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Ohm’s Law, Kirchhoff’s Loop Rule (Resistors in Series and Parallel)

Joseph Brown

Joseph Brown

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Study Guide Overview

This AP Physics 1 study guide covers kinematics (motion in one and two dimensions), dynamics (Newton's Laws and their applications), circular motion and gravitation, energy (work, energy conservation, and power), momentum (linear momentum and collisions), simple harmonic motion (SHM systems), and torque and rotational motion. It includes key equations, practice questions, and exam tips for each unit. The guide emphasizes core concepts like Newton's Laws, energy and momentum conservation, and free-body diagrams.

AP Physics 1: Ultimate Study Guide 🚀

Hey, future physicist! Let's get you prepped for the AP Physics 1 exam. Remember, you've got this! 💪

Table of Contents

  1. Unit 1: Kinematics
  2. Unit 2: Dynamics
  3. Unit 3: Circular Motion and Gravitation
  4. Unit 4: Energy
  5. Unit 5: Momentum
  6. Unit 6: Simple Harmonic Motion
  7. Unit 7: Torque and Rotational Motion
  8. Final Exam Focus
  9. Practice Questions

Unit 1: Kinematics 🚗

1.1 Motion in One Dimension

  • Displacement, Velocity, and Acceleration:
    • Displacement (Δx) is the change in position. It's a vector!
    • Velocity (v) is the rate of change of displacement. Also a vector!
    • Acceleration (a) is the rate of change of velocity. Yep, it's a vector too!
  • Key Equations:
    • v=ΔxΔtv = \frac{\Delta x}{\Delta t} (average velocity)
    • a=ΔvΔta = \frac{\Delta v}{\Delta t} (average acceleration)
    • Kinematic Equations (for constant acceleration):
      • vf=vi+atv_f = v_i + at
      • Δx=vit+12at2\Delta x = v_i t + \frac{1}{2}at^2
      • vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a\Delta x
      • Δx=12(vi+vf)t\Delta x = \frac{1}{2}(v_i + v_f)t
Key Concept

Remember to use the correct sign conventions for direction (e.g., right/up is positive, left/down is negative).

  • Graphs of Motion:
    • Position vs. Time: Slope = velocity
    • Velocity vs. Time: Slope = acceleration, Area = displacement
    • Acceleration vs. Time: Area = change in velocity

1.2 Motion in Two Dimensions

  • Projectile Motion:
    • Analyze horizontal and vertical motion separately.
    • Horizontal motion: constant velocity (ax=0a_x = 0)
    • Vertical motion: constant acceleration due to gravity (ay=g=9.8m/s2a_y = -g = -9.8 m/s^2)
    • Initial velocity components: vix=vicos(θ)v_{ix} = v_i \cos(\theta), viy=visin(θ)v_{iy} = v_i \sin(\theta)
Exam Tip

Remember that time is the same for both horizontal and vertical motion.

  • Relative Motion:
    • Velocity of an object relative to a frame of reference.
    • Use vector addition to find relative velocities.
Memory Aid

Think of projectile motion as a combination of a horizontal car moving at constant speed and a ball being thrown straight up and down. They happen at the same time, but are independent of each other.

Practice Question

Multiple Choice:

  1. A ball is thrown horizontally from the top of a building. If air resistance is negligible, what is the shape of the ball's trajectory? (A) A straight line (B) A parabola (C) A circle (D) A hyperbola

  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/s² (B) 4 m/s² (C) 5 m/s² (D) 10 m/s²

Free Response: A projectile is launched with an initial velocity of 30 m/s at an angle of 60 degrees above the horizontal. Neglect air resistance.

(a) Calculate the horizontal and vertical components of the initial velocity. (2 points) (b) Calculate the time it takes for the projectile to reach its maximum height. (2 points) (c) Calculate the maximum height reached by the projectile. (2 points) (d) Calculate the total time the projectile is in the air. (2 points) (e) Calculate the horizontal range of the projectile. (2 points)

Answer Key:

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

Free Response:

(a) vix=30cos(60)=15m/sv_{ix} = 30 \cos(60) = 15 m/s, viy=30sin(60)=26m/sv_{iy} = 30 \sin(60) = 26 m/s (2 points) (b) vf=vi+atv_f = v_i + at, 0=269.8t0 = 26 - 9.8t, t=2.65st = 2.65 s (2 points) (c) Δy=vit+12at2\Delta y = v_i t + \frac{1}{2}at^2, Δy=26(2.65)4.9(2.65)2=34.4m\Delta y = 26(2.65) - 4.9(2.65)^2 = 34.4 m (2 points) (d) Total time = 2 * time to max height = 2 * 2.65 = 5.3 s (2 points) (e) x=vxt=155.3=79.5mx = v_x t = 15 * 5.3 = 79.5 m (2 points)

Unit 2: Dynamics 🏋️‍♀️

2.1 Forces and Newton's Laws

  • Newton's First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net force.

