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Inelastic SCATTERING

  1. Jun 20, 2008 #1
    Can Someone look over this and tell me if the work is correct.

    1. The problem statement, all variables and given/known data
    An object comes in with known velocity (v) and known mass (m) moving directly along the x-axis. It strikes another object of the same mass (m) which is stationary. Find the change in energy of the incoming object based on the angle at which it scattered ([tex]\theta[/tex]) and the velocity of the target after the collision (u). The collision only occurs in 2 dimensions. (Note: The angle at which the target object is moving can be removed algebraically.)

    [tex]E_{i}[/tex]=Energy of Incoming Object before collision
    [tex]E_{f}[/tex]=Energy of Incoming Object after collision
    [tex]E_{2}[/tex]=Energy of Target Object after collision

    [tex]p_{i}[/tex]= Momentum of incoming object before collision.
    [tex]p_{f}[/tex]=Momentum of Incoming object after collision.
    [tex]p_{2}[/tex]=Momentum of Target object after collion.

    [tex]\phi[/tex]= Arbitrary Angle of Target object scattering (should not matter based on the note.


    2. Relevant equations
    [tex]E=1/2*m*v^{2}[/tex]
    p=m*v
    [tex]p^{2}/(2*m)=E[/tex]

    3. The attempt at a solution

    Energy Balance:
    [tex]E_{i}=E_{f}+E_{2}[/tex]

    X-Momentum Balance:
    [tex]p_{i}=p_{f}*Cos(\theta) +p_{2}*Cos(\phi)[/tex]

    Y-Momentum Balance: (This should be a zero momentum system in y-direction)
    [tex]p_{f}*Sin(\theta)=p_{2}*Sin(\phi)
    [/tex]

    Squaring only the Momentum Equations and adding them together.

    Y-Balance:
    [tex]p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)[/tex]

    X-Balance: (after getting [tex]\phi[/tex] isolated on one side then squaring)
    [tex]p_{2}^{2}*Cos^{2}(\phi)=p_{i}^{2} - p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}*Cos^{2}(\theta)[/tex]

    Adding the two momentum Equations and using [tex]Sin^{2}+Cos^{2}=1[/tex]
    [tex]p_{2}^{2}=p_{i}^{2}-p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}[/tex]

    Relating back to energy, using the relationship defined in section 2.
    Since all masses are the same I divide the newly found momentum equation by 2m in order to get to energy:
    [tex]E_{2}=E_{i}+E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)[/tex]

    (Note: [tex]\sqrt{2m}*\sqrt{2m}=2m[/tex] and [tex]p/\sqrt{2m}=\sqrt{E}[/tex] )

    Combining with the original Energy Balance Equation to Eliminate [tex]E_{2}[/tex]. This involves subtracting the equation I just solved for and the original equation.
    [tex]0=2*E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)[/tex]

    Applying the Quadratic Equation [\b]
    [tex]\sqrt{E_{f}}=\sqrt{E_{i}}*Cos(\theta) +\- \frac{\sqrt{\sqrt{E_{i}}^{2}*Cos^{2}(\theta)-0}}{4}[/tex]


    Minus Sign Answer Leads to 0, so nontrivial answer is:
    [tex]E_{f}=\frac{E_{i}*Cos(\theta)}{2}[/tex]

    Is this the correct solution, or is there a step that I made a mistake or false assumption? It seems to me that this is wrong, because even a grazing trajectory decreases the initial energy by 1/2.
     
  2. jcsd
  3. Jun 20, 2008 #2
    I think your error is in assuming they both deflect at the same angle.

    Objects always deflect at right angles to each other. That's how you remove the resultant theta of the object that is originally stationary.

    [​IMG] Right there, look at the p2 term.
     
    Last edited: Jun 20, 2008
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