Download AP Exam Study Overview (Without Rotational Dynamics)

Advanced Placement Physics
Physics Overview
The following equations and facts are not given on the exam, but are essential for your success.
v
x
t
v
2 r
T
g G
mAvAi  mBvBi  mAvA f  mBvB f
mAvAi  mB vB i   mA  mB  vAB f
1
1
mghi  mv 2i  mgh f  mv 2 f
2
2
mghtop 
1 2
mv bottom
2
m
r2
 mA  mB  vABi  mAvA f
 mB vB f
1 2 1 2
kx  mv
2
2
Klost / dissipated   K i   K f
qV 
1 2
mv
2
And any combination of Work Energy Theorem and Conservation of Energy that might be applicable to the problem.
Work is the area under Force distance.
Series circuits: current stays the same
Parallel circuits: current adds
Ek
q
r2
V    Ir (when there is internal
resistance)
Study Guide and Review
Series circuits: voltage adds
P  I 2R
Parallel circuits: voltage stays the same
P
V2
R
Right Hand Rule
A changing flux induces a current in a wire. The direction of the current is the opposite that specified by the right hand rule.
The current induced in the wire generates its own magnetic field, which is opposite to the field that caused the emf.
Physics Overview
1
Energy Overview
Energy is a central concept that connects the various strands of physics. Through the Work Energy Theorem
and the Principle of Conservation of Energy a host of equations and possibilities help generate solutions.
Work Energy Theorem
Total work equals the change in kinetic energy
Newtonian Mechanics
1
K  mv 2
2
Us 
Work, in the center, is equal to any change in energy around the circle. For
example:
U g  mgh
1 2
kx
2
Ug 
Gm1m2
r
1
2
m  v 
2
F  d cos 
W  U g  mg h
so
F  d cos  mg h
W  U E  qV
so
F  d cos  qV
Conservation of Energy
Energy can change forms
W  F  d cos
U= ½
LI2
Total Energyi  Total Energy f
Any energy around the circle can change into another energy around the
circle. For example:
1
2
m  v   mg h
2
qV 
U E  qV
Uc 
1
QV
2
1
2
m  v 
2
so
W  K 
1
U c  CV 2
2
1
2
m  v 
2
so
1
1
mghi  mv 2i  mgh f  mv 2 f
2
2
A charge accelerated by charged plates
Electricity & Magnetism
Power: If work and energy are important, then any variable that has work / energy in its equation is equally important.
Power is the rate that work is done or that energy is delivered, expended, or used. So from power we can get work and energy,
and from work and energy we can get power. One easy way is to set time to 1.0 s. Then power and work / energy have the same
value, but different units. If you find the time later in the problem just multiply or divide by the time to solve as necessary.
Study Guide and Review
Physics Overview
2
MOTION OF A SINGLE OBJECT: Relevant Kinematics, Force, and Energy
Start a problem by asking “What is the object doing?”, then “What is causing it to do that?”.
What direction is it moving in (if two find x and y components)? Is it moving at constant v (this includes v = 0)? Is it accelerating? Force? Energy change?
See if energy solves the problem first. Then think force and kinematics.
Most common usage is boxed. But, the most common usage is often a special case. Knowing the overall equations and logic will allow you to solve any scenario.
Situation
Kinematics
Constant Velocity
FN
v
v
Fg
Need constant velocity
Energy
F  0
x
t
Forces are vertical, while
motion is horizontal
or
d or x
Force
Always think sum of force
Fg  mg
x  vt
FN  mg cos
Inertia only. No force. No energy needed.
W  F  d cos
W  F  d cos 90o
Where  is the angle between F and d vectors.
