ap review.notebook - massillon middle school review slides.pdfap review.notebook 1 april 24, 2015...

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AP Review.notebook 1 April 24, 2015 May 66:23 AM AP Physics Review 2 D Motion Kenimatic Equations for Constant Acceleration The overall velocity of the projectile at any point of its motion is the vector sum of its x and y components at that point. Problem Solving Strategy: 1. Sketch the path of the projectile, including initial and final position, velocities, and accelerations. 2. Resolve the initial velocity into x and y components 3. Treat horizontal and vertical motion independently ~ Time is the only variable that can be used in both the x and y directions ~ a = 0 in the x direction unless otherwise stated. ~ a = g = 9.8 m/s2 = 10 m/s2 in the y direction. vi(x & y) = 0 m/s yi = y (height above ground) y = 0 so Δy = 0 yi = yi vf = y direction only viy =viy vfy = viy vtop = 0 m/s viy =viy vfy = viy vtop =vx vix =vx vfx =vx On the dished line, Δy = 0 m On the ground, Δy = yi Newton's Laws of Motion: 1. Law of Inertia = An object at rest stays at rest unless acted upon by an unbalanced force. An object in constant straight line motion stays in that motion unless acted upon by an unbalanced force. 2. Relates Force, mass and acceleration by the equation: FNet = ma 3. Action / Reaction. Every force has an equal and opposite force. Types of Forces: FNet = Sum of all forces = Σ F1 +F2 = ma Fg = mg = Force of Gravity = Weight FT = T = Force Tension in a rope or string FN = Force Normal = Surface Push (Perpendicular) Fs = Spring Force Fc = Centripital Force = Perpendicular to Velocity v vx = v cosθ viy = v sinθ θ May 66:23 AM AP Physics Review Forces Combine to get Fc Fc = mv2 r Physics Book 40 0 F N =F F F ll F ll 40 0 F g Problem Solving Strategy: 1. Sketch a free body diagram of the problem. Label all forces. 2. Resolve forces at angles into x and y components (unless inclined plane) 3. Treat horizontal and vertical forces independently ~ Clearly label +x and x, +y and y forces. ~ Sum all forces in the x direction. Same with y. 4. Balanced Forces means a = 0 and v = constant!!! Incline Plane = Parallel and Perpendicular Components = mg cosθ = mg sinθ mg Friction can be up or down the incline plane opposite motion. Pulley / Elevator Problems 3 kg 5 kg To find a: ma = F g5 F g3 a sign match F Big To find T: 1 Look at 1 Block ma = F g5 F T ma = F T F g3 On Elevator ma= F T F g a sign match direction In Elevator ma= F N F g a sign match direction Torque = F is perpendicular r is length between rotation and F AP Physics Review Momentum Impulse changes momentum Problem Solving Strategy: Use for Collisions or Explosions 1. Sketch a diagram of the problem. Show before collision and after collision. 2. Resolve momentum's at angles into x and y components (independent) 3. Momentum in x is conserved. Momentum in y is conserved. 4. Elastic Collisions = Energy Conserved Inelastic Collision = Energy Lost m 1 v 1x m 1 v 1y m 1 v' 1x m 1 v' 1y p 2 =p x 2 +p y 2 p x =m 1 v 1 p y =0 p' x =m 1 v 1 =p 1x +p 2x m 1 m 2 m 1 m 2 p' y =0=p 1y p 2y AP Physics Review Energy Problem Solving Strategy: Do not use for Collisions or Explosions 1. Sketch a diagram of the problem. 2. Direction does not matter for energy. U can convert directly to K. 3. Only time direction matters is when an object has an x velocity. 4. Set U and K equation to get the short cut equation. 5. W = Change in Energy. So W = U = K v i = 0 m/s Δh 1 v 1 = √ 2 g Δh 1 v 1 v 4 v 3 v 2 Δh 4 =0 Δh 3 Δh 2 = total h i v 4 = √ 2 g Δh 4 =0 v 3 = √ 2 g Δh 3 v 2 = √ 2 g Δh 2 May 66:23 AM AP Physics Review Other Equations in Newtonian Mechanics Problem Solving Strategy: Things you should know... 1. P0 is pressure at the top of the fluid in interest. Usually air pressure 1 x 105. 