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Electric Charge

Joseph Brown

Joseph Brown

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Study Guide Overview

This AP Physics 1 study guide covers kinematics (1D & 2D motion), dynamics (Newton's Laws, forces, FBDs), circular motion & rotation, work & energy (conservation of energy, power), linear momentum (impulse, collisions), simple harmonic motion (SHM systems, energy), and wave properties & phenomena. It includes key equations, practice questions, exam tips, and focuses on high-value topics like Newton's Laws, energy and momentum conservation, SHM, and waves.

AP Physics 1: Ultimate Study Guide ๐Ÿš€

Hey future physicist! ๐Ÿ‘‹ This guide is designed to be your go-to resource for acing the AP Physics 1 exam. Let's break down the key concepts and make sure you're feeling confident and ready to go! Remember, you've got this! ๐Ÿ’ช

โšก๏ธ Unit 1: Kinematics

1.1 Motion in One Dimension

  • Displacement vs. Distance:

    • Displacement is the change in position (a vector). ๐Ÿ“
    • Distance is the total path length (a scalar). ๐Ÿ›ค๏ธ
  • Velocity vs. Speed:

    • Velocity is the rate of change of displacement (a vector). ๐Ÿš—๐Ÿ’จ
    • Speed is the rate of change of distance (a scalar). ๐Ÿƒ
  • Acceleration: The rate of change of velocity. ๐Ÿ“ˆ

  • Key Equations:

    • v=ฮ”xฮ”tv = \frac{\Delta x}{\Delta t} (average velocity)
    • a=ฮ”vฮ”ta = \frac{\Delta v}{\Delta t} (average acceleration)
    • Kinematic Equations (for constant acceleration):
      • v=v0+atv = v_0 + at

      • ฮ”x=v0t+12at2\Delta x = v_0t + \frac{1}{2}at^2

      • v2=v02+2aฮ”xv^2 = v_0^2 + 2a\Delta x

Key Concept

Kinematic equations are only valid for constant acceleration. Make sure to check the conditions before applying them!

Memory Aid

SUVAT helps remember the variables in kinematic equations: S (displacement), U (initial velocity), V (final velocity), A (acceleration), T (time).

1.2 Motion in Two Dimensions

  • Projectile Motion:

    • Horizontal motion has constant velocity (no acceleration). โžก๏ธ
    • Vertical motion has constant acceleration due to gravity (g=9.8m/s2g = 9.8 m/s^2). โฌ‡๏ธ
    • Analyze horizontal and vertical motion separately. ๐Ÿงฎ
  • Key Concepts:

    • Initial velocity can be broken into horizontal (v0x=v0cosโกฮธv_{0x} = v_0 \cos\theta) and vertical (v0y=v0sinโกฮธv_{0y} = v_0 \sin\theta) components. ๐Ÿ“

    • Time of flight is determined by vertical motion. โฐ

    • Range is determined by horizontal motion. ๐ŸŽฏ

Common Mistake

Don't mix horizontal and vertical components in your equations! Keep them separate.

Practice Questions

Practice Question

Multiple Choice:

  1. A ball is thrown horizontally from the top of a building with an initial speed of 15 m/s. If the ball hits the ground 2.0 s later, how far horizontally from the base of the building does the ball land? (A) 7.5 m (B) 15 m (C) 30 m (D) 60 m

  2. A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. What is the average acceleration of the car? (A) 2 m/sยฒ (B) 4 m/sยฒ (C) 5 m/sยฒ (D) 10 m/sยฒ

Free Response:

A projectile is launched from the ground with an initial velocity of 30 m/s at an angle of 60 degrees above the horizontal. Assume air resistance is negligible.

