Translational motion test

Hi I have questions on the physics Translation Motion Test: 

Please note that I answered some of these questions elsewhere in the blog, but I'll re-answer them here.

2. Terminal velocity? how is it we can assume A=0 and no net forces. ?? So every object has a terminal velocity where A=O = gf so it does not accelerate further???
Terminal velocity, by definition, is the point when the drag force is equal to the weight.  With two equal and opposite forces, that object will experience no acceleration, as perNewton's Second Law (F=ma).  Every object that has a drag force acting on it will eventually reach terminal velocity if it keeps falling.  Once it does, the acceleration is zero.  That doesn't mean it's not moving; it just means the velocity is constant.

4. q = (KLsquared) / (Kprime L cubed) = ((K) / K prime) (1/L)
UM don't get the creation of K and Kprime, are we making the values of the proportionality constant up? And why wouldn't they be the same
The invention of K as a proportionality constant is a mathematical convention, but you really don't need it here.  If q is surface area/volume, well, they tell us that surface area is proportional to L² and volume is proportional to L³.  So, q would be proportional to L²/L³, or 1/L.  Adding in a proportionality constant is just a convention.  From there, q-ant/q-human would be (1/L_ant)/(1/L_human).  Ugly fraction, so let’s invert and multiply:  q-ant/q-human = L_human/L_ant.  L_human it says is 1 m and L_ant is 0.5 cm, or 5x10^-3m.  L_human/L_ant is then 1 m/(5x10^-3 m) = 0.2x10³ = 200.  The ratio is therefore 200:1.

5. In Q5 we drop the assumed K we used in 4 - How/why/ When is it appropriate to create these assumptions? - then we assume the length of the daughter cell keeps the same L squared and L cubed relationships with out the proportionality constant....
Then they create the formula L squared = (2 raised the the 2/3) wavelength squared ( I know its not a wavelength but thats what the symbol looks like :) ) where the second portion of the equation = surface area of the original cell, since that is the case and its what we are starting with then HOW can you go forth and say 2wavelenght squared = surface area of daughter cells and NOT use what I thought it would be which is (2 raised to 2/3 ) times wavelength squared.   Then my inability to do math pops up (despite reviewing the math foundation)
2:2 raised 2/3 = (2) / (2 raised 2/3) = which apparently = (2)/(2 raised 2/3) times (2 raised to the negative 2/3) / (2 raised to the negative 2/3) = 2 raised to the 1/3 / 1 = which they then changed back into a ratio? = 2 raised 1/3 : 1
After this question I was ready to freak out. I just don't get it.

This is one of the most asked about questions in the whole Kaplan curriculum.  Here’s my explanation:

This question reads:  “Cells can only absorb nutrients through their surfaces.  In order for a cell to absorb nutrients most efficiently, it will grow to a maximum size and then divide.  If the total volume remains constant when a cell divides, then the ratio of the total surface area of the daughter cells to that of the original cell is:”

Let’s start with defining a few variables here.  According to our passage, “each animal has a characteristic length L…its area is proportional to L²…and its volume is proportional to L³.”

Let’s also define the subscript “1” to mean the cell before it’s divided, and “2” to represent each cell after it’s divided.

The total volume here is constant – the one cell before division has the same volume as adding together the two cells after division.  This gives us the equation that L₁³=2L₂³.  Let’s take the cube-root of both sides.  Well, the cube root is the same as taking each side to the 1/3 power.  So we would have:

(L₁³)^1/3=(2L₂³)^1/3.  When you have an exponent raised to an exponent, we multiply their values.  This means that this simplifies out to L₁=2^1/3L₂.  The 2^1/3 is because we have to distribute the exponent to both values in the parentheses.

From here, we can figure out the answer to our question.  It asks us to find the total surface area of the daughter cells, as compared to the area of the original cell, or

2L₂²:L₁².  Remember that there are TWO daughter cells, so that’s why we have the two here.  Well, we know how L₁ and L₂ are related, so we can say that

2L₂²:L₁² can be represented as 2L₂²:(2^1/3L₂)² – we just substituted in our value for L₁ from that volume calculation before.  Distributing our exponent on the right-hand side, we have 2L₂²:2^2/3L₂².  Let’s divide both sides by L₂².  That gives use 2:2^2/3.  But… that doesn’t quite look like any of the answers here.  Since most of the answers have 1 on the right-hand side, let’s divide both sides by 2^2/3.  That gives us:

[2/2^2/3]:[2^2/3/2^2/3].  When we divide two items that have the same root and different exponents, we simply subtract the exponents.  That means that we have [2^1-2/3]:1, or 2^1/3:1.  That corresponds to answer choice (C).  If you’re having difficulty with exponents, I encourage you to go on to Google and type in “Exponent Practice.”  You want to be able to work with exponents very easily on the MCAT.  The four big rules are:

(X^A)(X^B) = X^A+B

(X^A)/(X^B) = X^A-B

(X^A)^B = X^AB

nth root of X = X^1/n

6. How is it A=mg in the Y direction or is this only when calc  Force = (M) (-g) 

Acceleration will always be in the same direction as the net force on the object (remember F = ma).  So, the ONLY force on this object is gravity.  Since gravity points downwards, our acceleration is downwards.  Also, if that’s the only force, then F = mg = ma.  So a = g.  But we know that.  “g” is the acceleration due to gravity by definition.

14. I doubled the M in Ke = mv squared /2 but that was wrong since I did adjust for delta V with less mass BUt V = DELTA X / DELTA T
right? but when V initial = F delta T / M

This is what most people do at first with this question.  But you need to keep in mind when more than one variable may get affected.  If you have a heavier clown being pushed by the same force, he’ll move slower.  The method they give for answering this question – what you allude to at the end of your question here – I think is a bad way of approaching it. Where do they talk in the passage about the relationship between mass and velocity?  The graph.  Now, the y-axis on the graph is v_y and they’re asking in the question about v₀. But v_y and v₀ are proportional (it’s always the same angle that the clown gets fired at, so v_y/v₀ is a constant (it’s equal to sinθ here – opposite over hypotenuse).  So, let’s look at what happens when you double the mass.

Starting with the 40 kg line, it shows us that the initial velocity is 42 m/s.  If we double the mass to 80 kg, initial velocity is 21 m/s.  In other words, doubling the mass halves the velocity.  Since K = ½ mv², doubling the mass will cause m to double and v to get cut in half.  The net effect?  We get cut in half for our kinetic energy.