2. h is depth in the fluid. 3. Buoyant Force on an object is equal to the weight of the displaced fluid. ρVg ~ Volume of displaced fluid = volume of object under water. ~ ρ = density of fluid V = volume of object submerged. 4. The Volume Flow Rate, FR = Av, has to be constant for a moving fluid. 5. Bernoulli's Equation, P + ρgy + ½ρv2 is a conservation of energy equation. ~ If fluid has no height, ρgy term = 0 ~ If fluid area is so big v = 0 (Top of water tank), then ½ρv2 =0 k = sp constant x = stretch / compress ignore () sign T = period = time for 1 cycle Tspring depends on mass & k Tpendulumdepends on length & g Lenght increases, T increases G = gravitational constant r= center of mass separation Ignore the negative sign L Δh vi = 0 m/s Velocity at the Bottom = √ 2 g Δh FT at Bottom = Fc =FT Fg Time to Swing across and back. T T = 2π √ L / g Time to swing across. ½T Spring is at equilibrium position. F = 0 N. Spring is at xmax. U is max. K and v = 0 Spring is back at x = 0 U=0J K and v = max Spring is at xmax. U is max. K and v = 0 AP Physics Review Fluid Mechanics For stationary fluids A v is Flow Rate. m3/s PP0 = ΔP AKA Gauge Pressure May 66:23 AM AP Physics Review Thermal Mechanics Problem Solving Strategy: Things you should know... 1. The ideal gas law is commonly used. They will give you PV graphs and ask for the temperature. ~ You can use this equation to solve for changes. ~ If you have number of molecules, use the kB equation. 2. Temperature comes from the motion of a substances molecules. ~ Motion is K... Kave is directly related to Temp! 3. U is internal energy, found by the KE of all molecules, or N Kave. ~ U = NKave = (3/2) N KB T= (3/2) n R T ~ It temperature increases, U increases. 4. Work is always work done "On the gas." ~ Contract = + W Expand = W ~ If pressure is not constant, than W = Area under the PV Curve 5. Q is the transfer of thermal energy, govern by the specific heat eq. N0 = Avogadro's Number = 6.02 x 1023 particles/mole n= = Number of Moles N = Number of Molecules or Atoms m molar mass PfVf nRTf PiVi nRTi = N N0 n= Thermal Expansion Rate of Heat Transfer ΔU = (3/2) nRΔT Not given but need! ΔQ = mcΔT Specific Heat Only use if P is constant Q Positive if energy is transferredinto the system Negative with energy isremoved fromthe system W Positive if work is doneon the system Negative if work is doneby the system ΔU Positive if the temperatureincreases Negative if the temperaturedecreases Thermal Processes Isobaric Pressure stays constant. This creates a horizontal line on a PV diagram. For an isobaric possess, W = PΔV ΔU=Q+W or ΔU=QPΔV Adiabatic No heat is exchanged with the surroundings. This means that Q = 0. Understand that this does not mean that ΔT=0 ΔU=0+W or ΔU=W Isovolumetric ΔU=Q+0 or ΔU=Q Isothermal Temperature stays constant. This means that there will be no chan in internal energy. Any work done on the system is accompanied by a loss of thermal energy and visa versa. 0=Q+W or Q = W Isovolumetric Volume stays constant. This creates a vertical line on a PV diagram. Since you cannot take the integral (area under the curve) for a vertical line, W = 0. ΔU=Q+W Understanding the Signs The gas is being compressed, so work is done on the gas. This causes more collisions, increasing molecular motion, increasing temperature, increasing U. Pressure is not constant, so you must find the area under the curve to calculate work. It is expanding so work is done by the gas ( W). The gases internal energy, U, must decrease. The area inside the process A B C A is the net work done. Because you start at A and end at A, PV is constant, so temperature returns to initial value. If ΔT = 0, then ΔU = 0. May 66:23 AM AP Physics Review Electricity Problem Solving Strategy: Things you should know... 1. Like charges repel, opposites attack. 