W 0
0o slope
FN  mg
Use Kinematic Equations
Accelerating in x
FN
a
Fret
Fg
vx  vo x  at
 F  ma
vx  vo x  2a  x  xo 
 Fx  Fp x  Fretarding x
2
FP
d or x
There is a sum of force
2
1
x  xo  vo xt  at 2
2
example
 Fx  Fp x  Ffr
In projectile motion
v y  vo y  2 g  y  yo 
Fret
2
2
y  yo  vo y t 
g
 Fy  Fg  Fretarding y
v y  vo y  gt
h or y
Fg
Horizontal projectile or
dropped object
y
Study Guide and Review
1 2
gt
2
1 2
gt
2
Wx   Fx  d cos
Where  is the angle between F and d vectors.
W  F  d cos 0o
No retarding forces present
W  F d
This time work is done, so there is a  energy.
W  Energy
W  K 
x  vo xt
Accelerating in y
A force through a distance.
example
 Fy  Fg  Fair resistance
1
2
m  v  The only thing changing is velocity, so K is changing.
2
Wy   Fy  h
Includes retarding forces
W  Fg  h
No retarding forces
W  U g  mg h Height is changing, so U is changing.
W  K 
mg h 
Physics Overview
1
2
m  v  Velocity is changing, so K is changing.
2
1
2
m  v 
2
1
1
mghi  mv 2i  mgh f  mv 2 f Energy conserved.
2
2
3
Situation
Kinematics
Kinematic Equations Apply
Inclines
a
y
vx  vo x  2a  x  xo 
2
2
1
x  xo  vo xt  at 2
2
x
Fret
vx  vo x  at
FN
vy  vo y  gt
v y  vo y  2 g  y  yo 
2
Fg
2
y  yo  vo y t 
Down a curve
Steep slope, large a
Less steep, a
decreaseing
1 2
gt
2
Force
Energy
Motion is parallel to slope
Work and energy can work parallel, in the x, and in the y
Acceleration down the slope
is caused by the addition of
Fg and FN. The resultant of
these two vectors is Fgsin.
Since the natural motion is
down the slope set that
direction as +. In some
problems it is useful to
reverse this (if the object is
going up hill).
W  F d
Force and distance vectors form similar triangles.
W  Fx  d x
W  Fy  d y
Work depends on F and d being parallel.
 F  Fg sin   Fretarding
FN  mg cos
Kinematics Fail
Force Fails
The net force is changing
as the vectors Fg and FN
change. In addition the
direction is changing.
The net force is changing as
the vectors Fg and FN
change.
W  K 
Pendulum or Swing
R
y
x
h
h R y
y  R  x  R cos
2
2
mg h 
Study Guide and Review
1
2
m  v 
2
1
1
mghi  mv 2i  mgh f  mv 2 f
2
2
Energy is directionless.
W  K 
1
2
m  v 
2
W  U g  mg h
mg h 
The Kinematic Equations
are designed for changing
velocity, but only work for
uniform (constant)
acceleration.
1
2
m  v 
2
1
1
mghi  mv 2i  mgh f  mv 2 f
2
2
A very important case.
mghtop 
Kinematics Fail
Force Fails
See accelerating down a
curve above.
See accelerating down a
curve above.
The velocity is zero at
either end.
The restoring force is greatest
at the ends, as is the
acceleration.
The velocity is greatest at
the lowest point
1
2
m  v 
2
W  U g  mg h
Acceleration is changing.
a=0
constant
velocity from
this point on
So any pair of parallel vector will solve the problem
1 2
mv bottom
2
No initial velocity moving to a height of zero.
Energy is directionless.
W  K 
1
2
m  v 
2
W  U g  mg h
The restoring force is zero in
the middle, and so is the
acceleration.
Physics Overview
mghtop 
1 2
mv bottom If it has no initial velocity and goes all the way down.
2
At the ends it is all potential no motion, in the middle it
is all motion no potential.
4
Situation
Kinematics
Object on string,
Vertical loop
v
FT
Center seeking.
Unlike previous scenarios, the object definitely has velocity at the top.
Instantaneous velocity is
tangent to the circlular
motion.
Force is centripetal, Fc, is the
sum of force in circular
motion. Toward center is +.
