AP Physics 1 Review Notes

Quick Info and Constants

• AP Physics 1 focuses on Mechanics, Waves and Sound, Electrostatics, and DC Circuits.

• Important Constants & Conversion Factors:

  • Universal Gravitational Constant: G = 6.67 \times 10^{-11} \frac{m^3}{kg \cdot s^2} = 6.67 \times 10^{-11} \frac{N \cdot m^2}{kg^2}

  • Atmospheric Pressure: 1 atm = 1.0 \times 10^5 \frac{N}{m^2} = 1.0 \times 10^5 Pa

  • Acceleration due to Gravity at Earth's Surface: g = 9.8 \frac{m}{s^2}

  • Gravitational Field Strength at Earth's Surface: g = 9.8 \frac{N}{kg}

• Prefixes:

  • Tera (T): 10^{12}

  • Giga (G): 10^9

  • Mega (M): 10^6

  • Kilo (k): 10^3

  • Centi (c): 10^{-2}

  • Milli (m): 10^{-3}

  • Micro (\mu): 10^{-6}

  • Nano (n): 10^{-9}

  • Pico (p): 10^{-12}

Units

• Hertz (Hz)

• Newton (N)

• Joule (J)

• Pascal (Pa)

• Kilogram (kg)

• Second (s)

• Meter (m)

• Watt (W)

Trigonometric Values for Common Angles

• \theta : 0°, 30°, 37°, 45°, 53°, 60°, 90°

• sin \theta: 0, 1/2, 3/5, (\sqrt{2})/2, 4/5, (\sqrt{3})/2, 1

• cos \theta: 1, (\sqrt{3})/2, 4/5, (\sqrt{2})/2, 3/5, 1/2, 0

• tan \theta: 0, (\sqrt{3})/3, 3/4, 1, 4/3, (\sqrt{3}), ∞

Conventions

• Frames of reference are inertial unless stated otherwise.

• Air resistance is negligible unless stated otherwise.

• Springs and strings are ideal unless stated otherwise.

• Fluids are ideal, and pipes are completely filled unless stated otherwise.

Geometry and Trigonometry

• Rectangle Area: A = bh

• Rectangular Solid Volume: V = lwh

• Triangle Area: A = (1/2)bh

• Cylinder Volume: V = \pi r^2 h

• Right Triangle: a^2 + b^2 = c^2, sin \theta = a/c, cos \theta = b/c, tan \theta = a/b

• Circle: A = \pi r^2, C = 2 \pi r, s = r\theta

• Sphere: S = 4\pi r^2, V = (4/3)\pi r^3

Mechanics and Fluids Formulas

• Kinematics:

  • vf = vi + at

  • x = x0 + vi t + (1/2)at^2

  • vf^2 = vi^2 + 2a(x - x_0)

• Center of Mass: x{cm} = (\Sigma mi xi) / (\Sigma mi)

• Net Force: F_{net} = \Sigma F

• Impulse: J = \int F dt

• Gravitational Force: |Fg| = G (m1 m_2) / r^2

• Friction: |Ff| \le \mu |FN|

• Spring Force: F_s = -k \Delta x

• Kinetic Energy: K = (1/2) m v^2

• Work: W = Fd = Fd cos\theta

• Work-Energy Theorem: \Delta K = \Sigma W = \Sigma F_i d

• Spring Potential Energy: U_s = (1/2) k (\Delta x)^2

• Gravitational Potential Energy: UG = -G (m1 m_2) / r

• Potential Energy: \Delta U = mg \Delta y

• Average Power: P_{avg} = (\Delta E) / (\Delta t)

• Instantaneous Power: P = Fv = Fv cos\theta

• Momentum: p = mv

• Impulse-Momentum Theorem: J = F \Delta t = \Delta p

• Average Velocity:v{avg} = (\Sigma vi) / (\Sigma m_i)