2. Electric Field lines show where (+) charges will go. ~ Point away from () charges and toward (+) charges. ~ E Field inside a conductor is 0 N/C. Must be at surface. 3. Electric Potential = V Potential (Stored) Energy = U 4. Forces are vectors, so you must think about x and y components. 5. Charge, q, is quantized. Comes in clumps of qe = 1.6 x 1019 C. 6. Voltage, V can be shown as ε in a circuit diagram. 7. Positive side of battery is the long plate. Current is always positive! 8. The voltage drop over any enclosed loop must be 0Ω. ~ Resistors and Capacitors in parallel have equal ΔV. 9. When hooked up in series, Capacitors have equal charge, q. ~ In series resistors have equal current, I. 10. VBat =Req Itot VBat Ceq=Qtot 11. Light bulb brightness is measured in Power (W). Increase P, increase brightness. Only equation where the sign of the charge matters! Potential Energy Stored in a Capacitor R for a wire is equal to ρ, resistivity of wire, length of wire, and Area Combine to get P = I2R=V2/R P also = W/t = E/t Could be K = qV or W = qEd 5V 5V 5V 5V C1 + C2 Series Connection Equal Charge or Current Parallel Connection Equal Voltage Drop over all loops 12 V 5 μF 4 μF 9 μF 12 V 12 V 5 μF 4 μF 9 μF When the switch first closes, the capacitors are still uncharged. With Q = 0F, V also must be 0, due to C = Q/V. If the voltage over the capacitor is 0, it acts like a wire. If the switch is open, everything is uncharged. C = Uncharged = Wire C = Charged = Unhooked If the switch is open for a long time, the capacitors are fully charged (Steady state). The now receive all of their possible voltage. Also, since they are fully charged, no more charge or current can flow through them and they act as if they are unhooked. RC Circuits How to get Req. Work your way down, then work back up solving for current (charge for capacitors) and voltage. May 66:23 AM AP Physics Review Magnetism 1. A Charge MOVING PERPENDICULAR through a magnetic field feels a force qvB acting on it. ~ The force felt is perpendicular to both velocty and BField ~ Determined by Right hand Rule (Positive Particles / Current) A B C A segment of a currentcarrying wire in a magnetic field B. The magnetic force exerted on each charge making up the current is qv x B and the net force on the segment of length L is IL x B F v B (A) In this rule, the fingers point in the direction of v, withBcoming out of your palm, so that you can curl your fingers in the direction of B. The direction of v x B, and the force on a positive charge, is the direction in which the thumb points. (B) In this rule, the vectorvis in the direction of your thumb andBin the direction of your fingers. The force FB on a positive charge is in the direction of your palm, as if you are pushing the particle with your hand. BField In BField Out Charges move in Circular Motion in a BField. Set FB =FC + vi X X X X XX X X X X X XX X Velocity Selectors combine E and B to allow charges with a specific velocity to pass through un deflected. FB = qvB, FE = qV, Set equal and solve for v. qE = qvB so v = E/B Velocity Selector Second Right Hand Rule for a long Current Carrying Wire Two long parallel current carrying wires can attract or repel each other. Each creates its own BField. Their created BFlield's can interact with the current in the other wire and exert a force on the other wire. Third Right Hand Rule for a coil of wire acting like a Bar Magnet Electromagnetic Induction Pulling a wire through BField creates induced ε (V) and I. Electromagnetic Induction Moving a magnet toward a coil of wire will induce a current that create a magnetic field that opposes the incoming / outgoing magnet. F = qE Used for induced voltage and current. ε = V, so BLv = IR