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

  • Newton's Third Law: For every action, there is an equal and opposite reaction.

    Newton's Second Law is the foundation of dynamics. Make sure you understand how to apply it in various scenarios.

  • Types of Forces:

    • Gravity: Fg=mgF_g = mg (weight)
    • Normal Force: Force exerted by a surface perpendicular to the surface.
    • Tension: Force exerted by a string or rope.
    • Friction: Force that opposes motion. Ff=μFNF_f = \mu F_N (static or kinetic)
Common Mistake

Don't forget to draw free-body diagrams to visualize all forces acting on an object.

2.2 Applications of Newton's Laws

  • Inclined Planes:
    • Resolve forces into components parallel and perpendicular to the plane.
    • Fg=mgsin(θ)F_{g\parallel} = mg \sin(\theta), Fg=mgcos(θ)F_{g\perp} = mg \cos(\theta)
  • Connected Objects:
    • Treat the system as a whole when finding acceleration.
    • Consider individual objects when finding tension or other internal forces.
Memory Aid

Remember "Free Body Diagrams are your BFFs" - always start with a FBD when solving force problems!

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 is 0.2, what is the net force acting on the block? (A) 10 N (B) 20 N (C) 30 N (D) 50 N

  2. Two blocks, one with mass m and the other with mass 2m, are connected by a string and pulled across a frictionless surface with a force F. What is the tension in the string connecting the two blocks? (A) F/3 (B) F/2 (C) 2F/3 (D) F

Free Response: A 5 kg block is placed on a 30-degree incline. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2. (a) Draw a free body diagram of all the forces acting on the block. (2 points) (b) Calculate the component of the gravitational force acting parallel to the incline. (2 points) (c) Calculate the component of the gravitational force acting perpendicular to the incline. (2 points) (d) Calculate the minimum static frictional force required to keep the block at rest. (2 points) (e) If the block starts to slide, calculate the acceleration of the block down the incline. (2 points)

Answer Key:

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

Free Response:

(a) Diagram should show weight (mg), normal force (N), and static friction (fs) (2 points) (b) Fg=59.8sin(30)=24.5NF_{g\parallel} = 5 * 9.8 * \sin(30) = 24.5 N (2 points) (c) Fg=59.8cos(30)=42.4NF_{g\perp} = 5 * 9.8 * \cos(30) = 42.4 N (2 points) (d) Fstatic=Fg=24.5NF_{static} = F_{g\parallel} = 24.5 N (2 points) (e) Fnet=FgFkineticF_{net} = F_{g\parallel} - F_{kinetic}, Fkinetic=μkFN=0.242.4=8.48NF_{kinetic} = \mu_k * F_N = 0.2 * 42.4 = 8.48N, a=(24.58.48)/5=3.2m/s2a = (24.5 - 8.48) / 5 = 3.2 m/s^2 (2 points)

Unit 3: Circular Motion and Gravitation 🎡

3.1 Uniform Circular Motion

  • Centripetal Acceleration: Acceleration directed towards the center of the circle. ac=v2ra_c = \frac{v^2}{r}
  • Centripetal Force: Net force causing circular motion. Fc=mv2rF_c = m\frac{v^2}{r}
  • Period (T) and Frequency (f):
    • T=1fT = \frac{1}{f}
    • v=2πrTv = \frac{2\pi r}{T}
Quick Fact

Centripetal force is not a new type of force; it's the net force that causes circular motion (e.g., tension, friction, gravity).

3.2 Law of Universal Gravitation

  • Gravitational Force: Fg=Gm1m2r2F_g = G\frac{m_1m_2}{r^2}, where G is the gravitational constant.
  • Gravitational Field: g=GMr2g = \frac{GM}{r^2}
  • Orbital Motion:
    • Set gravitational force equal to centripetal force to analyze orbits.
    • GMmr2=mv2rG\frac{Mm}{r^2} = m\frac{v^2}{r}
Exam Tip

Remember that the gravitational force is always attractive and acts along the line connecting the centers of the two masses.

