Physics 21 Study Guide

What Physics Is About

Physics is the science of how things move and why. PHYS 21 covers two big domains: mechanics (point particles and extended solids that translate, rotate, and collide) and fluids (liquids and gases that flow, push, and float things). Both are governed by a small handful of conservation laws — momentum, energy, angular momentum — applied with care to model the real world.

§1.1–1.2 — SI Units & Powers of Ten

Every physical quantity has a numerical value and a unit. Skipping units is the #1 source of wrong answers in physics.

The Seven SI Base Units

In PHYS 21 we use kg, m, s. Every other mechanical quantity is built from these (force = kg·m/s², energy = kg·m²/s², pressure = kg/(m·s²), etc.).

SI Prefixes You Should Know Cold

§2.1 — Dimensional Analysis

Every equation in physics must be dimensionally consistent: every term separated by + or − or = must have the same units. This is your free error checker.

Example: Sanity-Check a Formula

Suppose you wrote   v² = v₀² + 2a·t. Check: left side L²/T². Right side: L²/T² + (L/T²)(T) = L²/T² + L/T. Inconsistent — the second term is wrong (it should be 2a·Δx with units L²/T²).

Strategy for End-of-Chapter Problems

• Convert everything to SI at the start; never mix units.
• Use scientific notation for numbers < 10⁻³ or > 10⁴.
• Round only at the very end; carry 1 extra significant figure during work.
• End each problem by saying "the answer is ___ units, which makes physical sense because ___."

§2.3 — Position, Displacement, Velocity

Position x is a coordinate on a number line; choose an origin and a positive direction first. Then:

Displacement is a signed quantity (a vector in 1-D). It is not the same as distance traveled — a runner who returns to the start has displacement 0 but distance > 0.

§2.4 — Acceleration

Sign matters: positive a means the velocity is increasing in the +x direction (could be speeding up if v > 0 or slowing down if v < 0). Decelerating just means a and v point in opposite directions.

§2.5 — Constant-Acceleration Kinematics ("the Big Four")

If a is constant, integrating once gives v(t) and twice gives x(t). Memorize all four:

§2.7 — Free Fall

Near Earth's surface (and ignoring air resistance), every object falls with g = 9.80 m/s² directed downward, regardless of mass. Substitute a = ±g into the Big Four with a sign convention you choose at the start.

Worked Example

A ball is thrown straight up from y = 0 with v₀ = +20 m/s. Take +y up, so a = −g = −9.8 m/s². Find the maximum height.

At the top, v = 0. Use v² = v₀² + 2a·Δy:

Time to reach the top: t = (0 − 20)/(−9.8) ≈ 2.04 s. By symmetry, the round trip takes ~4.08 s.

Scalars vs. Vectors

A scalar has only magnitude (mass, time, temperature, energy). A vector has both magnitude and direction (displacement, velocity, force).

Vector Notation

We write vectors with arrows: v⃗, or in bold, v. Magnitude is |v⃗| or just v (no arrow). In 2-D, a vector decomposes into components along chosen axes:

§3.2 — Vector Addition (Two Methods)

Graphical (Tip-to-Tail)

Draw the first vector. Place the tail of the second at the tip of the first. The resultant R⃗ = A⃗ + B⃗ goes from the tail of the first to the tip of the last. (Order doesn't matter — vector addition is commutative.)

Component (Always Use This for Calculation)

Then |R| = √(R_x² + R_y²) and θ = tan⁻¹(R_y/R_x) (watching the quadrant).

§3.3 — Subtraction & Scalar Multiplication

Subtraction: A⃗ − B⃗ = A⃗ + (−B⃗), where −B⃗ is B⃗ flipped. Scalar multiplication: cA⃗ stretches or shrinks A⃗ by |c|, and reverses it if c < 0. Components scale: (cA)_x = cA_x.

The Dot Product (You'll Need This Soon)

The scalar product picks out the component of one vector along another:

• Parallel vectors (φ = 0): A·B = AB.
• Perpendicular (φ = 90°): A·B = 0. Used to test orthogonality.
• Anti-parallel (φ = 180°): A·B = −AB.

