AP Physics C Mechanics Quick Review

Velocity and acceleration are formally defined as instantaneous rates of change of position, so calculus --- not just algebraic kinematics formulas --- is the primary tool for one-dimensional motion with non-constant acceleration.

$v(t)=/dxdt$, $a(t)=/dvdt=/d^2xdt^2$. On a position-time graph, slope $=$ velocity; on a velocity-time graph, slope $=$ acceleration. The object turns around when $v(t)=0$ while $a(t)≠ 0$ at that instant. Speed is $|v(t)|$.

Instantaneous velocity ($dx/dt$ at one instant) vs. average velocity ($ x/ t$ over an interval) --- equal only at special instants or when velocity is constant.

Setting $a(t)=0$ to find maximum speed instead of finding where $v(t)$ is extremal --- extrema of $v(t)$ occur where $dv/dt=0$ with a sign change, not where $v(t)=0$ (that condition locates a turning point of POSITION, not maximum velocity).

$v=dx/dt$, $a=dv/dt=d^2x/dt^2$ $arrow$ slopes of $x$-$t$ and $v$-$t$ graphs.

Since velocity and acceleration are derivatives of position, recovering position or velocity from $a(t)$ requires integrating, with initial conditions fixing the constant of integration.

$v(t)=v_0+_0^t a(t') dt'$; $ x(t)=x_0+_0^t v(t') dt'$. Displacement $=$ signed area under a $v$-$t$ graph; $ v=$ signed area under an $a$-$t$ graph. For constant $a$: $v=v_0+at$, $x=x_0+v_0t+/12 at^2$, $v^2=v_0^2+2a x$.

A definite integral (area between two times, gives a specific $ x$ or $ v$) vs. an antiderivative (gives $x(t)$ or $v(t)$ as a function, with a constant fixed by initial conditions).

Applying the constant-acceleration formulas ($v=v_0+at$, etc.) when $a(t)$ is explicitly given as a non-constant function of time --- those shortcuts require CONSTANT $a$; with varying $a(t)$ you must integrate directly.

$ v= a dt$, $ x= v dt$ $arrow$ area under the curve, not slope.

When acceleration depends on velocity or position rather than time (drag forces, springs), Newton's second law becomes a differential equation solved by separation of variables or the identity $a=v dv/dx$.

Chain-rule trick: $a=/dvdt=/dvdx/dxdt=v/dvdx$, useful when $a$ is given as a function of $x$. Linear drag: $m/dvdt=-bv ⇒ /dvv=-/bm dt ⇒ v(t)=v_0e^-bt/m$ --- the signature exponential-decay solution of $dy/dt=-ky$.

$a(t)$ (acceleration as an explicit function of time --- integrate directly) vs. $a(x)$ or $a(v)$ (depends on position or velocity --- requires separation of variables or the $v dv/dx$ substitution first).

Trying to integrate $a(x)$ directly with respect to $t$ as if it were $a(t)$ --- you must first convert using $v dv=a(x) dx$ before integrating.

$a=v(dv/dx)$ when $a$ depends on $x$; separate variables to solve velocity- or position-dependent force problems.

Two-dimensional projectile motion is two independent one-dimensional kinematics problems happening simultaneously --- constant velocity horizontally, constant acceleration $g$ downward vertically --- linked only by a shared time variable.

$x(t)=x_0+v_0 t$; $y(t)=y_0+v_0 t-/12 gt^2$. Time of flight (level ground): $t=2v_0/g$. Range: $R=/v_0^2(2)g$, maximized at $=45^$. At the peak, $v_y=0$ but $v_x=v_0≠0$, and $a_y=-g$ throughout.

$R=v_0^2(2)/g$ (valid only when launch and landing heights are EQUAL) vs. unequal launch/landing height problems, which require solving the full $y(t)=0$ quadratic for time.

Assuming acceleration is zero at the top of the trajectory because $v_y=0$ there --- acceleration is $g$ downward for the ENTIRE flight, including the peak.