(a) Calculate the initial horizontal and vertical components of the velocity. (2 points) (b) Calculate the time the projectile is in the air. (3 points) (c) Calculate the horizontal distance the projectile travels before hitting the ground. (3 points) (d) Calculate the maximum height the projectile reaches. (2 points)

Answer Key:

Multiple Choice: 1. (C), 2. (B)

Free Response:

(a) v0x=30cosโก(60)=15m/sv_{0x} = 30 \cos(60) = 15 m/s, v0y=30sinโก(60)=26m/sv_{0y} = 30 \sin(60) = 26 m/s (2 points: 1 for each component) (b) Time to reach max height: 0=26โˆ’9.8t0 = 26 - 9.8t, t=2.65st = 2.65 s. Total time: 2โˆ—2.65=5.3s2 * 2.65 = 5.3 s (3 points: 1 for using vy=0v_y = 0 at max height, 1 for correct time to max height, 1 for total time) (c) x=15โˆ—5.3=79.5mx = 15 * 5.3 = 79.5 m (3 points: 1 for correct horizontal velocity, 1 for correct time, 1 for correct distance) (d) vf2=vi2+2aฮ”yv_f^2 = v_i^2 + 2 a \Delta y, 0=262โˆ’2โˆ—9.8โˆ—ฮ”y0 = 26^2 - 2 * 9.8 * \Delta y, ฮ”y=34.4m\Delta y = 34.4 m (2 points: 1 for correct formula, 1 for correct answer)

โš–๏ธ Unit 2: Dynamics

2.1 Forces and Newton's Laws

  • Newton's First Law (Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same velocity unless acted upon by a net force. โžก๏ธ

  • Newton's Second Law: Fnet=maF_{net} = ma. The net force on an object is equal to its mass times its acceleration. ๐Ÿ’ก

  • Newton's Third Law: For every action, there is an equal and opposite reaction. โ†”๏ธ

  • Types of Forces:

    • Gravity (Fg=mgF_g = mg)

    • Normal Force (FNF_N)

    • Tension (TT)

    • Friction (Ff=ฮผFNF_f = \mu F_N)

    • Spring Force (Fs=โˆ’kxF_s = -kx)

Memory Aid

FAN helps remember the types of forces: Friction, Applied, Normal, and Gravity.

2.2 Free-Body Diagrams (FBDs)

  • Draw a dot to represent the object.

  • Draw arrows representing all forces acting on the object.

  • Label each force with its symbol (e.g., FgF_g, FNF_N, TT). โœ๏ธ

  • Choose a coordinate system and resolve forces into components if needed. ๐Ÿ“

Exam Tip

Always draw a free-body diagram before applying Newton's Second Law. It helps visualize the forces and avoid mistakes.

2.3 Applications of Newton's Laws

  • Equilibrium: When the net force is zero, the object is in equilibrium (either at rest or moving with constant velocity). โš–๏ธ
  • Non-Equilibrium: When the net force is not zero, the object accelerates. ๐Ÿš€
  • Inclined Planes: Resolve forces into components parallel and perpendicular to the plane. โ›ฐ๏ธ

Practice Questions

Practice Question

Multiple Choice:

  1. A 10 kg block is pulled across a horizontal surface with a force of 50 N. If the coefficient of kinetic friction between the block and the surface is 0.2, what is the acceleration of the block? (A) 1 m/sยฒ (B) 3 m/sยฒ (C) 5 m/sยฒ (D) 7 m/sยฒ

  2. A 2 kg object is suspended from a string. What is the tension in the string? (A) 2 N (B) 9.8 N (C) 19.6 N (D) 20 N

Free Response:

A 5 kg block is placed on a 30-degree incline. The coefficient of static friction between the block and the incline is 0.4. (a) Draw a free-body diagram of the block. (2 points) (b) Calculate the component of the gravitational force parallel to the incline. (2 points) (c) Calculate the component of the gravitational force perpendicular to the incline. (2 points) (d) Determine if the block will slide down the incline. (4 points)

Answer Key:

Multiple Choice: 1. (B), 2. (C)

Free Response:

