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Physics Notes

Every section's key ideas, formulas and worked examples. Tap a section to jump to it.

Chapter 4 · Dynamics: Force and Newton's Laws
Chapter 5 · Friction
Chapter 6 · Uniform Circular Motion and Gravitation
Chapter 7 · Work and Energy
Chapter 8 · Linear Momentum and Collisions
Chapter 11 · Fluid Statics
Chapter 12 · Fluid Dynamics
Chapter 13 · Temperature, Kinetic Theory, and the Gas Laws
Chapter 14 · Heat and Heat Transfer Methods
Chapter 15 · Thermodynamics
Chapter 16 · Oscillatory Motion and Waves
Chapter 17 · Physics of Hearing

Chapter 4: Dynamics: Force and Newton's Laws

4.1 Development of Force Concept

A force is a push or pull with both size and direction, so it's a vector and adds like any other vector.

Practice 4.1 in the quiz → · Top ↑

4.2 Newton's First Law: Inertia

An object keeps doing what it's doing (staying still, or moving in a straight line at constant speed) unless a net external force acts on it.

Practice 4.2 in the quiz → · Top ↑

4.3 Newton's Second Law: Concept of a System

Acceleration equals net external force divided by mass: push harder and it speeds up more, add mass and it speeds up less.

a=Fnetm\displaystyle a = \frac{F_{net}}{m}Fnet=ma\displaystyle F_{net} = ma

Key terms

Weight vs. mass

w=mg\displaystyle w = mgg=9.80 m/s2 on Earth\displaystyle g = 9.80\ \text{m/s}^2 \text{ on Earth}
Mass Weight
What it is Amount of matter (inertia) Force of gravity on the object
Unit kilogram (kg) newton (N)
Changes with location? No Yes
1.0 kg object on Earth 1.0 kg 9.8 N
1.0 kg object on the Moon (g = 1.625 m/s²) 1.0 kg about 1.6 N

Worked examples

Problem Given Setup Answer
Lawn mower (Ex. 4.1) F_net = 51 N, m = 24 kg a = F_net / m 2.1 m/s²
Rocket sled (Ex. 4.2) 4 rockets, m = 2100 kg, a = 49 m/s², f = 650 N 4T − f = ma, so 4T = ma + f = 1.0×10⁵ N Each rocket: 2.6×10⁴ N

Common mistakes

Practice 4.3 in the quiz → · Top ↑

4.4 Newton's Third Law: Symmetry in Forces

Forces always come in pairs: if A pushes on B, B pushes back on A with the same size force in the opposite direction.

Everyday examples

You push on... ...which pushes back on you Result
Floor, backward (walking) You, forward You walk forward
Road, backward (car tires) Car, forward Car accelerates
Exhaust gas, backward (rocket) Rocket, forward (thrust) Rocket accelerates, even in a vacuum
Air, downward (helicopter, bird wings) Aircraft, upward (lift) It flies
Ball, forward (bat swing) Bat, backward Bats can break, as with Mariano Rivera's cutter

Choosing the system matters

The same situation gives different free-body diagrams depending on what you call the system. Pick the system so the force you want is external to it.

Worked examples

A professor (65.0 kg) pushes a cart (12.0 kg) of equipment (7.0 kg). She pushes back on the floor with 150 N; friction totals 24.0 N.

Problem System Setup Answer
Acceleration (Ex. 4.3) Professor + cart + equipment (84 kg) Floor pushes her forward 150 N; F_net = 150 − 24.0 = 126 N; a = 126 / 84 1.5 m/s²
Force on the cart (Ex. 4.4) Cart + equipment (19.0 kg) F_net = ma = 19.0 × 1.5 = 29 N; F_prof = F_net + f = 29 + 24.0 53 N

Only 53 N of her effort reaches the cart; the rest accelerates her own body.

Common mistakes

Practice 4.4 in the quiz → · Top ↑

4.5 Normal, Tension, and Other Forces

A supporting surface pushes back perpendicular to itself (normal force), a rope or cable pulls along its length (tension), and on a slope you split weight into parallel and perpendicular parts.

Key terms

Normal force rules

Inclines: resolving weight

Tilt the axes so one runs along the slope and one runs perpendicular to it. Then weight splits into two parts:

w∥=mgsin⁡θ\displaystyle w_{\parallel} = mg\sin\thetaw⊥=mgcos⁡θ\displaystyle w_{\perp} = mg\cos\theta

Acceleration down an incline:

a=mgsin⁡θ−fm\displaystyle a = \frac{mg\sin\theta - f}{m}

With no friction this becomes a = g sinθ, the same for any mass.

Tension rules

T=F⊥2sin⁡θ\displaystyle T = \frac{F_{\perp}}{2\sin\theta}

Worked examples

Problem Given Setup Answer
Skier, no friction (Ex. 4.5a) m = 60.0 kg, θ = 25° a = g sin25° = 9.80 × 0.4226 4.14 m/s²
Skier, with friction (Ex. 4.5b) f = 45.0 N a = (mg sin25° − f) / m = (248.5 − 45.0) / 60.0 3.39 m/s²
Skier, normal force same N = mg cos25° = 60 × 9.8 × 0.906 532 N
Hanging mass m = 5.00 kg T = mg 49.0 N
Tightrope walker (Ex. 4.6) m = 70.0 kg, sag 5.0° T = mg / (2 sin5.0°) = 686 / (2 × 0.0872) 3,900 N (about 6× his weight)

Problem-solving recipe

  1. Draw a free-body diagram: one dot, every force as an arrow.
  2. Pick axes that line up with the motion (along/perpendicular to a slope; horizontal/vertical for ropes).
  3. Break any force not on an axis into components with sin and cos.
  4. Write ΣF = ma separately for each axis. If nothing moves along an axis, ΣF = 0 there.
  5. Solve, then sanity-check units and size.

Common mistakes

Practice 4.5 in the quiz → · Top ↑

4.6 Problem-Solving Strategies

Every Newton's-laws problem follows the same four steps; the step students skip most is choosing the system and leaving out forces that don't act on it.

The four steps

  1. Sketch and identify forces. Draw the situation, then add an arrow for every force. Label each one, and make lengths and directions match the forces when you can.
  2. List knowns and unknowns, then define the system of interest. Draw a boundary around the object(s) you care about. Only external forces (from outside that boundary) go on the free-body diagram.
  3. Apply Newton's second law. Turn the free-body diagram into equations and solve.
    • All forces parallel (1-D): add them like ordinary numbers with + and − signs.
    • Forces at angles (2-D): split into two 1-D problems by projecting forces onto axes.
    • Choose axes that make it easy: one axis along the direction of motion; on a ramp, along and perpendicular to the slope.
  4. Check the answer. Is it reasonable (e.g. friction should make an object slide slower, not faster)? Do the units match (a force should come out in newtons, not m/s)?

Free-body diagram rules

Tarzan example (Figure 4.20)

Tarzan hangs still from a vine. Three forces appear in the sketch: the vine's tension T pulling up on him, his weight w pulling down, and F_T, the force he exerts down on the vine. With Tarzan as the system, F_T acts on the vine, not on him, so it's left off the diagram. Since he's stationary, T and w balance: T = −w (same size, opposite direction).

Decide acceleration first, axis by axis

Before writing any equation, ask: is the system accelerating along this axis?

Axis Accelerating? Equation
Direction of motion (say x) Yes F_net,x = ma
Perpendicular (say y) No F_net,y = 0

The zero-acceleration axis is usually where you find an unknown force like N or T.

Common mistakes

Practice 4.6 in the quiz → · Top ↑

4.7 Further Applications of Newton's Laws

This section stacks the earlier tools into harder problems: combine perpendicular forces, solve two unknown tensions with two equations, predict an elevator scale reading, and link F = ma to kinematics.

Combining perpendicular forces

When two forces are at right angles, find the total with the Pythagorean theorem and its direction with inverse tangent:

F=Fx2+Fy2\displaystyle F = \sqrt{F_x^2 + F_y^2}θ=tan⁡−1(FyFx)\displaystyle \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)

Drag (friction from a fluid like water or air) points opposite the motion. Once you know the direction of motion, the problem becomes 1-D: F_net = F_applied − F_drag = ma.

Unequal tensions

If a hanging object is held by two wires at different angles, the tensions differ. Use two equations (zero acceleration on both axes):

Solve the first for one tension, substitute into the second. The steeper (more vertical) wire carries more tension.

Elevator and bathroom scale

A scale reads the normal force it pushes up with. With the person as the system, F_s − mg = ma, so:

Fs=m(g+a)\displaystyle F_s = m(g + a)
Elevator motion a (up = +) Scale reads
Speeding up going up, or slowing down going down positive More than weight
At rest or constant speed (up or down) 0 Exactly weight
Speeding up going down, or slowing down going up negative Less than weight
Free fall −g Zero ("weightless")

Key idea: the scale responds to acceleration, not velocity. Constant speed feels the same as standing still.

Integrated problems: kinematics + forces

  1. List knowns and unknowns to spot which principles apply.
  2. Use kinematics (a = Δv / Δt, etc.) to find acceleration, or forces to find acceleration and then motion.
  3. Connect them through F_net = ma.