Horizontal: constant $v$. Vertical: constant $a=-g$. Same $t$ links them; peak has $v_y=0$, $a_y=-g$ still.

Newton's second law, $F=ma$, is a vector equation applied component-by-component after isolating each object with a correct free-body diagram; the first law is the special case $F=0$.

$ F_x=ma_x$, $ F_y=ma_y$, applied independently per axis (often rotated along an incline). Equilibrium (rest or constant velocity) $⇒ F=0$. Draw only forces acting ON the object (gravity, normal, tension, applied, friction) --- never forces it exerts on something else.

Mass (scalar, resistance to acceleration, constant) vs. weight (vector force $mg$, varies with the local gravitational field).

Drawing a "force of motion" in the direction of velocity on a free-body diagram --- velocity is NOT a force; net force can point anywhere relative to velocity, even opposite to it (deceleration).

$ F=ma$ per axis; draw ONLY real forces on the object, never velocity or "$ma$" itself.

Every force comes in an action-reaction pair acting on two DIFFERENT objects, while friction opposes relative sliding (kinetic) or impending sliding (static), bounded by $ N$.

$F_A on B=-F_B on A$ (equal magnitude, opposite direction, same force type, act on different objects). Static friction: $f_s≤q _s N$ (self-adjusting up to the max needed to prevent sliding). Kinetic friction: $f_k=_k N$ (constant once sliding, opposes relative motion).

A third-law pair (equal/opposite forces on TWO different objects, never both on the same free-body diagram) vs. two forces that happen to balance on ONE object (e.g., normal force and gravity on a resting block --- NOT a third-law pair).

Using $f_s=_s N$ as an equation instead of an inequality --- static friction takes WHATEVER value (up to $_s N$) is needed to prevent sliding; plugging in $_s N$ when the object isn't on the verge of slipping gives the wrong force.

3rd-law pairs act on different objects; $f_s≤q_sN$ (inequality), $f_k=_kN$ (equality).

A system of particles (or extended rigid body) moves, on average, as if all its mass were concentrated at the center of mass --- a mass-weighted average position found by a sum for discrete masses or an integral for continuous ones.

Discrete: $x_cm=/ m_ix_i m_i$. Continuous: $x_cm=/1M x dm$, with $dm$ expressed via a density function (e.g., $dm= dx$, $=M/L$ for a uniform rod). A uniform symmetric object's center of mass lies at its geometric center.

Center of mass (mass-weighted average position, depends on mass distribution) vs. geometric center/centroid (shape-based average) --- these coincide only when density is uniform.

Forgetting to rewrite $dm$ in terms of the integration variable (e.g., $dm=(M/L) dx$) before integrating $x_cm=(1/M) x dm$ --- the integral cannot be evaluated until $dm$ is expressed in terms of $dx$.

$x_cm=(1/M) x dm$; uniform density $arrow dm=(M/L) dx$ (or the area/volume analog).

Objects connected by an ideal (massless, inextensible) string over an ideal (massless, frictionless) pulley share a common magnitude of acceleration, letting you write one $ F=ma$ equation per object and solve the system simultaneously.

Ideal string/pulley $⇒$ same tension throughout and same $|a|$ for all connected objects. Atwood machine ($m_1$, $m_2$ over a pulley): $a=/(m_1-m_2)gm_1+m_2$, $T=/2m_1m_2gm_1+m_2$. Frictionless incline: $a=g$ along the surface; rotate axes parallel/perpendicular to the incline.

Tension in an ideal string over a pulley (same magnitude throughout) vs. normal force on an incline (perpendicular to the surface, generally $≠ mg$ if there is also acceleration perpendicular to the surface).

Using different symbols/values for the acceleration on each side of a pulley system and forgetting they must share the same MAGNITUDE --- fix one positive direction for the whole system before writing equations.

Ideal string+pulley $arrow$ same $T$, same $|a|$; rotate axes along the incline to separate components.

