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Finding final speed with two unknowns (Vf1 & Vf2)

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 0.540 kg moving east (+x direction) with a speed of 3.70 m/s collides head-on with a 0.440 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

    ball originally at rest m/s east
    ball originally moving east m/s east

    Known/Unknowns:
    m1= .540 kg
    Vi1= 3.70 m/s
    Vf1= ?
    m2= .440 kg
    Vi2= 0 m/s
    Vf2= ?


    2. Relevant equations
    [tex]
    \frac{1}{2}m_1vi_1^2 + \frac{1}{2}m_2vi_2^2 = \frac{1}{2}m_1vf_1^2 + \frac{1}{2}m_2vf_2^2
    [/tex]

    [tex]
    m_1\vec{vi}_1 + m_2\vec{vi}_2 = m_1\vec{vf}_1 + m_2\vec{vf}_2
    [/tex]

    Did I miss any?

    3. The attempt at a solution
    I've tried plugging everything into the above equations, but the two unknowns have me stuck. I know I'm supposed to use basic Algebra to give each unknown it's own equation and solve from there, but I have honestly forgotten how. Any help on how to solve for one of the unknowns is greatly appreciated. Thank you!
     
    Last edited: Oct 7, 2009
  2. jcsd
  3. Oct 7, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Rewrite the relevant equations by using v1i, v1f, v2i and v2f.
    Collect the terms congaing m1 in one side and m2 on the other side in both the equations.
    Take the ratio of left hand side and right hand side and equate.
    Now see whether you can proceed further.
     
  4. Oct 7, 2009 #3
    No. I tried to create separate equations and attempted to solve for Vf1 by substituting Vf2 with an equation for Vf2 (sorry if it doesn't make sense) but I only came up with zero...
     
  5. Oct 7, 2009 #4

    rl.bhat

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    Homework Helper

    It is always better to show your calculations, so that we can check where you went wrong. After rearranging the two equations , you get
    v1i^2 -v1f^2 = v2f^2 - v2i^2...(1)
    v1i -v1f = v2f - v2i........(2)
    Factorize both the sides in (1) and take the ratios of left hand side and right hand side of the equations.
     
  6. Oct 7, 2009 #5
    Oh okay I will. I just thought it would've been a little extreme as they took up the whole page in my binder. 8l

    Anyways, I constantly end up getting this and it doesn't work:

    (1/2m1vi1 - 1/2m1vf1)/(1/2m2)=Vf2

    I have a feeling it's completely wrong

    EDIT: Turns out I WAS completely wrong. I solved the problem finally. :D

    I did it by doing exactly what rl.bhat gave me and then solved for both v2f and v1f by getting each one alone.
    I eventually got v1f=v2f + 3.7 and v2f=3.7 - v1f.
    Then I plugged v1f into the original equation (second one listed under my "relevant equations" and solved for V1f. Then, I plugged that answer into the V2f=3.7 - v1f and found the answer.
     
    Last edited: Oct 7, 2009
  7. Oct 7, 2009 #6

    I must be missing something. Shouldn't you take into account the masses in to that equation? I think that it would be correct if m1=m2, but they are not.
     
  8. Oct 7, 2009 #7
    Don't feel bad, I thought the same thing. Here's why you wouldn't:

    You start out with the two equations I have listed as relevant equations (first is kinetic energy, second is momentum). Both of them will be used.

    Next, you rewrite the momentum equation by doing m1Vi1 - m1Vf1 = m2Vf2 - m2Vi2 and factoring out the mass. This leaves you with m1(Vi1 - Vf1) = m2(Vi2 - Vf2)

    Then you do the same thing for the kinetic energy equation and get m1(Vi1^2 - Vf1^2) = m2(Vi2^2 - Vf2^2)
    Which becomes m1(Vi1 - Vf1)(Vi1 + Vf1) = m2(Vi2 - Vf2)(Vi2 + Vf2)

    Now that you have the two equations, you divide the momentum equation by the kinetic equation.

    (m1(Vi1 - Vf1) = m2(Vi2 - Vf2)) / (m1(Vi1 - Vf1)(Vi1 + Vf1) = m2(Vi2 - Vf2)(Vi2 + Vf2))

    Both masses cancel out and so do the similar equations. This leaves you with Vi1 + Vf1 = Vi2 + Vf2

    Then just solve from there.

    It looks complicated but it really isn't that bad. : )
     
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