Physics FAQs and Topic Writeups

Shrike · Jun 15, 2005

This thread is a reference work, not a discussion. It comprises answers to Frequently Asked Questions (FAQs) about the physics portion of the MCAT, and write-ups of topics that give many students trouble on the test. Some of the posts are detailed, because the topics are non-trivial. Those in-depth posts are intended to guide your studying, not to substitute for other materials. We hope that this thread becomes a permanent resource for the SDN community.

It will be some time before this thread is completed. If you want to suggest a topic for inclusion (which request I will happily consider), please PM me. Meanwhile, readers and contributors should expect stuff to move around on a daily basis while construction continues.

Please do not post in this thread without permission. If you have permission, fire away, but be willing to have your posts edited, for format and content.

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics and economics, minored in physics, and spent several years accumulating unused school experience (in law).

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Table of Contents

Post 13: Buoyancy: Solving Problems
Post 03: Conservation Laws
Post 14: Conversion Units and Dimensional Analysis
Post 16: Conversion Units II
Post 17: Coulomb's Law
Post 07: How to Read Physical Science Passages
Post 08: Newton's Third Law
Post 02: Optics Problems: How to Solve Them
Post 12: Physics Constants Necessary for the MCAT and OAT
Post 09: Physics Equations
Post 10: Physics Equations II
Post 11: Physics Equations III
Post 06: Right Hand Rule for Magnetic Fields
Post 15: Simple Machines
Post 05: Tension: Ropes, Cables, Wires, Strings
Post 04: Waves: Transverse and Longitudinal

Shrike · Jun 16, 2005

The easiest way to solve MCAT optics problems -- all of them -- is to know three equations (which must be memorized):

1/o + 1/i = 1/f
m = -i/o
p = 1/f

... nine easy definitions (try them; you already know most of them):

o = distance (from the lens/mirror) to the object
i = distance to the image
f = focal length (sometimes hiding in the problem, as Radius of Curvature/2)
m = magnification (which, oddly, includes being right side up and upside down)
p = power
converging = bringing together
diverging = spreading apart
real = light goes there
virtual = light does not go there

... three things you already know about the world, for translating MCAT problems and figures into the proper form:

magnifying glasses, which are convex, make light come together
mirrors are different from lenses (they have the opposite effect on light)
concave is different from convex (it has the opposite effect on light)

... four easy, but perhaps novel, principles for setting up and interpreting the equations, based on the idea that positive is good:

object o --> always + (because it's always real, and real is good)
image i --> real is + (because real is good)
focal length f --> converging is + (because it's good to come together)
magnification m --> upright is + (because it's good to be upright)

... and one sensible method:

to find the effect of mutiple lenses or mirrors, add their powers to get the power of the combination.

Shrike · Jun 16, 2005

In theory, momentum is always conserved, regardless of what happens. On MCAT problems, momentum is conserved if the system, i.e., all the objects that are mentioned, includes everything that exerts force on anything else. The most common trouble here is the Earth -- if gravity causes objects to fall, but the Earth is not mentioned in the problem, then AAMC would claim, confusingly, that momentum is not conserved.

Momentum and its conservation is usually mentioned in collision problems.
Everything listed above for momentum is also true of angular momentum: it's always conserved in theory, but might not be according to AAMC if something is missing from the system. Angular momentum appears infrequently on the MCAT.
Energy is not conserved, except in special cases: it is conserved during perfectly elastic collisions (you know that it's perfectly elastic because they tell you so), and when the only forces involved in a problem are "conservative."

Conservative forces on the MCAT are: gravity, electrostatic, springs; also, any physical pushing or pulling that doesn't involve dynamic friction. Dynamic friction, electrical resistance, radioactive decay, and any collision that is not perfectly elastic, are not conservative. If in doubt, it's not.
Mass is conserved on the MCAT, except possibly in radioactive decay. If anyone has seen an exception to this principle in AAMC materials or tests, please advise.
Net charge is conserved, always.
The number of baryons (for us, protons and neutrons) is conserved, always. On the MCAT, we can just add them because we don't encounter antiprotons.

Shrike · Jun 17, 2005

There are two kinds of waves, longitudinal and transverse. Longitudinal waves are waves that vibrate in the same direction that they travel (propagate). Transverse waves vibrate across their direction of travel.

Every physical material -- solid, liquid, and gas -- can conduct a longitudinal wave -- that's a compression wave, also known as sound (though often we call it sound only if it's of a frequency near the range of human hearing) -- in any direction. There are no longitudinal waves other than compression, i.e., sound, waves that matter on the MCAT.