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

$a_c=/v^2r=^2r$, directed toward the center. $F_net,c=/mv^2r=m^2r$ --- not a new force, but whatever combination of real forces (tension, gravity, normal, friction) supplies the inward net force. $v=2 r/T$.

Centripetal force (the net inward force required for circular motion, $=mv^2/r$, supplied by real forces) vs. "centrifugal force" (fictitious, appears only in a rotating non-inertial frame --- never draw it on an inertial free-body diagram).

Adding an extra outward "centrifugal" force to an inertial-frame free-body diagram, or double-counting by drawing $F_c$ separately from the real forces that already sum to it --- $ F_real=mv^2/r$ IS the equation, not an additional force.

$a_c=v^2/r$ toward center; $ F_real=ma_c$, never a separate "centrifugal" force.

Newton's universal law of gravitation gives the inverse-square attraction between two masses, and this same force supplies the centripetal force for orbits.

$F_g=/Gm_1m_2r^2$, attractive, along the line joining the masses. Gravitational field: $g(r)=/GMr^2$ (reduces to $g≈ 9.8 m/s^2$ at Earth's surface, $r=R_E$). Circular orbit: $/GMmr^2=/mv^2r⇒ v=sqrtGM/r$.

$g≈9.8 m/s^2$ near Earth's surface (a local constant, short-range problems) vs. $g(r)=GM/r^2$ (the true distance-dependent field, required whenever altitude is comparable to $R_E$ or for any orbital problem).

Using altitude above the surface instead of distance from the CENTER of the planet in $F_g=Gm_1m_2/r^2$ --- $r$ is always measured from the center of mass of the attracting body; add the planet's radius to the altitude.

$F_g=Gm_1m_2/r^2$, $g(r)=GM/r^2$; $r$ measured from the planet's CENTER, always.

Work generalizes "force times distance" to variable forces via a line integral, so a "mathematical routines" FRQ often asks you to integrate a given $F(x)$ directly for work.

$W=_x_i^x_fF(x) dx$ along a line (more generally $W=F· dr$). Constant force: $W=Fd$, $$ the angle between force and displacement. Graphically, work $=$ signed area under an $F$-$x$ graph.

Work done BY a specific force (use only that force in the integral) vs. NET work (use $ F$, or sum the work done by each individual force).

Forgetting the $$ factor when force and displacement aren't parallel --- a force perpendicular to displacement (normal force during horizontal motion, tension in circular motion) does ZERO work, a very common exam setup.

$W= F dx$ (variable force) $=Fd$ (constant force); perpendicular force $arrow$ zero work.

The net work done on an object equals its change in kinetic energy --- a direct calculus consequence of integrating Newton's second law along the path of motion.

$W_net= KE=/12mv_f^2-/12mv_i^2$. Derivation: $W_net= F_net dx= ma dx= m/dvdtdx= mv dv=/12mv_f^2-/12mv_i^2$ (using $dx/dt=v$). Holds even for non-constant acceleration, since it is built directly from integration.

The work-energy theorem (relates NET work to change in KINETIC energy only, always true) vs. conservation of mechanical energy (relates non-conservative work to change in KE$+U$ --- a different, more restrictive bookkeeping).

Applying $W_net= KE$ using only ONE force's work (e.g., just gravity) while calling it "net work" --- every force acting on the object must be included, unless the problem explicitly asks for the work of just one force.

$W_net= F dx= KE$ --- derived from $F=ma$ via $a dx=v dv$.

A conservative force can be written as minus the derivative of a potential energy function, since its work depends only on endpoints, not path --- trading a force calculation for a simpler energy calculation.

$F=-/dUdx$ (1D); force points toward decreasing $U$. $U_spring=/12kx^2$ (from Hooke's law $F=-kx$). $U_grav=mgy$ near a surface; general: $U_grav=-/GMmr$ (defined so $U→0$ as $r→∞$). Conservative: gravity, spring, electrostatic force; non-conservative: friction, air resistance, applied pushes.

