Download MCAT Physics Review Historical Areas of Emphasis

Historical Areas of Emphasis
MCAT Physics Review
Grant Hart
[email protected]
Important Ideas about the
Physical Science part of the MCAT
• The problems are not complicated. They
usually involve just one or two concepts,
but you may have to dig a little in the
reading material to find what you need.
– You may also have to apply some common
sense to what you read.
– The majority of what you read is probably not
going to be relevant to the questions.
Mechanics
35%
Fluid Mechanics
15%
Waves, Optics, Sound
20%
Electricity & Magnetism
10%
Nuclear & Atomic Physics
10%
Tools
10%
Important Ideas about the MCAT
• The problems are almost all conceptual
and can be answered with fairly basic
physics. The reading may involve more
complicated ideas, but the questions are
based evaluating based on simple
physics.
1
Important Ideas about the MCAT
Suggestions for doing well
• Most of the time if you have to do more
than add or multiply a couple of numbers
together, you are probably on the wrong
track.
1. Read everything carefully.
2. There is a lot of unused information in
the reading. Don’t worry if you don’t use
it.
3. If you are weak in a topic, don’t just pass
it over. There are several techniques to
improve your chances when you guess.
If you aren’t familiar with a topic:
If you are familiar with the topic:
1. Use your common sense.
2. Eliminate unreasonable answers.
3. Guess, but mark the problem number so
that you can come back to it if you have
time.
1. Simplify.
2. Round your numbers.
3. Calculate.
You cannot use a calculator, so any
calculations will necessarily be simple. You
can use scratch paper if you need to.
4. Check for reasonableness.
2
How to Prepare
• Study the prime areas:
– Mechanics/E&M,circuits/Fluids/Radioactivity/
Waves/Optics
• Understand the concepts – complicated
problems are not the MCAT way.
How to Prepare
• Know the important equations. They are
generally closely related to the basic
concepts.
– Don’t try to memorize everything, particularly
equations. It will just confuse you and won’t
help you.
• Know how to read graphs and tables.
There will be a number of them on the
exam!
From the MCAT instructions:
Format of Physical Science Section
• Neither the passage-based questions nor
the independent questions test your ability
to memorize scientific facts. Rather, both
types of questions assess knowledge of
basic physical and biological science
concepts and your facility at problem
solving at using these concepts.
• 70 Minutes
• 52 questions. About half will be on
Physics and half on Chemistry. They may
be mixed together in the same reading.
• 7 readings of about 250 words each with
4-7 questions about each one.
• 13 questions unrelated to any reading.
3
How to approach a Physics
problem
Exam Preparation
• The purpose of this class is not to teach you
physics – you should know most of what you
need to know already.
• The purpose of this class is to help you organize
that material in your mind so you can get more
points on the exam.
• It is essential that you practice thinking physics,
that is the only way to recognize when the
principles come up in the reading.
1. Read
•
•
•
2. Organize your thoughts
•
•
•
How to approach a Physics
problem
3. Simplify the problem
•
Ignore extraneous information. The important
principles in step 2 will help recognize this.
4. Solve
•
•
Concepts used to select method.
Equations
•
Equations are only useful in two ways:
•
•
•
•
They organize the concepts – a good summary. This often
shows up as ratio problems.
You need them when you need a numerical answer.
Be careful – make sure your units are compatible and
watch the signs of things.
Be quick – most of the time you can round to 1 figure and
do a quick calculation.
Passage
Problems
Answers – are they reasonable?
Visualize and sketch it.
Decide what physics principles are
important.
Note given any needed information.
How to approach a Physics
problem
5. Think
•
•
Reasonable?
Units match?
6. After about 1 minute
•
•
•
Eliminate the unlikely answers
Guess
Write down the problem number if there is
hope.
4
What if the answer is bad?
- Wrong units
- Check multiply/divide, powers, bad equation
- Wrong number
Paradigms
• A paradigm is a model or typical pattern
that can be followed, particularly to solve
problems.
– Signs, algebra, bad equation, bad calculation
─ Quickly check your work – then make your
best guess.
• I will talk about several paradigms that can
be used to solve various classes of
problems in physics.
Notes on the Web
Paradigms we will use
A printout of these notes can be found at the
following url:
http://www.physics.byu.edu/faculty/hart/MCAT/
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
5
Paradigms we will use
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
Block on Inclined Plane Paradigm
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
• This is a paradigm for conservation of energy.
This is the easiest way to work a problem – if it
works.
Einitial  E final
• Energy and work:
E  KE  PE
PE  mgh
1
KE  mv 2
2
Block on Inclined Plane Paradigm
Block on Inclined Plane Paradigm
Einitial  0  mgh
initial
final
h
• There is no friction.
h
E final 
1 2
mv f  0
2
1 2
mv f  mgh
2
v f  2 gh
6
Block on Inclined Plane Paradigm
• As long as there is no friction, the path
between start and finish doesn’t matter.
– Free fall is the same as sliding down
something without friction in terms of what the
final velocity will be.
• For springs the PE is ½ k x2
Block on Inclined Plane Paradigm
• Use this technique whenever possible.
Key things to look for:
– Only conservative forces involved (usually
gravity, electric forces, and springs.)
– Time is not involved in the problem, you have
just an initial state and a final state.
– Usually just one object is moving.
Paradigms we will use
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
Porsche Paradigm
• Power:
P
 E W

