AP Physics 1 Ultimate Study Guide **NEW 2025 UNITS 1-8**

This guide provides a comprehensive overview of all 8 units covered in the AP Physics 1 curriculum. It includes key concepts, explanations, formulas, and AP-style example problems with detailed solutions to help you prepare for your exam in May.

General AP Physics 1 Exam Tips:

• Conceptual Understanding: Physics 1 is heavily conceptual. Don't just memorize formulas; understand the principles behind them and when to apply them.

• Free-Body Diagrams (FBDs): Master them. They are essential for almost any problem involving forces.

• Energy Conservation: This is a powerful tool. Learn to identify when mechanical energy is conserved and when work is done by non-conservative forces (like friction).

• Momentum Conservation: Understand when linear and angular momentum are conserved (absence of external net force/torque).

• Connections: Recognize the links between linear and rotational motion (analogous quantities and relationships).

• Task Verbs: Pay attention to verbs in FRQs: "Calculate," "Derive," "Explain," "Justify," "Describe." Each requires a different type of response. "Justify" and "Explain" often require referencing physics principles and connecting them to the specific scenario.

• Formula Sheet: Familiarize yourself with the AP Physics 1 formula sheet provided during the exam, but know that understanding when and how to use the formulas is more important than just finding them.

• Practice FRQs: Work through released FRQs from the College Board website. This is the best way to understand the expected depth and style of answers.

Unit 1: Kinematics

Overview: Kinematics is the description of motion without considering its causes (forces). It involves quantities like position, displacement, velocity, and acceleration.

Key Concepts & Explanations:

1. Scalars and Vectors:

• Scalar: A quantity with only magnitude (size). Examples: distance, speed, time, mass, energy.

• Vector: A quantity with both magnitude and direction. Examples: displacement, velocity, acceleration, force, momentum. Vectors are often represented by arrows.

2. Displacement, Velocity, and Acceleration (1D):

• Position (x or y): Location relative to an origin. Vector (implicitly in 1D by sign).

• Displacement (Δx or Δy): Change in position (final position - initial position). Vector. It's not the same as distance traveled.

• Formula: Δx = x_f - x_i

• Average Velocity (v_avg): Rate of change of displacement. Vector.

• Formula: v_avg = Δx / Δt

• Instantaneous Velocity (v): Velocity at a specific moment in time. Vector. It's the slope of the position-time graph.

• Average Acceleration (a_avg): Rate of change of velocity. Vector.

• Formula: a_avg = Δv / Δt

• Instantaneous Acceleration (a): Acceleration at a specific moment. Vector. It's the slope of the velocity-time graph.

3. Representing Motion:

• Motion Diagrams: Dots representing position at equal time intervals. Spacing shows speed; change in spacing shows acceleration.

• Graphs:

• Position vs. Time (x-t): Slope = instantaneous velocity. Curvature indicates acceleration.

• Velocity vs. Time (v-t): Slope = instantaneous acceleration. Area under the curve = displacement (Δx).

• Acceleration vs. Time (a-t): Area under the curve = change in velocity (Δv).

• Kinematic Equations (Constant Acceleration Only): These are crucial for solving many problems.

• v_f = v_i + at (Definition of acceleration rearranged)

• Δx = v_i*t + ½at² (Position as a function of time)

• v_f² = v_i² + 2aΔx (Velocity as a function of position - time not needed)

• Δx = ½(v_i + v_f)t (Average velocity definition combined)

• Remember: v_i (or v₀) is initial velocity, v_f (or v) is final velocity, a is constant acceleration, t is time, Δx is displacement. Choose a consistent positive direction!

4. Reference Frames and Relative Motion:

• Motion depends on the observer's frame of reference.

• Example: If you walk forward on a moving train, your velocity relative to the ground is the train's velocity plus your velocity relative to the train.

• Vector Notation: v_AG = v_AT + v_TG (Velocity of A relative to Ground = Velocity of A relative to Train + Velocity of Train relative to Ground).

5. Vectors and Motion in Two Dimensions (2D):

• Vector Components: Break vectors into perpendicular components (usually x and y). Use trigonometry: Ax = A cos θ, Ay = A sin θ. A = √(Ax² + Ay²), θ = tan⁻¹(Ay/Ax).

• Vector Addition: Add vectors by adding their corresponding components: Rx = Ax + Bx, Ry = Ay + By.

• Projectile Motion: Motion under gravity only (neglecting air resistance).

• Key Idea: Horizontal (x) and vertical (y) motions are independent.

• Horizontal: Constant velocity (ax = 0). Use Δx = vx*t.

• Vertical: Constant acceleration (ay = -g = -9.8 m/s² near Earth, assuming up is positive). Use the 1D kinematic equations for the y-motion.

• Time (t) is the link connecting the x and y motions.

AP Exam Style Examples (Unit 1):

Example 1 (MCQ Style - 1D Kinematics):

A car starts from rest and accelerates uniformly at 3.0 m/s² for 5.0 seconds. What is the car's displacement during this time?

(A) 15.0 m

(B) 37.5 m

(C) 75.0 m

(D) 150.0 m

Solution:

1. Identify Knowns:

• Initial velocity (v_i) = 0 m/s (starts from rest)

• Acceleration (a) = 3.0 m/s² (uniform)

• Time (t) = 5.0 s

2. Identify Unknown: Displacement (Δx)

3. Choose Equation: We need an equation relating v_i, a, t, and Δx. The best fit is: Δx = v_i*t + ½at²

4. Substitute and Solve:
   Δx = (0 m/s)(5.0 s) + ½(3.0 m/s²)(5.0 s)²
   Δx = 0 + ½(3.0 m/s²)(25.0 s²)
   Δx = ½(75.0 m)
   Δx = 37.5 m

5. Answer: (B)

In-depth Explanation: This problem tests direct application of kinematic equations under constant acceleration. We list the given variables and the target variable. By comparing these with the standard kinematic equations, we select the one that contains all knowns and the one unknown (Δx = v_i*t + ½at²). Substituting the values directly yields the displacement. It's crucial to recognize "starts from rest" means v_i = 0 and "accelerates uniformly" means a is constant, allowing the use of these specific equations.

Example 2 (FRQ Style Snippet - 2D Projectile Motion):

A ball is launched from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal. Neglect air resistance.

(a) Calculate the initial horizontal (vx) and vertical (vy) components of the velocity.

(b) Calculate the time it takes for the ball to reach its maximum height.

Solution:

(a) Initial Velocity Components:

1. Concept: Resolve the initial velocity vector into its x and y components using trigonometry.

2. Formulas: vx = v cos(θ), vy = v sin(θ)

3. Given: v = 20 m/s, θ = 30°

4. Calculate:

• vx = (20 m/s) cos(30°) = (20 m/s) (√3 / 2) ≈ 17.3 m/s

• vy_initial = (20 m/s) sin(30°) = (20 m/s) (1 / 2) = 10.0 m/s

(b) Time to Maximum Height:

1. Concept: At the maximum height of its trajectory, the vertical component of the ball's velocity (vy) is momentarily zero. The horizontal velocity (vx) remains constant. We can use the vertical motion information with kinematic equations.

2. Vertical Motion Knowns:

• Initial vertical velocity (v_yi) = +10.0 m/s (from part a, assuming up is positive)

• Final vertical velocity at max height (v_yf) = 0 m/s

• Vertical acceleration (ay) = -g = -9.8 m/s² (gravity acts downwards)

3. Identify Unknown: Time (t)

4. Choose Equation: We need an equation relating v_yi, v_yf, ay, and t. The best fit is: v_yf = v_yi + ay*t

5. Substitute and Solve:
   0 m/s = 10.0 m/s + (-9.8 m/s²) t
   -10.0 m/s = (-9.8 m/s²) t
   t = (-10.0 m/s) / (-9.8 m/s²) ≈ 1.02 s

In-depth Explanation:

Part (a) requires basic vector decomposition using sine and cosine, fundamental for any 2D problem.