Worked examples

Problem Given Setup Answer
Barge drag (Ex. 4.7) Tugs 2.7×10⁵ N (x) and 3.6×10⁵ N (y); m = 5.0×10⁶ kg; a = 0.075 m/s² F_app = √(2.7² + 3.6²)×10⁵ = 4.5×10⁵ N at 53°; F_D = F_app − ma 7.5×10⁴ N, opposite the motion
Traffic light (Ex. 4.8) m = 15.0 kg; wires at 30° and 45° T₂ = 1.225 T₁; 1.366 T₁ = 147 N T₁ = 108 N, T₂ = 132 N
Scale, accelerating up (Ex. 4.9a) m = 75.0 kg; a = 1.20 m/s² F_s = 75.0(9.80 + 1.20) 825 N (about 185 lb)
Scale, constant speed (Ex. 4.9b) v = 1 m/s, so a = 0 F_s = mg 735 N (his weight)
Soccer sprint (Ex. 4.10) m = 70.0 kg; 0 → 8.00 m/s in 2.50 s a = 8.00 / 2.50 = 3.20 m/s²; F = ma 224 N (about 50 lb)

Common mistakes

Practice 4.7 in the quiz → · Top ↑

Chapter 5: Friction

5.1 Friction

Friction opposes sliding between two surfaces and equals a coefficient times the normal force; it's stronger before things start moving (static) than once they slide (kinetic).

Two kinds of friction

fs≤μsN\displaystyle f_s \le \mu_s Nfk=μkN\displaystyle f_k = \mu_k N

Selected coefficients (Table 5.1)

Surfaces μ_s μ_k
Rubber on dry concrete 1.0 0.7
Shoes on wood 0.9 0.7
Rubber on wet concrete 0.7 0.5
Steel on steel (dry) 0.6 0.3
Wood on wood 0.5 0.3
Waxed wood on wet snow 0.14 0.1
Shoes on ice 0.1 0.05
Steel on steel (oiled) 0.05 0.03
Bone joint (synovial fluid) 0.016 0.015

Friction on a slope

On an incline, N = mg cosθ, so f_k = μ_k mg cosθ. If an object slides down at constant velocity, friction exactly balances the downhill pull:

μk=tan⁡θ\displaystyle \mu_k = \tan\theta

That gives a quick way to measure μ_k: tilt a surface until something slides at steady speed and take tan of the angle.

Worked examples

Problem Given Setup Answer
Crate, to start moving m = 100 kg, μ_s = 0.45 N = mg = 980 N; f_s(max) = 0.45 × 980 Push over 440 N
Crate, to keep moving μ_k = 0.30 f_k = 0.30 × 980 290 N at constant speed
Skier on 25° slope (Ex. 5.1) m = 62 kg, f_k = 45.0 N μ_k = f_k / (mg cos25°) = 45.0 / (62 × 9.80 × 0.906) μ_k = 0.082

Why friction happens

Surfaces are rough up close, so only the high spots touch. Molecules at those spots also stick to each other (adhesion). Pressing harder increases the true contact area, which is why friction depends on normal force, not on the overall surface area.

Common mistakes

Practice 5.1 in the quiz → · Top ↑

Chapter 6: Uniform Circular Motion and Gravitation

6.1 Rotation Angle and Angular Velocity

Angles in rotation are measured in radians, and angular velocity (ω) tells how fast the angle changes; a point's speed is v = rω.

Δθ=Δsr\displaystyle \Delta\theta = \frac{\Delta s}{r}ω=ΔθΔt\displaystyle \omega = \frac{\Delta\theta}{\Delta t}v=rω\displaystyle v = r\omega

Conversions

From To Multiply by
Revolutions Radians 2π
Degrees Radians π / 180
Radians Degrees 180 / π (1 rad ≈ 57.3°)
rev/min (rpm) rad/s 2π / 60

Common angles: 30° = π/6, 60° = π/3, 90° = π/2, 120° = 2π/3, 135° = 3π/4, 180° = π, 360° = 2π.

Key ideas

Worked examples

Problem Setup Answer
Car tire, r = 0.300 m, car at 15.0 m/s (Ex. 6.1) ω = v/r = 15.0 / 0.300 50.0 rad/s
Earth-mover tire, r = 1.20 m, same speed ω = 15.0 / 1.20 12.5 rad/s

Common mistakes: using degrees in v = rω or Δs = rΔθ (must be radians); forgetting the 2π when converting revolutions; thinking the outer edge has a bigger ω (it has a bigger v, same ω); calculator in the wrong mode.

rpm conversion isn't in 6.1 itself, but it shows up in problems, so it's included above.

Practice 6.1 in the quiz → · Top ↑

6.2 Centripetal Acceleration

An object moving in a circle at constant speed is still accelerating, because its direction keeps changing; that acceleration points to the center.

ac=v2r=rω2\displaystyle a_c = \frac{v^2}{r} = r\omega^2
Problem Setup Answer
Car on a 500 m curve at 25.0 m/s (Ex. 6.2) a_c = 25.0² / 500 1.25 m/s² (0.128g)
Ultracentrifuge, r = 0.0750 m, 7.50 × 10⁴ rpm (Ex. 6.3) ω = 7854 rad/s; a_c = 0.0750 × 7854² 4.63 × 10⁶ m/s² (472,000g)

Common mistakes: thinking constant speed means no acceleration; pointing a_c outward; forgetting to square v or ω; not converting rpm to rad/s.

Practice 6.2 in the quiz → · Top ↑

6.3 Centripetal Force

Whatever net force points to the center is the centripetal force; it isn't a new kind of force.

Fc=mv2r=mrω2\displaystyle F_c = m\frac{v^2}{r} = mr\omega^2
Situation What supplies F_c
Car on a flat curve Static friction from the road
Car on a banked curve (ideal) Horizontal part of the normal force
Tetherball, ball on a string Tension
Moon or satellite in orbit Gravity
Skater turning Friction on the skates

Flat curve: friction

μs=v2rg\displaystyle \mu_s = \frac{v^2}{rg}

This is the minimum coefficient needed. Mass cancels, so a loaded truck and a car need the same μ at the same speed.

Banked curve with no friction

tan⁡θ=v2rg\displaystyle \tan\theta = \frac{v^2}{rg}v=rgtan⁡θ\displaystyle v = \sqrt{rg\tan\theta}

Faster speeds and tighter curves need steeper banking. Mass cancels again. Friction lets cars go faster or slower than the ideal speed.

Problem Setup Answer
900 kg car, r = 500 m, 25.0 m/s: force (Ex. 6.4a) F_c = 900 × 25.0² / 500 1125 N
Minimum μ_s for that curve (Ex. 6.4b) μ_s = 25.0² / (500 × 9.80) 0.13
Ideal speed on a 65.0° bank, r = 100 m (Ex. 6.5) v = √(100 × 9.80 × tan65°) 45.8 m/s (165 km/h)

Common mistakes: drawing "centripetal force" as an extra force on a free-body diagram; thinking an object flies straight outward when released.

Practice 6.3 in the quiz → · Top ↑

6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force

Forces you seem to feel in a turning or accelerating vehicle aren't real; they come from your own inertia.

Common mistakes: saying a "centrifugal force" throws you outward in a turning car; mixing up the hemisphere directions.

Practice 6.4 in the quiz → · Top ↑

6.5 Newton's Universal Law of Gravitation

Every mass attracts every other mass with a force that weakens with the square of the distance between their centers.

F=GmMr2\displaystyle F = G\frac{mM}{r^2}G=6.674×10−11 N⋅m2/kg2\displaystyle G = 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2g=GMr2\displaystyle g = \frac{GM}{r^2}

Useful values: Earth M = 5.98 × 10²⁴ kg, R = 6.38 × 10⁶ m; Earth–Moon distance 3.84 × 10⁸ m; Moon's period 27.3 days.

Problem Setup Answer
Earth's gravity at the Moon's distance (Ex. 6.6a) g = GM / r² with r = 3.84 × 10⁸ m 2.70 × 10⁻³ m/s²
Moon's centripetal acceleration (Ex. 6.6b) ω = 2π / 27.3 days = 2.66 × 10⁻⁶ rad/s; a_c = rω² 2.72 × 10⁻³ m/s² (they match, so gravity holds the Moon in orbit)
Two 1.000 kg masses 1.000 m apart F = G × 1 × 1 / 1² 6.674 × 10⁻¹¹ N (tiny)

Common mistakes: measuring r from the surface instead of the center; forgetting to square r; saying there's no gravity in space.

Practice 6.5 in the quiz → · Top ↑

6.6 Satellites and Kepler's Laws

Kepler described how planets move; Newton's gravity explains why.

  1. First law: every orbit is an ellipse with the central body at one focus.
  2. Second law: the line from the Sun to a planet sweeps out equal areas in equal times, so the planet moves fastest when closest.
  3. Third law: T² is proportional to r³ for everything orbiting the same body.
T12T22=r13r23\displaystyle \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}v=GMr\displaystyle v = \sqrt{\frac{GM}{r}}T2=4π2GMr3\displaystyle T^2 = \frac{4\pi^2}{GM}r^3M=4π2r3GT2\displaystyle M = \frac{4\pi^2 r^3}{GT^2}
Problem Setup Answer
Satellite 1500 km up, compared with the Moon (Ex. 6.7) r₂ = 1500 + 6380 = 7880 km; T₂ = 27.3 d × (7880 / 3.84 × 10⁵)^1.5 1.93 h

Planets around the Sun all share r³/T² ≈ 3.35 × 10²⁴ km³/y².

Common mistakes: using altitude as r; comparing a moon of Jupiter with a planet around the Sun in Kepler's third law; thinking heavier satellites orbit faster.