(a) FBD should show FgF_g pointing down, FNF_N perpendicular to incline, FfF_f pointing up the incline (2 points: 1 for each correct force) (b) Fgโˆฅ=mgsinโก(30)=5โˆ—9.8โˆ—0.5=24.5NF_{g\parallel} = mg \sin(30) = 5 * 9.8 * 0.5 = 24.5 N (2 points: 1 for correct formula, 1 for correct answer) (c) FgโŠฅ=mgcosโก(30)=5โˆ—9.8โˆ—0.866=42.4NF_{g\perp} = mg \cos(30) = 5 * 9.8 * 0.866 = 42.4 N (2 points: 1 for correct formula, 1 for correct answer) (d) Ffmax=ฮผFN=0.4โˆ—42.4=16.96NF_{fmax} = \mu F_N = 0.4 * 42.4 = 16.96 N. Since Fgโˆฅ>FfmaxF_{g\parallel} > F_{fmax}, the block will slide (4 points: 1 for correct FNF_N, 1 for correct FfmaxF_{fmax}, 1 for comparison, 1 for correct conclusion)

โš™๏ธ Unit 3: Circular Motion and Rotation

3.1 Uniform Circular Motion

  • Centripetal Acceleration: Acceleration directed towards the center of the circle (ac=v2ra_c = \frac{v^2}{r}). ๐Ÿ”„
  • Centripetal Force: The net force that causes centripetal acceleration (Fc=mv2rF_c = m\frac{v^2}{r}). ๐ŸŽฏ
  • Key Concepts:
    • Velocity is always tangent to the circle. ๐Ÿš—

    • Centripetal force is not a new force, it's the net force causing circular motion. ๐Ÿ’ก

Quick Fact

Remember that centripetal force is always perpendicular to the velocity.

3.2 Rotational Motion

  • Angular Displacement (ฮธ\theta): The angle through which an object rotates. ๐Ÿ“

  • Angular Velocity (ฯ‰\omega): The rate of change of angular displacement (ฯ‰=ฮ”ฮธฮ”t\omega = \frac{\Delta\theta}{\Delta t}). ๐ŸŒ€

  • Angular Acceleration (ฮฑ\alpha): The rate of change of angular velocity (ฮฑ=ฮ”ฯ‰ฮ”t\alpha = \frac{\Delta\omega}{\Delta t}). ๐Ÿ“ˆ

  • Key Concepts:

    • Relationship between linear and angular quantities: v=rฯ‰v = r\omega, a=rฮฑa = r\alpha. ๐Ÿ”—

3.3 Torque

  • Torque (ฯ„\tau): A force that causes rotation (ฯ„=rFsinโกฮธ\tau = rF\sin\theta). ๐Ÿ”ฉ
  • Moment of Inertia (II): Resistance to rotational motion (analogous to mass in linear motion). ๐Ÿ‹๏ธ
  • Rotational Inertia: I=โˆ‘mr2I = \sum mr^2 for point masses.
  • Newton's Second Law for Rotation: ฯ„net=Iฮฑ\tau_{net} = I\alpha. ๐Ÿ’ก

Practice Questions

Practice Question

Multiple Choice:

  1. A car is moving around a circular track with a constant speed. Which of the following is true about the car's acceleration? (A) It is zero. (B) It is constant and directed tangent to the circle. (C) It is constant and directed toward the center of the circle. (D) It is changing and directed toward the center of the circle.

  2. A rotating disc has an angular velocity of 10 rad/s. If the disc's moment of inertia is 2 kgยทmยฒ, what is its rotational kinetic energy? (A) 20 J (B) 50 J (C) 100 J (D) 200 J

Free Response:

A 0.5 kg ball is attached to a 1 m long string and swung in a horizontal circle. The ball makes 2 revolutions per second.

(a) Calculate the angular velocity of the ball. (2 points) (b) Calculate the linear speed of the ball. (2 points) (c) Calculate the centripetal acceleration of the ball. (2 points) (d) Calculate the tension in the string. (4 points)

Answer Key:

Multiple Choice: 1. (C), 2. (C)

Free Response:

