AP Physics 1 Quick Review

Kinematics describes motion without regard to its cause; displacement, velocity, and acceleration are all vectors defined relative to a chosen coordinate system and instant of time.

Displacement $ x = x_f - x_i$ (not total distance traveled). Average velocity $ v = x/ t$; instantaneous velocity is the slope of the $x$-$t$ graph at a point. Average acceleration $ a = v/ t$; instantaneous acceleration is the slope of the $v$-$t$ graph. Speed is the magnitude of velocity; sign indicates direction along the chosen axis.

Distance (total path length, always $≥ 0$) vs. displacement (net vector change in position, which can be negative or even zero after motion has occurred).

Treating negative velocity as automatically ``slowing down'' --- a negative velocity only means motion in the negative direction; the object is slowing down only when velocity and acceleration have OPPOSITE signs, and speeding up when they share the SAME sign.

Displacement = net change; compare signs of $v$ and $a$ to know speeding up vs. slowing down.

The three motion graphs are linked by slope and area: the slope of one graph gives the next, and the area under one graph gives the change in the other.

Slope of $x$-$t$ graph $=$ velocity; slope of $v$-$t$ graph $=$ acceleration. Area under $v$-$t$ graph $=$ displacement; area under $a$-$t$ graph $=$ change in velocity. A parabolic $x$-$t$ graph indicates constant nonzero acceleration; a horizontal $v$-$t$ graph indicates zero acceleration.

A $v$-$t$ graph crossing the time axis ($v=0$, object momentarily at rest but may still be accelerating) vs. an $a$-$t$ graph crossing the time axis ($a=0$, an inflection point on the $x$-$t$ graph).

Reading the steepness of a $v$-$t$ graph as ``speed,'' or interpreting a curved $x$-$t$ graph as if it were a $v$-$t$ graph --- always identify which quantity is on each axis before extracting slope or area.

$x$-$t$ slope $→ v$; $v$-$t$ slope $→ a$; $v$-$t$ area $→ x$; $a$-$t$ area $→ v$.

For constant acceleration only, four linked equations connect displacement, initial/final velocity, acceleration, and time --- pick the one that omits the variable not given.

$v = v_0 + at$; $ x = v_0 t + /12 at^2$; $v^2 = v_0^2 + 2a x$; $ x = /v_0+v2t$. These hold only when acceleration is constant in both magnitude and direction over the interval.

The kinematic ``big four'' equations (valid ONLY for constant acceleration) vs. graphical slope/area methods (needed whenever acceleration changes with time).

Plugging into a kinematic equation across an interval where acceleration changes (e.g., across a bounce, an incline change, or a friction-direction switch) --- always split the motion into separate constant-acceleration segments first.

Big four equations apply only while $a$ is constant --- split the motion at any point $a$ changes.

Projectile motion is two independent 1-D motions happening at once: constant-velocity horizontal motion and constant-acceleration ($g$) vertical motion, linked only by a shared time.

Horizontal: $v_x = v_x0$ constant; $x = v_x0t$. Vertical: $v_y = v_y0-gt$; $y = v_y0t - /12 gt^2$; $g=9.8\ m/s^2$ downward. At the top of the arc, $v_y=0$ (not $v=0$, unless launched purely vertically). Maximum range (equal launch/landing height) occurs at $45^$.

Velocity at the peak of the arc (only the vertical component is zero; horizontal component is unchanged) vs. velocity at launch/landing (full vector, both components present).

Using the full initial speed instead of the vertical component $v_0$ in the vertical equations, or reusing a time value solved from a different sub-motion --- always resolve into components first and solve the vertical equation for $t$ before applying it horizontally.

Horizontal $=$ constant velocity; vertical $=$ free fall; time is the only shared link between them.

Newton's first law defines equilibrium (constant velocity, including rest) as the natural state absent a net force; the third law guarantees every force is paired with an equal, opposite reaction force on a DIFFERENT object.

First law: if $ F = 0$, an object's velocity does not change. Third law: $ F_A→ B = - F_B→ A$ --- equal magnitude, opposite direction, same force type, acting on two different objects.

Newton's third-law pairs (act on two DIFFERENT objects, never cancel each other, exist even at rest) vs. balanced forces in equilibrium (act on the SAME object, from different sources, and can sum to zero).

