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Momentum Question

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A curling stone thrown takes 4.8 s to travel 60 m. The stone collides with another stone. The collision is a glancing one. If the second stone is deflected 25° and travels 1.5 m/s, calculate the angle of deflection of the first stone after collision. Omit any effects due to friction.

    2. Relevant equations

    ρbefore = ρafter

    3. The attempt at a solution

    I looked at splitting it up into the x and y directions, but couldn't get anywhere. I looked in the answer key, and its 3.3 degrees, but I have no idea how to get there. Thanks!
     
  2. jcsd
  3. Dec 13, 2012 #2

    TSny

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    Hello, monkeygrif. Welcome to Physics Forums.

    You have the right idea of setting up x and y components of momentum conservation.

    Can you show more detail of your attempt?
     
  4. Dec 13, 2012 #3
    Thanks!

    x dir:

    0 = mass(a)v2(a) + m(b)v2(b)

    because of the wording of the question, i made the assumption that they had the same masses, and thus the masses were irrelevant, so:

    0 = v2(a) + v2(b)

    0 = v2(a)cosx - 1.5cos25

    v2(a)cosx = 1.5cos25

    y-dir:

    same setup here, which ends up with:

    v1(a) = v2(a) + v2(b)
    12.5 = v2(a)sinx + 1.5sin25
    v2(a)sinx = 11.87

    from there i used the tangent ratio, tanx = opposite/adjacent, which gave me the x value of 83.4 degrees
     
  5. Dec 13, 2012 #4

    TSny

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    OK. Looks pretty good. But note that you are taking the initial direction of motion of the first stone to be in the y-direction. That's fine. But then that means the deflection angles are measured with respect to the y-axis (not the x-axis). So, you'll need to think about whether you should use cosine or sine to get x-components. Similarly for y-components.
     
  6. Dec 13, 2012 #5
    Thanks so much! Its all worked out now. :)
     
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