(a) ฯ‰=2ฯ€f=2ฯ€โˆ—2=4ฯ€rad/s\omega = 2 \pi f = 2 \pi * 2 = 4 \pi rad/s (2 points: 1 for correct formula, 1 for correct answer) (b) v=rฯ‰=1โˆ—4ฯ€=4ฯ€m/sv = r \omega = 1 * 4 \pi = 4\pi m/s (2 points: 1 for correct formula, 1 for correct answer) (c) ac=v2r=(4ฯ€)21=16ฯ€2m/s2a_c = \frac{v^2}{r} = \frac{(4\pi)^2}{1} = 16\pi^2 m/s^2 (2 points: 1 for correct formula, 1 for correct answer) (d) T=Fc=mac=0.5โˆ—16ฯ€2=8ฯ€2NT = F_c = ma_c = 0.5 * 16\pi^2 = 8\pi^2 N (4 points: 1 for T=FcT = F_c, 1 for correct formula, 1 for correct substitution, 1 for correct answer)

๐Ÿ’ฅ Unit 4: Energy

4.1 Work and Energy

  • Work: The transfer of energy by a force (W=FdcosโกฮธW = Fd\cos\theta). ๐Ÿ‹๏ธ
  • Kinetic Energy: Energy of motion (KE=12mv2KE = \frac{1}{2}mv^2). ๐Ÿƒ
  • Potential Energy: Stored energy due to position or configuration.
    • Gravitational Potential Energy (PEg=mghPE_g = mgh). โ›ฐ๏ธ

    • Elastic Potential Energy (PEs=12kx2PE_s = \frac{1}{2}kx^2). ๐Ÿช€

Key Concept

Work-energy theorem: The net work done on an object is equal to the change in its kinetic energy (Wnet=ฮ”KEW_{net} = \Delta KE).

4.2 Conservation of Energy

  • Conservation of Mechanical Energy: In the absence of non-conservative forces (like friction), the total mechanical energy (KE + PE) remains constant. ๐Ÿ’ก

  • Non-Conservative Forces: Forces that dissipate energy (e.g., friction, air resistance). ๐Ÿ’จ

  • Work-Energy Principle: The change in total energy of a system is equal to the work done by non-conservative forces (Wnc=ฮ”EW_{nc} = \Delta E). ๐Ÿงฎ

Exam Tip

Always identify if non-conservative forces are present before applying conservation of energy. If friction is present, energy is not conserved.

4.3 Power

  • Power: The rate at which work is done or energy is transferred (P=Wฮ”t=FvP = \frac{W}{\Delta t} = Fv). โšก

Memory Aid

Power is like 'work per second'. Think of a powerful machine doing a lot of work quickly.

Practice Questions

Practice Question

Multiple Choice:

  1. A 2 kg object is lifted vertically 3 meters. How much work is done against gravity? (A) 6 J (B) 19.6 J (C) 58.8 J (D) 117.6 J

  2. A 1000 kg car is moving at 20 m/s. If the car's speed is doubled, what happens to its kinetic energy? (A) It doubles (B) It quadruples (C) It is halved (D) It remains the same

Free Response:

A 0.5 kg block is released from rest at the top of a frictionless ramp that is 2 m high. At the bottom of the ramp, the block slides onto a horizontal surface with a coefficient of kinetic friction of 0.3. (a) Calculate the potential energy of the block at the top of the ramp. (2 points) (b) Calculate the kinetic energy of the block at the bottom of the ramp. (2 points) (c) Calculate the speed of the block at the bottom of the ramp. (2 points) (d) Calculate the distance the block travels on the horizontal surface before stopping. (4 points)

Answer Key:

Multiple Choice: 1. (C), 2. (B)

Free Response:

(a) PE=mgh=0.5โˆ—9.8โˆ—2=9.8JPE = mgh = 0.5 * 9.8 * 2 = 9.8 J (2 points: 1 for correct formula, 1 for correct answer) (b) KE=PE=9.8JKE = PE = 9.8 J (2 points: 1 for conservation of energy, 1 for correct answer) (c) KE=12mv2KE = \frac{1}{2}mv^2, 9.8=12โˆ—0.5โˆ—v29.8 = \frac{1}{2} * 0.5 * v^2, v=6.26m/sv = 6.26 m/s (2 points: 1 for correct formula, 1 for correct answer) (d) Wfriction=ฮ”KEW_{friction} = \Delta KE, Ffd=KEF_f d = KE, ฮผmgd=KE\mu mgd = KE, 0.3โˆ—0.5โˆ—9.8โˆ—d=9.80.3 * 0.5 * 9.8 * d = 9.8, d=6.67md = 6.67 m (4 points: 1 for W=ฮ”KEW = \Delta KE, 1 for correct friction force, 1 for correct substitution, 1 for correct answer)

