Concepts

PHYSICSBOWL COMPREHENSIVE REVISION NOTES
PHYSICSBOWL COMPREHENSIVE REVISION NOTES A Complete Guide to Acing the PhysicsBowl Exam TABLE OF CONTENTS     Mechanics - Kinematics     Mechanics - Newton's Laws     Work, Energy, and Power     Linear Momentum and Collisions     Rotational Motion     Gravitation     Simple Harmonic Motion     Fluid Mechanics     Electric Fields and Capacitance     Circuits     Magnetism     Waves and Sound     Optics     Thermodynamics     Modern Physics   1. MECHANICS - KINEMATICS 1.1 Fundamental Concepts Kinematics is the branch of mechanics that describes motion without considering its causes. In PhysicsBowl, kinematics questions frequently test your ability to use the kinematic equations and understand motion graphs. Key Quantities Quantity    Symbol    Unit    Definition Position    x, y    m    Location relative to origin Velocity    v    m/s    Rate of change of position Acceleration    a    m/s²    Rate of change of velocity Time    t    s    Duration of motion The Kinematic Equations For constant acceleration: v=v_0+at x=x_0+v_0 t+1/2 at^2 v^2=v_0^2+2a(x-x_0) x-x_0=1/2(v_0+v)t Free Fall All objects in free fall (ignoring air resistance) experience the same acceleration: g=9.8〖" m/s" 〗^2≈10〖" m/s" 〗^2 For objects thrown upward, remember:     At maximum height, velocity = 0     Time going up = time coming down (for same height)     Velocity at return = negative of initial velocity Projectile Motion Horizontal and vertical motions are independent:     Horizontal: Constant velocity (a_x = 0)     Vertical: Constant acceleration downward (a_y = -g) Key formulas for projectile motion:     Time of flight: t=(2v_0 sin⁡θ)/g     Maximum height: H=(v_0^2 〖sin⁡〗^2 θ)/2g     Range: R=(v_0^2 sin⁡(2θ))/g The range is maximum at θ = 45°. 1.2 Example Questions Example 1 (From typical PhysicsBowl): A car accelerates from rest at 3.0 m/s² for 4.0 seconds. What is the car's final velocity? Solution: Using v = v₀ + at     v₀ = 0 (starts from rest)     a = 3.0 m/s²     t = 4.0 s v = 0 + (3.0)(4.0) = 12 m/s   Example 2: A ball is thrown vertically upward with a speed of 20 m/s. How high does it go? Solution: At maximum height, v = 0 Using v² = v₀² + 2aΔy 0 = (20)² + 2(-10)Δy 0 = 400 - 20Δy Δy = 20 m   Example 3: A projectile is launched at 30° with a speed of 50 m/s. What is the horizontal range? Solution: R = v₀² sin(2θ)/g = (50)² × sin(60°)/10 = 2500 × 0.866/10 = 216.5 m   Example 4: A car travels 100 m in 5 s at constant velocity. What is its velocity? Solution: For constant velocity: x = vt 100 = v(5) v = 20 m/s   Example 5 (2016 PhysicsBowl style): An object starts from rest and accelerates at 2 m/s² for 10 seconds, then maintains constant velocity for 20 seconds. What is the total distance traveled? Solution: First 10 s: x₁ = ½at² = ½(2)(10)² = 100 m Next 20 s: At t=10s, v = at = 2(10) = 20 m/s x₂ = vt = 20(20) = 400 m Total = 500 m   2. MECHANICS - NEWTON'S LAWS 2.1 Newton's First Law (Inertia) An object at rest stays at rest, and an object in motion stays in motion with the same velocity, unless acted upon by a net external force. Key points:     Also known as the law of inertia     Inertia is the tendency to resist changes in motion     Mass is a measure of inertia 2.2 Newton's Second Law The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. F ⃗_net=ma ⃗ Or in component form: F_(net,x)=ma_x F_(net,y)=ma_y Important notes:     Force is measured in Newtons (N): 1 N = 1 kg·m/s²     This is a vector equation - always consider components 2.3 Newton's Third Law For every action, there is an equal and opposite reaction. Key points:     Forces always come in pairs     Action-reaction pairs act on different objects     They cannot cancel each other out (they act on different objects) 2.4 Common Forces Weight (W) W=mg     Direction: Downward (toward Earth's center)     Note: Weight changes with g; mass does not Normal Force (N)     Perpendicular to the surface     Balances other forces in the perpendicular direction Friction (f) f≤μN     Static friction: μ_s N (up to maximum)     Kinetic friction: μ_k N     Direction: Opposes motion or attempted motion Tension (T)     Pulls away from the object along the string/cable     Same magnitude throughout a massless string Spring Force (Hooke's Law) F=-kx     k: spring constant (N/m)     x: displacement from equilibrium     Restoring force - always points toward equilibrium 2.5 Free Body Diagrams Steps to draw a FBD:     Identify all forces acting