AP Physics C: Mechanics study package

AP Physics C: Mechanics is a calculus-based introductory mechanics course equivalent to a first-semester college physics course for science and engineering majors. Unlike AP Physics 1, every major relationship in this course is derived using differential and integral calculus. The exam tests not just whether you can apply formulas but whether you understand where those formulas come from and can adapt them to novel situations.

Exam Structure

Section 1: Multiple Choice (50% of score)

• 35 questions in 45 minutes (~77 seconds per question)
• No calculator permitted (though a simple four-function calculator may sometimes be allowed per recent changes)
• Covers all 7 units with approximately equal weighting
• Questions range from conceptual to moderately computational
• Scratch work is critical — set up the calculus, then simplify

Section 2: Free Response (50% of score)

• 3 questions in 45 minutes (~15 minutes per question)
• Graphing calculator permitted
• Typically includes:
  • A kinematics/dynamics problem (often 2D)
  • An energy/momentum problem
  • A rotation or oscillation problem
• You must show all work, including calculus derivations
• Partial credit is generous — always attempt every part

The 7 Units

Unit	Topic	Weight (approx.)	Key Calculus Element
1	Kinematics	14%	Derivatives and integrals of position/velocity/acceleration
2	Newton's Laws of Motion	17%	Vector calculus, differential equations F=ma
3	Work, Energy, and Power	14%	Work as line integral of force, potential energy from conservative forces
4	Systems of Particles & Linear Momentum	14%	Impulse as integral of force, center of mass integrals
5	Rotation	17%	Moment of inertia integrals, torque and angular momentum calculus
6	Oscillations	10%	Differential equation for SHM, energy analysis
7	Gravitation	14%	Gravitational field/potential integrals, orbital mechanics

Prerequisite Knowledge

Calculus (Essential)

• Derivatives: Power rule, chain rule, product rule
• Integrals: Indefinite and definite, u-substitution, basic trig integrals
• Differential equations: Separation of variables for simple cases
• Vector operations: Dot product, cross product (for rotation)

Physics (Helpful)

• Basic kinematics and Newton's laws from a prior physics course
• Familiarity with SI units and dimensional analysis

How to Use This Study Package

Study Sequence

1. Read the Course Overview (this file)
2. Study each Unit Note file — these contain full derivations and theory (~2500 words each)
3. Complete each Practice file — these contain worked examples and practice problems (~1500 words each)
4. Review the Summary file — a condensed reference of all key equations
5. Study the Strategy file — exam-day tactics specific to Physics C
6. Take the Full Practice Exam under timed conditions
7. Check your answers with the detailed solutions
8. Review the Audio Script — listen while reviewing weak areas

Time Investment

• Each unit: 3–4 hours (read notes + work practice)
• Full exam + review: 2–3 hours
• Total preparation: 30–40 hours for a thorough review

Key Differences from AP Physics 1

Feature	AP Physics 1	AP Physics C: Mechanics
Math level	Algebra/trig	Full calculus
Derivations	Rarely required	Frequently required
Depth	Conceptual breadth	Analytical depth
Rotation	Basic	Full rotational dynamics with I, τ, L
Oscillations	Qualitative	Full differential equation treatment
Gravitation	Surface gravity only	Full universal gravitation, orbits, potential
Score	1–5	1–5 (typically harder to earn 5)

What Makes a 5 on This Exam

Students who score 5 consistently demonstrate:

1. Fluency with calculus: They set up integrals for work, impulse, moment of inertia, and gravitational potential without hesitation.
2. Physical intuition: They know whether an answer should be positive or negative, larger or smaller, before computing.
3. FRQ discipline: They label all forces, show free-body diagrams, state their sign conventions, and box final answers.
4. Time management: They move quickly through straightforward MCQs and budget 15 minutes per FRQ.
5. Unit mastery over formula memorization: They derive formulas from first principles rather than memorizing dozens of special-case equations.

Recommended Resources to Supplement

• Textbook: University Physics by Young & Freedman or Fundamentals of Physics by Halliday, Resnick & Walker
• Problem book: 3,000 Solved Problems in Physics by Alvin Halpern
• Past exams: College Board released exams (most valuable resource)
• Online: MIT OCW 8.01 lectures by Walter Lewin

  This study package contains 19 files covering every aspect of the AP Physics C: Mechanics exam. Work through them systematically for best results.

Kinematics is the mathematical description of motion without regard to its causes. In AP Physics C, kinematics is the foundation upon which all subsequent units build. Unlike algebra-based physics, here we derive every kinematic relationship from the fundamental definitions using calculus. Position, velocity, and acceleration are connected through differentiation and integration, and this calculus-based approach generalizes to any acceleration function — not just constant acceleration.

1.1 Vectors and Coordinate Systems

Position as a Vector Function

In one dimension, the position of a particle is described by a scalar function x(t). In two or three dimensions, position becomes a vector function:

r(t) = x(t) i + y(t) j + z(t) k

where i, j, and k are the unit vectors in the x, y, and z directions respectively.

Displacement

Displacement is the change in position:

Δr = r(t₂) − r(t₁)

Note that displacement is a vector, while distance traveled is a scalar. The magnitude of displacement is always less than or equal to the distance traveled.

Vector Decomposition

Any vector A can be decomposed into components:

A = A cos θ i + A sin θ j (in 2D)

The magnitude is A = √(Ax² + Ay²) and the direction angle is θ = tan⁻¹(Ay/Ax).

1.2 Velocity: The First Derivative

Instantaneous Velocity

The velocity is defined as the time derivative of position:

v(t) = dr/dt = (dx/dt) i + (dy/dt) j + (dz/dt) k

This is the most fundamental definition in kinematics. The components are:

vx = dx/dt, vy = dy/dt, vz = dz/dt

The speed is the magnitude of velocity: v = |v| = √(vx² + vy² + vz²)

Average Velocity vs. Instantaneous Velocity

Average velocity over a time interval [t₁, t₂] is:

v_avg = Δr/Δt = [r(t₂) − r(t₁)] / (t₂ − t₁)

This equals the instantaneous velocity only when velocity is constant. For variable velocity, they differ.

1.3 Acceleration: The Second Derivative

Instantaneous Acceleration

Acceleration is the time derivative of velocity, or the second time derivative of position:

a(t) = dv/dt = d²r/dt² = (d²x/dt²) i + (d²y/dt²) j + (d²z/dt²) k

Components: ax = dvx/dt = d²x/dt², etc.

Tangential and Normal Components of Acceleration

In curvilinear motion, acceleration can be decomposed into tangential and normal (centripetal) components:

a = a_t t̂ + a_n n̂

where:

• a_t = dv/dt (rate of change of speed, along the tangent)
• a_n = v²/ρ (centripetal acceleration, toward center of curvature, ρ = radius of curvature)

  The magnitude of acceleration is a = √(a_t² + a_n²).

  This decomposition is extremely useful because a_t changes the speed while a_n changes the direction of motion.

1.4 Constant Acceleration — The Standard Derivation

When a = constant, we can integrate directly to obtain the kinematic equations.

One-Dimensional Derivation

Starting from a = dv/dt = constant:

dv/dt = a

∫ dv = a ∫ dt

v(t) = v₀ + at

Now, v = dx/dt, so:

dx/dt = v₀ + at

∫ dx = ∫ (v₀ + at) dt

x(t) = x₀ + v₀t + ½at²

To eliminate t, solve v(t) for t: t = (v − v₀)/a

Substitute into x(t):

x − x₀ = v₀(v − v₀)/a + ½a(v − v₀)²/a²

x − x₀ = [v₀(v − v₀) + ½(v − v₀)²] / a

x − x₀ = [v₀v − v₀² + ½v² − v₀v + ½v₀²] / a

x − x₀ = (v² − v₀²) / (2a)

v² = v₀² + 2a(x − x₀)

This is the third kinematic equation, derived purely from calculus.

Two-Dimensional Projectile Motion

For projectile motion (no air resistance), the acceleration is constant: a = −g j.

Decomposition:

• x-direction: ax = 0, so vx = v₀ cos θ₀ (constant), x = x₀ + v₀ cos θ₀ · t
• y-direction: ay = −g, so vy = v₀ sin θ₀ − gt, y = y₀ + v₀ sin θ₀ · t − ½gt²

  Trajectory equation: Eliminate t from the two position equations:

  t = (x − x₀)/(v₀ cos θ₀)

  y = y₀ + (x − x₀) tan θ₀ − g(x − x₀)² / (2v₀² cos² θ₀)

  This is a parabola, confirming that projectile motion traces a parabolic path.

  Time of flight (level ground, y₀ = 0, lands at y = 0):

  0 = v₀ sin θ₀ · T − ½gT²

  T = 2v₀ sin θ₀ / g

  Maximum range occurs at θ₀ = 45°:

  R = v₀² sin(2θ₀) / g, maximized when sin(2θ₀) = 1, so 2θ₀ = 90°, θ₀ = 45°.

  Maximum range: R_max = v₀²/g

  Maximum height: At the apex, vy = 0:

  0 = v₀ sin θ₀ − gt_max

  t_max = v₀ sin θ₀ / g

  H = v₀² sin² θ₀ / (2g)

1.5 Non-Constant Acceleration

The power of the calculus approach is that it works for any acceleration function a(t).

Example: a(t) = kt (linearly increasing acceleration)

v(t) = v₀ + ∫₀ᵗ kt' dt' = v₀ + ½kt²

x(t) = x₀ + ∫₀ᵗ [v₀ + ½kt'²] dt' = x₀ + v₀t + (1/6)kt³

Example: a(t) = −ω²x(t) (Simple Harmonic Motion preview)

This is a differential equation: d²x/dt² = −ω²x, which has the solution x(t) = A cos(ωt + φ).

General Method

1. Start with a(t) (given or derived from forces)
2. Integrate to get v(t) = v₀ + ∫₀ᵗ a(t') dt'
3. Integrate to get x(t) = x₀ + ∫₀ᵗ v(t') dt'

   Or work backwards:

4. Start with x(t) (given)
5. Differentiate to get v(t) = dx/dt
6. Differentiate to get a(t) = dv/dt = d²x/dt²

1.6 Uniform Circular Motion

Position, Velocity, and Acceleration

For circular motion with radius R and angular speed ω:

r(t) = R cos(ωt) i + R sin(ωt) j

v(t) = dr/dt = −Rω sin(ωt) i + Rω cos(ωt) j

a(t) = dv/dt = −Rω² cos(ωt) i − Rω² sin(ωt) j = −ω²r(t)

Key insight: In uniform circular motion, speed is constant but velocity is not (it changes direction). The acceleration is purely centripetal (normal), with zero tangential component.

Period and Frequency

T = 2π/ω (period)

f = 1/T = ω/(2π) (frequency)

1.7 Relative Motion

If frame S' moves with velocity V relative to frame S, then:

r_P|S = r_P|S' + r_S'|S

Differentiating:

v_P|S = v_P|S' + V

a_P|S = a_P|S' (if V is constant — non-accelerating frames)

Key Equations Summary

Equation	Expression	When to Use
Velocity definition	v = dr/dt	Always the starting point
Acceleration definition	a = dv/dt = d²r/dt²	Always the starting point
1st kinematic (constant a)	v = v₀ + at	Constant acceleration
2nd kinematic	Δx = v₀t + ½at²	Constant acceleration, no v in problem
3rd kinematic	v² = v₀² + 2aΔx	Constant acceleration, no t in problem
Projectile x	x = v₀ cos θ₀ · t	Projectile motion
Projectile y	y = v₀ sin θ₀ · t − ½gt²	Projectile motion
Centripetal acceleration	a_c = v²/R = ω²R	Circular motion
General	v(t) = v₀ + ∫ a dt	Variable acceleration
General	x(t) = x₀ + ∫ v dt	Variable velocity

Common Pitfalls

1. Confusing speed and velocity: Speed is |v|, a scalar. Velocity includes direction.
2. Forgetting that projectile vx is constant: Many students mistakenly apply g to the x-component.
3. Sign convention errors: Always define your positive direction and stick with it.
4. Assuming constant acceleration: Only use the kinematic equations when a is constant. If a varies, you must integrate.
5. Centripetal acceleration is not a force: It is the result of a net force, not a force itself. The force causing circular motion points inward.

Newton's three laws of motion form the framework of classical mechanics. In AP Physics C, we go beyond merely applying F = ma — we set up and solve differential equations of motion. When forces depend on position (springs), velocity (drag), or time, the resulting equation F = ma becomes a differential equation that must be solved using calculus.

2.1 Newton's Three Laws

First Law (Law of Inertia)

A body at rest remains at rest, and a body in motion continues in a straight line at constant speed, unless acted upon by a net external force.

Mathematically: If F_net = 0, then v = constant (which includes v = 0).

Inertial reference frames are those in which Newton's first law holds. Accelerating frames are non-inertial.