Trying to cancel a third-law pair on a single free-body diagram (e.g., canceling a table's normal force on a book with the book's weight) --- these are two different forces on the SAME object, not a third-law pair; gravity-on-book's true partner is gravity-on-Earth.

3rd-law pairs: same magnitude, opposite direction, different objects, never cancel on one diagram.

A free-body diagram isolates ONE object and shows every external force acting ON it, which is the essential first step to correctly applying $ F = m a$.

$ F = m a$, applied along each axis separately: $ F_x = ma_x$, $ F_y = ma_y$. Common forces: weight ($F_g=mg$, straight down), normal force ($F_N$, perpendicular to the surface), tension ($F_T$, along a string, pulling away from the object), applied force, and friction (parallel to the surface, opposing relative sliding).

Mass (intrinsic scalar property, constant everywhere, kg) vs. weight ($F_g=mg$, a force that varies with local gravitational field, N).

Drawing the net force or an internal, non-existent ``force of motion'' on a free-body diagram instead of only the real external forces --- if a force has no identifiable agent (rope, surface, field) touching or pulling on the object, it does not belong on the diagram.

FBD = only real external forces on ONE object; then split $ F=ma$ into $x$ and $y$ separately.

Friction opposes relative sliding (kinetic) or impending sliding (static), and both are proportional to the normal force through a coefficient that depends on the two surfaces in contact.

Kinetic friction: $f_k = _k F_N$ (fixed once sliding). Static friction: $f_s ≤ _s F_N$, adjusting up to this maximum to prevent sliding --- it is NOT always at its maximum. Typically $_s > _k$ for the same surfaces.

Static friction (variable, matches the applied force up to $_sF_N$, present with NO relative sliding) vs. kinetic friction (constant at $_kF_N$, present only DURING relative sliding).

Automatically computing friction as $_sF_N$ or $_kF_N$ even when the object is in equilibrium and not on the verge of sliding --- static friction then equals whatever value keeps $ F=0$, which may be less than $_sF_N$.

Kinetic friction is fixed ($_kF_N$); static friction varies, capped at $_sF_N$.

An object moving at constant speed in a circle still accelerates because its velocity direction constantly changes; this centripetal acceleration always points toward the circle's center and requires a net inward force.

$a_c = /v^2r = ^2r$; $F_c = /mv^2r$, directed toward the center. ``Centripetal force'' is not a new force --- it names whichever real net force (tension, gravity, normal force, friction) happens to point toward the center.

Centripetal force (the real, center-pointing NET force required for circular motion) vs. ``centrifugal force'' (a fictitious outward effect felt only in the rotating, non-inertial frame --- it never appears on an inertial-frame free-body diagram).

Adding a separate outward ``centrifugal force'' arrow to a free-body diagram, or forgetting that at the top of a vertical circle gravity AND normal/tension force both point toward the center and ADD, rather than subtract --- always sum real forces along the radial direction toward the center.

Centripetal force = net inward real force; never add a fictitious outward force to an inertial-frame FBD.

Every pair of masses attracts each other with a force that depends on both masses and falls off with the square of the distance between their centers.

$F_g = /Gm_1m_2r^2$, where $G=6.67×10^-11\ N·m^2/kg^2$ and $r$ is measured center-to-center. Near a planet's surface this reduces to the local gravitational field $g = /GMr^2$, which is why $g$ decreases with altitude and with a planet's radius.

$g$ (local gravitational field strength, $9.8\ m/s^2$ near Earth's surface, changes with distance) vs. $G$ (the universal gravitational constant, the same everywhere).

Using an object's altitude above the surface instead of its distance from the planet's CENTER in $F_g=Gm_1m_2/r^2$, or doubling $r$ and forgetting the force drops by a factor of 4, not 2 --- $r$ always means center-to-center distance in an inverse-square relationship.

$F_g=Gm_1m_2/r^2$; $r$ = center-to-center; doubling $r$ quarters the force.

An ideal spring exerts a restoring force directly proportional to its displacement from natural length, always pointing back toward equilibrium.

$F_s = -kx$ (Hooke's Law), where $k$ is the spring constant (N/m) and $x$ is displacement from the spring's natural (unstretched) length. The negative sign shows the force always opposes the displacement, restoring the spring toward $x=0$.

Spring force magnitude $kx$ used in $ F=ma$ (a force, in newtons) vs. elastic potential energy $/12kx^2$ (an energy, in joules) --- do not substitute one where the other is required.