Compression waves propagate in all directions through their media (i.e., sound diffuses).

Every physical material with a surface (i.e., every solid object, and every liquid/gas and liquid/liquid interface) can conduct a transverse wave in any direction along the surface, with displacement normal to the surface. Note that a thin wire is a special case, essentially a surface with normals pointing out of the wire in all directions, so it can vibrate in any direction perpendicular to it.

Though there have been MCAT passages with waves traveling in multiple directions on surfaces, these were unusual; in almost all problems, there is only one wave. Waves on surfaces, and especially liquid surfaces, can be very complex to model, so they don't appear very often.

Other waves are possible, including transverse waves far from the surface of objects. Though these waves might appear, their precise properties shouldn't matter on the MCAT.

In addition to waves in substances, light can be modeled as a transverse wave. Never mind what it's a wave of.

Shrike · Jun 17, 2005

On the MCAT:

Ropes, cables, wires, and strings are just ways to apply forces (except when they're vibrating...).
The force on a rope etc. is always applied along the line of the rope, and it is the same in both directions. It is equal to, by definition, the tension on the rope.
The amount of force, i.e., the tension, is the same throughout the rope. In particular, it is the same on each side of a pulley.

Shrike · Jun 17, 2005

This is everything you need to know about the Right Hand Rule (RHR), or rather the two versions thereof. One or the other of them applies any time you are finding a direction (whether of field, force, acceleration, or even something else), and charges are moving.

I detail the flat hand method -- no sticking fingers in different directions perpendicular to each other.

There are two related right hand rules. Let's say: Right Hand Rule 1 (RHR1) -- used to determine the direction a charged particle, moving in a magnetic field, will be pushed; Right Hand Rule 2 (RHR2) -- used to determine the direction of a magnetic field created by the movement of charges.

There are three parts of your hand to remember, and I try to make it is easy as possible:

Fingers = Field: stick your fingers (of your right hand) in the direction of the magnetic field; if you aren't told the field (i.e., you are doing RHR2), the fingers are eventually going to tell you its direction. In other words, fingers = the first field you encounter in the problem. Remember, this is only for magnetic fields; electric fields have nothing to do with magnetism problems.
Thumb = hitchhike, in the direction of motion. Motion of what? Of the thing in the problem that's moving (the charged particle or particles; if there's just a current given, not particles, then that current would be the movement, by convention in the positive direction even though it's usually the movement of negative particles in the opposite direction). In other words, the first motion you encounter in the problem. (Note: the direction it's going now, not the direction of its acceleration.)
Palm = Push. Your palm will face in the direction that a positively-charged particle will be pushed by the magnetic field. For RHR1 problems, this is usually what the problem asked you to figure out. For RHR2 problems, it doesn't really apply but it still works: a positively-charged particle moving in the same direction as the current will be pushed toward the wire by the magnetic field.

So, if asked to figure out the direction a particle is pushed, stick your fingers in the direction of the magnetic field, and hitchhike in the direction the particle is moving. Your palm is now trying to push the charge, in the appropriate direction. You just used RHR1.

If asked to make a magnetic field, hitchhike in the direction of the moving charges (or the current); your fingers, which now are allowed to curl, show the direction of the field, which curls around the path of the charges. You just used RHR2.

Related issues:

The magnetic field does zero work, always. This is because it is always perpendicular to motion, turning the particle rather than speeding it up or slowing it down. That's why we can kind of throw it in there at the end of electrostatics: it has no effect on the energy of anything.
It is possible to use the analogous left hand rule to solve problems for negative charges moving in magnetic fields, but I recommend against it. Use RHR, of whichever type, and for negative charges or currents the answer is the opposite direction.
Electric fields do not create magnetic fields; these two are unrelated for your purposes. You'll know this if you try to apply RHR inappropriately -- you can't stick your fingers in two different directions.

Shrike · Jun 17, 2005

None. Or very close to none.

Time is of the essence on the Physical Sciences section; nearly everyone needs to finsh the section, but for most people that isn't going to happen if they spend too much time reading unnecessary material. And it turns out that most of the text in physics passage is unnecessary for answering the questions.