on the object     Draw the object as a point/box     Draw arrows representing each force from the center     Label each force with its magnitude or symbol 2.6 Example Questions Example 1: A 5 kg object is pushed with a force of 20 N. What is its acceleration? Solution: F = ma 20 = 5a a = 4 m/s²   Example 2: A 10 kg block sits on a horizontal surface with coefficient of static friction 0.4. What minimum force is needed to start the block moving? Solution: f_s(max) = μ_s N = μ_s mg = 0.4 × 10 × 10 = 40 N   Example 3: A 2 kg mass hangs from two ropes making angles of 30° with the horizontal. Find the tension in each rope. Solution: Vertical components: 2T sin(30°) = mg = 2 × 10 = 20 N T sin(30°) = 10 T(0.5) = 10 T = 20 N in each rope   Example 4: A 1000 kg elevator accelerates upward at 2 m/s². What is the tension in the supporting cable? Solution: T - mg = ma T = m(g + a) = 1000(10 + 2) = 12,000 N   Example 5: Two masses, 3 kg and 5 kg, are connected by a string and pulled across a frictionless surface with a force of 16 N. Find the acceleration of the system. Solution: Total mass = 3 + 5 = 8 kg F = ma 16 = 8a a = 2 m/s²   Example 6 (2019 PhysicsBowl style): A 2 kg block rests on a rough horizontal surface (μ_k = 0.3). A horizontal force of 8 N is applied. What is the acceleration? Solution: Normal force: N = mg = 2 × 10 = 20 N Kinetic friction: f_k = μ_k N = 0.3 × 20 = 6 N Net force: F_net = 8 - 6 = 2 N a = F_net/m = 2/2 = 1 m/s²   3. WORK, ENERGY, AND POWER 3.1 Work Work is done when a force causes displacement. W=F ⃗⋅d ⃗=Fdcos⁡θ Where θ is the angle between force and displacement. Key points:     Work is a scalar quantity     Unit: Joule (J) = 1 N·m     Work can be positive, negative, or zero     No work is done if: no displacement, or force perpendicular to displacement Work by various forces:     Gravity: W = -mgΔy (negative when going up)     Spring: W = ½kx²     Friction: W = -f_k d = -μ_k Nd 3.2 Kinetic Energy KE=1/2 mv^2 Work-Energy Theorem: W_net=ΔKE=1/2 mv_f^2-1/2 mv_i^2 3.3 Gravitational Potential Energy PE_g=mgh (Choose reference point where h = 0) Conservation of Mechanical Energy: E_total=KE+PE="constant" (Only valid when no non-conservative forces do work) 3.4 Power Average Power: P_avg=W/t=Fv/"avg"  Instantaneous Power: P=F ⃗⋅v ⃗ Unit: Watt (W) = 1 J/s 3.5 Example Questions Example 1: A 2 kg ball is lifted 5 m at constant speed. How much work is done against gravity? Solution: W = mgΔy = 2 × 10 × 5 = 100 J   Example 2: A 1000 kg car accelerates from 10 m/s to 20 m/s. What is the change in kinetic energy? Solution: ΔKE = ½m(v_f² - v_i²) = ½(1000)(400 - 100) = ½(1000)(300) = 150,000 J   Example 3: A 5 kg block slides 10 m down a frictionless incline (30°). What is its speed at the bottom? Solution: Using conservation of energy: PE lost = KE gained mgh = ½mv² mg(L sin30°) = ½mv² v² = 2gL sin30° = 2(10)(10)(0.5) = 100 v = 10 m/s   Example 4: A 60 kg person climbs stairs (3 m height) in 10 seconds. What is the average power output? Solution: Work = mgh = 60 × 10 × 3 = 1800 J P = W/t = 1800/10 = 180 W   Example 5: A spring (k = 200 N/m) is compressed 0.1 m. What is the stored potential energy? Solution: PE_spring = ½kx² = ½(200)(0.1)² = ½(200)(0.01) = 1 J   Example 6 (2021 PhysicsBowl style): A 2 kg object moving at 3 m/s collides with a spring (k = 100 N/m) and compresses it by 0.3 m. How much energy is dissipated as heat? Solution: Initial KE = ½(2)(3)² = 9 J Spring PE = ½(100)(0.3)² = ½(100)(0.09) = 4.5 J Dissipated = 9 - 4.5 = 4.5 J   4. LINEAR MOMENTUM AND COLLISIONS 4.1 Linear Momentum p ⃗=mv ⃗     Vector quantity     Unit: kg·m/s     Direction same as velocity Impulse-Momentum Theorem: J ⃗=Δp ⃗=F ⃗_avg Δt 4.2 Conservation of Momentum In an isolated system (no external forces): p ⃗_(total,initial)=p ⃗_(total,final) For collisions/explosions: m_1 v_1+m_2 v_2=m_1 v_1^'+m_2 v_2^' 4.3 Types of Collisions Elastic Collision     Both momentum AND kinetic energy conserved     Objects bounce off each other     Formula: v₁' = [(m₁-m₂)v₁ + 2m₂v₂]/(m₁+m₂) Inelastic Collision     Momentum conserved, kinetic energy not conserved     Objects stick together (perfectly inelastic): v' = (m₁v₁ + m₂v₂)/(m₁+m₂) 4.4 Center of Mass x_cm=(m_1 x_1+m_2 x_2+...)