

Linear and rotational partners: x ↔ θ, v ↔ ω, a ↔ α, m ↔ I, F ↔ τ, p ↔ L and ½mv² ↔ ½Iω². The rotational equations look like the linear ones with each quantity swapped for its partner.
Constant angular acceleration: ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt² and ω² = ω₀² + 2αΔθ, used exactly like the linear versions and only when α is constant.
Linking a point to the rotation: s = rθ, v = rω, a_t = rα: These need angles in radians. Every point on a rigid object has the same θ, ω and α, but points farther from the axis move faster; points on the axis don't move.
Two kinds of acceleration for a point on a spinning object: Tangential acceleration a_t = rα changes its speed; centripetal acceleration a_c = ω²r = v²/r changes its direction. At constant ω only the centripetal part remains.
Torque: τ = rF sin θ: r runs from the axis to where the force acts, and θ is the angle between r and F. A force pointing straight toward or away from the axis makes no torque, and a force perpendicular to r makes the most.
Lever arm: Torque also equals force times the lever arm, the perpendicular distance from the axis to the force's line of action. A longer wrench gives a longer lever arm and more torque for the same push.
Weight acts at the center of mass: To find the torque from an object's weight, put the whole weight at its center of mass, which is the middle of a uniform beam.
Rotational inertia of point masses: I = Σmr²: r is each mass's distance from the axis, so I depends on which axis you choose. Moving a mass twice as far from the axis makes its share of I four times as large.
Shapes to recognize: Hoop about its center MR²; solid disk or cylinder ½MR²; solid sphere (2/5)MR²; thin rod (1/12)ML² about its center and (1/3)ML² about one end. Problems give these when needed, and the key idea is that mass farther out means a bigger I.
Parallel axis theorem: I = I_cm + Md²: d is the distance between the axis through the center of mass and the parallel axis you want. Of all axes pointing the same way, the one through the center of mass gives the smallest I.
Newton's second law for rotation: α = τ_net / I: More net torque means more angular acceleration; more rotational inertia means less. Use the same clockwise or counterclockwise sign choice for τ and α.
Pulley with mass: The string tensions on the two sides are no longer equal: (T₂ − T₁)R = Iα, with a = Rα if the string doesn't slip. For two hanging masses this gives a = (m₂ − m₁)g/(m₁ + m₂ + I/R²).

Static equilibrium and balanced beams

Units 2, 5

Two conditions: ΣF = 0 and Στ = 0: An object stays in static equilibrium only if both the net force and the net torque are zero. One doesn't guarantee the other: two equal, opposite forces along different lines make zero net force but a nonzero torque.
Choose the axis to eliminate an unknown: For an object in equilibrium, net torque is zero about every point, so put the axis where an unknown force acts and that force drops out of the torque equation.
Seesaw balance: m₁r₁ = m₂r₂: The torques from the two weights about the pivot must cancel, so the heavier person sits closer to the pivot.
Beam on two supports: With vertical supports and loads, the support forces add up to the total weight, and the support closer to the center of mass (or to a heavy load) carries more of it.
About to tip: When a beam or block is just about to tip over an edge, the force from the other support or edge drops to zero, and all the support acts at the tipping edge.

Rotational energy, angular momentum and rolling

Unit 6

Rotational kinetic energy: K = ½Iω²: A spinning object has kinetic energy even if its center stays put. An object that moves and spins has K = ½Mv_cm² + ½Iω².
Work done by a torque: W = τΔθ: For a constant torque with Δθ in radians; for a changing torque, use the area under the torque vs. angle graph. Net work by torques equals the change in rotational kinetic energy.
Angular momentum: L = Iω, or L = mvr sin θ for a moving point: Units are kg·m²/s. Even an object moving in a straight line has angular momentum about a point not on its path, equal to mv times the perpendicular distance from the point to the path.
Angular impulse: τΔt = ΔL: The area under a torque vs. time graph is the change in angular momentum.
Conservation of angular momentum: I₁ω₁ = I₂ω₂: With no net external torque, L stays constant. A skater who pulls in their arms lowers I and spins faster, and their kinetic energy rises (K = L²/(2I)) because their muscles do work.
Rotational collisions: When clay sticks to a spinning disk or a person jumps onto a merry-go-round, angular momentum about the axis is conserved but kinetic energy is not. Someone moving at speed v along the rim's tangent brings L = mvR.
Rolling without slipping: v_cm = Rω and a_cm = Rα: The contact point is momentarily at rest and the top moves at 2v_cm. Static friction removes no energy here, so mechanical energy is conserved.
Rolling down a ramp: v = √(2gh / (1 + I/(MR²))): From Mgh = ½Mv² + ½Iω². The smaller I/(MR²) is, the faster it rolls, so a solid sphere beats a solid cylinder, which beats a hoop, whatever their masses and radii. All roll slower than a frictionless slider, which reaches √(2gh).
Rolling while slipping: Kinetic friction acts, v_cm and Rω are no longer equal and mechanical energy turns into thermal energy. AP Physics 1 asks only for a description in words.
Elliptical orbits: faster when closer: Gravity exerts no torque about the central body, so the satellite's angular momentum is constant; at the nearest and farthest points r₁v₁ = r₂v₂. Total energy ½mv² − GMm/r is also constant.
Escape velocity: v = √(2GM/r): The speed at distance r that makes the total energy exactly zero. It's √2 times the circular orbit speed at the same r and doesn't depend on the escaping object's mass.
Circular orbit energy: Combining v = √(GM/r) with U = −GMm/r gives K = GMm/(2r) and total energy E = −GMm/(2r). A negative total energy means the satellite is bound to the planet.