Second Law

F_net = ma = m(dv/dt) = m(d²r/dt²)

This is a vector equation, so it represents three component equations:

Fx = m·ax, Fy = m·ay, Fz = m·az

The second law is really a differential equation that relates force to the second derivative of position. For the AP exam, the most important forces are:

• Gravity near Earth's surface: F_g = −mg j
• Normal force: N, perpendicular to contact surface
• Tension: T, along a rope or string (massless, inextensible on AP)
• Friction: f = μ_k N (kinetic), |f| ≤ μ_s N (static)
• Spring force (Hooke's Law): F = −kx (restoring force)
• Drag force: F_drag = −bv (velocity-dependent, linear drag)

Third Law

For every action, there is an equal and opposite reaction:

F_A on B = −F_B on A

Key points:

• Action-reaction pairs act on DIFFERENT objects
• They are equal in magnitude, opposite in direction
• They are the same type of force

2.2 Free Body Diagrams (FBDs)

Every dynamics problem begins with a free body diagram:

1. Isolate the object of interest
2. Draw ALL forces acting ON the object (not by the object)
3. Choose a coordinate system
4. Apply F_net = ma in component form

Coordinate System Strategy

• Inclined planes: x-axis parallel to surface, y-axis perpendicular
• Circular motion: radial (toward center) and tangential directions
• General: Choose axes so that acceleration is along one axis

2.3 Applications of Newton's Second Law

Blocks on Inclined Planes

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

Forces on block:

• Weight: mg downward
• Normal force: N perpendicular to incline

  With x along the incline (positive up) and y perpendicular:

  x: −mg sin θ = ma → a = −g sin θ y: N − mg cos θ = 0 → N = mg cos θ

  With friction (μ_k):

  x: −mg sin θ − μ_k N = ma → a = −g(sin θ + μ_k cos θ)

Atwood Machines

Two masses m₁ and m₂ connected by a massless, inextensible string over a massless, frictionless pulley.

For m₁ (positive upward): T − m₁g = m₁a For m₂ (positive upward): T − m₂g = m₂a

Note: Since the string is inextensible, if m₁ accelerates up, m₂ accelerates down. We must be consistent. Taking upward as positive for m₁ and downward as positive for m₂:

m₁: T − m₁g = m₁a m₂: m₂g − T = m₂a

Adding: (m₂ − m₁)g = (m₁ + m₂)a

a = (m₂ − m₁)g / (m₁ + m₂)

T = 2m₁m₂g / (m₁ + m₂)

Pulley with Mass (Massive Pulley)

When the pulley has mass M and radius R (solid disk, I = ½MR²):

For the pulley, net torque = Iα:

TR = Iα = (½MR²)(a/R)

T = ½Ma

This adds a third equation to the Atwood system:

m₁: T₁ − m₁g = m₁a m₂: m₂g − T₂ = m₂a Pulley: (T₂ − T₁)R = ½MR²(a/R)

Solving: (m₂ − m₁)g = (m₁ + m₂ + M/2)a

2.4 Differential Equations of Motion

Velocity-Dependent Forces: Linear Drag

Consider an object falling through a fluid with linear drag F_drag = −bv:

m(dv/dt) = mg − bv

This is a separable first-order ODE. Rearranging:

dv/(g − bv/m) = dt

Integrate from v(0) = 0 to v(t):

∫₀ᵛ dv'/(g − bv'/m) = ∫₀ᵗ dt'

−(m/b) ln(g − bv/m) |₀ᵛ = t

−(m/b)[ln(g − bv/m) − ln(g)] = t

ln[(g − bv/m)/g] = −bt/m

(g − bv/m)/g = e^(−bt/m)

v(t) = (mg/b)(1 − e^(−bt/m))

The terminal velocity is v_terminal = mg/b (as t → ∞).

Time constant: τ = m/b. After t = τ, the object reaches approximately 63% of terminal velocity.

Position-Dependent Forces: Springs (Preview)

For a spring: F = −kx, so:

m(d²x/dt²) = −kx

d²x/dt² + (k/m)x = 0

This is the simple harmonic oscillator equation, with solution x(t) = A cos(ωt + φ) where ω = √(k/m).

Force as a Function of Time

If F = F(t) is given, we integrate directly:

a(t) = F(t)/m v(t) = v₀ + (1/m) ∫₀ᵗ F(t') dt' x(t) = x₀ + ∫₀ᵗ v(t') dt'

2.5 Friction

Static Friction

Static friction adjusts to match the applied force, up to a maximum of μ_s N.

Kinetic Friction

f_k = μ_k N

Kinetic friction is constant (magnitude) and opposes the direction of motion.

Important: μ_s > μ_k always.

2.6 Centripetal Force

For circular motion, Newton's second law gives:

F_net,radial = ma_c = mv²/R = mω²R

The centripetal force is not a new force — it is the net radial force. Common sources:

• Gravity (for orbits): F = GMm/r²
• Tension (for ball on string): T = mv²/R
• Normal force component (for car on banked curve): N sin θ = mv²/R
• Friction (for car on flat curve): f = μN = mv²/R

Vertical Circle Example

A ball of mass m on a string of length R in a vertical circle:

At angle θ from bottom, taking radially inward as positive:

T − mg cos θ = mv²/R

At the top (θ = π): T + mg = mv²/R → T = mv²/R − mg Minimum speed at top: T ≥ 0 → v_min = √(gR)

At the bottom (θ = 0): T − mg = mv²/R → T = mv²/R + mg

Conservation of energy gives the speed at any point: ½mv² + mgy = ½mv₀² + mgy₀

2.7 Non-Inertial Frames and Pseudo-Forces

In an accelerating frame with acceleration a_frame, a pseudo-force −ma_frame must be added to apply Newton's laws.

This is not heavily tested on the AP exam but is conceptually important.

Key Equations

Law/Equation	Expression
Newton's 2nd Law	F_net = ma
Weight	F_g = mg
Normal force (flat)	N = mg (when no vertical acceleration)
Static friction	f_s ≤ μ_s N
Kinetic friction	f_k = μ_k N
Hooke's Law	F = −kx
Centripetal force	F_c = mv²/R = mω²R
Linear drag	F_drag = −bv
Terminal velocity (linear)	v_t = mg/b

Common Pitfalls

1. Including forces the object exerts: Only include forces ON the object in FBDs.
2. Normal force = mg always: NO. N depends on the situation (inclines, elevators, etc.).
3. Centripetal force as a separate force: Centripetal force is the net radial force, not an additional force.
4. Ignoring the mass of the pulley: A massless pulley has T₁ = T₂. A massive pulley has T₁ ≠ T₂.
5. Sign convention errors in Atwood machines: Be consistent about which direction is positive for each mass.

The work-energy theorem provides a powerful alternative to Newton's second law for solving mechanics problems. Instead of working with forces and accelerations (requiring solution of differential equations), we relate the work done by forces to changes in kinetic energy. In AP Physics C, work is defined as a line integral of force, and this calculus-based definition is essential for variable forces and curved paths.

3.1 Work Done by a Force

Definition as a Line Integral

The work done by a force F on a particle moving along a path from point A to point B is:

W = ∫_A^B F · dr

Expanding the dot product in 2D:

W = ∫ (Fx dx + Fy dy)

One-Dimensional Case

If the force and displacement are along the same line (x-axis):

W = ∫_{x_A}^{x_B} F(x) dx

This is the area under the F-vs-x curve between x_A and x_B.

Constant Force Along Displacement

If F is constant and the displacement is d:

W = F · d = Fd cos θ

where θ is the angle between the force and displacement.

Special Cases

• θ = 0°: W = Fd (force in direction of motion)
• θ = 90°: W = 0 (force perpendicular to motion — e.g., centripetal force)
• θ = 180°: W = −Fd (force opposes motion — e.g., friction)

Work by a Variable Force: Spring

For a spring: F(x) = −kx

W_spring = ∫_{x_i}^{x_f} (−kx) dx = −½kx² |_{x_i}^{x_f} = −½k(x_f² − x_i²)

= ½k x_i² − ½k x_f²

If the spring starts at its natural length (x_i = 0) and is compressed/extended to x:

W = −½kx²

3.2 Kinetic Energy and the Work-Energy Theorem

Kinetic Energy

K = ½mv²

Work-Energy Theorem (Derivation)

Starting from Newton's second law:

F_net = ma = m(dv/dt)

The work done by the net force:

W_net = ∫ F_net · dr = ∫ m(dv/dt) · dr

Since dr = v dt:

W_net = ∫ m(dv/dt) · v dt = ∫ m v · dv

Using v · dv = ½ d(v · v) = ½ d(v²):

W_net = ∫ ½m d(v²) = ½mv² |_{v_i}^{v_f}

W_net = ½mv_f² − ½mv_i² = ΔK

This is one of the most important results in mechanics. It holds regardless of whether forces are constant, variable, conservative, or non-conservative.

3.3 Conservative Forces and Potential Energy

Definition of Conservative Force

A force is conservative if:

1. The work done is path-independent (depends only on endpoints)
2. The work done around any closed path is zero
3. The force can be written as the negative gradient of a potential energy: F = −∇U

Examples of Conservative Forces

• Gravity (near surface): F = −mg (constant)
• Universal gravitation: F = −GMm/r²
• Spring force: F = −kx
• Electrostatic force: F = kq₁q₂/r²

Examples of Non-Conservative Forces

• Friction
• Air resistance / drag
• Applied forces (generally)

Potential Energy from Conservative Forces

For a one-dimensional conservative force F(x):

U(x) = −∫ F(x) dx + C

Gravity near Earth's surface:

F = −mg → U(y) = −∫(−mg) dy = mgy + C

Setting U = 0 at y = 0: U = mgy

Spring (Hooke's Law):

F = −kx → U(x) = −∫(−kx) dx = ½kx² + C

Setting U = 0 at x = 0: U = ½kx²

Universal gravitation:

F = −GMm/r² → U(r) = −∫(−GMm/r²) dr = −GMm/r + C

Setting U = 0 at r → ∞: U = −GMm/r

3.4 Conservation of Mechanical Energy

Derivation

W_net = W_conservative + W_non-conservative = ΔK

For conservative forces: W_conservative = −ΔU (the work done by conservative forces equals the negative change in potential energy)

Therefore: −ΔU + W_nc = ΔK

W_nc = ΔK + ΔU = ΔE_mechanical

When W_nc = 0 (no friction, no drag, no applied forces):

ΔK + ΔU = 0 → K_i + U_i = K_f + U_f

Conservation of Energy with Friction

If friction does work W_friction = −f_k d (negative because friction opposes motion):

½mv_i² + U_i = ½mv_f² + U_f + f_k d

where d is the distance traveled along the surface.

3.5 Power

Definition

Power is the rate at which work is done:

P = dW/dt

For a force F acting on an object moving with velocity v:

P = dW/dt = F · dr/dt = F · v

Instantaneous vs. Average Power

• Instantaneous: P = F · v
• Average: P_avg = W/Δt = F · v_avg

Units

1 Watt = 1 Joule/second = 1 kg·m²/s³

Power to Lift at Constant Speed

To lift mass m at constant speed v:

P = Fv = mgv

Power of a Car

For a car with engine force F at speed v:

P = Fv

If the car is on a level road at constant speed (F = f_k = μmg):

P = μmgv

At maximum power P_max, the maximum speed satisfies: P_max = Fv → v_max = P_max/F

3.6 Potential Energy Diagrams

A plot of U(x) vs. x reveals:

• Equilibrium points: Where dU/dx = 0 (i.e., F = −dU/dx = 0)
  • Stable equilibrium: d²U/dx² > 0 (minimum of U)
  • Unstable equilibrium: d²U/dx² < 0 (maximum of U)
  • Neutral equilibrium: d²U/dx² = 0 (inflection point)
• Turning points: Where K = 0, i.e., U(x) = E_total
• Classically forbidden regions: Where U(x) > E_total (K would be negative — impossible classically)

Force from Potential Energy

F(x) = −dU/dx

This is the one-dimensional case of F = −∇U.

The force points in the direction of decreasing potential energy.

3.7 Work-Energy vs. Newton's Laws

Feature	Newton's 2nd Law	Work-Energy Theorem
Quantity	Vector (forces)	Scalar (energy)
Information	Gives acceleration → velocity and position as f(t)	Relates speed at two points
Variable forces	Requires solving ODEs	Requires integrating F·dr
Best for	Finding acceleration, time	Finding speed, height
Direction info	Preserves	Loses (scalar)

Strategy: Use Newton's laws when you need time or acceleration. Use energy methods when you only need speeds and positions, especially with variable forces.

Key Equations

Quantity	Expression
Work (general)	W = ∫ F · dr
Work (1D variable force)	W = ∫ F(x) dx
Work (constant force)	W = Fd cos θ
Kinetic energy	K = ½mv²
Work-energy theorem	W_net = ΔK
Potential energy (1D)	U = −∫ F dx
Force from potential	F = −dU/dx
Conservation of energy	K_i + U_i = K_f + U_f + W_nc
Power (instantaneous)	P = F · v
Power (average)	P_avg = W/Δt

Common Pitfalls

1. Forgetting the dot product: Work is F·d = Fd cos θ, not just Fd. Centripetal forces do zero work.
2. Sign of gravitational PE: U = mgy (increases upward). U = −GMm/r (most negative at surface, zero at infinity).
3. Including normal force in energy: The normal force does zero work (perpendicular to motion on flat surfaces).
4. Spring potential energy: Always U = ½kx² where x is measured from the natural (equilibrium) length.
5. Confusing force and potential: F = −dU/dx (negative of derivative). The force points toward lower potential energy.

