Physics 1
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AP Physics 1: Algebra-Based — Cheatsheet

Formulas, exam-day tips, and key terms on one page.

Formulas & relationships

Displacement
Δx = x_f − x_i
Final minus initial. A negative result simply means motion in the negative direction, not "less than zero distance".
Average velocity
v_avg = Δx / Δt = (x_f − x_i) / (t_f − t_i)
Units: meters per second, m/s. The sign of v_avg is the sign of Δx.
Velocity–time (no Δx)
v = v₀ + at
Use when you have or want time but not displacement.
Position–time (no v)
Δx = v₀t + ½at²
The ½ is not optional — dropping it is the single most common kinematics mistake.
Timeless equation (no t)
v² = v₀² + 2aΔx
The rescue equation when time is neither given nor asked for.
Slope and area
slope of x–t = v · slope of v–t = a · area under v–t = Δx
Two different graphs, three different meanings — keep straight which graph you are reading.
Projectile equations (horizontal launch, taking down as +)
horizontal: Δx = v_x·t · vertical: Δy = ½g·t² · v_y = g·t
No acceleration horizontally (v_x constant); a = g ≈ 10 m/s² vertically.
Newton’s second law
ΣF = ma
ΣF is the vector sum of all forces (newtons, N). Solve component by component: ΣF_x = ma_x and ΣF_y = ma_y.
Weight
W = mg
g = 10 m/s² here (AP tables use 9.8). Weight is a force in newtons; it is never measured in kilograms.
Weight components on an incline (angle θ)
along incline: mg sinθ · perpendicular: mg cosθ
On a ramp, gravity splits into a part that pulls the object down the slope (mg sinθ) and a part pressing it into the surface (mg cosθ).
Friction force
f_k = μ_k N · f_s ≤ μ_s N
On flat ground with no vertical push, N = mg. Static friction is an inequality: it only reaches its maximum right at the verge of slipping.
Centripetal acceleration and force
a_c = v² / r · F_c = mv² / r
Both point toward the center of the circle. Doubling the speed quadruples both, because v is squared.
Newton’s law of universal gravitation
F = G·m₁·m₂ / r²
G = 6.67 × 10⁻¹¹ N·m²/kg². The force is an inverse-square law: triple the separation and the force drops to one-ninth.
Work
W = F·d·cosθ
θ is the angle between the force and the displacement. Force along the motion → cosθ = 1; opposite → cosθ = −1; perpendicular → cosθ = 0.
Work–energy theorem
W_net = ΔKE = ½mv_f² − ½mv_i²
The total work done by all forces equals the change in kinetic energy. Speeding up means positive net work; slowing down means negative.
Kinetic energy
KE = ½mv²
v is speed. The ½ and the square are both essential — dropping either is a common error.
Potential energy
PE_grav = mgh · PE_spring = ½kx²
h is height above your chosen reference; x is the spring’s stretch or compression from equilibrium.
Conservation of mechanical energy
KE_i + PE_i = KE_f + PE_f
Valid when no friction or drag removes energy. With friction, add the dissipated thermal energy to the final side.
Energy with friction
KE_i + PE_i = KE_f + PE_f + E_thermal
Friction converts mechanical energy to heat. Total energy is still conserved — it just leaves the mechanical account.
Power
P = W / t · P = F·v
Units: watts (W) = joules per second. Use W/t when you know the total work and time; use Fv when you know a steady force and speed.
Momentum
p = mv
A vector in kg·m/s. Its direction is the direction of motion; reverse the motion and the sign flips.
Impulse–momentum theorem
J = F·Δt = Δp = m·v_f − m·v_i
Impulse (N·s) equals the change in momentum. For a bounce, remember v_f and v_i point in opposite directions.
Conservation of momentum
m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
Total momentum before = total momentum after. Keep signs consistent: opposite directions get opposite signs.
Perfectly inelastic collision
m₁v₁ + m₂v₂ = (m₁ + m₂)·v′
The objects stick, so they share one final velocity v′. Solve for v′ by dividing the total momentum by the total mass.
Center of mass (two objects)
x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
A mass-weighted average of positions. Placing your origin at one mass simplifies the arithmetic.
Linear–angular link
v = rω · a_t = rα
r is the distance from the rotation axis. Angular quantities use radians; v and a_t come out in m/s and m/s².
Rotational kinematics (constant α)
ω = ω₀ + αt · θ = ω₀t + ½αt²
Identical in form to the linear equations, with θ→x, ω→v, α→a. Every trick you learned in kinematics carries over.
Torque
τ = r·F·sinθ
θ is the angle between the position vector r and the force. Maximum torque when θ = 90° (force perpendicular); zero when θ = 0° (force along r).
Rotational inertia and Newton’s second law
I_point = mr² · τ_net = Iα
I is measured in kg·m². Mass far from the axis raises I steeply because of the square on r.
Static equilibrium conditions
ΣF = 0 and Στ = 0
Both must hold. For a balanced beam this reduces to (clockwise torques) = (counterclockwise torques) about any chosen pivot.
Rotational and total kinetic energy
KE_rot = ½Iω² · KE_total = ½mv² + ½Iω²
For a rolling object, add both terms. Energy conservation from a height h then reads mgh = ½mv² + ½Iω².
Angular momentum
L = Iω · L_point = mvr
Units kg·m²/s. For a point mass on a circle, r is its distance from the axis. Angular momentum is a vector along the spin axis.
Conservation of angular momentum
I₁ω₁ = I₂ω₂
Valid when net external torque is zero. If I drops to half, ω doubles; if I doubles, ω halves.
Rolling condition
v = rω
Ties center speed to spin rate for rolling without slipping. Also a = rα for the accelerations.
Hooke’s law and spring period
F = −kx · T = 2π√(m/k)
k is in N/m. A stiffer spring (larger k) gives a shorter period; a heavier mass gives a longer one.