Practice 6.6 in the quiz → · Top ↑

Chapter 7: Work and Energy

7.1 Work: The Scientific Definition

Work is done only when a force moves something, and only the part of the force along the motion counts.

W=Fdcos⁡θ\displaystyle W = Fd\cos\theta
Situation Angle Work
Force along the motion (pushing a cart forward) 0° Positive, W = Fd
Force at an angle (pushing a mower handle 35° below horizontal) θ W = Fd cosθ
Force perpendicular to motion (carrying a briefcase across a level floor) 90° Zero
No displacement (holding a heavy box still) — Zero
Force opposite the motion (friction, lowering a load) 180° Negative, W = −Fd

Worked example (Ex. 7.1): pushing a mower with 75.0 N at 35° below horizontal for 25.0 m: W = 75.0 × 25.0 × cos35° = 1.54 × 10³ J (0.367 kcal).

Common mistakes: using sin instead of cos; calling it work when the force is perpendicular to the motion; forgetting the negative sign for friction.

Practice 7.1 in the quiz → · Top ↑

7.2 Kinetic Energy and the Work-Energy Theorem

The net work done on an object equals its change in kinetic energy.

KE=12mv2\displaystyle KE = \tfrac{1}{2}mv^2Wnet=12mv2−12mv02=ΔKE\displaystyle W_{net} = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 = \Delta KE

Worked examples (30.0 kg package on rollers)

Problem Setup Answer
KE at 0.500 m/s (Ex. 7.2) ½ × 30.0 × 0.500² 3.75 J
Net work: 120 N push, 5.00 N friction, 0.800 m (Ex. 7.3) (120 − 5.00) × 0.800, or 96.0 J − 4.00 J 92.0 J
Final speed (Ex. 7.4) ½mv² = 92.0 + 3.75 = 95.75 J; v = √(2 × 95.75 / 30.0) 2.53 m/s
Coasting distance after the push (Ex. 7.5) Friction removes 95.75 J: d = 95.75 / 5.00 19.2 m

Common mistakes: forgetting to square v; using the applied force instead of the net force; leaving out the initial KE.

Practice 7.2 in the quiz → · Top ↑

7.3 Gravitational Potential Energy

Lifting something stores energy that comes back as kinetic energy when it falls.

ΔPEg=mgh\displaystyle \Delta PE_g = mghv=2g∣h∣\displaystyle v = \sqrt{2g|h|}v=2g∣h∣+v02\displaystyle v = \sqrt{2g|h| + v_0^2}

Worked examples

Problem Setup Answer
Raise a 0.500 kg clock weight 1.00 m 0.500 × 9.80 × 1.00 4.90 J
Roller coaster from rest, 20.0 m drop (Ex. 7.7a) √(2 × 9.80 × 20.0) 19.8 m/s
Same coaster starting at 5.00 m/s (Ex. 7.7b) √(392 + 25.0) 20.4 m/s
60.0 kg person lands stiffly from 3.00 m, knees bend 0.500 cm (Ex. 7.6) F = mgh / d = 1764 / 0.00500 3.53 × 10⁵ N (500× his weight)

Note in 7.7b: starting at 5 m/s only raises the final speed from 19.8 to 20.4 m/s, not by 5 m/s, because energies add, not speeds.

Common mistakes: adding speeds instead of energies; forgetting h is the vertical drop, not the track length; keeping m when it cancels.

Practice 7.3 in the quiz → · Top ↑

7.4 Conservative Forces and Potential Energy

If only conservative forces act, kinetic plus potential energy stays constant.

PEs=12kx2\displaystyle PE_s = \tfrac{1}{2}kx^2KEi+PEi=KEf+PEf\displaystyle KE_i + PE_i = KE_f + PE_f

Worked example (Ex. 7.8): a 0.100 kg toy car, spring with k = 250.0 N/m compressed 4.00 cm, track rises 0.180 m.

Point Energy equation Speed
Leaving the spring ½kx² = ½mv² → v = x√(k/m) = 0.0400 × √2500 2.00 m/s
Top of the 0.180 m rise ½kx² = ½mv² + mgh → v = √(kx²/m − 2gh) 0.687 m/s

Common mistakes: using cm instead of m for x; forgetting to square x; using KE_i + PE_i = KE_f + PE_f when friction is significant.

Practice 7.4 in the quiz → · Top ↑

7.6 Conservation of Energy

Total energy never disappears; it only changes form or moves between systems.

KEi+PEi+Wnc+OEi=KEf+PEf+OEf\displaystyle KE_i + PE_i + W_{nc} + OE_i = KE_f + PE_f + OE_f
Efficiency=WoutEin\displaystyle \text{Efficiency} = \frac{W_{out}}{E_{in}}

Energy chains

Process Conversion
Eating and moving Chemical → kinetic + potential + thermal
Photosynthesis Light → chemical
Coal power plant Chemical → thermal → mechanical (turbine) → electrical
Solar-powered motor Radiant → electrical → mechanical

Typical efficiencies (Table 7.2)

Device or activity Efficiency
Electric motor 98%
Gas heater 90%
Coal power plant 42%
Diesel engine 35%
Gasoline engine 30%
Cycling and climbing 20%
Solar cell 10%
Weightlifting 9%
Shoveling 3%
Surface swimming 2%

Energy problem-solving steps

  1. Define the system, knowns, and unknown. Sketch it.
  2. List the forces. Do you know a potential energy for each?
  3. All conservative → KE_i + PE_i = KE_f + PE_f.
  4. Friction or other energy involved → use the full equation with W_nc and OE.
  5. Zero out terms: set h = 0 at the start or end; cross out KE when something is at rest.
  6. Check: friction work should be negative, PE lower at the bottom, speeds reasonable.

Common mistakes: saying energy is "lost" (it's converted, usually to thermal); efficiency over 100%; calculating work by gravity separately and also counting PE (that double-counts it).

Section 7.5 (Nonconservative Forces) wasn't included in this batch. 7.6 uses W_nc from it, so it's worth adding if it's assigned.

Practice 7.6 in the quiz → · Top ↑

Chapter 8: Linear Momentum and Collisions

8.1 Linear Momentum and Force

Momentum is mass times velocity, and a net force is what changes it.

p⃗=mv⃗\displaystyle \vec{p} = m\vec{v}Fnet=ΔpΔt\displaystyle F_{net} = \frac{\Delta p}{\Delta t}Δp=m(vf−vi)\displaystyle \Delta p = m(v_f - v_i)
Problem Setup Answer
110 kg player at 8.00 m/s (Ex. 8.1a) p = 110 × 8.00 880 kg·m/s
0.410 kg football at 25.0 m/s (Ex. 8.1b) p = 0.410 × 25.0 10.3 kg·m/s (player has about 85× more)
Tennis serve: 0.057 kg, 0 → 58 m/s in 5.0 ms (Ex. 8.2) Δp = 0.057 × 58 = 3.3 kg·m/s; F = 3.3 / 0.0050 660 N

Common mistakes: forgetting momentum's direction (sign); using m × final speed instead of m × change in velocity; ms not converted to s.

Practice 8.1 in the quiz → · Top ↑

8.2 Impulse

Impulse is force times time, and it equals the change in momentum.

Impulse=Fnet Δt=Δp\displaystyle \text{Impulse} = F_{net}\,\Delta t = \Delta p

Common mistakes: saying a cushion reduces impulse (it reduces force; the impulse is the same); forgetting the factor of 2 for a bounce.

Practice 8.2 in the quiz → · Top ↑

8.3 Conservation of Momentum

If no net external force acts on a system, its total momentum never changes.

p1+p2=p1′+p2′\displaystyle p_1 + p_2 = p_1' + p_2'm1v1+m2v2=m1v1′+m2v2′\displaystyle m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'

Common mistakes: dropping the sign of a velocity going the other way; including an object in the system that has an outside force on it (like a falling ball with gravity).

Practice 8.3 in the quiz → · Top ↑

8.4 Elastic Collisions in One Dimension

In an elastic collision both momentum and kinetic energy are conserved.

m1v1+m2v2=m1v1′+m2v2′\displaystyle m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'12m1v12+12m2v22=12m1v1′2+12m2v2′2\displaystyle \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 = \tfrac{1}{2}m_1v_1'^2 + \tfrac{1}{2}m_2v_2'^2
v1′=m1−m2m1+m2 v1\displaystyle v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1v2′=2m1m1+m2 v1\displaystyle v_2' = \frac{2m_1}{m_1 + m_2}\,v_1
Case (target at rest) What happens
Equal masses The first object stops; the second leaves at the first one's speed (Newton's cradle)
Light hits heavy Light one bounces back; heavy one moves forward slowly
Heavy hits light Heavy one keeps going slightly slower; light one shoots forward at up to 2v

Worked example (Ex. 8.4): 0.500 kg at 4.00 m/s hits 3.50 kg at rest. v₁′ = (0.500 − 3.50)/4.00 × 4.00 = −3.00 m/s and v₂′ = 2(0.500)/4.00 × 4.00 = 1.00 m/s. Check: KE = 4.00 J and p = 2.00 kg·m/s both before and after.

