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Homework Help: Linear momentum

  1. Jul 6, 2010 #1
    Linear momentum 2d collision

    1. The problem statement, all variables and given/known data
    Two particles A and B of mass [tex]m[/tex] and [tex]3m[/tex] respectively, A collides with B. Find the coefficient of restitution [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction.

    Velocity of each particle before collision.
    [tex]\dot{\textbf{r}}_{A}=9 \textbf{i}+5\textbf{j}[/tex]
    [tex]\dot{\textbf{r}}_{B}=2 \textbf{i}+2\textbf{j}[/tex]

    The x and y velocity components before collision
    [tex]\dot{x}_{A}[/tex], [tex]\dot{y}_{A}[/tex], [tex]\dot{x}_{B}[/tex] and [tex]\dot{y}_{B}[/tex]

    The x and y velocity components after collision
    [tex]\dot{X}_{A}[/tex], [tex]\dot{Y}_{A}[/tex], [tex]\dot{X}_{B}[/tex] and [tex]\dot{Y}_{B}[/tex]

    2. Relevant equations
    The common tangent is the [tex]\textbf{j}[/tex] axis.

    [tex]\dot{y}_{A}\textbf{j}=\dot{Y}_{A}\textbf{j}[/tex] and [tex]\dot{y}_{B}\textbf{j}=\dot{Y}_{B}\textbf{j}[/tex]

    [tex](\dot{X}_{A}-\dot{X}_{B})\textbf{i}=-e(\dot{x}_{A}-\dot{x}_{B})\textbf{i}[/tex]

    [tex]m\dot{\textbf{r}}=m\textbf{v}[/tex]

    3. The attempt at a solution
    With the values above I find the velocities after collision

    [tex]\textbf{v}_{A}=(-\frac{21}{4}e+\frac{15}{4})\textbf{i}+5\textbf{j}[/tex]
    [tex]\textbf{v}_{B}=(\frac{7}{4}e+\frac{15}{4})\textbf{i}+2\textbf{j}[/tex]

    How do I find [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction?

    If I use(which is in the i-direction)

    [tex](\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})[/tex]

    I get [tex]e=0[/tex], a totally inelastic collision.

    Thanks in advance.
     
    Last edited: Jul 6, 2010
  2. jcsd
  3. Jul 6, 2010 #2

    kuruman

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    What do you think is the correct expression for the coefficient of restitution when you have a collision in 2-d? How did you conclude that the y-component of each particle's momentum is the same before and after the collision? Finally, what is a "common tangent" and why is it the y-axis?
     
  4. Jul 6, 2010 #3
    Hi

    The common tangent is the plane perpendicular to the common normal plane. We are told the common normal is in the i-direction. In the course text we are told

    'In an (instantaneous) collision between two smooth non-rotating objects, where the area of contact at the moment of impact lies on a common tangent plane, the velocities parallel to the tangent plane remain unchanged before and after impact.'

    Thanks
     
  5. Jul 6, 2010 #4

    kuruman

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    OK, then, the coefficient of restitution is given by

    [tex]e=\left| \frac{\dot{X_A}-\dot{X}_B}{\dot{x_A}-\dot{x}_B} \right| [/tex]
    You know both terms in the denominator and it is given that

    [tex]\dot{X}_A=0[/tex]

    Use momentum conservation in the x-direction to find the remaining term.
     
  6. Jul 6, 2010 #5
    Hi

    Thanks for the reply, in explaining to you the coordinate system it got me thinking again and obviously there is no [tex]\dot{X}_A[/tex] so obviously I need to solve

    [tex]
    (\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})
    [/tex]

    for e

    Thanks
     
  7. Jul 7, 2010 #6

    kuruman

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    I hope you can finish from this point.
     
  8. Jul 7, 2010 #7
    Yes, all done, thanks again
     
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