There is a height difference from top to bottom, but the object has speed at
the top as well. And the bottom may not necessarily be the lowest point in
the problem
2 r
T
Acceleration is toward
center, centripetal.
ac 
Object on string,
Horizontal loop
FT
Fg
FT
Object turning on
flat surface
Ffr
Fg
Roller Coaster
v
r
Fc  FT  Fg
Find tension at the bottom.
Fc  FT  Fg
Center seeking.
No work is done
Instantaneous velocity is
tangent to the circlular
motion.
Force is centripetal, Fc, is the
sum of force in circular
motion. Toward center is +.
W  F  d cos
W  F  d cos 90o
2 r
T
Acceleration is toward
center, centripetal.
ac 
Fc
2
1
1
2
2
mghtop i  mvtop i  mghbottom f  mvbottom f
2
2
Find tension at the top.
Tangential Velocity
v
Fg
Energy
Tangential Velocity
v
ac
Force
2 r
T
ac 
FT  Fc  Fg
2
2
v2
r
Tangential Velocity
v
Find Fc by adding vectors (tip
to tail). Then solve for FT.
2
Why is there circular motion?
Object not sliding off disk, or
car turning on a road.
Where  is the angle between F and d vectors.
W 0
At any instant the direction of motion (tangent to the circle) is perpendicular
to the center seeking Fc, the FT, and the Fg.
And for one revolution there is no total displacement from the origin, since a
single revolution brings you back to the starting point.
No work is done
See above.
Fc  Ffr
v
r
Need uniform slope
Force centripetal in the loop.
Energy works everywhere, with its directionless advantage.
Kinematics only work on
sections that have
constant slope.
To find the speed needed to
have passengers feel
weightless at the top of the
loop
You can solve for any point A using any other point B. Use the complete
equation. The car will usually have both speed and height at every point.
An exception is the lowest point on the track, or if it starts with zero velocity
at the top of a hill (this is unlikely since roller-coasters don’t stop at the top of
each hill). If it has velocity and height at the top you need to include both.
If the track is curved try
energy.
Fc  Fg
1
1
2
2
mghA  mvA  mghB  mvB
2
2
Study Guide and Review
Physics Overview
5
Situation
Spring
-x
x=0
+x
m
m
Kinematics
Force
Force Fails
See Down a curve, and
Pendulum above.
See Down a curve, and
Pendulum above.
W  K 
The velocity is zero at
maximum +/- x
(amplitude)
The restoring force is greatest
at maximum +/- x
(amplitude).
W  U S 
The velocity is greatest at
x=0
The restoring force is zero at
x = 0, and so is the
acceleration.
Potential difference.
The force of the electric field
m
Particle accelerated
by electric field
v
Energy is directionless.
Positive charges go in the
opposite direction.
It acts like a projectile
x  vx o t
y
Charged particle in a
magnetic field
Path curved by field.
2 r
T
ac 
v2
r
1
2
k  x 
2
If it starts at maximum x (amplitude) and it converts all the springs energy
into speed of the object pushed / pulled by the spring
1 2 1 2
kx  mv
2
2
1 2 1 2 1 2
1
kx i  mv i  kx f  mv 2 f
2
2
2
2
Electromagnetism: New forms of energy, but energy is still conserved.
W  K 
1 2
mv
2
W  U E  qV
qV 
1 2
mv
2
1
1
qVi  mv 2i  qV f  mv 2 f
2
2
Electric field is perpendicular
Work is done in the direction of the electric field.
Particle is forced toward the
plate with opposite sign.
W  F d
F
q
F  qE
 F  ma
ma  qE
E
Forced to center by field.
If the field is large enough
the particle will follow a
circular path.
v
Study Guide and Review
1 2
at get a from F
2
1
2
m  v 
2
W  Energy
F  qE
Velocity increases.
But, + particles are more
massive, don’t accelerate
as quickly, and have lower
final velocities.