Practice Question

Multiple Choice:

  1. A car travels around a circular track at a constant speed. What is the direction of the centripetal acceleration of the car? (A) Tangent to the circle (B) Toward the center of the circle (C) Away from the center of the circle (D) In the direction of motion

  2. If the distance between two objects is doubled, how does the gravitational force between them change? (A) It is halved (B) It is doubled (C) It is quadrupled (D) It is quartered

Free Response: A satellite of mass m orbits a planet of mass M at a radius r.

(a) Derive an expression for the orbital speed of the satellite in terms of G, M, and r. (3 points) (b) Derive an expression for the period of the satellite in terms of G, M, and r. (3 points) (c) If the orbital radius is doubled, how does the orbital speed change? (2 points) (d) If the orbital radius is doubled, how does the period change? (2 points)

Answer Key:

*Multiple Choice: 1. (B), 2. (D)

Free Response:

(a) GMmr2=mv2rG\frac{Mm}{r^2} = m\frac{v^2}{r}, v=GMrv = \sqrt{\frac{GM}{r}} (3 points) (b) v=2πrTv = \frac{2\pi r}{T}, T=2πrv=2πrGMr=2πr3GMT = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} = 2\pi \sqrt{\frac{r^3}{GM}} (3 points) (c) If r is doubled, v becomes v/2v/\sqrt{2} (2 points) (d) If r is doubled, T becomes T8T\sqrt{8} (2 points)

Unit 4: Energy ⚡

4.1 Work and Energy

  • Work: W=Fdcos(θ)W = Fd\cos(\theta). Work is the transfer of energy by a force.
  • Kinetic Energy: KE=12mv2KE = \frac{1}{2}mv^2. Energy of motion.
  • Potential Energy:
    • Gravitational: PEg=mghPE_g = mgh
    • Elastic: PEs=12kx2PE_s = \frac{1}{2}kx^2
  • Work-Energy Theorem: Wnet=ΔKEW_{net} = \Delta KE. The net work done on an object equals the change in its kinetic energy.
Key Concept

Work is a scalar quantity and can be positive or negative depending on the direction of the force relative to the displacement.

4.2 Conservation of Energy

  • Total Mechanical Energy: E=KE+PEE = KE + PE
  • Conservation of Mechanical Energy: In the absence of non-conservative forces (like friction), the total mechanical energy of a system remains constant. KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_f
  • Power: P=WtP = \frac{W}{t}. Rate at which work is done or energy is transferred.
Memory Aid

"Energy is never created or destroyed, just transformed!" - Remember the law of conservation of energy.

Practice Question

Multiple Choice:

  1. A 2 kg ball is dropped from a height of 10 m. What is its kinetic energy just before it hits the ground? (Neglect air resistance). (A) 100 J (B) 196 J (C) 200 J (D) 392 J

  2. A spring with a spring constant of 200 N/m is compressed by 0.2 m. What is the potential energy stored in the spring? (A) 4 J (B) 8 J (C) 16 J (D) 20 J

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, it slides along a horizontal surface and compresses a spring with a spring constant of 100 N/m.

(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 maximum compression of the spring. (2 points) (e) If friction is present on the horizontal surface, how would the maximum compression of the spring change? Explain. (2 points)

Answer Key:

*Multiple Choice: 1. (B), 2. (A)

Free Response:

(a) PE=mgh=0.59.82=9.8JPE = mgh = 0.5 * 9.8 * 2 = 9.8 J (2 points) (b) KE=PE=9.8JKE = PE = 9.8 J (2 points) (c) KE=12mv2KE = \frac{1}{2}mv^2, 9.8=120.5v29.8 = \frac{1}{2} * 0.5 * v^2, v=6.26m/sv = 6.26 m/s (2 points) (d) PEspring=12kx2PE_{spring} = \frac{1}{2}kx^2, 9.8=12100x29.8 = \frac{1}{2} * 100 * x^2, x=0.44mx = 0.44 m (2 points) (e) The maximum compression would be less because some energy would be lost due to friction as thermal energy. (2 points)

Unit 5: Momentum ⚽

5.1 Linear Momentum

  • Momentum: p=mvp = mv. Momentum is a vector quantity.
  • Impulse: J=FΔt=ΔpJ = F\Delta t = \Delta p. Impulse is the change in momentum.
Common Mistake

Remember that impulse is equal to the area under the force vs. time graph.

