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When and how do I use sin and cos regarding 2-D collisions?

  1. Dec 15, 2009 #1
    I think what I'm doing is overkill because all I want to know is when to use sin and cos in the components regarding (isolated) two-dimensional collisions. I'm just showing more content because it might help for context. Also, there are more steps in order to find the final velocity which is why you can see some steps cut off, but I didn't bother posting more pictures of it because I thought it was irrelevant from then on as I am only looking for why sin and cos are used.

    1. The problem statement, all variables and given/known data
    On QUESTION #1 III., why is sin used in the East component and cos used in the North component, when in QUESTION #2 II., a different but ultimately similar question, the sin and cos are switched? Please feel free to click through the images in order for context. Also, how do I know if I should label the components East and North? Why not other compass directions? I hope I'm making some sense :P

    I. http://bayimg.com/image/oajpoaaca.jpg [Broken]
    II. http://bayimg.com/image/aajajaacb.jpg [Broken]
    III. http://bayimg.com/image/bajadaacb.jpg [Broken]
    IV. http://bayimg.com/image/cajaaaacb.jpg [Broken]
    V. http://bayimg.com/image/cajaoaacb.jpg [Broken]

    I. http://bayimg.com/image/cajcpaacb.jpg [Broken]
    II. http://bayimg.com/image/dajcmaacb.jpg [Broken]
    III. http://bayimg.com/image/eajcjaacb.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution
    Looked at sine laws, properties of triangles, and so far I have found nothing I can apply it to. Please forgive me if I'm asking such a simple question.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 15, 2009 #2


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    Homework Helper

    it is because of how the angle is measured.

    sin =opp/hyp


    in 3, the east is opposite to the angle, so the east component is hyp*sin, similarly the north is adjacent to the angle. Understand?
  4. Dec 15, 2009 #3
    I would like to say that I understand what you mean, but sadly I don't. Although I think I found out a way how and when to use sin and cos. Wordings such as [12.0 E of N] use East for sin and North for cos; while [12.0 N or E] use East for cos and North for sin. Was this what you meant by "because of the how angle is measured"? My apologies for failing to understand
  5. Dec 15, 2009 #4


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    Homework Helper

    See this diagram:

    http://img16.imageshack.us/img16/3596/diagrm.jpg [Broken]

    Lets call DAC=α (alpha) and CAB=β

    sin=opp/hyp and cos=adj/hyp.

    Consider triangle ADC,

    [tex]cos \alpha =\frac{adjacent}{hypotenuse}= \frac{AD}{AC}[/tex]

    [tex]sin\alpha = \frac{opposite}{hypotenuse}= \frac{DC}{AC}[/tex]

    Consider triangle ABC,

    [tex]sin \beta = \frac{opposite}{hypotenuse}=\frac{BC}{AC}[/tex]

    [tex]cos \beta=\frac{adjacent}{hypotenuse}= \frac{AB}{AC}[/tex]

    Do you understand what I meant by how the angle is measured now?
    Last edited by a moderator: May 4, 2017
  6. Dec 15, 2009 #5
    Thanks so much! I already knew the properties of sin and cos as shown in your second post, but it was just the thought of using East and North as a property of a triangle in 2-D collisions and matching it with sin and cos that confused me. I read your first post again and I finally figured it out. Again, thank you :)
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