Part (b) relies on understanding the key characteristic of projectile motion's peak: vertical velocity is zero. We isolate the vertical motion (since ay is constant) and apply a 1D kinematic equation. We identify the initial vertical velocity (from part a), the final vertical velocity (0 m/s at the peak), and the acceleration due to gravity (-9.8 m/s²). The equation v_f = v_i + at allows us to solve for the time t it takes to reach this point. This highlights the independence of horizontal and vertical motion – we only used vertical quantities to find the time to the peak.

Unit 2: Force and Translational Dynamics

Overview: This unit introduces forces as interactions that cause changes in motion (acceleration). Newton's Laws of Motion form the foundation.

Key Concepts & Explanations:

1. Systems and Center of Mass:

• System: A defined collection of objects. Forces can be internal (within the system) or external (from outside the system).

• Center of Mass (CM): The average position of the mass in a system. For a system of particles, x_cm = (m₁x₁ + m₂x₂ + ...)/(m₁ + m₂ + ...). The CM of a system moves as if all the system's mass were concentrated there and all external forces acted there.

2. Forces and Free-Body Diagrams (FBDs):

• Force (F): A push or pull on an object resulting from interaction with another object. Vector quantity. Unit: Newton (N). 1 N = 1 kg⋅m/s².

• Free-Body Diagram (FBD): A diagram showing only one object (or system) and all the external forces acting on that object. Arrows represent forces, drawn originating from the object and pointing in the correct direction. ESSENTIAL for solving force problems.

3. Newton's Third Law (Action-Reaction):

• For every action force, there is an equal and opposite reaction force.

• F_AB = - F_BA (The force object A exerts on B is equal in magnitude and opposite in direction to the force object B exerts on A).

• Crucial Point: Action-reaction pairs act on different objects. They never cancel each other out when considering the motion of a single object.

4. Newton's First Law (Inertia):

• An object at rest stays at rest, and an object in motion stays in motion with constant velocity (constant speed and direction) unless acted upon by a net external force.

• Inertia: The natural tendency of an object to resist changes in its state of motion. Mass is a measure of inertia.

• Equilibrium: When the net external force on an object is zero (ΣF = 0), the object is in equilibrium (either at rest or moving with constant velocity). Acceleration is zero.

5. Newton's Second Law (ΣF = ma):

• The acceleration (a) of an object is directly proportional to the net external force (ΣF) acting on it and inversely proportional to its mass (m).

• Formula: ΣF = ma (This is the cornerstone of dynamics).

• ΣF means the vector sum of all external forces acting on the object.

• Often applied in component form: ΣFx = max, ΣFy = may.

6. Specific Forces:

• Gravitational Force (Weight, Fg): Force exerted by a large body (like Earth) on an object. Near Earth's surface, Fg = mg, directed downwards. g ≈ 9.8 N/kg (or 9.8 m/s²) is the gravitational field strength.

• Normal Force (FN or N): Contact force acting perpendicular to a surface, preventing an object from penetrating it. Its magnitude adjusts to satisfy Newton's laws (often balances perpendicular components of other forces). It's not always equal to mg.

• Tension (FT or T): Pulling force exerted by a string, rope, or cable. Acts along the rope, away from the object.

• Friction (Ff or f): Contact force that opposes relative motion or attempted relative motion between surfaces. Acts parallel to the surface.

• Static Friction (Fs): Prevents motion from starting. Adjusts its magnitude up to a maximum. 0 ≤ Fs ≤ Fs_max, where Fs_max = μs * FN. (μs = coefficient of static friction).

• Kinetic Friction (Fk): Opposes motion when surfaces are sliding. Usually constant magnitude. Fk = μk * FN. (μk = coefficient of kinetic friction). Typically, μk < μs.

• Spring Force (Fs): Force exerted by a stretched or compressed spring. Acts to restore the spring to its equilibrium length.

• Hooke's Law: Fs = -kx. k is the spring constant (stiffness), x is the displacement from equilibrium. The minus sign indicates it's a restoring force (opposite to displacement).

7. Circular Motion (Uniform):

• An object moving in a circle at constant speed still has changing velocity (because direction changes), meaning it is accelerating.

• Centripetal Acceleration (ac): Acceleration directed towards the center of the circle. Responsible for changing the direction of velocity.

• Formula: ac = v²/r = ω²r (r is radius, v is tangential speed, ω is angular velocity).

• Centripetal Force (Fc): The net force directed towards the center of the circle that causes the centripetal acceleration. It's NOT a new type of force. It's the result of other forces (gravity, tension, normal force, friction, etc.).

• Formula (from Newton's 2nd Law): ΣF_radial = mac = mv²/r.

AP Exam Style Examples (Unit 2):

Example 1 (MCQ Style - Newton's Laws/Friction):

A 10 kg box rests on a horizontal surface. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. A horizontal force F is applied to the box. The box will start to move when F is slightly greater than:

(A) 3.0 N

(B) 5.0 N

(C) 29.4 N

(D) 49.0 N

Solution:

1. Concept: The box will start to move when the applied horizontal force F overcomes the maximum static friction Fs_max.

2. FBD (Vertical): Forces are Gravity (Fg down) and Normal Force (FN up). Since the surface is horizontal and there's no vertical acceleration, ΣFy = FN - Fg = 0, so FN = Fg.

3. Calculate Fg: Fg = mg = (10 kg)(9.8 m/s²) = 98 N.

4. Calculate FN: FN = Fg = 98 N.

5. Calculate Fs_max: Use the formula Fs_max = μs * FN.

• Fs_max = (0.50) * (98 N) = 49 N.

6. Conclusion: The applied force F must be slightly greater than Fs_max to initiate motion.

7. Answer: (D)

In-depth Explanation: This problem tests understanding of static friction. The key is realizing that static friction prevents motion up to a maximum value. To make the box move, the applied force must exceed this maximum. We first draw an FBD (or visualize it) to find the Normal Force. On a simple horizontal surface with only vertical forces being gravity and normal force, FN = mg. Then, we apply the formula for maximum static friction, Fs_max = μs * FN. The coefficient of kinetic friction (μk) is irrelevant for finding the force needed to start the motion.

Example 2 (FRQ Style Snippet - Circular Motion):

A 0.50 kg ball is attached to the end of a cord of length 1.5 m. The ball is swung in a vertical circle. At the very top of the circle, the speed of the ball is 4.0 m/s.

(a) Draw a free-body diagram for the ball when it is at the top of the circle.

(b) Calculate the tension in the cord at the top of the circle.

Solution:

(a) Free-Body Diagram (at the top):

1. Identify Object: The ball.

2. Identify Forces:

• Gravity (Fg): Acts downwards, towards the center of the Earth. Fg = mg.

• Tension (FT): Acts along the cord, pulling downwards on the ball when it's at the top of the vertical circle (since the cord is above the ball).

3. Draw Diagram: Draw a dot representing the ball. Draw an arrow pointing down labeled Fg. Draw another arrow pointing down from the ball, along the cord direction, labeled FT.

(b) Calculate Tension (FT):

1. Concept: The ball is undergoing uniform circular motion (at this instant). The net force directed towards the center of the circle provides the necessary centripetal force (Fc = mac = mv²/r). At the top of the circle, the center is directly below the ball.

2. Apply Newton's Second Law (Radial Direction): Choose "down" (towards the center) as the positive direction for this analysis.

• ΣF_radial = mac

• The forces acting radially (towards the center) are Fg and FT. Both point down.