Practice 8.4 in the quiz → · Top ↑

8.5 Inelastic Collisions in One Dimension

In an inelastic collision momentum is still conserved, but kinetic energy is not.

v′=m1v1+m2v2m1+m2(objects stick)\displaystyle v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \quad (\text{objects stick})ΔKE=KE′−KE\displaystyle \Delta KE = KE' - KE
Problem Setup Answer
0.150 kg puck at 35.0 m/s caught by 70.0 kg goalie (Ex. 8.5a) v′ = 0.150 × 35.0 / 70.15 0.0748 m/s
KE lost (Ex. 8.5b) 91.9 J before, 0.196 J after −91.7 J
Spring carts: 0.350 kg at 2.00 → −4.00 m/s; 0.500 kg at −0.500 m/s (Ex. 8.6) v₂′ = (0.700 − 0.250 + 1.40) / 0.500 3.70 m/s; KE gained 5.46 J from the spring

Common mistakes: thinking KE is conserved in every collision; forgetting to add the masses when objects stick; sign errors when objects move toward each other.

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8.6 Collisions of Point Masses in Two Dimensions

In 2D, momentum is conserved separately along x and along y.

m1v1=m1v1′cos⁡θ1+m2v2′cos⁡θ2\displaystyle m_1v_1 = m_1v_1'\cos\theta_1 + m_2v_2'\cos\theta_20=m1v1′sin⁡θ1+m2v2′sin⁡θ2\displaystyle 0 = m_1v_1'\sin\theta_1 + m_2v_2'\sin\theta_2

Worked example (Ex. 8.7): 0.250 kg at 2.00 m/s hits 0.400 kg at rest; afterward m₁ moves 1.50 m/s at 45.0°. Then tanθ₂ = (1.50 × 0.7071) / (1.50 × 0.7071 − 2.00) → θ₂ = 312° (48° below the x-axis) and v₂′ = 0.886 m/s. KE drops, so it's inelastic.

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8.7 Introduction to Rocket Propulsion

A rocket speeds up by throwing mass out the back; the exhaust pushes the rocket forward (Newton's third law).

Thrust=veΔmΔt\displaystyle \text{Thrust} = v_e\frac{\Delta m}{\Delta t}a=vemΔmΔt−g\displaystyle a = \frac{v_e}{m}\frac{\Delta m}{\Delta t} - gv=veln⁡m0mr\displaystyle v = v_e \ln\frac{m_0}{m_r}
Problem Setup Answer
Moon rocket liftoff: m = 2.80×10⁶ kg, burn 1.40×10⁴ kg/s, vₑ = 2.40×10³ m/s (Ex. 8.8) Thrust = 3.36×10⁷ N; 3.36×10⁷ / 2.80×10⁶ = 12.0 m/s²; minus 9.80 2.20 m/s²
Mass ratio to reach 11.2 km/s with vₑ = 2.5 km/s ln(m₀/m_r) = 4.48 → m₀/m_r = e^4.48 88 (98.9% of the rocket is fuel)

Common mistakes: forgetting to subtract g at liftoff; using the burn rate as a force; thinking rockets need air to push on.

Practice 8.7 in the quiz → · Top ↑

Chapter 11: Fluid Statics

11.1 What Is a Fluid?

Liquids, gases and plasmas are fluids because they flow when pushed sideways (shear); solids resist.

Phase Shape Volume Compress easily?
Solid Definite Definite No
Liquid Takes container's Definite No
Gas Takes container's Fills container Yes
Plasma Takes container's Fills container Yes (charged particles)

Practice 11.1 in the quiz → · Top ↑

11.2 Density

ρ=mV\displaystyle \rho = \frac{m}{V}
Substance ρ (kg/m³)
Air 1.29
Ice 917
Water 1000
Seawater 1025
Blood 1050
Aluminum 2700
Iron / steel 7800
Mercury 13,600
Gold 19,320

Ex. 11.1: a reservoir 50.0 km² × 40.0 m deep holds V = 2.00 × 10⁹ m³, so m = ρV = 2.00 × 10¹² kg.

Practice 11.2 in the quiz → · Top ↑

11.3 Pressure

P=FA\displaystyle P = \frac{F}{A}F=PA\displaystyle F = PA

Ex. 11.2: a tank at 6.90 × 10⁶ Pa with a 0.150 m diameter end: F = PA = 6.90 × 10⁶ × π(0.0750)² = 1.22 × 10⁵ N.

Practice 11.3 in the quiz → · Top ↑

11.4 Variation of Pressure with Depth

P=hρg\displaystyle P = h\rho gPtotal=Patm+hρg\displaystyle P_{total} = P_{atm} + h\rho g
Problem Setup Answer
Dam 500 m wide, 80.0 m deep (Ex. 11.3) P̄ = 40.0 × 1000 × 9.80 = 3.92 × 10⁵ Pa; F = P̄ × (80.0 × 500) 1.57 × 10¹⁰ N
Depth of water for 1 atm (Ex. 11.5) h = 1.01 × 10⁵ / (1000 × 9.80) 10.3 m

Practice 11.4 in the quiz → · Top ↑

11.5 Pascal's Principle

A pressure change in an enclosed fluid is passed on undiminished to every part of it.

F1A1=F2A2\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}F2=A2A1F1\displaystyle F_2 = \frac{A_2}{A_1}F_1

Ex. 11.6 (brakes): 500 N on a 0.500 cm cylinder drives 2.50 cm wheel cylinders: F₂ = (2.50/0.500)² × 500 = 1.25 × 10⁴ N each.

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11.6 Gauge Pressure, Absolute Pressure, and Measurement

Pabs=Pg+Patm\displaystyle P_{abs} = P_g + P_{atm}
Unit Equals
1 atm 1.013 × 10⁵ Pa = 760 mm Hg = 14.7 psi = 10.3 m water
1 mm Hg 133 Pa
1 psi 6.90 × 10³ Pa
Problem Setup Answer
Tire gauge reads 34 psi P_abs = 34 + 14.7 48.7 psi
IV bag height for 18 mm Hg (Ex. 11.7) 18 × 133 = 2400 Pa; h = 2400 / (1000 × 9.80) 0.24 m

Practice 11.6 in the quiz → · Top ↑

11.7 Archimedes' Principle

The buoyant force on an object equals the weight of the fluid it displaces.

FB=ρfluidVsub g\displaystyle F_B = \rho_{fluid} V_{sub}\, gfraction submerged=ρobjρfluid\displaystyle \text{fraction submerged} = \frac{\rho_{obj}}{\rho_{fluid}}SG=ρobjρwater\displaystyle SG = \frac{\rho_{obj}}{\rho_{water}}
Problem Setup Answer
Floating woman, 97.0% submerged (Ex. 11.9) ρ = 0.970 × 1000 970 kg/m³
Coin: 8.630 g in air, 7.800 g in water (Ex. 11.10) V = 0.830 cm³; ρ = 8.630 / 0.830 10.4 g/cm³ (silver)
1.00 × 10⁷ kg of steel as a solid block (Ex. 11.8) V = 1.28 × 10³ m³; F_B = 1.3 × 10⁷ N vs. weight 9.8 × 10⁷ N Sinks; as a hull displacing 1.00 × 10⁵ m³ it floats

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11.8 Surface Tension and Capillary Action

γ=FL\displaystyle \gamma = \frac{F}{L}P=4γr (bubble)\displaystyle P = \frac{4\gamma}{r}\ (\text{bubble})h=2γcos⁡θρgr\displaystyle h = \frac{2\gamma\cos\theta}{\rho g r}
Problem Setup Answer
Soap bubble, r = 2.00 × 10⁻⁴ m (Ex. 11.11) P = 4 × 0.037 / 2.00 × 10⁻⁴ 740 Pa (5.56 mm Hg)
Tube radius to lift sap 100 m (Ex. 11.12) r = 2(0.0728) / (1050 × 9.80 × 100) 1.41 × 10⁻⁷ m (capillary action alone can't do it)

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11.9 Pressures in the Body

Ex. 11.13: an eardrum (1.00 cm²) tears at 3.00 N: P = 3.00 × 10⁴ Pa (226 mm Hg), reached at h = P/ρg = 3.06 m of water.

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Chapter 12: Fluid Dynamics

12.1 Flow Rate and Its Relation to Velocity

Q=Vt=Avˉ\displaystyle Q = \frac{V}{t} = A\bar{v}A1v1=A2v2\displaystyle A_1v_1 = A_2v_2
Problem Setup Answer
Hose r = 0.900 cm, Q = 0.500 L/s (Ex. 12.2a) v = 5.00 × 10⁻⁴ / π(0.00900)² 1.96 m/s
Nozzle r = 0.250 cm (Ex. 12.2b) v = (0.900/0.250)² × 1.96 25.5 m/s
Aorta r = 10 mm, Q = 5.0 L/min (Ex. 12.3) v = Q / πr² 0.27 m/s; about 5 × 10⁹ capillaries

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12.2 Bernoulli's Equation

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2\displaystyle P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2

Ex. 12.4: water speeds from 1.96 to 25.5 m/s at the nozzle (P₂ = 1.01 × 10⁵ Pa): P₁ = P₂ + ½ρ(v₂² − v₁²) = 4.24 × 10⁵ Pa.

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12.3 General Applications of Bernoulli's Equation

v=2gh (Torricelli)\displaystyle v = \sqrt{2gh}\ (\text{Torricelli})Power=(P+12ρv2+ρgh)Q\displaystyle \text{Power} = \left(P + \tfrac{1}{2}\rho v^2 + \rho g h\right)Q

Ex. 12.6: pump raises 40.0 L/s from 0.700 × 10⁶ to 1.62 × 10⁶ Pa: power = 0.920 × 10⁶ × 0.0400 = 36.8 kW (about 50 hp).