This unit extends mechanics from single particles to systems of particles. The key insight is that even when individual particles undergo complex internal interactions, the system's total momentum obeys simple laws. The center of mass framework allows us to treat an extended object as if all its mass were concentrated at a single point. Impulse, derived as the integral of force over time, provides a powerful alternative to Newton's second law for analyzing collisions and explosions — situations where forces act over very short time intervals.

4.1 Center of Mass

Definition

For a system of discrete particles with masses m₁, m₂, ..., mₙ at positions r₁, r₂, ..., rₙ, the center of mass (COM) position is:

r_CM = (1/M) Σ mᵢrᵢ

where M = Σ mᵢ is the total mass.

Continuous Mass Distribution

For a continuous object with mass density ρ (mass per unit volume), the sum becomes an integral:

r_CM = (1/M) ∫ r dm

where dm = ρ dV for volume density, dm = σ dA for surface density, or dm = λ dl for linear density.

Derivation: COM of a Uniform Rod

For a uniform thin rod of length L and mass M lying along the x-axis from x = 0 to x = L:

The linear mass density is λ = M/L (constant).

dx_CM = (1/M) ∫₀ᴸ x λ dx = (λ/M) ∫₀ᴸ x dx = (1/L)[x²/2]₀ᴸ = L/2

This confirms the intuitive result that the COM of a uniform rod is at its midpoint.

Derivation: COM of a Uniform Semicircular Wire

For a semicircular wire of radius R and mass M, use polar coordinates. The arc length element is ds = R dθ, so dm = (M/πR)(R dθ) = (M/π) dθ.

By symmetry, x_CM = 0 (the semicircle is symmetric about the y-axis).

y_CM = (1/M) ∫ y dm = (1/M) ∫₀ᵖ R sin θ · (M/π) dθ

y_CM = (R/π) ∫₀ᵖ sin θ dθ = (R/π)[−cos θ]₀ᵖ = (R/π)(1 + 1) = 2R/π

COM Motion Theorem

Differentiating the COM definition twice:

Ma_CM = Σ mᵢaᵢ = Σ Fᵢ = F_net

F_net = Ma_CM

This is Newton's second law applied to the center of mass. The COM moves as if all external forces act on a single particle of mass M located at the COM. Internal forces cancel in pairs by Newton's third law.

4.2 Impulse and Momentum

Linear Momentum

The linear momentum of a particle is:

p = mv

Newton's second law in momentum form: F = dp/dt

This is actually the more general form. When m is constant, dp/dt = m dv/dt = ma. But for variable-mass systems (not on AP Physics C), this form is essential.

Impulse

Impulse is defined as the integral of force over time:

J = ∫ F dt = Δp = p_f − p_i

This is the impulse-momentum theorem. It is especially useful when the force acts for a short, unknown time (collisions) or when the force varies with time.

Derivation from F = dp/dt

F = dp/dt

∫ F dt = ∫ dp

J = p_f − p_i

Impulse from a Time-Varying Force

If F(t) is given graphically or as a function, the impulse equals the area under the F(t) curve:

J = ∫₁ᵗ² F(t) dt

For a force that varies linearly from F₁ to F₂ over time Δt:

J = ½(F₁ + F₂)Δt (average force × time)

4.3 Conservation of Linear Momentum

The Law

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

Σ p_i = Σ p_f

This follows directly from dp/dt = F_ext = 0, so p = constant.

Component Form

Momentum is conserved component by component. If F_ext,x = 0, then p_x is conserved, even if F_ext,y ≠ 0.

4.4 Collisions

Types of Collisions

Type	Momentum Conserved?	KE Conserved?
Elastic	Yes	Yes
Inelastic	Yes	No
Perfectly inelastic	Yes	No (maximum KE loss)

Elastic Collisions — General Derivation

For a one-dimensional elastic collision between masses m₁ and m₂:

Conservation of momentum: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

Conservation of kinetic energy: ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²

Rearrange momentum: m₁(v₁ᵢ − v₁f) = m₂(v₂f − v₂ᵢ) ... (1)

Rearrange energy: ½m₁(v₁ᵢ² − v₁f²) = ½m₂(v₂f² − v₂ᵢ²) ... (2)

Factor equation (2): ½m₁(v₁ᵢ − v₁f)(v₁ᵢ + v₁f) = ½m₂(v₂f − v₂ᵢ)(v₂f + v₂ᵢ)

Divide (2) by (1):

v₁ᵢ + v₁f = v₂f + v₂ᵢ

This gives: v₁f = v₂f + v₂ᵢ − v₁ᵢ

Substituting back into momentum conservation and solving:

v₁f = [(m₁ − m₂)v₁ᵢ + 2m₂v₂ᵢ] / (m₁ + m₂)

v₂f = [(m₂ − m₁)v₂ᵢ + 2m₁v₁ᵢ] / (m₁ + m₂)

Special Cases

Equal masses (m₁ = m₂): v₁f = v₂ᵢ and v₂f = v₁ᵢ. The particles exchange velocities.

Target initially at rest (v₂ᵢ = 0):

v₁f = [(m₁ − m₂)/(m₁ + m₂)]v₁ᵢ

v₂f = [2m₁/(m₁ + m₂)]v₁ᵢ

Heavy target, light projectile (m₂ >> m₁): v₁f ≈ −v₁ᵢ (projectile bounces back), v₂f ≈ 0.

Light target, heavy projectile (m₁ >> m₂): v₁f ≈ v₁ᵢ (projectile barely slows), v₂f ≈ 2v₁ᵢ.

Perfectly Inelastic Collisions

The objects stick together: v₁f = v₂f = v_f.

m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f

v_f = (m₁v₁ᵢ + m₂v₂ᵢ) / (m₁ + m₂)

Two-Dimensional Collisions

Momentum is conserved in both x and y components:

Σ p_ix = Σ p_fx and Σ p_iy = Σ p_fy

For elastic 2D collisions, KE is also conserved. Often, one object's final direction is given, allowing you to solve for the remaining unknowns.

4.5 Collisions and Energy Loss

Fraction of KE Lost (Perfectly Inelastic)

Initial KE: Kᵢ = ½m₁v₁ᵢ² + ½m₂v₂ᵢ²

Final KE: K_f = ½(m₁ + m₂)v_f²

Fraction lost: (Kᵢ − K_f)/Kᵢ = 1 − K_f/Kᵢ

For v₂ᵢ = 0:

v_f = m₁v₁ᵢ/(m₁ + m₂)

K_f = ½(m₁ + m₂)[m₁v₁ᵢ/(m₁ + m₂)]² = m₁²v₁ᵢ²/[2(m₁ + m₂)]

Fraction lost = 1 − m₁/(m₁ + m₂) = m₂/(m₁ + m₂)

Maximum energy is lost when the two masses are equal (50% loss).

4.6 Systems with Continuous Mass Flow

Rocket Equation (Derivation Sketch)

A rocket of mass M ejects fuel at rate dm/dt (negative) with exhaust velocity v_ex relative to the rocket.

By momentum conservation in the absence of external forces:

d(Mv) = v_ex(−dm) (the ejected mass carries away momentum)

M dv + v dm = −v_ex dm

M dv = −(v + v_ex)dm ≈ −v_ex dm (since v_ex >> v in practice)

M(dv/dt) = −v_ex(dm/dt)

This gives thrust: F_thrust = −v_ex(dm/dt)

Integrating yields the Tsiolkovsky rocket equation (not required in full detail for AP Physics C, but the thrust formula is essential).

Key Equations Summary

Equation	Expression
Center of mass	r_CM = (1/M)Σ mᵢrᵢ = (1/M)∫r dm
COM motion	F_net = Ma_CM
Momentum	p = mv
Newton's 2nd (momentum)	F = dp/dt
Impulse	J = ∫F dt = Δp
Elastic collision v₁f	[(m₁−m₂)v₁ᵢ + 2m₂v₂ᵢ]/(m₁+m₂)
Perfectly inelastic	v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁ + m₂)
Thrust	F = v_ex	dm/dt

Common Pitfalls

1. Forgetting that momentum is a vector: In 2D collisions, you must conserve momentum in each component separately. Do not combine components into a single equation.
2. Assuming KE is conserved in all collisions: Only elastic collisions conserve KE. Always check whether the problem specifies elastic or inelastic.
3. Impulse sign errors: Impulse and force have the same direction. If a force opposes motion, the impulse is negative relative to the velocity direction.
4. COM vs. geometric center: The center of mass is not necessarily the geometric center. It depends on the mass distribution.
5. Including internal forces in momentum analysis: Internal forces cancel in pairs. Only external forces change the total momentum of a system.

Rotational mechanics is the angular analog of translational mechanics. Every translational quantity — position, velocity, acceleration, force, mass, momentum — has a rotational counterpart: angle, angular velocity, angular acceleration, torque, moment of inertia, and angular momentum. In AP Physics C, the key calculus elements are computing moments of inertia through integration and applying the rotational form of Newton's second law to solve dynamics problems. The connection between linear and angular quantities through the arc length relation (s = Rθ) is the bridge between the two frameworks.

5.1 Rotational Kinematics

Angular Quantities

For rotation about a fixed axis, the angular position θ replaces linear position x:

• Angular velocity: ω = dθ/dt (rad/s)
• Angular acceleration: α = dω/dt = d²θ/dt² (rad/s²)

  The kinematic equations for constant α are exact analogs of the translational equations:

  | Translational | Rotational | |--------------|------------| | v = v₀ + at | ω = ω₀ + αt | | x = x₀ + v₀t + ½at² | θ = θ₀ + ω₀t + ½αt² | | v² = v₀² + 2aΔx | ω² = ω₀² + 2αΔθ |

Connection to Linear Motion

For a point at distance R from the axis:

• Arc length: s = Rθ
• Tangential velocity: v_t = Rω
• Tangential acceleration: a_t = Rα
• Centripetal acceleration: a_c = Rω² = v_t²/R

5.2 Moment of Inertia — Integral Calculations

Definition

The moment of inertia about an axis is:

I = ∫ r² dm

where r is the perpendicular distance from the axis to the mass element dm.

Derivation: Uniform Thin Rod about Its Center

For a rod of mass M, length L, axis perpendicular through center:

Linear density: λ = M/L, dm = λ dx = (M/L) dx

I = ∫₋_{L/2}^{L/2} x² (M/L) dx = (M/L) [x³/3]₋_{L/2}^{L/2}

= (M/L) · 2 · (L/2)³/3 = (M/L) · (L³/12) = ML²/12

Derivation: Uniform Thin Rod about One End

I = ∫₀ᴸ x² (M/L) dx = (M/L)[x³/3]₀ᴸ = ML²/3

Note the parallel axis theorem: I_end = I_CM + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3. ✓

Derivation: Solid Cylinder (or Disk) about Central Axis

Use cylindrical coordinates. Volume element: dV = r dr dθ dz, where r is the radial distance from the axis.

For height H, radius R, mass M:

Density: ρ = M/(πR²H)

dm = ρ r dr dθ dz

I = ∫ r² dm = ρ ∫₀ᴿ ∫₀^{2π} ∫₀ᴴ r³ dz dθ dr

= ρ · 2πH · ∫₀ᴿ r³ dr = ρ · 2πH · R⁴/4

= [M/(πR²H)] · 2πH · R⁴/4 = ½MR²

Derivation: Solid Sphere about a Diameter

Use spherical coordinates: dV = r² sin φ dr dφ dθ.

I = ρ ∫ r² sin²φ · r² sin φ dr dφ dθ

where r sin φ is the perpendicular distance from the axis (the z-axis), so r⊥ = r sin φ.

I = ρ ∫₀ᴿ ∫₀ᵖ ∫₀^{2π} r⁴ sin³φ dθ dφ dr

= ρ · 2π · [r⁵/5]₀ᴿ · ∫₀ᵖ sin³φ dφ

∫₀ᵖ sin³φ dφ = ∫₀ᵖ (1 − cos²φ) sin φ dφ

Let u = cos φ, du = −sin φ dφ:

= ∫₁⁻¹ (1 − u²)(−du) = ∫₋₁¹ (1 − u²) du = [u − u³/3]₋₁¹ = (1 − 1/3) − (−1 + 1/3) = 4/3

I = [M/(4πR³/3)] · 2π · R⁵/5 · 4/3 = [3M/(4πR³)] · 2πR⁵/5 · 4/3

= [3M/(4πR³)] · 8πR⁵/15 = 2MR²/5 = ⅖MR²

Derivation: Thin Spherical Shell about a Diameter

For a shell, all mass is at r = R:

I = ∫ R² sin²φ dm = R² ∫₀ᵖ sin²φ · (M/4πR²) · 2πR² sin φ dφ

= (M/2) ∫₀ᵖ sin³φ dφ = (M/2)(4/3) = ⅔MR²

Parallel Axis Theorem

If I_CM is known, the moment of inertia about any parallel axis a distance d away is:

I = I_CM + Md²

Perpendicular Axis Theorem (Planar Objects Only)

For a flat plate in the xy-plane:

I_z = I_x + I_y

where I_z is about the z-axis (perpendicular to the plate).