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Page 1: AP Review.notebook - Massillon Middle School Review Slides.pdfAP Review.notebook 1 April 24, 2015 May 66:23 AM AP Physics Review 2 D Motion Kenimatic Equations for Constant Acceleration

AP Review.notebook

1

April 24, 2015

May 6­6:23 AM

AP Physics Review ­ 2 D Motion

Kenimatic Equations for Constant Acceleration

The overall velocity of the projectile at any point of its motion is the vector sum of its x and y components at that point.

Problem Solving Strategy:  1.  Sketch the path of the projectile, including initial and final position, velocities,  

and accelerations.   2.  Resolve the initial velocity into x and y components   3.  Treat horizontal and vertical motion independently  

~ Time is the only variable that can be used in both the x and y directions ~ a = 0 in the x direction unless otherwise stated. ~ a = g = ­9.8 m/s2 = ­10 m/s2 in the y direction.

vi(x & y) = 0 m/syi = y (height above ground)

y = 0  so Δy = 0 ­ yi = ­yivf = y direction only

viy = viy vfy = ­viy

vtop = 0 m/s

viy = viy vfy = ­viy

vtop = vx

vix = vx vfx = vxOn the dished line, Δy = 0 m

On the ground, Δy = ­yi

Newton's Laws of Motion:   1.  Law of Inertia = An object at rest stays at rest unless acted upon by an unbalanced force. An object in 

constant straight line motion stays in that motion unless acted upon by an unbalanced force.

  2.  Relates Force, mass and acceleration by the equation:   FNet = ma 

  3.  Action / Reaction. Every force has an equal and opposite force. 

Types of Forces: FNet = Sum of all forces  =  Σ F1 + F2 = ma Fg = mg = Force of Gravity = Weight

FT = T = Force Tension in a rope or string FN = Force Normal = Surface Push (Perpendicular)

Fs = Spring Force Fc = Centripital Force = Perpendicular to Velocity

v

vx = v cosθ

viy = v sinθ

θ

May 6­6:23 AM

AP Physics Review ­ Forces

Combine to get Fc

Fc =mv2

r

Physics

 Book 

400

FN = F

 FF ll

F ll

400

 Fg

Problem Solving Strategy:  1.  Sketch a free body diagram of the problem. Label all forces.   2.  Resolve forces at angles into x and y components (unless inclined plane)  3.  Treat horizontal and vertical forces independently  

~ Clearly label +x and ­x, +y and ­y forces.

~ Sum all forces in the x direction. Same with y. 

  4.  Balanced Forces means a = 0 and v = constant!!!

Incline Plane = Parallel and Perpendicular Components

= mg cosθ

= mg sinθmg

Friction can be up or down the incline plane opposite motion.

Pulley / Elevator Problems

3 kg

5 kg

To find a:ma = Fg5 ­ Fg3

a sign match FBig

To find T: 1 Look at 1 Block

ma = Fg5 ­ FT

ma = FT ­ Fg3

On Elevatorma =  FT ­ Fg

a sign match direction

In Elevatorma =  FN ­ Fg

a sign match direction

Torque = F is perpendicularr is length between rotation and F

AP Physics Review ­ Momentum 

Impulse changes momentum

Problem Solving Strategy: Use for Collisions or Explosions   1.  Sketch a diagram of the problem. Show before collision and after collision.   2.  Resolve momentum's at angles into x and y components (independent)

  3.  Momentum in x is conserved. Momentum in y is conserved.  

  4.  Elastic Collisions = Energy Conserved Inelastic Collision = Energy Lost   

m1v1x

m1v1y

m1v'1x

m1v'1yp2  = p

x2  + 

p y2

px = m1v1

py = 0

p'x = m1v1 = p1x + p2xm1

m2

m1

m2

p'y = 0 = p1y ­ p2y

AP Physics Review ­ Energy 

Problem Solving Strategy: Do not use for Collisions or Explosions   1.  Sketch a diagram of the problem.    2.  Direction does not matter for energy. 

U can convert directly to K.

  3.  Only time direction matters is when an object has an x velocity.

  4.  Set U and K equation to get the short cut equation.  

  5.  W = Change in Energy. So W = U = K

vi = 0 m/s

Δh1v1 = √ 2 g Δh1

v1

v4

v3

v2

Δh4 = 0

Δh3

Δh2 = total hi v4 = √ 2 g Δh4 = 0

v3 = √ 2 g Δh3

v2 = √ 2 g Δh2

May 6­6:23 AM

AP Physics Review ­ Other Equations in Newtonian Mechanics

Problem Solving Strategy: Things you should know...

  1.  P0 is pressure at the top of the fluid in interest. Usually air pressure  1 x 105.  

  2.  h is depth in the fluid.  3.  Buoyant Force on an object is equal to the weight of the displaced fluid. ρVg 

~ Volume of displaced fluid = volume of object under water. 