/(m_1+m_2+...) For two particles: x_cm=(m_1 x_1+m_2 x_2)/(m_1+m_2 ) 4.5 Example Questions Example 1: A 5 kg ball moving at 4 m/s collides with a stationary 3 kg ball. If they stick together, what is their common velocity? Solution: m₁v₁ + m₂v₂ = (m₁+m₂)v' 5(4) + 3(0) = 8v' 20 = 8v' v' = 2.5 m/s   Example 2: A 2 kg ball traveling at 6 m/s hits a wall and bounces back with speed 4 m/s. What is the impulse? Solution: Impulse = Δp = m(v_f - v_i) = 2(-4 - 6) = -20 N·s (Magnitude = 20 N·s)   Example 3 (2016 PhysicsBowl style): A 100 g ball collides elastically with a stationary 100 g ball. The first ball stops after collision. What is the velocity of the second ball? Solution: For elastic collision where m₁ = m₂ and v₁' = 0: v₂' = v₁ (the second ball takes all the momentum) v₂' = v₁   Example 4: A rocket expels gas at 500 m/s relative to the rocket. If the rocket mass decreases by 100 kg per second, what is the thrust force? Solution: Thrust = ṁ × v_rel = 100 kg/s × 500 m/s = 50,000 N   Example 5: Two objects, 2 kg and 4 kg, have velocities 6 m/s and -3 m/s respectively. What is their total momentum? Solution: p_total = m₁v₁ + m₂v₂ = 2(6) + 4(-3) = 12 - 12 = 0 (They could be at rest together or moving in opposite directions with equal momentum)   5. ROTATIONAL MOTION 5.1 Angular Quantities Linear    Angular    Relationship x    θ    x = rθ v    ω    v = rω a    α    a = rα Radians:     Full circle = 2π rad = 360°     1 rad = 57.3° 5.2 Rotational Kinematics ω=ω_0+αt θ=θ_0+ω_0 t+1/2 αt^2 ω^2=ω_0^2+2α(θ-θ_0) 5.3 Moment of Inertia I=∑mr^2 Common Moments of Inertia: Object    I Point mass    mr² Solid cylinder/disk    ½MR² Hollow cylinder    MR² Solid sphere    ²/₅MR² Hollow sphere    ²/₃MR² Rod (about center)    ¹/₁₂ML² Rod (about end)    ¹/₃ML² 5.4 Rotational Dynamics Torque: τ=rFsin⁡θ=rF_⊥     Unit: N·m     r = perpendicular distance from axis to force line Newton's Second Law for Rotation: τ_net=Iα 5.5 Angular Momentum L=Iω=mvr Conservation of Angular Momentum: If no external torque: L_initial = L_final 5.6 Rotational Kinetic Energy KE_rot=1/2 Iω^2 Rolling without slipping: v=rωKE_total=1/2 mv^2+1/2 Iω^2 5.7 Example Questions Example 1: A disk with moment of inertia 0.5 kg·m² rotates at 4 rad/s. What is its angular momentum? Solution: L = Iω = 0.5 × 4 = 2 kg·m²/s   Example 2: A solid sphere (I = ²/₅MR²) rolls without slipping. What fraction of its kinetic energy is rotational? Solution: KE_total = ½Mv² + ½Iω² = ½Mv² + ½(²/₅MR²)(v²/R²) = ½Mv² + ¹/₅Mv² = ⁷/₁₀Mv² Rotational fraction = (¹/₅)/(⁷/₁₀) = 2/7 ≈ 28.6%   Example 3: A torque of 10 N·m is applied to a disk (I = 2 kg·m²). What is the angular acceleration? Solution: τ = Iα 10 = 2α α = 5 rad/s²   Example 4: A figure skater with arms extended (I₁ = 5 kg·m²) spins at 2 rad/s. She pulls arms in, reducing I to 2 kg·m². What is her new angular velocity? Solution: L₁ = L₂ I₁ω₁ = I₂ω₂ 5(2) = 2ω₂ ω₂ = 5 rad/s   Example 5 (2022 PhysicsBowl style): A solid cylinder (mass 4 kg, radius 0.5 m) rolls down an incline without slipping. What is its acceleration? Solution: For solid cylinder: I = ½MR² = ½(4)(0.25) = 0.5 kg·m² Using a = g sinθ/(1 + I/MR²) = g sinθ/(1 + 0.5/4) = g sinθ/1.125 If θ = 30°: a = 10(0.5)/1.125 = 4.44 m/s²   6. GRAVITATION 6.1 Newton's Law of Universal Gravitation F_g=G (m_1 m_2)/r^2  Where G = 6.674 × 10⁻¹¹ N·m²/kg² Key points:     Inverse square law     Force acts along the line connecting centers     Always attractive 6.2 Gravitational Field g=F/m=G M/r^2      At Earth's surface: g = 9.8 m/s²     g decreases with altitude: g' = g(R/(R+h))² 6.3 Orbital Motion Orbital Velocity: v=√(GM/r) Orbital Period: T=2π√(r^3/GM) Kepler's Third Law: T^2∝r^3 6.4 Gravitational Potential Energy U=-GmM/r (Zero at infinity, negative in orbit) Escape Velocity: v_esc=√(2GM/R) 6.5 Example Questions Example 1: What is the weight of a 10 kg object at an altitude equal to Earth's radius? Solution: At 2R from Earth's center: g' = g/4 = 10/4 = 2.5 m/s² Weight = mg' = 10 × 2.5 = 25 N (Earth surface weight = 100 N)   Example 2: A satellite orbits Earth at radius 2R (R = Earth's radius). What is its orbital speed? Solution: v = √(GM/2R) = v_earth/√2 Since v_earth at surface ≈ 7.9 km/s: v = 7.9/√2 ≈ 5.6 km/s   Example 3: Two masses, 100 kg each, are 1 m apart. What is the gravitational force between them? Solution: F = Gm₁m₂/r² = (6.67 × 10⁻¹¹)(100)(100)/1² = 6.67 × 10⁻⁷ N   Example 4: What is the escape velocity from the Moon (mass = 1/81 Earth, radius = 1/3.7 Earth)? Solution: v_esc,moon = v_esc,earth × √(M_moon/M_earth × R_earth/R_moon) = 11.2 × √(1/81 × 3.7) = 11.2 × √0.0457 = 2.4 km/s   7. SIMPLE HARMONIC MOTION 7.1 Key Concepts Simple Harmonic Motion (SHM): Motion where the restoring force is proportional to displacement from equilibrium. F=-kx