The AP Physics 1 equation sheet, explained line by line

v_x

velocity at time t (m/s)

v_x0

velocity at t = 0 (m/s)

a_x

acceleration, constant (m/s²)

t

time (s)

Use it when: Speed changes at a steady rate (constant acceleration) and you know or want the time.

Do not use it when: Acceleration changes (a spring, a pendulum, air drag). Then the three kinematics lines are all wrong.

Example: A car leaves a light from rest with a = 3.0 m/s². After 4.0 s: v = 0 + (3.0)(4.0) = 12 m/s.

Trap: Signs. Pick a positive direction first. A ball thrown up has a = −9.8 m/s² on the way up AND on the way down.

Unit 1: Kinematics Topic 1.2, 1.3

x

position at time t (m)

x₀

starting position (m)

v_x0

starting velocity (m/s)

a_x

constant acceleration (m/s²)

t

time (s)

Use it when: Constant acceleration and the problem asks where or how far, with time known (or asked for).

Do not use it when: Acceleration is not constant. Also not for the total distance if the object turns around: it gives displacement.

Example: A cart starts with 2.0 m/s and speeds up at 1.5 m/s² for 4.0 s: Δx = (2.0)(4.0) + ½(1.5)(4.0)² = 8.0 + 12 = 20 m.

Trap: Forgetting the ½, or squaring before multiplying by a in the wrong order. Projectiles: use it twice, once for x (with a_x = 0) and once for y (with a_y = −9.8 m/s²).

Unit 1: Kinematics Topic 1.2, 1.5

v_x

final velocity (m/s)

v_x0

starting velocity (m/s)

a_x

constant acceleration (m/s²)

x − x₀

displacement (m)

Use it when: Constant acceleration and time is NOT given and NOT asked (braking distance, a dropped ball's speed after falling some height).

Do not use it when: Acceleration changes. And it cannot tell you the direction of v: the square root has a ± you must choose.

Example: A car going 20 m/s brakes at 5.0 m/s². Stopping distance: 0 = 20² + 2(−5.0)Δx, so Δx = 400 / 10 = 40 m.

Trap: Doubling the speed makes the stopping distance FOUR times bigger (it goes with v²). AP loves that ratio question.

Unit 1: Kinematics Topic 1.2

x_cm

position of the center of mass (m)

m_i

mass of each piece (kg)

x_i

position of each piece (m)

Use it when: Finding the balance point of a system of objects, or tracking a system as one particle.

Do not use it when: A single object with a non-uniform shape you cannot split into pieces with known centers.

Example: 2.0 kg at x = 0 and 3.0 kg at x = 5.0 m: x_cm = (0 + 15) / 5.0 = 3.0 m (closer to the heavier mass).

Trap: The center of mass is always closer to the heavier mass. If your answer is closer to the lighter one, check the arithmetic. It need not be inside any object (a ring, a boomerang).

Unit 2: Force and Translational Dynamics Topic 2.1

a_sys

acceleration of the system (m/s²), a vector

ΣF

sum of all EXTERNAL forces on the system (N)

m_sys

total mass of the system (kg)

Use it when: Any time you relate forces to motion: draw a free-body diagram, add forces along each axis, divide by mass.

Do not use it when: Do not include internal forces (one part of the system pushing another) and do not add ma as a force on the diagram.

Example: A 4.0 kg box is pushed with 30 N against 10 N of friction: a = (30 − 10) / 4.0 = 5.0 m/s² in the push direction.

Trap: Zero net force means constant velocity, not necessarily at rest. Work one axis at a time.

Unit 2: Force and Translational Dynamics Topic 2.4, 2.5

F_g

gravitational force each mass exerts on the other (N)

G

6.67 × 10⁻¹¹ N·m²/kg²

m₁, m₂

the two masses (kg)

r

distance between their CENTERS (m)

Use it when: Planets, moons, satellites, or when the distance from a planet changes a lot. Near Earth's surface just use F_g = mg.

Do not use it when: Do not use the height above the ground as r. Use radius plus height.