t
t
(Porsche speeding up)
7
Porsche Paradigm
• It can go from 0 to 60 in 3 seconds, what is the power?
P
 E K f  Ki


t
t
1
2
mv 2
t
• Divide whatever change in energy you have by the time
interval. That is the power, the rate at which energy
changes. You don’t use this for electrical power in
circuits, although it works at the microscopic level.
Braking Car Paradigm
• This paradigm is for kinematics – description of
motion.
• This is used when the following quantities are
involved:
- Position
- Velocity
- Force
- Time
- Acceleration
Paradigms we will use
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
Braking Car Paradigm
• Basic Equations:
v  v0  a t
1
x  x0  v0t  a t 2
2
2
2
v  v0  2a  x  x0 


F  ma  F  ma
W  mg
F f  N
8
Braking Car Paradigm
Possible questions:
• What is the acceleration?
v 2  v02  2a  x  x0 
• Typical Problem:
v0
t=0
x
 v02  2ad
 v02
2d
• What is the coefficient of friction?
a
v=0
t = tf
F f  ma
d
  F 
 mg  ma
  ag
Possible questions:
• How big is the frictional force?
v  v0    v0
a
t
f
 t0 
tf
F  ma
• Given μ and d, what was v0?
 mg  ma   g  a
 v02  2ad
v0  2  g d
Braking Car Paradigm
• Remember – this is for anything speeding
up or slowing down, whether horizontally
or vertically.
• Use whichever equations have the right
variables in them.
• Make sure that conservation of energy is
not the easier way to do it.
9
Aside – Newton’s Laws
• A number of conceptual questions address
Newton’s Laws directly, not in the context of
kinematics.
• In many ways Newton’s first law is conceptually
the hardest.
Example:
• A car goes around a corner. The bag of
groceries on the seat slides and hits the
door. What force caused that?
– None! The bag just traveled in a straight line
and the car turned under it.
– When an object has no net force acting on it, then it
moves at a constant speed in a straight line.
– It does not take a force to keep something moving!
Another Example:
Another Example:
• A skydiver jumps out of a plane. His
speed increases until he reaches terminal
velocity. How big is the force of air
resistance on him at first?
• A skydiver jumps out of a plane. His
speed increases until he reaches terminal
velocity. How big is the force of air
resistance on him after he reaches
terminal velocity?
– Greater than mg.
– Equal to mg.
– Less than mg.
– Greater than mg.
– Equal to mg.
– Less than mg.
10
Circular Motion
• Circular motion is the 2-D motion that you
are most likely to see.
• The a in F=ma is a vector acceleration – if
the direction of the velocity changes but
the magnitude does not, you still have an
acceleration.
• In circular motion the acceleration is
toward the center (centripetal) and of
magnitude v2 / r.
Equilibrium
• When an object is at rest, then all the
forces add to zero. This is a vector sum!
• Occasionally they may ask a problem
where the torques need to sum to zero as
well.
Gravity and Orbital Motion
• Newton’s law of gravity:
– Only attractive
– Inverse square law
F
Gm1m2
r2
• In orbit gravity supplies the centripetal force
necessary to keep you tied to the Earth.
Path without gravity
Actual path
Paradigms we will use
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
11
Ball on an Inclined Plane Paradigm
Ball on an Inclined Plane Paradigm
• Rotation:
– This is a lot like ordinary kinematics and energy
conservation. Make substitutions for the variables
and all the equations are the same:
v 
a 
mI
x 
F 
p  mv  L  I
KE  12 mv 2  KE  12 I 2
add the relationships
v  r and a  r
Ball on an Inclined Plane Paradigm
initial
h
Ball has radius r
final
h
• Equating the two:
• At the bottom
E  12 mv 2  12 I 2
v
E  12 mv 2  12 I  
r
I 