5.2 Conservation of Momentum

  • Conservation of Linear Momentum: In a closed system, the total momentum remains constant. m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}
  • Types of Collisions:
    • Elastic: Both momentum and kinetic energy are conserved.
    • Inelastic: Momentum is conserved, but kinetic energy is not.
    • Perfectly Inelastic: Objects stick together after the collision.
Memory Aid

"Momentum is always conserved in collisions, but energy is only conserved in elastic collisions!"

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) 3.33 m/s (C) 5 m/s (D) 10 m/s

  2. A force of 10 N acts on an object for 2 seconds. What is the impulse exerted on the object? (A) 5 Ns (B) 10 Ns (C) 20 Ns (D) 40 Ns

Free Response: A 0.5 kg cart is moving to the right at 2 m/s and collides with a 1 kg cart that is initially at rest. After the collision, the 0.5 kg cart moves to the left at 1 m/s.

(a) Calculate the initial momentum of the system. (2 points) (b) Calculate the final momentum of the 0.5 kg cart. (2 points) (c) Calculate the final velocity of the 1 kg cart. (2 points) (d) Is this collision elastic or inelastic? Justify your answer. (2 points) (e) Calculate the change in kinetic energy of the system. (2 points)

Answer Key:

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

Free Response:

(a) pi=(0.52)+(10)=1kgm/sp_i = (0.5 * 2) + (1 * 0) = 1 kg m/s (2 points) (b) p1f=0.51=0.5kgm/sp_{1f} = 0.5 * -1 = -0.5 kg m/s (2 points) (c) pi=pfp_i = p_f, 1=0.5+1v2f1 = -0.5 + 1 * v_{2f}, v2f=1.5m/sv_{2f} = 1.5 m/s (2 points) (d) Inelastic. KEi=0.522/2=1JKE_i = 0.5 * 2^2 / 2 = 1J, KEf=(0.5(1)2/2)+(11.52/2)=1.375JKE_f = (0.5 * (-1)^2 / 2) + (1 * 1.5^2 / 2) = 1.375J Kinetic energy is not conserved. (2 points) (e) ΔKE=1.3751=0.375J\Delta KE = 1.375 - 1 = 0.375J (2 points)

Unit 6: Simple Harmonic Motion 🎵

6.1 Simple Harmonic Motion (SHM)

  • SHM Definition: Periodic motion where the restoring force is proportional to the displacement from equilibrium.
  • Period (T) and Frequency (f):
    • T=1fT = \frac{1}{f}
  • Amplitude (A): Maximum displacement from equilibrium.
Quick Fact

SHM is characterized by sinusoidal motion.

6.2 SHM Systems

  • Mass-Spring System:
    • T=2πmkT = 2\pi\sqrt{\frac{m}{k}}
  • Simple Pendulum:
    • T=2πLgT = 2\pi\sqrt{\frac{L}{g}}
Exam Tip

Note that the period of a pendulum is independent of mass and amplitude (for small angles).

Practice Question

Multiple Choice:

  1. A mass-spring system oscillates with a period T. If the mass is doubled, what is the new period of oscillation? (A) T/2 (B) T/\sqrt{2} (C) T\sqrt{2} (D) 2T

  2. A simple pendulum has a length L and a period T. If the length of the pendulum is quadrupled, what is the new period of oscillation? (A) T/2 (B) T (C) 2T (D) 4T

Free Response: A 0.2 kg mass is attached to a spring with a spring constant of 50 N/m. The mass is pulled 0.1 m from its equilibrium position and released.

(a) Calculate the period of oscillation of the mass-spring system. (2 points) (b) Calculate the maximum speed of the mass during its oscillation. (2 points) (c) Calculate the total energy of the system. (2 points) (d) If the amplitude of oscillation is doubled, how does the maximum speed change? (2 points) (e) If the amplitude of oscillation is doubled, how does the period change? (2 points)

Answer Key:

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

Free Response:

(a) T=2πmk=2π0.250=0.4sT = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.2}{50}} = 0.4 s (2 points) (b) vmax=Akm=0.1500.2=1.58m/sv_{max} = A\sqrt{\frac{k}{m}} = 0.1\sqrt{\frac{50}{0.2}} = 1.58 m/s (2 points) (c) E=12kA2=12500.12=0.25JE = \frac{1}{2}kA^2 = \frac{1}{2} * 50 * 0.1^2 = 0.25 J (2 points) (d) Maximum speed doubles (2 points) (e) Period does not change (2 points)

Unit 7: Torque and Rotational Motion 🔄

7.1 Torque

  • Torque: τ=rFsin(θ)\tau = rF\sin(\theta). Torque is the rotational analog of force.
  • Lever Arm: The perpendicular distance from the axis of rotation to the line of action of the force.
Key Concept

Torque causes rotational motion. It's a vector quantity, and its direction is determined by the right-hand rule.

