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Placements of vector diagrams and angles in 2-D collisions

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Regarding two-dimensional collisions, I just don't understand why my book suddenly changes vector positions and theta angles at different places than what I was accustomed to. I understand that they're getting me to think by changing questions and answers, but I'm confused WHY they placed it there. My important queries and comments are in the photo links that are bolded and the other links are just for context. Also, I have decided not to post more pictures for continuation in Question #1 because the next steps are irrelevant.

    STEP I. http://img686.imageshack.us/img686/938/41145645.jpg [Broken]
    STEP II. http://img198.imageshack.us/img198/9696/65034028.jpg [Broken]
    STEP III. http://img39.imageshack.us/img39/677/66356214.jpg [Broken]
    STEP IV. http://img198.imageshack.us/img198/4056/91707666.jpg [Broken]

    STEP I. http://img64.imageshack.us/img64/4439/20331196.jpg [Broken]
    STEP II. http://img198.imageshack.us/img198/1593/58858513f.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    Attempted to "decipher" the question in hopes of why the book placed it there, but to no avail. It could be the wording, though, but I don't see anything. Forgive me for posting because it's hard being taught only from the book with no other sources of help other than physicsforums.com
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 17, 2009 #2


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    Your arrangement (Step IV) is not different from the book's arrangement in Step III.

    Step III says that
    -pCf = pAf + pBf

    Step IV says that
    -pCf = pBf + pAf

    In other words, if you start at the tip of vector pCf and go first to the left and then down (Step III) you end up at the origin the same as if you went down first and then to the left (Step IV). Vector addition is commutative, it makes no difference which term comes first and which comes second when you add vectors.

    However, the book's way of showing the diagram (Step III) makes it necessary to place the angle outside the triangle so that it becomes obvious to the reader that the direction of vector is 50o (or whatever) "North of East". If they placed the angle inside the triangle, it would not be as obvious. Of course, you way of drawing the same thing (Step IV) takes care of the problem automatically. So, in retrospect, I would say that your way of showing what is going on is less confusing.

    I am not sure why the book asks you to use a ruler and a protractor. Maybe to reinforce the idea that trig functions can be obtained by taking ratios of sides of a right triangle instead of pushing buttons on one's calculator. Just a guess. :uhh:

    Although you don't show exactly what the question is, it obviously has to do with momentum conservation in two dimensions. You need to write all vectors involved as vectors and them add them vectorially to get the answer. That's the purpose of the part you think you don't need.

    I don't see how the vector diagram can be placed in the third quadrant. Because vector pBf points "East", the vector diagram must be placed in the first quadrant or the fourth, if you displaced the whole thing down by an amount equal to the magnitude of pAf.

    I don't think anyone is trying to confuse you. Each situation needs to be analyzed on its own merits and there are many different ways to approach a physics problem, all of them correct. The trick is to be able to see that a solution is correct even if it is not the way you would do it.
    Last edited by a moderator: May 4, 2017
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