Forgetting the restoring direction of $F_s=-kx$ and drawing it the wrong way on a free-body diagram, or measuring $x$ from the wrong origin --- $x$ must be measured from the spring's own natural length, never from the ground or another fixed point.

$F_s=-kx$ always points back toward the spring's natural length --- restoring, not repelling.

Work measures the energy transferred to or from an object by a force acting over a displacement, and only the force component PARALLEL to the displacement does work.

$W = Fd$, where $$ is the angle between the force and displacement vectors. Work is a scalar (positive, negative, or zero), measured in joules. $W>0$ if the force has a component along the motion; $W<0$ if it opposes motion; $W=0$ if perpendicular to displacement.

Work done by one individual force (can be positive, negative, or zero for that force alone) vs. net work (the sum of all individual works, which equals the change in kinetic energy).

Computing $W=Fd$ without the $$ factor for an angled force, or assuming perpendicular-to-motion forces like normal force always do zero work in every scenario --- normal force does zero work on a level floor but can do nonzero work when the direction of motion is not perpendicular to it (e.g., on a curving path).

$W=Fd$ --- only the force component along the direction of motion counts.

The net work done on an object equals its change in kinetic energy --- this lets you skip force/acceleration analysis entirely when only speeds are needed.

$KE = /12mv^2$. Work-energy theorem: $W_net = KE = KE_f-KE_i$. This holds for the NET work (the sum of every force's work), not the work of any single force alone.

The work-energy theorem ($W_net= KE$, always true, involves ALL forces) vs. conservation of mechanical energy ($KE_i+PE_i=KE_f+PE_f$, true only when no non-conservative force does work).

Applying $W_net= KE$ using only ONE force's work (e.g., only gravity) when friction or an applied force also does work --- $W_net$ must include the work of every force acting on the object.

$W_net= KE$ --- always true, but ``net'' means every force's work, summed.

Potential energy stores the capacity to do work based on an object's position in a force field (gravity) or a system's configuration (a stretched spring), defined relative to a chosen reference point.

Gravitational PE (near Earth's surface): $PE_g = mgh$, with $h$ measured above a chosen reference level. Elastic PE: $PE_s = /12kx^2$, with $x$ measured from the spring's natural length. Only CHANGES in $PE$ are physically meaningful; the zero reference is arbitrary.

$PE_g=mgh$ (linear in $h$, near-surface approximation) vs. $PE_s=/12kx^2$ (quadratic in $x$) --- doubling height doubles gravitational PE, but doubling spring stretch quadruples elastic PE.

Measuring $h$ from an inconsistent reference level between an object's initial and final states (e.g., from the table for $PE_i$ but from the floor for $PE_f$) --- fix ONE reference height and use it throughout the whole problem.

$PE_g=mgh$ (linear in $h$); $PE_s=/12kx^2$ (quadratic in $x$) --- always fix one reference point.

When only conservative forces (gravity, ideal springs) do work on a system, total mechanical energy is conserved; a non-conservative force like friction converts mechanical energy into other forms.

$KE_i+PE_i = KE_f+PE_f$ when no non-conservative work is done. With friction/applied forces present: $KE_i+PE_i+W_nc = KE_f+PE_f$, where $W_nc$ is typically negative, often $W_nc=-f_kd$ for kinetic friction.

Conservative forces (gravity, spring force --- work is path-independent, mechanical energy conserved) vs. non-conservative forces (friction, applied pushes --- work is path-dependent, mechanical energy changes).

Applying $KE_i+PE_i=KE_f+PE_f$ on a surface with friction, or assuming energy is simply ``lost'' rather than converted to thermal energy --- always check for friction or an applied force before assuming mechanical energy alone is conserved.

No friction/applied force $→ KE_i+PE_i=KE_f+PE_f$; with friction, subtract $f_kd$ from the total.

Power is the rate at which energy is transferred or work is done, distinguishing ``how much work'' from ``how fast that work is done.''

$P = /Wt = / Et$; for a constant force along the direction of motion, instantaneous power $P = Fv$ (or $Fv$ at angle $$). Units: watts (W) $=$ J/s.

Work (total energy transferred over an entire process, in joules) vs. power (the RATE of that transfer, in watts) --- two engines can do identical total work at very different powers.

