Password to access protected areas of this website

To access any protected areas of this website with physics content please look at the right-hand side of the graphic at the top of the page.  To access copyrighted UIL materials please contact me for your password.

Why have I done this?

In an effort to continue to make my content freely available to my students, parents, colleagues and other educators, but to keep “web-bots” from stealing content from this domain, I have embedded the password information into the graphic at the top of the webpage.  The password should open all publicly accessible portions of this site and you should only have to input it once and your browser will store it (unless you have extremely tight security settings on your web browser).  If you have any problems accessing course content or if the password does not appear to work, please contact me as soon as possible.


AP Physics: Tension Knots lab example

Data is based on system 1, 2011-2013 school year (last couple years) so the final answer is different than the masses this year.  Also the angles will be somewhat different just due to the new set-up.

Tension knots system 1

Tension knots system 3 2012



AP-1: Test review materials

In addition to completing your HW and the practice problems from class, here are any documents that I generate for you this weekend, starting with the “quiz” that I didn’t give you in class today.  3rd period you didn’t receive this document on paper (sorry), 4th you got your copy, and B-day you will get yours on Tuesday.  The answer key is on the second page.

AP-1–Linear Motion problems with graphing KEY


AP-1: Linear motion 2 HW helps

#18 & 19:

As discussed in some periods, webassign may score you incorrectly when you try to find the slopes and values of the graphs as indicated in the questions.  Eyeball it to the best of your ability and make sure you know HOW to answer the questions, but I will ensure that you receive credit for items like that that webassign marks incorrect that you actually have right.


This problem is a common issue for students because (if you read carefully) you will notice that you have two distinct motions represented here.  The knee-jerk approach to this problem is to list out your variables and try to stuff numbers into an equation.  This one is not that easy.  You actually have two separate physics problems here: one is constant velocity motion, one is accelerated (slowing down) motion.


Your final answer to the question must add the distance the car travels during both time periods, so you first need to find those distances.

Constant Velocity part
If an object is moving at a constant speed, only 1 of our kinematics equations describes its motion: the defn. of velocity–V=(delta)x/t
Based on the information provided in the problem this should be simple to solve for the distance traveled during the reaction time.

Accelerated part
Since this is not constant velocity, the equation we used above goes out the window.  You CANNOT use it because it does not account for the acceleration of the car.  What this does mean, however, is that you can use any of the other 3 equations, provided you have the requisite information.  Here’s what we know:

x = ?          Vo = same speed as first part, since you have not yet started braking              V = 0 (zero, you have come to a stop)

a = given in problem          t = unknown

There are several ways to approach this solution.  You can go directly to the displacement (x) if you select the “right” euqation, but you could also find the time first, then use that information to find the displacement.  It makes no difference.  One important point of caution:  Your acceleration opposes the velocity, so you must make sure it is _________?____________

If you answered “negative” you get a cookie.
Don’t forget to add the 2 distances from the 2 parts of the problem when you finish.

I encourage you to approach this question as two separate motions (in some ways similar to number 5, above).  While it can be done as an entire trip (up and then down) it may be easier for you to think about it in two parts.  Use the information provided to find the height to which the ball goes up–don’t forget to account for the fact that gravitational acceleration is down.  Then use the total height to find the impact speed when it falls all the way back to the ground.

Now, on the way down something interesting happens with your problem solving.  If you work this from the max height to the ground then your

displacement is ____?____
velocity is ____?____
acceleration is ____?____

Since everything is negative (you did answer negative or downward for all of those items, right?), you might say that concerns about direction cancel out…they really sort of do mathematically.  In any event you  can solve this using positive numbers provided you recognize that your final answer is downward.  This is important because of the way the question is asked in the problem.  your final answer for your

velocity is ____?____
speed is ____?____

and since the question asks “how fast” your answer should probably be speed.


Number 8 is your challenge problem for this assignment.