Charged particle
parallel to plates
Energy
Kinematics Fail
W  qE  d
No work is done
Fc  FB
W  F  d cos 90o
v2
m  qvB
r
W 0
At any instant the direction of motion (tangent to the circle) is perpendicular
to the center seeking Fc, and the FB.
Physics Overview
6
Collisions
Circular Motion
Momentum is always conserved in a collision
The key to circular motion is to ask:
“What is causing it to stay in a circle?”
Know the following equations
mAvAi  mBvBi  mAvA f  mBvB f
2 objects before, 2 objects after
Centripetal means center seeking
mAvAi  mB vB i   mA  mB  vAB f
2 objects before, 1 objects after
The direction of motion is toward the center
 mA  mB  vABi  mAvA f  mBvB f
1 objects before, 2 objects after
If there are more than two objects add
Any force pointing to the center is a positive force.
Any force pointing away from the center is a negative force.
Force Centripetal is the sum of forces in these problems
mC vC , mD vD , etc.
If the collisions happen in two dimensions, x and y, turn all mv vectors into x
and y components. Solve for the result in the x direction and then solve for the
result in the y direction. Take the final x and y and use Pythagorean Theorem
to find the overall resultants.
If the object has momentum it also has kinetic energy.
Total energy is also conserved, but energy changes forms.
Perfectly elastic collision (An idealized unrealistic case)
It is not drawn on free body diagrams since it is the net force.
Any force (gravity, tension, friction, normal, magnetic, etc.) can contribute to Fc
Possible equations
Fc  Fg
To find minimum speed at the top of a roller coaster loop.
Fc  Ffr
Object is revolving on a horizontal surface, or a car turning.
Fc  FT
If a (horizontal) string spins the object in a horizontal circle.
Fc  FT  Fg
2
Kinetic energy is conserved in this rare case.
Use these two equations together (System of 2 equations & 2 varibles).
mAvAi  mBvBi  mAvA f  mBvB f
1
1
1
1
2
2
mAv A i  mB vB i  mAvA 2 f  mB vB 2 f
2
2
2
2
Inelastic Collision (The common type)
Kinetic energy is lost or dissipated.
2
If an object at the end of a string is spinning through the air
Fc  FT  Fg
and gravity pulls the string down from the horizontal.
An object (at the top) spinning at the end of a string in a
Fc  FT  Fg
vertical circle.
An object (at the bottom) spinning at the end of a string in a
Fc  FN
vertical circle.
Inside an amusement park ride (Gravitron)
Fc  FB
For a charged particle in a magnetic field.
There is less kinetic energy after the collision.
Substitute and solve.
Klost / dissipated   K i   K f
v2
r
Velocity is Tangential
(Note this is the opposite of change in kinetic energy. Change in
kinetic energy is Kf – Ki. The value is the same, but the sign is
reversed.)
1
1
1

2
2  1
Klost / dissipated   mAv A i  mB vB i    mAv A 2 f  mB vB 2 f 
2
2
2
 2

The kinetic energy lost often turns into heat from the impact
K lost / dissipated  mcT
Study Guide and Review
Fc  m
The instantaneous velocity is tangent to the circular motion.
v
2 r
T
T is the period, the time for one revolution.
If the object is released (the force stops working) then the object will move at
this velocity in a direction tangent to the circle at the time of the release.
Physics Overview
7
Rates and Graphing
Rates and Graphing
A change in a variable as a function of time (in seconds).
In the graphed examples the y intercepts and slopes would depend on
where the problem started and on how fast the rate is changing.
d
(or v = dx/dt)
t
Velocity is the slope (derivative) of distance time graph
v
v
v0
v
Constant Velocity: change in position
d
d
t
d
t
Acceleration:
a
v
(or a = dv/dt)
t
Distance increases (or decreases) in an exponential manner.
a
a
d
d
t
t
Acceleration is the slope (derivative) of velocity time graph
a
a
v
Q
(or I = dQ/dt)
t
Current stays the same in a series circuit. All the resistors are in
line. It’s like a traffic jam on one road with no alternate routes. All the cars are
going the same speed on the entire road, so the amount of cars passing any
point in a certain time interval is the same everywhere.
