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Homework Help: Collision between two particles.

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Question: First image
    My working: Second image

    I used Va+Vb for the speed of separation. I got the answer for (a), but couldn't get the answer for (b).

    I know the proper expression is "Vb-Va" with Vb being negative.

    But if I use Vb-Va, (a) would be wrong.

    How come?

    2. Relevant equations
    Principle of conservation of momentum, Newton's Experimental Law.

    3. The attempt at a solution

    As shown in the second image.

    Attached Files:

  2. jcsd
  3. Apr 14, 2013 #2

    Doc Al

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    Staff: Mentor

    Realize that when you post an image of your work, instead of typing it in, that it's difficult to make comments. (You'll get quicker help if you make it easy to comment. :wink:)

    In any case, the speed of approach of the particles is (2u) - (-u) = 3u, not 2u. (Wrong!)

    And the correct expression is Vb - Va, not Vb + Va.

    Edit: My statement about speed of approach was incorrect; it should be (u) - (-u) = 2u, just as you had it. I misread the attachment.
    Last edited: Apr 14, 2013
  4. Apr 14, 2013 #3
    Coconut (:tongue:) just lucked out. Coconut did not use the correct equation for velocity of separation and still got the correct answer. Coconut should redo the first part.

    Coconut may assign right as positive and left as negative.
    Its all the game of two equations: conservation of linear momentum and that velocity of separation one.

    BTW, are you a coconut fan ? :tongue:
  5. Apr 14, 2013 #4
    I thought both particles are coming together with the same speed?

    What ._.
    No, coconut is derived from a word in a language directly translated from some english words which have the same pronunciation as the abbreviation of my real name being directly translated into english.

    (wow, that was fun)
  6. Apr 14, 2013 #5

    Doc Al

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    Staff: Mentor

    Oops! I misread your post. My bad! :redface:

    In any case, the best way to approach this is algebraically without plugging in numbers for e until a later step.

    And the relative velocity is Va - Vb, so the key relation would be Vb' - Va' = e(Va - Vb).
  7. Apr 16, 2013 #6
    u = 2Va + Vb (from principle of conservation of momentum)

    Vb-Va = e (Ua-Ub)

    Vb-Va = e怀(u- -u)

    Vb-Va = e (2u)

    Vb-Va = e (4Va + 2Vb)

    When e > 1/2,

    Vb-Va < 1/2 (4Va + 2Vb)

    < 2Va+Vb

    3Va > 0

    Still not reversed :uhh:
  8. Apr 16, 2013 #7

    Doc Al

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    Staff: Mentor

    Good. Call this equation 1.

    Call this equation 2.

    Now use equation 1 to eliminate Vb from equation 2, thus expressing Va in terms of u and e only.

    Then you'll be able to draw some conclusions about Va.

    (Then do the same for Vb.)
  9. Apr 16, 2013 #8
    u = 2Va + Vb--1
    Vb-Va = e(2u)--2


    u-2Va-Va = e(2u)
    u-3Va= 2eu

    Va= (u-2eu)/3

    =u(1-2e)/3 (is this sufficient to jump to the next statement?)

    when e> 1/2, Va<0
  10. Apr 16, 2013 #9

    Doc Al

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    Staff: Mentor

    Good. The sign of 1-2e changes when e > 1/2.
  11. Apr 17, 2013 #10
    Ok thanks.
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