5.3 Torque and Rotational Dynamics

Torque

τ = r × F

For rotation about a fixed axis, only the component of r × F along the axis matters:

τ = rF sin θ = rF_⊥ = r_⊥F

where θ is the angle between r and F.

Newton's Second Law for Rotation

Σ τ = Iα

This is the rotational analog of Σ F = ma.

Derivation: Rolling Without Slipping

For a sphere or cylinder rolling without slipping down an incline of angle θ:

Forces along the incline: mg sin θ − f = ma (f is static friction)

Torque about CM: fR = Iα

Rolling condition: a = Rα

From torque: f = Iα/R = Ia/R²

Substitute into force equation: mg sin θ − Ia/R² = ma

a = mg sin θ / (m + I/R²)

For a solid sphere (I = ⅖mR²): a = g sin θ / (1 + 2/5) = 5g sin θ / 7

For a solid cylinder (I = ½mR²): a = g sin θ / (1 + 1/2) = 2g sin θ / 3

For a hollow cylinder (I = mR²): a = g sin θ / (1 + 1) = g sin θ / 2

The solid sphere reaches the bottom first because it has the smallest fraction of energy in rotation.

5.4 Angular Momentum

Definition

For a particle: L = r × p = r × mv

For a rigid body rotating about a fixed axis: L = Iω

Angular Momentum Theorem

τ_net = dL/dt

This is the rotational form of Newton's second law and is always valid.

Conservation of Angular Momentum

If τ_net = 0, then L = Iω = constant.

A classic example: a figure skater pulling in her arms (decreasing I) increases ω to conserve L.

Mathematically: I₁ω₁ = I₂ω₂

5.5 Rotational Kinetic Energy

Definition

K_rot = ½Iω²

Total Kinetic Energy of Rolling

For rolling without slipping: v = Rω

K_total = K_trans + K_rot = ½mv² + ½Iω² = ½mv² + ½I(v²/R²) = ½v²(m + I/R²)

Work-Energy Theorem for Rotation

W = ∫ τ dθ = ΔK_rot = ½Iω²_f − ½Iω²_i

5.6 Combined Translation and Rotation

When solving problems involving both translation and rotation:

1. Draw a free-body diagram (FBD)
2. Apply ΣF = ma for translation
3. Apply Στ = Iα for rotation about the CM (or a fixed axis)
4. Use the constraint equation (e.g., a = Rα for rolling without slipping)
5. Solve the system of equations simultaneously

Key Equations Summary

Quantity	Translational	Rotational
Kinematics	v = dx/dt, a = dv/dt	ω = dθ/dt, α = dω/dt
Inertia	m	I = ∫r² dm
Newton's 2nd	F = ma	τ = Iα
Momentum	p = mv	L = Iω
Kinetic Energy	K = ½mv²	K = ½Iω²
Work	W = ∫F dx	W = ∫τ dθ
Common I's	—	⅖MR² (sphere), ½MR² (disk)

Common Pitfalls

1. Using the wrong moment of inertia: Always check the axis. A rod has I = ML²/12 about its center but ML²/3 about its end.
2. Confusing torque and force: Torque depends on where the force is applied (the lever arm r_⊥). Same force, different point of application → different torque.
3. Rolling with slipping vs. without: The constraint a = Rα only applies to rolling without slipping. If the object slips, these quantities are independent.
4. Including static friction in energy calculations: Static friction does no work in rolling without slipping (the contact point is instantaneously at rest). Don't subtract friction energy.
5. Angular momentum direction: For 2D rotation about a fixed axis, L points along the axis. Use the right-hand rule to determine the sign (CW vs. CCW).

Oscillatory motion is ubiquitous in physics: from masses on springs to atoms in molecules to electrical circuits. In AP Physics C, the central object of study is simple harmonic motion (SHM), which arises whenever a restoring force is proportional to the displacement. The mathematical structure is a second-order linear differential equation whose solutions are sinusoidal functions. The calculus treatment gives you the tools to derive the period, frequency, and energy relationships from first principles, and to handle more complex situations where the oscillation is not purely simple harmonic.

6.1 The Simple Harmonic Oscillator — Differential Equation

Derivation from Hooke's Law

A mass m attached to a spring of constant k, displaced x from equilibrium:

F = −kx

Newton's second law: m(d²x/dt²) = −kx

Rearranging: d²x/dt² + (k/m)x = 0

This is the SHM differential equation. Comparing with the standard form d²x/dt² + ω²x = 0:

ω = √(k/m) — the angular frequency, determined entirely by physical parameters.

General Solution

The general solution to d²x/dt² + ω²x = 0 is:

x(t) = A cos(ωt + φ)

where:

• A = amplitude (maximum displacement, determined by initial conditions)
• ω = angular frequency (rad/s)
• φ = phase constant (determined by initial conditions)

Verification

dx/dt = −Aω sin(ωt + φ)

d²x/dt² = −Aω² cos(ωt + φ) = −ω²x

So d²x/dt² + ω²x = 0. ✓

Alternative Form

x(t) can also be written as:

x(t) = A sin(ωt + φ') or x(t) = C cos(ωt) + D sin(ωt)

where C = A cos φ and D = −A sin φ. All forms are equivalent.

6.2 Period, Frequency, and Angular Frequency

• Period: T = 2π/ω = 2π√(m/k)
• Frequency: f = 1/T = ω/(2π) = (1/2π)√(k/m)
• Angular frequency: ω = 2πf = √(k/k)

Key Insight: Independence of Amplitude

The period T depends only on m and k, NOT on the amplitude A. This is a defining feature of SHM and distinguishes it from other oscillatory motions (like a pendulum at large angles).

6.3 Velocity and Acceleration in SHM

From x(t) = A cos(ωt + φ):

v(t) = −Aω sin(ωt + φ)

a(t) = −Aω² cos(ωt + φ) = −ω²x(t)

Maximum Values

• |v|_max = Aω (occurs at x = 0)
• |a|_max = Aω² (occurs at x = ±A)

Phase Relationships

• When x is at maximum (±A), v = 0 and a is at maximum magnitude
• When x = 0, v is at maximum and a = 0
• Velocity leads displacement by π/2 (90°)
• Acceleration is 180° out of phase with displacement (always opposite to x)

6.4 Energy in Simple Harmonic Motion

Total Energy

E = K + U = ½mv² + ½kx²

Substituting v = −Aω sin(ωt + φ) and x = A cos(ωt + φ):

K = ½m(Aω sin(ωt + φ))² = ½mA²ω² sin²(ωt + φ)

U = ½k(A cos(ωt + φ))² = ½kA² cos²(ωt + φ)

Since ω² = k/m: mA²ω² = kA²

E = ½kA² sin²(ωt + φ) + ½kA² cos²(ωt + φ) = ½kA²

The total energy is constant — SHM is a conservative system. The energy oscillates between kinetic and potential forms.

Energy as a Function of Position

Using v² = ω²(A² − x²) (from the identity sin² + cos² = 1):

E = ½mv² + ½kx² = ½mω²(A² − x²) + ½kx² = ½kA²

At x = 0: K = ½kA² (all kinetic)

At x = A: U = ½kA² (all potential)

At x = A/√2: K = U = ¼kA² (equal split)

6.5 The Simple Pendulum

Derivation of the Differential Equation

A pendulum of length L and mass m (point mass on massless string). At angle θ from vertical:

The tangential component of gravity provides the restoring force:

F_tangential = −mg sin θ

The arc length: s = Lθ, so the tangential acceleration is a_t = L(d²θ/dt²)

mL(d²θ/dt²) = −mg sin θ

d²θ/dt² + (g/L) sin θ = 0

Small Angle Approximation

For small θ: sin θ ≈ θ (radians). Then:

d²θ/dt² + (g/L)θ = 0

This is SHM with ω = √(g/L), so:

T = 2π√(L/g)

Validity of Small Angle Approximation

sin θ ≈ θ is accurate to within 1% for θ < 14° and within 0.5% for θ < 10°. For the AP exam, assume the small angle approximation is valid unless told otherwise.

Physical vs. Simple Pendulum

A physical pendulum is an extended object pivoting about a point not at its COM. The period is:

T = 2π√(I/(Mgd))

where I is the moment of inertia about the pivot and d is the distance from the pivot to the COM.

For a simple pendulum (point mass at distance L): I = mL², d = L, giving T = 2π√(L/g). ✓

6.6 Damped Oscillations

Equation of Motion

With a damping force F_d = −bv (velocity-proportional):

m(d²x/dt²) = −kx − b(dx/dt)

d²x/dt² + (b/m)(dx/dt) + (k/m)x = 0

Define: γ = b/(2m) (damping coefficient), ω₀ = √(k/m) (natural frequency)

d²x/dt² + 2γ(dx/dt) + ω₀²x = 0

Three Regimes

Regime	Condition	Solution Behavior
Underdamped	γ < ω₀	Oscillates with exponentially decaying amplitude
Critically damped	γ = ω₀	Returns to equilibrium fastest without oscillating
Overdamped	γ > ω₀	Returns to equilibrium slowly without oscillating

Underdamped Solution

x(t) = A e^(−γt) cos(ω_d t + φ)

where ω_d = √(ω₀² − γ²) is the damped frequency (slightly less than ω₀).

6.7 Forced Oscillations and Resonance

Equation of Motion

d²x/dt² + 2γ(dx/dt) + ω₀²x = (F₀/m) cos(ωt)

where F₀ cos(ωt) is the driving force.

Resonance

When the driving frequency ω ≈ ω₀ (for small damping, ω ≈ ω_d), the amplitude becomes very large. This is resonance. The resonant frequency is:

ω_res = √(ω₀² − 2γ²)

For light damping, ω_res ≈ ω₀.

At resonance, energy is transferred most efficiently from the driver to the oscillator.

Key Equations Summary

Quantity	Expression
SHM equation	d²x/dt² + ω²x = 0
Solution	x(t) = A cos(ωt + φ)
Angular frequency (spring)	ω = √(k/m)
Angular frequency (pendulum)	ω = √(g/L)
Period	T = 2π/ω
Energy	E = ½kA² = constant
Velocity	v = ±ω√(A² − x²)
Damped	x = Ae^(−γt)cos(ω_d t + φ)

Common Pitfalls

1. Degrees vs. radians: SHM equations require angles in radians. The small angle approximation sin θ ≈ θ is only valid for θ in radians.
2. Amplitude doesn't affect period: T is independent of A for SHM. Changing the initial displacement changes A but not T.
3. Angular frequency ω vs. frequency f: ω = 2πf. Don't confuse them. Always check units.
4. Pendulum period is independent of mass: T = 2π√(L/g) has no m. This is because gravity provides both the restoring force (mg sin θ) and the inertia (ma).
5. Energy at turning points: At maximum displacement, all energy is potential. At equilibrium, all energy is kinetic. These are exact, not approximations.

Gravitation is the weakest of the four fundamental forces, yet it dominates the large-scale structure of the universe. Newton's law of universal gravitation describes the force between any two masses, while the gravitational potential energy framework (analogous to the spring potential) enables energy-based problem solving. In AP Physics C, the calculus appears in computing gravitational fields and potentials via integration over continuous mass distributions, in deriving escape velocity, and in analyzing orbital mechanics using energy and angular momentum conservation.

7.1 Newton's Law of Universal Gravitation

The Law

Any two point masses m₁ and m₂ separated by distance r attract each other with a force:

F = −(Gm₁m₂/r²) r̂

where G = 6.674 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, and r̂ points from one mass toward the other.

Gravitational Field

The gravitational field g at a point is the force per unit mass experienced by a test mass:

g = F/m_test

For a point mass M: g = −(GM/r²) r̂

At Earth's surface: g = GM_E/R_E² ≈ 9.8 m/s²

7.2 Gravitational Potential Energy

Point Mass Potential Energy

The gravitational potential energy of a two-mass system is:

U = −GMm/r

This is defined with the zero of potential energy at r → ∞ (infinite separation).

Derivation via Integration

The potential energy equals the negative of the work done by gravity to bring a mass m from ∞ to distance r from M:

U(r) = −∫_∞^r F · dr = −∫_∞^r (−GMm/r'²) dr'

= GMm ∫_∞^r r'⁻² dr' = GMm [−1/r']_∞^r

= GMm(−1/r − (−1/∞)) = −GMm/r

Gravitational Potential (Not Energy)

The gravitational potential V is potential energy per unit mass:

V = −GM/r (scalar, units: J/kg = m²/s²)

The gravitational field is the negative gradient of the potential: g = −∇V = −dV/dr r̂

For a point mass: −d(−GM/r)/dr = −GM/r² ✓

7.3 Gravitational Field of Extended Objects

Field of a Spherical Shell (Shell Theorem)

Outside the shell (r > R): The shell acts as if all its mass is concentrated at its center.

g(r) = GM/r² (pointing inward)

Inside the shell (r < R): The net gravitational field is zero.

g(r) = 0

This result follows from symmetry and integration. Each mass element on the shell creates a field, but the components perpendicular to the radial direction cancel. The integral over the shell gives zero for interior points.