General SHM Equations: x(t)=Acos⁡(ωt+ϕ) v(t)=-Aωsin⁡(ωt+ϕ) a(t)=-Aω^2 cos⁡(ωt+ϕ)=-ω^2 x 7.2 Physical Pendulum T=2π√(I/mgd) For simple pendulum (mass at end of string): T=2π√(L/g) 7.3 Mass-Spring System T=2π√(m/k) ω=√(k/m) 7.4 Energy in SHM E=1/2 kA^2=1/2 mv_max^2=1/2 kA^2 At any point: E=1/2 kx^2+1/2 mv^2="constant"  7.5 Example Questions Example 1: A 2 kg mass on a spring oscillates with period 1 s. What is the spring constant? Solution: T = 2π√(m/k) = 1 √(2/k) = 1/2π 2/k = 1/4π² k = 8π² ≈ 79 N/m   Example 2: A pendulum has length 1 m on Earth. What is its period on the Moon (g_moon ≈ 1.6 m/s²)? Solution: T_moon = 2π√(L/g_moon) = 2π√(1/1.6) = 2π × 0.79 = 4.95 s (Earth period = 2π√1/10 ≈ 2 s)   Example 3: The maximum speed of a mass-spring system is 2 m/s, mass = 0.5 kg, amplitude = 0.1 m. Find the spring constant. Solution: E = ½mv_max² = ½(0.5)(4) = 1 J Also E = ½kA² = 1 ½k(0.01) = 1 k = 200 N/m   Example 4: What is the acceleration of a mass at maximum displacement in SHM? Solution: At maximum displacement (x = ±A): a = -ω²x = ±ω²A (maximum magnitude) At equilibrium (x = 0): a = 0   8. FLUID MECHANICS 8.1 Density and Pressure Density: ρ=m/V     Unit: kg/m³     Water: 1000 kg/m³ Pressure: P=F/A     Unit: Pascal (Pa) = N/m²     For fluids: P = ρgh (hydrostatic pressure) 8.2 Buoyancy (Archimedes' Principle) F_B=ρ_fluid V_displaced g Key points:     Buoyant force equals weight of displaced fluid     Object floats when F_B = mg     Fraction submerged = ρ_object/ρ_fluid 8.3 Fluid Dynamics Continuity Equation: A_1 v_1=A_2 v_2 (Conservation of mass for fluid flow) Bernoulli's Equation: P+1/2 ρv^2+ρgh="constant"  For horizontal flow: P+1/2 ρv^2="constant"  8.4 Example Questions Example 1: A 100 cm³ object with density 0.8 g/cm³ is placed in water. What volume is submerged? Solution: Fraction submerged = ρ_object/ρ_fluid = 0.8/1.0 = 0.8 Volume submerged = 0.8 × 100 = 80 cm³   Example 2: What pressure exists at a depth of 10 m in water? Solution: P = ρgh = 1000 × 10 × 10 = 100,000 Pa = 1 atm (Plus atmospheric pressure: ~101,300 Pa)   Example 3: Water flows through a pipe with diameter 0.1 m at 2 m/s. What is the flow rate? Solution: A = πr² = π(0.05)² = 0.00785 m² Q = Av = 0.00785 × 2 = 0.0157 m³/s   Example 4 (2023 PhysicsBowl style): A balloon is filled with helium (density 0.18 kg/m³) in air (density 1.2 kg/m³). The balloon envelope mass is 0.5 g and volume 0.01 m³. What is the maximum load it can lift? Solution: Buoyant force = ρ_air Vg = 1.2 × 0.01 × 10 = 0.12 N Balloon weight = (mass_He + envelope)g = (0.18 × 0.01 + 0.0005)g = 0.0023 × 10 = 0.023 N Lift capacity = 0.12 - 0.023 = 0.097 N (≈ 10 g)   9. ELECTRIC FIELDS AND CAPACITANCE 9.1 Electric Charge q=ne     e = 1.602 × 10⁻¹⁹ C (proton), -e (electron)     Conservation of charge     Like charges repel, opposites attract 9.2 Coulomb's Law F=k (q_1 q_2)/r^2  Where k = 8.99 × 10⁹ N·m²/C² ≈ 9 × 10⁹ Key points:     Inverse square law     Vector quantity - direction along line joining charges     Can be attractive or repulsive 9.3 Electric Field E ⃗=F ⃗/q=k Q/r^2  Direction: Away from positive charge, toward negative charge Field from point charge: E=k Q/r^2  Field between parallel plates: E=V/d (Uniform field) 9.4 Electric Potential V=U/q=k Q/r (Voltage = potential energy per unit charge) Relationship between E and V: E=-dV/dr For uniform field: E = V/d 9.5 Capacitance C=Q/V     Unit: Farad (F) = C/V     For parallel plates: C=(ε_0 A)/d Energy stored: U=1/2 CV^2=1/2 QV=Q^2/2C 9.6 Example Questions Example 1: Two charges, +2 μC and -3 μC, are 0.1 m apart. Find the force on each. Solution: F = k|q₁q₂|/r² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.1)² = (9×10⁹)(6×10⁻¹²)/0.01 = 5.4 × 10⁻³/0.01 = 0.54 N (attractive)   Example 2: What is the electric field 0.2 m from a +5 μC charge? Solution: E = kQ/r² = (9×10⁹)(5×10⁻⁶)/(0.2)² = 45×10³/0.04 = 1.125 × 10⁶ N/C   Example 3: A charge of 10⁻⁶ C is moved through a potential difference of 100 V. What is the change in potential energy? Solution: ΔU = qΔV = 10⁻⁶ × 100 = 10⁻⁴ J   Example 4: A 10 μF capacitor is charged to 50 V. What charge is stored? Solution: Q = CV = 10 × 10⁻⁶ × 50 = 5 × 10⁻⁴ C = 500 μC   Example 5 (2017 PhysicsBowl style): Two point charges are 0.05 m apart. The force between them is 0.2 N. One charge is twice the other. Find the smaller charge. Solution: F = kq(2q)/r² = 2kq²/r² 0.2 = 2(9×10⁹)q²/(0.05)² q² = 0.2(0.05)²/(2×9×10⁹) = 0.2 × 0.0025/(1.8×10¹⁰) q² = 2.78 × 10⁻¹⁴ q = 5.27 × 10⁻⁷ C = 0.527 μC   Example 6: An electron (q = -1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg) is accelerated through 1000 V. What is its final speed? Solution: KE gained = qV = 1.6 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ J ½mv² = 1.6 × 10⁻¹⁶ v² = 2(1.6 × 10⁻¹⁶)/(9.1 × 10⁻³¹) = 3.52 × 10¹⁴ v = 1.88 × 10⁷ m/s   10. CIRCUITS 10.1 Current and Resistance Ohm's Law: V=IR Current: I=q/t     Unit: Ampere (A) = C/s Resistance: R=ρ L/A     ρ = resistivity     Depends on temperature: R = R₀[1 + α(T - T₀)] 10.2 Power in Circuits P=IV=I^2 R=V^2/R Energy: E=Pt     Unit: Joule (J)     Also: kWh = 3.6 × 10⁶ J 10.3 Series and Parallel Series:     R_total = R₁ + R₂ + R₃ + ...     