7.2 Rotational Kinematics and Dynamics

  • Rotational Kinematic Equations: Similar to linear kinematics, but with angular variables.
  • Rotational Inertia (Moment of Inertia): I=mr2I = \sum mr^2. Resistance to changes in rotational motion.
  • Newton's Second Law for Rotation: τnet=Iα\tau_{net} = I\alpha
  • Rotational Kinetic Energy: KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2
  • Angular Momentum: L=IωL = I\omega. Conserved in the absence of external torques.
Memory Aid

Think of rotational motion as the circular version of linear motion. Many concepts have direct analogs!

Practice Question

Multiple Choice:

  1. A force is applied to a door handle. At which location will the force create the most torque? (A) Near the hinge (B) In the middle of the door (C) At the edge of the door, farthest from the hinge (D) At any point, torque is the same

  2. A solid cylinder and a hollow cylinder with the same mass and radius are released from the top of an incline. Which cylinder will reach the bottom first? (A) The solid cylinder (B) The hollow cylinder (C) They will reach the bottom at the same time (D) It depends on the incline angle

Free Response: A 2 kg solid disk with a radius of 0.5 m is rotating at 10 rad/s. A frictional force is applied at the edge of the disk, bringing it to rest in 5 seconds.

(a) Calculate the initial rotational kinetic energy of the disk. (2 points) (b) Calculate the angular acceleration of the disk. (2 points) (c) Calculate the torque applied by the frictional force. (2 points) (d) Calculate the work done by the frictional force to stop the disk. (2 points) (e) If the radius of the disk is doubled, how does the rotational inertia change? (2 points)

Answer Key:

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

Free Response:

(a) I=12mr2=1220.52=0.25kgm2I = \frac{1}{2}mr^2 = \frac{1}{2} * 2 * 0.5^2 = 0.25 kgm^2, KErot=12Iω2=120.25102=12.5JKE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} * 0.25 * 10^2 = 12.5 J (2 points) (b) α=ΔωΔt=0105=2rad/s2\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 - 10}{5} = -2 rad/s^2 (2 points) (c) τ=Iα=0.252=0.5Nm\tau = I\alpha = 0.25 * -2 = -0.5 Nm (2 points) (d) W=ΔKErot=012.5=12.5JW = \Delta KE_{rot} = 0 - 12.5 = -12.5 J (2 points) (e) Rotational inertia is quadrupled (2 points)

Final Exam Focus 🎯

High-Priority Topics

  • Forces and Motion: Newton's Laws, Free-Body Diagrams, Inclined Planes, Friction

  • Energy Conservation: Work-Energy Theorem, Potential and Kinetic Energy, Power

  • Momentum Conservation: Collisions, Impulse

  • Circular Motion and Gravitation: Centripetal Force, Universal Gravitation

  • Simple Harmonic Motion: Mass-Spring Systems, Pendulums

    Focus on understanding the fundamental principles and how they connect to each other. AP questions often combine multiple concepts.

Common Question Types

  • Multiple Choice: Conceptual questions, calculations, graph analysis
  • Free Response: Experimental design, derivations, multi-step problem solving

Last-Minute Tips

  • Time Management: Don't spend too long on any one question. Move on and come back if time allows.
  • Free Body Diagrams: Always start with a free-body diagram when dealing with forces.
  • Units: Always include units in your calculations and answers.
  • Show Your Work: Even if you don't get the final answer, you can get partial credit for showing your work.
  • Review Formulas: Make sure you are familiar with all the formulas on the formula sheet.
  • Stay Calm: Take deep breaths and remember that you are prepared! You've got this! 🧘‍♀️

Practice Questions

See practice questions at the end of each unit.

Good luck on your exam! You've worked hard and you're ready to rock this! 🌟

Question 1 of 9

A 10-ohm resistor has a current of 2 amps flowing through it. What is the voltage drop across the resistor? ⚡

5 V

12 V

20 V

8 V