AP Assignment

Read the notes on Circular and Harmonic Motion and Read Sections 13.1 –13.4
Do Worksheet 1 1-6

http://h1.ripway.com/DrCherdack/NotesonCircHarmonic2-09.doc

http://h1.ripway.com/DrCherdack/unit9worksheet1r1RNC.doc

Topics and Research

Please read pages 99-118 in The Principle of Relativity for Monday so you can gain some value from our class.

Honors Schedule

Here is the beginning of this Unit's schedule and some more reading on rotational motion.

http://h1.ripway.com/DrCherdack/HonorsUnit7Rotsch09.doc

http://h1.ripway.com/DrCherdack/NotesonTorquerev12-30.doc

Honors Research Topics in Physics

Here's a nice problem. Solve it tonight and bring it in tomorrow. By the way this is a requirement, not a suggestion.

Relativity problem 1-09

A train is 300 m long in and has a mass ( without fuel) of 1 million kg bothe as measured its own frame. It leaves a station and accelerates to and maintains a velocity of .6 c ( its an express train). It then enters a tunnel that is 400m long. The conductor on the train at the back of the train notes the time on his watch as noon or zero just as the back of the train enters the tunnel He also notices that the clock on the tunnel wall also says noon. The tunnel is exactly 300 m long in its own frame.


The usual “paradox” mentioned here is that the train sees the tunnel as shorter and therefore cannot fit entirely inside the tunnel. meanwhile the tunnel sees the train as shorter so it fits easily inside. Who is right?

For convenience assume that the tunnel frame and train frame are both lined with tape measures and clocks so we can always find t, t x, and x’. Assume the train tape =0 at back of train and tunnel tape = 0 at entrance to tunnel. To deal with this “paradox” we wiil answer the following. What is x at the front of the train ( x’ = 300m) when t’=0. What is t when t’=0 and x’=300m? What is x’ when x =300m and t =0 and more to the point , what is x when x’=300m and t=0? When you find all this, you’ll know who is right.

Also, how much fuel had to be converted to KE to get the train up to this speed?

Group Meeting Reports

MOst of you have not turned in enough reports. Let me have them by Wednesday!

Honors Addl Practice Correction

Ben has pointed out the solution to KE initial for problem 22 is 8.7 E-15 J. The solution I published missed the divide by two in M vsqrd/2 so it is twice the correct answer given by Ben.

I would guess there are a couple more errors out there

AP MIdyear

Bring your official equation and information sheets. Get adequate sleep, and good luck tomorrow.

Honors Equation and Definition sheet and one comment

1)Here it is

http://h1.ripway.com/DrCherdack/midyr%20DefsEqns-08-9.doc

If you end up with any silly boxes, highlight them, convert to symbol font and read the Greek letters, and then ask yourself why Bill Gates is a success.

2) I received one question about problem 10 about the elevator. Note g is  a number not a vector ; it is +10 ( 9.8 for labs etc.) N/kg  or m/s2. 

Fg is  a vector; it points downward and equals - Mg if up is +. Therefore when subtracting Fg from both sides of Fscale + Fg = Ma we get Fscale = Ma+ Mg and again g is 10 NOT -10.  a may be negative of course remember sign is direction not necessarily an indication of getting bigger or smaller. [vectors, F and a, are boldface]

Honors Solutions to first set of practice problems

Here are solutions and a few corrections to the first set. Once again, these have not been proof read yet.
http://h1.ripway.com/DrCherdack/Midyrrev08-9hnrssln.doc

Honors Solutions to Additional Practice probs

Here are solutions to the second set of problems ( additional problems). There are also some minor corrections to the text of the problems. Note that I generated these solutions in a small house with 4 children around and 3 other adults discussing the days events. The physics is probably okay but there are probably a couple of arithmetic or algebraic errors. My next task is to proof read your exam and then develop the solutions to the first set, although we did many of these in class. After that its enter the latest labs received and homework, if I can get to it, and try to contact those of you who are still missing stuff. 

