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Homework Help: Two Puck Colliding

  1. Oct 4, 2006 #1
    Suppose there are two puck of equal mass. That the first puck moves to the east at a velocity of 3.6 m/s while the second one is at rest. Then there is an elastic collision. The first puck moves at an angle theta north of east with a velocity of 1.3 m/s. The second puck moves south of east at angle phi. What is the angle theta?

    I have attempted the problem but haven't gotten it to work out yet. Here is what I did.

    Momentum in the x direction:
    [tex]v_{1i}=v_{1f}*cos(\theta )+v_{2f}cos(\phi )[/tex]

    Momentum in the y direction:
    [tex]0=v_{1f}*sin(\theta )+v_{2f}sin(\phi )[/tex]

    Consevation of Energy:
    [tex]v_{1i}^2=v_{1f}^2+v_{2f}^2[/tex]
     
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2

    berkeman

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    Staff: Mentor

    It looks like you can get v2 from the conservation of energy equation, and then you have 2 equations and the two unknown angles....
     
  4. Oct 4, 2006 #3
    I have a hard time solving the equations once I get them.

    From energy we can say:
    [tex]v_{2f}=\sqrt{v_{1i}^2-v_{1f}^2[/tex]

    From y momentum we get:
    [tex]sin(\phi )=\frac{-v_{1f}*sin(\theta )}{v_{2f}}[/tex]

    From the x momentum we get:
    [tex]cos(\phi )=\frac{v_{1i}-v_{1f}*cos(\theta )}{v_{2f}}[/tex]

    Here is the trick. I draw the triangles then I say that the adjacent legs are the same value since both have the same hypotnuse v2f and the same angle phi.
    [tex]v_{1i}-v_{1f}*cos(\theta )=\sqrt{v_{2f}^2-(v_{1f}*sin(\phi )^2}[/tex]

    From here it is possible to solve I just wondered if anyone knew any easier way.
     
  5. Oct 4, 2006 #4
    Is the angle between the final velocity of the pucks supposed to be 90 degrees because when I worked it out I got something like 88.6 degrees???
     
  6. Oct 4, 2006 #5
    I read that it is always 90 degrees so I guess I can say:

    90= theta - phi

    since phi is going in the negative direction.

    Then I could use the double angle thereom which might be a little easier. The final expression I ended up with was:

    [tex]cos(\theta )=\frac{V_{1f}}{V_{1i}}[/tex]

    Which seems so simple, yet I had to go through so many hoops to get it. I feel like there must be some short cut.
     
  7. Oct 5, 2006 #6
    yep it is much easier to say that:
    sin(Phi)=sin(pheta-90)=-cos(pheta)
     
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