๐ŸŽข Unit 5: Momentum

5.1 Linear Momentum

  • Momentum: A measure of an object's mass in motion (p=mvp = mv). ๐Ÿš—

  • Impulse: The change in momentum (J=ฮ”p=Fฮ”tJ = \Delta p = F\Delta t). ๐Ÿ’ฅ

Quick Fact

Impulse is the area under the force-time graph.

5.2 Conservation of Momentum

  • Conservation of Momentum: In a closed system, the total momentum remains constant. ๐Ÿ’ก
  • Collisions:
    • Elastic Collisions: Both momentum and kinetic energy are conserved. ๐Ÿ“

    • Inelastic Collisions: Momentum is conserved, but kinetic energy is not. ๐Ÿš—๐Ÿ’ฅ

    • Perfectly Inelastic Collisions: Objects stick together after the collision. ๐Ÿš‚

Common Mistake

Remember that momentum is a vector quantity, so direction matters!

Practice Questions

Practice Question

Multiple Choice:

  1. A 2 kg ball moving at 5 m/s collides head-on with a 1 kg ball at rest. If the collision is perfectly inelastic, what is the final velocity of the combined mass? (A) 2.5 m/s (B) 3.3 m/s (C) 5 m/s (D) 10 m/s

  2. A 5 kg object experiences a net force of 10 N for 2 seconds. What is the change in momentum of the object? (A) 5 kg m/s (B) 10 kg m/s (C) 20 kg m/s (D) 40 kg m/s

Free Response:

A 1 kg cart moving at 4 m/s collides with a stationary 2 kg cart. After the collision, the 1 kg cart bounces back with a velocity of 1 m/s.

(a) Calculate the initial momentum of the system. (2 points) (b) Calculate the final velocity of the 2 kg cart. (4 points) (c) Determine if the collision is elastic or inelastic. (4 points)

Answer Key:

Multiple Choice: 1. (B), 2. (C)

Free Response:

(a) pinitial=m1v1+m2v2=1โˆ—4+2โˆ—0=4kgm/sp_{initial} = m_1v_1 + m_2v_2 = 1 * 4 + 2 * 0 = 4 kg m/s (2 points: 1 for correct formula, 1 for correct answer) (b) pinitial=pfinalp_{initial} = p_{final}, 4=1โˆ—(โˆ’1)+2โˆ—v24 = 1 * (-1) + 2 * v_2, v2=2.5m/sv_2 = 2.5 m/s (4 points: 1 for conservation of momentum, 1 for correct substitution, 1 for correct sign, 1 for correct answer) (c) KEinitial=12โˆ—1โˆ—42=8JKE_{initial} = \frac{1}{2} * 1 * 4^2 = 8 J, KEfinal=12โˆ—1โˆ—(โˆ’1)2+12โˆ—2โˆ—2.52=7.25JKE_{final} = \frac{1}{2} * 1 * (-1)^2 + \frac{1}{2} * 2 * 2.5^2 = 7.25 J. Since KEinitialโ‰ KEfinalKE_{initial} \ne KE_{final}, the collision is inelastic (4 points: 1 for correct KEinitialKE_{initial}, 1 for correct KEfinalKE_{final}, 1 for comparison, 1 for correct conclusion)

๐Ÿ“ณ Unit 6: Simple Harmonic Motion

6.1 Simple Harmonic Motion (SHM)

  • SHM: Periodic motion where the restoring force is proportional to the displacement. ๐Ÿช€
  • Key Concepts:
    • Amplitude (AA): Maximum displacement from equilibrium. ๐Ÿ“
    • Period (TT): Time for one complete oscillation. โฐ
    • Frequency (ff): Number of oscillations per second (f=1Tf = \frac{1}{T}). ๐ŸŒ€
    • Angular Frequency (ฯ‰\omega): ฯ‰=2ฯ€f=2ฯ€T\omega = 2\pi f = \frac{2\pi}{T}. ๐Ÿงฎ