Example: Earth (5.97 × 10²⁴ kg, radius 6.37 × 10⁶ m) on a 70 kg person: F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(70) / (6.37 × 10⁶)² ≈ 690 N, the same as mg = 70 × 9.8 = 686 N.

Trap: Inverse square: double r and the force becomes ¼. Both masses feel the SAME size force (Newton's third law).

Unit 2: Force and Translational Dynamics Topic 2.6, 6.6

F_f

friction force (N)

μ

coefficient of friction (no units): μ_s static, μ_k kinetic

F_N

normal force (N)

Use it when: Sliding: kinetic friction EQUALS μ_kF_N. Not sliding: static friction is whatever is needed, UP TO μ_sF_N (that is the ≤).

Do not use it when: Do not set static friction equal to μ_sF_N unless the object is just about to slip.

Example: F_N = 50 N, μ_s = 0.50: static friction can be up to 25 N. Push with 10 N and the box stays put; friction is 10 N, not 25 N.

Trap: F_N is not always mg (ramps, pulling at an angle, elevators). Find F_N from the perpendicular axis first.

Unit 2: Force and Translational Dynamics Topic 2.7

F_s

force the spring exerts (N)

k

spring constant (N/m)

Δx

stretch or compression from the spring's natural length (m)

Use it when: Hooke's law: an ideal spring stretched or compressed by Δx. The minus sign means the force points back toward natural length.

Do not use it when: Δx is NOT the spring's total length. Do not use it past the spring's elastic limit.

Example: k = 200 N/m stretched 0.050 m: |F| = 200 × 0.050 = 10 N, pulling back.

Trap: On a graph of F vs Δx, the slope is k and the area under the line is the stored energy.

Unit 2: Force and Translational Dynamics Topic 2.8, 7.1

a_c

centripetal acceleration, toward the center (m/s²)

v

speed (m/s)

r

radius of the circle (m)

Use it when: Anything moving in a circle (or part of one): cars on curves, the top of a loop, satellites.

Do not use it when: Centripetal force is not a new force on the diagram. It is the net inward force made by real forces (tension, gravity, normal, friction).

Example: A car takes a 25 m radius curve at 10 m/s: a_c = 10² / 25 = 4.0 m/s², toward the center.

Trap: Write ΣF_{toward center} = mv²/r. Forces pointing toward the center are positive.

Unit 2: Force and Translational Dynamics Topic 2.9, 6.6

K

translational kinetic energy (J)

m

mass (kg)

v

speed (m/s)

Use it when: Energy of motion. Any energy bar chart or conservation of energy problem.

Do not use it when: It is a scalar: never give it a direction and never split it into x and y kinetic energies as vectors.

Example: A 1000 kg car at 20 m/s: K = ½(1000)(20)² = 2.0 × 10⁵ J.

Trap: Double speed = four times K. K can never be negative.

Unit 3: Work, Energy, and Power Topic 3.1

W

work done by one force (J)

F_∥

part of the force along the motion (N)

d

distance the point of contact moves (m)

θ

angle between the force and the motion

Use it when: A constant force acts while the object moves. Positive work adds energy, negative work takes it away.

Do not use it when: A force perpendicular to the motion does zero work (normal force on flat ground, tension in a circular swing). Changing forces need the area under an F-x graph.

Example: Pulling a sled 10 m with 50 N at 60° above the ground: W = 50 × 10 × cos 60° = 250 J.

Trap: θ is between the force and the DISPLACEMENT, not between the force and the ground in every case. Friction on a sliding box: θ = 180°, so W is negative.

Unit 3: Work, Energy, and Power Topic 3.2

ΔK

change in kinetic energy (J)

ΣW_i

total work by all forces (J)

Use it when: The work-energy theorem: net work changes kinetic energy. Great when you know forces and distances but not time.

Do not use it when: If you already counted gravity as potential energy, do not also count its work here (no double counting).

Example: A 2.0 kg box from rest gets 36 J of net work: ½(2.0)v² = 36, so v = 6.0 m/s.

Trap: It is the NET work, from all forces together, that equals ΔK.

Unit 3: Work, Energy, and Power Topic 3.2, 3.4

U_s

elastic potential energy in the spring (J)

k

spring constant (N/m)

Δx

stretch or compression (m)

Use it when: Energy stored in a stretched or squashed spring (launchers, bouncing blocks, oscillators).