This is the one problem I will not help you complete fully, but I will give you the following hint:  You need to draw a picture and set up two separate equations for the displacement of the truck (constant velocity) and the displacement of the car (slowing down so need the equation with acceleration).  Write one of the displacements in terms of the other (i.e., 500 – x).  Both of your equations will be missing a displacement and time term.  Solve for time (when they meet the same amount of time must have elapsed).  You will have to either graph it or use the quadratic formula.  Use time to solve for displacement.

Last hint:  Because the car is slowing down fairly rapidly, it will travel the lesser distance.  This may help you narrow down your answer.

If you miss it, don’t worry about the points.  I always factor some participation points into your grades that will offset these type of challenge problems :)

Good luck.

HW #’s 33 & 34 (and notes about 35)

33. A helicopter is rising at X m/s when a bag is dropped from it.

(a) After Y s, what is the bag’s velocity?

(b) How far has the bag fallen?

(c) How far below the helicopter is the bag?

34. A helicopter descends at A m/s when a bag is dropped from it.

(a) After B s, what is the bag’s velocity?

(b) How far has the bag fallen?

(c) How far below the helicopter is the bag?

OK, so these are basically the same problem.  In each problem you know the same information at the beginning.  Since everybody has different numbers I have given variables in the problems here.  For each problem you know:

a = 9.81 m/s^2 down, t and Vo.  In 23 you will want to make your initial velocity positive while the acceleration of gravity is negative.  This means the bag keeps rising for a bit as a result of its inertia but eventually changes direction and falls. In 24 you will want to make everything positive since everything (displacement, velocity, acceleration) is in the downwards direction.

Use the definition of acceleration (a=(delta)V/t) to find the final velocity for both questions part a).  Remember that delta = final – initial.  Your sign changes are very important in this problem!  Your answer may be positive or negative depending on the initial velocity YOU have been given by webassign.  Postive would mean it is still going ___________, but negative means it is going _________.

On part b) I suggest using x=Vot + .5at^2.  In 23 make sure you put in a positive initial velocity and a negative acceleration.  Since x s the displacement this positive and negative discrepancy will take care of the up/down issue and just give you the position of the bag (i.e., “how far it has fallen”) in your result.  If the bag is on its way down (see note from part a) and has fallen past the starting point you will get a negative answer.  The question asked “how far.” This means it wants the absolute value.  In 24 this point is moot since everything is going downward already.  What you get is what you are looking for.

Part c)  These are really your challenge problems on this assignment.  I will give you a big hint, but I am not going to explain completely.  You have two motions here: The helicopter moving at a constant speed (given in your problem), and the falling bag, accelerating due to gravity.  You already know how far the bag has fallen relative to where it started (answer b).  You must find how far the helicopter has gone either up or down (in #23 or 24 respectively) using constant velocity and either add to or subtract from your value from b.  Which one you do depends on the direction of motion.  Draw a before and after picture for each if you are having a hard time seeing it.  Good luck.

For question 35, the scenario is the same as 34, but since the helo is descending the initial velocity of the bag will be downward (negative) along with the acceleration of gravity and the displacement (all negative).  Like part of #7, above, you can drop these negative signs (or keep them as you wish) since all aspects of your motion are down, in the same direction.  Again, take care with part a to ensure that your answer is reflective of direction (down or negative) since the problem specifically asks for “velocity.”


AP-1: Problem solving stations solutions from class

AP-1: Here are the problems and solutions from the practice word problems that we did in class.

Worksheet–stations problem solving solutions with problems

And here are the practice problems from the power point notes worked out for you:

Linear motion practice problems with solutions (from ppt notes)


AP-C: Your “study guide” for the multi-dimensional motion exam

Let’s think about what we have learned/relearned during this unit of study together.  We began with a review of basic vectors: Resolving into components and expressing as a magnitude and direction, essentially converting between Cartesian and Polar notation respectively.  We reviewed vector addition and vector “subtraction,” the latter of which is merely the addition of a “negative” vector.  We learned to multiply vectors by taking both a dot (scalar) and cross (vector) product.  For the cross product we learned to use the Right-Hand-Rule to determine the direction of the resultant.  Remember that dot products just return a number/amount/magnitude.