You'll see this in Lecture 9 as work: W = F⃗·d⃗ = Fd cos φ.

The Big Idea: Independence of x and y

In 2-D motion under constant acceleration (e.g., gravity), the x- and y-axes evolve independently. Apply the Big Four kinematic equations to each axis separately and link them only through time t.

§3.4 — Projectile Equations

Initial speed v₀ at angle θ above horizontal: v_0x = v₀ cos θ, v_0y = v₀ sin θ. Take +y up, +x in launch direction; a_x = 0, a_y = −g.

Range, Max Height, Time of Flight (level ground, y_final = y_initial)

Worked Example: Cliff Dive

A diver leaves a 30-m cliff horizontally at v₀ = 6 m/s. How far from the base does she hit the water?

Vertical: y = h − ½ g t² = 0 ⇒ t = √(2h/g) = √(60/9.8) ≈ 2.47 s. Horizontal: x = v₀ t = 6(2.47) ≈ 14.8 m.

Notice you never need to mix x and y in the algebra — they're independent until you read off "at the same time."

The Three Laws

§4.5 — Free-Body Diagrams (FBDs)

The single most useful tool in mechanics. Steps:

1. Pick the system (one object, or a connected group treated as one).
2. Replace it with a dot.
3. Draw every force acting on the dot (no internal forces).
4. Choose axes — usually one along expected motion or along a surface.
5. Decompose forces into components.
6. Apply ΣF_x = ma_x and ΣF_y = ma_y.

§4.6 — Common Forces in Mechanics

§5.1 — Friction

• Static friction adjusts up to a maximum (μ_sN) to prevent slip.
• Once slip starts, kinetic friction kicks in. Almost always μ_k < μ_s — that's why a stationary car can hold a hill but starts skidding once it begins to slide.
• μ depends on the pair of surfaces, not on contact area or normal force individually.

The Inclined-Plane Recipe

For a block of mass m on a frictionless incline of angle θ:

With friction (block sliding down): a = g(sin θ − μ_k cos θ). The block won't slide at all if μ_s ≥ tan θ.

Tension & Pulleys

A massless, frictionless pulley simply redirects the tension — magnitude is the same on both sides. A massless, inextensible rope means both ends share the same |a|.

Atwood Machine

Two masses m₁ and m₂ (m₂ > m₁) over a pulley. Let a be the acceleration of m₂ downward. Newton on each mass:

Add: a = (m₂ − m₁)g/(m₁ + m₂). Then T = 2 m₁ m₂ g / (m₁ + m₂). Limits: m₁ = m₂ → a = 0 ✓. m₂ ≫ m₁ → a → g ✓.

§6.1 — Angle & Angular Velocity

Measure angles in radians: a full circle is 2π rad, and arc length s = r θ only in radians. Angular velocity ω is rate of change of angle:

§6.2 — Centripetal Acceleration

An object moving in a circle at constant speed is still accelerating — its velocity vector keeps turning. The acceleration points toward the center (centripetal):

Worked Example: Conical Pendulum / Banked Curve

A car of mass m rounds a curve of radius r at speed v on a road banked at angle θ. Without friction: tan θ = v²/(rg). For larger v, friction must supply the extra inward force.

Topics You Must Own

• Units & Dimensions: SI base units, dimensional consistency as an error check.
• 1-D Kinematics: all four constant-acceleration equations; signed velocity and acceleration; free fall with chosen sign convention.
• Vectors: decomposition into components, magnitude/direction, tip-to-tail addition, dot product.
• 2-D & Projectiles: independent x and y motion; range/height/time formulas for level launch; complementary-angle symmetry.
• Newton's Laws: the meaning of each, free-body diagrams, action–reaction pairs, apparent weight in elevators.
• Friction: static (inequality) vs. kinetic (equality), critical-angle relation tan θ_c = μ_s.
• Inclines, tension, pulleys: tilted-axes recipe; Atwood machine.
• Circular Motion: a_c = v²/r = rω², "centripetal force" as the net inward component.