Do not use it when: Δx is from the natural length, not from where the block was hung.

Example: k = 200 N/m compressed 0.10 m: U_s = ½(200)(0.10)² = 1.0 J.

Trap: Double the compression = four times the energy.

Unit 3: Work, Energy, and Power Topic 3.3, 7.4

U_G

gravitational potential energy of two masses (J)

r

center-to-center distance (m)

Use it when: Satellites and planets, far from the surface, where g is not constant. Zero is set at infinite separation.

Do not use it when: Near the surface use ΔU_g = mgΔy instead.

Example: Earth-Moon: U = −(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²) / (3.84 × 10⁸) ≈ −7.6 × 10²⁸ J.

Trap: It is negative and goes as 1/r (not 1/r²). Moving apart makes it LESS negative, so it increases.

Unit 3: Work, Energy, and Power Topic 3.3, 6.6

ΔU_g

change in gravitational potential energy (J)

m

mass (kg)

g

9.8 m/s²

Δy

change in height (m), up positive

Use it when: Objects near Earth's surface going up or down: ramps, roller coasters, dropped balls.

Do not use it when: Far from Earth (satellites). Only the CHANGE in height matters, not the path.

Example: Lifting a 2.0 kg book 1.5 m: ΔU_g = 2.0 × 9.8 × 1.5 = 29 J.

Trap: Potential energy belongs to the object-Earth SYSTEM. If Earth is not in your system, use work done by gravity instead.

Unit 3: Work, Energy, and Power Topic 3.3, 3.4

P_avg

average power (W = J/s)

W, ΔE

work done or energy changed (J)

Δt

time taken (s)

Use it when: How fast energy is transferred: motors, climbing stairs, lifting loads.

Do not use it when: Power is not energy. A small power for a long time can do a lot of work.

Example: A motor does 600 J of work in 3.0 s: P = 600 / 3.0 = 200 W.

Trap: Watch kilo: 1 kW = 1000 W. Time must be in seconds.

Unit 3: Work, Energy, and Power Topic 3.5

P_inst

power at one instant (W)

F_∥

force along the velocity (N)

v

speed at that instant (m/s)

Use it when: A force acts on something moving at a known speed (a car engine at cruising speed, towing).

Do not use it when: If the speed changes, the power at one instant is not the average.

Example: Engine force 2000 N at 25 m/s: P = 2000 × 25 = 5.0 × 10⁴ W.

Trap: At constant speed the driving force equals the resistive force, so power = (resistive force) × v.

Unit 3: Work, Energy, and Power Topic 3.5

p

momentum (kg·m/s), a vector

m

mass (kg)

v

velocity (m/s)

Use it when: Collisions, explosions, pushes. Momentum has a direction, so signs matter.

Do not use it when: Do not mix it up with kinetic energy: momentum is a vector, kinetic energy is a scalar.

Example: A 0.15 kg ball at 40 m/s: p = 0.15 × 40 = 6.0 kg·m/s in the direction of motion.

Trap: Two objects with equal and opposite momenta have total momentum zero but lots of kinetic energy.

Unit 4: Linear Momentum Topic 4.1

F_net

net force (N)

Δp

change in momentum (kg·m/s)

Δt

time (s)

Use it when: Newton's second law in momentum form: how big a force is needed to change momentum in a time.

Do not use it when: It gives the AVERAGE force over Δt if the force varies.

Example: A 0.15 kg ball goes from +40 m/s to −30 m/s in 0.010 s: F = 0.15(−30 − 40) / 0.010 = −1050 N (about 1.1 × 10³ N backward).

Trap: A bounce changes momentum MORE than a stop (the change includes reversing), so the force is bigger.

Unit 4: Linear Momentum Topic 4.2

J

impulse (N·s = kg·m/s)

F_avg

average force (N)

Δt

contact time (s)

Use it when: Short hits: bats, crashes, landing. Impulse is also the area under a force-time graph.

Do not use it when: Do not use the peak force as the average force.