Then we progressed into two dimensional motion which included the following topics:

  • Projectiles, which I would further break down into
    • –Simple “PreAP” projectiles inlcuding horizontal and full trajectory (flat ground) launches.
    • –More complicated projectiles like those requiring a system of equations to solve (the baseball fence problem), inturupted trajectory (throwing something at a wall), and arcing over the cliff problems
  • Uniform circular motion focusing on
    • –finding the centripetal acceleration or the tangential and angular velocity of a rotating, revolving, orbiting system
    • –recognizing that “uniform” circular motion is constant speed and this fact ties the speed to the circumference and the orbital period
    • –understanding that the force and acceleration vectors must point inside the curve, directly towards the center of the arc if an object is to rotate with a uniform speed, thus placing the velocity and acceleration vectors at right angles to each other.  This independence allows for the acceleration to change on the direction of the velocity vector, but not the magnitude of the velocity (a.k.a. the speed)
  • General principles of 2-D motion: applying concepts of 1-D motion and calculus to 2 or 3 D systems to determine displacement, velocity, and/or acceleration from given information.

For your exam tomorrow, plan on 12-15 multiple choice items (remember no calcs, no eqn sheets) and 3 large free response items, of which these will be broken down into sub questions.  I would estimate spending 30-ish minutes on the MC portion of the exam and 45 on the FR portion.  You will have the entire period to work, however, so you should not be rushed in any way.

I will be available for tutorial in the morning, please stop by if you need assistance.


AP HW: Projectiles 1, numbers 8, 9 & 10

Repost from past year

AP Projectiles #1

Problem 8, 9, 10


Several of you have inquired this evening about questions 8, 9 and 10.  Here are your helps:



In this problem you are given a velocity function which is initially in the x-direction only and an acceleration that is in the negative x-direction as well as the negative y-direction (at least on my version).  In my opinion, the easiest way to do this problem is to break apart the equation into x- and y- directions and do kinematics.  We can do this because a) vectors at right angles are independent (that is to say the y-acceleration does not affect the x-direction of motion) and b) because the acceleration in both directions is constant…note no time dependency in the accel. Equation.


OK, so in the x direction we know

Starting position (origin = 0)

Initial velocity (your value)

Final velocity (0…where it stops moving positively and starts moving negatively)

Acceleration (your value)

Final position (going to solve for this)

Time (can solve for this using a=deltaV/t)


In the y-direction

Starting position (origin = 0)

Initial velocity (0)

Final velocity (Going to be solving for this)

Acceleration (your value)

Final position (going to solve for this)

Time (same as x-direction)



Nothing happening (unless your vector has a k-hat component and I don’t think any should)



Very similar to number 8 except you have a time dependent acceleration in the x-direction, so we will need to use some calculus.

a)      Take derivative, evaluate for t=4s

b)      Set derivative = 0, solve for time

c)       Set velocity (the function you are given) = 0, solve for time.  Do x- and y-directions separately.  Both must be zero at the same time to satisfy question.  If all answers are not real then never. My answer with my numbers is NEVER, but some of you may have a value.

d)      Set v=10, solve for t



Two of you have asked about number 10.  I checked both of your work.  WebAssign is correct here.  You have probably rounded too much.  If you take your answers to 2 or 3 sig figs, rounding only at the end of your solutions, you will probably get it.  Also, be sure you take into account the bullet drop in meters, not centimeters as stated.


AP-C: Linear Motion 3 HW #8, 16, 15

AP C Linear Motion3,

Problem 8

This is a pretty touch problem.  I began working on it this morning and had to stop and try again to find an easier solution using alternative kinematics equations than what I first tried.  The key is recognizing that you can use a system of equations–multiple equations each describing a different aspect of the motion, with multiple unknowns.  If you can find the right set of equations you can end up with a fairly simple solution.  Here’s the one I found that worked best.

To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.0 km/h, and 24.4 m when the initial speed is 47.8 km/h.  These are the numbers from my problem, of course.

(a) What is your reaction time?
(b) Acceleration?