Strategy

• Always draw a picture first. Pictures cost nothing and prevent sign errors.
• Pick axes early. For inclines, tilt them. For projectiles, +y up.
• Solve algebraically; substitute numbers last. Catches more mistakes and makes limit-checking possible.
• Check a limit. "What if μ → 0? What if θ → 0?" — your answer should reduce to something familiar.
• Watch units. Convert km→m, g→kg, °→rad before plugging in.

§6.5 — Universal Gravitation

Every pair of point masses attracts along the line joining them with a force:

• G = 6.674 × 10⁻¹¹ N·m²/kg² (Cavendish, 1798).
• r is the center-to-center distance.
• For a spherical body (uniform density), gravity acts as if all the mass were at its center (shell theorem).

Surface Gravity from First Principles

§6.6 — Satellites & Kepler's Laws

For a circular orbit at radius r around a body of mass M, gravity supplies the centripetal force:

Worked Example: ISS

Altitude 400 km above Earth, so r = 6.37 × 10⁶ + 4 × 10⁵ = 6.77 × 10⁶ m. v ≈ √[(6.67×10⁻¹¹)(5.97×10²⁴)/(6.77×10⁶)] ≈ 7.67 km/s; T ≈ 5550 s ≈ 92 minutes. Matches reality.

Kepler in Words

• 1st Law: Planets orbit on ellipses with the Sun at one focus.
• 2nd Law: Equal areas swept in equal times (a consequence of angular momentum conservation, Lecture 14).
• 3rd Law: T² ∝ r³. Constant of proportionality only depends on the central mass.

Non-Uniform Circular Motion (Discussion 4 Topic)

If speed is changing in addition to direction, total acceleration has two components:

|a| = √(a_c² + a_t²), and a points inward and forward/backward by the angle tan⁻¹(a_t/a_c).

§7.1 — Work Done by a Constant Force

• Work is a scalar measured in joules (J = N·m = kg·m²/s²).
• φ is the angle between F⃗ and d⃗. F ⊥ d → W = 0 (e.g., gravity does no work on horizontal motion; centripetal force does no work, ever).
• Sign of W matches the sign of cos φ: pushing along motion does + work; friction does − work.

§7.2 — Kinetic Energy & the Work-Energy Theorem

The total work done by every force adds up to the change in kinetic energy. This is one of the most important and reliable problem-solving tools in mechanics.

§7.3 — Variable Force & "Area Under F-x"

For a position-dependent force, work is the area under the F-vs-x curve:

For a spring (Hooke's law F = −kx) stretched from 0 to x:

Power

A 1-horsepower motor delivers 746 W. A bright LED bulb consumes ~10 W; a treadmill at jogging speed dissipates ~600 W; a Tesla Model S can pull ~500 kW for a few seconds.

§7.4 — Conservative Forces & Potential Energy

A force is conservative if the work it does between two points is path-independent (equivalently: the work around any closed loop is zero). Gravity and springs are conservative; friction and air drag are not.

The zero of potential energy is your choice; only differences matter.

§7.5–7.6 — Mechanical Energy & Conservation

If only conservative forces act:

If non-conservative forces (friction, drag, applied push) also act:

Worked Example: Loop-the-Loop

A ball starts at rest at height h on a frictionless track that curves into a vertical loop of radius R. Minimum h for the ball to complete the loop: at the top, gravity supplies the entire centripetal force, so v_top² = gR. Energy conservation between start and top:

§8.1 — Linear Momentum & Impulse

For constant mass this reduces to F⃗ = ma⃗. But the momentum form survives even when mass changes (rockets, raindrops collecting water).

§8.3 — Conservation of Momentum

If the net external force on a system is zero, the total momentum of the system is conserved:

Momentum conservation holds even when energy isn't conserved. Collisions, explosions, and even sticky bumps all conserve momentum as long as you can ignore external forces during the brief impact.

§8.5 — Types of Collisions

1-D Elastic Collision (Eq. 8.32)

m₁ hits m₂ (initially at rest). Conservation of p and KE gives:

• Equal masses: v_1f = 0, v_2f = v_1i — they swap velocities (Newton's cradle).
• Big hits small (m₁ ≫ m₂): v_1f ≈ v_1i, v_2f ≈ 2v_1i (gravitational slingshot).
• Small hits big (m₁ ≪ m₂): v_1f ≈ −v_1i, v_2f ≈ 0 (ball bouncing off a wall).