Pendulum period
T = 2π√(L / g)
L is the string length; g = 10 m/s². Mass and (small) amplitude do not appear — only length changes the period on a given planet.
Period, frequency, angular frequency
f = 1 / T · ω = 2πf = 2π / T
T in seconds, f in hertz (Hz), ω in rad/s. Doubling the frequency halves the period.
Energy in SHM
E_total = ½kA² = ½mv_max²
All potential at the extremes, all kinetic at equilibrium. Setting the two equal gives v_max = A√(k/m).
Density, pressure, and pressure with depth
ρ = m/V · P = F/A · P = P₀ + ρgh
ρgh is the gauge pressure — the extra pressure from a depth h of fluid. P₀ is the pressure at the surface (often atmospheric).
Buoyant force
F_b = ρ_fluid · V_displaced · g
Uses the *fluid’s* density and the volume of fluid pushed aside. For a fully submerged object, V_displaced equals the object’s own volume.
Continuity equation
A₁v₁ = A₂v₂
For incompressible, steady flow. Area and speed are inversely related: halve the area and the speed doubles.
Bernoulli’s equation
P + ½ρv² + ρgh = constant (along a streamline)
For flow at constant height the ρgh terms cancel, leaving P + ½ρv² constant — higher speed forces lower pressure.
The translation chain
slope of x–t = v · slope of v–t = a · area under a–t = Δv · area under v–t = Δx
Slopes going down, areas going up. Area below the time axis is negative in both directions.
Angled projectile setup
v₀ₓ = v₀ cos θ · v₀ᵧ = v₀ sin θ · x: Δx = v₀ₓt · y: Δy = v₀ᵧt − ½gt², vᵧ = v₀ᵧ − gt
Taking up as positive, so g enters with a minus sign. Flipping that convention is fine as long as you flip it everywhere.
Range on level ground
R = v₀² sin(2θ) / g
Only valid when landing height equals launch height. sin(2θ) peaks at 2θ = 90°, i.e. θ = 45°.
The two-step method
Step 1 (system): a = ΣF_external / Σm · Step 2 (one body): ΣF on that body = m_that body · a
Step 1 gives the acceleration. Step 2 uses that acceleration to solve for the internal force.
Block on a frictionless incline
along: mg sin θ = ma → a = g sin θ · perpendicular: N = mg cos θ
The acceleration down a frictionless incline is independent of mass — the same reason all objects fall together.
Newton's second law for circular motion
ΣF_toward center = m v² / r
Sum the *real* forces, taking toward-the-center as positive. Then set that sum equal to mv²/r.
Circular orbit
GMm/r² = mv²/r → v = √(GM/r) · T = 2π√(r³/GM)
The orbiting mass m cancels: orbital speed and period depend only on the central mass and the radius, never on the satellite.
Energy accounting
W_external = ΔK + ΔU + ΔE_thermal
Everything crossing the boundary on the left; everything stored or dissipated inside on the right.
Work from a graph
W = area under the F-vs-x curve · For a spring: W = ½kx² (the triangle under F = kx)
The ½ in the spring energy is not a separate rule — it is the area of the triangle under a straight line through the origin.
Force from a potential-energy curve
F = −dU/dx (the negative slope)
The minus sign means force always points *downhill* on the U curve, toward lower potential energy.
Two-dimensional conservation
Σm v_x before = Σm v_x after · Σm v_y before = Σm v_y after
Two separate scalar equations. Recombine with |p| = √(pₓ² + pᵧ²) and θ = tan⁻¹(pᵧ/pₓ).
Which law applies
ALL collisions: momentum conserved · ELASTIC only: kinetic energy also conserved
Momentum is the reliable one. Kinetic energy is the special case, so never assume it.
Rotational inertia
I = Σ m r² (units kg·m²)
r is measured from the axis of rotation, not from the center of mass, unless the two coincide.
Newton's second law for rotation
Στ = I α
Torque plays the role of force, rotational inertia the role of mass, angular acceleration the role of acceleration.
Static equilibrium
ΣF_x = 0 · ΣF_y = 0 · Στ = 0 about ANY point
The torque condition holds about every point when the body is in equilibrium, which is what makes the pivot a free choice.
Rolling without slipping
v_cm = ωR · a_cm = αR · distance traveled = Rθ
Only valid when there is no slipping. A spinning wheel on ice violates all three.
Total kinetic energy of a rolling object
K = ½mv² + ½Iω² and with I = cmR² plus ω = v/R: K = ½(1 + c) mv²
c is the shape coefficient: 2/5 for a solid sphere, ½ for a disk, 1 for a hoop.
Conservation of angular momentum
I₁ω₁ = I₂ω₂ (when Στ_external = 0)
Reduce I and ω must rise to compensate. This is the entire content of the skater, diver and neutron-star examples.
Angular impulse–momentum theorem
τ Δt = ΔL = I ω_f − I ω_i
The exact analogue of F Δt = Δp. For a varying torque, the angular impulse is the area under a torque–time graph.
The defining relation
F = −kx → a = −(k/m)x
The minus sign is the physics: force and acceleration always point back toward equilibrium, opposite the displacement.
Maxima in SHM
v_max = ωA = A√(k/m) · a_max = ω²A = (k/m)A
Both scale with amplitude, which is why a wider swing is both faster at the bottom and harder pulled at the ends.
Periods
Spring: T = 2π√(m/k) · Pendulum: T = 2π√(L/g)
Spring period depends on mass but not g; pendulum period depends on g but not mass. Exactly reversed.
Archimedes' principle
F_b = ρ_fluid · V_displaced · g
V_displaced is the submerged volume, which equals the object's total volume only when it is fully underwater.
Continuity equation
A₁v₁ = A₂v₂, with A = πr² for a circular pipe
Volume per second is conserved. Halving the radius quadruples the speed, not doubles it.
Bernoulli's equation
P + ½ρv² + ρgh = constant along a streamline
Pressure work + kinetic energy + potential energy, each per unit volume. Every term has units of pascals.

