A useful physics equation sheet helps you recognize a relationship, identify the variables and check whether its assumptions fit the problem. It cannot replace a diagram or an explanation of why the relationship applies.
Start with a simple motion problem
A cart starts from rest and accelerates uniformly at 2 m/s² for 3 seconds. The question asks for its final velocity. “Uniformly” tells us acceleration is constant, so v = v₀ + at applies.
- Choose forward as positive.
- Identify v₀ = 0 m/s, a = 2 m/s² and t = 3 s.
- Calculate v = 0 + (2)(3) = 6 m/s.
- Interpret the result: the cart is moving forward at 6 meters per second after 3 seconds.
Know the symbols before substituting
The symbol Δ means “final minus initial.” Position x is measured in meters (m), velocity v in m/s, acceleration a in m/s², time t in seconds (s), mass m in kilograms (kg), force F in newtons (N), and energy or work in joules (J). A subscript 0 denotes an initial value.
Common relationships and their conditions
| Idea | Relationship | When to use it |
|---|---|---|
| Constant acceleration | v = v₀ + at Δx = v₀t + ½at² v² = v₀² + 2aΔx | One axis with constant acceleration. Keep signs consistent. |
| Net force | ΣF = ma | Add all force components along the chosen axis; F means net force here. |
| Friction | fₖ = μₖN 0 ≤ fₛ ≤ μₛN | Kinetic friction for sliding; static friction adjusts up to a maximum. N is normal-force magnitude. |
| Energy | K = ½mv² ΔUᵍ = mgΔh Uˢ = ½kx² | Near Earth use approximately constant g; for an ideal spring x is displacement from equilibrium. |
| Work | W = Fd cos θ Wnet = ΔK | The first expression assumes a constant force. θ is the angle between force and displacement. |
| Momentum and impulse | p = mv J = Δp = Fnet,avg Δt | Momentum is a vector. A system’s total momentum stays constant only when external impulse is zero or negligible. |
| Circular motion | aᶜ = v²/r | Radial acceleration points toward the center. “Centripetal force” describes the net inward force, not an extra force to add. |
| Rotation | τ = rF sin θ Στ = Iα Krot = ½Iω² | Define the axis. I is rotational inertia, α angular acceleration and ω angular speed. |
| Oscillations | T = 2π√(m/k) T = 2π√(L/g) | Ideal mass-spring system; simple pendulum at small angles, respectively. |
| Fluids | ρ = m/V P = P₀ + ρgh Fᵇ = ρfluid Vdisplaced g | Density, pressure at depth in a uniform stationary fluid, and buoyant force. Current course coverage includes fluids. |
Three common mistakes
- Mixing force pairs: Newton’s third-law forces act on different objects; they do not cancel on one object’s force diagram.
- Calling energy “lost”: in an inelastic collision, some kinetic energy becomes internal energy or other forms. Total energy is still conserved.
- Choosing a formula before the system: first decide which objects belong to your system and what crosses its boundary.
Practice with a reason
A 2 kg cart’s velocity changes from 1 m/s to 4 m/s in the positive direction. Find its change in momentum.
Check the reasoning
Δp = m(vfinal − vinitial) = 2(4 − 1) = 6 kg·m/s, positive. The net impulse has the same value, 6 N·s.
Next, explain why knowing the time interval would let you calculate the average net force. Continue through our AP Physics 1 lessons and practice.
Recovered original PDF
Revised study edition
Download the corrected study guide (PDF). The revised edition reflects the corrections on this page. Independent teacher review is pending.
Download the original archived handout (PDF). This preserves the original download for existing links. It predates the corrections above and is not the current official exam reference. Use the revised explanation on this page; the revised edition is linked above.