E  12  m  2 v 2
r 

• At the top
E  mgh
2
• No slipping or sliding.
Notes on the Web (Week 2)
A printout of these notes can be found at the
following url:
http://www.physics.byu.edu/faculty/hart/MCAT/
I 

mgh  12  m  2 v 2
r


v
2 gh
I 

1  2 
 mr 
12
Sample MCAT physics problems
• sampleitems.pdf
Paradigms we will use
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
Collisions
• In collisions, momentum is always
conserved!
Colliding Blocks Paradigm
Before:
m1
• If kinetic energy is conserved, it is called
elastic.
• If kinetic energy is not conserved, it is
inelastic.
• If the two stick together it is called totally
inelastic. (maximum energy loss)
v1
v2
m2
Note: v2 < 0
x
After:
v1’
m1
m2
v2’
Note: now v1’ < 0
x
13
Colliding Blocks Paradigm
• Momentum conservation:
m1v1  m2v2  m1v1  m2v2
2 unknowns, 1 equation. Need more information…
Colliding Blocks Paradigm
• However, totally inelastic, v1’ = v2’
Before:
m1
v1
x
m1
m2
v
x
– KE conserved:
m1v12  12 m2 v22  12 m1v12  12 m2v22
• Lots of nasty algebra, not likely to see on MCAT.
Paradigms we will use
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
m2
After:
• Elastic Collision:
1
2
v2
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
m1v1  m2v2  m1  m2 v
v
m1v1  m2v2 
m1  m2 
Electric Circuits
• Definition of Current: I=∆q/ ∆t … flow of charge
• Ohm’s Law: V=IR
• Series and Parallel combinations of resistors:
– Series: Req = R1 + R2
– Parallel: 1/Req = 1/R1 + 1/R2
• Electrical Power: P = I V
• Kirchoff’s rules (only conceptually)
– Loop rule: Sum of voltage changes around a loop = 0
– Junction rule: Current into a junction = current out of that junction
14
Circuit Paradigm
Circuit Paradigm
R1 = 2 Ω
I2
2Ω
R2 = 1 Ω
1Ω
1Ω
I3 R3 = 1 Ω
I1
6V
6V
1) V across R1 = 6V
so
I2 = V1 / R1 = 6 V / 2 Ω = 3 A
2) V across R2 and R3 = 6V, so
6 V = V2 + V3 = I3 R2 + I3 R3 = I3 ( R2 + R3 ) = I3 ( 2 Ω)
I3 = 6 V / 2 Ω = 3 A
3) I1 = I2 + I3 = 3 A + 3 A = 6 A
Circuit Paradigm
Circuit Paradigm
2Ω
I2
1Ω
I3
• Power dissipated in R1
1Ω
– P = I V = (3 A) (6 V) = 18 W
– P = I2 R = (3 A)2 (2 Ω) = 18 W
– P = V2 / R = (6 V)2 / (2 Ω) = 18 W
I1
Quick solution:
2Ω
1Ω
6V
2Ω
1Ω
1Ω
• Similar expressions for R2 and R3.
2Ω
6V
6V
6V
I1 = 6 V / 1 Ω = 6 A.
Because the equivalent resistance of I2 is the same as I3,
I2 = I3 = 3 A.
V2 = V3 = 3 V.
– P2 = (3 A) (3 V) = 9 W
– P3 = (3 A) (3 V) = 9 W
15
Paradigms we will use
• Charge in Capacitor
(Electric Forces)
• Water Tank (Fluids)
• Wave (Waves and
Sound)
• Ball hitting wall (Optics –
reflection)
• Cart going into sand
(Optics – refraction)
• 14C (Radioactivity and
Half-life)
• Block on Inclined Plane
(Energy Conservation)
• Porsche (Power)
• Braking Car (Kinematics)
• Ball on Inclined Plane
(Rotation)
• Colliding Blocks
(Momentum
Conservation)
• Circuit (Resistance,
Current and Voltage)
Charge in Capacitor Paradigm
Electric Forces
• Coulomb’s Law:
– Opposites attract, likes
repel
– Force proportional to
charge
– Inverse-square law, like
gravity
• Electric Field:
– Force per unit charge
• Electric Potential:
– Work per unit charge
d
k q1q2
r2
F  qE
V  E d
Charge in Capacitor Paradigm
• Possible questions:
∆V
F
∆V
d
+q
E
– What is the force on the charge?
+
q
E
• F=qE
• E = ∆V / d
– How much work is done in moving a charge
from one plate to the other?
• W = q ∆V = q E d
16