Perfectly Inelastic (Stick Together)

Maximum KE is lost (consistent with momentum conservation). The lost KE goes to deformation, heat, and sound.

2-D Collisions & Center of Mass

In 2-D, momentum is conserved component-wise (two scalar equations). For a system of particles, define:

The CM moves as if all the mass were concentrated there and all external forces acted on it: M_tot a⃗_CM = F⃗_{ext, net}. Internal forces shift particles relative to the CM but never move the CM itself.

§9.1 — The First Condition for Equilibrium

For a rigid body to be in translational equilibrium, the net external force must vanish: ΣF⃗ = 0. This gives two scalar equations in 2-D.

§9.2 — Torque & the Second Condition

A force can cause an object to rotate. The rotational analog of force is torque τ:

• r: distance from pivot to point of application.
• φ: angle between r⃗ and F⃗.
• r_⊥ = r sin φ is the "lever arm" — perpendicular distance from pivot to the line of action of F.
• Sign convention: counterclockwise (+), clockwise (−). Or pick whatever you want — just stay consistent.

§9.3–9.4 — Static Equilibrium Recipe

1. Draw a free-body diagram showing every force and where it's applied.
2. Write ΣF_x = 0 and ΣF_y = 0.
3. Choose a clever pivot — usually where the most unknowns act — and write Στ = 0.
4. Solve the resulting 3 equations for up to 3 unknowns.

Worked Example: Seesaw Balance (OpenStax 9.1 generalized)

A child of mass m₁ = 25 kg sits 1.6 m left of the pivot. A larger child of mass m₂ sits 1.0 m right. Find m₂ for balance.

Στ = 0 about pivot: m₁ g (1.6) − m₂ g (1.0) = 0 ⇒ m₂ = 1.6 × 25 = 40 kg.

Worked Example: Ladder Against a Wall

Uniform ladder of length L, mass M, leaning at angle θ against a frictionless vertical wall. The ground has coefficient of static friction μ_s. Find the minimum θ before slipping.

Forces: weight Mg (down at center), normal N from ground (up), normal N_w from wall (horizontal toward ladder), friction f_s at base (horizontal toward wall).

ΣF_x: f_s = N_w. ΣF_y: N = Mg. Στ about base: N_w L sin θ − Mg (L/2) cos θ = 0 ⇒ N_w = (Mg/2) cot θ. At slip: f_s = μ_s N = μ_s Mg, so:

§9.5–9.6 — Stability & Levers (Discussion 6)

An object is stable if its center of gravity, when slightly displaced, returns toward the original position (lower CM). A simple machine (lever, pulley, wheel-and-axle) trades distance for force; ideal mechanical advantage = (effort arm)/(load arm).

§10.1 — Angular Quantities & Their Linear Analogs

§10.2 — Constant Angular Acceleration ("the Big Four, Rotated")

Identical structure to the linear Big Four — same problem-solving recipe, different letters.

§10.3 — Newton's 2nd Law for Rotation & Moment of Inertia

The moment of inertia I measures how the mass is distributed about the rotation axis. More mass farther from the axis → larger I → harder to spin up. Same mass closer in → smaller I → easier to spin.

Moments of Inertia You Should Know Cold

§10.4 — Rotational Kinetic Energy

Same form as ½ m v², with mass replaced by moment of inertia and speed replaced by angular speed.

Rolling Without Slipping

An object rolling without slipping has total kinetic energy equal to translational + rotational:

This explains the racing-down-a-ramp puzzle of Lec 13: more I means more rotational KE for the same total energy, so less goes into translation, so smaller v_CM.

Worked Example (OpenStax 10.13/10.14)

A solid sphere (I = (2/5)MR²) rolls without slipping down a ramp of height h. Find v_CM at the bottom.