• Fg + FT = mac

• Substitute Fg = mg and ac = v²/r:

• mg + FT = mv²/r

3. Knowns:

• Mass (m) = 0.50 kg

• Speed (v) = 4.0 m/s

• Radius (r) = Cord length = 1.5 m

• g = 9.8 m/s²

4. Solve for FT:

• FT = mv²/r - mg

• FT = (0.50 kg)(4.0 m/s)² / (1.5 m) - (0.50 kg)(9.8 m/s²)

• FT = (0.50 kg)(16.0 m²/s²) / (1.5 m) - 4.9 N

• FT = (8.0 kg⋅m²/s²) / (1.5 m) - 4.9 N

• FT ≈ 5.33 N - 4.9 N

• FT ≈ 0.43 N

In-depth Explanation:

Part (a) requires correctly identifying the forces acting on the ball at the specified location. Gravity always acts down. Tension acts along the cord, away from the object it's attached to, so from the ball's perspective, the cord pulls it downwards at the top.

Part (b) applies Newton's Second Law specifically for circular motion. The key is recognizing that the net force towards the center equals mv²/r. At the top, both tension and gravity contribute to this net centripetal force. We set up the ΣF_radial = mac equation, sum the forces pointing towards the center (Fg and FT), and set this sum equal to mv²/r. Then, we solve algebraically for the unknown tension FT. It's important to note that if the speed were lower, the tension could become zero (the minimum speed condition to complete the circle).

Unit 3: Work, Energy, and Power

Overview: This unit introduces energy as a conserved quantity and work as the mechanism for transferring energy. Power describes the rate of energy transfer.

Key Concepts & Explanations:

1. Translational Kinetic Energy (KE):

• The energy an object possesses due to its motion (translation = moving from one point to another). Scalar quantity. Unit: Joule (J).

• Formula: KE = ½mv² (m = mass, v = speed).

2. Work (W):

• The energy transferred to or from an object or system by a force acting over a distance. Scalar quantity. Unit: Joule (J).

• Formula (Constant Force): W = F d cos(θ)

• F = magnitude of the force

• d = magnitude of the displacement

• θ = angle between the force vector and the displacement vector.

• Work is positive if the force component is in the direction of displacement (energy added).

• Work is negative if the force component is opposite to displacement (energy removed).

• Work is zero if the force is perpendicular to displacement (cos(90°) = 0).

• Work by a Variable Force: The work done is the area under the Force vs. Position (F-x) graph.

3. Work-Energy Theorem:

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

• Formula: W_net = ΔKE = KE_f - KE_i

• W_net is the work done by the net force (sum of work done by all individual forces).

4. Potential Energy (PE or U):

• Stored energy associated with the position or configuration of an object or system. Depends on the interaction involved. Scalar quantity. Unit: Joule (J).

• Change in potential energy (ΔPE) is defined as the negative of the work done by a conservative force. ΔPE = -W_conservative.

• Gravitational Potential Energy (PEg or Ug): Energy stored due to an object's vertical position relative to a reference level (where PEg = 0). Associated with the conservative force of gravity.

• Formula (near Earth's surface): ΔPEg = mgΔh (where Δh is the change in vertical height). Often written as PEg = mgh if h=0 is the reference level.

• Elastic (Spring) Potential Energy (PEs or Us): Energy stored in a spring when it's stretched or compressed from its equilibrium position. Associated with the conservative spring force.

• Formula: PEs = ½kx² (k = spring constant, x = displacement from equilibrium).

5. Conservative vs. Non-Conservative Forces:

• Conservative Force: Work done by this force depends only on the initial and final positions, not the path taken (e.g., gravity, spring force). Mechanical energy can be conserved if only conservative forces do work. Potential energy can be defined for conservative forces.

• Non-Conservative Force: Work done depends on the path taken (e.g., friction, air resistance, applied push/pull). These forces dissipate mechanical energy (often as heat).

6. Conservation of Mechanical Energy:

• Mechanical Energy (E): The sum of kinetic and potential energies of a system. E = KE + PE.

• Principle: If the net work done by non-conservative forces is zero (W_nc = 0), then the total mechanical energy of the system is conserved (remains constant).

• Formula: Ei = Ef or KEi + PEgi + PEsi = KEf + PEgf + PEef

• Modified Form (with non-conservative work): If non-conservative forces do perform net work:

• W_nc = ΔE = Ef - Ei = (KEf + PEf) - (KEi + PEi)

7. Power (P):

• The rate at which work is done or energy is transferred. Scalar quantity. Unit: Watt (W). 1 W = 1 J/s.

• Formulas:

• P_avg = W / Δt = ΔE / Δt (Average power)

• P = F v cos(θ) (Instantaneous power delivered by force F to an object moving at velocity v, where θ is the angle between F and v).

AP Exam Style Examples (Unit 3):

Example 1 (MCQ Style - Work-Energy Theorem):

A 2.0 kg object is pushed horizontally by a constant force of 10 N across a frictionless surface for a distance of 5.0 m. If the object starts from rest, what is its final kinetic energy?

(A) 10 J

(B) 25 J

(C) 50 J

(D) 100 J

Solution:

1. Concept: Use the Work-Energy Theorem: W_net = ΔKE. Since the surface is frictionless and the force is horizontal, the net force in the direction of motion is the applied force.

2. Calculate Net Work (W_net): The applied force is constant and in the same direction as the displacement (θ = 0°).

• W_net = W_applied = F d cos(0°) = (10 N) (5.0 m) 1 = 50 J

3. Apply Work-Energy Theorem: W_net = ΔKE = KE_f - KE_i

4. Initial KE: The object starts from rest (v_i = 0), so KE_i = ½m(v_i)² = 0.

5. Solve for Final KE:

• 50 J = KE_f - 0

• KE_f = 50 J

6. Answer: (C)

In-depth Explanation: This problem directly tests the Work-Energy Theorem. The key is to calculate the net work done. Since friction is absent and the normal force/gravity do no work (perpendicular to displacement), the net work is just the work done by the applied force. The angle between the force and displacement is 0°, so cos(θ)=1. Once the net work is found (50 J), the Work-Energy Theorem states this equals the change in kinetic energy. Since the initial kinetic energy is zero, the final kinetic energy must be equal to the net work done. We didn't even need to calculate the final velocity.

Example 2 (FRQ Style Snippet - Conservation of Energy):

A 0.5 kg block is placed on a vertical spring of constant k = 1000 N/m, compressing it by 0.1 m from its equilibrium position. The block is then released from rest. Assume no energy loss due to friction.

(a) Calculate the total mechanical energy of the block-spring system when the block is released.

(b) Calculate the maximum height the block reaches above the release point.

Solution:

(a) Total Mechanical Energy at Release:

1. Concept: Total mechanical energy E = KE + PEg + PEs. At the release point (maximum compression), the block is momentarily at rest. Let's choose the release point (the position of the block when the spring is compressed by 0.1 m) as h=0.

2. Energy Components at Release (Initial State):

• KEi = 0 (released from rest)

• PEgi = 0 (at the chosen reference height h=0)

• PEsi = ½kx², where x is the compression distance (0.1 m).

3. Calculate PEs:

• PEsi = ½ (1000 N/m) (0.1 m)²

• PEsi = ½ (1000 N/m) (0.01 m²)

• PEsi = 5.0 J

4. Calculate Total Energy:

• Ei = KEi + PEgi + PEsi = 0 + 0 + 5.0 J = 5.0 J

(b) Maximum Height Above Release Point:

1. Concept: As the block moves upwards, spring potential energy converts to kinetic and gravitational potential energy. At max height (h_max relative to release point), the block is momentarily at rest (KEf=0). Since energy is conserved (Ei = Ef), the initial energy (all spring PE) equals the final energy (all gravitational PE relative to the release point).

2. Energy States:

• Initial (Release): h=0, v=0, x=0.1m. Ei = PEsi = 5.0 J.

• Final (Max Height): h=h_max, v=0. Block has left the spring, so PEsf = 0. Ef = PEgf = mg*h_max.