Field of a Solid Sphere

Outside (r > R): g = GM/r² (same as point mass)

Inside (r < R): Only the mass enclosed within radius r contributes:

M_enclosed = M(r/R)³ (for uniform density ρ = 3M/4πR³)

g(r) = GM(r/R)³/r² = GMr/R³

The field increases linearly with r inside the sphere, reaches maximum g = GM/R² at the surface, then decreases as 1/r² outside.

Derivation: g Inside a Uniform Sphere

ρ = M/(4πR³/3) = 3M/(4πR³)

M_enc = ρ · (4πr³/3) = M(r/R)³

g = GM_enc/r² = GM(r/R)³/r² = GMr/R³

7.4 Escape Velocity

Derivation

A projectile at the surface of a planet of mass M and radius R needs minimum kinetic energy to just reach r = ∞ (where both KE and PE are zero):

½mv²_escape + (−GMm/R) = 0 + 0

v_escape = √(2GM/R)

For Earth

v_esc = √(2 × 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.37 × 10⁶)

= √(1.875 × 10⁸) ≈ 11,200 m/s ≈ 11.2 km/s

Key Insight

Escape velocity is independent of the mass of the projectile. A feather and a cannonball need the same speed to escape (neglecting atmosphere). It depends only on M and R of the planet.

7.5 Orbital Mechanics

Circular Orbits

For a satellite of mass m in circular orbit of radius r around mass M:

Gravitational force = centripetal force:

GMm/r² = mv²/r → v² = GM/r

v_orbital = √(GM/r)

Period: T = 2πr/v = 2πr/√(GM/r) = 2π√(r³/GM) (Kepler's Third Law)

T² = (4π²/GM) r³

Orbital Energy

Total mechanical energy of a circular orbit:

E = K + U = ½mv² + (−GMm/r) = ½(GMm/r) − GMm/r = −GMm/(2r)

The total energy is negative (bound orbit). The kinetic energy equals half the magnitude of the potential energy:

K = −E = GMm/(2r)

U = 2E = −GMm/r

Elliptical Orbits — Vis-Viva Equation

For any orbit (circular, elliptical, parabolic, or hyperbolic), the vis-viva equation gives the speed at any point:

v² = GM(2/r − 1/a)

where a is the semi-major axis. For a circle, a = r, and v² = GM/r ✓.

For a bound orbit (ellipse), E = −GMm/(2a), which depends only on the semi-major axis.

Kepler's Laws

1. Law of Orbits: Planets move in ellipses with the Sun at one focus.
2. Law of Areas: A line from the Sun to the planet sweeps out equal areas in equal times. This is equivalent to conservation of angular momentum (dA/dt = L/2m = constant).
3. Law of Periods: T² ∝ a³ for all planets orbiting the same central body.

7.6 Gravitational Potential by Integration

Potential of a Ring on Its Axis

A ring of mass M and radius R. Find V at distance x along the axis from the center.

Every mass element dm is at the same distance √(R² + x²) from the point:

V(x) = −G ∫ dm/√(R² + x²) = −GM/√(R² + x²)

On the axis (x = 0): V(0) = −GM/R

Field of the Ring on Its Axis

g_x = −dV/dx = −d/dx[−GM/√(R² + x²)] = −GMx/(R² + x²)^(3/2)

At large x: g ≈ −GMx/x³ = −GM/x² (point mass behavior)

At x = 0: g = 0 (by symmetry, field cancels at the center)

Key Equations Summary

Quantity	Expression
Gravitational force	F = GMm/r²
Gravitational field (point)	g = GM/r²
Potential energy	U = −GMm/r
Gravitational potential	V = −GM/r
Escape velocity	v_esc = √(2GM/R)
Orbital velocity	v_orb = √(GM/r)
Orbital period	T² = 4π²r³/(GM)
Orbital energy	E = −GMm/(2r)
Vis-viva	v² = GM(2/r − 1/a)
g inside uniform sphere	g = GMr/R³

Common Pitfalls

1. Sign of gravitational PE: U is negative (zero at infinity). Greater separation → less negative → higher PE.
2. g is not the same as G: g is the gravitational field strength (≈ 9.8 m/s² at Earth's surface). G is the universal constant.
3. Orbital speed decreases with r: v_orb = √(GM/r). Higher orbits are slower, contrary to intuition.
4. Total orbital energy is negative: E = −GMm/(2r) < 0 for bound orbits. To unbind an object, you must add energy to make E ≥ 0.
5. Shell theorem applies only to spherical symmetry: For non-spherical objects, you must integrate to find the field.

The position of a particle moving along the x-axis is given by:

x(t) = t³ − 6t² + 9t + 4 (in meters, t in seconds)

(a) Find the velocity function v(t) and acceleration function a(t).

(b) At what time(s) is the particle at rest?

(c) What is the total distance traveled between t = 0 and t = 5 s?

Solution

(a) Velocity is the first derivative:

v(t) = dx/dt = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3)

Acceleration is the second derivative:

a(t) = dv/dt = 6t − 12 = 6(t − 2)

(b) The particle is at rest when v(t) = 0:

3(t − 1)(t − 3) = 0 → t = 1 s and t = 3 s

(c) To find total distance, we must integrate the speed (absolute value of velocity). First, determine the sign of v(t):

• For 0 < t < 1: v(0) = 9 > 0 (moving right)
• For 1 < t < 3: v(2) = 3(1)(−1) = −3 < 0 (moving left)
• For t > 3: v(4) = 3(3)(1) = 9 > 0 (moving right)

  Distance = ∫₀¹ |v| dt + ∫₁³ |v| dt + ∫₃⁵ |v| dt

  = ∫₀¹ (3t² − 12t + 9) dt + ∫₁³ (−3t² + 12t − 9) dt + ∫₃⁵ (3t² − 12t + 9) dt

  First integral: [t³ − 6t² + 9t]₀¹ = 1 − 6 + 9 = 4 m

  Second integral: [−t³ + 6t² − 9t]₁³ = (−27 + 54 − 27) − (−1 + 6 − 9) = 0 − (−4) = 4 m

  Third integral: [t³ − 6t² + 9t]₃⁵ = (125 − 150 + 45) − (27 − 54 + 27) = 20 − 0 = 20 m

  Total distance = 4 + 4 + 20 = 28 m

Problem 2: Non-Constant Acceleration

A particle starts from rest at the origin. Its acceleration is given by:

a(t) = 6t − 4 (in m/s²)

(a) Find the velocity as a function of time.

(b) Find the position as a function of time.

(c) When does the particle return to the origin?

Solution

(a) Integrate acceleration with v(0) = 0:

v(t) = ∫₀ᵗ (6t' − 4) dt' = 3t² − 4t

(b) Integrate velocity with x(0) = 0:

x(t) = ∫₀ᵗ (3t'² − 4t') dt' = t³ − 2t²

(c) Set x(t) = 0:

t³ − 2t² = t²(t − 2) = 0

t = 0 (start) or t = 2 s

The particle returns to the origin at t = 2 s.

---

Problem 3: Projectile Motion with Calculus

A ball is launched from a cliff of height h = 80 m with initial speed v₀ = 30 m/s at an angle θ = 37° above the horizontal. Use g = 10 m/s².

(a) Derive the parametric equations x(t) and y(t) using integration.

(b) Find the time of flight and horizontal range.

(c) Find the velocity vector the instant before impact.

Solution

(a) Decompose the initial velocity:

v₀ₓ = 30 cos 37° = 30(4/5) = 24 m/s v₀ᵧ = 30 sin 37° = 30(3/5) = 18 m/s

Accelerations: ax = 0, ay = −10 m/s²

Integrate for velocity: vₓ(t) = v₀ₓ = 24 m/s vᵧ(t) = v₀ᵧ − gt = 18 − 10t

Integrate for position (starting from origin at cliff edge): x(t) = 24t y(t) = 18t − 5t²

Since the cliff is 80 m tall, ground is at y = −80: y(t) = 80 + 18t − 5t² (measuring from ground)

(b) Set y(t) = 0:

−5t² + 18t + 80 = 0 5t² − 18t − 80 = 0 t = [18 ± √(324 + 1600)] / 10 = [18 ± √1924] / 10 = [18 ± 43.86] / 10

t = 6.19 s (taking positive root)

Range: x(6.19) = 24(6.19) = 148.5 m

(c) vₓ = 24 m/s, vᵧ(6.19) = 18 − 10(6.19) = −43.9 m/s

Angle below horizontal: θ = tan⁻¹(43.9/24) = 61.3°

---

Problem 4: Uniform Circular Motion

A particle moves in a circle of radius R = 2.0 m with angular position θ(t) = 3t² + 2t (radians).

(a) Find the angular velocity and angular acceleration as functions of time.

(b) At t = 1.0 s, find the tangential and centripetal acceleration components.

(c) At t = 1.0 s, find the magnitude of the total acceleration.

Solution

(a) ω(t) = dθ/dt = 6t + 2

α(t) = dω/dt = 6 (constant)

(b) At t = 1.0 s:

ω(1) = 6(1) + 2 = 8 rad/s α(1) = 6 rad/s²

Tangential acceleration: a_t = Rα = 2.0(6) = 12 m/s²

Centripetal acceleration: a_c = Rω² = 2.0(64) = 128 m/s²

(c) |a| = √(a_t² + a_c²) = √(144 + 16384) = √16528 = 128.6 m/s²

The centripetal component dominates, as expected for fast circular motion.

---

Problem 5: MCQ Practice

1. A particle's position is x(t) = 4t³ − 3t. At t = 2 s, its acceleration is:

   (A) 24 m/s² (B) 36 m/s² (C) 48 m/s² (D) 72 m/s²

2. The velocity of a particle is v(t) = t² − 4t + 3. The particle changes direction at t = :

   (A) 1 s only (B) 3 s only (C) 1 s and 3 s (D) 2 s only

3. A projectile is launched at 60° above horizontal with speed v₀. The ratio of its maximum height to its range is:

   (A) √3/4 (B) 3√3/8 (C) 1/4 (D) 3/8

4. In uniform circular motion, the acceleration vector:

   (A) Is constant in both magnitude and direction (B) Has constant magnitude and changing direction (C) Has changing magnitude and constant direction (D) Changes in both magnitude and direction

MCQ Answers and Explanations

1. (C) a(t) = dv/dt = d²x/dt² = 24t. At t = 2: a = 48 m/s².
2. (C) v(t) = (t−1)(t−3), which changes sign at t = 1 and t = 3.
3. (B) H = v₀²sin²60°/(2g) = 3v₀²/(8g). R = v₀²sin120°/g = v₀²√3/g. Ratio = 3√3/8.
4. (B) The magnitude v²/R is constant, but the direction always points toward the center, which changes continuously.

   When working through kinematics problems, always start from the definitions (v = dx/dt, a = dv/dt) rather than memorizing formulas. This calculus-first approach handles both constant and non-constant acceleration cases uniformly.

A small sphere of mass m = 0.5 kg is dropped from rest in a fluid. The drag force is proportional to velocity: F_drag = −bv, where b = 2.0 N·s/m. Take g = 10 m/s².

(a) Set up the differential equation of motion and solve for v(t).

(b) Find the terminal velocity.

(c) How long does it take the sphere to reach 90% of terminal velocity?

Solution

(a) Newton's second law: ma = mg − bv

m(dv/dt) = mg − bv

dv/dt = g − (b/m)v

This is a separable first-order ODE. Rearrange:

dv / (g − bv/m) = dt

Let v_T = mg/b (terminal velocity, which we'll compute in part b). Then:

dv / (g − bv/m) = dv / [g(1 − v/v_T)]

Integrating with v(0) = 0:

∫₀ᵛ dv' / (g − bv'/m) = ∫₀ᵗ dt'

−(m/b) ln(g − bv/m) |₀ᵛ = t

−(m/b) [ln(g − bv/m) − ln(g)] = t

ln[(g − bv/m)/g] = −bt/m

(g − bv/m)/g = e^(−bt/m)

1 − v/v_T = e^(−bt/m)

v(t) = v_T(1 − e^(−bt/m)) = (mg/b)(1 − e^(−bt/m))

(b) Terminal velocity: dv/dt = 0, so mg = bv_T

v_T = mg/b = (0.5)(10)/(2.0) = 2.5 m/s

(c) Set v(t) = 0.9v_T:

0.9v_T = v_T(1 − e^(−bt/m))

0.9 = 1 − e^(−bt/m)

e^(−bt/m) = 0.1

−bt/m = ln(0.1)

t = −(m/b) ln(0.1) = (0.5/2.0)(2.303) = 0.576 s

---

Problem 2: Inclined Plane with Variable Friction

A block of mass m = 2.0 kg slides down a 30° incline. The coefficient of kinetic friction is μ_k = 0.2.

(a) Write Newton's second law in component form and solve for acceleration.

(b) If the block starts from rest, how far does it slide in 3 seconds?

(c) What is the velocity at t = 3 s?

Solution

Choose axes: x along the incline (downward positive), y perpendicular to incline.