6.2 SHM Systems

  • Mass-Spring System:
    • Period: T=2ฯ€mkT = 2\pi \sqrt{\frac{m}{k}}.
    • Frequency: f=12ฯ€kmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}.
  • Simple Pendulum:
    • Period: T=2ฯ€LgT = 2\pi \sqrt{\frac{L}{g}}.

    • Frequency: f=12ฯ€gLf = \frac{1}{2\pi} \sqrt{\frac{g}{L}}.

Memory Aid

Remember the 'Mass over Spring' for the spring period and 'Length over Gravity' for the pendulum period.

6.3 Energy in SHM

  • Energy Conservation: Total energy in SHM is constant and alternates between kinetic and potential energy. ๐Ÿ’ก
  • Maximum Velocity: Occurs at the equilibrium position.
  • Maximum Displacement: Occurs when velocity is zero.

Practice Questions

Practice Question

Multiple Choice:

  1. A mass-spring system oscillates with a period of 2 seconds. If the mass is doubled, what is the new period? (A) 1 s (B) โˆš2 s (C) 2โˆš2 s (D) 4 s

  2. A pendulum has a length of 1 meter. What is its approximate period? (A) 0.5 s (B) 1 s (C) 2 s (D) 4 s

Free Response:

A 0.2 kg mass is attached to a spring with a spring constant of 20 N/m. The mass is pulled 0.1 m from equilibrium and released.

(a) Calculate the period of oscillation. (2 points) (b) Calculate the maximum speed of the mass. (4 points) (c) Calculate the total energy of the system. (4 points)

Answer Key:

Multiple Choice: 1. (C), 2. (C)

Free Response:

(a) T=2ฯ€mk=2ฯ€0.220=0.628sT = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2}{20}} = 0.628 s (2 points: 1 for correct formula, 1 for correct answer) (b) KEmax=PEmaxKE_{max} = PE_{max}, 12mv2=12kA2\frac{1}{2}mv^2 = \frac{1}{2}kA^2, v=Akm=0.1200.2=1m/sv = A\sqrt{\frac{k}{m}} = 0.1\sqrt{\frac{20}{0.2}} = 1 m/s (4 points: 1 for energy conservation, 1 for correct formula, 1 for correct substitution, 1 for correct answer) (c) E=12kA2=12โˆ—20โˆ—0.12=0.1JE = \frac{1}{2}kA^2 = \frac{1}{2} * 20 * 0.1^2 = 0.1 J (4 points: 1 for correct formula, 1 for correct substitution, 1 for correct answer)

๐ŸŒŠ Unit 7: Waves

7.1 Wave Properties

  • Transverse Waves: Particles oscillate perpendicular to the wave's direction (e.g., light waves). ใ€ฐ๏ธ
  • Longitudinal Waves: Particles oscillate parallel to the wave's direction (e.g., sound waves). ๐Ÿ”Š
  • Key Concepts:
    • Wavelength (ฮป\lambda): Distance between two consecutive crests or troughs. ๐Ÿ“
    • Amplitude (AA): Maximum displacement from equilibrium. โฌ†๏ธ
    • Frequency (ff): Number of oscillations per second. ๐ŸŒ€
    • Wave Speed (vv): v=fฮปv = f\lambda. ๐Ÿš—

7.2 Wave Phenomena

  • Superposition: When two or more waves overlap, their displacements add together. โž•

  • Interference:

    • Constructive Interference: Waves add to produce a larger amplitude. โฌ†๏ธ
    • Destructive Interference: Waves cancel to produce a smaller amplitude. โฌ‡๏ธ
  • Diffraction: The bending of waves around obstacles. ๐Ÿšง

  • Reflection: The bouncing of waves off a surface. ๐Ÿชž

  • Refraction: The bending of waves as they pass from one medium to another. ๐Ÿ‘“

Exam Tip

Understand the difference between reflection and refraction. Reflection is bouncing off, refraction is bending through.