$U_grav=mgy$ (valid only near a planet's surface, roughly constant $g$) vs. $U_grav=-GMm/r$ (general form for large-altitude or orbital problems --- note the sign and the $1/r$, not $1/r^2$, dependence).

Dropping the negative sign when going between $U(x)$ and $F=-dU/dx$ --- writing $F=+dU/dx$ reverses the force direction relative to the potential slope.

$F=-dU/dx$; $U_spring=/12kx^2$; $U_grav$ (surface) $=mgy$; $U_grav$ (general) $=-GMm/r$.

When only conservative forces do work on a system, total mechanical energy (KE$+U$) is conserved; any non-conservative work changes that total by exactly the amount of work it does.

Only conservative forces act: $KE_i+U_i=KE_f+U_f$. General bookkeeping: $KE_i+U_i+W_nc=KE_f+U_f$, where $W_nc$ is the work done by non-conservative forces (often negative for friction). Energy sloshes between KE and $U$ in a spring/pendulum system while the total stays fixed absent friction/drag.

Conservation of ENERGY (always true overall --- energy transforms, e.g., into heat) vs. conservation of MECHANICAL energy (KE$+U$ specifically --- true only when non-conservative forces do zero net work).

Setting $KE_i+U_i=KE_f+U_f$ on a problem that explicitly includes friction or air resistance --- the $W_nc$ term (typically negative) must be included, or the answer overstates the final speed.

$KE+U=$const only if $W_nc=0$; otherwise $KE_i+U_i+W_nc=KE_f+U_f$.

Power measures the RATE at which work is done or energy is transferred, and for a force acting on a moving object it reduces to a clean dot product of force and velocity.

$P_avg=W/ t$; instantaneous $P=/dWdt=F·v=Fv$. For a motor delivering constant power against a resistive force, $P=Fv$ lets you solve for the changing force or terminal velocity as speed changes.

Work (a scalar total, in joules, accumulated over a process) vs. power (a rate, in watts $=$ J/s, describing how FAST that work is done --- the same work done twice as fast is twice the power, not twice the work).

Assuming constant force when a problem states constant POWER --- with $P$ constant, $F=P/v$ changes as $v$ changes, so constant-acceleration kinematics do not apply; set up $P=Fv=mv(dv/dt)$ and integrate instead.

$P=dW/dt=F·v$; constant power $⇒$ changing force ($F=P/v$), not constant acceleration.

Impulse --- the time-integral of net force --- equals the change in momentum, making it the calculus generalization of Newton's second law for forces that vary over an interaction time.

$J= F dt= p=m v$ (constant mass). Graphically, impulse $=$ area under an $F$-$t$ graph. For a large force over a short time (impacts), $J≈ F_avg t$ is useful when average force is given/asked for.

Impulse ($ F dt$, a vector, changes momentum, units N$·$s) vs. work ($ F dx$, a scalar, changes kinetic energy, units J) --- same-looking integrals with respect to different variables, producing different quantities.

Dropping direction/sign in a collision or bounce problem --- velocity (and hence momentum and impulse) reversing direction is a SIGN CHANGE, not just a magnitude difference; $ p$ for a bouncing ball must treat "before" and "after" as vectors.

$J= F dt= p$; area under $F$-$t$ graph $=$ impulse; mind the sign on reversed velocities.

For an isolated system (zero net external force), total momentum is conserved --- a direct consequence of Newton's third law, since internal forces between system members always cancel in pairs.

$ p_i,initial= p_i,final$ per component, whenever $ F_ext=0$ (or $≈0$ during a brief collision, even with gravity/friction present, since impulsive internal forces dominate). Holds regardless of whether kinetic energy is conserved.

Conservation of momentum (requires $ F_ext=0$, holds in ALL collisions) vs. conservation of kinetic energy (an EXTRA condition, true only in elastic collisions).