• Density: ρ = m / V

• Pressure: P = F / A

• Hydrostatic Pressure: P = P_0 + ρgh

• Gauge Pressure: P_{gauge} = ρgh

• Buoyant Force: F_b = ρVg

• Continuity Equation: A1 v1 = A2 v2

Rotational Motion

• Angular Velocity: ω = ω_0 + αt

• Angular Displacement: θ = θ0 + ω0 t + (1/2) α t^2

• Velocity and Angular Velocity: v = rω

• Tangential Acceleration: a_t = rα

• Torque: τ = rF = rF sinθ

• Rotational Inertia: I = \Sigma mi ri^2

• Parallel Axis Theorem: I' = I + Md^2

• Net Torque: τ_{net} = Iα

• Rotational Kinetic Energy: K = (1/2) Iω^2

• Work (Rotational): W = τ \Delta θ

• Angular Momentum: L = Iω

• τ = \Delta L / \Delta t

• L = rmv sinθ

• Arc Length: \Delta r = r \Delta θ

• Period and Frequency: T = 1/f, f = 1/T

• Simple Harmonic Motion (SHM): T = 2\pi \sqrt{m/k}, T = 2\pi \sqrt{L/g}

• Position in SHM: x = A cos(2\pi ft), x = A sin(2\pi ft)

Definitions

• Kinematics: Study of motion using words, numbers, graphs, and equations.

• Scalar: Quantity with only magnitude (e.g., 5 m/s).

• Vector: Quantity with magnitude and direction (e.g., 5 m/s West).

• Distance: Scalar quantity representing ground covered during motion.

• Displacement: Vector quantity representing the change in position ($\Delta x = x{final} - x{initial}$).

• Speed: Ratio of distance and time (always positive).

  • Average Speed = d/t

• Velocity: Ratio of displacement and time (can be + or -).

  • Average Velocity = \Delta x / t

• Instantaneous Velocity: Speed at a specific instant in time (e.g., speedometer reading).

• Relative Motion: Motion relative to a reference point (e.g., ground).

Relative Motion

• Velocity with Respect to the Ground: The overall speed and direction of an object as measured by a stationary observer.

  • Relative concept: Always relative to a reference point.

  • Vector quantity: Has magnitude and direction.

  • Calculation: Add the velocity of the object in still water to the velocity of the river current.

Example Problems

• 2D Relative Motion:

  • Van velocity: 30 m/s at 20° S of E.

  • Wind velocity: 20 m/s due East.

Important Steps

1. Break down initial velocity components:

   • V{vanx} = 30 cos(20°) = 28.19 m/s

   • V{vany} = -30 sin(20°) = -10.26 m/s

2. Wind velocity components:

   • V{windx} = 20 m/s

   • V{windy} = 0 m/s

3. Resultant velocity components:

   • V{rx} = V{vanx} + V{windx} = 28.19 + 20 = 48.19 m/s

   • V{ry} = V{vany} + V{windy} = -10.26 + 0 = -10.26 m/s

4. Magnitude of resultant velocity:

   • |V| = \sqrt{V{rx}^2 + V{ry}^2}

   • |V| = \sqrt{48.19^2 + (-10.26)^2} = \sqrt{2322.28 + 105.27} = \sqrt{2427.55} = 49.3 m/s

Projectile Motion

• Horizontal velocity remains constant.

• Vertical motion is under constant acceleration due to gravity.

Example Problems

• Horizontal Launch:
  Given: Soccer ball kicked horizontally off a 22m high hill, landing 35m from the edge.
  Find the initial horizontal velocity?

Steps:

1. Find the time the ball spends in the air using vertical motion:

• y = v_{iy} t + (1/2)gt^2

• 22 = 0 + (1/2)(9.8)t^2

• t = \sqrt{2(22)/9.8} = \sqrt{4.49} = 2.12 s

1. Calculate initial horizontal velocity:

• x = v_x t

• 35 = v_x (2.12)

• v_x = 35 / 2.12 ≈ 16.5 m/s

• Vertical Launch:
  Given: Rock thrown horizontally at 23.7 m/s off a hill, reaches the ground in 5.7 seconds.
  Find the height of the hill?