Using an object's average velocity when the problem asks for instantaneous power at a specific moment, or vice versa --- $P=Fv$ requires the velocity AT that instant, not the average velocity over the whole trip.

$P=W/t=Fv$ --- power is the rate, not the total amount, of energy transfer.

Momentum quantifies ``quantity of motion'' as the product of mass and velocity, and impulse is the mechanism (force acting over time) that changes it.

$ p = m v$ (a vector, same direction as velocity). Impulse $ J = F t$ (also the area under an $F$-$t$ graph for varying force). Units of both: kg$·$m/s $=$ N$·$s.

Momentum (a property an object HAS at an instant, $p=mv$) vs. impulse (something DONE TO an object over a time interval, $J=F t$, causing a change in momentum).

Using the maximum or initial force instead of the average force in $J=F t$ for a non-constant force (e.g., a collision) --- for varying force, use the average force over $ t$, or the area under the full $F$-$t$ graph.

$p=mv$ (a snapshot); $J=F t$ (a cause) $= p$ (the effect).

The impulse-momentum theorem is the momentum analog of the work-energy theorem: net impulse on an object equals its change in momentum.

$ J_net = p = m v_f - m v_i$. This is a vector equation --- direction (sign) matters, especially when velocity reverses, as in a bounce off a wall.

A bounce (velocity REVERSES, so $ p=m(v_f-(-v_i))=m(v_f+v_i)$, a LARGE impulse) vs. a stop (velocity goes to zero, $ p=-mv_i$, a smaller-magnitude impulse) --- bouncing back requires more impulse than simply stopping.

Forgetting to assign opposite signs to ``before'' and ``after'' velocities when an object reverses direction, which understates $ p$ --- always define a positive direction first and keep every velocity's sign consistent with it.

$J= p$; a full bounce-back changes momentum by roughly DOUBLE a simple stop.

In any isolated system (zero net EXTERNAL force), total momentum is conserved even though individual objects' momenta can change dramatically, such as during a collision.

$ p_i = p_f$, i.e., $m_1v_1i+m_2v_2i = m_1v_1f+m_2v_2f$ for a two-object system, applied along each axis independently. Conservation holds regardless of whether the collision is elastic or inelastic.

Conservation of momentum (holds in EVERY collision/explosion with no net external force, elastic or not) vs. conservation of kinetic energy (holds ONLY in perfectly elastic collisions).

Ignoring an external force (friction from the ground, an external push) that makes the SYSTEM non-isolated, or applying momentum conservation to just one object instead of the total system --- momentum conservation applies to TOTAL system momentum only when net external force $≈ 0$ during the interaction.

Total system $p$ is always conserved (no net external force); total $KE$ is conserved ONLY if elastic.

Collisions are classified by whether kinetic energy is conserved, and a system's center of mass moves at constant velocity throughout any collision, since internal forces cannot change total momentum.

Elastic: momentum AND kinetic energy conserved. Inelastic: momentum conserved, kinetic energy is NOT. Perfectly inelastic: objects stick together, $v_f = /m_1v_1i+m_2v_2im_1+m_2$. Center of mass: $x_cm = / m_ix_i m_i$; its velocity is unaffected by internal collision forces.

Perfectly inelastic (objects stick, maximum kinetic-energy loss for that momentum) vs. ``just inelastic'' (objects separate after colliding but still lose some kinetic energy) --- ``inelastic'' alone does NOT mean the objects stick together.

Assuming kinetic energy is conserved by default, or computing final KE with the wrong combined mass after a perfectly inelastic collision --- always verify the collision type before assuming elastic behavior.

Elastic: $p$ AND $KE$ conserved. Inelastic: only $p$ conserved. Perfectly inelastic: objects stick, one common $v_f$.

Torque is the rotational analog of force --- it measures a force's effectiveness at causing angular acceleration, depending on the force's magnitude and how far from (and at what angle to) the axis it is applied.

$ = rF = r_ F$, where $r$ is the distance from the pivot to the point of application and $$ is the angle between $ r$ and $ F$. Rotational equilibrium: $=0$ (combined with $ F=0$ for full static equilibrium). Convention: counterclockwise torques are typically positive.

The lever arm $r_$ (perpendicular distance from the axis to the LINE OF ACTION of the force) vs. the full distance $r$ to the point of application --- torque uses only the perpendicular/effective component, not simply $rF$.