Mgh = ½ M v² + ½ (2/5)MR² (v/R)² = ½ M v²(1 + 2/5) = (7/10) M v². So v = √(10gh/7) ≈ 0.845 √(2gh) — slower than a frictionless slide (v = √(2gh)).

§10.5 — Angular Momentum & Its Conservation

Classic Example: Spinning Skater

An ice skater spins with arms extended at ω₀, then pulls them in. With no external torque, L is conserved: I₀ ω₀ = I_f ω_f. Since I drops, ω rises.

Where does the KE come from? KE_rot = ½ I ω² = ½ L²/I. Smaller I → larger KE. The skater does work pulling the arms in against the centrifugal pseudo-force; that work shows up as added rotational KE.

Other Examples

• Diver tucks to spin faster, opens up to slow before entering water.
• A neutron star spinning at 700 Hz is a stellar core that collapsed from R ~ 10⁶ km to ~10 km — I dropped by 10¹⁰, so ω rose by 10¹⁰.
• Helicopter tail rotor is a torque counter-balance: spinning the main rotor would otherwise spin the body the opposite way (L conservation).

Topics You Must Own

• Universal Gravitation: F = G m₁m₂/r², g_surface derivation, satellite v = √(GM/r), Kepler's 3rd Law T² ∝ r³.
• Work & Energy: W = F⃗·d⃗, KE = ½mv², work-energy theorem W_net = ΔKE, spring W = ½kx².
• Energy Conservation: KE + U_grav + U_spring constant when only conservative forces act; W_nc = ΔE_mech.
• Momentum & Impulse: p = mv, J = ∫F dt = Δp, conservation when ΣF_ext = 0.
• Collisions: elastic (both p & KE), inelastic (only p), perfectly inelastic (stick); 1-D elastic formulas.
• Center of Mass: M a_CM = F_ext; CM moves like a point particle.
• Torque & Equilibrium: τ = rF sin φ, ΣF = 0 and Στ = 0 about any pivot.
• Rotational Kinematics & Dynamics: Big Four with θ, ω, α; Στ = Iα; moments of inertia for hoop/disk/sphere/rod; parallel-axis theorem.
• Rotational Energy & Rolling: KE_rot = ½Iω², total KE = ½Mv² + ½Iω², v = Rω for rolling without slipping.
• Angular Momentum: L = Iω, conserved when Στ_ext = 0; skater spins, neutron stars.

Strategy

• Energy method beats kinematics when path or force is messy and you only care about start/end.
• Momentum method for collisions (always conserved) — then use energy conservation only if elastic.
• Pick a pivot to kill unknowns when computing torques. Often the point where the most unknown forces act.
• Linear ↔ angular dictionary: if a linear-mechanics intuition works, the angular analog likely does too.
• Check rolling-without-slipping: the contact point is instantaneously at rest. v_CM = Rω, a_CM = Rα.

§10.6 — Extended-Body Collisions & Rotation

When a free particle strikes an extended object, both linear and angular momentum are conserved (assuming no external impulsive forces).

Worked Example: Bullet Embeds in a Door

A bullet of mass m at velocity v hits a vertical door (mass M, width L, hinged at one side) at distance d from the hinge and embeds. Find the angular velocity of door + bullet just after impact.

Conservation of L about the hinge: m v d = (I_door + m d²) ω. With I_door = (1/3) M L²:

Linear momentum is not conserved because the hinge exerts an impulsive horizontal force. Angular momentum about the hinge is conserved because the hinge force has zero torque about itself.

§11.1 — What Is a Fluid?

A fluid is a substance that flows — it has no fixed shape and conforms to its container. Liquids (nearly incompressible) and gases (highly compressible) are both fluids.

§11.2 — Density

Water: ρ ≈ 1000 kg/m³ at 4 °C. Air at sea level: ρ ≈ 1.29 kg/m³. Mercury: ρ ≈ 13,600 kg/m³. Specific gravity = ρ/ρ_water (a dimensionless ratio).

§11.3–11.4 — Pressure

• Atmospheric: 1 atm = 101,325 Pa = 14.7 psi = 760 mmHg.
• Pressure is a scalar — it pushes equally in all directions on a fluid element.