Simple harmonic motion

Unit 7

When motion is simple harmonic: SHM needs a restoring force proportional to the displacement and pointing back toward equilibrium, F = −kx. A pendulum is close to SHM only for small angles, roughly under 15°.
Period and frequency: T = 1/f: Frequency is in hertz (cycles per second). The angular frequency ω = 2πf, in rad/s, appears in SHM equations.
Mass on a spring: T = 2π√(m/k): More mass means a slower oscillation and a stiffer spring a faster one. The period doesn't depend on amplitude or g, so it's the same for a vertical spring, whose equilibrium is simply shifted down by mg/k.
Simple pendulum: T = 2π√(L/g): The period doesn't depend on the bob's mass or (for small angles) the amplitude. Four times the length doubles the period, and with a weaker g, as on the Moon, the period is longer.
Position over time: x = A cos(2πft) or A sin(2πft): Use cosine if the object is released from x = A at t = 0 and sine if it starts at equilibrium moving in the positive direction.
Where the extremes happen: At equilibrium: greatest speed and kinetic energy, zero acceleration and force, lowest potential energy. At the turning points (±A): zero speed, greatest acceleration and force (pointing back toward equilibrium), greatest potential energy.
Maximum speed and acceleration: v_max = 2πfA = A√(k/m) and a_max = (2πf)²A = kA/m. Doubling the amplitude doubles both but leaves the period unchanged.
Energy of a mass–spring oscillator: E = ½kA²: E = ½kx² + ½mv² at every point, so E also equals ½mv_max². Doubling the amplitude makes the total energy four times as large.
SHM graphs: x, v and a are all sinusoidal with the same period. v is zero wherever x is at a peak, and a is always the mirror image of x (a = −(k/m)x). K and U swing between zero and E twice per period and never go negative.
Pendulum energy: The bob trades gravitational potential energy for kinetic energy. Its height above the lowest point is h = L(1 − cos θ), so its speed at the bottom is √(2gh).

Fluids

Unit 8

Density: ρ = m/V: Measured in kg/m³; water is 1000 kg/m³. AP Physics 1 treats fluids as ideal: incompressible (the density never changes) and with no viscosity.
Pressure: P = F/A: The perpendicular force per unit area, a scalar measured in pascals (1 Pa = 1 N/m²). The same force on a smaller area makes a bigger pressure.
Pressure at depth: P = P₀ + ρgh: h is measured down from the surface, where the pressure is P₀. Pressure depends only on depth and the fluid's density, not on the container's shape or width, so points at the same depth in the same connected fluid have the same pressure.
Absolute vs. gauge pressure: Absolute pressure includes the atmosphere; gauge pressure is the extra amount, P − P_atm, which is ρgh in an open container. Tire gauges read gauge pressure.
10 m of water ≈ 1 atm: ρgh = (1000 kg/m³)(9.8 m/s²)(10 m) ≈ 1 × 10⁵ Pa, so a diver 10 m down feels about twice atmospheric pressure in absolute terms.
Buoyant force: F_b = ρ_fluid V_sub g: The fluid's upward push equals the weight of the fluid displaced. Use the fluid's density and the submerged volume, not the object's density; it comes from the pressure on the bottom being greater than the pressure on the top.
Floating: A floating object has F_b = mg, so the fraction underwater equals ρ_object/ρ_fluid. An object floats if its average density is less than the fluid's.
Fully submerged objects: Apparent weight = mg − ρ_fluid V g, so a scale reads less underwater. A fully submerged object in an incompressible fluid feels the same buoyant force at any depth.
Continuity equation: A₁v₁ = A₂v₂: In a full pipe the volume flow rate Av (m³/s) is the same everywhere, so fluid moves faster where the pipe is narrower. For a round pipe A = πr², so halving the diameter makes the speed four times as large.
Bernoulli's equation: P + ρgy + ½ρv² is the same along the flow: It's energy conservation per unit volume, and each term is in pascals (J/m³). It holds for an ideal fluid in steady flow, comparing two points in the same stream.
Faster flow, lower pressure: At the same height, the fluid moves faster where the pressure is lower. A pressure difference is what pushes fluid to speed up, just as Newton's second law says.
Torricelli's theorem: v = √(2gh): The speed of fluid leaving a hole a depth h below the surface of a large open tank, the same as dropping h. It assumes the top and the hole are both at atmospheric pressure and the top surface barely moves.
Bernoulli setup tips: Any point open to the air has P = P_atm, and the surface of a large tank has v ≈ 0. Use continuity to find a second speed before you use Bernoulli.