~ ρ = density of fluid V = volume of object submerged. 

  4.  The Volume Flow Rate, FR = Av, has to be constant for a moving fluid.

  5.  Bernoulli's Equation, P + ρgy + ½ρv2 is a conservation of energy equation. ~ If fluid has no height, ρgy term = 0 

~ If fluid area is so big v = 0 (Top of water tank), then ½ρv2 = 0

k = sp constantx = stretch / compress   

 ignore (­) sign

T = period = time for 1 cycle

Tspring depends on mass & k 

Tpendulum depends on length & g

Lenght increases, T increases

G = gravitational constant

r = center of mass separation 

Ignore the negative sign

L

Δh

vi = 0 m/s

Velocity at the Bottom = √ 2 g Δh

FT at Bottom =   Fc = FT ­ Fg

Time to Swing across and back. T

T = 2π √ L / g 

Time to swing across.    ½ T

Spring is at equilibrium position. F = 0 N.

Spring is at xmax. U is max. 

K and v = 0

Spring is back at x = 0

U = 0 J    K and v = max

Spring is at xmax. U is max. 

K and v = 0

AP Physics Review ­ Fluid MechanicsFor stationary 

fluids

A v  is Flow Rate.  m3/s

P ­ P0 = ΔP AKA Gauge Pressure

May 6­6:23 AM

AP Physics Review ­ Thermal Mechanics

Problem Solving Strategy: Things you should know...

  1.  The ideal gas law is commonly used. They will give you PV graphs and ask for the temperature. 

~ You can use this equation to solve for changes. 

~ If you have number of molecules, use the kB equation. 

  2.    Temperature comes from the motion of a substances molecules. 

~ Motion is K...    Kave is directly related to Temp!  

  3.   U is internal energy, found by the KE of all molecules, or N Kave.

~  U = NKave = (3/2) N KB T = (3/2) n R T 

~  It temperature increases, U increases.

  4.  Work is always work done "On the gas."    

~ Contract = + W Expand = ­ W 

~ If pressure is not constant, than W = Area under the PV Curve

  5.  Q is the transfer of thermal energy, govern by the specific heat eq.

N0 = Avogadro's Number = 6.02 x 1023 particles/mole

n =  = Number of Moles

N = Number of Molecules or Atoms

m

molar mass

PfVf          nRTfPiVi     nRTi

=

NN0

n =

Thermal Expansion

Rate of Heat Transfer

ΔU = (3/2) nRΔT

Not given but need!

ΔQ = mcΔTSpecific Heat

Only use if P is constant

Q  Positive if energy is transferred into the system

Negative with energy is removed from the system

   W  Positive if work is done on the system

Negative if work is done by the system

  ΔU  Positive if the temperature increasesNegative if the temperature decreases

Thermal ProcessesIsobaric ­ Pressure stays constant. This creates a horizontal line on a PV diagram. For an isobaric possess, W = PΔV

ΔU = Q + W    or    ΔU = Q ­ PΔV

Adiabatic ­No heat is exchanged with the surroundings. This means that Q = 0. Understand that this does not mean that ΔT = 0

    ΔU = 0 + W    or    ΔU = W

Isovolumetric ­

ΔU = Q + 0   or   ΔU = Q

Isothermal ­ Temperature stays constant. This means that there will be no change in internal energy. Any work done on the system is accompanied by a loss of thermal energy and visa versa.

0 = Q + W   or   Q = ­W

Isovolumetric ­ Volume stays constant. This creates a vertical line on a PV diagram. Since you cannot take the integral (area under the curve) for a vertical line, W = 0.

ΔU = Q + W        Understanding the Signs

The gas is being compressed, so work is done on the gas. This causes more collisions, increasing molecular motion, increasing temperature, increasing U.

Pressure is not constant, so you must find the area under the curve to calculate work. It is expanding so work is done by the gas ( ­ W). The gases internal energy, U, must decrease.

The area inside the process A B C A is the net work done. 

Because you start at A and end at A, PV is constant, so temperature returns to initial value. If ΔT = 0, then ΔU = 0.