3. Apply Conservation of Energy: Ei = Ef

• 5.0 J = mg*h_max

4. Solve for h_max:

• h_max = 5.0 J / (mg) = 5.0 J / (0.5 kg * 9.8 m/s²) = 5.0 J / 4.9 N ≈ 1.02 m

In-depth Explanation:

Part (a) calculates the initial stored energy. At release, all energy is stored as elastic potential energy (PEs = ½kx²), as KE is zero and we set PEg=0 at this level.

Part (b) uses conservation of mechanical energy. The total initial energy (5.0 J) must equal the total final energy. At maximum height, speed is zero (KEf = 0), and the block has left the spring (PEsf = 0). All initial energy converts to gravitational potential energy (PEgf = mgh_max) relative to the release point. Setting Ei = Ef allows solving for h_max.

Unit 4: Linear Momentum

Overview: This unit focuses on momentum, a measure of mass in motion, and impulse, the change in momentum. Conservation of linear momentum is a fundamental principle, especially useful for analyzing collisions and explosions.

Key Concepts & Explanations:

1. Linear Momentum (p):

• A measure of an object's "quantity of motion." Vector quantity. Unit: kg⋅m/s.

• Formula: p = mv (m = mass, v = velocity).

• Direction of momentum is the same as the direction of velocity.

• System Momentum: The total momentum of a system is the vector sum of the momenta of individual objects within the system: P_total = p₁ + p₂ + ... = m₁v₁ + m₂v₂ + ...

2. Impulse (J):

• The change in momentum of an object. Vector quantity. Unit: N⋅s (or kg⋅m/s).

• Formula (Definition): J = Δp = p_f - p_i = m*v_f - m*v_i

• Impulse is also related to the force causing the change in momentum and the time over which the force acts.

• Formula (Impulse-Momentum Theorem): J = F_avg * Δt

• F_avg is the average net force acting on the object.

• Δt is the time interval over which the force acts.

• Combining these: F_avg * Δt = Δp. This shows that a force applied over time changes an object's momentum.

• Variable Force: Impulse is the area under the Force vs. Time (F-t) graph.

3. Conservation of Linear Momentum:

• Principle: If the net external force acting on a system is zero, then the total linear momentum of the system remains constant (is conserved).

• Formula (for an isolated system): Σp_initial = Σp_final

• m₁v₁i + m₂v₂i + ... = m₁v₁f + m₂v₂f + ...

• Applies to collisions (objects hitting each other) and explosions (objects breaking apart).

• Momentum is conserved in each direction independently (e.g., Σpx_initial = Σpx_final and Σpy_initial = Σpy_final).

4. Collisions:

• Interactions where objects exert strong forces on each other over a short time interval. Momentum of the system is conserved if external forces (like friction during the brief collision) are negligible.

• Elastic Collision: A collision where the total kinetic energy of the system is conserved (ΣKE_initial = ΣKE_final). Objects bounce off each other perfectly.

• Inelastic Collision: A collision where the total kinetic energy of the system is not conserved (ΣKE_initial > ΣKE_final). Some KE is transformed into other forms (heat, sound, deformation). Objects may bounce off, but not perfectly.

• Perfectly Inelastic Collision: A type of inelastic collision where the objects stick together after colliding and move with a common final velocity. Maximum kinetic energy is lost (consistent with momentum conservation).

AP Exam Style Examples (Unit 4):

Example 1 (MCQ Style - Impulse-Momentum):

A 0.2 kg softball is pitched horizontally at 30 m/s. The batter hits it, and the ball leaves the bat horizontally in the opposite direction at 40 m/s. If the contact time between bat and ball is 0.005 s, what is the magnitude of the average force exerted by the bat on the ball?

(A) 280 N

(B) 400 N

(C) 1200 N

(D) 2800 N

Solution:

1. Concept: Use the Impulse-Momentum Theorem: F_avg * Δt = Δp = p_f - p_i.

2. Define Direction: Let the initial direction (pitch) be negative (-) and the final direction (hit) be positive (+).

3. Calculate Initial Momentum (pi):

• v_i = -30 m/s

• p_i = m v_i = (0.2 kg) (-30 m/s) = -6.0 kg⋅m/s

4. Calculate Final Momentum (pf):

• v_f = +40 m/s

• p_f = m v_f = (0.2 kg) (+40 m/s) = +8.0 kg⋅m/s

5. Calculate Change in Momentum (Δp):

• Δp = p_f - p_i = (+8.0 kg⋅m/s) - (-6.0 kg⋅m/s) = +14.0 kg⋅m/s

6. Apply Impulse-Momentum Theorem to find F_avg:

• F_avg * Δt = Δp

• F_avg * (0.005 s) = 14.0 kg⋅m/s

• F_avg = (14.0 kg⋅m/s) / (0.005 s) = 2800 N

7. Answer: (D)

In-depth Explanation: This problem connects force, time, and the change in momentum (impulse). The crucial first step is setting a clear positive/negative direction because momentum is a vector. The initial velocity is in one direction, and the final velocity is in the opposite direction. We calculate initial and final momentum using p=mv. The change in momentum (Δp) is the final minus the initial value – paying close attention to the signs is vital here. The impulse (J = Δp) is then related to the average force and contact time by J = F_avg * Δt. Solving for F_avg gives the magnitude of the average force during the collision.

Example 2 (FRQ Style Snippet - Conservation of Momentum/Collisions):

Block A of mass 2.0 kg moves to the right at 5.0 m/s and collides with Block B of mass 3.0 kg, which is initially at rest. After the collision, Block A moves to the left at 1.0 m/s.

(a) Calculate the velocity (magnitude and direction) of Block B immediately after the collision.

(b) Determine whether the collision was elastic or inelastic. Justify your answer with calculations.

Solution:

(a) Velocity of Block B after Collision:

1. Concept: Since there are no significant external horizontal forces mentioned during the brief collision, the total linear momentum of the two-block system is conserved.

2. Define System: Blocks A and B.

3. Define Direction: Let right be positive (+), left be negative (-).

4. Apply Conservation of Momentum: Σp_initial = Σp_final

• m_A*v_Ai + m_B*v_Bi = m_A*v_Af + m_B*v_Bf

5. Knowns:

• m_A = 2.0 kg, v_Ai = +5.0 m/s

• m_B = 3.0 kg, v_Bi = 0 m/s

• v_Af = -1.0 m/s (moves left)

6. Unknown: v_Bf

7. Substitute and Solve:

• (2.0 kg)(+5.0 m/s) + (3.0 kg)(0 m/s) = (2.0 kg)(-1.0 m/s) + (3.0 kg)*v_Bf

• +10.0 kg⋅m/s + 0 = -2.0 kg⋅m/s + (3.0 kg)*v_Bf

• +12.0 kg⋅m/s = (3.0 kg)*v_Bf

• v_Bf = (+12.0 kg⋅m/s) / (3.0 kg) = +4.0 m/s

8. Result: The velocity of Block B after the collision is 4.0 m/s to the right.

(b) Elastic or Inelastic Collision?

1. Concept: Compare the total kinetic energy of the system before the collision (ΣKE_initial) with the total kinetic energy after the collision (ΣKE_final). If KE_initial = KE_final, it's elastic. If KE_initial > KE_final, it's inelastic.

2. Calculate Initial KE:

• KE_Ai = ½ m_A (v_Ai)² = ½ (2.0 kg) (5.0 m/s)² = ½ 2.0 25 = 25 J

• KE_Bi = ½ m_B (v_Bi)² = ½ (3.0 kg) (0 m/s)² = 0 J

• ΣKE_initial = KE_Ai + KE_Bi = 25 J + 0 J = 25 J

3. Calculate Final KE:

• KE_Af = ½ m_A (v_Af)² = ½ (2.0 kg) (-1.0 m/s)² = ½ 2.0 1 = 1 J

• KE_Bf = ½ m_B (v_Bf)² = ½ (3.0 kg) (+4.0 m/s)² = ½ 3.0 16 = 24 J

• ΣKE_final = KE_Af + KE_Bf = 1 J + 24 J = 25 J

4. Compare: ΣKE_initial = 25 J and ΣKE_final = 25 J.

5. Conclusion: Since the total kinetic energy before the collision equals the total kinetic energy after the collision (ΣKE_initial = ΣKE_final), the collision was elastic.