Problem-solving checklists

Units 1, 2, 3, 4, 5, 6, 7, 8

Which tool?: Use forces and Newton's laws when you need an acceleration or a force at an instant, energy when the question links speeds and positions without time, momentum for collisions and explosions, and impulse when force and time are involved. Use the rotational versions when things spin.
What the exam assumes unless it says otherwise: The reference frame is inertial, air resistance is negligible, springs and strings are ideal (massless), and fluids are ideal and fill their pipes completely. If a question drops one of these, it will say so.
Force problems: Pick the system, draw the free-body diagram, put one axis along the acceleration and write ΣF = ma for each axis. Solve with letters first, then plug in numbers.
Energy problems: Pick the system and the starting and ending moments, then list which kinds of energy exist at each (K, U_g, U_s, thermal). Add any work by outside forces, write the energy equation and solve.
Collision problems: Check that the system is isolated during the interaction, choose a positive direction and set total momentum before equal to total momentum after. Use energy conservation only before or after the collision, unless it's elastic.
Torque and rotation problems: Choose the axis, find each force's lever arm and whether it turns things clockwise or counterclockwise, and write Στ = Iα (or Στ = 0). If something also moves in a line, combine this with ΣF = ma and a = rα.
Derivations: Start from a basic law (Newton's second law, energy or momentum conservation) and write it for this situation. Use only quantities the problem gives and physical constants, and finish with the requested quantity alone on one side.
'What if it doubles?' questions: Write the relationship, cancel everything that stays the same, then scale. Example: in F = Gm₁m₂/r², doubling r makes F one-fourth as large.
Justifying a claim: Name the law or principle, apply it to the specific objects in the problem, then state the conclusion. Make sure your words agree with your equations, diagrams and graphs.
Final check: Check that the units are right, that every vector answer has a direction and that the size is reasonable (even a world-class sprinter tops out around 12 m/s). Then check that a limiting case makes sense.

Experiments and linearizing data

Units 1, 2, 3, 4, 5, 6, 7, 8

Designing an experiment: Name what you'll change (independent variable), what you'll measure (dependent variable) and what you'll keep the same. Say which equipment measures each quantity and how, test a wide range of values and repeat trials.
Common equipment: Meterstick or measuring tape, stopwatch, motion detector (position and velocity vs. time), force sensor or spring scale, photogate, balance, protractor and video analysis.
Reducing uncertainty: Repeat and average, and time many cycles and divide (for example, time 10 oscillations to find one period). Larger distances and times make measuring errors a smaller fraction of the result.
Linearize: rearrange into y = (slope)x + b: Choose what to plot on each axis so the equation becomes a straight line, then find the unknown from the slope or intercept.
Motion and force examples: Dropped from rest: Δy against t² has slope g/2; spring: F against Δx has slope k; Newton's second law: a against F_net has slope 1/m. Sliding from rest down a frictionless ramp: v² against drop height h has slope 2g; circular motion: net inward force against v² has slope m/r.
Oscillation examples: Pendulum: T² against L has slope 4π²/g, so g = 4π²/slope; mass–spring: T² against m has slope 4π²/k, so k = 4π²/slope.
Fluid examples: Absolute pressure against depth has slope ρg and intercept equal to the surface pressure; buoyant force against submerged volume has slope ρ_fluid g. For Torricelli's theorem, v² against depth h has slope 2g.
Drawing and using a best-fit line: Label each axis with the quantity and unit, and choose a scale that uses most of the grid. Draw one straight line with points scattered evenly on both sides, and find the slope from two points on the line far apart, not from data points; give the slope its units.
Reading the intercept: If theory predicts the line goes through the origin but it doesn't, something systematic is going on, such as friction, a measuring offset or a mass you ignored.