May 6­6:23 AM

AP Physics Review ­ ElectricityProblem Solving Strategy: Things you should know...

  1.  Like charges repel, opposites attack.   2.  Electric Field lines show where (+) charges will go. 

~ Point away from (­) charges and toward (+) charges. ~ E Field inside a conductor is 0 N/C. Must be at surface.

  3.  Electric Potential = V        Potential (Stored) Energy = U 

  4.  Forces are vectors, so you must think about x and y components.   5.  Charge, q, is quantized. Comes in clumps of qe = 1.6 x 10­19 C. 

  6.  Voltage, V can be shown as ε in a circuit diagram.   7.  Positive side of battery is the long plate. Current is always positive!   8.  The voltage drop over any enclosed loop must be 0Ω. 

~ Resistors and Capacitors in parallel have equal ΔV.   9.  When hooked up in series, Capacitors have equal charge, q. 

~ In series resistors have equal current, I.

  10.  VBat = Req Itot VBat  Ceq = Qtot 

  11.  Light bulb brightness is measured in Power (W). Increase P,      increase brightness.

Only equation where the sign of the charge matters!

Potential Energy Stored in a Capacitor

R for a wire is equal to ρ, resistivity of wire, 

length of wire, and Area

Combine to get P = I2R = V2/R

P also = W/t = E/t

Could be   K = qV  

            or       W = qEd

5 V 5 V 5 V 5 V

C1

­+

C2

Series ConnectionEqual Charge or Current

Parallel ConnectionEqual Voltage Drop over all loops

12 V

2 Ω

4 Ω 5 μF4 μF

9 μF

12 V

2 Ω

4 Ω 12 V

2 Ω

4 Ω 5 μF4 μF

9 μF

When the switch first closes, the  

capacitors are still uncharged. With  

Q = 0F, V also must be 0, due to  

C = Q/V. If the voltage over the  

capacitor is 0, it acts like a wire.

If the switch is open, everything is uncharged. 

C = Uncharged = Wire C = Charged = Unhooked

If the switch is open for a long time, 

the capacitors are fully charged 

(Steady state). The now receive all of 

their possible voltage. Also, since they 

are fully charged, no more charge or 

current can flow through them and 

they act as if they are unhooked.

RC ­ Circuits

How to get Req.

Work your way down, then work back up solving for current (charge for capacitors) and voltage.

May 6­6:23 AM

AP Physics Review ­ Magnetism  1.  A Charge MOVING PERPENDICULAR through a magnetic field   feels a force qvB acting on it. 

~ The force felt is perpendicular to both velocty and BField~ Determined by Right hand Rule (Positive Particles / Current)

  

A B C

A segment of a current­carrying wire in a magnetic field B. The magnetic force exerted on each charge making up the current is qv x B and the net force on the segment of length L is IL x B

F

v

B

(A) In this rule, the fingers point in the direction of v, with B coming out of your palm, so that you can curl your fingers in the direction of B. The direction of v x B, and the force on a positive charge, is the direction in which the thumb points.

(B) In this rule, the vector v is in the direction of your thumb and B in the direction of your fingers. The force FB on a positive charge is in the direction of your palm, as if you are pushing the particle with your hand.

BField In BField Out

Charges move in Circular Motion in a BField. Set FB = FC

+vi

X    X   X   X   X  X   X  

X    X   X   X   X  X   X  

Velocity Selectors combine E and B to allow charges with a specific velocity to pass through un deflected. FB = qvB,    FE = qV, Set equal and solve for v.

qE = qvB      so      v = E/B

Velocity Selector

Second Right Hand Rule for a long Current Carrying Wire Two long parallel current carrying wires 

can attract or repel each other. Each creates its own BField. Their created BFlield's can interact with the current in the other wire and exert a force on the other wire.

Third Right Hand Rule for a coil of wire acting like a Bar Magnet

Electromagnetic Induction 

Pulling a wire through BField 

creates induced ε (V) and I.

Electromagnetic Induction 

Moving a magnet toward a coil of wire will induce a current that create a magnetic field that opposes the incoming / outgoing magnet.

F = qE

Used for induced voltage and current.