When you first turn to a physics passage, it is fine to glance at the first few lines to see what it's about, though even this isn't necessary. What you must do is check out every picture, table, graph, and block-quoted (offset) equation. You have a particular task with each of these items:

Pictures: Often these show an experimental setup, or the system being studied. Examine it to see what's going on. If it's confusing, do not spend too much time here; sometimes, you don't really need to understand the entire experiment to answer all of the questions.
Tables: These represent the most important part of many passages, because they show experimental results and AAMC loves experiments. For the moment (before you've read the questions), you have two tasks: (1) understand what the table is talking about, i.e., what experiment was performed -- you may have to read a few lines of text above the table in order to do this; and (2) see what was measured -- in other words, read and understand the column headings.
Graphs: Do not analyze the graphs, just see what they are graphs of, by checking the labels on the axes and by reading a couple of lines of text above them if that's necessary.
Equations: Do not attempt to understand any equation that is offset from the rest of the text, but do see what it's an equation for; again, this may necessitate reading a bit of text.

You then proceed to the questions, referring back to the pictures, tables, graphs, and equations as necessary.

About one question per passage, maybe not even that many, will require reading some more of the text, but by now you'll have a good idea where to look and what you're looking for, so you'll be able to find it quickly. Many people worry about these questions, but they're rarely a problem. If you have practiced enough passages and studied the right material, you'll know immediately when they are asking you something that requires reference to the passage, because it's not amazingly simple but it's also not something that you saw in the pictures, tables, graphs, or equations. For such not-so-simple questions, look in the passage.

You might think it would make sense to read the passage first if you're going to have to go back to it later. Not true. For one thing, some passages have no such questions. For another, for most test-takers reading the passage is a time sink. If you had the discipline to limit yourself to a quick skimming it would be fine, but under testing conditions most people hate to miss anything, so skimming becomes reading, and reading becomes time expiring before you're done or having to spend insufficient time on questions to compensate.

This method will be uncomfortable at first, and maybe forever. To those who are still reading the passages because they feel more comfortable when they do that, ask yourself whether the discomfiture is actually costing you points, and whether the time lost is worth it. If you're not 100 percent sure, test it both ways, but only after you've practiced the new method enough to get good at it. Remember, the point is getting questions right in the time allotted, not liking it.

xanthines · Jun 17, 2005

In the classes I teach, a disconcerting number of students don't really understand Newton's third law of motion and get confused by the statement:

"For every action, there is an equal and opposite reaction."

In fact, the law would be more accurately expressed this way:

"For every force, there is an equal and opposite force."

Or, in mathematical terms:

"F1 on 2 = -F2 on 1"

For a book lying on a table (i.e., stationary), it is commonly thought that Newton's 3rd Law is when the Normal Force and the weight of the book (m*g) cancel each other out and therefore the book goes nowhere. Actually, these two forces are not related by the Third Law. Given a (gravitational) force between the Earth and the book, the Third Law implies a force between the book and Earth, not the not the book and the table. Consider the math:

FG = G(MEarth)(Mbook)/ r^2

You usually learn that acceleration due to gravity, g, is

g = GMEarth/r^2

and the force due to gravity, aka weight, is mg. Newton's 3rd Law states that the book must exert the same amount of force on the Earth... and it does!

acceleration (a) of due to the book = G(Mbook)/r^2.

Since in each case you're using masses of the book and earth for accelerations, the forces come out the same when you multiply by the mass of the other body. Forces are masses * accelerations, which means you will arrive at the same force whether or not you're using g or the acceleration due to the book. We've gotten used to just thinking about acceleration as g (which is immensely useful, no doubt), but it puts up blinders when thinking about forces, most notably weight.

Using the gravitational equation, you can figure out the force on the book due to the earth, as well as the force on the Earth due to the book. Each time, you will find the magnitude to be the same (although different directions). Using mg incorporates the mass of the book (or whatever) into the gravitational equation. Because of this, whenever you figure out the forces on the book or on the Earth, you end up using all the same variables. Try it out! This is what Newton meant by the Third Law.

-X

psiyung · Jun 17, 2005

These are some common Physics equations that will probably need to be memorized and understood for the MCAT

vectors

Pythagorean Theorem: a^2 + b^2 = c^2
**Remember the common 3-4-5 and 5-12-13 triangles**

sin(theta) = (opposite / hypotenuse)
cos(theta) = (adjacent / hypotenuse)
tan(theta) = (opposite / adjacent)

Distance, Displacement, Velocity, and Acceleration

Displacement d = change in position
Speed = (distance) / (time)

Velocity v = (displacement) / (time) -- this is a vector
**Note that displacement is the NET distance from the starting point**

Acceleration a = (change in velocity) / (time) -- this is a vector
**Because acceleration is a vector, a change in velocity or a change in direction means the object has accelerated**