Example: A 200 N average force for 0.050 s: J = 200 × 0.050 = 10 N·s, so the momentum changes by 10 kg·m/s.

Trap: Air bags and bent knees do not reduce the impulse, they make Δt longer so the force is smaller.

Unit 4: Linear Momentum Topic 4.2

v_cm

velocity of the system's center of mass (m/s)

p_i

momentum of each object

Use it when: Collisions and explosions: with no external force, v_cm stays the same before, during and after.

Do not use it when: If a net external force acts (friction, a push), v_cm changes.

Example: 2.0 kg at +3.0 m/s and 1.0 kg at −3.0 m/s: v_cm = (6.0 − 3.0) / 3.0 = +1.0 m/s. After they stick together, they move at 1.0 m/s.

Trap: Use signs for directions. Momentum is conserved in ALL collisions with no external force; kinetic energy only in elastic ones.

Unit 4: Linear Momentum Topic 4.3, 4.4

ω

angular velocity (rad/s)

ω₀

starting angular velocity (rad/s)

α

constant angular acceleration (rad/s²)

t

time (s)

Use it when: Something spinning up or down at a steady rate: wheels, fans, turntables. Same as v = v₀ + at.

Do not use it when: Angular acceleration not constant.

Example: A wheel from rest with α = 2.0 rad/s² for 5.0 s: ω = 0 + 2.0 × 5.0 = 10 rad/s.

Trap: Use radians, not degrees or revolutions. 1 rev = 2π rad; rpm ÷ 60 × 2π gives rad/s.

Unit 5: Torque and Rotational Dynamics Topic 5.1

θ

angular position (rad)

θ₀

starting angle (rad)

Use it when: Constant angular acceleration: how far round something turns in a time.

Do not use it when: Angular acceleration changes.

Example: Same wheel, after 5.0 s: θ = 0 + 0 + ½(2.0)(5.0)² = 25 rad ≈ 4.0 revolutions.

Trap: Divide radians by 2π to get revolutions.

Unit 5: Torque and Rotational Dynamics Topic 5.1

θ − θ₀

angular displacement (rad)

Use it when: Constant angular acceleration with no time given or asked.

Do not use it when: Angular acceleration changes.

Example: A fan at 10 rad/s slows at 2.0 rad/s² to rest: 0 = 10² + 2(−2.0)Δθ, so Δθ = 25 rad.

Trap: Slowing down means α has the opposite sign to ω.

Unit 5: Torque and Rotational Dynamics Topic 5.1

v

linear speed of a point (m/s)

r

distance of that point from the axis (m)

ω

angular speed (rad/s)

Use it when: Linking spinning to moving: the edge of a wheel, a point on a merry-go-round, a rope on a pulley.

Do not use it when: ω must be in rad/s, not rpm.

Example: A point 0.30 m from the axis of a wheel turning at 10 rad/s: v = 0.30 × 10 = 3.0 m/s.

Trap: Every point on a rigid object has the same ω, but points farther out have bigger v.

Unit 5: Torque and Rotational Dynamics Topic 5.2

a_T

tangential acceleration (m/s²)

α

angular acceleration (rad/s²)

Use it when: Linking angular acceleration to the speeding up of a point on the rim (or a rope wrapped on a pulley).

Do not use it when: It is not the centripetal acceleration (that is v²/r, toward the center).

Example: r = 0.30 m, α = 2.0 rad/s²: a_T = 0.60 m/s².

Trap: A point at constant ω has a_T = 0 but still has centripetal acceleration.

Unit 5: Torque and Rotational Dynamics Topic 5.2

τ

torque (N·m)

r

distance from the axis to where the force acts (m)

r_⊥

lever arm: perpendicular distance from axis to the force's line (m)

θ

angle between r and F

Use it when: Turning effect of a force: wrenches, doors, seesaws, beams on supports.

Do not use it when: A force through the axis (or along r) has zero torque. Do not use the distance along the beam if the force is slanted without the sin θ.

Example: 40 N at the end of a 0.25 m wrench at 90°: τ = 0.25 × 40 = 10 N·m. At 30°: 10 × sin 30° = 5.0 N·m.

Trap: Pick a sign: counterclockwise positive (state it). For balance, Στ = 0 about ANY point.

