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Gen Phys 1: First class assignment and webassign information

Don’t forget to set up your webassign account.  The instructions I gave you in class can be found here.

Frequently asked questions regarding webassign and online homework:

There may be a glitch in assignment 1.2, problem 13.  If you encounter an issue, please email me.

Finally, I realized that I accidentally copied an older and somewhat less clear version of the “Mechanics Discovery” assignment at the front of your practice packet that I asked you to do for next class.  The content is generally the same, but the instructions are slightly less clear than the version linked in the unit page section of the website which you can find HERE.  Below, please find solutions worked out for that introductory assignment.
Mechanics Discovery–solutions

Also on the unit page section of the website, you will find the calculus notes from the first class, plus some practice problems involving some basic derivatives and integrals.  If you have never taken any calculus before this semester, or if you wish to brush up, you might want to do this practice.  We will being applying predominately power-rule derivatives and integrals as we dive into kinematics in our next classes.

Looking forward to our next meeting!


PreAP: Electrostatics 1 HW helps

Several folks have inquired about HW questions, particularly #’s 19-22.  These are certainly the more difficult problems on the HW assignment.  I will provide you with both some general information and some specific helps on each.

First, 19 & 20 are essentially the same.  19 has a North-South orientation, while 20 has an East-West orientation.  21 & 22 are similar except that you have more complicated 2-D vectors to consider.  Secondly, and perhaps the most important help I can give you: Ignore the signs when you use Coulomb’s Law.  Ignore the signs while you are doing your calculations and when you are done ask yourself, “What is the direction of this force?”  The remainder of the problem will depend on how you utilize that info.

Some background: If you keep the signs in the calculation when doing Coulomb’s Law Q1*Q2 give some interesting results.  If you have like charges (either + &+ or – & -) you will always end up with a positive force.  What does this mean?  Well, like charges repel, so a positive result must mean the charges push each other away.  Does the opposite hold?  Well if Q1 & Q2 have opposite signs (e.g., + & -) and you apply Coulomb’s Law your result will be negative.  This means an attractive force, or a force along a direction toward the second charge.

Let’s begin with the easier scenarios (19 & 20).

Three particles are placed in a line. The left particle has a charge of -67 µC, the middle, qm = +42 µC, and the right, qr = –87 µC. The middle particle is 72 cm from each of the others, as shown in Figure 20-13.

Figure 20-13.

(a) Find the net force on the middle particle.
Let’s get some labels here.  Call the Left particle A, middle B, and right C.

In order to find the force on B we only need to calculate the force of A on B and of C on B.  The force of A on C is irrelevant to this problem.

So Fnet = FAB + FCB

FAB = ?    B is pulled to the LEFT

FCB = ?     B is pulled to the RIGHT

Since these forces are in opposite directions you need to treat them so, so when you calculate the net or total force you must SUBTRACT.

Part B)

Essentially the same idea, but a little different.  TO find the force on the right particle, you want

Fnet = FAC + FBC.

You already know the value of FBC from part A, but you must reconsider that the direction will be to the LEFT, relative to charge C.

FAC will need to be calculated.  It’s direction is repulsive, so to the RIGHT on charge C.  Again, subtraction will be called for.  A final caution: When finding the magnitude of FAC, be sure to consider the whole distance between particles A and C.

#’s 21 & 22:

We will look at number 21.  Like 19 and 20 you have a couple forces to calculate and consider.  The only difference here is that you must apply your rules for adding vectors at funky angles (find the x- and y-components for each vector, add them, find the resultant).  As always, one must consider direction whenever we “add” vectors.  Looking closely at one of your classmates versions of 21:

Three charged spheres are located at the positions shown in Figure 20-14. Find the total force on sphere B. (qB = -7.4 µC, and qC = +5.5 µC.)

First use Coulomb’s Law to find the two forces: The force of A on B, and the force of C on B.  Both are attractive, so charge B is pulled generally toward the left.
When I solved with your numbers, I found the following vectors:
187 N to the Left
146 N @ 36.9 degrees South of West

Breaking into components and adding the x- and y-directions respectively I get:
X: -304 (left)
Y: -87.7 (South)

Using trig an Pythagorean, I conclude with: 316 N @ 16.1 degrees South of West (take care in putting that into Webassign as you need to refernce the whole angle from 0-degrees).


AP-C: Magnetic field practice from SMU

Loads of Practice problems:


AP-C: B-fields old quiz for practice (With KEY)

Below is a quiz that I have assigned as a take home quiz several times in recent years.  It focuses on the B-force only, so more the first half of our most recent unit.  Solutions are worked out in the pdf document.

Quiz 19.2–AP–Magnetic Fields

B-fields quiz 1–KEY


AP-1: E-fields and forces practice answer KEY

Worksheet–KEY–PreAP Efields Forces


AP-1 Waves & Sound HW #2 problem 15

A rock band plays at a(n) 85 dB sound level. (a) How many times greater is the sound pressure from another rock band playing at 100 dB? etc.

So, this is your “challenge problem” on this assignment.  As such I will not help you complete it, but I will help you get started.