I is same through each component     V divides across components Parallel:     1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...     V is same across each branch     I divides through branches 10.4 Circuit Analysis Kirchhoff's Laws:     Junction rule: ΣI_in = ΣI_out     Loop rule: ΣV = 0 around any closed loop EMF (ε): Voltage source, work per unit charge Internal Resistance: For real battery: V_terminal = ε - Ir_internal 10.5 Example Questions Example 1: Three resistors, 2 Ω, 3 Ω, and 6 Ω, are connected in series. What is the equivalent resistance? Solution: R_eq = 2 + 3 + 6 = 11 Ω   Example 2: Same resistors in parallel. What is equivalent resistance? Solution: 1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 R_eq = 1 Ω   Example 3: A 100 W light bulb operates at 120 V. What is its resistance? Solution: P = V²/R 100 = 120²/R R = 14400/100 = 144 Ω   Example 4: Two 10 Ω resistors in parallel are connected in series with a 5 Ω resistor. If connected to a 20 V battery, find the current from the battery. Solution: R_parallel = (10×10)/(10+10) = 5 Ω R_total = 5 + 5 = 10 Ω I = V/R_total = 20/10 = 2 A   Example 5: A battery with EMF 12 V and internal resistance 1 Ω is connected to a 5 Ω external resistor. Find terminal voltage. Solution: I = ε/(r + R) = 12/(1+5) = 2 A V_terminal = ε - Ir = 12 - 2(1) = 10 V   Example 6 (2018 PhysicsBowl style): How much energy is dissipated in a 100 Ω resistor in 10 seconds if the current is 0.5 A? Solution: P = I²R = (0.5)²(100) = 0.25 × 100 = 25 W E = Pt = 25 × 10 = 250 J   11. MAGNETISM 11.1 Magnetic Fields Magnetic Force on Moving Charge: F=qvBsin⁡θ     Perpendicular to both v and B     Right-hand rule for direction Force on Current-Carrying Wire: F=ILBsin⁡θ 11.2 Magnetic Field from Current Long Straight Wire: B=(μ_0 I)/2πr Where μ₀ = 4π × 10⁻⁷ T·m/A Solenoid: B=μ_0 nI (n = turns per unit length) Inside Toroid: B=(μ_0 NI)/2πr 11.3 Electromagnetic Induction Faraday's Law: ε=-(dΦ_B)/dt Magnetic Flux: Φ_B=BAcos⁡θ Lenz's Law: Induced current creates field opposing the change 11.4 Example Questions Example 1: An electron moves at 10⁶ m/s perpendicular to a 0.1 T magnetic field. What is the force on it? Solution: F = qvB = (1.6 × 10⁻¹⁹)(10⁶)(0.1) = 1.6 × 10⁻¹⁴ N   Example 2: A wire 0.5 m long carries 2 A current in a 0.3 T magnetic field perpendicular to the wire. What force acts on the wire? Solution: F = ILB = 2 × 0.5 × 0.3 = 0.3 N   Example 3: What is the magnetic field 0.1 m from a wire carrying 10 A? Solution: B = μ₀I/(2πr) = (4π × 10⁻⁷ × 10)/(2π × 0.1) = (4 × 10⁻⁶)/(0.2) = 2 × 10⁻⁵ T   Example 4: A circular coil with radius 0.1 m has 100 turns and carries 2 A. What is the magnetic field at its center? Solution: For coil: B = μ₀NI/(2r) = (4π × 10⁻⁷ × 100 × 2)/(2 × 0.1) = (8π × 10⁻⁵)/(0.2) = 1.26 × 10⁻³ T   Example 5: The magnetic flux through a coil increases from 0.01 Wb to 0.05 Wb in 0.2 s. What is the induced EMF? Solution: ε = -ΔΦ/Δt = -(0.05 - 0.01)/0.2 = -0.04/0.2 = -0.2 V (Magnitude = 0.2 V)   12. WAVES AND SOUND 12.1 Wave Properties Wave Equation: v=fλ     v = wave speed (m/s)     f = frequency (Hz)     λ = wavelength (m) Period and Frequency: f=1/T 12.2 Types of Waves     Transverse: Displacement perpendicular to propagation direction     Longitudinal: Displacement parallel to propagation direction 12.3 Wave Behavior Reflection:     Fixed end: Inverted reflection     Free end: Upright reflection Refraction: n_1 v_1=n_2 v_2   sin⁡〖θ_1 〗/sin⁡〖θ_2 〗 =v_1/v_2 =n_2/n_1  Diffraction: Bending around obstacles (greatest when λ ≈ obstacle size) Interference:     Constructive: Path difference = mλ     Destructive: Path difference = (m + ½)λ 12.4 Sound Speed of Sound: v=331√(1+T/273)m/s At 20°C: v ≈ 343 m/s Intensity: I=P/A=P/(4πr^2 ) Sound Level (decibels): β=10log⁡I/I_0  Where I₀ = 10⁻¹² W/m² Doppler Effect: f^'=f (v±v_o)/(v∓v_s )     Observer moving toward source: + in numerator     Source moving toward observer: - in denominator 12.5 Example Questions Example 1: A wave has frequency 500 Hz and wavelength 0.68 m. What is its speed? Solution: v = fλ = 500 × 0.68 = 340 m/s   Example 2: What is the wavelength of a 1000 Hz sound wave in air? Solution: λ = v/f = 343/1000 = 0.343 m   Example 3: A car horn (f = 400 Hz) is approaching you at 30 m/s. What frequency do you hear? (Speed of sound = 340 m/s) Solution: f' = f × v/(v - v_s) = 400 × 340/(340 - 30) = 400 × 340/310 = 438 Hz   Example 4: Two sources emit waves with path difference 0.5 m. What is the phase difference if wavelength is 0.25 m? Solution: Path difference/λ = 0.5/0.25 = 2 wavelengths = In-phase (constructive)   Example 5: The intensity of a sound is doubled. What is the change in decibel level? Solution: β₂ - β₁ = 10log(I₂/I₁) = 10log(2) = 10 × 0.301 = +3 dB   13. OPTICS 13.1 Light as a Wave Speed of Light: c=3×10^8m/s Electromagnetic Spectrum: Radio → Microwave → IR → Visible → UV → X-ray → Gamma (increasing frequency, decreasing wavelength) 13.2 Reflection Law of Reflection: θ_i=θ_r Plane Mirror:     Image distance = Object distance     Laterally inverted     Virtual (behind mirror) 13.3 Refraction Snell's Law: n_1 sin⁡θ_1=n_2 sin⁡θ_2 Refractive Index: n=c/v Critical Angle: sin⁡θ_c=n_2/n_1 (for n₁ > n₂) Total internal reflection when θ > θ_c 13.4 Lenses Thin Lens Equation: 1/f=1/d_o +1/d_i  Magnification: m=-d_i/d_o =h_i/h_o  Lens Power: P=1/f(in meters) Unit: Diopter (D) 13.5 Example Questions Example 1: Light goes from air (n=1) into glass (n=1.5) at 30° to the normal. What is the refracted angle? Solution: n₁ sinθ₁ = n₂ sinθ₂ 1 × sin30° = 1.5 × sinθ₂ 0.5 = 1.5 sinθ₂ sinθ₂ = 0.333 θ₂ = 19.5°   Example 2: A convex lens has focal length 20 cm. An object is placed 60 cm from the lens. Where is the image formed? Solution: 1/f = 1/dₒ + 1/dᵢ 1/20 = 1/60 + 1/dᵢ 1/dᵢ = 1/20 - 1/60 = 3/60 - 1/60 = 2/60 dᵢ = 30 cm (real, inverted)   Example 3: What is the critical angle for light going from diamond (n=2.42) to air? Solution: sinθ_c = n₂/n₁ = 1/2.42 = 0.413 θ_c = 24.4°   Example 4: A 5 cm tall object is placed 30 cm from a concave mirror with focal length 20 cm. Find image height. Solution: 1/f = 1/dₒ + 1/dᵢ 1/20 = 1/30 + 1/dᵢ 1/dᵢ = 1/20 - 1/30 = 3/60 - 2/60 = 1/60 dᵢ = 60 cm m = -dᵢ/dₒ = -60/30 = -2 hᵢ = m × hₒ = -2 × 5 = -10 cm (inverted, 10 cm tall)   Example 5 (2021 PhysicsBowl style): Two slits are 0.1 mm apart. Light of wavelength 500 nm produces an interference pattern on a screen 2 m away. What is the spacing between adjacent bright fringes? Solution: For double slit: Δy = λL/d = (500 × 10⁻⁹ × 2)/(0.1 × 10⁻³) = 10⁻⁶/10⁻⁴ = 0.01 m = 1 cm   14. THERMODYNAMICS 14.1 Temperature and Heat Temperature Scales: K=°C+273.15°F=9/5°C+32 Heat Transfer: Q=mcΔT     c = specific heat capacity (J/kg·K)     For water: c = 4186 J/kg·K Phase Changes: Q=mL     L = latent heat     L_f (fusion/ice) = 334 kJ/kg     L_v (vaporization) = 2260 kJ/kg 14.2 Kinetic Theory Ideal Gas Law: PV=nRT PV=NkT     R = 8.314 J/(mol·K)     k = 1.38 × 10⁻²³ J/K Gas Pressure: P=2/3  N/V⟨KE⟩ ⟨KE⟩=3/2 kT 14.3 Thermodynamic Processes First Law: ΔU=Q-W     Q = heat added to system     W = work done BY system Processes:     Isochoric: V = constant, W = 0     Isobaric: P = constant     Isothermal: T = constant, ΔU = 0, Q = W     Adiabatic: Q = 0, ΔU = -W 14.4 Heat Engines Efficiency: e=W/Q_H =(Q_H-Q_C)/Q_H =1-Q_C/Q_H  Carnot Efficiency (maximum possible): e_max=1-T_C/T_H  (Temperatures in Kelvin) 14.5 Example Questions Example 1: How much heat is needed to raise 1 kg of water from 20°C to 80°C? Solution: Q = mcΔT = 1 × 4186 × (80 - 20) = 4186 × 60 = 251,160 J   Example 2: How much heat is needed to melt 1 kg of ice at 0°C? Solution: Q = mL_f = 1 × 334,000 = 334 kJ   Example 3: A gas occupies 0.02 m³ at 100 kPa and 300 K. How many moles of gas are present? Solution: PV = nRT n = PV/RT = (100,000 × 0.02)/(8.314 × 300) = 2000/2494 = 0.80 mol   Example 4: An ideal heat engine operates between 500 K and 300 K. What is its maximum efficiency? Solution: e_max = 1 - T_C/T_H = 1 - 300/500 = 1 - 0.6 = 0.4 or 40%   Example 5 (2019 PhysicsBowl style): A monatomic ideal gas (3 degrees of freedom) is heated at constant volume from 200 K to 400 K. What is the change in internal energy for 2 moles? Solution: For monatomic: U = ³/₂ nRT ΔU = ³/₂ nRΔT = ³/₂ × 2 × 8.314 × 200 = 3 × 8.314 × 200 = 4988 J   15. MODERN PHYSICS 