Steps:

1. Calculate height using vertical motion:

• y = (1/2)gt^2

• y = (1/2)(9.8)(5.7)^2 = 4.9 \times 32.49 ≈ 159.2 m

Dropping Supplies from a Plane - Example

Given: Plane flying at 358.0 km/hr at a height of 230.0m.
Find how far ahead of the landing site the plane should drop supplies?

Steps:

1. Convert velocity to m/s: 358.0 km/hr = 99.44 m/s.

2. Calculate time the supplies are in the air:

• y = (1/2)gt^2

• t = \sqrt{2y/g} = \sqrt{2(230.0)/9.8} ≈ 6.85 s

1. Calculate horizontal distance:

• x = v_x t = (99.44)(6.85) ≈ 681.2 m

Projectile Motion Problems

• Ball thrown at 60 degrees above the horizontal with an initial velocity of 40 m/s from a 50-m-high building:

  1. Time to reach the highest point:

     • v_{0y} = 40 sin(60°) = 34.64 m/s

     • 0 = 34.64 - 9.81t

     • t = 34.64/9.81 ≈ 3.53 seconds

  2. Maximum height from the base of the building:

     • h = 34.64 \times 3.53 + (1/2)(-9.81)(3.53)^2

     • h = 122.28 - 61.14 ≈ 61.14 m

     • Total height = 61.14 + 50 = 111.14 m ≈ 111 m

• Projectile fired horizontally at 8 m/s from an 80m-high cliff:

  • Time to hit the ground:

    • 80 = 0 + (1/2)(9.81)t^2

    • t = \sqrt{80/4.905} ≈ 4.04 seconds

• Final velocities:

  • V_{fx} = 8 m/s

  • V_{fy} = 0 + 9.81(4.04) = 39.63 m/s

  • V = \sqrt{8^2 + 39.63^2} ≈ 40.4 m/s

  • Angle of impact: θ = arctan(39.63/8) ≈ 78.6° ≈ 79°

Initial Velocity Components

• Ball kicked at 15m/s at an angle of 37 degrees:

  • Horizontal component: V_{ix} = 15 \times cos(37°) ≈ 11.98 m/s

  • Vertical component: V_{iy} = 15 \times sin(37°) ≈ 9.03 m/s
    Total time in air:

• 0 = 9.03t - 4.905t^2

• t = (9.03 \pm \sqrt{9.03^2 + 0})/9.81 ≈ 1.84 seconds
  Range (horizontal displacement):

• Range = 11.98 \times 1.84 ≈ 22 meters

Stone Thrown Example

• Horizontal component = 25m/s, total time = 3s:

  • Range = 25 m/s * 3 s = 75 m

  • Initial vertical component: 0 = V_{iy}(3) - (1/2)(9.81)(3)^2

  • V_{iy} = 14.7 m/s

  • Initial velocity: V_0 = \sqrt{25^2 + 14.7^2} ≈ 29 m/s

  • Angle of throw: θ = arctan(14.7/25) ≈ 30°

Relative Motion Examples

• Topic 1.1 Video 1: Scalar vs. Vector

  • Vector: Magnitude and direction.

  • Scalar: Only magnitude.

  • Speedometer measures speed (scalar).

• Initial position: 20th floor. After 1s: 19th floor. Total fall time: 3s. Air resistance negligible. Floors evenly spaced.

  • Distance fallen is proportional to t^2.

  • After 3s: (3²/1²) × 1 = 9 floors.

• Initial velocity v_0 = 10 m/s upward. Bowling pin rotates. Find max height of center of mass.