Linear Motion

v = v0 + at

d = v0t + 0.5at^2

v^2 = V0^2 +2ad

vavg = 0.5(v + v0)

**Note that these equations are under the assumption that the objects are moving with constant acceleration**

Objects shot into projectile motion

dx = v0xt

V0x = V0xcos(theta) V0y = V0sin(theta)

ax = 0 ay = -10 m/s^2 (acceleration due to gravity)

Peak height of an object: V0 x sin(theta) = sqrt(2gh)

Range of object = Vicos(theta) x (time)

Newtons Law and the Law of Universal Gravitation

Force = mass x acceleration or F = ma

F = (Gm1m2) / r^2 where G = 6.67 x 10^-11 m^3 kg^-1 s^-2 (do not memorize this constant!)
**m1 and m2 are the masses of the objects while r is the distance between the center of masses of the two objects**

**Note that both objects described feel forces of the same magnitude**

**Note that combind these two equations we get**:
F = (Gm1m2) / r^2 = ma --> from here we can find the acceleration of either object just by plugging in the mass of the other object

Inclined Planes

F = mgsin(theta)
**this is the force causing an object to move up or down the incline plane**

Fn = mgcos(theta)
**this is the force being applied on the object that continuously changes the direction of velocity by creating a continuous centripetal acceleration**

Circular Motion

Centripetal Acceleration = ac = (v^2) / (r) *r = radius of circle*
**Note that this acceleration ALWAYS points toward the center of a circle**

Centripetal Force = F(c) = mv^2 / r

Frictional Forces

**Static friction -- Force opposing motion when two objects are not moving relative to each other**

f(s) = mus x Fn where mu(s) is the fraction of the Normal Force (Fn)

**Kinetic Friction -- Force opposing motion when two objects are moving relative to each other**

f(k) = mu(k) x F

**mu(s) > mu(k)**

Hooke's Law

F = -k(x)
k is a constant for a particular object
x is the deformation or the change in position of the object

Equilibrium

Means no linear or angular acceleration applied the an object (object moves and rotates at a constant velocity or zero velocity). The sum of all forces acting on such objects is zero

F(upward) = F(downward)
F(rightward) = F(leftward)

psiyung · Jun 17, 2005

Torque
Tau = (force) x (moment arm)

Tau(clockwise) = Tae (counterclockwise) for static problems

Center of mass = x = (m1x1+m2x2+....) / (m1+m2+....)
**Note the equation is dependent on the number of objects**

Energy

Kinetic Energy = K = .5(mv^2)

Gravitational Potential Energy = U(g) = mgh
**this equation is used for objects near the earth's surface**

Elastic Potential Energy = U(e) = .5(k)(x^2)
**this shows the potential energy for objects following Hookes Law**

Work

W = Fdcos(theta)
**Work is done by all forces except that of FRICTION**

W = (delta)K + (delta)U
**this equations can also be written:
.5(mv(f)^2) + mgh(f) = .5(mv(i)^2) + mgh(i)**

Power
P = (work) / (time) -- measured in watts
P = Fdcos(theta) / (time)
P = Fvcos(theta)
**sometimes power is defined by transferred energy (E) which is Work + heat(q)** P = (delta)E / (time)

momentum

p = mv -- mass and velocity of an object
**Momentum is always conserved**

Collisions

Elastic collisions: U(i) + K(i) = U(f) + K(f)
** Mechanical energies before and after collision are equal

Inelastic collision: Use the conservation of momentum to solve inelastic collision problems since mechanical energy is converted internal energy
**p(i) = p(f)

Impulse

Is the change in momentum: J = (delta)p

F(avg) = m(vf-vi) / (change in time)

mass defect
Nucleus of an atom has less mass than the sum of its individual parts. The difference of the mass is the mass defect. Binding energy holding protons and neutrons can be found using the defect

E = mc^2 where m is the mass defect

Density of and Specific Gravity

Density = mass / volume (kg/m^3)

S.G. = (density of a particular substance) / (density of water)

Fluid Pressure on objects

P = Force / area

P = (density)(g)

**where g is gravity and density and y is the density and depth of the fluid the object can be found respectively**

if a fluid is open to the atmosphere, the pressure is:
**Pressure = (density)(g)

+ P(atm)

Pascal's Principle

Any change in pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls
**Principal seen in the Hydraulic Lift where: F(1)/A(2) = F(2)/A(2)

Archimedes Principle
Any fluid applies a buoyant force to an object that is partially or completely submerged in it -- this force is equal to the weight of the fluid that the object has displaced.