In-depth Explanation:

Part (a) uses the principle of conservation of linear momentum, the cornerstone for analyzing collisions. We define the system (both blocks) and apply Σp_initial = Σp_final, carefully substituting the known masses and initial velocities, and the known final velocity of block A (remembering the sign change for direction). This allows us to solve for the single unknown, v_Bf.

Part (b) tests the definition of an elastic collision. We must calculate the total kinetic energy of the system both before and after the collision using KE = ½mv² for each block and summing them. Comparing the initial and final total KE determines the type of collision. Because KE is conserved here (25 J = 25 J), the collision is elastic. If KE had decreased, it would have been inelastic. Remember that momentum is conserved in both elastic and inelastic collisions (assuming an isolated system), but kinetic energy is only conserved in elastic collisions.

Unit 5: Torque and Rotational Dynamics

Overview: This unit extends the principles of dynamics (forces and motion) to objects that rotate. We introduce rotational analogs for displacement, velocity, acceleration, force (torque), and mass (rotational inertia).

Key Concepts & Explanations:

1. Rotational Kinematics: Description of rotational motion.

• Angular Position (θ): Angle of rotation relative to a reference axis. Unit: radians (rad).

• Angular Displacement (Δθ): Change in angular position (Δθ = θf - θi). Vector (direction via right-hand rule, but often treated as scalar +/- for fixed axis).

• Average Angular Velocity (ω_avg): Rate of change of angular displacement. ω_avg = Δθ / Δt. Unit: rad/s.

• Instantaneous Angular Velocity (ω): Angular velocity at a specific moment. Slope of θ-t graph.

• Average Angular Acceleration (α_avg): Rate of change of angular velocity. α_avg = Δω / Δt. Unit: rad/s².

• Instantaneous Angular Acceleration (α): Angular acceleration at a specific moment. Slope of ω-t graph.

• Rotational Kinematic Equations (Constant α Only): Analogous to linear equations.

• ωf = ωi + αt

• Δθ = ωi*t + ½αt²

• ωf² = ωi² + 2αΔθ

• Δθ = ½(ωi + ωf)t

2. Connecting Linear and Rotational Motion: For a point on a rotating rigid body at radius r from the axis:

• Tangential Speed (vt): vt = rω (ω must be in rad/s)

• Tangential Acceleration (at): Component of acceleration parallel to motion, due to changing angular speed. at = rα

• Centripetal (Radial) Acceleration (ac or ar): Component of acceleration directed towards the center, due to changing direction of velocity. ac = vt²/r = rω²

• Total linear acceleration is the vector sum of at and ac.

3. Torque (τ):

• The rotational equivalent of force; the "twist" or "turn" caused by a force. Vector quantity. Unit: Newton-meter (N⋅m).

• Formula: τ = r F sin(φ)

• r = distance from the pivot point (axis of rotation) to where the force is applied (lever arm distance).

• F = magnitude of the force.

• φ = angle between the lever arm vector r (pointing from pivot to force application point) and the force vector F.

• Alternatively: τ = r F⊥ (lever arm distance times perpendicular component of force) or τ = r⊥ F (force times perpendicular distance from pivot to line of action of force - the "lever arm").

• Torque causes angular acceleration. Positive/negative sign indicates direction (e.g., counter-clockwise +, clockwise -).

4. Rotational Inertia (I):

• Also called Moment of Inertia. The rotational equivalent of mass; resistance of an object to changes in its rotational motion (resistance to angular acceleration). Scalar quantity. Unit: kg⋅m².

• Depends on the object's mass and how that mass is distributed relative to the axis of rotation. Mass farther from the axis contributes more to rotational inertia.

• Formula (Conceptual/Point Masses): I = Σ(mi * ri²) (Sum of mass times radius squared for all particles in the object).

• Formulas for I for common shapes (solid sphere, hoop, disk, rod) are provided on the AP formula sheet. You usually don't derive them, but you need to know when to use which one. Example: Hoop I = MR², Solid Disk I = ½MR².

5. Rotational Equilibrium:

• An object is in rotational equilibrium if its angular velocity is constant (including zero). This requires the net external torque acting on it to be zero.

• Conditions for Static Equilibrium (at rest):

• ΣF = 0 (Translational equilibrium - no linear acceleration)

• Στ = 0 (Rotational equilibrium - no angular acceleration)

• Torques can be calculated about any pivot point. Choosing a pivot where an unknown force acts often simplifies calculations.

6. Newton's First Law in Rotational Form:

• An object at rest rotationally or rotating with constant angular velocity will continue to do so unless acted upon by a net external torque.

7. Newton's Second Law in Rotational Form:

• The angular acceleration (α) of an object is directly proportional to the net external torque (Στ) acting on it and inversely proportional to its rotational inertia (I).

• Formula: Στ = Iα (The rotational analog of ΣF = ma).

AP Exam Style Examples (Unit 5):

Example 1 (MCQ Style - Torque):

A light rod of length 2.0 m is pivoted at its center. A force F1 = 10 N is applied perpendicularly downwards at the left end, and a force F2 = 5 N is applied perpendicularly downwards at the right end. What is the magnitude of the net torque about the pivot?

(A) 0 N⋅m

(B) 5 N⋅m

(C) 10 N⋅m

(D) 15 N⋅m

Solution:

1. Concept: Net torque is the vector sum of individual torques. Define a sign convention (e.g., counter-clockwise CCW as +, clockwise CW as -). The pivot is at the center, so the distance r from the pivot to each end is half the length (1.0 m).

2. Torque due to F1 (τ1):

• Force F1 = 10 N is at r1 = 1.0 m from the pivot.

• The force is perpendicular (sin(φ) = sin(90°) = 1).

• This force would cause a CCW rotation if acting alone. So, τ1 is positive.

• τ1 = r1 F1 sin(90°) = (1.0 m) (10 N) 1 = +10 N⋅m

3. Torque due to F2 (τ2):

• Force F2 = 5 N is at r2 = 1.0 m from the pivot.

• The force is perpendicular (sin(φ) = sin(90°) = 1).

• This force would cause a CW rotation if acting alone. So, τ2 is negative.

• τ2 = - r2 F2 sin(90°) = - (1.0 m) (5 N) 1 = -5 N⋅m

4. Net Torque (Στ):

• Στ = τ1 + τ2 = (+10 N⋅m) + (-5 N⋅m) = +5 N⋅m

5. Magnitude: The magnitude is 5 N⋅m.

6. Answer: (B)

In-depth Explanation: This problem requires calculating individual torques and summing them, paying close attention to the sign convention representing the direction of rotation each torque would cause. Torque is calculated as rFsin(φ). Here, φ=90° for both forces. The distance r is measured from the pivot. F1 causes a counter-clockwise (+) torque, while F2 causes a clockwise (-) torque. The net torque is the algebraic sum.

Example 2 (FRQ Style Snippet - Rotational Dynamics):

A uniform solid disk of mass M = 2.0 kg and radius R = 0.5 m is mounted on a frictionless axle through its center. A light cord is wrapped around the rim, and a constant force T = 10 N is applied tangentially to the cord.

(a) Calculate the rotational inertia (I) of the disk. (Formula for solid disk: I = ½MR²)

(b) Calculate the angular acceleration (α) of the disk.

