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 400
F
N
= F F F
l l
Fll
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 = m g s i n θ mg 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 Elevator ma = FT Fg
a sign match direction
In Elevator ma = FN Fg
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 m1v1x m1v1y m1v'1x m1v'1y p2 = px2 + py2 px = m1v1 py = 0 p'x = m1v1 = p1x + p2x m1 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 Δh1 v1 = √ 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 constant x = 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 Mechanics
For 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 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 transferred into the system Negative with energy is removed from the system W Positive if work is done
- n the system
Negative if work is done by the system ΔU Positive if the temperature increases Negative if the temperature decreases
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 = 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
- n 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
- r W = qEd
5 V 5 V 5 V 5 V C1 + C2
Series Connection Equal Charge or Current Parallel Connection Equal Voltage Drop over all loops 12 V 2 Ω 4 Ω 5 μF 4 μF 9 μF 12 V 2 Ω 4 Ω 12 V 2 Ω 4 Ω 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, 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
- f 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