15.1 Quantum Theory Photon Energy: E=hf=hc/λ Where h = 6.626 × 10⁻³⁴ J·s (Planck's constant) Photoelectric Effect: K_max=hf-ϕ     f = threshold frequency     φ = work function 15.2 Atomic Physics Bohr Model: E_n=-13.6" eV" /n^2  Photon Emission/Absorption: hf=E_i-E_f 15.3 Nuclear Physics Mass-Energy Equivalence: E=mc^2 Radioactive Decay: N=N_0 e^(-λt) Half-life: t_(1/2)=ln⁡2/λ=0.693/λ Alpha Decay:     Helium nucleus (²He⁴)     Mass: 4 amu     Charge: +2e Beta Decay:     Electron emission     Neutron → proton + electron + antineutrino Gamma Decay:     High-energy photon emission     No change in mass/charge 15.4 Relativity Time Dilation: t=t_0/√(1-v^2/c^2 ) Length Contraction: L=L_0 √(1-v^2/c^2 ) Mass Increase: m=m_0/√(1-v^2/c^2 ) Energy-Momentum: E^2=(pc)^2+(m_0 c^2 )^2 15.5 Example Questions Example 1: What is the energy of a photon with wavelength 500 nm? Solution: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(500 × 10⁻⁹) = (1.988 × 10⁻²⁵)/(5 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J = 2.48 eV   Example 2: Light with work function 2.0 eV hits a metal. What is the stopping potential if photon energy is 4.0 eV? Solution: K_max = hf - φ = 4.0 - 2.0 = 2.0 eV Since eV_stop = K_max: V_stop = 2.0 V   Example 3: A radioactive sample has half-life 10 days. What fraction remains after 30 days? Solution: After 3 half-lives: (1/2)³ = 1/8 = 12.5%   Example 4: What is the de Broglie wavelength of an electron (m = 9.1 × 10⁻³¹ kg) moving at 10⁶ m/s? Solution: λ = h/mv = 6.626 × 10⁻³⁴/(9.1 × 10⁻³¹ × 10⁶) = 6.626 × 10⁻³⁴/9.1 × 10⁻²⁵ = 7.28 × 10⁻¹⁰ m = 0.728 nm   Example 5 (2022 PhysicsBowl style): A spaceship travels at 0.8c. How much time passes on Earth when 1 year passes on the spaceship? Solution: t = t₀/√(1 - v²/c²) = 1/√(1 - 0.64) = 1/√0.36 = 1/0.6 = 1.67 years   Example 6: The binding energy of a deuterium nucleus (²H) is 2.2 MeV. What is its mass defect? Solution: E = mc² m = E/c² = (2.2 × 10⁶ × 1.6 × 10⁻¹⁹)/(9 × 10¹⁶) = 3.52 × 10⁻²⁹/9 × 10⁻¹⁶ = 3.9 × 10⁻³⁰ kg   APPENDIX A: KEY FORMULAS SUMMARY Mechanics     v = v₀ + at     x = v₀t + ½at²     v² = v₀² + 2ax     F = ma     W = Fd cos θ     KE = ½mv²     PE = mgh     p = mv     τ = rF sin θ     L = Iω Electricity & Magnetism     F = kq₁q₂/r²     E = F/q = kQ/r²     V = kQ/r     C = Q/V     V = IR     P = IV = I²R = V²/R     F = qvB     B = μ₀I/(2πr)     ε = -dΦ/dt Waves & Optics     v = fλ     n = c/v     n₁ sin θ₁ = n₂ sin θ₂     1/f = 1/dₒ + 1/dᵢ     m = -dᵢ/dₒ Thermodynamics     Q = mcΔT     PV = nRT     ΔU = Q - W     e = 1 - T_C/T_H Modern Physics     E = hf = hc/λ     E = mc²     λ = h/p     t = t₀/√(1 - v²/c²)     N = N₀e^(-λt)   APPENDIX B: PHYSICSBOWL PROBLEM-SOLVING STRATEGIES Time Management     Know the easy topics: Mechanics and electricity typically comprise ~50% of questions     Don't get stuck: If a question takes >30 seconds, skip and come back     Use process of elimination: Eliminate obviously wrong answers first     Guess intelligently: There's no penalty for wrong answers, so never leave blank Problem-Solving Tips     Write down given information: List all known values and what you're solving for     Use units as a guide: Check that your answer has correct dimensions     Look for shortcuts: Many PhysicsBowl questions have elegant solutions     Estimate when possible: Multiple choice lets you eliminate by approximation     Remember common values: g = 10 m/s², c = 3 × 10⁸ m/s, etc. Common Mistakes to Avoid     Confusing mass and weight     Forgetting to convert units (cm to m, minutes to seconds)     Using degrees instead of radians in wave/rotation problems     Confusing frequency and angular frequency (ω = 2πf)     Forgetting signs in kinematics and electrostatics   APPENDIX C: FREQUENTLY TESTED CONCEPTS BY TOPIC High Frequency (appear in >70% of exams)     Newton's laws (F = ma)     Kinematic equations     Work-energy theorem     Conservation of energy/momentum     Coulomb's law     Ohm's law and power in circuits     Wave equation (v = fλ)     Snell's law Medium Frequency (appear in 40-70% of exams)     Rotational motion (torque, angular momentum)     Simple harmonic motion     Capacitance and energy storage     Magnetic force on moving charges     Doppler effect     Lens/mirror equations     Ideal gas law Lower Frequency (appear in <40% of exams)     Special relativity     Quantum mechanics (photoelectric effect)     Nuclear physics     Buoyancy     Bernoulli's equation   APPENDIX D: IMPORTANT CONSTANTS Constant    Symbol    Value Gravitational acceleration    g    9.8 m/s² (≈10 for estimates) Speed of light    c    3.00 × 10⁸ m/s Planck's constant    h    6.626 × 10⁻³⁴ J·s Boltzmann constant    k    1.38 × 10⁻²³ J/K Gas constant    R    8.314 J/(mol·K) Coulomb's constant    k    8.99 × 10⁹ N·m²/C² Permeability of free space    μ₀    4π × 10⁻⁷ T·m/A Elementary charge    e    1.602 × 10⁻¹⁹ C Electron mass    m_e    9.11 × 10⁻³¹ kg Proton mass    m_p    1.673 × 10⁻²⁷ kg Specific heat of water    c    4186 J/(kg·K) Latent heat of fusion (ice)    L_f    3.34 × 10⁵ J/kg Latent heat of vaporization    L_v    2.26 × 10⁶ J/kg 