Solution:

(a) Rotational Inertia (I):

1. Concept: Use the given formula for the rotational inertia of a uniform solid disk about its center.

2. Formula: I = ½MR²

3. Knowns: M = 2.0 kg, R = 0.5 m

4. Calculate:

• I = ½ (2.0 kg) (0.5 m)²

• I = ½ (2.0 kg) (0.25 m²)

• I = 0.25 kg⋅m²

(b) Angular Acceleration (α):

1. Concept: Apply Newton's Second Law for Rotation: Στ = Iα. We need to find the net torque acting on the disk.

2. Identify Torques: The only force causing a torque about the central axle is the tension T applied tangentially at the rim. Gravity and the axle force act at the pivot (r=0) or are balanced, producing no torque.

3. Calculate Torque (τ): The tension T acts at radius R and is tangential (perpendicular to the radius, φ=90°).

• τ = R T sin(90°) = R * T

• τ = (0.5 m) * (10 N) = 5.0 N⋅m

4. Apply Newton's Second Law for Rotation:

• Στ = Iα

• 5.0 N⋅m = (0.25 kg⋅m²) * α

5. Solve for α:

• α = (5.0 N⋅m) / (0.25 kg⋅m²) = 20 rad/s²

In-depth Explanation:

Part (a) is a direct application of the provided formula for rotational inertia of a disk.

Part (b) applies the rotational equivalent of Newton's Second Law (Στ = Iα). First, identify the forces creating torque about the axis of rotation. The tension T applied at the rim (r=R) and perpendicular to the radius creates a torque τ = R*T. This is the net torque since other forces (gravity, axle support) exert no torque about the center. Using the rotational inertia I calculated in part (a) and the calculated net torque τ, we solve Στ = Iα for the angular acceleration α.

Unit 6: Energy and Momentum of Rotating Systems

Overview: This unit applies the concepts of energy and momentum to rotating systems, introducing rotational kinetic energy and angular momentum, along with their conservation principles.

Key Concepts & Explanations:

1. Rotational Kinetic Energy (KE_rot or K_rot):

• The kinetic energy an object possesses due to its rotation. Scalar quantity. Unit: Joule (J).

• Formula: KE_rot = ½Iω² (I = rotational inertia, ω = angular speed in rad/s).

• Total Kinetic Energy: For an object that is both translating and rotating (like a rolling ball), the total KE is the sum of translational and rotational KE:

• KE_total = KE_trans + KE_rot = ½mv_cm² + ½I_cmω² (v_cm is velocity of center of mass, I_cm is rotational inertia about center of mass).

2. Torque and Work:

• Work done by a constant torque τ rotating an object through an angular displacement Δθ.

• Formula: W = τΔθ (Analogous to W = FΔx). Δθ must be in radians.

• Work-Energy Theorem for Rotation: The net work done by torques equals the change in rotational kinetic energy: W_net_torque = ΔKE_rot.

3. Angular Momentum (L):

• The rotational equivalent of linear momentum; a measure of an object's "quantity of rotation." Vector quantity. Unit: kg⋅m²/s.

• Formula (for a rigid body rotating about a fixed axis): L = Iω

• Direction is given by the right-hand rule (curl fingers in direction of rotation, thumb points in direction of L and ω).

• Formula (for a point particle): L = r p sin(φ) = mvr*sin(φ), where r is the position vector from the reference point, p=mv is linear momentum, and φ is the angle between r and p.

4. Angular Impulse:

• The change in angular momentum.

• Formula: ΔL = Lf - Li = Iωf - Iωi

• Related to the average net external torque and time interval:

• Formula (Angular Impulse-Momentum Theorem): τ_avg_net * Δt = ΔL

5. Conservation of Angular Momentum:

• Principle: If the net external torque acting on a system is zero, then the total angular momentum of the system remains constant (is conserved).

• Formula (for an isolated system): ΣL_initial = ΣL_final

• I_i ω_i = I_f ω_f (for a single object changing its I or ω)

• I₁ω₁i + I₂ω₂i + ... = I₁ω₁f + I₂ω₂f + ... (for a system of objects)

• Examples: Ice skater pulling arms in (I decreases, ω increases), rotating platform collisions, planetary orbits (approximately).

6. Rolling Motion (Without Slipping):

• A combination of translation of the center of mass and rotation about the center of mass.

• Condition for rolling without slipping: The linear speed of the center of mass v_cm is related to the angular speed ω by v_cm = Rω (where R is the radius).

• Energy: Use KE_total = ½mv_cm² + ½I_cmω². Often used in conservation of energy problems (e.g., ball rolling down a ramp PEg_initial = KE_total_final).

• Friction: Static friction is typically the force that causes rolling, providing the necessary torque. It does no work if the object rolls without slipping because the point of contact is instantaneously at rest relative to the surface.

7. Motion of Orbiting Satellites: (Conceptual overlap with gravity)

• Gravity provides the centripetal force (GMm/r² = mv²/r).

• Mechanical energy (E = KE + PEg = ½mv² - GMm/r) is conserved for a satellite in a stable orbit (assuming no air resistance or external forces).

• Angular momentum (L = mvr*sin(90°) = mvr for circular orbits, or L=Iω) is conserved because the gravitational force exerts no torque about the central body (force is radial, r is radial, φ=180° or 0°, sin(φ)=0).

AP Exam Style Examples (Unit 6):

Example 1 (MCQ Style - Conservation of Angular Momentum):

An ice skater is spinning with her arms outstretched. Her rotational inertia is I and her angular speed is ω. She pulls her arms in, reducing her rotational inertia to I/3. Assuming no external torque acts on her, what is her new angular speed?

(A) ω/9

(B) ω/3

(C) ω

(D) 3ω

Solution:

1. Concept: Since there is no net external torque (friction is assumed negligible for this ideal case), the skater's angular momentum L is conserved.

2. Apply Conservation of Angular Momentum: Li = Lf

3. Formula: L = Iω

• Ii ωi = If ωf

4. Knowns/Given:

• Ii = I

• ωi = ω

• If = I/3

5. Unknown: ωf

6. Substitute and Solve:

• (I) (ω) = (I/3) ωf

• Divide both sides by I: ω = (1/3) * ωf

• Multiply both sides by 3: ωf = 3ω

7. Answer: (D)

In-depth Explanation: This is a classic application of conservation of angular momentum (L=Iω). The key principle is that if no external torque acts on the system (the skater), its angular momentum cannot change. When the skater pulls her arms in, she changes the distribution of her mass relative to the axis of rotation, decreasing her rotational inertia (I). Because L must remain constant and L = Iω, if I decreases, ω must increase proportionally. Here, I becomes 1/3 of its original value, so ω must become 3 times its original value.

Example 2 (FRQ Style Snippet - Rolling Motion and Energy):

A solid sphere of mass M and radius R starts from rest at the top of an incline of vertical height h. It rolls down the incline without slipping. (Rotational inertia of a solid sphere: I = (2/5)MR²)

(a) Determine the total kinetic energy of the sphere at the bottom of the incline in terms of M, g, and h.

(b) Determine the translational speed (v_cm) of the sphere's center of mass at the bottom of the incline in terms of g and h.

Solution:

(a) Total Kinetic Energy at Bottom:

1. Concept: Use conservation of mechanical energy. The system is the sphere and the Earth. Assume the incline provides no non-conservative work (no friction work since it rolls without slipping, normal force does no work).

2. Energy States:

• Initial (Top): h_i = h, v_i = 0, ω_i = 0. Choose PEg=0 at the bottom. Ei = KEi_trans + KEi_rot + PEgi = 0 + 0 + Mgh.

• Final (Bottom): h_f = 0, v_f = v_cm, ω_f = ω. Ef = KEf_trans + KEf_rot + PEgf = ½Mv_cm² + ½Iω² + 0.

3. Apply Conservation of Energy: Ei = Ef

• Mgh = ½Mv_cm² + ½Iω²

4. Result: The total kinetic energy at the bottom (KE_total_final = ½Mv_cm² + ½Iω²) is equal to the initial potential energy.