§11.5 — Pressure with Depth in a Static Fluid

Where h is depth below a free surface at pressure P₀. Two consequences:

• Pascal's principle: a pressure change applied anywhere in an enclosed incompressible fluid is transmitted undiminished to every point — basis of hydraulic lifts.
• Pressure depends only on depth, not container shape ("hydrostatic paradox"): a thin tall tube and a wide shallow tank can give the same pressure at the bottom.

Hydraulic Lift

Volume conservation requires F₁ · d₁ = F₂ · d₂: you trade distance for force, with no free lunch in energy.

§11.6 — Gauge vs. Absolute Pressure

Most pressure gauges (tire, blood pressure cuff) read pressure above atmospheric:

A "32 psi" tire has absolute pressure ~46.7 psi. A flat tire has gauge 0, absolute 1 atm.

Manometers & Barometers

A U-tube manometer measures gauge pressure as a height difference: ΔP = ρ g Δh. A mercury barometer reads atmospheric pressure: 1 atm pushes up a column of Hg by 760 mm.

§11.7 — Archimedes' Principle

Mathematically, F_B arises because the pressure on the bottom of the submerged volume is greater than the pressure on the top — that pressure difference times the cross-section area equals ρ V g.

Floats vs. Sinks

• Object density < fluid density → floats; submerged fraction = ρ_obj/ρ_fluid.
• Object density = fluid density → neutrally buoyant.
• Object density > fluid density → sinks (apparent weight = (ρ_obj − ρ_fluid) V g).

Worked Example: Iceberg Above Water

ρ_ice = 920 kg/m³, ρ_seawater = 1025 kg/m³. Submerged fraction = 920/1025 ≈ 0.898 → about 10% above water. (Hence "tip of the iceberg.")

§12.1 — Flow Rate & Continuity

For an incompressible fluid in steady flow, the same mass passes every cross-section per unit time. With Q = volume flow rate (m³/s):

Pinch a hose: smaller A → larger v. The water comes out faster.

§12.2 — Bernoulli's Equation

Conservation of energy per unit volume along a streamline of an incompressible, non-viscous, steady flow:

Three energy densities, each with units of pressure (Pa = J/m³):

• P — static (pressure) energy density.
• ½ ρ v² — kinetic energy density.
• ρ g h — gravitational potential energy density.

§12.3 — Applications

Torricelli's Theorem (Tank Drain)

A large tank has a small hole at depth h below the water surface. Apply Bernoulli between the surface (large area, v ≈ 0, P = P_atm) and the exit (P = P_atm):

Same as a free-fall from height h. Deep water exits faster.

Venturi Meter (Constricted Pipe)

Horizontal pipe narrows from A₁ to A₂. Continuity: v₂ = (A₁/A₂) v₁. Bernoulli (h₁ = h₂):

Pressure drops at the constriction. Carburetors and atomizers exploit this to draw fuel into the airflow.

Why Wings Lift

A cambered (curved) airfoil shape forces air over the top to travel a longer path, increasing v above the wing. Bernoulli says P drops there, and the resulting pressure imbalance lifts the wing. (Reality is more subtle than the "equal-time" simplification, but Bernoulli + Newton's 3rd law is the right starting picture.)

§12.4 — Viscosity

Real fluids resist relative motion of adjacent layers — that resistance is viscosity η, measured in Pa·s. Honey is more viscous than water; air is even less viscous.

Two parallel plates separated by fluid: the force per unit area to slide one plate relative to the other is proportional to the velocity gradient. Honey is η ≈ 10 Pa·s; water is ~0.001 Pa·s; air is ~10⁻⁵ Pa·s.

§12.5 — Laminar vs. Turbulent Flow & Reynolds Number

At low speeds, fluid layers slide smoothly past each other (laminar); at high speeds, swirling chaos sets in (turbulent). The dimensionless Reynolds number tells you which regime:

• N_R < ~2000 → laminar.
• N_R ~ 2000–3000 → unstable transition.
• N_R > ~3000 → turbulent.