**F(b) = W(fluid)** -- can also be written

**F(b) = (density of fluid)(g)(Volume of fluid displaced)

Ideal Fluids

The equation of continuity: The mass flow rate of a fluid has the same value at every position along a tube that has a single entry and a single exit point for fluid flow --> for two positions along a tube:
**(density)(A1)(v1) = (density)(A2)(v2)**
A= cross sectional area of tube
v = fluid velocity

Since density is constant in an imcompressible fluid, it does not change during flow: A1v1 = A2v2
volume flow rate = Q = Av

Bernoulli's Equation
In a steady flow of nonviscous, imcompressible fluid of a cetain density, the pressure, the speed, and the elevation at two points are related by the following equation:

P1 + .5(density)(v1^2) + (density)(g)(y1) = P2 + .5(density)(v2^2) + (density)(g)(y2)

psiyung · Jun 17, 2005

Solids -- Included for completeness; DO NOT LEARN THESE! [according to Shrike]

Stress = (Force) / (area) -- Stress is what is done to an object
**The maximum stress applied to an object that allows the object to regain its original dimensions is called the yield point -- beyond this point, an object will not regain its shape**

Strain is the fractional change in the objects shape
**it equals the (change in dimension) / (original dimension)**

Always note that stress is proportional to strain:

The modulus of elasticity for a certain objects = (stress) / (strain)

Young's Modulus - deals with length changes in objects with an applied force
**F = (Y)(change in length)(Area) / (Original Length)**

Shear Modulus -- deals with shear deformation of objects with an applied force
**F = (S)(amount of shear)(Area) / (original length)**

Bulk Modulus -- deals with volume changes of objects by an applied pressure
** P = (-B)(change in volume) / (original volume)**
**Note that the negative sign denotes that anytime there is an increase in pressure, there is a decrease on volume**

Waves

V(wave) = (frequency)(wavelength)
**frequency is measured in hertz or cycles/second**

The period of a wave is the time required for one up and down cycle: On an x-axis of time, it is the point from any point of the wave function, to that exact point where the wave function has repeated itself
**T = (1/frequency)**

Intensity

Is measured in decibels -- β = 10log(I/Ii)
**Note that for every factor of 10 that intensity increases, the decibels increase an additional 10 decibels**

beats
Beats occur when two waves of slightly different frequencies are superimposed. The beat frequncy produced will be the difference between the frequencies of the two original waves:

f(beat) = |f1 - f2|

Doppler Effect

fo = fs [(1)/(1 - v(s)/v)]
**This is for a source (s) moving towards a stationary observer**
**fo stands for the frequency head by the observer**
**v(s) is the speed of object the source is coming from while v is the speed of sounds (340 m/s)**

fo = fs [(1) / (1+v(s)/v) ]
**for a source moving away from an observer**

Electricity

F = (k)(q1)(q2) / ( r^2)
**Coulomb's Law -- remember that opposite charges attract each other and like charges repel each other**

Electric Field's due to a point charge

The electric field that exists at a point in space is the electrostatic force felt by a test charge at the point divided by the charge itself

E = (force) / (test charge)
**if we plug the Coulumbs equations we get E = k(q1)/(r^2)**
**test charge also shown as qo

Constant electrical fields

(force) = (qo)(E)

The electrical potential V at any given point is the electrical potential energy of a test charge qo, divided by the charge itself:

V = (EPE) / (qo)
**EPE (electrical potential energy) = (E)(q)(d)**
**plugging in these two equations we get V = Ed -- note that Voltage is the potential for work by an electric field to move any charge from one point to another**

Volatge by a point charge:
V = (k)(q1) / (r)

circuits

Voltage = IR (Ohm's Law)
R = resistance
I = current

Capacitance = Q / V Q = CV
Q = charge
V = voltage
**Capacitance is the ability to store charge per unit voltage**

The energy stored within a capacitor can be found using these equations:
U = .5(QV)
U = .5(C)(V^2)
U = .5(Q^2) / (C)

Resistors

Total resistance for resistors in series: R(eff) = R1 + R2 + R3 .......

Total resistance for resistance in parallel = 1/R(eff) = 1/R1 + 1/R2 + 1/R3......

Capacitors

Total capacitance of capacitors in series: 1/C(eff) = 1/C1 + 1/C2 + 1/C3 ...

Total capacitance of capacitors in parallel: C(eff) = C1 + C2 + C3 .....