Study, but remember to get adequate sleep. It's vital to be alert when taking exams.

http://h1.ripway.com/DrCherdack/HnrMidyrAddlslns08-9.xls

Honors Additional and Highly Recommended Review

Here are some additional questions well worthy of your time. Most this stuff should look pretty familiar from our work during this semester. Study hard, but get some rest as well.

http://h1.ripway.com/DrCherdack/Physics%20Midyear%20AddPract08-9.doc

AP Review Again

Further apologies, the water leaving the tank does so at 10.8 m/s and the elevator poweris 8.5 E4 watts. In both cases I misread the questions in one case I think I reversed the height and in the other case I read the elevator mass as a more realistic 9500 kg.

AP Review

For some reason I divided by 2 more than was correct on two problems. Number 22 a 240 kg cart moving at 2.2 m/s has a KE of 240/2 x2.2 sqrd + 580 J and so eff = 580/ (100x12) = .48 or 48%

#34 The force 1200x 20sqrd/85 = 5650 N . mu = 5650/12000 = .47

These are examples of the dangers of inadequate sleep.

Honors Review

Correct answer for #16 is 1 sec. It is missing. The helicopter problem is missing the velocity. It is rising at 15 m/s.

AP Thermo and problem 63

We have emphasized the pure processes in our thermo work, but it is certainly possible for a process to be none of these. Heat in or out with a change in volume and pressure and temperature is the norm. The delta Us can be found from the end point temperatures, using c and n, but the work is best found from the area under the P vs V curve. In problem 63 process C-A, this area is RECTANGLE 20,000 X (40-80) plus a TRIANGLE (200,000 -20,000) high with a base of (40-80) m3. Given the work and the change i n U , you can figure out Q.

AP Solutions

Here is a draft set of solutions for the review version 2

http://h1.ripway.com/DrCherdack/APMidyrRev%20ver2sln.xls

Honors Assignment

As agreed, here are tonight's problems: 5,7,12 ,14,20

All students Study Group Reports

By next Monday,  each student should have turned in one study group report for each unit and one more for overall review of the semester. For Honors this means Units 5 and 6 plus one. For AP this means Units 6,7,8 plus one. Again, the report must say where, when, who, and subject covered. Just a few lines will do.

AP Review The new and improved model

After conferring with our marketing staff and trying a few prototypes out on selected customers, we have come up with the new improved review sheet. Changes and additions are marked with asterisks.

http://h1.ripway.com/DrCherdack/APImidyrRev08-9Ver2.doc

Honors Practice Questions

Here are the first 50 questions do 20 per night. There will be more on our more recent work soon, however reviewing collision, reaction rate and circular motion homework of the last couple of weeks along with the break review should put you in good shape - provided you manage to squeeze in some sleep as well.

http://h1.ripway.com/DrCherdack/Midyrrev08-9hnrsdrft.doc

AP Midyear review

Having had the weekend off, I hope you are ready to bear down and work. Here is your review package for the midyear exam. Do 20 questions per night and you should be in good shape.

http://h1.ripway.com/DrCherdack/AP%20I%20%20Midyear%20REV%2008-9.doc

ALL AP and HONORS Grades

I have posted grades on TAC, your HAC. They are up to last Thursday and do not include the AP thermo test. Several of you have outstanding labs due for initial or resubmission. These are noted in the two lab categories labelled appropriately. In addition, several of you have "see me"s on tests or simply tests you did not take or ones I couldn't grade. You must clear these up before the end of this week.

Honors Homework for Sunday 1/18

Read this and do the problems as your homework for today ( Sunday). I 've given you a link but also printed it here. Note the ws are omegas; blogspot can't write Greek.

http://h1.ripway.com/DrCherdack/Rotational%20Dynamics%20Eqs.doc

Rotational Motion in Magnetic Fields Notes

Magnetic fields are peculiar in that the force they create is felt only by moving charged particles and the forces are perpendicular to the field and the velocity of the charged particles. We will explain all this next quarter. For now, just note that the direction of the forces makes them useful for making charged particles to go around in circles since the force to make something go in a circle is perpendicular to velocity ( the velocity is tangential to the circle and the force must be inward radially).