§12.6 — Poiseuille's Law (Laminar Pipe Flow)

Volume flow rate through a pipe of radius r and length L driven by pressure difference ΔP. The dramatic r⁴ dependence means a small change in pipe radius has a huge effect on flow.

Worked Examples

• OS 12.8: Water flow through a tap — picking r and L.
• OS 12.9: IV bag height needed to overcome venous pressure.
• OS 12.10: Estimating turbulence onset in arteries during exercise (when N_R crosses 2000).

The Three Master Conservation Laws of PHYS 21

1. Energy (KE + PE + thermal + …) — useful when forces are conservative or when you only care about start/end states.
2. Linear momentum (Σ p⃗) — useful for collisions and any system with no net external force during a brief interval.
3. Angular momentum (Σ L⃗) — useful for rotations and orbits with no net external torque about a chosen pivot.

Most "hard" problems on the final yield to one of these. Always ask: what is conserved here, and between which two instants?

Fun Synthesis Examples

1. Ballistic Pendulum (momentum + energy, in sequence)

A bullet (m, v) embeds in a hanging block (M, at rest). The combined mass swings up to height h. Find v.

During impact (very fast): momentum conservation, mv = (m+M)V.

After impact (slow swing): energy conservation, ½(m+M)V² = (m+M)gh, so V = √(2gh).

Combine: v = (1 + M/m) √(2gh). You can't use energy through the impact (some KE went to heat) and you can't use momentum through the swing (gravity is external) — pick the right tool for each phase.

2. Spinning Stool (angular momentum)

You sit on a frictionless rotating stool holding a spinning bicycle wheel horizontally with L_wheel = +L₀. You flip the wheel over so its L is now −L₀. Total system L is conserved → you and the stool spin in the +direction with L = 2L₀.

3. Hot-Air Balloon (Archimedes for gas)

Hot air at temperature T inside the balloon has lower density ρ' = ρ_air(T₀/T). The buoyant force (ρ_air V g) exceeds the gas weight (ρ' V g), and the balloon rises until ρ' equals the local atmospheric density.

4. Falling vs. Rolling Race

A solid sphere released from rest down a frictionless ramp of height h reaches v = √(2gh). Released as rolling-without-slipping down the same ramp, it reaches v = √(10gh/7) ≈ 0.85√(2gh) — slower because some PE went into rotational KE.

Common Final-Exam Pitfalls

• Forgetting that g is unsigned; chosen sign comes from your axis convention.
• Treating "centripetal force" as a separate force rather than the net inward component of real forces.
• Using energy conservation through inelastic collisions, where it doesn't hold.
• Computing torque about a moving point that isn't the CM — torque equation gets ugly.
• Confusing moment arm with distance — only the perpendicular component of r matters for τ.
• Mixing angular and linear quantities without converting via v = rω.
• Ignoring Bernoulli's altitude term ρgh when fluid moves vertically.
• Forgetting that Poiseuille goes as r⁴ — a 20% change in r is a 100% change in flow.

Coverage Map (by Topic Cluster)

3-Hour Strategy

• Read every problem first. Do the easy ones to bank points before time pressure mounts.
• Allocate roughly equal time per problem. If a problem is sucking up > 1.5× its share, move on and come back.
• Always draw a picture and label knowns/unknowns. Even a partial-credit picture earns points.
• Sanity-check signs and limits before submitting an answer.
• Use the formula sheet (if any) sparingly — knowing where formulas come from beats memorization under pressure.

Kinematics (1-D & 2-D)

Vectors

Newton's Laws & Forces

Circular Motion & Gravity

Work, Energy, Power

Momentum & Collisions

Rotation

Fluids

Physical Constants

Useful Conversions

Useful Approximations

• g ≈ 10 m/s² for fast estimates (introduces ~2% error).
• π² ≈ 10 for back-of-envelope (4π² ≈ 40, useful in Kepler's 3rd Law).
• √2 ≈ 1.414, √3 ≈ 1.732, √10 ≈ 3.162.
• sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ for small θ in radians.
• 1 m/s ≈ 2.24 mph — quick speed conversion.
• 1 N ≈ ¼ pound, so a 70-kg person weighs ~700 N ≈ 154 lb.