7.3 Sound Waves

  • Speed of Sound: Depends on the medium. ๐Ÿ’จ
  • Doppler Effect: The change in frequency due to relative motion between the source and observer. ๐Ÿš‘

Practice Questions

Practice Question

Multiple Choice:

  1. A wave has a frequency of 10 Hz and a wavelength of 2 meters. What is the speed of the wave? (A) 5 m/s (B) 10 m/s (C) 20 m/s (D) 40 m/s

  2. When two waves with the same amplitude and frequency meet in phase, what type of interference occurs? (A) Constructive (B) Destructive (C) No interference (D) Refraction

Free Response:

A sound wave with a frequency of 500 Hz is emitted from a stationary source. An observer is moving towards the source with a speed of 20 m/s. The speed of sound is 340 m/s.

(a) Calculate the wavelength of the sound wave. (2 points) (b) Calculate the frequency observed by the moving observer. (4 points) (c) Explain the Doppler effect. (4 points)

Answer Key:

Multiple Choice: 1. (C), 2. (A)

Free Response:

(a) v=fฮปv = f\lambda, 340=500ฮป340 = 500\lambda, ฮป=0.68m\lambda = 0.68 m (2 points: 1 for correct formula, 1 for correct answer) (b) fโ€ฒ=fv+vov=500340+20340=529.4Hzf' = f\frac{v + v_o}{v} = 500 \frac{340 + 20}{340} = 529.4 Hz (4 points: 1 for correct formula, 1 for correct substitution, 1 for correct sign, 1 for correct answer) (c) The Doppler effect is the change in frequency of a wave for an observer moving relative to its source. When the observer moves towards the source, the observed frequency increases, and when the observer moves away, the observed frequency decreases (4 points: 1 for definition, 1 for moving towards, 1 for moving away, 1 for complete explanation)

๐Ÿ’ก Unit 8: Electrostatics (Not on AP Physics 1 Exam)

8.1 Electric Charge

  • Two Types of Charge: Positive and negative. โž•โž–
  • Neutral Objects: Contain equal amounts of positive and negative charge. โš–๏ธ
  • Like Charges Repel, Opposites Attract. ๐Ÿงฒ

8.2 Electric Fields

  • Electric Field Lines: Show the direction a positive test charge would move. โžก๏ธ

  • Field Lines Point Away from Positive Charges and Towards Negative Charges. โšก

    Electric field lines
Exam Tip

Remember, electric field lines are a way to visualize the direction of the electric force on a positive test charge.

๐ŸŽฏ Final Exam Focus

High-Value Topics

  • Newton's Laws and Forces: A solid understanding is crucial for dynamics problems. ๐Ÿ‹๏ธ

  • Energy Conservation: Frequently tested in various scenarios. ๐Ÿ’ก

  • Momentum Conservation: Key for collision problems. ๐Ÿ’ฅ

  • Simple Harmonic Motion: Understanding the key parameters and equations is essential. ๐Ÿช€

  • Waves: Grasp the basic properties and wave phenomena. ๐ŸŒŠ

    Focus on mastering these areas as they form the backbone of the AP Physics 1 exam.

Common Question Types

  • Multiple Choice: Conceptual questions, calculations, graphical analysis.
  • Free Response: Experimental design, quantitative analysis, paragraph-length responses.

Last-Minute Tips

  • Time Management: Allocate time wisely for each question. Don't spend too long on one problem.
  • Read Carefully: Pay close attention to the wording of each question. ๐Ÿง
  • Show Your Work: Even if you don't get the final answer, you can earn partial credit for showing your steps. ๐Ÿ“
  • Stay Calm: Take deep breaths and approach each problem methodically. You've got this! ๐Ÿง˜

Good luck on your AP Physics 1 exam! You're well-prepared and ready to shine! โœจ

Question 1 of 5

What are the two types of electric charge? โšก

Positive and neutral

Negative and neutral

Positive and negative

Static and dynamic