• KE_total_final = Mgh

(b) Translational Speed at Bottom:

1. Concept: Use the energy conservation result from part (a) and the condition for rolling without slipping (v_cm = Rω, so ω = v_cm / R). Substitute I = (2/5)MR².

2. Substitute into Energy Equation:

• Mgh = ½Mv_cm² + ½ I ω²

• Mgh = ½Mv_cm² + ½ (²/₅MR²) (v_cm / R)²

3. Simplify:

• Mgh = ½Mv_cm² + ½ (²/₅MR²) (v_cm² / R²)

• Mgh = ½Mv_cm² + (¹/₅)Mv_cm² (Notice R² cancels)

4. Factor and Solve for v_cm:

• Mgh = Mv_cm² * (½ + ¹/₅)

• Cancel M: gh = v_cm² (⁵/₁₀ + ²/₁₀) = v_cm² (⁷/₁₀)

• v_cm² = (¹⁰/₇)gh

• v_cm = √((10/7)gh)

In-depth Explanation:

Part (a) applies conservation of energy between the top and bottom of the incline. The initial energy is purely gravitational potential energy (Mgh). The final energy (at the bottom) is purely kinetic, but because the sphere is rolling, it has both translational (½Mv_cm²) and rotational (½Iω²) kinetic energy. By energy conservation, this total final kinetic energy must equal the initial potential energy, Mgh.

Part (b) takes the energy conservation equation from (a) and uses the specifics of rolling motion. We substitute the formula for I for a solid sphere ((2/5)MR²) and the relationship between linear and angular speed for rolling (ω = v_cm / R). This allows us to express the total kinetic energy solely in terms of v_cm. Simplifying the equation and solving for v_cm gives the final translational speed. Note that the final speed is less than it would be if the object just slid without friction (v = √(2gh)), because some of the initial potential energy goes into rotational kinetic energy instead of just translational.

Unit 7: Oscillations

Overview: This unit focuses on a specific type of periodic motion called Simple Harmonic Motion (SHM), characterized by a restoring force proportional to displacement. Examples include mass-spring systems and pendulums (for small angles).

Key Concepts & Explanations:

1. Defining Simple Harmonic Motion (SHM):

• Periodic motion where the restoring force (F_restore) acting on the object is directly proportional to the object's displacement (x) from its equilibrium position and is directed opposite to the displacement.

• Condition for SHM: F_restore = -kx (Hooke's Law is the classic example). k is a proportionality constant (spring constant in the spring example).

• Equilibrium Position: The position where the net force on the object is zero.

2. Frequency and Period of SHM:

• Period (T): Time required to complete one full cycle of motion. Unit: seconds (s).

• Frequency (f): Number of cycles completed per unit time. Unit: Hertz (Hz) or s⁻¹.

• Relationship: f = 1/T and T = 1/f.

• Angular Frequency (ω): Related to frequency and period. Unit: rad/s.

• Formula: ω = 2πf = 2π / T

• Period Formulas (Important!):

• Mass-Spring System: T = 2π√(m/k) (m=mass, k=spring constant). Note: Period depends on mass and spring stiffness, not amplitude (for ideal SHM).

• Simple Pendulum (Small Angle Approximation): T = 2π√(L/g) (L=length of pendulum, g=acceleration due to gravity). Note: Period depends on length and gravity, not mass or amplitude (for small angles < ~15°).

3. Representing and Analyzing SHM:

• Motion is sinusoidal (described by sine or cosine functions).

• Position vs. Time: x(t) = A cos(ωt + φ) or x(t) = A sin(ωt + φ)

• A = Amplitude (maximum displacement from equilibrium).

• ω = Angular frequency (2πf).

• φ = Phase constant (determines initial position at t=0).

• Velocity vs. Time: Velocity is the time derivative of position. Max speed occurs at equilibrium (x=0), zero speed at max displacement (x=±A). v(t) = -Aω * sin(ωt + φ) (if x used cosine). v_max = Aω.

• Acceleration vs. Time: Acceleration is the time derivative of velocity. Max acceleration occurs at max displacement (x=±A), zero acceleration at equilibrium (x=0). a(t) = -Aω² * cos(ωt + φ) (if x used cosine). a_max = Aω². Note: a(t) = -ω²x(t), showing acceleration is proportional and opposite to displacement, confirming SHM.

4. Energy of Simple Harmonic Oscillators:

• Assume conservative forces (ideal spring, no friction/air resistance). Total mechanical energy E is conserved.

• Energy transforms between kinetic (KE = ½mv²) and potential (PEs = ½kx² for a spring).

• E_total = KE + PEs = ½mv² + ½kx² = constant

• At maximum displacement (x = ±A), v = 0, so energy is all potential: E_total = ½kA².

• At equilibrium position (x = 0), PEs = 0, speed is maximum (v = v_max), so energy is all kinetic: E_total = ½mv_max².

• Therefore: E_total = ½kA² = ½mv_max²

AP Exam Style Examples (Unit 7):

Example 1 (MCQ Style - Period):

A simple pendulum has a period T on Earth. If the pendulum is taken to the Moon, where the acceleration due to gravity is approximately g/6, what is its new period in terms of T?

(A) T/6

(B) T/√6

(C) T

(D) T√6

Solution:

1. Concept: The period of a simple pendulum depends on its length and the local acceleration due to gravity g.

2. Formula: T = 2π√(L/g)

3. On Earth: T_earth = T = 2π√(L/g_earth)

4. On Moon: g_moon = g_earth / 6. Let the new period be T_moon.

• T_moon = 2π√(L/g_moon) = 2π√(L / (g_earth / 6))

5. Simplify and Compare:

• T_moon = 2π√(6L / g_earth) = √6 * [2π√(L / g_earth)]

• T_moon = √6 * T_earth = T√6

6. Answer: (D)

In-depth Explanation: This problem tests the understanding of the formula for the period of a simple pendulum and how it depends on g. We write the formula for the period on Earth (T) and on the Moon (T_moon), substituting g_moon = g_earth / 6. By rearranging the expression for T_moon, we can factor out the expression for T_earth, showing that T_moon = T_earth * √6. The period increases on the Moon because the weaker gravity provides a weaker restoring force, making the oscillations slower.

Example 2 (FRQ Style Snippet - Energy in SHM):

A block of mass m = 0.5 kg is attached to an ideal horizontal spring with spring constant k = 50 N/m. The block is pulled to a position x = +0.2 m (amplitude A = 0.2 m) and released from rest.

(a) Calculate the total mechanical energy of the block-spring system.

(b) Calculate the maximum speed (v_max) of the block.

Solution:

(a) Total Mechanical Energy (E):

1. Concept: In SHM with no friction, total mechanical energy is conserved. It can be calculated at any point in the cycle. The easiest point is at maximum displacement (amplitude), where the block is momentarily at rest and all energy is potential.

2. Formula at Amplitude: E_total = KE + PEs = 0 + ½kA²

3. Knowns: k = 50 N/m, A = 0.2 m

4. Calculate:

• E_total = ½ (50 N/m) (0.2 m)²

• E_total = ½ (50 N/m) (0.04 m²)

• E_total = 1.0 J

(b) Maximum Speed (v_max):

1. Concept: Maximum speed occurs when the block passes through the equilibrium position (x = 0). At this point, all the energy is kinetic.

2. Formula at Equilibrium: E_total = KE_max + PEs = ½mv_max² + 0

3. Apply Energy Conservation: The total energy calculated in part (a) must equal the energy at equilibrium.

• E_total = ½mv_max²

• 1.0 J = ½ (0.5 kg) v_max²

4. Solve for v_max:

• 1.0 J = (0.25 kg) * v_max²

• v_max² = (1.0 J) / (0.25 kg) = 4.0 m²/s²

• v_max = √(4.0 m²/s²) = 2.0 m/s

In-depth Explanation:

Part (a) calculates the total energy of the oscillator. Since energy is conserved, we can calculate it at the most convenient point: the amplitude (x=A), where v=0. Here, E = PEs_max = ½kA².