The magnetic force is given by Fmag = Qv X B. Here B is the magnetic field strength ( I have no clue as to why its B)and it has a direction just as electric and gravity fields do. The X is not simple multiplication. A X B is equal yto AB sin angle between A and B. [The strength of B is measured in Tesla, T. If this makes it clear as to why its called B, let me know.] The X operation ends up with a vector which is perpendicular to both V and B. Let’s just focus on the strength of the force and the fact that it is cnetripetal for now.

We have demonstrated that a centripetal, ac, = wvtan or v2/r or w2r since vtan =wr.


Mac = QvB or Mvsqrd/r =QvB dividing both sides by v we get Mv/r = QB or r =Mv/QB

WE can also solve for w using v = wr. We have Mac = Mvsqrd2r=QwrB divide both sides by w and we have Mw=QB or w = QB/M

Now lets get some numbers. If we know the force and mass we can find r for say a deuteron moving at 2 x106 m/s in a field whose strength is 5 Tesla ( this is about 100,000 time the earth’s magnetic field the one that makes compasses work). Go ahead and do this. Also find w. Try it on your own and then check. If you need to find Q and M from notes go ahead and do it. I’m not telling you any more.

The answers are r = 8.53 x10-3 m or .85 cm (a third of an inch) and w= 2.4 x108 rad/sec.

Now you do it for tritium nucleus, T, traveling at 1x106 m/s and then for a T 2.5 x106 m/s.

Do it again for a field of strength B = 3 T to see how r and w change with B.

Honors Nuclear Reactions

Nearly all of you do not understand at least some fundamental part of the relations between quantities.

Print this out and study it.


If you put units in your equations and do dimensional analysis you will avoid some of the more inane mistakes you are currently making. Dimensional analysis means when you do the algebra on the units the result should yield the units you need for the quantity you are finding.
(m3 means cubic meter here) ( per sec is the same as /sec, etc.)
1) Power is energy transferred or transformed per unit of time or J/s
2) If you know the reactions per sec per m3 and you know the energy per reaction you should know how to find the energy per sec per m3. Think about it and make sure you can do it.
3) The energy per m3 per sec is the power per m3, i.e. J/s per m3 = W/m3. By the way, given the amount of fuel per m3 and the sigmav Whcih by the way has units m3/sec) the reactions per m3 have nothing to do with the volume of the plant. The total number of reactions in the plant of course = the reactions per m3 time the volume. You must think and understand rather than try to remember a jumble of formulae or how we did old problems.
4) If 1 MW is one million W than you must know that you must divide the number of W by 1 million to get the number of MW. After all, don't you divide the number of cents by 100 to get the number of dollars, or is this news to you?
5) The power output of a plant is the fusion ( or other microscopic) energy generated by the plant per sec times the efficiency. The fusion power generated by a fusion plant is of course the fusion power per cubic m times the volume of the reactor. Thus output = power per m3 x volume times efficiency
6) The mass of fuel in anything = number of particles per m3 x mass per particle times volume of the thing.
7)The kg of fusion fuel used per day is reactions per sec per m3 x volume x sec per day x the mass of the fuel particles per each reaction ( 5 x mp for DT)
8)To find kg of coal used per day, given J/kg of coal clearly you need to find the Joules needed per day and to do that you can take the output power ( W = J/s) times sec per day and then you figure out how to find out how much fuel you use because I won't tell you how to do things you should have learned 4 years ago. I will tell you that by now you should be able to convert MW to J/s. Aso, dimesional analysis may help if you do it right.
9) You must get some understanding of what numbers mean. You are not to write down whatever nonesense you get out of your calculator. A power plant does not use a mass of fuel greater that the mass of the entire galaxy. 10 to the 37th power kg is not correct! In the future answers displaying this level of not caring will result in loss of credit for the entire problem.

Honors Power ( Staircase) Lab

1) It should be absolutely clear to you that energy conservation can be expressed as thework put into a system = change in energy of that system less what ever leaks out. The energy change of the system includeds its KE, its PE, and its thermal ( microscopic) energy.