Forgetting momentum is a VECTOR --- in 2D collisions, the $x$-component and $y$-component must be conserved SEPARATELY; conserving only speed/magnitude loses information and gives wrong answers.

$ p$ before $= p$ after (vector, per component) whenever $F_ext≈0$; true for ALL collisions.

All collisions conserve momentum, but only ELASTIC collisions also conserve kinetic energy; perfectly inelastic collisions are the other extreme, where colliding objects stick together.

Perfectly inelastic: $m_1v_1+m_2v_2=(m_1+m_2)v_f$. Elastic (1D), from conserving both $p$ and KE: $v_1f=/m_1-m_2m_1+m_2v_1i+/2m_2m_1+m_2v_2i$ (symmetric for $v_2f$). Two-dimensional collisions require conserving $p_x$ and $p_y$ as separate equations.

Elastic collision (KE conserved, objects bounce apart) vs. perfectly inelastic collision (KE NOT conserved --- energy converts to heat/deformation, objects move together afterward).

Assuming KE is conserved by default in a collision problem that doesn't say "elastic" --- most AP collision problems are inelastic (or ask you to classify by comparing KE before/after); never assume elasticity unless stated or derivable.

Momentum conserved in ALL collisions; KE conserved ONLY if elastic; "stick together" $=$ perfectly inelastic.

The center of mass of any system of particles obeys Newton's second law using only the NET EXTERNAL force, moving as if all the system's mass were concentrated there --- regardless of internal complications.

$F_net,ext=Ma_cm$, where $a_cm=d^2x_cm/dt^2$. Total momentum $P=Mv_cm$, so if $F_ext=0$, $v_cm$ is constant (consistent with momentum conservation). Internal forces (collisions, explosions) never change the center of mass's trajectory, only how mass is distributed around it.

Motion of individual particles/fragments (can be complicated, involve internal forces) vs. motion of the system's center of mass (always simple --- follows $F_ext=Ma_cm$; a projectile that explodes mid-flight has its CM continue along the ORIGINAL parabola).

Assuming an internal explosion or collision changes the center of mass's path --- as long as no new external force appears, the CM continues exactly along its pre-existing trajectory, even as fragments scatter.

$F_ext=Ma_cm$; internal forces (explosions/collisions) never move the center of mass off its original path.

Torque is the rotational analog of force --- a force's effectiveness at producing angular acceleration about a pivot --- and an object is in rotational equilibrium when net torque about ANY axis is zero.

$=r×F$, magnitude $=rF=r_ F$ ($r_$ is the lever arm, the perpendicular distance from pivot to the force's line of action). Static equilibrium requires BOTH $F=0$ and $=0$ simultaneously. Convention: counterclockwise torque is typically positive.

Force (causes linear acceleration, $ F=ma$) vs. torque (causes angular acceleration, $=I$) --- an object can have zero net force but nonzero net torque (spins in place), or vice versa.

Using the full distance $r$ instead of the perpendicular lever arm $r_=r$ when the force isn't applied perpendicular to the position vector --- always resolve the perpendicular component before multiplying.

$=r_ F=rF$; equilibrium needs $ F=0$ AND $=0$, about any chosen pivot.

Rotational kinematics is a direct angular analog of linear kinematics, with $$, $$, $$ related by the same derivative/integral and constant-acceleration relationships as $x$, $v$, $a$.

$=d/dt$, $=d/dt=d^2/dt^2$. Constant-$$ analogs: $=_0+ t$, $=_0+_0t+/12 t^2$, $^2=_0^2+2$. Linear-angular link at radius $r$: $v= r$, $a_tan= r$, $a_c=^2r$ --- a rotating point generally has BOTH tangential and centripetal acceleration.

Tangential acceleration $a_tan= r$ (changes rotation SPEED, tangent to the path) vs. centripetal/radial acceleration $a_c=^2r=v^2/r$ (changes DIRECTION, always toward the center) --- total acceleration is their vector sum.