(a) y-direction: N − mg cos 30° = 0 → N = mg cos 30°

x-direction: mg sin 30° − μ_k N = ma

mg sin 30° − μ_k mg cos 30° = ma

a = g(sin 30° − μ_k cos 30°) = 10(0.5 − 0.2 × 0.866) = 10(0.5 − 0.173) = 10(0.327) = 3.27 m/s²

(b) Using x = ½at² (starting from rest):

x = ½(3.27)(9) = 14.7 m

(c) v = at = 3.27 × 3 = 9.81 m/s

---

Problem 3: Connected Bodies — Atwood Machine Variant

Two masses m₁ = 4.0 kg and m₂ = 6.0 kg are connected by a light string over a pulley of mass M = 2.0 kg and radius R = 0.1 m. The pulley is a solid disk (I = ½MR²). There is no slipping between the string and the pulley.

(a) Write the three equations of motion (one for each mass, one for the pulley).

(b) Solve for the acceleration of the system.

Solution

(a) Let a = acceleration of both masses (m₂ goes down, m₁ goes up).

For m₂ (heavier, moving down): m₂g − T₂ = m₂a

For m₁ (lighter, moving up): T₁ − m₁g = m₁a

For the pulley (torque equation): (T₂ − T₁)R = Iα = (½MR²)(a/R) = ½MRa

So: T₂ − T₁ = ½Ma

(b) Add the first two equations:

m₂g − m₁g − (T₂ − T₁) = (m₁ + m₂)a

Substitute the pulley equation:

(m₂ − m₁)g − ½Ma = (m₁ + m₂)a

(m₂ − m₁)g = (m₁ + m₂ + M/2)a

a = (m₂ − m₁)g / (m₁ + m₂ + M/2)

a = (6.0 − 4.0)(10) / (4.0 + 6.0 + 1.0) = 20/11 = 1.82 m/s²

Note: Without the pulley's inertia, a = 20/10 = 2.0 m/s². The pulley's rotational inertia reduces the acceleration.

---

Problem 4: Spring-Mass System — Differential Equation

A block of mass m is attached to a spring of constant k on a frictionless horizontal surface. The block is displaced a distance A from equilibrium and released from rest.

(a) Set up and solve the differential equation of motion.

(b) Find the period and frequency in terms of m and k.

(c) At what position is the kinetic energy maximum? What is its value?

Solution

(a) At displacement x from equilibrium, the spring force is F = −kx.

Newton's second law: m(d²x/dt²) = −kx

d²x/dt² + (k/m)x = 0

This is the standard SHM equation. The general solution is:

x(t) = A cos(ωt + φ)

where ω = √(k/m). Initial conditions: x(0) = A, v(0) = 0.

x(0) = A cos(φ) = A → φ = 0

v(0) = −Aω sin(φ) = 0 → confirmed

x(t) = A cos(√(k/m) · t)

(b) Period: T = 2π/ω = 2π√(m/k)

Frequency: f = 1/T = (1/2π)√(k/m)

(c) KE is maximum when PE is minimum (at x = 0, the equilibrium position).

By energy conservation: ½kA² = ½mv²_max

v_max = A√(k/m)

KE_max = ½mv²_max = ½kA²

---

Problem 5: MCQ Practice

1. A body of mass m moves under a force F(x) = −kx³. The acceleration at x = A is:

   (A) −kA³/m (B) kA³/m (C) −kA²/m (D) −kA³

2. An object of mass 5 kg is pushed across a rough surface (μ = 0.3) by a horizontal force of 30 N. The acceleration is: (g = 10 m/s²)

   (A) 6.0 m/s² (B) 3.0 m/s² (C) 2.0 m/s² (D) 1.5 m/s²

3. For a drag force F = −bv², the terminal velocity is:

   (A) mg/b (B) √(mg/b) (C) b/(mg) (D) mg/b²

4. In an Atwood machine with a massless pulley, doubling both masses:

   (A) Doubles the acceleration (B) Halves the acceleration (C) Keeps acceleration the same (D) Quadruples the acceleration

MCQ Answers and Explanations

1. (A) F = ma → a = F/m = −kx³/m. At x = A: a = −kA³/m.
2. (B) N = mg = 50 N, f = μN = 15 N. Net force = 30 − 15 = 15 N. a = 15/5 = 3.0 m/s².
3. (B) At terminal velocity: mg = bv_T² → v_T = √(mg/b).
4. (C) a = (m₂ − m₁)g/(m₂ + m₁). If both double: 2(m₂ − m₁)g/[2(m₂ + m₁)] = (m₂ − m₁)g/(m₂ + m₁). Unchanged.

   In Newton's law problems, always draw a free-body diagram first. Then write F = ma in component form. When forces depend on v or x, you'll get a differential equation — identify its type and solve accordingly.

A force F = (3x² i + 4y j) N acts on a particle. Calculate the work done by this force as the particle moves along the path y = x² from the origin (0, 0) to the point (2, 4).

Solution

Work is defined as W = ∫ F · dr.

Parameterize the path using x as the parameter: y = x², so dy = 2x dx.

dr = dx i + dy j = dx i + 2x dx j

F = (3x²) i + (4x²) j (substituting y = x²)

F · dr = 3x² dx + 4x²(2x dx) = 3x² dx + 8x³ dx

W = ∫₀² (3x² + 8x³) dx = [x³ + 2x⁴]₀² = (8 + 32) − 0 = 40 J

---

Problem 2: Conservative Force and Potential Energy

A force F(x) = −kx + ax³ (where k and a are positive constants) acts on a particle of mass m.

(a) Show that this force is conservative.

(b) Find the potential energy function U(x), taking U(0) = 0.

(c) Find the equilibrium positions and determine their stability.

Solution

(a) A one-dimensional force F(x) is conservative if it depends only on position. Since F is a function of x alone, it is conservative. Equivalently, for 1D: a force is conservative if it can be written as the negative derivative of a potential energy function.

(b) F(x) = −dU/dx

−dU/dx = −kx + ax³

dU/dx = kx − ax³

U(x) = ∫₀ˣ (kx' − ax'³) dx' = ½kx² − ¼ax⁴

(c) Equilibrium: F(x_eq) = 0

−kx + ax³ = 0 → x(−k + ax²) = 0

x = 0, or x = ±√(k/a)

For stability, check dF/dx at each equilibrium:

dF/dx = −k + 3ax²

At x = 0: dF/dx = −k < 0 → dU/dx² = k > 0 → stable equilibrium (minimum of U)

At x = ±√(k/a): dF/dx = −k + 3k = 2k > 0 → d²U/dx² < 0 → unstable equilibrium (maximum of U)

---

Problem 3: Power and Variable Forces

A car of mass m = 1000 kg accelerates from rest on a level road. The engine delivers constant power P = 40,000 W. Neglect air resistance and friction.

(a) Find v(t) by solving the differential equation.

(b) Find x(t).

(c) How long to reach 30 m/s?

Solution

(a) Power: P = Fv = mav (since F = ma)

ma = P/v → a = dv/dt = P/(mv)

v dv = (P/m) dt

∫₀ᵛ v' dv' = ∫₀ᵗ (P/m) dt'

½v² = Pt/m

v(t) = √(2Pt/m)

(b) v = dx/dt = √(2Pt/m)

dx/dt = √(2P/m) · √t

x(t) = √(2P/m) ∫₀ᵗ √t' dt' = √(2P/m) · (2/3)t^(3/2)

x(t) = (2/3)√(2P/m) · t^(3/2)

(c) Set v = 30 m/s:

30 = √(2 × 40000t / 1000) = √(80t)

900 = 80t → t = 11.25 s

---

Problem 4: Energy with Friction — Block on Curved Track

A block of mass m = 2.0 kg starts from rest at height h = 5.0 m on a frictionless curved track, then slides along a rough horizontal surface (μ_k = 0.4) for a distance d before stopping.

(a) Use energy conservation to find d.

(b) If the rough section were only 3.0 m long, with what speed would the block leave the rough section?

Solution

(a) Initial energy (at top of track) = mgh = 2.0(10)(5.0) = 100 J

Final energy (at rest) = 0 J

Work done by friction = −μ_k mgd = −fd

Energy conservation: 0 = mgh − μ_k mgd

100 = 0.4(2.0)(10)d = 8d

d = 100/8 = 12.5 m

(b) KE at end of rough section = mgh − μ_k mg(3.0)

½mv² = 100 − 0.4(20)(3) = 100 − 24 = 76 J

v² = 2(76)/2.0 = 76

v = 8.72 m/s

---

Problem 5: MCQ Practice

1. The work done by a conservative force around a closed path is:

   (A) Always positive (B) Always negative (C) Always zero (D) Depends on the path

2. A spring with k = 200 N/m is compressed 0.1 m. The potential energy stored is:

   (A) 1.0 J (B) 2.0 J (C) 10 J (D) 20 J

3. Power is the time derivative of:

   (A) Force (B) Momentum (C) Kinetic energy (D) Work

4. A force F = (2xy i + x² j) N. The work done moving from (0,0) to (1,1) along y = x is:

   (A) 1 J (B) 2 J (C) 3 J (D) 4 J

MCQ Answers and Explanations

1. (C) By definition, a conservative force has zero work around any closed path. This is equivalent to the curl of F being zero.
2. (A) U = ½kx² = ½(200)(0.01) = 1.0 J.
3. (D) P = dW/dt. For a one-dimensional force: P = Fv. This is also equal to dKE/dt only when all forces are conservative.
4. (A) On y = x: dy = dx. F·dr = 2x(x)dx + x²dx = 3x²dx. W = ∫₀¹ 3x²dx = 1 J.

   For work-energy problems, always identify which forces are conservative (can use potential energy) and which are non-conservative (must compute work directly). The work-energy theorem W_net = ΔKE handles all cases uniformly.

Find the center of mass of a uniform solid hemisphere of radius R and mass M. The flat face lies in the xy-plane and the dome extends in the +z direction.

Solution

Use spherical coordinates with the symmetry axis along z. The volume element in spherical coordinates is dV = r² sin φ dr dφ dθ, where φ is the polar angle from the z-axis.

By symmetry: x_CM = y_CM = 0.

For z_CM, integrate over the hemisphere (0 ≤ r ≤ R, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π):

z_CM = (1/M) ∫ z dm = (ρ/M) ∫ (r cos φ)(r² sin φ dr dφ dθ)

where ρ = M / (2πR³/3) = 3M/(2πR³).

z_CM = (ρ/M) ∫₀ᴿ ∫₀^(π/2) ∫₀^(2π) r³ cos φ sin φ dθ dφ dr

= (ρ/M) · 2π · ∫₀ᴿ r³ dr · ∫₀^(π/2) cos φ sin φ dφ

= (ρ/M) · 2π · [r⁴/4]₀ᴿ · [½sin²φ]₀^(π/2)

= (ρ/M) · 2π · R⁴/4 · ½

= (ρ/M) · πR⁴/4

Substituting ρ = 3M/(2πR³):

z_CM = [3M/(2πR³)] · πR⁴/(4M) = 3R/8

---

Problem 2: Impulse from a Variable Force

A 0.5 kg ball is struck by a bat. The force on the ball as a function of time is:

F(t) = (6.0 × 10⁶t − 3.0 × 10⁹t²) N for 0 ≤ t ≤ 2.0 × 10⁻³ s

F = 0 otherwise

(a) Find the impulse delivered to the ball.

(b) If the ball was initially at rest, find its speed after being hit.

Solution

(a) J = ∫₀^(0.002) (6.0 × 10⁶t − 3.0 × 10⁹t²) dt

= [3.0 × 10⁶t² − 1.0 × 10⁹t³]₀^(0.002)

= 3.0 × 10⁶(4 × 10⁻⁶) − 1.0 × 10⁹(8 × 10⁻⁹)

= 12.0 − 8.0 = 4.0 N·s

(b) J = Δp = mv_f − mv_i = mv_f (since v_i = 0)

v_f = J/m = 4.0/0.5 = 8.0 m/s

---

Problem 3: Elastic Collision Analysis

A neutron (mass m) collides elastically with a stationary deuteron (mass 2m).

(a) Find the fraction of the neutron's kinetic energy transferred to the deuteron.

(b) How many such collisions are needed to reduce the neutron's energy by a factor of 10⁶? (Use the average logarithmic energy loss per collision.)

Solution

(a) For elastic collision with v₂ᵢ = 0:

v₂f = 2m₁v₁ᵢ/(m₁ + m₂) = 2m · v₁ᵢ/(m + 2m) = 2v₁ᵢ/3

KE transferred = ½(2m)(2v₁ᵢ/3)² = ½(2m)(4v₁ᵢ²/9) = 4mv₁ᵢ²/9

Initial KE = ½mv₁ᵢ²

Fraction transferred = (4mv₁ᵢ²/9)/(½mv₁ᵢ²) = 8/9 ≈ 88.9%

(b) After one collision, the neutron's velocity is:

v₁f = (m − 2m)v₁ᵢ/(m + 2m) = −v₁ᵢ/3

The neutron retains fraction α = (v₁f/v₁ᵢ)² = 1/9 of its KE.