Electric Power

P = IV
P = I^2(R)
P = V^2/R

magnetism

F = qvBsin(theta)
F = force on the charge moving through the magnetic field
q = charge
B = magnetic field
**Note that the force is directed perpendicular to both the velocity and the magnetic field**

Alternating Circuits

V(max) = sqrt(2) rms
I(max) = sqrt(2) rms

Light and Optics

Index of refraction

n = c/v
**Compares the speed of light in a vacuum with the speed of light through a given medium**

angle of refraction -- Snell's Law

n1sin(theta) = n2sin(theta)

E = hf
**this equation shows the higher the freqeuncies of light have more energy**

Mirrors and lenses -- learn the optics laws.

Shrike · Jun 18, 2005

Not many. And those you need, you do not need with much precision at all.

pi = 3 (Not kidding; this should be good enough. 3.1 in a pinch)
sin(30) = 0.5; cos(30) = 0.9
sin(45) = 0.7; cos(45) = 0.7
sin(60) = 0.9; cos(60) = 0.5
the speed of light in a vacuum (and air): c = 3 x 10^8m/s
acceleration due to gravity near the surface of the Earth: g = 10 m/s^2
the density of water, in the following units: 1000kg/m^3 = 1g/cm^3

Also useful:

the speed of sound in air = 340 m/s
1 atmosphere = 100,000Pa = 10m of water = 760mmHg
the density of water, in the following units: 1kg/L

Shrike · Jun 18, 2005

First, you must understand this principle: the buoyant force, which is directed upward, is equal by definition to the weight of the fluid displaced.

Problem type 1: the object is floating, in equilibrium, on the surface of a fluid.

If the density of the fluid above the surface can be neglected (i.e., it's much lighter than the fluid below, usually a gas above a liquid), then you need to understand two things:

The buoyant force, which is equal to the weight, mg, of the fluid displaced, also equals the weight of the object (because the object is in equilibrium); it follows that the mass of the object is exactly equal to the mass of the fluid displaced.
The portion of the object that is below the fluid equals the ratio of the object's density to that of the fluid.

The second principle follows from the first. Notably, it means that the way an object floats does not depend on how strong gravity is.

If the density of the fluid above the surface cannot be neglected (i.e., it's a liquid or a very dense gas), then the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface. In practice, subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object. (If you're finding how much is below the surface of the heavier fluid, subtract the density of the lighter one from both the heavier fluid and the object, and find the ratio as above.) An example shows it better:

Question: An object is floating at the surface of two fluids; two thirds of its volume is below the surface. If the densities of the two fluids are 1.0 g/cm^3 and 1.6g/cm^3, what is the density of the object?

Answer: Subtract the density of the lighter fluid from that of the heavier, and we get 0.6 g/cm^3. Two thirds of the object is below the surface, so that yields a density of 2/3 * 0.6 = 0.4 g/cm^3. Then we need to add the density of the first fluid back in: 0.4 + 1.0 = 1.4 g/cm^3.

Problem type 2: the object is below the surface of the fluid.

The easiest way to do these problems quickly is to do everything in terms of water: figure out what the buoyant force would be if the fluid were water, and then convert at the end. You need to have the density of water memorized in three different units: kg/m^3, g/cm^3, and kg/L. Note that a liter (L) is equal to 1000 cc's; many MCAT objects are most easily considered in these units.

To solve the problem, you will want to find the buoyant force, which is the weight of the fluid dispaced. To do this, you find the mass of the fluid displaced, and at first you pretend it's water. After finding the buoyant force, you subtract from the weight of the object to find its apparent weight; use that to get acceleration if needed. An example:

Question: A 20 cm x 40 cm x 5 cm object of density 0.4g/cm^3 is submerged with specific gravity of 1.2g/cm^3. Find the apparent weight of the object, and its acceleration.

Answer: The object's volume is 4000cm^3, or 4L. If the fluid were water, the mass displaced would therefore be 4kg (because the density of water is 1kg/L), so the buoyant force would be 4kg x g = 40N. This fluid is 1.2 times as dense as water, so the buoyant force is 1.2 x 40N = 48N.

The object is 0.4 times as dense as water, so its weight is 40N x 0.4 = 16N (and its mass is therefore 1.6 kg; we use that below). (Note that again we do everything in terms of water, because it's the easiest way. Note also that the volume of the object is, by definition, the same as that of the fluid displaced.) The apparent weight is therefore 16N - 48N = -32N. The negative weight means the net force on the object is upward; this makes sense because it's lighter than the fluid, i.e., it wants to float.

To find the object's acceleration, use F = ma --> a = F/m = 32N/1.6kg = 20m/s^2.