If you don’t know what to do with the logarithmic equation, this may help a bit:

Let x = your variable and y = a number.

Log 10 x = y

Raise both sides as exponents of ten (the log base)

10 Log 10 x = 10 y

Raising the log expression to be the exponent of the log base effectively undoes the logarithmic operator, leaving behind the result

x = 10 y

You can solve the equation with any log base in exactly the same way.


AP Study Resource (All classes)

We had just gotten access to some AP study resources through the library’s Gale databases.  Since I have not yet seen them, I cannot guarantee the quality of the practice exams, they do seem like a good study option.  Use the following link to create your personal account and track your progress:

At the top hover over “High School Tools” –> Advanced Placement Exams (AP)

Then select the class you want.  If you are looking at physics, be sure you select AP-1 or AP-C respectively.  Other classes like AP-US History are available.


AP Circuits HW 2, problems 2 & 3, an Quiz KEY

I think you will find these much easier than first appearance.   For number 3, just be careful with your signs.  I have included the solutions for the currents below for a random sample problem.

AP Circuits HW 2, problem 2

AP Circuits HW 2, problem 3

Notes on problem 3:

Many of you have asked about the negative current.  Here you go.

So, about the currents…as I explained it is something of a guessing game.  I think that I misconstrued a question in class today about negative currents.  Is it OK to have a negative current?  Yes! As long as your algebra proves the validity of your solution.  A negative indicates that you chose the wrong direction for one or two branches of your loop rules.  If your algebra does NOT work out, then you have to go back and redo your direction in your loops.

So, to my example problem.  First, there are a huge range of values available on that problem so yours may be VERY different.  In fact, I just recycled the problem twice on my teacher view (I can do that with my version of the assignment) and the values of battery 1 and 2 flip flopped…something that would change the way I did my loops.  Before I get into details here is something to remember: You have two parallel branches with batteries and one with a resistor.  If it helps you model it against my work, you can rotate your circuit by 180 degrees out of the plane of the paper until your larger voltage is on the left like mine was.  As long as you keep the right resistors attached to the right batteries in the right order you have the same circuit!

1) I used the inside loops rather than the outside since that would give the the current through the central 4 ohm resistor in both equations…easier algebra.

2) I choose the direction of loop 1 to be consistent with my 5V battery

3) I choose the direction of loop 2 to be consistent with the larger voltage, in this case the 5V battery.  Basically when 5V meets 3V in the opposite direction, you net 2V in the direction of the 5V source.  Essentially the bigger battery wins, right?  (underlying big picture question: why?  More unbalanced charge at the terminals of course).  So, you are correct that loops appear to conflict in the center 4 ohm resistor but…

4) Look at equation 2–notice how the signs of the voltage losses and gains compliment the direction established in equation 1.  Consistency is the key!  The middle term in equation 1 (-I1(4)) indicates a loss through the center 4 ohm resistor.  Look at equation 2.  The middle term is positive reflecting a “backflow” of current in the opposite direction that I established in equation 1.  Again, it doesn’t matter which is correct.  Consistency wins.  Your math will tell you the direction (positive or negative).

5) Equation 3 represents the node rule for the junction at the top of the circuit, again using the larger battery for the positive direction.  Again, consistency.

I trust you can follow the algebra?  Let me know if you need assistance with that.  It is no bother.  I should ask you to check me as I actually did this problem last year and just re-posted the work today, so by all means scrutinize my algebra.  It is very easy to switch a sign when you are dealing with systems of equations, as you surely appreciate.

Now, back to my original point.  Is it OK that I got a negative value for I2?  You bet!  Why?  When I check it against equation 3 it returns a true solution.  What does it mean?  It means that the direction of current flow I established in I2 (to the right) is wrong.  Turns out this current should be flowing to the left.  Does it matter?  No.  The value​ of the current is correct.  Remember also that I2 represents the current in the 5 ohm resistor.  Why should it be to the left?  Well, the math really answers the question but lets go back to the problem.  Set aside the need for consistency of signs in your algebra and just study the figure.  We have two batteries working “against each other” in the outside loop, but complicated by the fact that both are connected by the central branch with the 4 ohm resistor.  Both batteries are similar in voltage (one is not way bigger than the other), so imagine a positive charge (conventional current) in the wires.  If you are by the first 4 ohm resistor you dont want to go toward that battery.  If you are by the 5 ohm resistor you don’t want to go towards that battery.  Imagine both batteries pushing out positive charge…where does it go?  Towards the middle and down the central branch, getting sucked away to the negative terminals on both sides.  Now does the negative answer for I2 make sense?

So why didn’t I do that in the first place?  Consistency.  I made a guess, that turned out to be wrong, but because I was consistent in establishing direction on all of my work, it still returned an algebraically true solution that also makes good physical, conceptual sense for the circuit in the problem.

Below please find the key to the back side of the recent quiz (that I handed back today).  I did not grade this problem since you were pressed for time.  Please be sure to review prior to your exam.

AP Circuit properties quiz back side KEY

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