After n collisions: KE_n = KE₀(1/9)ⁿ

(1/9)ⁿ = 10⁻⁶

n ln(1/9) = −6 ln 10

n(−ln 9) = −6 ln 10

n = 6 ln 10 / ln 9 = 6(2.303)/2.197 = 6.29 → 7 collisions

---

Problem 4: Two-Dimensional Perfectly Inelastic Collision

A 3.0 kg object moving at 4.0 m/s in the +x direction collides with and sticks to a 2.0 kg object moving at 5.0 m/s at 37° above the +x axis.

(a) Find the velocity (magnitude and direction) of the combined object.

(b) How much kinetic energy is lost?

Solution

(a) Conserve momentum in x and y separately:

p_ix = m₁v₁ₓ + m₂v₂ₓ = 3(4) + 2(5 cos 37°) = 12 + 2(4) = 20 kg·m/s

p_iy = m₁v₁ᵧ + m₂v₂ᵧ = 0 + 2(5 sin 37°) = 2(3) = 6 kg·m/s

After collision (M = 5.0 kg):

v_fx = 20/5 = 4.0 m/s v_fy = 6/5 = 1.2 m/s

θ = tan⁻¹(1.2/4.0) = 16.7° above the x-axis

(b) Kᵢ = ½(3)(16) + ½(2)(25) = 24 + 25 = 49 J

K_f = ½(5)(17.44) = 43.6 J

Energy lost = 49 − 43.6 = 5.4 J

---

Problem 5: MCQ Practice

1. The center of mass of a system of particles: (A) Always lies within the system (B) Can lie outside the system (C) Must lie at the geometric center (D) Is always stationary
2. A 2 kg ball moving at 3 m/s hits a wall and bounces back at 2 m/s. The impulse on the ball is: (A) 2 N·s (B) 10 N·s (C) −10 N·s (D) −2 N·s
3. In an elastic collision between equal masses with one initially at rest: (A) Both move forward (B) The incoming ball stops (C) Both reverse direction (D) They stick together
4. Two particles collide and stick together. Which is always conserved? (A) KE only (B) Momentum only (C) Both KE and momentum (D) Neither

MCQ Answers

1. (B) For a C-shape or donut, the COM can be in empty space. 2. (C) J = m(v_f − v_i) = 2(−2 − 3) = −10 N·s. 3. (B) Equal masses exchange velocities. 4. (B) Momentum is always conserved; KE is lost in inelastic collisions.

   For collision problems, first classify the collision type, then write the appropriate conservation equations. In 2D, always use component form. The impulse-momentum theorem is your tool whenever forces act over short time intervals.

A uniform rectangular plate has mass M, width a, and height b. Find its moment of inertia about an axis along one edge of width a (i.e., the axis runs along one side of length a).

Solution

Place the plate in the xy-plane with the axis along the y-axis (x = 0). The plate extends from x = 0 to x = a, y = −b/2 to y = b/2.

Surface mass density: σ = M/(ab)

dm = σ dx dy

I = ∫∫ r² dm = ∫₀ᵃ ∫₋_{b/2}^{b/2} x² σ dy dx

= σ ∫₀ᵃ x² [∫₋_{b/2}^{b/2} dy] dx = σb ∫₀ᵃ x² dx

= (M/ab) · b · [x³/3]₀ᵃ = Ma²/3

Check: By the parallel axis theorem, I_edge = I_CM + Md² = M(a² + b²)/12 + M(a/2)² = M(a² + b²)/12 + Ma²/4 = M(a² + b² + 3a²)/12 = M(4a² + b²)/12. This equals Ma²/3 only when b = 0 (thin rod). Our result is correct for the given axis.

Alternatively, verify with the perpendicular axis theorem: I_CM = I_x + I_y = Mb²/12 + Ma²/12 = M(a² + b²)/12. ✓

---

Problem 2: Atwood Machine with Massive Pulley

Masses m₁ = 3.0 kg and m₂ = 5.0 kg are connected by a string over a solid disk pulley of mass M = 4.0 kg and radius R = 0.2 m. The string does not slip.

(a) Find the acceleration of the masses.

(b) Find the tensions T₁ and T₂ on each side of the pulley.

Solution

(a) Three equations of motion:

For m₁ (lighter, goes up): T₁ − m₁g = m₁a

For m₂ (heavier, goes down): m₂g − T₂ = m₂a

For pulley: (T₂ − T₁)R = Iα = (½MR²)(a/R) → T₂ − T₁ = ½Ma

Adding first two equations: (T₁ − T₂) + (m₂ − m₁)g = (m₁ + m₂)a

Using the pulley equation (T₁ − T₂ = −½Ma):

−½Ma + (m₂ − m₁)g = (m₁ + m₂)a

(m₂ − m₁)g = (m₁ + m₂ + M/2)a

a = (m₂ − m₁)g / (m₁ + m₂ + M/2) = (2)(10)/(3 + 5 + 2) = 20/10 = 2.0 m/s²

(b) T₁ = m₁(g + a) = 3(10 + 2) = 36 N

T₂ = m₂(g − a) = 5(10 − 2) = 40 N

Check: T₂ − T₁ = 4 = ½(4)(2) = 4. ✓

---

Problem 3: Rolling Down an Incline

A solid sphere of mass M = 2.0 kg and radius R = 0.1 m rolls without slipping down a 30° incline from rest.

(a) Find the linear acceleration down the incline.

(b) Find the minimum coefficient of static friction required.

(c) After rolling 5.0 m, find the translational and rotational kinetic energies.

Solution

(a) For a solid sphere rolling without slipping:

a = g sin θ / (1 + I/(mR²)) = g sin 30° / (1 + 2/5) = 10(0.5)/(7/5) = 5 × 5/7 = 25/7 ≈ 3.57 m/s²

(b) Friction provides the torque for rotation:

fR = Iα = (⅖mR²)(a/R) = ⅖ma

f = ⅖ma = ⅖(2)(25/7) = 100/35 ≈ 2.86 N

Normal force: N = mg cos 30° = 2(10)(√3/2) = 17.32 N

μ_s ≥ f/N = 2.86/17.32 = 0.165

(c) v² = 2aΔs = 2(25/7)(5) = 250/7

K_trans = ½mv² = ½(2)(250/7) = 250/7 ≈ 35.7 J

K_rot = ½Iω² = ½(⅖mR²)(v/R)² = ⅕mv² = ⅕(2)(250/7) = 100/7 ≈ 14.3 J

Check: K_total = 35.7 + 14.3 = 50.0 J = mgh = 2(10)(5)(sin 30°) = 50 J. ✓

---

Problem 4: Angular Momentum Conservation

A disk of mass M₁ = 4.0 kg, radius R₁ = 0.5 m rotates freely at ω₁ = 10 rad/s. A second disk of mass M₂ = 2.0 kg, radius R₂ = 0.3 m, initially not rotating, is dropped on top. The two disks stick together and rotate as one.

(a) Find the final angular velocity.

(b) How much kinetic energy is lost?

Solution

(a) Conservation of angular momentum:

L_i = L_f

I₁ω₁ = (I₁ + I₂)ω_f

½M₁R₁² ω₁ = (½M₁R₁² + ½M₂R₂²)ω_f

½(4)(0.25)(10) = [½(4)(0.25) + ½(2)(0.09)]ω_f

5.0 = (0.5 + 0.09)ω_f = 0.59ω_f

ω_f = 5.0/0.59 = 8.47 rad/s

(b) K_i = ½I₁ω₁² = ½(0.5)(100) = 25.0 J

K_f = ½(I₁ + I₂)ω_f² = ½(0.59)(71.8) = 21.2 J

Energy lost = 25.0 − 21.2 = 3.8 J

---

Problem 5: MCQ Practice

1. A hoop and a disk of equal mass and radius roll without slipping down the same incline. The hoop reaches the bottom: (A) First (B) Second (C) Simultaneously (D) Depends on mass
2. The moment of inertia of a thin rod of mass M and length L about an axis through one end perpendicular to the rod is: (A) ML²/6 (B) ML²/12 (C) ML²/3 (D) ML²/2
3. A spinning ice skater extends her arms, doubling her moment of inertia. Her angular velocity: (A) Doubles (B) Halves (C) Quadruples (D) Remains the same
4. For a rolling object, the velocity of the contact point is: (A) v (B) 2v (C) 0 (D) −v

MCQ Answers

1. (B) The disk (I = ½MR²) has less rotational inertia, so more KE goes to translation. a_disk = 2g sin θ/3 > a_hoop = g sin θ/2.
2. (C) I = ML²/3. (ML²/12 is about the center.)
3. (B) L = Iω = constant. If I doubles, ω halves.
4. (C) The contact point is instantaneously at rest (the definition of rolling without slipping).

   Rotation problems require a systematic approach: compute I, set up τ = Iα and/or L = Iω, and use the rolling constraint when applicable. Always verify energy conservation as a check.

A 0.5 kg mass on a spring (k = 200 N/m) is pulled 0.1 m from equilibrium and released from rest at t = 0.

(a) Write the position function x(t).

(b) At what time does the mass first pass through x = −0.05 m?

(c) What is the speed at x = −0.05 m?

Solution

(a) ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s

x(t) = A cos(ωt + φ), with x(0) = 0.1 = A cos φ and v(0) = 0 = −Aω sin φ

From v(0) = 0: sin φ = 0, so φ = 0 or π. Since x(0) = 0.1 > 0 and A > 0: φ = 0, A = 0.1 m.

x(t) = 0.1 cos(20t) (in meters)

(b) Set x(t) = −0.05:

−0.05 = 0.1 cos(20t) cos(20t) = −0.5 20t = 2π/3 (first time cos θ = −0.5 after t = 0) t = 2π/(3 × 20) = π/30 ≈ 0.105 s

(c) v = ±ω√(A² − x²) = 20√(0.01 − 0.0025) = 20√0.0075 = 20(0.0866) = 1.73 m/s

(The sign is negative since the mass is moving in the −x direction at this point.)

---

Problem 2: Energy Analysis

A mass-spring system has total energy E = 8.0 J and spring constant k = 800 N/m.

(a) Find the amplitude.

(b) At what position is the kinetic energy equal to twice the potential energy?

(c) At what position is the speed half the maximum speed?

Solution

(a) E = ½kA²

8.0 = ½(800)A² → A² = 0.02 → A = 0.141 m

(b) K = 2U, and K + U = E = 8 J

So 2U + U = 8 → U = 8/3 J

½kx² = 8/3 → x² = 16/(3 × 800) = 1/150 → x = ±0.0816 m

(c) v = v_max/2, so v² = v²_max/4

½mv² = ½m(v²_max/4) = K_max/4 = E/4 = 2 J

U = E − K = 8 − 2 = 6 J

½kx² = 6 → x² = 12/800 = 0.015 → x = ±0.122 m

---

Problem 3: Physical Pendulum

A uniform rod of mass M = 1.0 kg and length L = 1.0 m is pivoted at one end and swings as a physical pendulum.

(a) Find the period of small oscillations.

(b) What is the length of an equivalent simple pendulum?

Solution

(a) For a rod pivoted at one end:

I = ML²/3, d = L/2 (distance from pivot to COM)

T = 2π√(I/(Mgd)) = 2π√(ML²/3)/(MgL/2)) = 2π√(2L/(3g))

T = 2π√(2 × 1.0/(3 × 9.8)) = 2π√(0.0680) = 2π(0.261) = 1.64 s

(b) For a simple pendulum: T = 2π√(L_eq/g)

2π√(L_eq/g) = 2π√(2L/3g)

L_eq = 2L/3 ≈ 0.667 m

The equivalent simple pendulum has length 2L/3, which is shorter than the rod. This makes sense because the mass is distributed, with some mass closer to the pivot than a simple pendulum would be.

---

Problem 4: Damped Oscillations

A 0.2 kg mass on a spring (k = 50 N/m) experiences damping with b = 1.0 N·s/m.

(a) Determine whether the system is underdamped, critically damped, or overdamped.

(b) If underdamped, find the time for the amplitude to decay to half its initial value.

Solution

(a) ω₀ = √(k/m) = √(50/0.2) = √250 = 15.81 rad/s

γ = b/(2m) = 1.0/(0.4) = 2.5 s⁻¹

Since γ = 2.5 < ω₀ = 15.81, the system is underdamped.

(b) The amplitude decays as Ae^(−γt). Set e^(−γt) = 0.5:

e^(−2.5t) = 0.5

−2.5t = ln(0.5) = −0.693 t = 0.693/2.5 = 0.277 s

---

Problem 5: MCQ Practice

1. If you double the mass on a spring, the period: (A) Doubles (B) Increases by √2 (C) Halves (D) Is unchanged
2. A pendulum on Earth has period T. On the Moon (g ≈ g_Earth/6), the period is: (A) T/6 (B) T/√6 (C) √6 T (D) 6T
3. At x = 0 in SHM, the acceleration is: (A) Maximum (B) Minimum (C) Zero (D) Depends on amplitude
4. For a mass-spring system, the maximum speed occurs: (A) At x = A (B) At x = 0 (C) At x = A/2 (D) At x = A/√2

MCQ Answers

1. (B) T = 2π√(m/k). Doubling m gives T' = 2π√(2m/k) = T√2.
2. (C) T = 2π√(L/g). If g → g/6, T → T√6.
3. (C) a = −ω²x. At x = 0, a = 0.
4. (B) v_max = Aω occurs at x = 0 (equilibrium).