Forgetting that a force applied AT the pivot ($r=0$) or directed straight through it ($=0^$ or $180^$) produces zero torque, or mixing up clockwise/counterclockwise signs when summing --- assign a consistent positive rotational direction before summing torques.

$=rF$; a force through the pivot, or at the pivot, contributes zero torque.

Moment of inertia is the rotational analog of mass --- it measures resistance to angular acceleration, and unlike mass, it depends on HOW that mass is distributed relative to the axis of rotation.

$I = m_ir_i^2$ for point masses; common rigid-body results (given on the AP equation sheet): solid disk/cylinder about center, $I=/12MR^2$; thin hoop about center, $I=MR^2$; solid sphere, $I=/25MR^2$; thin rod about center, $I=/112ML^2$; thin rod about end, $I=/13ML^2$.

Mass (resistance to LINEAR acceleration, one fixed number) vs. moment of inertia (resistance to ANGULAR acceleration, changes with the chosen axis, even for the same object).

Using an object's moment of inertia formula for rotation about its CENTER when the problem actually pivots it about a different axis (e.g., a rod pivoted at its end) --- always match the $I$ formula to the axis described in the problem.

$I$ depends on mass distribution AND axis choice --- same object, different axis, different $I$.

Just as unbalanced force causes linear acceleration, unbalanced torque causes angular acceleration, with moment of inertia playing the role mass plays in the linear version.

$_net = I$ (rotational analog of $F_net=ma$). Linear and angular quantities link, for a point at radius $r$ from the axis, via $v=r$ and $a_t=r$ (tangential acceleration).

$F=ma$ (linear dynamics, uses mass $m$) vs. $=I$ (rotational dynamics, uses moment of inertia $I$) --- never plug torque into $F=ma$ or force into $=I$.

Substituting an object's mass in place of its moment of inertia (or vice versa) when switching between linear and rotational motion, or confusing $$ (rad/s$^2$) with linear acceleration $a$ (m/s$^2$) --- link them only through $a=r$.

$_net=I$ is the rotational twin of $F_net=ma$; link them with $a=r$.

Rotational motion with constant angular acceleration follows the exact same equation structure as linear kinematics, with angular variables substituted in.

$ = _0+ t$; $ = _0t+/12 t^2$; $^2=_0^2+2$. Angular position $$ (rad), angular velocity $$ (rad/s), angular acceleration $$ (rad/s$^2$) are the rotational analogs of $x$, $v$, $a$.

Angular velocity $$ (the SAME for every point on a rotating rigid body) vs. tangential/linear velocity $v=r$ (DIFFERENT for every point, larger farther from the axis).

Mixing degrees and radians in $$, $$, or $v=r$ --- every rotational kinematic equation and $v=r$ requires angles in RADIANS.

Same 3 equations as linear kinematics, angular symbols swapped in --- always work in radians.

A rolling object without slipping has both translational and rotational kinetic energy simultaneously, so its total kinetic energy is the sum of both forms.

$KE_rot = /12I^2$. Rolling without slipping: $KE_total = /12mv_cm^2+/12I^2$, with rolling constraint $v_cm=R$. Comparing equal-mass, equal-radius objects rolling down an incline, the one with SMALLER $I$ (mass concentrated near the axis) accelerates faster.

Rolling without slipping (static friction acts but does zero work; $v_cm=R$ holds exactly) vs. sliding/skidding (kinetic friction does negative work; $v_cm R$).

Using only $/12mv^2$ for a rolling object, ignoring rotational KE --- a rolling object always arrives slower at the bottom of an incline than a frictionless sliding object of the same mass, because some energy goes into rotation.

Rolling KE $=/12mv^2+/12I^2$; smaller $I$ (compact shape) wins a rolling race down a ramp.

Angular momentum is the rotational analog of linear momentum, quantifying ``quantity of rotational motion'' for a spinning or orbiting object.

Rigid body rotating about a fixed axis: $L=I$. Point particle: $L=mvr = mv_ r$, using the velocity component perpendicular to $ r$. Units: kg$·$m$^2$/s.

Angular momentum of a rotating rigid body ($L=I$) vs. angular momentum of a moving point particle about an external axis ($L=mvr$) --- both are called ``angular momentum'' but apply to different physical setups.

Using $L=mvr$ without the $$ factor when velocity is not perpendicular to the position vector from the axis --- only the perpendicular component of velocity (or of $r$) contributes to angular momentum about that axis.

