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