Current adds in a parallel circuit. The electrons have multiple
pathways to choose from. If 100 C arrive at a fork in the circuit they must split
up. Due to conservation of charge, the amount of electrons in the parallel
paths must add up to the amount of electrons arriving at the fork.
emf: change in flux (magnetic field thru an area)  

(or ε = -dΦ/dt)
t
Remember, current can only be generated by a changing flux. So a closed
loop of wire must move through the field, or the loop must be getting larger, or
the loop must be rotating.
Loop moved thru
Bar moved, enlarging loop
Loop rotates
v
t
W
(or P = dW/dt)
t
Remember you can convert directly to work (or energy) from power if you solve
the problem using 1 second for time. (Example: 100 W, means 100 J in one
second). If you get information about time later in the problem, just multiply by
the amount of time to find the actual total work (or energy). Example: If the
proceeding 100 W was delivered for 1 minute, then 100 J were delivered each
second for 60 s. So 6000 J of work (energy) was done, used, or delivered.
Current: charge moving through a point in a circuit I 
t
change in velocity
Power: Work (and any form of energy) done in a time t P 
t
Areas Under Curves

Velocity is the area (integral) under the acceleration graph.
Displacement is the area (integral) under the velocity graph.
B  A cos 
t
 B v

B  A cos 
t
*** Work is the area under the force distance curve.



t
t
Study Guide and Review
Physics Overview
t
8
Capacitors
Graphs of exponential decays:
C = Q/V
Capacitance (measured in Farads) depends only on geometry/shape
Capacitors store charge and energy
Energy Stored: U = ½ CV2 = ½ QV = ½ Q2/C
Voltage across a capacitor in circuit: Q/C (so like a wire when uncharged)
Combinations of Capacitors:
(Series) 1/Ceq = 1/C1 + 1/C2 + …. (decreases capacitance)
(Parallel) Ceq = C1 + C2 + …. (increases capacitance)
Dielectrics: k is the dielectric constant; dielectrics decrease the electric field and
voltage (at a given Q) and therefore increase capacitance by the factor of k
Parallel Plate Capcitor: C = kεA/d
Graphs of exponential increase (Q during charging):
Charging a capacitor in an RC circuit:
The charge on the capacitor increases exponentially and the current to it
decreases exponentially, usually according to equations like,
Q(t) = Qmax(1-e-t/RC)
I(t) = dQ/dt = I0e-t/RC
When fully charged, current quits flowing to it
Discharging a capacitor in an RC circuit:
The charge on the capacitor decreases exponentially and the current decreases
exponentially, usually according to equations like,
Q(t) = Q0e-t/RC
I(t) = -dQ/dt = I0e-t/RC
Ampere’s Law
Inductors
Voltage: LdI/dt (usually a drop in voltage, so subtract)
Energy: U = ½ LI2
Current through inductor cannot change instantly
At steady-state, acts like wire (no voltage)
L-R circuits (current grows): I(t) = Imaxe-Rt/L
L-C circuits (current oscillates; charge on capacitor oscillates; current max
when Q = 0; Q max when I = 0)
Study Guide and Review
§Bdl = µ0Ienclosed
Biot-Savart
dB = (µ/4π) (I dl x r)/r3 (note: dl and r on top are vectors; this is inverse square
law with r; can be written with just direction of r (r “hat”) on top, then r squared is on
bottom)
Physics Overview
9
1. Charge
a. Fundamental charge is e = 1.602 x 10-19 C (electron charge = -e)
b. Continuous charge densities:
i. λ = charge per length (e.g., C/m)
ii. σ = charge per area (e.g., C/m 2)
iii. ρ = charge per volume (e.g., C/m 3)
2. Coulomb’s Law for force between point charges
a. F = kq1q2/r2 (this is a vector; direction: like charges repel;
opposites attract)
i. k = 8.99 x 109 Nm2/C2 = 1/4πε0
ii. ε0 = 8.85 x 10-12 C2/Nm2 = 1/4πk
b. For net force on a single charge Q due to multiple charges, sum
the forces due to each individual charge (but do it by component