  • Maximum height: h = v_0^2 / (2g)

• Distance at different times will be proportional to t^2
  At t=1s, distance = 1 floor
  At t=3s, distance = (3^2/1^2) * 1 = 9 floors

Q): Diver moving horizontally with speed v dives off the edge of a vertical cliff and lands in water a distance d from teh base of the cliff.How far from the base of teh cliff would the diver have landed if teh diver initially had been moving horizontally wwith speed 2v?
A): 2d, The time in the air for a horizontal projectile is dependent on the height and independent of the initial speed. Since the distance (d = vt)
will be twice as much at a speed of 2v

• Car moving in a straight line w an accleration given by accompanying graph. *What is teh speed of teh car at t = 3s?
  The area under accleration-time graph is teh change in speed. *Ans: 10.5 m/s*# UNIT 2 Pak

Center of mass

• On AP equation sheet
  X{cm} = \frac{\sum mixi}{\sum mi}
  V{cm}= \frac{\sum mivi}{\sum mi}
  a{cm} = \frac{\sum miai}{\sum mi}

• Center of mass is a collection of particles with little to no interaction.

• Objects with shape will have uniform density and obvious centers of mass.

• When objects are free to rotate, they rotate around their center of mass
  *Objects center of mass will follow projectile motion pattern.

Forces

*Forces are the result of an interaction between two objects.
*An object cannot exert force on itself.

5 Steps to solve Free Body Diagram

1. Draw the free body diagram

2. Break forces into components

3. Redraw FBD

4. Sum the forces

5. Sum the forces in a direction perpendicular to the direction in steep

Force Normal

• Caused by a surface.

• is Normal (perpendicular) to the surface

• pushes away from teh surface.

Force of tension is in a rope, string, cable, chain or something similar

• Ideal ropes have negligible mass and do not stretch

• Ideal roeps have same magnitude force of tension at all points in rope
  NOTE I force of tension is less at teh bottom if there is mass to consider.

• Force of Tension is always parallel to direction of rope, wire, string or cable.

CONTACT FORCES result from interaction of one object touching another object and result from electric forces between the atoms or teh objects

• ex tension, friction, force normal, force applied and teh srpign force.

Laws

Newtons First Law:An object at rest will remain at rest and an object in motion will remain at a constant velocity unless acted upon by a net, external force.

This is called teh law of inertia. Valid when measurements are taken from an inertial reference frame.

Newton's second law

\sum F = ma

• Identify object or system on which we are summing teh forces

• Identify direction in which we are summing teh forces.
  Acceleration is always int he same direction as net force
  Translational EQUILIBRIUM F = 0
  Object is either at rest or moving at a constant velocity -> a = 0If an object is at rest, there could be forces acting on the object form other objects, however, teh net force acting on the objects is 0

Newton's third law:

For every force object 1 exerts on object 2, object 2 exert an equal but opposiote force on object 1. F{12} = -F{21}

Gravitational force

The interaction between an object with mass and another object w mass F=mg
ie weight, gravitational force, all mean teh same thing.
Direction of fore of gravity ins always toward center of mass of planet; down.

Force of friction

*Parallel to teh surface
*Always opposes sliding motion
*is independent of teh direction of force applied
\Equation for force of friction: |F| <= \mu|F_N|
Coefficient of Friction, \mu : ratio of maximum force of friction and force normal.