Reporting only centripetal acceleration for an object whose angular speed is CHANGING ($≠0$) --- when rotation speeds up or slows down, total acceleration must include BOTH the tangential ($ r$) and centripetal ($^2r$) components.

$=d/dt$, $=d/dt$; $v= r$, $a_tan= r$, $a_c=^2r$ --- combine both when $≠0$.

Moment of inertia is the rotational analog of mass --- a measure of resistance to angular acceleration --- but unlike mass, it depends on HOW mass is distributed relative to the rotation axis, computed via an integral.

$I= r^2 dm$ (continuous) or $I= m_ir_i^2$ (discrete), $r$ = distance from the axis. Common shapes about center: solid sphere $I=/25MR^2$; solid disk/cylinder $I=/12MR^2$; hoop $I=MR^2$; thin rod about center $I=/112ML^2$; thin rod about end $I=/13ML^2$. Parallel axis theorem: $I=I_cm+Md^2$, $d$ = distance from the CM axis to the new parallel axis.

Moment of inertia about the CENTER OF MASS ($I_cm$, the minimum for that axis direction) vs. about a parallel off-center axis (always larger --- add $Md^2$ via the parallel axis theorem, never subtract).

Applying $I=I_cm+Md^2$ starting from an $I$ that is NOT already about the center of mass --- the theorem only works when the known $I$ is $I_cm$; you cannot shift directly between two arbitrary parallel axes.

$I= r^2 dm$; parallel axis: $I=I_cm+Md^2$ (only from the CM axis outward).

Just as $ F=ma$ governs linear acceleration, $=I$ governs angular acceleration --- the rotational equation of motion for pulleys with mass, rotating rods, and rolling objects.

$=I$ (about a fixed axis, or about the center of mass for a rigid body). For a falling rod pivoted at one end, or a massive pulley, combine $=I$ with the linear $ F=ma$ equations for attached masses, plus the constraint $a= r$ linking a string's linear acceleration to the pulley's angular acceleration.

A massless, frictionless pulley (same tension on both sides, no rotational inertia) vs. a pulley WITH mass/moment of inertia (tension differs on each side, since $=I$ requires a net torque).

Assuming equal string tension on both sides of a pulley with nonzero mass/moment of inertia --- with $I≠0$, $(T_1-T_2)R=I$ requires $T_1≠ T_2$ in general; only an idealized MASSLESS pulley has equal tension.

$=I$; massive pulley $arrow T_1≠ T_2$, linked to blocks via $a= R$.

A rotating object stores kinetic energy in its spin, with moment of inertia playing the role mass plays in translational KE, and a rolling object has BOTH translational and rotational kinetic energy simultaneously.

$KE_rot=/12I^2$. Rolling without slipping (radius $R$, moment of inertia $I$ about its own center): $KE_total=/12Mv_cm^2+/12I^2=/12Mv_cm^2≤ft(1+/IMR^2)$, using $v_cm= R$. Larger $I/(MR^2)$ (hoops) accelerates more slowly down an incline than smaller $I/(MR^2)$ (solid spheres), since more energy goes into rotation.

$KE_trans=/12Mv_cm^2$ (motion of the center of mass) vs. $KE_rot=/12I^2$ (spin about the center of mass) --- a rolling object's total KE is the SUM of both, not just one.

Using only $/12Mv^2$ for a rolling object's kinetic energy in an energy-conservation problem --- omitting the $/12I^2$ term overstates the final translational speed, since some PE also converts to spin.

$KE_rolling=/12Mv^2+/12I^2=/12Mv^2(1+I/MR^2)$; rolling constraint $v= R$ links them.

Angular momentum is the rotational analog of linear momentum, conserved whenever net external torque is zero --- the key tool for problems where a rotating system's shape (and moment of inertia) changes.