Unit 5: Torque and Rotational Dynamics Topic 5.3, 5.5

I

rotational inertia (kg·m²)

r_i

distance of each mass from the axis (m)

Use it when: Rotational inertia of point masses, or to compare shapes: mass far from the axis gives bigger I.

Do not use it when: For solid shapes (disk, rod, sphere) the problem gives you I; they are not on the sheet.

Example: Two 1.0 kg masses each 0.50 m from the axis: I = 2 × 1.0 × 0.50² = 0.50 kg·m².

Trap: I depends on WHERE the axis is. Distance is squared: double r = four times I.

Unit 5: Torque and Rotational Dynamics Topic 5.4

I′

rotational inertia about a new parallel axis (kg·m²)

I_cm

about the axis through the center of mass

M

total mass (kg)

d

distance between the two axes (m)

Use it when: Parallel axis theorem: you know I through the center and need it about another axis parallel to that one.

Do not use it when: The axes must be parallel and one must pass through the center of mass.

Example: A 2.0 kg, 1.0 m rod: I_cm = ML²/12 = 0.167 kg·m². About an end: 0.167 + 2.0 × 0.50² = 0.67 kg·m².

Trap: I is smallest about the center of mass. Any other parallel axis adds Md².

Unit 5: Torque and Rotational Dynamics Topic 5.4

α_sys

angular acceleration (rad/s²)

τ_net

net external torque (N·m)

I_sys

rotational inertia (kg·m²)

Use it when: Newton's second law for rotation: pulleys with mass, spinning disks, falling rods.

Do not use it when: Torques and I must be about the SAME axis.

Example: 10 N·m net torque on I = 0.50 kg·m²: α = 10 / 0.50 = 20 rad/s².

Trap: A pulley with mass has DIFFERENT tensions on its two sides; the difference makes the torque.

Unit 5: Torque and Rotational Dynamics Topic 5.6

K

rotational kinetic energy (J)

I

rotational inertia (kg·m²)

ω

angular speed (rad/s)

Use it when: Spinning objects. Rolling objects have BOTH: K = ½mv² + ½Iω².

Do not use it when: Do not forget the translational part for something rolling down a ramp.

Example: I = 0.50 kg·m² at 4.0 rad/s: K = ½(0.50)(4.0)² = 4.0 J.

Trap: Rolling races: the object with smaller I/(mR²) wins (solid sphere beats hoop), whatever its mass.

Unit 6: Energy and Momentum of Rotating Systems Topic 6.1, 6.5

W

work done by a torque (J)

Δθ

angle turned (rad)

Use it when: A constant torque turns something through an angle (winding, opening a valve, a motor spinning up).

Do not use it when: Δθ must be in radians.

Example: A 10 N·m torque turns a wheel 3.0 rad: W = 30 J, which becomes rotational kinetic energy if nothing else acts.

Trap: Area under a torque vs angle graph = work.

Unit 6: Energy and Momentum of Rotating Systems Topic 6.2

L

angular momentum (kg·m²/s)

I

rotational inertia (kg·m²)

ω

angular speed (rad/s)

Use it when: A rigid spinning object. With no external torque, L stays constant: a skater pulling in her arms spins faster.

Do not use it when: When I changes, L stays the same but rotational K does NOT stay the same.

Example: I = 0.50 kg·m² at 4.0 rad/s: L = 2.0 kg·m²/s. Halve I → ω = 8.0 rad/s.

Trap: Pulling arms in: L same, ω up, K up (the skater's muscles do work).

Unit 6: Energy and Momentum of Rotating Systems Topic 6.3, 6.4

r

distance from the chosen point to the object (m)

θ

angle between r and v

r sin θ

perpendicular distance from the point to the line of motion

Use it when: Angular momentum of a small object moving past a point (a ball hitting a rod, a kid jumping onto a merry-go-round).

Do not use it when: The point must be stated; the answer depends on it.

Example: A 0.20 kg puck at 3.0 m/s passes 2.0 m from a pivot (closest distance): L = 2.0 × 0.20 × 3.0 = 1.2 kg·m²/s.