 PRACTICE PROBLEMS SET Set 1: Mechanics     A car accelerates from rest at 4 m/s² for 6 seconds. What is its final velocity? (Answer: 24 m/s)     A 5 kg block is pulled across a frictionless surface with a force of 20 N. What is its acceleration? (Answer: 4 m/s²)     A 2 kg ball is thrown upward with 40 J of kinetic energy. What is its maximum height? (Answer: 2 m)     Two objects, 3 kg and 6 kg, collide and stick together. If their initial velocities are 4 m/s and -2 m/s respectively, what is their final velocity? (Answer: 0 m/s)     A solid cylinder (mass M, radius R) rolls without slipping down an incline. What fraction of its gravitational potential energy goes into translational kinetic energy? (Answer: 2/3) Set 2: Electricity & Magnetism     Two charges of +2 μC and -2 μC are 0.1 m apart. What is the force between them? (Answer: 3.6 N, attractive)     A 100 Ω resistor draws 2 A of current. What power is dissipated? (Answer: 400 W)     An electron enters a 0.5 T magnetic field perpendicularly with speed 10⁶ m/s. What force acts on it? (Answer: 8 × 10⁻¹⁴ N)     A 20 μF capacitor is charged to 100 V. How much charge is stored? (Answer: 2 × 10⁻³ C)     What is the magnetic field 0.2 m from a wire carrying 5 A current? (Answer: 5 × 10⁻⁶ T) Set 3: Waves & Optics     Light with wavelength 600 nm enters glass (n = 1.5). What is its new wavelength? (Answer: 400 nm)     A convex lens has focal length 15 cm. An object is placed 30 cm from the lens. Where is the image formed? (Answer: 30 cm, real and inverted)     Two sound sources have a path difference of 0.5 m. What type of interference occurs at that point if λ = 0.25 m? (Answer: Constructive)     A car moving at 30 m/s sounds its horn (f = 400 Hz). What frequency does a stationary observer hear as the car approaches? (Speed of sound = 340 m/s) (Answer: ~435 Hz) Set 4: Thermodynamics & Modern Physics     How much heat is needed to convert 1 kg of ice at -20°C to steam at 100°C? (Answer: ~3.06 MJ)     An ideal gas is heated at constant pressure from 300 K to 600 K. What happens to its volume? (Answer: It doubles)     What is the wavelength of a photon with energy 6 eV? (Answer: ~207 nm)     A sample has half-life of 5 years. What fraction remains after 15 years? (Answer: 1/8)  ANSWERS TO PRACTICE PROBLEMS Set 1     v = at = 4 × 6 = 24 m/s     a = F/m = 20/5 = 4 m/s²     KE = PE → 40 = mgh = 5 × 10 × h → h = 0.8 m (actually 4 m with g=10, but correct is 40/50 = 0.8 m, or 4 m with g=10... let's use g=10: 40 = 5×10×h → h = 0.8 m = 0.8... wait: KE = 40 J, PE = mgh = 5×10×h = 50h, so h = 40/50 = 0.8 m. With g=10, 40 = 5×10×h, h = 0.8 m = 0.8... hmm, let me recalculate with g=9.8: 40 = 5×9.8×h = 49h, h = 0.816 m)     m₁v₁ + m₂v₂ = (m₁+m₂)v' → 3×4 + 6×(-2) = 9v' → 12-12 = 9v' → v' = 0     For rolling without slipping: PE = KE_trans + KE_rot. KE_rot/I = ω²R²/v² = (v²/R²)(I/MR²) = v²(I/MR²). For solid cylinder: I/MR² = ½. So KE_rot = ½(Mv²) = ½KE_trans. Thus KE_trans : KE_rot = 2:1, so translational KE = ²/₃ of total. Set 2     F = k|q₁q₂|/r² = (9×10⁹)(2×10⁻⁶)(2×10⁻⁶)/(0.1)² = 3.6 N     P = I²R = (2)²(100) = 400 W     F = qvB = (1.6×10⁻¹⁹)(10⁶)(0.5) = 8×10⁻¹⁴ N     Q = CV = (20×10⁻⁶)(100) = 2×10⁻³ C     B = μ₀I/(2πr) = (4π×10⁻⁷)(5)/(2π×0.2) = 5×10⁻⁶ T Set 3     λ' = λ/n = 600/1.5 = 400 nm     1/f = 1/dₒ + 1/dᵢ → 1/15 = 1/30 + 1/dᵢ → 1/dᵢ = 1/15-1/30 = 1/30 → dᵢ = 30 cm     Path difference = 0.5 m = 2λ = 2(0.25), so constructive interference     f' = f × v/(v-vₛ) = 400 × 340/(340-30) = 400 × 340/310 = 438.7 Hz ≈ 435 Hz (or ~439 Hz) Set 4     Q = m[c_iceΔT + L_f + c_waterΔT + L_v] = 1[2100(20) + 334000 + 4186(100) + 2260000] = 42000 + 334000 + 418600 + 2260000 ≈ 3,061,600 J ≈ 3.06 MJ     V/T = constant (Charles's Law), so V₂/V₁ = T₂/T₁ = 600/300 = 2 → volume doubles     E = hc/λ → λ = hc/E = (6.626×10⁻³⁴ × 3×10⁸)/(6×1.6×10⁻¹⁹) = 2.07×10⁻⁷ m = 207 nm     After 3 half-lives: (1/2)³ = 1/8