Hooke's Law

The force required to stretch an elastic object is directly proporitonal to the extension of teh spring for small distances. The force exerted back by teh string is known as Hooke's law.
Fs = -kx Where Fs is teh force exerted by teh spring, x is the displacement relative to teh unstretched length of teh spring, and k is teh spring constant.
The spring force is called restoring forec because teh force exerted by teh spring is always in teh opposite direction of teh displacement. The force required to stretch or compress teh spring is equal and opposite to teh restoring force. K is the spring constant, it is measured in newtons pre meter (N/m). \,x is the change in teh length of teh srpign; it is measured in meters (m). it is not teh entire lenft hof teh spring, just teh displacement from equilibrium. slope = k. larger k means more force required to stretch or compress
Force of friction does not depend on the size of surface area of contact between two surfaces.
µ = tan(⁡𝜃). This formula can be used in two cases.
(a) This formula is used when an object is stationary on a ramp and the angle 𝜃 (theta) is such that the object just breaks free from static friction and begins to slide. Meaning any angle smaller than θ will result in teh object remaining stationary. The "mu" calculated will be teh coefficient of static friction.
(b) This formula can also be used to find teh coefficient of kinetic friction if teh object is sliding down teh hill at a constant speed.

Universal Gravity (Force of Gravity equation)

F = G \frac{Mm}{r^2}
Where G= 6.67 * 10^{-11} \frac{N m^2}{kg^2}

Centripetal Force

Fc = mac = \frac{mv^2}{r}

• is teh force that makes a moving object change direction

• Ceases to exist when an object stops moving in a circle

• Centripetal Accelaration a_c = \frac{v^2}{r}

A centrifugal force DOES NOT Exist

• occurs whenever a moving object changes direction

• does not change teh speed of an object

• acts at right angles to teh velocity at any instant

• is directed toward teh center of a circle (center seeking)

• Circular Motion
  is an illusion experienced in a rotation reference frame

Solving problems from the PAK

Question 4
1) Normal force, N = mg (since it's on a horizontal surface)
N = 15kg * 9.8 m/s^2 = 147 N.
1) When teh box is about to move, Ff(static) = \mus * N = Applied Force 75 N = \mus * 147 N -> \mus = 75/147 = 0.51
\to 0.51
2) Net Horiztonal Force, F{net} = ma F{net} = 15 kg * 1.5 m/s^2 = 22.5 N
\to 22.5N
3) Force of Kinetic friciton, F{net} = F{applied} = F{kinetic friction}, F{kinetic friction} = 75N - 22.5 N = 52.5 N
\to 52.5N
4) Coeeficnet of Kientic firction, F(kinetic friction) = \muk*N => 52.5N = \muk * 147N => \mu_k = 52.5/147 = 0.357

Spring problems

Question 2

Concider placing as spring in a verticla positon and putting a 100-g object on it. We see it compressed by 2 cm. g = 10 m/s^2
(a) If an additonal load of 300 grams is placed on top of te preivous object, how much more will teh spring compress this time? _ cm
(b) Find teh spring constant. _ N/m

Anwsers with explained Logic:

For springs the force is proportional to displacement (Hook's law). The ratio of additonal weight to initial weight will equal teh ratio of additonal compression to initial compression
Initial Force= 0.1kg10 m/s^2 = 1NAdditonal Focre = 0.3kg10 m/s^2 = 3NIf 1 N causes 2 cm Compression, then aditional 3 N will equal 6cm. Spring contant is F = kx. X = 2cm -> 0.02 m. 1/0.02 => 50N/m

Question 2.9

A block of 5kg sits on an inclined plane of 30 degrees above teh horizonta ant dbegins to slide dwon it. teh coefficeitn of Kienetic fictoin is 15%.
What's teh accerelraitno of teh block, how logn does it take for teh bclock to slightdwon tot eh btoomofethte ramp if it straits fomr rest at te top?

1. Find teh force of gravity (along teh ramp)
   Fg_{parallel} = mg sin theta = 5*10sin30 = 25N

2. If θ is 45 and the accelerateion rate is 5m/s
   Since force is mg sinθ

3. The passenger in a race car driving over a hill at a constant rate o 15 m/s feel weightless,
   What is teh raciduso f curvature fo teh hill?
   Answer When you are wegithless, teh noraml forc eis zero because you are essentially in freefall meaining teh only foce acting on you is vgravity and ether is no suface tot exert a contatc froc eto cuontercat teh froce fo gravity

4. A 1 kg ball is on a sring of length 2m.If teh ball is spinning in a vertical circle at a constant velocity with a period of 2s 9 g= 10m/s
   The highest tension At teh botom of teh circle, hwen tension will eb maicm, at this piont, two force asct pm teh ball. Cnetriptal force (due to tnesion, gavriitonal froce pilluing dwn.