Trap: Even an object moving in a straight line has angular momentum about a point not on that line.

Unit 6: Energy and Momentum of Rotating Systems Topic 6.3, 6.4

ΔL

change in angular momentum (kg·m²/s)

τ

net external torque (N·m)

Δt

time (s)

Use it when: Angular impulse: a torque acting for a time changes angular momentum (like J = FΔt).

Do not use it when: If the net external torque is zero, ΔL = 0.

Example: A 5.0 N·m torque for 2.0 s: ΔL = 10 kg·m²/s.

Trap: Area under a torque-time graph = change in angular momentum.

Unit 6: Energy and Momentum of Rotating Systems Topic 6.3

Δx_cm

distance the center moves (m)

r

radius (m)

Δθ

angle turned (rad)

Use it when: Rolling WITHOUT slipping. Then also v_cm = rω and a_cm = rα.

Do not use it when: Skidding or spinning in place (slipping): then the center does not move rΔθ.

Example: A 0.35 m radius wheel turns once (2π rad): Δx = 0.35 × 2π ≈ 2.2 m.

Trap: Rolling without slipping: static friction does no work, so mechanical energy is conserved.

Unit 6: Energy and Momentum of Rotating Systems Topic 6.5

T

period: time for one full cycle (s)

f

frequency: cycles per second (Hz)

Use it when: Any repeating motion: springs, pendulums, circular motion.

Do not use it when: Not the time to go from one side to the other (that is half a period).

Example: 4.0 vibrations each second: T = 1/4.0 = 0.25 s.

Trap: Also ω = 2πf = 2π/T for circles and oscillations (not printed this way on the sheet).

Unit 7: Oscillations Topic 7.2

T_s

period of a mass on a spring (s)

m

mass on the spring (kg)

k

spring constant (N/m)

Use it when: A mass oscillating on an ideal spring (horizontal or vertical).

Do not use it when: Does NOT depend on amplitude or on g. Not for pendulums.

Example: 0.50 kg on k = 50 N/m: T = 2π√(0.50/50) = 2π(0.10) = 0.63 s.

Trap: Four times the mass doubles the period (square root).

Unit 7: Oscillations Topic 7.2

T_p

period of a simple pendulum (s)

ℓ

length of the string to the bob's center (m)

g

9.8 m/s²

Use it when: A simple pendulum with SMALL swings (under about 15°).

Do not use it when: Large angles; a physical (solid) pendulum. Mass does not appear: changing mass does not change the period.

Example: ℓ = 1.0 m: T = 2π√(1.0/9.8) ≈ 2.0 s.

Trap: On the Moon (smaller g) a pendulum clock runs SLOW (longer period).

Unit 7: Oscillations Topic 7.2

x

position from equilibrium (m)

A

amplitude (m)

f

frequency (Hz)

t

time (s)

Use it when: Simple harmonic motion that STARTS at maximum displacement (released from rest at x = A).

Do not use it when: Calculator in DEGREES: 2πft is in radians.

Example: A = 0.10 m, f = 0.50 Hz, t = 1.0 s: x = 0.10 cos(π) = −0.10 m (other side).

Trap: Set your calculator to RADIANS for this line.

Unit 7: Oscillations Topic 7.3

x

position from equilibrium (m)

Use it when: SHM that STARTS at equilibrium moving in the + direction (a block hit while at rest at the center).

Do not use it when: Starting at the end: use cosine.

Example: A = 0.10 m, f = 0.50 Hz, t = 0.50 s: x = 0.10 sin(π/2) = +0.10 m.

Trap: Speed is greatest at x = 0; acceleration is greatest at x = ±A.

Unit 7: Oscillations Topic 7.3

ρ

density (kg/m³)

m

mass (kg)

V

volume (m³)

Use it when: Comparing materials; deciding sink or float; turning a volume of fluid into a mass.

Do not use it when: Density does not change when you cut an object in half (same material).

Example: A 0.54 kg block of volume 2.0 × 10⁻⁴ m³: ρ = 0.54 / 2.0 × 10⁻⁴ = 2700 kg/m³ (aluminum).