In other words: Work in = Delta KE + delta PE +losses (including all forms of "microscopic" energy such as heat, noise, light. Some of the heat stays in the system and some is lost. The heat that stays in yields a rise in system microscopic nery i.e. delta U).

In the stair lab, we are ignoring losses to heat etc. Thus Win = Mgdelta h + Mv sqrd/2.Win does not equal KE. While KE does equal Fnet dot distance most of the time, this fact is useless here since you cannot calculate the net force very well.
Note KEi = 0 so delta KE = the KE you calculated.

The proportion of average power used to create the KE is (KE/delta t)/(total delata E/delta t) or just the KE dic=vided by the total work done since the delta ts cancel.

AP Thermo

Here is a new corrected version of the Carnot Cycle. ( A few cell references were wrong on my original spreadsheet). This will be excellent practice for you even if you already took the test. ( If you turn it may you can fool me into thinking you actually get this stuff- oops that was too cynical even for me.) Destroy any copies of the old one. I am also posting solutions for this.
I deleted yesterday's post so I am also including the link to the thermo notes with the solution to the sample cycle in the notes. Use the R5 notes for everything else.

http://h1.ripway.com/DrCherdack/Cycle%20Carnot%201-09r1.doc

http://h1.ripway.com/DrCherdack/Cycle%20Carnot%201-09r1solns.doc


http://h1.ripway.com/DrCherdack/NotesonThermoR4soln.doc

Honors Rotational Motion

Folks,

Frequency, f, is the number of complete turns an object makes in a second. Therefore omega, teh angular speed is equal to 2pi x frequency.

Honors Next Topic Rotational Motion

By Thursday, Read this

http://h1.ripway.com/DrCherdack/RotMotionDefseetc-07.doc

and try these:

Rotational problems

Your are 3 m from the center of a carousel. The carousel turns half a turn, i.e. pi radians. How far did you move?

The carousel makes one complete rotation in 10 seconds, how far did you move in those ten seconds?

How fast were you going in m/s? This is your tangential velocity.

How many radians did it turn in those ten seconds? [ answer is 2 pi] How many radians did it turn in one second?

What was your angular speed (its called omega) in radians / second?

Was your tangential speed (m/s) equal to omega times 3 m?

Honors Reaction Problems

As unbelievable as it may seem, I left a number out of a problem. Problem 5 on reaction rates, T = 300 keV.

AP Thermo Notes,etc.

Here is a revised set of notes including a Carnot Cycle problem. Do this cycle tonight, we will try to go over it in class tomorrow.

http://h1.ripway.com/DrCherdack/NotesonThermoR5.doc

Also see the grain elevator Carnot Cycle notes which we covered today

http://h1.ripway.com/DrCherdack/CarnotCycleforGrainElevatorSheet.doc

Honors Review Make sure you get these right

There were many errors which showed up on several papers. Please make....actually never mind the please... just make sure you understand these items completely. The numbers are from the honors review sheets you did over the break.

All: learn the meanings of average and difference and final.

Learn how to take a definition and turn it into an equation for a quantity used in the definition

5) While vavg = (vf+vi)/2 when ( and only when) a is constant, this is not a definition of vavg. Definition of vavg in x direction is deltax/delta t same applies in any other direction

8) a DOES NOT = vavg/delta t. It is remarkable that this is still an issue. a = change in v divided by time it took change to occur or a = (vf-vi)/delta t = delta v/delta t AND delta v DOES NOT = vavg. By the way, the relation for a in the previous sentence is a definition not just an equation.

12) delta x = vavg delta t It is amazing how many of you correctly identified vavg as 15 m/s and the said delta x = 30 m/s x delta t anyway. vavg does not = vf or = vi unless a is zero. If a is constant vavg = (vf+vi)/2 as noted above (and about a hundred other times)

17) Vectors and Scalars:

Position is a vector. You have to give direction and distance, i.e. what direction and how far you are from the origin. 5 miles doesn't give your position. 5 miles at 15 degrees north of east from Brush's deli ( a sort of origin around here) does give your position. 5 miles is called the magnitude of this position vector. 15 deg N of E is the direction, obviously.
Displacement is the change in position so naturally it is a vector (position f - position i). In other words, its not enough to say I moved 2 miles, that's just the magnitude of the displacement, you also need to say what direction.