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12.4 Viscosity and Laminar Flow; Poiseuille's Law

Q=P2−P1R\displaystyle Q = \frac{P_2 - P_1}{R}R=8ηlπr4\displaystyle R = \frac{8\eta l}{\pi r^4}Q=(P2−P1)πr48ηl\displaystyle Q = \frac{(P_2 - P_1)\pi r^4}{8\eta l}

Ex. 12.8: an IV needle (r = 0.150 mm, l = 2.50 cm, Q = 0.120 cm³/s, vein at 8.00 mm Hg) needs P₂ = 8ηlQ/πr⁴ + P₁ = 1.62 × 10⁴ Pa.

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12.5 The Onset of Turbulence

NR=2ρvrη\displaystyle N_R = \frac{2\rho v r}{\eta}
Reynolds number (tube) Flow
Below about 2000 Laminar
2000–3000 Unstable, can switch
Above about 3000 Turbulent

Ex. 12.9: the IV needle's flow (ρ = 1025, v = 1.70 m/s, r = 0.150 mm, η = 1.00 × 10⁻³) has N_R = 523, laminar.

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12.6 Motion of an Object in a Viscous Fluid

NR′=ρvLη\displaystyle N_R' = \frac{\rho v L}{\eta}FS=6πRηv (Stokes’ law)\displaystyle F_S = 6\pi R\eta v\ (\text{Stokes' law})

Ex. 12.10: a 7.40 cm ball at 40.0 m/s in air: N_R′ = 1.29 × 40.0 × 0.0740 / 1.81 × 10⁻⁵ = 2.11 × 10⁵, so it has a turbulent wake.

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12.7 Diffusion, Osmosis and Related Processes

xrms=2Dt\displaystyle x_{rms} = \sqrt{2Dt}

Ex. 12.11: glucose spreading 1.0 cm in water: t = x²/2D = (0.010)² / (2 × 6.7 × 10⁻¹⁰) = 7.5 × 10⁴ s (21 h).

Common mistakes (Unit 6): using gauge pressure where absolute is needed (or the reverse); diameter instead of radius in A = πr²; forgetting r⁴ in Poiseuille's law; using the object's volume instead of the submerged volume for buoyancy; mixing L and m³.

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Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

13.1 Temperature

Temperature is what a thermometer measures. Later in the chapter it gets a deeper meaning: the average kinetic energy of the molecules.

T°F=95T°C+32\displaystyle T_{°F} = \tfrac{9}{5}T_{°C} + 32T°C=59(T°F−32)\displaystyle T_{°C} = \tfrac{5}{9}(T_{°F} - 32)TK=T°C+273.15\displaystyle T_K = T_{°C} + 273.15
Reference point °F °C K
Absolute zero −459.67 −273.15 0
Water freezes 32 0 273.15
Room temperature 77 25 298
Body temperature 98.6 37.0 310
Water boils 212 100 373.15

Ex. 13.1: room temperature 25 °C: T = (9/5)(25) + 32 = 77 °F, and T = 25 + 273.15 = 298 K.

Ex. 13.2 (Reaumur scale, water freezes at 0 °R and boils at 80 °R): T = 0.8 × 25 = 20 °R.

Practice 13.1 in the quiz → · Top ↑

13.2 Thermal Expansion of Solids and Liquids

Hotter atoms vibrate harder and push their neighbors a little farther away, so the whole object grows.

ΔL=αLΔT\displaystyle \Delta L = \alpha L \Delta TΔA=2αAΔT\displaystyle \Delta A = 2\alpha A \Delta TΔV=βVΔT\displaystyle \Delta V = \beta V \Delta Tβ≈3α\displaystyle \beta \approx 3\alpha
Material α (1/°C) β (1/°C)
Aluminum 25 × 10⁻⁶ 75 × 10⁻⁶
Brass 19 × 10⁻⁶ 56 × 10⁻⁶
Copper 17 × 10⁻⁶ 51 × 10⁻⁶
Iron or steel 12 × 10⁻⁶ 35 × 10⁻⁶
Concrete 12 × 10⁻⁶ 36 × 10⁻⁶
Glass (ordinary) 9 × 10⁻⁶ 27 × 10⁻⁶
Glass (Pyrex) 3 × 10⁻⁶ 9 × 10⁻⁶
Water — 210 × 10⁻⁶
Mercury — 180 × 10⁻⁶
Gasoline (petrol) — 950 × 10⁻⁶
Air (and most gases) — 3400 × 10⁻⁶
Problem Setup Answer
Golden Gate Bridge, 1275 m of steel, −15 °C to 40 °C (Ex. 13.3) ΔL = 12 × 10⁻⁶ × 1275 × 55 0.84 m
60.0 L steel tank full of gas, 15.0 °C → 35.0 °C (Ex. 13.4) V_spill = (950 − 35) × 10⁻⁶ × 60.0 × 20.0 1.10 L spills
Same gas, not allowed to expand, B = 1.00 × 10⁹ Pa (Ex. 13.5) P = (1.10 / 60.0) × 1.00 × 10⁹ 1.83 × 10⁷ Pa (about 2500 psi)

Practice 13.2 in the quiz → · Top ↑

13.3 The Ideal Gas Law

Gas molecules are far apart compared with their size, so all gases act almost the same way. An ideal gas has tiny molecules that only interact by bouncing (elastic collisions).

PV=NkT\displaystyle PV = NkTPV=nRT\displaystyle PV = nRTn=NNA\displaystyle n = \frac{N}{N_A}P1V1T1=P2V2T2\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
Constant Value
Boltzmann constant k 1.38 × 10⁻²³ J/K
Avogadro's number N_A 6.02 × 10²³ mol⁻¹
Gas constant R = N_A k 8.31 J/(mol·K) = 0.0821 L·atm/(mol·K) = 1.99 cal/(mol·K)
1 atm 1.01 × 10⁵ Pa
Problem Setup Answer
Bike tire at 7.00 × 10⁵ Pa, 18.0 °C → 35.0 °C (Ex. 13.6) P_f = 7.00 × 10⁵ × 308/291 7.41 × 10⁵ Pa
Molecules in 1.00 m³ at STP (Ex. 13.7) N = (1.01 × 10⁵)(1.00) / (1.38 × 10⁻²³ × 273) 2.68 × 10²⁵ molecules
Moles per m³ and liters per mole at STP (Ex. 13.8) 2.68 × 10²⁵ / 6.02 × 10²³; 1000 / 44.5 44.5 mol/m³; 22.5 L/mol
Moles in a 2.00 L tire at 7.00 × 10⁵ Pa, 18.0 °C (Ex. 13.9) n = (7.00 × 10⁵)(2.00 × 10⁻³) / (8.31 × 291) 0.579 mol

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13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

Kinetic theory — gas pressure comes from huge numbers of molecules bouncing off the walls. Each bounce flips the molecule's momentum (Δp = 2mv), which pushes on the wall.

PV=13Nmv2‾\displaystyle PV = \tfrac{1}{3}Nm\overline{v^2}KE‾=12mv2‾=32kT\displaystyle \overline{KE} = \tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}kTvrms=3kTm\displaystyle v_{rms} = \sqrt{\frac{3kT}{m}}
Problem Setup Answer
Average KE of a gas molecule at 20.0 °C (Ex. 13.10a) KE = 1.5 × 1.38 × 10⁻²³ × 293 6.07 × 10⁻²¹ J
v_rms of N₂ at 20.0 °C, m = 4.65 × 10⁻²⁶ kg (Ex. 13.10b) v = √(3 × 1.38 × 10⁻²³ × 293 / 4.65 × 10⁻²⁶) 511 m/s
T for helium (m = 6.65 × 10⁻²⁷ kg) to reach v_rms = 11.1 km/s (Ex. 13.11) T = m v² / 3k 1.98 × 10⁴ K

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13.5 Phase Changes

At low temperatures or high pressures, real gases stop acting ideal and turn into liquids and solids.

PV=constant (ideal gas, fixed T)\displaystyle PV = \text{constant}\ (\text{ideal gas, fixed } T)Ptotal=P1+P2+P3+⋯ (Dalton)\displaystyle P_{total} = P_1 + P_2 + P_3 + \cdots\ (\text{Dalton})
Substance Critical T (°C) Critical P (atm) Triple point T (K) Triple point P (atm)
Water 374.3 219.0 273.16 0.00600
Carbon dioxide 31.1 73.2 216.55 5.11
Nitrogen −146.9 33.6 63.18 0.124
Oxygen −118.4 50.3 54.36 0.00151

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13.6 Humidity, Evaporation, and Boiling

"It's not the heat, it's the humidity." Sweat cools you by evaporating, and humid air slows evaporation.