   For oscillation problems, always identify the type (spring, pendulum, physical pendulum), determine ω from physical parameters, and use initial conditions to find A and φ.

A uniform solid sphere has mass M = 4.0 × 10²⁴ kg and radius R = 6.0 × 10⁶ m.

(a) Find g at r = 3.0 × 10⁶ m (half the radius).

(b) Find g at r = 1.2 × 10⁷ m (twice the radius).

(c) At what radius does g equal half its surface value?

Solution

(a) Inside the sphere (r < R):

g = GMr/R³ = (6.67 × 10⁻¹¹)(4 × 10²⁴)(3 × 10⁶)/(6 × 10⁶)³

= 8.008 × 10¹⁷ / 2.16 × 10²⁰ = 3.71 m/s²

(b) Outside the sphere (r > R):

g = GM/r² = (6.67 × 10⁻¹¹)(4 × 10²⁴)/(1.2 × 10⁷)²

= 2.668 × 10¹⁴/1.44 × 10¹⁴ = 1.85 m/s²

(c) At the surface: g_surface = GM/R².

For r > R: g = GM/r² = GM/(2R²) when g = g_surface/2. So r = R√2 = 8.49 × 10⁶ m.

---

Problem 2: Escape Velocity and Orbital Energy

A satellite of mass m = 500 kg orbits Earth (M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m) at an altitude of 300 km.

(a) Find the orbital speed.

(b) Find the total orbital energy.

(c) What minimum speed boost is needed to escape?

Solution

(a) Orbital radius: r = R_E + h = 6.37 × 10⁶ + 3 × 10⁵ = 6.67 × 10⁶ m

v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.67 × 10⁶)

= √(5.97 × 10⁷) = 7,726 m/s

(b) E = −GMm/(2r) = −(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(500)/(2 × 6.67 × 10⁶)

= −(1.99 × 10¹⁷)/(1.334 × 10⁷) = −1.49 × 10¹⁰ J

(c) At infinity, E = 0. The energy needed is ΔE = 0 − E = 1.49 × 10¹⁰ J.

Since KE already equals ½mv² = ½(500)(7726²) = 1.493 × 10¹⁰ J, and E = K + U = K − GMm/r = K − 2K = −K:

The current KE is GMm/(2r). To escape, total energy must be ≥ 0:

½mv²_total = GMm/r → v_esc at this orbit = √(2GM/r) = √2 × v_orb

Δv needed = v_esc − v_orb = v_orb(√2 − 1) = 7726(0.414) = 3,199 m/s

---

Problem 3: Gravitational Potential of a Ring

A ring of mass M = 0.1 kg and radius R = 0.5 m lies in the xy-plane centered at the origin.

(a) Find the gravitational potential at a point on the axis at x = 1.0 m.

(b) Find the gravitational field at that point.

Solution

(a) V = −GM/√(R² + x²) = −(6.67 × 10⁻¹¹)(0.1)/√(0.25 + 1.0)

= −6.67 × 10⁻¹²/√1.25 = −6.67 × 10⁻¹²/1.118 = −5.97 × 10⁻¹² J/kg

(b) g_x = −GMx/(R² + x²)^(3/2) = −(6.67 × 10⁻¹¹)(0.1)(1.0)/(1.25)^{3/2}

= −6.67 × 10⁻¹²/1.398 = −4.77 × 10⁻¹² m/s² (toward the ring)

---

Problem 4: Satellite Transfer

A satellite is in a circular orbit of radius r₁ around Earth. It fires its engines to move to a circular orbit of radius r₂ = 2r₁.

(a) How does the orbital speed change? (Express as a ratio.)

(b) How does the total energy change? (Express as a ratio.)

(c) Does the satellite speed up or slow down?

Solution

(a) v = √(GM/r). If r doubles: v₂/v₁ = √(r₁/r₂) = √(1/2) = 1/√2 ≈ 0.707

(b) E = −GMm/(2r). If r doubles: E₂/E₁ = r₁/r₂ = 1/2

The energy is halved (less negative → higher, but still negative).

(c) The satellite slows down. Counterintuitively, moving to a higher orbit requires the satellite to slow down. The engine burn increases energy (makes it less negative), but most of that energy goes into gravitational potential.

---

Problem 5: MCQ Practice

1. If you double the radius of a planet (keeping density constant), the escape velocity: (A) Doubles (B) Increases by 2^(2/3) (C) Quadruples (D) Halves
2. A satellite in elliptical orbit is fastest at: (A) Aphelion (B) Perihelion (C) Semi-minor axis (D) Center of ellipse
3. The gravitational field inside a uniform spherical shell is: (A) GM/r² (B) Zero (C) GM r/R³ (D) GM/R²
4. For a circular orbit, the ratio of kinetic energy to potential energy is: (A) 1:1 (B) 1:−2 (C) 2:1 (D) −1:2

MCQ Answers

1. (B) v_esc = √(2GM/R). If R → 2R, M ∝ R³ → 8M. v_esc = √(2G·8M/2R) = √(8GM/R) = 2√2 × original. Ratio = 2^(3/2) = 2^(2/3) × 2^(5/6)... Actually: v ∝ √(M/R) = √(R³/R) = R. So v_esc doubles. The answer is (A). Wait — let me recheck. M = ρ(4/3)πR³. If R → 2R: M → 8M, v_esc = √(2G·8M/(2R)) = √(8GM/R) = 2√2 v_0. So (B) 2^(3/2) is the closest... actually 2^(3/2) = 2√2 ≈ 2.83. The answer is the factor 2^(3/2). Among the choices, (B) is intended.
2. (B) By conservation of angular momentum, L = mvr is constant. At perihelion (closest), r is minimum so v is maximum.
3. (B) Shell theorem: zero field inside.
4. (B) K = GMm/(2r), U = −GMm/r. K/U = −1/2.

   Gravitational problems often combine energy conservation with the 1/r potential. For orbital mechanics, remember that the vis-viva equation handles any orbit, and that higher orbits are slower, not faster.

• v = dr/dt, a = dv/dt = d²r/dt²
• v = v₀ + at, Δx = v₀t + ½at², v² = v₀² + 2aΔx
• a_c = v²/R = ω²R, a_t = dv/dt = Rα
• v(t) = v₀ + ∫₀ᵗ a(t') dt', x(t) = x₀ + ∫₀ᵗ v(t') dt'

Unit 2: Newton's Laws

• ΣF = ma (or ΣF = dp/dt)
• Weight: W = mg, Normal force: N = mg cos θ (on incline)
• Friction: f_s ≤ μ_sN, f_k = μ_kN
• Drag: F_drag = −bv (linear), F_drag = −cv² (quadratic)

Unit 3: Work, Energy, Power

• W = ∫F · dr, W = ∫ F dx (1D)
• K = ½mv², U_grav = mgh, U_spring = ½kx²
• W_nc = ΔK + ΔU (work-energy theorem)
• P = dW/dt = F · v, P_avg = W/Δt
• Conservative: ∮F · dr = 0

Unit 4: Systems and Momentum

• p = mv, F = dp/dt
• J = ∫F dt = Δp
• r_CM = (1/M)Σmᵢrᵢ = (1/M)∫r dm
• Σp_i = Σp_f (if F_ext = 0)
• Elastic: v₁f = [(m₁−m₂)v₁ᵢ + 2m₂v₂ᵢ]/(m₁+m₂)
• Perfectly inelastic: v_f = (m₁v₁ᵢ + m₂v₂ᵢ)/(m₁+m₂)

Unit 5: Rotation

• ω = dθ/dt, α = dω/dt
• v_t = Rω, a_t = Rα, a_c = Rω²
• τ = r × F, |τ| = rF sin θ
• Στ = Iα
• I = ∫r² dm
• K_rot = ½Iω², L = Iω
• τ = dL/dt, L conserved if τ_net = 0
• Parallel axis: I = I_CM + Md²

Common Moments of Inertia

Object	Axis	I
Thin rod	Through center, ⊥	ML²/12
Thin rod	Through end, ⊥	ML²/3
Solid cylinder	Central axis	½MR²
Hollow cylinder	Central axis	MR²
Solid sphere	Diameter	⅖MR²
Hollow sphere	Diameter	⅔MR²
Thin spherical shell	Diameter	⅔MR²

Unit 6: Oscillations

• d²x/dt² + ω²x = 0
• x(t) = A cos(ωt + φ)
• ω_spring = √(k/m), ω_pendulum = √(g/L)
• T = 2π/ω, f = 1/T
• E = ½kA² = ½mv²_max
• v = ±ω√(A² − x²)
• Physical pendulum: T = 2π√(I/Mgd)
• Damped: x = Ae^(−γt)cos(ω_d t + φ), γ = b/(2m)

Unit 7: Gravitation

• F = GMm/r²
• g = GM/r² (outside sphere), g = GMr/R³ (inside)
• U = −GMm/r, V = −GM/r
• v_escape = √(2GM/R)
• v_orbital = √(GM/r)
• T² = 4π²r³/(GM) (Kepler's 3rd)
• E_orbital = −GMm/(2r)
• Vis-viva: v² = GM(2/r − 1/a)

Constants

• G = 6.674 × 10⁻¹¹ N·m²/kg²
• g ≈ 9.8 m/s² (Earth surface)
• R_Earth ≈ 6.37 × 10⁶ m
• M_Earth ≈ 5.97 × 10²⁴ kg

Time Management

• You have approximately 77 seconds per question.
• Easy conceptual questions may take 20–30 seconds. Save that time for harder ones.
• If a question requires more than 2 minutes of computation, skip it and return.
• Aim to answer all 35 questions. There is no penalty for guessing.

MCQ Approach

1. Read the entire question first — identify what's being asked before starting calculations.
2. Identify the physics — which unit/concept applies? Write the relevant equation mentally.
3. Check dimensions — if an answer has units of m/s when you expect m/s², eliminate it.
4. Use limits — set a variable to 0 or ∞ and check which answers make physical sense.
5. Eliminate aggressively — even eliminating 2 of 5 choices improves your odds significantly.

Common MCQ Tricks

• "Which of the following is true?" — evaluate each statement individually.
• Graph questions — relate slopes and areas to derivatives and integrals.
• "Twice as far/ fast/heavy" — check whether the relationship is linear, quadratic, or inverse.
• Sign conventions — many wrong answers differ only by a sign.

Section 2: Free Response (3 questions, 45 minutes)

Time Budget

• 15 minutes per question — no more, no less.
• If stuck on one part, move to the next and come back.
• The first subpart is usually the easiest. Always attempt it.

FRQ Structure (Typical)

• Question 1: Kinematics/Dynamics (often 2D, possibly with calculus)
• Question 2: Energy/Momentum (conservation laws)
• Question 3: Rotation or Oscillations/Gravitation

Earning Maximum Points

1. Show every step — partial credit is generous. A correct setup earns points even if the arithmetic is wrong.
2. Define your variables and sign conventions — state "positive direction is to the right" or "upward is positive."
3. Draw free-body diagrams — even if not explicitly asked. Label every force with its name and expression.
4. Show the calculus — when integrating, write the integral, the limits, and the antiderivative. Don't jump to the answer.
5. Box your final answer — make it easy for the grader to find.
6. Always include units — a number without units is incomplete.
7. If you make an error, don't erase — cross it out and start over. Graders can award credit for correct work that follows from an earlier error ("carry-forward credit").

Calculus on the FRQ

• When asked to derive, start from first principles (F = ma, work integral, etc.).
• Show the integral setup explicitly: W = ∫F · dr.
• For differential equations, identify the type and method of solution.
• If you can't solve a differential equation analytically, at least set it up correctly — that's worth significant partial credit.

Common FRQ Mistakes

1. Forgetting to account for pulley mass — if a pulley has mass M, include Iα in the analysis.
2. Using the wrong moment of inertia — always state which I you're using and verify the axis.
3. Ignoring rotational KE in rolling problems — total KE = ½mv² + ½Iω².
4. Sign errors in potential energy — gravitational PE is −GMm/r (negative!).
5. Forgetting that impulse is a vector — in 2D, compute components separately.

General Exam-Day Tips

Before the Exam

• Sleep well the night before. Physics C requires clear analytical thinking.
• Review the equation summary sheet (this package's Equation Summary file).
• Do 2–3 practice FRQs the day before — not more. You want to be sharp, not exhausted.

During the Exam

• Breathe — if you hit a wall on a problem, take 5 seconds and reset.
• Trust your preparation — you've seen all these problem types.
• Don't leave anything blank — especially on MCQ. An educated guess is better than nothing.
• Check your work if time permits. On FRQ, verify that energy is conserved, units match, and signs are consistent.

Score Estimation

• 5: ~65–70% correct (raw score ~85–95 out of 150)
• 4: ~50–55% correct (raw score ~65–80)
• 3: ~35–40% correct (raw score ~45–60)
• The curve is generous for Physics C — a 5 typically requires ~60–65%.

  Strategy beats cramming. Knowing how to approach each question type is as important as knowing the physics.