Since velocity is the rate of displacement, i.e. rate of change of position, it too is a vector. It has a magnitude and a direction. Speed is just the magnitude of velocity and does not have a direction so its a scalar. Average speed is distance traveled over time and is non zero even when going in a complete circle. Average velocity going around a complete circle is zero since position f = position i and displacement is zero. At any one instant someone going in a circle has an instantaneous velocity, but the average is zero for a complete circle.

acceleration is the rate of change of velocity and has direction so it is a vector.

Energy is not a vector, it has no direction. You may be confused because one can calculate KE from velocity components. KE equals m vsquared/2. If one has orthogonal (i.e. at right angle) components such as x and y, then the v squared = vx squared + vy squared and
KE = m vx squared/2 + m vy squared/2 but that doesn't mean energy has x and y components, its just how Pythagoras works out. The energy is just there; it has no direction. KE depends only on total speed.

13) A - sign in front of a vector doesn't mean its making things smaller, it means the vector is pointed in the negative direction. Negative acceleration usually means accelerating down or left or south or west by convention. The acceleration due to gravity in free fall is always down or negative. It doesn't matter whether the ball is going up and slowing down, or stopped for an instant, or going down and speeding up, the acceleration is down and therefore negative.

Physics Olympics Results

Two Ridge High School teams took first and third place overall in addition to placing first in two of six events in the New Jersey Physics Olympics competition among 41 teams from over thirty high schools. Congratulations to Amy Chen, Sandra Korn, Helen Stolyar, Vanessa Tan, Dan Golding, Dennis Shih, Aakil Kadali, Chris D'Arcy, Dan Finnie, and Rob Kastner and our back home technical support Sydney Orthmann, Ben Hotz and Bobby Musso. Your skill, hard work, creativity, and team spirit really paid off.

All students

While I do not know who placed masking tape markers on the floor. I do know that after repeated requests, no one has removed them. Make sure I cannot say the same by next Tuesday.

AP Section 3

Congratulations, as a whole you did well to very well on the mechanics test. A few of you will need to attend review sessions before the midyear.

RPOT @ Monmouth

Good luck to the RPOTs =]. Remember to think about the thermodynamics of how snow is formed.

AP Section 3 A Major Blunder

There I was complaining that you were all asleep and my brain was also sleeping . All the U s we calculated in the cycle need to be multiplied by R. I just multiplied delta T by 3/2 n instead of 3/2 R n. This will also affect your Qs as well as the delta U s.

The cycle looks a lot less efficient now.

I hope you took notes in pencil.

Honors Review Package

I have been grading some of the break review packages and

1) There is still an absolutely appalling level of confusion about averages and differences. About half of you still think vavg = (vf-vi)/2. Get over this averages have nothing ( or almost nothing) to do with differences. Some of you seem to think you can find acceleration by taking the average velocity and dividing by the time. This of course could give you a substantial acceleration while v is constant. THIS IS NONESENSE. Delta does not mean average; average does not mean difference! YOU MUST GET THIS RIGHT IMMEDIATELY.

2) In trying to generate solution sets for three packages in one night ( while dealing with a tablet that insisted in scrolling past the page I was working on) a few mistakes crept in. For #20 the answer is b, 1 sec. For #30.b the answer is 2.4E6 m/s ( I didn't take the sqrt of the 5.76). For #44 the KE total after the collsion should be 4.5 E-15 andthe difference in KE should be 4.5 E-15 - 6.68 E-15 = -2.2E-15 J. Iwill try o upload the new solutions so the existing link posted Monday will give you corrected set.

Honors Nuclear Work Sheets

Here are the sheets that are overdue. Do nuclear reaction worksheet 2 tonight and then do the problems on the revised reaction rates notes for Monday. It should make you relative experts on the subject.