since vector quantities)
2
Discrete:
∑kQqi/ri2 i
Continuous:
3. Electric field
a. Due to a point charge q at a point a distance r away: E = kq/r2 (this
is a vector; direction is away from positive charges & toward
negative ones)
b. For the total E field due to multiple charges, sum the E field vectors
due to each individual charge (but do it by component since vector
quantities)
2
Discrete:
∑kqi/ri2 i
Continuous:
4. Electric field lines
a. Point away from positive charges and toward negative ones
b. The density of the lines is proportional to the strength of the E field
c. Direction of the field lines is the same as the direction of the E field
d. Lines start on + or at infinity and end on – or at infinity
e. The number of fields lines starting on a + charge or ending on a –
charge is proportional to the magnitude of the charge
Field lines are radial as the approach a point charge; at large distances,
they are again approximately radial, but now they point toward/away from
the approximate center of the charges
5. Electric Flux/Gauss’s Law
a. Electric Flux = φ =
surface, reduces to EA)
(if E is constant & perpendicular to
b. Electric Flux over a closed surface φ
c. Gauss’s Law: Electric flux over a closed surface is equal to the
charge enclosed by the surface divided by ε0 (or multiplied by 4πk)
i. φ
= Qinside/ε0 = 4πkQinside
ii. Only charges enclosed by a surface contribute to the net
flux through it
Study Guide and Review
6. Conductors
d. Any net charge is located on the surface of a conductor; the
charge density on the interior of a conductor is zero everywhere
e. The electric field is zero everyone on the inside of a conductor
i. The electric field on the surface of a conductor points
perpendicular to the surface and has a value given by, E =
σ/ε0
f. If a conductor has a hollow region, the total charge on the inner
surface of the hollow region will be the opposite of the total charge
inside the hollow region
g. Shielding:
i. Charges on the outer surface of a conductor arrange
themselves so as to create an electric field inside of the
conductor that cancels out any electric field external to the
conductor.
ii. Charges on the inner surface of a hollow region of a
conductor arrange themselves so as to create an electric
field outside of the hollow that cancels out any electric field
due to the charges in the hollow region.
1. So the field inside a hollow region depends only
on the inner surface charge and any charges in
the hollow region
iii. Faraday’s cage: Inside a hollow conductor (with no
enclosed charges in the hollow region), the electric field
will be zero.
h. The electric potential (V) is the same everywhere in a conductor
7. Electric Potential (V) (NOTE: This is a scalar; energy per charge)
i. Discrete: ∑kqi/rii Continuous:
ii. E = -dV/dx or ΔV = -SEdl
8. Specifics you should know (both the result and how to find them)
i. E of an infinite line of charge: E(r) = 2kλ/r (radial in direction)
j. E of an infinite plane of charge: E = σ/2ε0 (perpendicular to plane)
k. For a uniform spherical shell of radius R of charge:
i. E(r) = 0, V = kQ/R (inside shell)
ii. E = kq/r2 and V = kQ/r (outside shell)
l. For a uniform, solid sphere of total charge Q and radius R:
i. E(r) = kQ/r2 and V = kQ/r (outside of sphere)
ii. E(r) = kQinside/r2 = kQr/R3 (inside of sphere)
m. E of a uniform, hollow cylinder of total charge Q and radius R
i. E(r) = 2kλ/r in the radial direction (outside of the cylinder)
ii. E = 0 (inside of the cylinder)
iii. Solid cylinder: use Gauss’s Law with cylinders as surfaces
n. There is a discontinuity in the perpendicular component of E field
on either side of a surface charge of density σ given by, |ΔE| = σ/ε 0
Physics Overview
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Study Guide and Review
Physics Overview
11