On the exam

How to get a 5

Key terms

Kinematic equations (constant a)v = v₀ + at; x = x₀ + v₀t + ½at²; v² = v₀² + 2aΔx.
Newton's second lawΣF = ma, applied separately along each axis. Acceleration is in the direction of the NET force, not of any single force.
Weight vs massMass is the amount of matter and is frame-independent; weight is the gravitational force mg and changes with location.
Work–energy theoremW_net = ΔK = ½mv_f² − ½mv_i². The net work done on an object equals its change in kinetic energy.
Conservation of mechanical energyKE + PE is constant when only conservative forces do work. The condition is what makes the statement true, and omitting it is what loses the point.
Impulse–momentum theoremJ = FΔt = Δp = mv_f − mv_i. On a force–time graph, the impulse is the area under the curve.
Elastic vs inelastic collisionBoth conserve momentum. Elastic conserves KE; inelastic does not. Perfectly inelastic → objects stick.
Centripetal acceleration and forcea_c = v²/r toward center; F_c = mv²/r. It is a net-force requirement, not a new force.
Hooke's law and spring PEF = −kx (restoring); elastic PE = ½kx². k = spring constant (N/m).
Conditions for static equilibriumΣF = 0 AND Στ = 0. Torque τ = rF sinθ.
SHM period (pendulum & spring)Pendulum: T = 2π√(L/g). Spring: T = 2π√(m/k). Period independent of amplitude.
Newton's third lawForces come in equal and opposite pairs acting on DIFFERENT objects. The two never cancel, because they are not on the same body — which is why a horse can pull a cart.
Displacement vs. distanceDisplacement is the signed change in position, Δx = x_f − x_i (a vector). Distance is the total path length traveled (a scalar) and is never negative.
Slope and area on a v–t graphSlope of a velocity–time graph = acceleration; signed area between the curve and the time axis = displacement.
Key idea of projectile motionHorizontal and vertical motions are independent: a_x = 0 with constant v_x, while a_y = −g. The two axes share only the time of flight.
Newton’s second lawΣF = ma. Acceleration points along the net force and is inversely proportional to inertial mass.
Newton’s third-law pairTwo forces equal in magnitude, opposite in direction, of the same type, acting on different objects. They never cancel on a single free-body diagram.
Static vs. kinetic frictionStatic friction adjusts up to a maximum μsN to prevent sliding; kinetic friction has fixed magnitude μkN and opposes relative sliding. Usually μs > μk.
Centripetal accelerationa_c = v²/r directed toward the center of the circular path, caused by whatever real force points inward — tension, friction, gravity, or the normal force.
Newton’s law of universal gravitationF = Gm₁m₂/r², attractive and along the line joining the centers. The field strength a distance r from mass M is g = GM/r².
Work done by a constant forceW = Fd cos θ, where θ is the angle between force and displacement. A force perpendicular to the displacement, such as the normal force on a slope, does zero work.
PowerP = W/Δt = Fv cos θ, measured in watts. At constant speed the drive force equals the resistive force, so P = Fv.
When is momentum conserved?Whenever the net external force on the chosen system is zero, or negligible during a brief collision. Internal forces always cancel in third-law pairs.
Elastic vs. inelastic collisionMomentum is conserved in both; kinetic energy is conserved only in elastic collisions. A perfectly inelastic collision leaves the objects moving together and loses the maximum possible kinetic energy.