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Vector subtraction with momentum

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    solve this change in velocity when the V2` = 19.04 m/s [S59W]
    and V2=16.32 m/s [S36W]



    2. Relevant equations



    3. The attempt at a solution
    I solved this using the components method first, for which i got the answer 7.536 m/s [W 26.78 N]

    And now for the vector subtraction piece:

    I did this diagram:
    http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dffdfd.jpg.html

    So this is what i did
    I used cosine law like this:
    Δv^2 = (19.04)^2 + (16.32)^2 -2(19.04)(16.32)cos23
    Δv= 7.536

    then i used sine law like this:

    7.536/sin23 = 19.04/sinx
    for x = 81°

    and then when i place it into the diagram and subtract 36 from it, i get 45°. So obviously it would be [W 45 N] but by using the components method i get a totally different angle why?
     
  2. jcsd
  3. Nov 11, 2012 #2

    TSny

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    There's more than one solution for x.
     
  4. Nov 11, 2012 #3
    Umm what do you mean by one solution..
    according to the diagram i made... this is the x we were looking for.,..
     
  5. Nov 11, 2012 #4

    TSny

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    For example, the inverse sine of 0.5 has more than one answer, not just 30 degrees.
     
  6. Nov 11, 2012 #5
    I am confused.. where are you getting the 0.5 and 30 degress from?
     
  7. Nov 11, 2012 #6

    TSny

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    I just made up an example to illustrate an important point. What is the value of sin-1(0.5)? Your calculator will probably give you 30o. But that's not the only answer.
     
  8. Nov 11, 2012 #7
    so whats the other answer then?
     
  9. Nov 11, 2012 #8

    TSny

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    Are you familiar with the unit circle representation of the sine function? See http://maine.edc.org/file.php/1/tools/TrigUnitCircle2.html [Broken] for example and click the "y = sinθ" box and the "show coordinates" box. Then click and drag the green dot on the horizontal axis of the graph. Can you find another angle besides 30o where the sine function has a value of 0.5?
     
    Last edited by a moderator: May 6, 2017
  10. Nov 11, 2012 #9
    so 149 degrees?????
     
    Last edited by a moderator: May 6, 2017
  11. Nov 11, 2012 #10

    TSny

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    Yes, actually 150 degrees. So, both 30 degrees and 150 degrees would be solutions to sin x = 0.5. Note that if you add 30 and 150 you get 180 degrees. The sine function has the property that if two angles add to 180 degrees, then the sine of the angles will be equal. So, what would be another angle that would have the same sine as 81 degrees?
     
  12. Nov 11, 2012 #11
    I am guessing. 99°
     
  13. Nov 11, 2012 #12

    TSny

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    No need to guess. :smile: Grab the calculator and see if sin 99o = sin 81o.

    From your diagram, does the angle "x" look like its greater than or less than 90 degrees?
     
  14. Nov 11, 2012 #13
    It looks like its greater
     
  15. Nov 11, 2012 #14
    Well these velocities are used for finding the change in velocity for the second ball. the first ball had a velocity change of 18.02 m/s [E 32 S] and had a mass of 0.165 kg

    I need to find the mass for the second ball.
    And this equation was given.. m2 = (v1`-v1 / v2 - v2`) m1
    So i changed the diagram so that it would be v2 - v2` and i get 27 as the angle and the angle i get is [E 27 E] and the velocity of 7.536 m/s. I used that information to cancel out the [E32S] and [E27E] and divided the velocityand multiplied by the mass of the first ball. I got the final mass to be 0.39 kg. Was my process correct?
     
  16. Nov 11, 2012 #15

    TSny

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    Did you mean to type [E 27 S]?
    I don't understand how you are cancelling angles.

    But something's wrong. Do you see why the direction of the change in velocity of m2 should be exactly in the opposite direction of the change in velocity of m1? But that's not what you're getting. I don't see any error in how you got the change in velocity of m2. So, either v2 or v2' individually is wrong, or the change in v1 is wrong.
     
  17. Nov 11, 2012 #16
    Ok well lets continue on with finding the angle for Change in velocity for the second ball so it is 99 degrees im i correct? but how would i represent that .. because in my personal notes i showed the whole process of finding the angle "x" to be 81 degrees. should i just write beside the answer for x that .... x can have two angle and then continue on with that?

    So i do use 99 degrees as for my x....
    I would have a diagram like this... http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dffdfd-1.jpg.html

    So it would be [W 27 N] so i think now i can cancel out the [ E 32 S] yes the angles are a bit off but i think that would the mistake in the experiment data collected..
     
  18. Nov 11, 2012 #17

    TSny

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    Ah, these are experimental values. Good.
    So, if you chalk up the difference between 27 degrees and 32 degrees as experimental error, I believe your calculation for m2 (.39 kg) is correct.
     
  19. Nov 11, 2012 #18
    Oh so i guess the angles have to be completely opposite like we just found to cancel out correct?? BUT HOW sould i explain to my teacher that the 81 is same as 99... i mean what should i right in my notes?
     
  20. Nov 11, 2012 #19

    TSny

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    Well, 81 is not the same as 99. Only 99o is the correct answer for your problem. It's just that the sine of 81o is the same as the sine of 99o. You're going to run into that problem when you use the law of sines to find an angle larger than 90 degrees. I think the component method is the best way to subtract vectors. You got the correct answer that way.

    Alternately, you could use the law of sines to find the other unknown angle in the triangle, and then find the angle x from the fact that the three angles must add up to 180 degrees.
     
  21. Nov 11, 2012 #20
    On my sheet with this question they give the equation

    m2 = (v1`-v1 / v2 - v2`)m1

    So does that mean the change in velocity should be with v2-v2`
    not v2`-v2 ?????? that how we did it
     
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