% relative humidity=vapor densitysaturation vapor density×100\displaystyle \%\ \text{relative humidity} = \frac{\text{vapor density}}{\text{saturation vapor density}} \times 100nV=PRT\displaystyle \frac{n}{V} = \frac{P}{RT}ρ=nV×M\displaystyle \rho = \frac{n}{V} \times M
T (°C) Vapor pressure (Pa) Saturation vapor density (g/m³)
−10 2.60 × 10² 2.36
0 6.10 × 10² 4.84
10 1.19 × 10³ 9.40
15 1.69 × 10³ 12.8
20 2.33 × 10³ 17.2
25 3.17 × 10³ 23.0
30 4.24 × 10³ 30.4
37 6.31 × 10³ 44.0
40 7.34 × 10³ 51.1
100 1.01 × 10⁵ 598
Problem Setup Answer
Vapor density at 20.0 °C from P = 2.33 × 10³ Pa (Ex. 13.12) n/V = 2.33 × 10³ / (8.31 × 293) = 0.957 mol/m³; × 18.0 17.2 g/m³ (matches the table)
25.0 °C air with 9.40 g/m³ (Ex. 13.13a, b) 9.40 / 23.0 × 100; dew point where saturation = 9.40 40.9%; dew point 10.0 °C
25.0 °C, dew point −10.0 °C (Ex. 13.13c) 2.36 / 23.0 × 100 10.3%

Common mistakes (Chapter 13): using °C instead of K in the gas law or in KE = (3/2)kT (ΔT is fine in either); using gauge pressure instead of absolute pressure; mixing up N (molecules) with n (moles), or k with R; using grams per mole instead of kg per mole when finding a molecule's mass; forgetting the square root in v_rms; using α instead of β for volume expansion; thinking air "holds" water vapor.

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Chapter 14: Heat and Heat Transfer Methods

14.1 Heat

Heat is energy that moves on its own because of a temperature difference. It always flows from hot to cold and stops at thermal equilibrium (same temperature).

1.000 kcal=4186 J\displaystyle 1.000\ \text{kcal} = 4186\ \text{J}1 kcal=1000 cal=1 Cal (food)\displaystyle 1\ \text{kcal} = 1000\ \text{cal} = 1\ \text{Cal (food)}

Common mistake: mixing kcal and J. Multiply kcal by 4186 to get J; a 250 Cal snack is 250 kcal = 1.05 × 10⁶ J.

Practice 14.1 in the quiz → · Top ↑

14.2 Temperature Change and Heat Capacity

Q=mcΔT\displaystyle Q = mc\Delta TΔT=Qmc\displaystyle \Delta T = \frac{Q}{mc}Tf=m1c1T1+m2c2T2m1c1+m2c2\displaystyle T_f = \frac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1 + m_2c_2}
Substance c (J/kg·°C)
Water (15 °C) 4186
Human body (avg) 3500
Ethanol 2450
Ice (−50 to 0 °C) 2090
Steam (100 °C, constant P) 2020
Wood 1700
Air (dry, constant P) 1015
Aluminum 900
Glass, concrete 840
Iron, steel 452
Copper 387
Silver 235
Mercury 139
Gold 129
Lead 128

Ex. 14.1: 0.500 kg aluminum pan + 0.250 kg water, heated from 20.0 to 80.0 °C: Q_w = 0.250 × 4186 × 60.0 = 62.8 kJ, Q_Al = 0.500 × 900 × 60.0 = 27.0 kJ → 89.8 kJ total (69.9% to the water).

Ex. 14.2: a 10,000 kg truck drops 75.0 m; 10% of Mgh = 7.35 × 10⁵ J goes into 100 kg of brake material (c = 800): ΔT = 7.35 × 10⁵ / (100 × 800) → 9.2 °C.

Ex. 14.3: 0.250 kg of 20.0 °C water poured into a 0.500 kg aluminum pan at 150 °C: T_f = (0.500·900·150 + 0.250·4186·20.0) / (0.500·900 + 0.250·4186) → 59.1 °C (close to the water's start, because water's c is big).

Common mistakes: using the final temperature instead of ΔT; forgetting the container (the pan) also soaks up heat; using liters as kilograms for anything but water.

Practice 14.2 in the quiz → · Top ↑

14.3 Phase Change and Latent Heat

Q=mLf (melt/freeze)\displaystyle Q = mL_f\ (\text{melt/freeze})Q=mLv (boil/condense)\displaystyle Q = mL_v\ (\text{boil/condense})Q=mLs (sublimate)\displaystyle Q = mL_s\ (\text{sublimate})
Substance Melting pt (°C) L_f (kJ/kg) Boiling pt (°C) L_v (kJ/kg)
Water 0.00 334 100.0 2256 (2430 at 37 °C)
Ethanol −114 104 78.3 854
Nitrogen −210.0 25.5 −195.8 201
Mercury −38.9 11.8 357 272
Lead 327 24.5 1750 871
Aluminum 660 380 2450 11,400
Copper 1083 134 2595 5069
Gold 1063 64.5 2660 1578

Ex. 14.4: three 6.0 g ice cubes at 0 °C chill 0.25 kg of soda at 20 °C. Ice first melts (0.018 × 334,000 = 6012 J), then that meltwater warms up: T_f = (20,930 − 6012) / 1122 → 13 °C.

Problem strategy: check whether a melting or boiling point is crossed. If so, work in steps: temperature change → phase change → temperature change, then add the Q's.

Common mistakes: skipping the plateau (forgetting mL_f or mL_v); using water's c for ice or steam; using kJ/kg values as J/kg (334 kJ/kg = 334,000 J/kg).

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14.4 Heat Transfer Methods

Every heat transfer happens by one of three methods, and all three are driven only by a temperature difference.

Method How it works Example
Conduction Collisions between molecules in matter that isn't flowing; needs contact Stove burner → bottom of a pan; metal bat handle on a cold night
Convection A fluid (liquid or gas) moves and carries the energy with it Forced-air furnace, weather, boiling pot, car cooling system
Radiation Electromagnetic waves (mostly infrared) are emitted and absorbed; no medium needed Sun warming Earth, feeling a campfire from across the yard

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14.5 Conduction

Qt=kA(T2−T1)d\displaystyle \frac{Q}{t} = \frac{kA(T_2 - T_1)}{d}R=dk\displaystyle R = \frac{d}{k}
Substance k (J/s·m·°C)
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel (iron) 80
Stainless steel 14
Ice 2.2
Glass, concrete brick 0.84
Water 0.6
Fatty tissue 0.2
Wood 0.08–0.16
Snow (dry) 0.10
Wool 0.04
Down feathers 0.025
Air 0.023
Styrofoam 0.010

Ex. 14.5: a Styrofoam cooler (A = 0.950 m², d = 2.50 cm) at 0 °C inside, 35.0 °C outside: Q/t = 0.010 × 0.950 × 35.0 / 0.0250 = 13.3 W; in one day Q = 1.15 × 10⁶ J, melting m = Q/L_f → 3.44 kg of ice.

Ex. 14.6: water boils off at 1.00 g/s (2256 W) through an aluminum pan bottom 0.800 cm thick and 14.0 cm across: ΔT = (Q/t)·d/(kA) = 2256 × 0.00800 / (220 × 0.0154) → 5.33 °C.

Common mistakes: leaving d in cm (convert to m); using diameter for the radius when finding A; forgetting Q/t is a rate, so multiply by time (in seconds) to get Q.

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14.6 Convection

Qt=mcΔTt\displaystyle \frac{Q}{t} = \frac{mc\Delta T}{t}Qt=mLvt (sweat)\displaystyle \frac{Q}{t} = \frac{mL_v}{t}\ (\text{sweat})

Ex. 14.7: a 12.0 × 18.0 × 3.00 m house (m = 1.29 × 648 = 836 kg of air, c ≈ 1000) has its air replaced every 30.0 min, warmed by 10.0 °C: Q/t = 836 × 1000 × 10.0 / 1800 → 4.64 kW.

Ex. 14.8: a resting person makes 120 W. To shed it all by sweating (L_v = 2430 J/g at body temperature): m/t = 120 / 2430 = 0.0494 g/s ≈ 2.96 g/min.

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14.7 Radiation

Qt=σeAT4\displaystyle \frac{Q}{t} = \sigma e A T^4Qnett=σeA(T24−T14)\displaystyle \frac{Q_{net}}{t} = \sigma e A\left(T_2^4 - T_1^4\right)σ=5.67×10−8 J/(s⋅m2⋅K4)\displaystyle \sigma = 5.67\times10^{-8}\ \text{J/(s}\cdot\text{m}^2\cdot\text{K}^4\text{)}
Surface Emissivity e
Ideal blackbody 1
Carbon black 0.99
Human skin (infrared) 0.97
Earth (average) 0.65
Tungsten filament about 0.5
Perfect reflector 0

Ex. 14.9: unclothed person, skin 33.0 °C (306 K), room 22.0 °C (295 K), A = 1.50 m², e = 0.97: Q/t = 5.67 × 10⁻⁸ × 0.97 × 1.50 × (295⁴ − 306⁴) → −99 W (losing about 99 W).

Common mistakes: plugging °C into T⁴ (always add 273); computing (T₂ − T₁)⁴ instead of T₂⁴ − T₁⁴; forgetting e.

Common mistakes (Chapter 14): kcal vs J (× 4186); skipping a phase-change plateau; using water's c for ice or steam; ΔT is the same in K and °C but T⁴ needs kelvin; cm instead of m in conduction thickness; forgetting Q/t is a rate (watts).

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Chapter 15: Thermodynamics

15.1 The First Law of Thermodynamics

The first law is conservation of energy for a system where energy moves in or out by heat and work.

ΔEint=Q−W\displaystyle \Delta E_{int} = Q - W
Quantity Positive (+) means Negative (−) means
Q Heat flows into the system Heat flows out of the system
W Work done by the system (gas expands) Work done on the system (gas is compressed)
ΔE_int Internal energy goes up Internal energy goes down

Ex. 15.1a: 40.00 J of heat in and 10.00 J of work done by the system, then 25.00 J of heat out and 4.00 J of work done on it: Q = 15.00 J, W = 6.00 J, so ΔE_int = 9.00 J.