The reliance on water makes many calculations easier to do quickly, and also provides a running sanity check -- you know it immediately when you are getting a nonsensical result -- but it actually doesn't change the method in a technical sense.

Shrike · Jun 19, 2005

The only conversion you will have to do is of prefixes, e.g., micro-whatsis to kilo-whatsis. No kilograms to slugs, or liters to hogsheads, or meters to chains, or...

For those who have trouble converting units, I suggest the following method:

First, forget the conversion factors listed in whatever reference material you're using. They're correct, of course, but if you're having trouble they're what're mixing you up. Instead, just think in terms of big and small, as shown here. Also, eschew multiplying by negative exponents -- divide instead. Finally, think of the units as variables, and cancel as you would with variables.

An example problem will show the method best. Lets say you want to know the speed of light, c = 3x10^8 m/s, but in cm/ms.

Write down 3x10^8 m/s on the left. Write down the units you want, cm/ms, on the right. Now multiple and divide to get rid of the m and the s, and to get cm and ms in the right spots:

Code:

3x10^8 [U]m[/U] x _____[U]cm[/U] x _____[U]s[/U]_ = _______[U]cm[/U]
       s        m         ms          ms

Now fill in the numbers, which you should know, just focusing on which unit is bigger, and remembering that it takes more of the smaller ones to make a big one:

Code:

3x10^8 [U]m[/U] x [U]10^2[/U]_[U]cm[/U] x ___[U]1[/U]_[U]s[/U]_ = _______[U]cm[/U]
       s      1 m    10^3 ms          ms

Check to ensure that the old units cancel out, that you wind up with the new units, that you have more of the smaller units in each fraction (the fractions must each equal 1); now do the math:

Code:

3x10^8 [U]m[/U] x [U]10^2[/U]_[U]cm[/U] x ___[U]1[/U]_[U]s[/U]_ = [U]3x10^7[/U]_[U]cm[/U]
       s      1 m    10^3 ms          ms

Finally, never use a squared or cubed conversion factor: use the normal conversion factor, and square or cube it at the end. As another example, find the density of osmium in kg/m^3:

Code:

[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x (____[U]cm[/U])^3 = _______[U]kg[/U]_
     cm^3        g    (    m )^3          m^3

Code:

[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x ([U]100[/U]_[U]cm[/U])^3 = [U]3x10^7[/U]_[U]kg[/U]_
     cm^3   1000 g    (  1 m )^3          m^3

Code:

[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x [U]10^6[/U]_[U]cm^3[/U] = [U]2.24x10^4[/U]_[U]kg[/U]_
     cm^3   1000 g       1 m^3              m^3

Notice that we filled in the cm/m conversion, which we know well, and squared at the end. Notice also that there's no need to use scientific notation for many numbers -- the notation is a tool, not an end in itself.

If this method doesn't solve your conversion problems, I'll be surprised.

Shrike · Jun 22, 2005

Recent MCATs have included several questions about simple machines -- devices that alter the strength of the input force and the distance through which force is applied. You should be familiar with the force multiplication concept, and somewhat familiar with some of the machines that use it.

The basic concept is this: you can change force, but you can't get something for nothing, i.e., energy is not created or destroyed by any of these machines. The energy of a force is, more properly stated, the work done by that force; it follows that the work done by an input force is the same as that done on the object that is being moved (unless there's friction, in which case energy is lost; we'll neglect friction for the rest of this discussion). As work = force x distance x cos(theta), if we increase force we have to decrease distance equivalently, and vice versa. (For this discussion we'll consider forces that do work in the direction in which they are applied, i.e., cos(theta) = 1, and W = Fd.) The amount by which force is increased by a system, and therefore distance decreased, is the force multiplier.

Because of the fixed relationship between force and distance, it is often easier to consider distance (which we can see) than force, and then work backwards. For example, if you can see that the thing you're moving travels, say, three times farther than the force that moves it, then the force on the object is three times less than the input; if you can see that it travels only a quarter as far, then the force exerted on it is four times as great, and so on.

The simple machines (and other devices with force multipliers) follow, including all six classic simple machines (even those that aren't likely to appear in the test; I note which these are). Do not memorize these; instead, be moderately familiar with them, especially the more common ones.

