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Another Momentum one, sorry

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A .5kg hockey puck moving at 2 m/sec strikes another puck with a mass of 1kg. The puck then travels at an angle of 60 degrees with a speed of .75m/sec with respect to the motion of the original puck.

    a) What is the final speed of the first puck?
    b) What is the final direction of the first puck?

    m1=.5kg
    m2=1kg
    v1=?
    v2=.75 m/sec

    2. Relevant equations
    p=mv
    ΔP1 + ΔP2 = (F1 + F2)(Δt)


    3. The attempt at a solution

    m1u1 = (m1+m2)v cos60
    m2u2 = (m1+m2)v sin60

    This is all I can get. I think I'm right so far lol. I need help badly! This momentum stuff is killing me.
     
  2. jcsd
  3. Jun 2, 2010 #2

    rock.freak667

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    right, so the initial x-direction momentum is m1u1, the initial y-direction momentum is?

    But they did not say the masses stick together, so you cannot assume that. Instead assume mass 1's final velocity is v1 (vx1 and vy1 components) and mass 2's is v2, they said v2 is 0.75 m/s at 60°.

    So what is the final x-direction momentum and y-direction momentum?
     
  4. Jun 4, 2010 #3
    Well I've been working on this question and this is what I came up with. Please take the time to read it and give your input. Thank you very much :):)

    m1=.5kg
    m2=1kg
    v1I=2 m/sec
    v2I=0 m/sec
    v2F=.75 m/sec

    P=mvcos
    =(1)(.75)
    =.75 kgm/sec

    PTotal=m1v1I+m2v2I
    PTotal=m1v1F+m2v2F

    PTotal=(m1+m2)v
    =(.5)(1)(2)
    =1kgm/sec

    PTotal=m1vF1+m2vF2
    1=(.5)vF1 + (1)cos30
    1=vF1 + 0.866
    .134=vF1 = final speed
     
  5. Jun 4, 2010 #4

    rock.freak667

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    Why did you neglect the angle?

    Your answer is correct, but why do you keep assuming that the masses stick together?

    You have another component to deal with, the vertical component.
     
  6. Jun 4, 2010 #5
    Would the angle be 90 degrees which would make the whole thing 0?


    I just don't know how to do it singly lol.


    Alright I'll work on this
     
  7. Jun 4, 2010 #6

    rock.freak667

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    The question said at 60°

    Just write it out as single components

    m1u1+m2u2=m1v1+m2v2


    Remember consider the x and y directions separately at first.
     
  8. Jun 14, 2010 #7
    ok i think i got it, hopefully lol:

    p=mv

    m1u1+m2u2=m1v1+m2v2cos60

    (.5)(2)+(1)(0)=.5v1+(1)(.75)cos60

    1=.5v1+.375

    .625=.5v1

    1.25=v1

    b) The final direction would just be forward because the value of the speed is positive. Or do you use a formula to find the direction. I'm guessing you use v2=u2+2as, but we don't know acceleration unless it's the gravitational acceleration.

    Thank you for any help you give :D
     
  9. Jun 14, 2010 #8

    rock.freak667

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    This gives you the horizontal component.

    Remember you need to apply conservation of momentum vertically as well.

    When you get vx and vy for the puck, the direction is just the angle to the horizontal.
     
  10. Jun 15, 2010 #9
    Would the vertical component be the same as the first equation except we would use sin60 instead of cos60?
     
  11. Jun 15, 2010 #10

    rock.freak667

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    Yes but your initial vertical momentum would be zero.
     
  12. Jun 15, 2010 #11
    Alright I got it, thanks alot :D
     
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