Ex. 15.1b: 150.00 J of heat out (Q = −150.00 J) and 159.00 J of work done on the system (W = −159.00 J): ΔE_int = −150.00 − (−159.00) = 9.00 J. Different path, same change.

Human metabolism

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15.2 The First Law of Thermodynamics and Some Simple Processes

W=PΔV (constant pressure)\displaystyle W = P\Delta V\ (\text{constant pressure})W=area under the PV curve\displaystyle W = \text{area under the } PV \text{ curve}Wcycle=area inside the loop\displaystyle W_{cycle} = \text{area inside the loop}
Process What stays constant Key fact First law becomes
Isobaric Pressure W = PΔV ΔE_int = Q − PΔV
Isochoric Volume W = 0 (vertical line on PV) ΔE_int = Q
Isothermal Temperature ΔE_int = 0 (ideal gas) Q = W
Adiabatic No heat flow Q = 0 ΔE_int = −W
Eint=32NkT=32nRT\displaystyle E_{int} = \tfrac{3}{2}NkT = \tfrac{3}{2}nRTΔEint=32nR ΔT (monatomic)\displaystyle \Delta E_{int} = \tfrac{3}{2}nR\,\Delta T\ (\text{monatomic})

Ex. 15.2: a rectangle cycle ABCDA with P_AB = 1.50 × 10⁶ Pa, P_CD = 2.00 × 10⁵ Pa, ΔV = 5.00 × 10⁻⁴ m³: W_AB = 750 J, W_CD = −100 J, the two vertical legs do 0, so W = (1.50 × 10⁶ − 2.00 × 10⁵)(5.00 × 10⁻⁴) = 650 J.

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15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency

Second law (first form): heat flows on its own from hot to cold, never from cold to hot.

Second law (second form): in a cycle, no engine can turn heat from a reservoir completely into work. Some Q_c must go out.

W=Qh−Qc\displaystyle W = Q_h - Q_cEff=WQh=1−QcQh\displaystyle Eff = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}

Ex. 15.3: a coal plant takes in Q_h = 2.50 × 10¹⁴ J per day and dumps Q_c = 1.48 × 10¹⁴ J. W = 1.02 × 10¹⁴ J; Eff = 1.02/2.50 = 0.408 (40.8%). At 2.5 × 10⁶ J/kg it burns 1.0 × 10⁸ kg of coal, making 1.0 × 10⁸ × 44/12 = 3.7 × 10⁸ kg of CO₂ per day.

The Otto cycle (gasoline engine)

Path Process Engine stroke
A→B Adiabatic compression (work done on gas) Compression stroke
B→C Isochoric, Q_h added Spark ignites fuel
C→D Adiabatic expansion (gas does work) Power stroke
D→A Isochoric, Q_c released Exhaust + intake

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15.4 Carnot's Perfect Heat Engine: The Second Law of Thermodynamics Restated

EffC=1−TcTh\displaystyle Eff_C = 1 - \frac{T_c}{T_h}(T in kelvin: TK=T°C+273)\displaystyle (T \text{ in kelvin: } T_K = T_{°C} + 273)

Ex. 15.4: a nuclear reactor with water at 300 °C (573 K) and a condenser at 27 °C (300 K): Eff_C = 1 − 300/573 = 0.476 (47.6%). The real plant gets about 35%.

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15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators

Heat pumps, air conditioners and refrigerators are heat engines run backward: work input W moves heat from cold to hot.

Qh=Qc+W\displaystyle Q_h = Q_c + WCOPhp=QhW=1Eff\displaystyle COP_{hp} = \frac{Q_h}{W} = \frac{1}{Eff}COPref=QcW=COPhp−1\displaystyle COP_{ref} = \frac{Q_c}{W} = COP_{hp} - 1

Ex. 15.5: a heat pump with T_h = 45.0 °C (318 K) and T_c = −15.0 °C (258 K): Eff_C = 1 − 258/318 = 0.1887, so COP_hp = 1/0.1887 = 5.30. Each 1 J of electricity delivers 5.30 J to the house (4.3 J of it from the cold outside air).

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15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy

ΔS=QT (reversible, T in K)\displaystyle \Delta S = \frac{Q}{T}\ (\text{reversible, } T \text{ in K})ΔStot=ΔSh+ΔSc\displaystyle \Delta S_{tot} = \Delta S_h + \Delta S_cWunavail=ΔS⋅T0\displaystyle W_{unavail} = \Delta S \cdot T_0
Problem Setup Answer
4000 J flows from 600 K to 250 K (Ex. 15.6) ΔS_h = −4000/600 = −6.67 J/K; ΔS_c = 4000/250 = 16.0 J/K ΔS_tot = 9.33 J/K
Carnot engine 600 K → 100 K with 4000 J (Ex. 15.7a) Eff = 1 − 100/600 = 0.833 W = 3333 J
Same 4000 J first drops to 250 K (Ex. 15.7b) Eff = 1 − 100/250 = 0.600 W = 2400 J; 933 J lost = 9.33 × 100
Melt 1.00 kg of ice at 0 °C (Ex. 15.8) Q = mL_f = 3.34 × 10⁵ J; ΔS = Q/273 1.22 × 10³ J/K

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15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

The second law happens because disorder is overwhelmingly more likely than order.

S=kln⁡W\displaystyle S = k\ln WΔS=kln⁡Wf−kln⁡Wi=kln⁡WfWi\displaystyle \Delta S = k\ln W_f - k\ln W_i = k\ln\frac{W_f}{W_i}k=1.38×10−23 J/K\displaystyle k = 1.38\times10^{-23}\ \text{J/K}
5 coins 5H 4H 1T 3H 2T 2H 3T 1H 4T 5T Total
Microstates W 1 5 10 10 5 1 32

Ex. 15.9: 100 coins go from 60 H/40 T (W = 1.4 × 10²⁸) to 50 H/50 T (W = 1.0 × 10²⁹): ΔS = (1.38 × 10⁻²³)[ln(1.0 × 10²⁹) − ln(1.4 × 10²⁸)] = 2.7 × 10⁻²³ J/K.

Common mistakes (Chapter 15): sign of W (positive when the gas expands and does work; negative when compressed); sign of Q (positive into the system); forgetting ΔE_int = 0 for a full cycle or an isothermal ideal gas; using °C instead of kelvin in Eff_C, COP or ΔS = Q/T; mixing up efficiency (W/Q_h, always < 1) with **COP** (Q_h/W or Q_c/W, usually > 1); using COP_hp for a fridge (fridge uses Q_c); confusing W for work with W for number of microstates.

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Chapter 16: Oscillatory Motion and Waves

16.1 Hooke's Law: Stress and Strain Revisited

F=−kx\displaystyle F = -kxk=−Fx\displaystyle k = -\frac{F}{x}PEel=12kx2\displaystyle PE_{el} = \tfrac{1}{2}kx^2

Ex. 16.1: a car settles 1.20 cm when an 80.0 kg person gets in: F = 80.0 × 9.80 = 784 N, k = 784 / 0.0120 = 6.53 × 10⁴ N/m.

Ex. 16.2: a toy gun spring (k = 50.0 N/m) compressed 0.150 m stores PE = ½(50.0)(0.150)² = 0.563 J, and launches a 2.00 g dart at v = √(2 × 0.563 / 0.00200) = 23.7 m/s.

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16.2 Period and Frequency in Oscillations

f=1T\displaystyle f = \frac{1}{T}T=1f\displaystyle T = \frac{1}{f}1 Hz=1 cycle/s=1 s−1\displaystyle 1\ \text{Hz} = 1\ \text{cycle/s} = 1\ \text{s}^{-1}
Problem Setup Answer
Ultrasound, T = 0.400 µs (Ex. 16.3a) f = 1 / 0.400 × 10⁻⁶ s 2.50 × 10⁶ Hz
Middle C, f = 264 Hz (Ex. 16.3b) T = 1 / 264 3.79 × 10⁻³ s = 3.79 ms

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16.3 Simple Harmonic Motion: A Special Periodic Motion

T=2πmk\displaystyle T = 2\pi\sqrt{\frac{m}{k}}f=12πkm\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}
x(t)=Xcos⁡2πtT\displaystyle x(t) = X\cos\frac{2\pi t}{T}v(t)=−vmaxsin⁡2πtT\displaystyle v(t) = -v_{max}\sin\frac{2\pi t}{T}a(t)=−kXmcos⁡2πtT\displaystyle a(t) = -\frac{kX}{m}\cos\frac{2\pi t}{T}

Ex. 16.4: a 900 kg car with k = 6.53 × 10⁴ N/m (bad shocks): f = (1/2π)√(6.53 × 10⁴ / 900) = 1.36 Hz, T = 1/f = 0.738 s.

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16.4 The Simple Pendulum

T=2πLg\displaystyle T = 2\pi\sqrt{\frac{L}{g}}g=4π2LT2\displaystyle g = \frac{4\pi^2 L}{T^2}L=gT24π2\displaystyle L = \frac{gT^2}{4\pi^2}

Ex. 16.5: a 75.000 cm pendulum has a period of 1.7357 s: g = 4π²(0.75000) / (1.7357)² = 9.8281 m/s².

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16.5 Energy and the Simple Harmonic Oscillator

12mv2+12kx2=12kX2\displaystyle \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kX^2vmax=Xkm\displaystyle v_{max} = X\sqrt{\frac{k}{m}}v=±vmax1−x2X2\displaystyle v = \pm v_{max}\sqrt{1 - \frac{x^2}{X^2}}

Ex. 16.6: the 900 kg car (k = 6.53 × 10⁴ N/m) bounces with amplitude 0.100 m: v_max = √(6.53 × 10⁴ / 900) × 0.100 = 0.852 m/s.