Rigid body: $L=I$. Point particle: $L=mvr$ --- only the perpendicular component counts.

When no net external torque acts on a system, its total angular momentum stays constant even as its moment of inertia and angular velocity individually change --- the classic example is a spinning skater pulling in her arms.

$L_i = L_f ⇒ I_i_i = I_f_f$, when $_net,ext=0$. Because $I$ can change while $L$ stays fixed, $$ must change inversely: pulling mass IN (decreasing $I$) makes $$ INCREASE.

Conservation of angular momentum ($L$ conserved when $_net,ext=0$, even as $I$ and $$ both change) vs. conservation of linear momentum ($p$ conserved when $F_net,ext=0$) --- two SEPARATE laws that can each hold or fail independently in the same scenario.

Assuming $$ (or rotational KE) is conserved instead of $L$ when a skater changes body shape --- $$ changes precisely BECAUSE $L$, not $$, stays constant; rotational KE actually increases when $I$ decreases, from internal muscular work.

$I_i_i=I_f_f$ (no external torque) --- pull mass in, $I$ drops, $$ rises.

Simple harmonic motion (SHM) occurs whenever the restoring force (or torque) is directly proportional to displacement from equilibrium and points back toward it, producing sinusoidal position, velocity, and acceleration over time.

$x(t)=A( t+)$; $v(t)=-A( t+)$; $a(t)=-A^2( t+)=-^2x$. $ = 2 f = /2T$. $|v|$ is maximum at $x=0$; $|a|$ is maximum at $x=± A$.

Amplitude $A$ (maximum displacement, fixed by initial conditions, does NOT affect the period) vs. angular frequency $$ (set entirely by the system's physical properties, $k/m$ or $g/L$) --- changing amplitude does not change the period in ideal SHM.

Assuming maximum speed occurs at the extremes of motion, or maximum acceleration occurs at equilibrium --- it is the reverse: $v_max$ at $x=0$, $a_max$ at $x=± A$.

$v_max$ at $x=0$; $a_max$ at $x=± A$ --- always opposite ends of the cycle.

A mass on an ideal spring undergoes SHM because Hooke's law force is exactly the restoring-force condition, and the resulting period depends only on mass and spring stiffness --- never on amplitude.

$T = 2/mk$; $=sqrtk/m$. Total mechanical energy: $E = /12kA^2 = /12mv_max^2$, constant, continuously trading between kinetic and elastic potential energy.

Period of a mass-spring system ($T=2m/k$, depends on mass) vs. period of a simple pendulum ($T=2L/g$, independent of mass) --- doubling mass on a spring increases its period; doubling mass on a pendulum changes nothing.

Assuming a heavier mass on a spring speeds up the oscillation --- more mass means MORE inertia to overcome, so $T$ increases with $m$, it does not decrease.

Spring: $T=2m/k$ (mass matters). Total energy $E=/12kA^2$, constant throughout the cycle.

For small angles, a simple pendulum's restoring force is approximately proportional to its angular displacement, making it approximately simple harmonic with a period that depends only on length and local gravity.

$T = 2/Lg$, valid only for small angular amplitudes (typically $<15^$). $T$ does NOT depend on mass or (for small angles) on amplitude --- only on length $L$ and gravitational field strength $g$.

Pendulum period ($T=2L/g$, no mass dependence) vs. spring period ($T=2m/k$, no length/gravity dependence but does depend on mass) --- know which variables each system's period depends on.

Assuming a heavier pendulum bob swings with a different period, or applying $T=2L/g$ at large swing angles where the small-angle approximation breaks down --- period is mass-independent, and the formula loses accuracy at large amplitudes.

Pendulum: $T=2L/g$ --- mass-independent; only $L$ and $g$ matter (small angles only).

Density describes how tightly mass is packed into a volume, while pressure describes how force is distributed over an area --- both are essential to describing how fluids behave and exert forces.

$ = /mV$ (kg/m$^3$). Pressure: $P = /F_A$ (Pa $=$ N/m$^2$). Pressure at depth $h$ in a static fluid: $P = P_0+ gh$, where $P_0$ is the pressure at the surface.

Pressure (a scalar, acts equally in ALL directions at a given point, independent of a surface's orientation) vs. force (a vector, depends on the area over which it's distributed, $F=PA$).