What 's the formula. At the top of teh circle. T+Mg = mv^2
The formula is in this test, study and practice those question you should be prepared for anything!!!!
THE EQUATION BECOMES: T + Mg = \frac{m_v^2}{r} Both forces cuotnribtoe to the centipetal force and not against each toher. This meaisn:
At te top of te circle. Mg = ceitrpetyal FOCE - This is where some trickey, and or teh queson asj for a tricky answer.

Colligeboard

Applyinewtons secnd alw, T-mg= am
Solving for Tension T = (Mg + mg)Mashed - And mass;less = Mashed,
Mashed / Mg = m(g+a).

Pulley and FRQ

TEhc ey psitcalr picnipel here,s newtosn thiird law- fre every atcitn, tehre is an eqaulnad oppoiset reatction. THINKBOYA TO SMA SEGMENT IF ROPE. EACH SESGMETNTIS PUYLLED BY te segmnet berfore it. If these forecs weren't eaqul, then teh segments would acceelerate. - tehrope would ebar or streytgch. Therefore to mantan equalbra, eahc semgmetn t must tnanisntite sames forct to nex tsement. This fore transtmisocn contruoues trhourgo ut entire orpe

Free body diagrams

*Draw the object as a point or simple shape
Show all external forces as arrows
Make arew lenthts approximately proportional to forec magitudes
Label all forces clearly and include coordinated axes when relevant
Use consisnt colour coding
FORCES MUST BE CONTINUOSLY applied to eb considered in an FBD EX A force must be cintiously appliwed to an object of orte be inclusded in its Free body diagarm because an FBD only shows to teh forces acting on a object ata a specific mment in time. Forces muts amek contaatc ecentp for nin-cocntracrt force sin dore tot eb concuddeed on a FBD EXAMPLE An pbject cinnected to another pbjetct that has a tesncon toe d DOSE not have Froce of t uppin its
1Gravititon forcw
Always points dfown
Mamg
Acts ata he object s cejrtu7fof mass
Noraml fotc
PREVENT OBJETS FROM PSSING TRHO7UGH SUFRACES!!!!
Forctios
Tensin acts alon groeps strings. Saem amgutide rthgout an ideal stsring Massls
All appucatuins and force

Unit 3

Formlas:
*KE, POTENTRial Enreugy
*Me=pe + ke PE = mgh V = sqrt2gH
This alie for mormla, where the height formj teh mroomti9nn
If W +ESys W and KE PE, then e + w = Ef, If W - Essys orm re esues 5es

springSystem

*T =2 * sqrtk/m where is ter peropd mas of pbjet ans k i is the sprtin constant
*Gravitional mass:
TE= 2 * *r V + FGmM E ^2

Eruiclbirim

Equblrim is when thee su of all forcest = 0 sum of all tourqw = 0. No accerretaion

TYPES

y
Subtle rerterns to orgnaial poition if t disurve, unstaubl emives wat Form position, and netyral STtas ys in new podtion if disterbded.
motion is contripal froec = mv r

Analysis Piints

For Spting. Dhw eqtlibriim poition, indcule wieght , and rpsibg irc er . Conerdsisde ddamopmg if orers
Eciullibtrum draw fsd

Trison Probslem

*Is=f/3 - tehfoce fmyt aacvceeelratre ootgh asses

Poteintial Enrrgy Explained:

-teh higerhte pbject on teh hill, the mmoregravatiibonal pitenal enre ttyu ha
The amount f p[tenal enrgt pbjet
ecams is diretclt related ti its icrase ihght.