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Velocity after collision

  1. Jan 17, 2010 #1
    1. A solid ball A of mass 10kg travelling to the right at 2m/s collides with a solid ball B of mass 5kg travelling to the left at 5m/s. The collision is elastic. Find the velocities of the balls after the collision

    2. Momentum before collision = Momentum after collision

    3. (10kg x x2m/s) - (5kg x 5m/s) = 10v + 5v
    -5 = 15v
    v = -5/15 = -0.333m/s

    Could somebody check this please?
  2. jcsd
  3. Jan 17, 2010 #2
    No, it is not the case that the balls have identical velocities after the collision - this only happens when they coalesce together in a perfectly inelastic collision. So all you can conclude is that
    [tex](10)(2) + (5)(-5) = 10(v_{1}) + 5(v_{2})[/tex]
    To get further, you have to appeal to either conservation of energy or use the shortcut notion that for elastic collisions, relative speed of approach = relative speed of separation (which itself is a result obtained when invoking COE).
  4. Jan 17, 2010 #3
    So if it's an elastic collision some kinetic energy has been lost - transfered to other forms - like heat and deformation.
    1/2mv2 + -1/2mv2 = 1/2mv2 + 1/2mv2

  5. Jan 17, 2010 #4
    the second -1/2mv should be normal size
  6. Jan 17, 2010 #5
    No! That's an inelastic collision. An elastic collision is one in which no kinetic energy is lost at all. So, total initial kinetic energy = total final kinetic energy. This gives:
    [tex]\frac{1}{2}(10)(2)^{2} + \frac{1}{2}(5)(-5)^{2} = \frac{1}{2}(10)(v_{1})^{2} + \frac{1}{2}(5)(v_{2})^{2}[/tex]

    Solve simultaneously with the COLM equation for v1 and v2.
  7. Jan 17, 2010 #6
    In an elastic collision there is no net change in K.E.
    Why don't you just try to equate initial and final K.E and momentum using the two unknown velocities and the two known ones?
  8. Jan 17, 2010 #7
    yes of course - elastic.
    20+62.5 = 5v12 + 2.5v22
    82.5 = 5v12 + 2.5v22

    but I have two velocities in the same equation
    how can this be solved for each one?
  9. Jan 17, 2010 #8
    You still have [tex] -5 = 10v_{1} + 5v_{2} [/tex] from conservation of momentum
  10. Jan 17, 2010 #9
    Yeah Fightfish's right...
  11. Jan 17, 2010 #10
    I'm really lost now
  12. Jan 17, 2010 #11
    Why don't you try the equating relative velocities before and after collision?
    It is a two variable and two eqn. problem?
  13. Jan 17, 2010 #12
    You have two equations:
    [tex]82.5 = 5v_{1}^{2} + 2.5v_{2}^{2}[/tex]
    [tex]-5 = 10v_{1} + 5v_{2}[/tex]
    Solve them simultaneously to obtain v1 and v2.
  14. Jan 17, 2010 #13
    I can't solve it simultaneously. If I double the first I loose both variables. If I half the second I loose both also. If I solve by substitution the same happens
  15. Jan 17, 2010 #14
    [tex]-5 = 10v_{1} + 5v_{2}[/tex]
    [tex]- 2v_{1} - 1 = v_{2}[/tex]

    [tex]82.5 = 5v_{1}^{2} + 2.5v_{2}^{2}[/tex]
    [tex]33 = 2v_{1}^{2} + v_{2}^{2} = 2v_{1}^{2} + (- 2v_{1} - 1)^2 = 6v_{1}^{2} + 4v_{1} + 1[/tex]
    [tex]3v_{1}^{2} + 2v_{1} - 16 = 0[/tex]

    I guess that is sufficient for you to carry on?
    Last edited: Jan 18, 2010
  16. Jan 17, 2010 #15
    v1=8/3 or -2

    v2=-19/3 or -6.3


    when -6.3

    when 3

    How do I know which to choose, as the direction the balls went could be any?
  17. Jan 18, 2010 #16
    I apologise. I made a slight careless mistake above. The mistake in that post has since been corrected, and you may carry on.
  18. Jan 18, 2010 #17
    I don't understand. How can I finish this problem? anyone?
  19. Jan 19, 2010 #18
    Do the same thing as you did again in #15 except using the corrected version of the equation that I gave in #14 (edited in the post itself). You should take this opportunity as well to see how to approach such questions.

    For your reference, the corrected version of the equation to solve is
    [tex]3v_{1}^{2} + 2v_{1} - 16 = 0[/tex]
    I hope you will try to understand how that equation is arrived at.
  20. Jan 19, 2010 #19
    I understand how the equation was derived - by simultaneous equations - substitution.
    I worked it myself after you showed me. It was late and I guess I was tired and rushing.
    All good :cool:
    Thanks for your support.
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