Rigid body about a fixed axis: $L=I$. Point particle: $L=r×p$, magnitude $L=rp$. Rotational Newton's second law: $_net=dL/dt$. If $_net,ext=0$: $L$ conserved, $I_1_1=I_2_2$ (e.g., a spinning skater or rotating platform whose moment of inertia changes as mass redistributes).

Conservation of angular momentum ($I_1_1=I_2_2$, requires $_ext=0$, holds even though $KE_rot$ generally CHANGES when $I$ changes) vs. conservation of rotational kinetic energy (a separate, stronger condition NOT automatic even when $L$ is conserved).

Assuming rotational kinetic energy is conserved along with angular momentum when $I$ changes --- $KE_rot=/12I^2=L^2/(2I)$ actually INCREASES as $I$ decreases at fixed $L$ (e.g., a skater pulling in their arms), since internal work is done.

$L=I=rp$; $_ext=0arrow L$ conserved ($I_1_1=I_2_2$), even though $KE_rot$ often changes.

"Rolling without slipping" is a kinematic CONSTRAINT linking translational and rotational motion --- the contact point is instantaneously at rest relative to the surface --- and it is what allows static friction to act without doing work.

Rolling constraint: $v_cm= R$, $a_cm= R$. The contact point has zero instantaneous velocity relative to the ground; static friction supplies the torque needed to maintain this constraint (and does zero work). Object rolling down an incline from rest: $a_cm=/g1+I_cm/(MR^2)$ --- smaller $I/(MR^2)$ reaches the bottom faster.

Rolling WITHOUT slipping ($v_cm= R$ exactly, STATIC friction, no energy loss) vs. rolling WITH slipping ($v_cm≠ R$, KINETIC friction acts, energy lost to heat) --- $v= R$ is valid only in the no-slip case.

Applying $v_cm= R$ to an object that starts out slipping (e.g., a bowling ball thrown with no initial spin) --- the constraint only holds once/if the object transitions to pure rolling; before that, treat $v_cm$ and $$ as independent.

No slip: $v= R$, $a= R$ (static friction, no loss); slipping: $v≠ R$ (kinetic friction, energy lost).

A bound orbit is governed by the same gravitational force law as any two-body attraction, but becomes tractable by combining Newton's law of gravitation with conservation of angular momentum and energy, especially for elliptical orbits.

Circular orbit: $/GMmr^2=/mv^2r⇒ v=sqrtGM/r$, period $T=2r^3/GM$ (Kepler's third law: $T^2 r^3$). Circular orbit energy: $E=KE+U=/12mv^2-/GMmr=-/GMm2r$ (negative $=$ bound). Elliptical orbits: $L=mvr$ is conserved at every point (Kepler's second law: equal areas in equal times), and $E=-GMm/(2a)$ ($a=$ semi-major axis) is conserved throughout.

Kepler's second law (equal areas swept in equal times --- angular momentum conservation, so speed is NOT constant on an ellipse, fastest at perigee) vs. uniform circular orbital motion (constant speed, true only for perfectly circular orbits).

Assuming orbital speed is constant for an elliptical orbit --- only circular orbits have constant speed; elliptical orbits speed up near perigee and slow down near apogee, governed by $L=mvr=$const, not $v=$const.

Circular orbit: $v=sqrtGM/r$, $E=-GMm/2r$. Elliptical: $L$ and $E$ conserved, but speed varies (fastest at closest approach).

Simple harmonic motion is defined by a differential equation --- acceleration proportional to and opposite in sign from displacement --- whose solution is always sinusoidal, regardless of the physical system producing it.

Defining equation: $/d^2xdt^2=-^2x$ (from $F=-kx=ma$ with $^2=k/m$ for a spring). General solution: $x(t)=A( t+)$, $A$ = amplitude, $$ = phase constant from initial conditions. Then $v(t)=-A( t+)$, $a(t)=-A^2( t+)=-^2x(t)$. Period $T=2/$; frequency $f=1/T=/2$.