Pulley: A single pulley or a system thereof is not necessarily a simple machine, but it can be. For MCAT-style pulley systems, output force is input force multiplied by the number of ropes that are effectively pulling on the object to be moved; equivalently, by the number of ropes that must (not just might) shorten in order for the resistance to move. Another hint: if none of the pulleys is free to move, then the force multiplier is 1: force and distance remain the same. Of course, if the force multiplier is not 1, i.e., you get greater force on the object than you put in, you will get less distance by an equivalent factor. Appears on the MCAT.
Inclined plane: the force multiplier (relative to straight lifting) is the sine of the angle of the plane with the horizontal. Distance is divided by the same factor. Consider the problem another way: If you push a block up an incline, you push along the hypotenuse, but lift only the amount of its height; the ratio of these two is sin(theta) Because distance is less than what you input, force must be greater by the same factor -- the force multiplier. Common on the MCAT.
Lever: the force multiplier is the ratio of distance from fulcrum to object, to distance from fulcrum to applied force. (The "fulcrum" of a lever is its pivot point.) Look at the picture to make sure you know whether distance goes up or down (hence force down or up). This treatment works even when the force and the object are on the same side of the fulcrum. I saw a lever -- a claw hammer pulling a nail -- and a question about its force multiplier and another about its fulcrum, on the August 2004 MCAT, so you can't ignore this one.
Hydraulic jack: not usually considered to be a simple machine like the aforementioned, but effectively the same. Force multiplier is the ratio of the areas of the tops of the two pistons. Recall that force changes, not pressure -- pressure is the same (at the same height) everywhere in a body of liquid. Assuming that the change in height is negligible, it also doesn't matter what liquid is used. Appears on the MCAT.
Wedge: similar to an inclined plane; the force multiplier is length divided by width; distance is divided by the same factor. Unlikely on the MCAT.
Wheel and axle: the force multiplier is ratio of radii of the wheel and the axle. To figure whether to divide or multiply the two, look at the picture to see whether the distance the object travels is greater or less than the distance over which the force is applied; that will tell you whether the force is decreased or increased, respectively. Unlikely on the MCAT.
Gears (technically not a separate type of simple machine, but they work the same as the others): the force multiplier is the ratio of the numbers of teeth on the two wheels. Again, look at the picture to see which increases, force or distance. Unlikely on the MCAT.
Screw: an inclined plane wrapped around a rod. Unlikely on the MCAT. For completeness: the force multiplier is 2 x pi x radius x (turns/unit length). Don't worry about this one at all.

It is possible to combine two or more of the types: for example, a cam is an inclined plane combined with a wheel and axle. But don't worry: such a problem is very unlikely to appear on the MCAT, and if it did it would necessarily be a very simple setup.

psiyung · Jul 1, 2005

Some people still have problems converting cm^3 and mm^3 to m^3 for problems such as density and volume.

1) When converting from cm^3 to m^3 move 6 decimal places to the left. For example if the volume of a box is 120 cm^3, then converting it, we will get .000120 m^3 or 1.2 x 10^-4 m^3

2) When converting from mm^3 to m^3 move 9 decimal places to the left. For example, if the volume is 120 mm^3, the conveting it, we will get .000000120 m^3.

**REASONING**

1 m^3 = 1m x 1m x 1m = 100 cm x 100 cm x 100cm = 1000mm x 1000mm x 1000 mm

QofQuimica · Aug 12, 2005

Coulomb's Law describes what happens when two charges are placed close to one another. The magnitude of the force between the two charges, called the electrostatic force, depends on two variables: the magnitudes of the two charges, and the distance between the two charges. Whether the force is attractive or repulsive depends on the signs of the two charges: like charges will repel, and unlike charges will attract. Mathematically, two charges that repel will have a positive value for the electrostatic force, while two charges that attract will have a negative value for the electrostatic force.

Dependence on the Magnitudes of the Charges:
The electrostatic force is directly dependent on the magnitude of each of the two charges. If one charge's magnitude is doubled, the electrostatic force between them will be doubled. If the second charge's magnitude is doubled, again, this will double the force between them. If the magnitudes of both charges are doubled, the force between them will be quadrupled.

F

q1
F

q2

Dependence on the Distance Between the Charges:
The electrostatic force is indirectly dependent on the square of the distance between the two charges. This is why Coulomb's Law is known as an inverse square law, like Newton's Law of Universal Gravitation. If the distance between the two charges is doubled, the force is cut by one-fourth. If the distance between the two charges is halved, the force between them is quadrupled.

F

1/(r^2)

Putting all of these variables together with a permittivity constant gives us Coulomb's Law:

F = k(q1)(q2)/(r^2)

You should understand and memorize Coulomb's Law, but you do NOT need to memorize the value of k for the MCAT.

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QofQuimica

Seriously, dude, I think you're overreacting....

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