Treating pressure as if it only pushes downward like weight, or forgetting that pressure at a given depth is the SAME in every direction --- fluid pressure at a point pushes equally on a surface no matter how that surface is tilted at that depth.

$=m/V$; $P=P_0+ gh$ --- pressure increases with depth and pushes equally in all directions.

Pascal's principle explains how enclosed fluids transmit pressure changes undiminished throughout, enabling hydraulic force multiplication, while Archimedes's principle explains the upward force on submerged or floating objects.

Pascal's principle: $/F_1A_1 = /F_2A_2$ (pressure is transmitted equally throughout an enclosed fluid). Archimedes's principle: $F_B = _fluidV_displaced g$, always upward, equal to the weight of fluid displaced by the submerged volume.

Buoyant force (depends on the FLUID's density and the volume DISPLACED, not the object's own density) vs. the object's weight (depends on the object's own mass) --- an object floats when $F_B$ from partial submersion equals its full weight; it sinks when even full submersion cannot generate enough $F_B$.

Using the object's total volume instead of only the SUBMERGED volume for a partially floating object, or using the object's own density instead of the fluid's in the buoyancy formula --- $V_displaced$ is only the submerged portion, and $$ in $F_B= Vg$ is the FLUID's density.

$F_B=_fluidV_displacedg$ (upward) --- use fluid density and submerged volume only, never the object's.

For an incompressible fluid flowing through a pipe with no leaks, the same volume of fluid must pass every cross-section per second, so the fluid speeds up where the pipe narrows.

Continuity equation: $A_1v_1 = A_2v_2$ (volume flow rate $Q=Av$ is constant along a tube of flow for an incompressible fluid). A smaller cross-sectional area REQUIRES a larger fluid speed to maintain the same flow rate.

Volume flow rate $Q=Av$ (constant throughout a single, non-branching, incompressible pipe) vs. fluid speed $v$ (changes inversely with area --- NOT constant along the pipe).

Assuming fluid speed stays the same when a pipe's cross-sectional area changes, or applying $A_1v_1=A_2v_2$ across a branching pipe without summing the flow into every branch --- the continuity equation must account for every opening the fluid can flow through.

$A_1v_1=A_2v_2$ --- narrower pipe $→$ faster flow, same volume rate.

Bernoulli's equation is a statement of energy conservation for an ideal (incompressible, non-viscous) flowing fluid, relating pressure, height, and speed along a streamline.

$P+ gy+/12 v^2 = constant along a streamline$, i.e., $P_1+ gy_1+/12 v_1^2 = P_2+ gy_2+/12 v_2^2$. At constant height, an increase in fluid speed corresponds to a DECREASE in pressure, and vice versa.

Bernoulli's equation (an ENERGY relationship for a moving, ideal fluid) vs. the static-fluid equation $P=P_0+ gh$ (a special case of Bernoulli's equation valid only when $v=0$).

Assuming pressure and speed are independent, or that higher speed always means higher pressure --- combined with the continuity equation, Bernoulli's equation shows that where a pipe narrows (higher $v$), pressure must DROP to keep the sum constant.

$P+ gy+/12 v^2=$ const --- faster flow (narrower pipe) means LOWER pressure at the same height.

One formula sheet: kinematics, $F=ma$, energy, momentum, $=I$, SHM, fluids --- know all cold.

When two terms sound alike, ask: vector or scalar? constant or variable? conserved or not?

Every force on a diagram must trace back to a real agent: gravity, a surface, a rope, a spring, a fluid, or a push/pull.

The sheet gives you equations; it never tells you WHICH one to use or HOW to set up the diagram.

Isolate $→$ align axes $→$ resolve components $→$ split into $x$/$y$ equations $→$ solve $→$ sanity-check.

42 MCQ (50%, 85 min, single-select) + 4 FRQ --- routines, translation, experiment, qual/quant (50%, 95 min).

Define states $→$ fix one reference height $→$ classify forces $→$ write the energy equation $→$ solve $→$ check.

Incline: $mg$ along, $mg$ into --- new tilted axes, and radians for every rotational formula.

If a claim ``sounds right'' but skips a condition (mass, angle, direction, elastic vs.\ inelastic), check the condition before trusting it.

Kinematics $→$ dynamics $→$ energy $→$ momentum $→$ rotation $→$ SHM $→$ fluids --- one connected toolkit, not eight isolated topics.