2 parts to each motion:

–Constant velocity before break is applied (during reaction time)
–During acceleration (slowing)

For constant velocity portion: v = x/t
For slowing motion: 2ax = v^2 – v0^2

Now, this is where it gets tricky.  You will use 2ax etc. twice, once for each situation, but, and this is the big key here, you must substitute for x in that equation since the total x given in the problem is BOTH the constant speed distance traveled AND the slowing down distance traveled.  As a result, you will need to subtract the constant velocity distance from the total stopping distance.

Basically you work will look like this:
Scenario #1:
2a(x1-v01*t) = V1^2 – vo1^2

Scenario #2:
2a(x2-v02*t) = V2^2 – vo2^2

X1 & X2 – given distances
V01 & V02 – given enitial velocities
t – reaction time (unknown)
a – acceleration (unknown)
V1 & V2 – final velocities, in this case both = 0

Notice the terms x – V0*t: These are you taking the distance given, and subtracting the pre-braking, constant velocity distance out of this formula, essentially leaving behind just the distance traveled during the negative acceleration.
Now, at this point you have 2 equations with 2 unknowns.  Solve however you wish, but I prefer substitution.  The algebra gets nasty.  I actually had to redo my problem as I made one simple arithmetic mistake…not fun.  Take it step by step.
Have fun!



This is one of my favorite problems on the assignment.  It has elements of number 8 but is much easier.  You know a couple pieces of information.  In my version of the problem I know it falls from height h, and falls 0.49h in the last second.  The easiest way to solve this is to not focus on the last second, but to focus on the first part of the fall instead.  During whatever time that is (one second less than the total fall time) we know it falls 0.51h, the difference between 1 and whatever value of h you were given in your problem.  This lets us set up two equations:

Whole fall: h = (1/2)gt^2
First part: 0.51h = (1/2)g(t-1)^2  <– this is the key.  notice how we used the info from the problem, but we changed it up.

The reason I like to focus on the first part of the fall is because when Vo is zero, you get to cancel that variable.  Makes the algebra MUCH easier.
So, we are left with 2 equations with 2 unknowns.  Solve as you wish.  I generally prefer substitution.  In this case I simply take the first equation and plug in for h in the second, then used my graphing calculator to find the roots of the resulting quadratic to solve for time.  But wait, you say!  It’s quadratic so it has two solutions!  You are absolutely correct, but one of them is impossible.  I will eave it to you to figure out why, but I will say that one solution is outside the domain of the thing you are solving for.


Most common problem in dealing with this question is that the time given in your problem in part A–50ms–is 0.050s, not 0.50s.  Remember the things on the stopwatch everyone calls milliseconds are in fact NOT MILLISECONDS.  This is why a few of you have gotten answers with are simply 10x too big.  Solving this question then is a simple application of a standard displacement function for constant acceleration, that kinematics equation that you know and love:

x=Vot + .5at^2, of course Vo = ?

Also note the units on the answer.

As for the rest of the problem you can either calculate each one individually, or recognize that because of the constant acceleration of gravity you are dealing with a quadratic relationship between distance fallen and time.  That is to say, in two time the amount of time, the object will fall 4 times as far, or in three times the time, nine times as far, etc.  Work it out if you are confused and look for a quadratic pattern in your answers.


AP-1: Dimensional analysis practice and upcoming quiz notes

Solutions to some of the practice exercises from your notes:

AP-1- Symbolic alg and dimensional analysis practice

Relevant power point notes from class:

Metric system and conversions…you will remember this one from the summer packet if you used it.

All about matter…this is what we were looking at in class Friday (A-day).  These and a number of other items can be found on the unit page for this unit: AP-1 –> Fall semester –> Unit 1

Remember that you will have a quiz on Wednesday (A) or Thursday (B) over material that we have covered since the start of school, specifically:

  1. Metric system and conversions between metric prefixes
  2. General dimensional analysis and conversions between units
  3. Applying orders of magnitude estimation
  4. The fundamental structure of matter

I planning the quiz to be about 20-25 minutes in length and I will call time.  Be ready, work diligently.

If you have not already done so, please set up your webassign account!  We will get some practice using webassign and learning about that system after the quiz.

Older posts «

Switch to mobile version