We will tackle one more topic related to this. We will compare the masses of fuel to masses of reaction products to see if it matches the amount of energy made available by the reaction i.e. does the reaction energy = c squared x the mass difference ?

http://h1.ripway.com/DrCherdack/Nuclear%20Reaction%20Rates%20WS%202.doc

http://h1.ripway.com/DrCherdack/Reaction%20Rates.doc

Honors Heating Groups

To my heating groups:

Try these links that provide some information on beam design including the negative source beams that are more efficient at higher particle energies.

http://h1.ripway.com/DrCherdack/Neutral%20Beam%20Heating%20Devi.doc

All students

We are collecting change in 506 for KIVA. Be prepared to shell out generously.

AP Quiz

I am continuing my campaign for parity between my AP sections. In recognition that Section3 will not have a lab period tomorrow, their quiz will be on Wednesday.

AP and Honors

Here are the solution sets. Not 100% guaranteed but should be okay. Good Luck

1)AP http://h1.ripway.com/DrCherdack/AddlMechRevQuest08-9sln.doc

2) AP http://h1.ripway.com/DrCherdack/2008WntrFRAsgnsln.doc


3) Honors http://h1.ripway.com/DrCherdack/Break%20Rev%20Pkg%2008-9sln.doc

Honors Review package

29 has an errror: top is at 0V bottom at 16000V

Momentum

I am still seeing errors where people calculate delta p incorrectly.  AGAIN delta means final - initial and AGAIN direction matters. If something bounces back it has a reversed direction and reversed sign. For instance a 1.5 kg ball hitting a wall at 10 m/s and bouncing straight back at 8 m/s has a change in velocity of (  -8 -+10) = -18 m/s and a change in momentum of -27 kg m/s.

Archimedes Lab

The main purpose of the lab was to demonstrate that FB = density of fluid x g x volume submerged.  FB was simply the change in mass reading x g and could be found to .001 N with some care. We confirmed that our equation was correct by using the equation and the MEASURED FB from delta M x g  to find the volume of the submerged object and then using that to find the object's density. If the density found this way proved to be correct by comparison with standard values, then our equation relating FB and volume is correct.

The comparison of FB found exactly from the balance mass reading  increase following from density x g x change in fluid height in the beaker x beaker area was a sidelight and the measurements were crude and could yield  large ( 30% or so errors.)

Projectile Motion Lab

Many of you missed the fact that the lab actually demonstrated two principles.

1) Energy ( i.e. really Large scale KE + PE) was conserved in the pendulum type motion that followed the collision. Delta PE = - delta KE and in this instance PE final = KE right after the collision

2) Momentum was conserved in the inelastic collision (even while some KE was converted to microscopic energy such as distortions in the rubber of the catcher and noise.   IT IS NOT the fact that the mass of the catcher and ball is different from the mass of the ball alone that means KE is not preserved. It is the fact that they stuck together and there were losses of KE into hard to measure forms of energy and transfers of energy outside the system ( sound is an example of both) that kept KE from being conserved.

Energy conservation after the collision and momentum conservation during the collision were demonstrated to be true. This is true because the initial velocity ( launch speed) of the ball determined by a two stage calculation relying first on E conservation to find the v' after the collision and then using that v' and conservation of p to find the v of the ball before the collision i.e. its launch speed was confirmed by the previously proven projectile motion methods  relating vx to delta x.

AP Additional Problems

Actually these are mostly MC questions rather than full blown problems. Try to answer them without using the equation portion of the sheets from the front of 2008. Also, see how many you can answer without a calculator.

http://h1.ripway.com/DrCherdack/AddlMechRevQuest08-9.doc

AP Additional Problems

I hope to post them by about noon tomorrow. Stay tuned

Honors Momentum and Collision Test

It is okay to print out the answer sheet for you so you can place you answers on it.

Power ( Staircase) Lab

The purpose of the lab was to show how the amount of work done and the rate at which is done are both reflected in the power[use your own words]. I think a table would be good way to present the data and the calculation results. I would like to see the calculations themselves as well. All in all you should be calculating the delta PE, the delta KE and the power and the proportion of the total power that is used for the KE for each run.

Then write a conclusion about how load ( work done) and rate (or speed) affect power.

Enjoy the remainder of the break.