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16.6 Uniform Circular Motion and Simple Harmonic Motion

x=Xcos⁡ωt\displaystyle x = X\cos\omega tω=2πT\displaystyle \omega = \frac{2\pi}{T}T=2πXvmax\displaystyle T = \frac{2\pi X}{v_{max}}

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16.7 Damped Harmonic Motion

Wnc=Δ(KE+PE)\displaystyle W_{nc} = \Delta(KE + PE)f=μkmg\displaystyle f = \mu_k mg
Type What happens Example
Underdamped Oscillates, crossing equilibrium many times while the amplitude shrinks Guitar string, playground swing
Critically damped Returns to equilibrium as fast as possible without oscillating (overshoots at most once) Good car shocks, bathroom scale needle
Overdamped Creeps slowly back to equilibrium without oscillating Door closer set too stiff

Ex. 16.7: a 0.200 kg object on a spring (k = 50.0 N/m) with μ_k = 0.0800 feels friction f = 0.0800 × 0.200 × 9.80 = 0.157 N. Released 0.100 m from equilibrium, it travels a total of 1.59 m before stopping.

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16.8 Forced Oscillations and Resonance

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16.9 Waves

vw=λT\displaystyle v_w = \frac{\lambda}{T}vw=fλ\displaystyle v_w = f\lambda

Ex. 16.8: ocean wave crests 10.0 m apart; a gull bobs once every 5.00 s: v_w = λ/T = 10.0 / 5.00 = 2.00 m/s.

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16.10 Superposition and Interference

f1=vw2L\displaystyle f_1 = \frac{v_w}{2L}fn=nf1\displaystyle f_n = n f_1λn=2Ln\displaystyle \lambda_n = \frac{2L}{n}fB=∣f1−f2∣\displaystyle f_B = |f_1 - f_2|

Ex. (beats): a 256 Hz tuning fork and a string give 2 beats per second, so the string is 254 Hz or 258 Hz.

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16.11 Energy in Waves: Intensity

I=PA\displaystyle I = \frac{P}{A}E=IAt\displaystyle E = IAtI∝X2\displaystyle I \propto X^2
Problem Setup Answer
700 W/m² on 0.500 m² for 4.00 h (Ex. 16.9a) E = 700 × 0.500 × (4.00 × 3600) 5.04 × 10⁶ J
Same sunlight focused to 1/200 the area (Ex. 16.9b) I′ = 200 × 700 1.40 × 10⁵ W/m²
Two 1.00 W/m² waves, perfect constructive (Ex. 16.10) amplitude 2X → intensity (2)² = 4× 4.00 W/m²

Common mistakes (Chapter 16): mixing up period and frequency (T = 1/f; seconds vs. Hz); thinking a bigger amplitude changes the period (it doesn't, for springs or small-angle pendulums); thinking a heavier bob changes a pendulum's period (only L and g matter); forgetting the square root in T = 2π√(m/k) or 2π√(L/g); forgetting 2π; using cm instead of m for x or L; thinking intensity ∝ amplitude (it's ∝ amplitude²); thinking the water moves along with a wave (only the energy does); using λ = L instead of λ = 2L for a string's fundamental.

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Chapter 17: Physics of Hearing

17.1 Sound

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17.2 Speed of Sound, Frequency, and Wavelength

vw=fλ\displaystyle v_w = f\lambdavw=(331 m/s)T273 K\displaystyle v_w = (331\ \text{m/s})\sqrt{\frac{T}{273\ \text{K}}}
Medium v (m/s)
Air (0°C) 331
Air (20°C) 343
Carbon dioxide (0°C) 259
Helium (0°C) 965
Hydrogen (0°C) 1290
Fresh water (20°C) 1480
Seawater / human tissue 1540
Vulcanized rubber 54
Lead 1960
Marble 3810
Aluminum 5120
Pyrex glass 5640
Steel 5960

Ex. 17.1: at 30.0°C, v = 331√(303/273) = 348.7 m/s. Wavelengths: 348.7/20 = 17 m and 348.7/20,000 = 1.7 cm.

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17.3 Sound Intensity and Sound Level

I=PA\displaystyle I = \frac{P}{A}I=(Δp)22ρvw\displaystyle I = \frac{(\Delta p)^2}{2\rho v_w}β (dB)=10log⁡10 ⁣(II0)\displaystyle \beta\ (\text{dB}) = 10\log_{10}\!\left(\frac{I}{I_0}\right)I0=10−12 W/m2\displaystyle I_0 = 10^{-12}\ \text{W/m}^2
β (dB) I (W/m²) Example
0 10⁻¹² Threshold of hearing at 1000 Hz
10 10⁻¹¹ Rustle of leaves
20 10⁻¹⁰ Whisper at 1 m
40 10⁻⁸ Average home
60 10⁻⁶ Normal conversation
70 10⁻⁵ Busy traffic
80 10⁻⁴ Loud radio, classroom lecture
90 10⁻³ Inside a heavy truck
100 10⁻² Noisy factory, siren at 30 m
120 1 Loud rock concert; threshold of pain
140 10² Jet at 30 m; damage in seconds
160 10⁴ Eardrums burst
I₂/I₁ β₂ − β₁
2 3 dB
5 7 dB
10 10 dB

Ex. 17.2: Δp = 0.656 Pa in 0°C air: I = 0.656² / (2 × 1.29 × 331) = 5.04 × 10⁻⁴ W/m², so β = 10 log(5.04 × 10⁸) = 87 dB.

Ex. 17.3: twice the intensity: β₂ − β₁ = 10 log(2.00) = 3.01 dB.

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17.4 Doppler Effect and Sonic Booms

fobs=fs(vwvw±vs) (moving source)\displaystyle f_{obs} = f_s\left(\frac{v_w}{v_w \pm v_s}\right)\ (\text{moving source})fobs=fs(vw±vobsvw) (moving observer)\displaystyle f_{obs} = f_s\left(\frac{v_w \pm v_{obs}}{v_w}\right)\ (\text{moving observer})

Ex. 17.4: a 150 Hz horn on a train at 35.0 m/s (v = 340 m/s): approaching 150 × 340/305 = 167 Hz; leaving 150 × 340/375 = 136 Hz. The engineer hears 150 Hz.

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17.5 Sound Interference and Resonance: Standing Waves in Air Columns

fn=nvw4L, n=1,3,5,… (closed one end)\displaystyle f_n = n\frac{v_w}{4L},\ n = 1, 3, 5, \dots\ (\text{closed one end})fn=nvw2L, n=1,2,3,… (open both ends)\displaystyle f_n = n\frac{v_w}{2L},\ n = 1, 2, 3, \dots\ (\text{open both ends})fB=∣f1−f2∣\displaystyle f_B = |f_1 - f_2|
Tube Ends Fundamental λ Harmonics present Examples
Closed at one end Node + antinode 4L Odd only: f, 3f, 5f… Clarinet, trumpet, your throat, a bottle
Open at both ends Antinode + antinode 2L All: f, 2f, 3f… Flute, some organ pipes, oboe

Ex. 17.5: a closed tube with a 128 Hz fundamental at 22.0°C: v = 331√(295/273) = 344 m/s, L = 344 / (4 × 128) = 0.672 m. Its 4th overtone is n = 9: 9 × 128 = 1.15 kHz.

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17.6 Hearing

Perception Physical quantity
Pitch Frequency
Loudness Intensity and frequency
Timbre / tone Number and relative intensity of the frequencies

Ex. 17.6: from the equal-loudness graph, a 100 Hz sound at 80 dB has a loudness of 75 phons; a 4000 Hz sound at 70 phons is only 67 dB.

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17.7 Ultrasound

Z=ρv\displaystyle Z = \rho va=(Z2−Z1)2(Z1+Z2)2\displaystyle a = \frac{(Z_2 - Z_1)^2}{(Z_1 + Z_2)^2}λ=vwf\displaystyle \lambda = \frac{v_w}{f}
Medium ρ (kg/m³) v (m/s) Z (kg/(m²·s))
Air 1.3 330 429
Water 1000 1500 1.5 × 10⁶
Blood 1060 1570 1.66 × 10⁶
Fat 925 1450 1.34 × 10⁶
Muscle 1075 1590 1.70 × 10⁶
Bone 1400–1900 4080 5.7–7.8 × 10⁶
Problem Setup Answer
Fat to muscle (Ex. 17.7) Z_fat = 925 × 1450; a = (1.34 − 1.70)² / (1.34 + 1.70)² 1.34 × 10⁶; a = 0.014 (1.4% reflected)
7 MHz scan in tissue λ = 1540 / 7 × 10⁶; depth = 500λ 0.22 mm detail; 0.11 m deep
2.50 MHz on blood at 20.0 cm/s toward probe (Ex. 17.8) Blood receives 2,500,325 Hz; echo returns 2,500,649 Hz Beat frequency 649 Hz

Common mistakes (Chapter 17): treating decibels as linear (doubling intensity adds 3 dB, it doesn't double the dB); using ln instead of log₁₀; using °C instead of kelvin in v = 331√(T/273); thinking higher frequency travels faster (speed depends on the medium only); using 2L for a closed tube (it's 4L) or allowing even harmonics in a closed tube; flipping the Doppler sign (approaching must give a higher frequency); forgetting that an echo travels there and back.

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