Angular frequency $$ of SHM ($=sqrtk/m$ for a spring, a mathematical parameter of the oscillation) vs. angular velocity $$ of circular motion --- same symbol/units, but SHM's $$ is not literally a rotation rate (though SHM can be visualized as the projection of uniform circular motion).

Defaulting to $x=A( t)$ regardless of initial conditions --- always determine $A$ and $$ (or choose sin vs.\ cos) from the GIVEN initial position and velocity, since the object may not start at maximum displacement.

$d^2x/dt^2=-^2x arrow x(t)=A( t+)$; $a=-^2x$ always --- the defining signature of SHM.

For an ideal spring-mass system, Newton's second law directly produces the SHM equation, and total mechanical energy sloshes continuously between spring potential energy and kinetic energy while its SUM stays constant.

$=sqrtk/m$, independent of amplitude. $T=2m/k$. Total energy: $E=/12kA^2=/12mv_max^2$ (constant); at position $x$: $/12kx^2+/12mv^2=/12kA^2⇒ v=±A^2-x^2$. Max speed at $x=0$; max acceleration/force at $x=± A$ (turning points, $v=0$ there).

Period of a spring-mass system ($T=2m/k$ --- depends on mass and spring constant, NOT amplitude) --- a key SHM feature is that period/frequency is amplitude-independent for an ideal spring obeying Hooke's law.

Assuming a larger amplitude changes the period for an ideal spring --- $T=2m/k$ contains no $A$ at all; changing amplitude changes maximum speed and energy, never the period, as long as $F=-kx$ holds exactly.

$=sqrtk/m$, $T=2m/k$ --- independent of amplitude; $E=/12kA^2=/12kx^2+/12mv^2$ throughout.

Both a simple pendulum (point mass on a massless string) and a physical pendulum (any rigid body swinging about a pivot) undergo SHM only in the SMALL-ANGLE approximation, where the restoring torque becomes proportional to angular displacement.

Simple pendulum: for small $$, $=I$ becomes $-mgL≈-mgL=mL^2/d^2dt^2$, giving $=sqrtg/L$, $T=2L/g$ (independent of mass and amplitude, small angles). Physical pendulum (rigid body pivoted a distance $d$ from its center of mass, moment of inertia $I$ about the pivot): $=sqrtmgd/I$, $T=2I/(mgd)$ --- reduces to the simple-pendulum result when $I=mL^2$ and $d=L$.

Simple pendulum $T=2L/g$ (point mass, depends only on string length $L$ and $g$) vs. physical pendulum $T=2I/mgd$ (extended rigid body, depends on its full moment of inertia $I$ about the pivot AND the distance $d$ to its center of mass).

Using $T=2L/g$ for a physical/extended pendulum (a swinging rod or meter stick) instead of $T=2I/mgd$ --- the simple-pendulum formula is only valid for an idealized point mass on a massless string.

Simple pendulum: $T=2L/g$. Physical pendulum: $T=2I/mgd$ --- both require the SMALL-ANGLE approximation to be SHM at all.

Every formula above traces back to $F=ma$ (or $=I$) via a derivative or integral --- know the derivations, not just the results.

When two formulas look alike, ask: constant or variable? conservative or not? ideal (massless) or not?

Match the KEYWORDS in a problem to the governing equation before touching a calculator.

Constants are given; every derived formula and derivation is NOT --- know how to get from $F=ma$ to everything else.

Diagram $arrow$ identify what's conserved/changing $arrow$ symbolic setup $arrow$ substitute $arrow$ justify in words.

42 MCQ (50%, 85 min) + 4 FRQ --- routines, representations, experiment, qualitative/quantitative (50%, 95 min).

Separate FBDs $arrow F=ma$ per object $arrow =I$ for anything with mass/$I$ $arrow$ link with $a= R$ $arrow$ solve simultaneously.

Isolated + no ext.\ force $arrow p$ conserved. No ext.\ torque $arrow L$ conserved. No non-conservative work $arrow$ mechanical $E$ conserved. Otherwise, track $W_nc$.