ε = V, so  BLv = IR

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May 6­6:23 AM

AP Physics Review ­ Waves & Optics

n = λλn

Frequency does not change in n.

θIncidence  θReflection 

θRefraction 

Amplitude

Rest Line

Govern by Snell's Law.

More Dense to Less Dense~ Toward the Normal

More Dense to Less Dense~ Away from the Normal

so

si

EyeEye Eye

I

O

Principle Axis         Vertex

Center of Curvature

FocalPoint (f)

Principle Axis Vertex

Center of Curvature

FocalPoint (f)

FocalPoint (f)

Vertex(V)

FocalPoint (f)

FocalPoint (f)

Vertex(V)

FocalPoint (f)

Concave Mirror

Convex Mirror

Convex Lens

Concave Lens

Converging Devises

Images can be upright or inverted, enlarged or reduced, real or virtual

Diverging Devises

Images will be upright, reduced, and real.

+ + +

­ ­ ­

Total Internal Reflection

sisohihof

M

real

real

upright

upright

converging

upright

virtual

virtual

inverted

inverted

diverging

inverted

+ When ­ When

M > 1 = Enlarged

M = 1 =

1 > M > 0 = Reduced  

Mirror Directions:

1. In parallel, out through/away focus.

2. In through/toward focus, out parallel

3. In through / toward CofC, back through/ away from CofC

Lens Directions:

1. In parallel, out through/away focus.

2. In through/toward focus, out parallel

3. Straight through the vertex 

Young's Double Slit Experiment 

Proved light is a wave

L >>>>>> d

If that distance is nλ, then constructive interference

If it's ½nλ, then destructive interference

m = 1

m = 2

m = 0

m = ­1

m = ­2

1 Slit... Then d = slit width not slit separation. 

This can happen with sound as well.

May 6­6:23 AM

AP Physics Review ­ Atomic and Nuclear Physics

Kmax = qVs

E = hc

λ

For a Photon

Use for DeBrogli Wavelength questions. Can substitute mv

Stopping Potential / Voltage

Problem Solving Strategy: Things you should know...

  1.  Atomic Mass (energy) and Charge must be conserved.    2.  Look for units and use the correct planks constant h.  3.  Forms of atomic radiation (I did not cover...)

~ Alpha (α) Decay = Release 2P and 2N. Atomic mass ­ 4.

~ Beta (β) Decay = N becomes a P and an e­ is released. 

~ Fisson = split nuecleus (mass/energy and charge    conserved).

~ Fusion = fuse nuclei (mass/energy and charge conserved).  

 4.  Photon energy related to f.    E = hf.

~ High f, low λ, high energy.

 5.   Electrons absorb (release) energy and jump up (fall down) an energy level with a  difference of energy equal to the photon.

K = hf ­ φKinetic Energy of ejected Electron

Energy of the photon

Work Function"Toll"

hfc = φ

E2 ­ E1 = ΔE = hf

Cuttoff Frequency 

Only works for light at the violent end of the spectrum. Violet light has a higher frequency so it has high energy. They love to ask in terms of wavelength.

Compton Scattering proves light (massless) has momentum. 

E = hf = pc.

De Broglie took it further, said all matter moves in wave motion govern by λ = h/p = h / (mv)

n1 = Ground State

n2 = 1st Excited State (2nd E Level)

n3 = 2nd Excited State

May 10­7:27 AM May 10­7:27 AM

May 10­7:26 AM May 10­6:13 AM

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May 10­5:58 AM May 6­8:50 AM

2000

May 6­10:07 AM May 6­10:08 AM

2000

May 6­8:49 AM

03b

May 6­10:11 AM

08b

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May 6­10:18 AM

10b

May 7­9:19 PM

10

May 7­9:20 PM

07b

May 7­9:31 PM

08

May 7­9:23 PM

06

May 7­9:40 PM

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May 9­8:34 AM May 9­8:36 AM

May 9­8:38 AM May 7­9:40 PM

12

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11b

May 10­8:32 AM

May 10­8:35 AM May 10­8:39 AM

08b

May 10­8:39 AM

06

May 10­8:44 AM

11

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May 10­8:52 AM