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Collision of 2 balls of different masses traveling opposite direction

  1. Feb 11, 2008 #1
    Question

    Two small balls A and B have masses 0.5kg and 0.2kg respectively. They are moving towards each other in opposited directions on a smooth horizontal table when they collide directly. Immediately before the collision, the spped of A is 3m/s and the speed of B is 2m/s. The speed of A immediately after the collision is 1.5m/s. The direction of the motion of A is unchanged as a result of the collision.

    By modelling the balls as particles, find

    (a) The speed of B immediately after the collision


    (b) The magnitude of the impulse exerted on B after the collision

    Equations

    [tex]Ft=mv-mu[/tex]

    My Attempt

    (a) Conservation of momentum would mean that the initial momentum must be equal to the final momentum.

    [tex]Ft=mv-mu[/tex]

    [tex](0.5\times3.0)-(0.2\times2.0)=(0.5\times1.5)-(0.2\times x)[/tex]

    [tex]1.1=0.75-(0.2\timesx)[/tex]

    [tex]x=11.3m/s[/tex]

    (b) I have no Idea where to start here, I think it's more a question of now knowing what it is asking for. Is it asking for my value for [tex]Ft[/tex]?


    Any help would be great :cool:
     
  2. jcsd
  3. Feb 11, 2008 #2

    Hootenanny

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    You might want to check your arithmetic on question (a). Yes, (b) is just asking you for the impulse which is the quantity Ft.
     
  4. Feb 11, 2008 #3
    Sorry let me do that

    first bit again.

    (a)

    [tex]1.1=0.75-0.2x[/tex]

    [tex]1\frac{7}{15} = 0.26[/tex]

    [tex]x = 7.3m/s[/tex]

    I hope thats a bit better o:)

    (b)

    [tex]Ft=mu-mv[/tex]

    [tex]Ft=0.2\times2-0.2\times7.3 = 1.06Ns[/tex]

    Does that look a bit better?
     
  5. Feb 11, 2008 #4

    Hootenanny

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    Good :smile:
    Not so good, where's your x gone?
    Take a sanity check, does this seem reasonable to you?
     
  6. Feb 11, 2008 #5
    Sorry let me do that

    first bit again.

    (a)

    [tex]1.1=0.75-0.2x[/tex]

    [tex]\frac{1.1}{0.75} = 1\frac{7}{15} = 0.2x[/tex]

    [tex] x = \frac{1\frac{7}{15}}{0.2} = 7\frac{1}{3}[/tex]

    I would have thought th answer to be around 3.5 as that would make all the initial and final figures equal...
     
  7. Feb 11, 2008 #6

    Hootenanny

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    Your making your mistake between this step,
    And this one,
    Let me show you step by step,

    1.1=0.75-0.2x

    1.1-0.75 = -0.2x

    [tex]x = \frac{1.1-0.75}{-0.2}[/tex]

    Do you follow?
     
  8. Feb 11, 2008 #7
    Oh my word! I can't believe I did that! Thats a shocker haha. Cheers I follow that, and now with that I will get the correct answer for (b) thanks =]
     
  9. Feb 11, 2008 #8

    Hootenanny

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    A pleasure as always :smile:
     
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