Trap: Units! 1 cm³ = 10⁻⁶ m³ and 1 L = 10⁻³ m³. Water is 1000 kg/m³ = 1 g/cm³ (not on the sheet).

Unit 8: Fluids Topic 8.1

P

pressure (Pa = N/m²)

F_⊥

force perpendicular to the surface (N)

A

area (m²)

Use it when: Force spread over an area: snowshoes, nails, hydraulic lifts, the force of a fluid on a wall or window.

Do not use it when: Pressure is a scalar; it has no direction. Use only the perpendicular part of a slanted force.

Example: A 600 N person on two feet with 0.040 m² total: P = 600 / 0.040 = 1.5 × 10⁴ Pa.

Trap: cm² to m²: divide by 10 000. A small area means a BIG pressure.

Unit 8: Fluids Topic 8.2

P

absolute pressure at depth h (Pa)

P₀

pressure at the surface, often 1 atm = 1.0 × 10⁵ Pa

ρ

density of the FLUID (kg/m³)

h

depth below the surface (m)

Use it when: Pressure in a still fluid at a depth: divers, dams, submarines, U-tubes.

Do not use it when: Moving fluids (use Bernoulli). The object's density does not belong here, only the fluid's.

Example: 10 m down in fresh water: P = 1.0 × 10⁵ + 1000 × 9.8 × 10 = 1.98 × 10⁵ Pa ≈ 2 atm.

Trap: h is depth (measured down from the surface), not height. Container shape does not matter: same depth, same pressure.

Unit 8: Fluids Topic 8.2

P_gauge

pressure above atmospheric (Pa)

Use it when: A tire gauge, a blood pressure cuff, "how much extra pressure" questions. Absolute = gauge + atmospheric.

Do not use it when: If asked for absolute (total) pressure, add P₀.

Example: 10 m down in fresh water: P_gauge = 1000 × 9.8 × 10 = 9.8 × 10⁴ Pa.

Trap: Read the question for the word "absolute" or "gauge". They differ by 1.0 × 10⁵ Pa.

Unit 8: Fluids Topic 8.2

F_b

buoyant force, upward (N)

ρ

density of the FLUID (kg/m³)

V

volume of fluid displaced = submerged volume of the object (m³)

Use it when: Archimedes: anything in a fluid (fully or partly under) gets pushed up by the weight of the fluid it pushes aside.

Do not use it when: Do not use the object's density or the object's whole volume when it only floats partly under.

Example: A 1.0 × 10⁻³ m³ (1 L) rock fully under water: F_b = 1000 × 1.0 × 10⁻³ × 9.8 = 9.8 N.

Trap: Floating means F_b = mg. Fully under, F_b does not change with depth.

Unit 8: Fluids Topic 8.3

A

cross-sectional area of the pipe (m²)

v

flow speed there (m/s)

Use it when: Continuity: an incompressible fluid in a full pipe. Narrower pipe → faster flow.

Do not use it when: Gases being compressed, or a pipe that splits (then the total flow in = total flow out).

Example: A hose with A = 2.0 × 10⁻⁴ m² at 1.5 m/s into a nozzle of 0.50 × 10⁻⁴ m²: v₂ = (2.0 × 1.5) / 0.50 = 6.0 m/s.

Trap: Area of a circle is πr²: halving the DIAMETER makes the area ¼ and the speed four times bigger.

Unit 8: Fluids Topic 8.4

P

pressure (Pa)

ρ

fluid density (kg/m³)

y

height of each point (m)

v

flow speed (m/s)

Use it when: Bernoulli: energy conservation for an ideal fluid moving along a streamline (pipes, tanks with holes, nozzles).

Do not use it when: Viscous (sticky) fluids, turbulence, or two points not on the same flow.

Example: Horizontal pipe, water: P₁ = 1.5 × 10⁵ Pa, v₁ = 2.0 m/s, v₂ = 6.0 m/s: P₂ = 1.5 × 10⁵ + ½(1000)(2.0² − 6.0²) = 1.34 × 10⁵ Pa.

Trap: Faster flow at the same height means LOWER pressure. Open to air: P = P_atm. A big tank's top surface: v ≈ 0.

Unit 8: Fluids Topic 8.4