Part (b) finds the maximum speed. This occurs at the equilibrium position (x=0), where PEs=0. At this point, all the system's energy (which we found in part a) is kinetic energy (KE_max = ½mv_max²). By setting E_total = KE_max, we can solve for v_max. This demonstrates the continuous conversion between potential and kinetic energy in SHM while the total energy remains constant.

Unit 8: Fluids

Overview: This unit introduces the properties of fluids (liquids and gases) and the principles governing their behavior at rest (hydrostatics) and in motion (hydrodynamics - basic concepts).

Key Concepts & Explanations:

1. Internal Structure and Density:

• Fluids: Substances that can flow and take the shape of their container (liquids and gases).

• Density (ρ): Mass per unit volume. Scalar quantity. Unit: kg/m³.

• Formula: ρ = m / V (m=mass, V=volume).

• Density is an intrinsic property of a substance (at a given temperature and pressure). Water ≈ 1000 kg/m³.

2. Pressure (P):

• Force exerted perpendicularly per unit area. Scalar quantity. Unit: Pascal (Pa). 1 Pa = 1 N/m².

• Formula: P = F⊥ / A (F⊥ = perpendicular force, A = area).

• Pressure in a fluid acts equally in all directions at a given depth.

• Pressure vs. Depth (Hydrostatic Pressure): Pressure increases with depth in a fluid due to the weight of the fluid above.

• Formula: P = P₀ + ρgh

• P = Absolute pressure at depth h.

• P₀ = Pressure at the surface (often atmospheric pressure, P_atm ≈ 1.01 x 10⁵ Pa).

• ρ = Density of the fluid.

• g = Acceleration due to gravity.

• h = Depth below the surface.

• Gauge Pressure: Pressure relative to atmospheric pressure (P_gauge = P - P_atm = ρgh).

3. Fluids and Newton's Laws:

• Pascal's Principle: A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. (Basis for hydraulic lifts: F₁/A₁ = F₂/A₂).

• Archimedes' Principle and Buoyancy: An object wholly or partially submerged in a fluid experiences an upward buoyant force (FB) equal in magnitude to the weight of the fluid displaced by the object.

• Formula: FB = ρ_fluid V_submerged g

• ρ_fluid = Density of the fluid.

• V_submerged = Volume of the object that is submerged in the fluid.

• The buoyant force arises from the pressure difference between the bottom and top of the submerged object.

• Floating: An object floats if its average density is less than the fluid density. When floating in equilibrium, FB = Fg_object (Buoyant force equals the object's weight). ρ_fluid V_submerged g = ρ_object V_object g. This means V_submerged / V_object = ρ_object / ρ_fluid.

• Sinking: An object sinks if its average density is greater than the fluid density. When submerged, FB < Fg_object. The apparent weight when submerged is Fg_apparent = Fg_object - FB.

4. Fluids and Conservation Laws (Introduction):

• Equation of Continuity (Conservation of Mass for Incompressible Fluids): For fluid flowing through a pipe of varying cross-sectional area, the volume flow rate is constant.

• Formula: A₁v₁ = A₂v₂ (A=cross-sectional area, v=fluid speed). Where the pipe narrows, the fluid speeds up.

• Bernoulli's Principle (Conservation of Energy for Fluids): Relates pressure, speed, and height in a moving fluid. For horizontal flow, pressure is lower where speed is higher.

• Formula: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ = constant (along a streamline).

• Often applied qualitatively in Physics 1: Faster flow corresponds to lower pressure (e.g., airplane wing lift).

AP Exam Style Examples (Unit 8):

Example 1 (MCQ Style - Buoyancy):

An object has a weight of 100 N in air. When fully submerged in water (density 1000 kg/m³), its apparent weight is 60 N. What is the volume of the object? (Use g = 10 m/s²)

(A) 0.004 m³

(B) 0.006 m³

(C) 0.010 m³

(D) 0.040 m³

Solution:

1. Concept: The apparent weight is the actual weight minus the buoyant force (Fg_apparent = Fg_object - FB). The buoyant force depends on the volume of the object (since it's fully submerged) and the density of the fluid (water).

2. Find Buoyant Force (FB):

• FB = Fg_object - Fg_apparent

• FB = 100 N - 60 N = 40 N

3. Apply Archimedes' Principle: FB = ρ_fluid V_submerged g. Since the object is fully submerged, V_submerged = V_object.

• FB = ρ_water V_object g

4. Knowns:

• FB = 40 N

• ρ_water = 1000 kg/m³

• g = 10 m/s² (use provided value)

5. Solve for V_object:

• 40 N = (1000 kg/m³) V_object (10 m/s²)

• 40 N = (10000 kg/(m²⋅s²)) * V_object (Note: kg⋅m/s² = N)

• 40 N = (10000 N/m³) * V_object

• V_object = (40 N) / (10000 N/m³) = 0.004 m³

6. Answer: (A)

In-depth Explanation: This problem connects weight, apparent weight, and buoyancy. The difference between the weight in air and the apparent weight in water is the magnitude of the upward buoyant force exerted by the water. We calculate FB = 100N - 60N = 40N. Then, we use Archimedes' Principle, FB = ρ_fluid V_submerged g. Since the object is fully submerged, its volume V_object is equal to the volume of water displaced V_submerged. We know FB, ρ_fluid (water density), and g, allowing us to solve for V_object.

Example 2 (FRQ Style Snippet - Pressure):

A submarine is cruising at a depth of 200 m below the surface of the ocean. The density of seawater is approximately 1025 kg/m³. Atmospheric pressure at the surface is 1.01 x 10⁵ Pa.

(a) Calculate the gauge pressure on the submarine's hull.

(b) Calculate the absolute pressure on the submarine's hull.

Solution:

(a) Gauge Pressure:

1. Concept: Gauge pressure is the pressure relative to atmospheric pressure, caused only by the fluid column above.

2. Formula: P_gauge = ρgh

3. Knowns:

• ρ = 1025 kg/m³ (density of seawater)

• g ≈ 9.8 m/s² (standard gravity)

• h = 200 m (depth)

4. Calculate:

• P_gauge = (1025 kg/m³) (9.8 m/s²) (200 m)

• P_gauge ≈ 2,009,000 kg/(m⋅s²) = 2,009,000 Pa

• P_gauge ≈ 2.01 x 10⁶ Pa

(b) Absolute Pressure:

1. Concept: Absolute pressure is the total pressure, including the pressure at the surface (atmospheric pressure) plus the gauge pressure.

2. Formula: P_absolute = P₀ + P_gauge = P₀ + ρgh

3. Knowns:

• P₀ = P_atm = 1.01 x 10⁵ Pa

• P_gauge ≈ 2.01 x 10⁶ Pa (from part a)

4. Calculate:

• P_absolute = (1.01 x 10⁵ Pa) + (2.01 x 10⁶ Pa)

• P_absolute = (0.101 x 10⁶ Pa) + (2.01 x 10⁶ Pa)

• P_absolute = 2.111 x 10⁶ Pa

• P_absolute ≈ 2.11 x 10⁶ Pa

In-depth Explanation:

Part (a) asks for gauge pressure, which is simply the pressure due to the weight of the fluid column (ρgh). We use the density of seawater and the given depth.

Part (b) asks for absolute pressure. This is the total pressure at that depth, which includes the atmospheric pressure acting on the surface (P₀) plus the gauge pressure (ρgh). We add the atmospheric pressure to the gauge pressure calculated in part (a). This highlights the difference between gauge pressure (relative to atmosphere) and absolute pressure (total pressure relative to a vacuum).