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Conservation of Momentum?

  1. Oct 24, 2012 #1
    Conservation of Momentum???

    An unstable nucleus of mass 2.7 x 10-26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m1 = 1.0 x 10-26 kg, moves in the positive y-direction with speed v1 = 4.8 x 106 m/s. Another particle, of mass m2 = 1.2 x 10-26 kg, moves in the positive x-direction with speed v2 = 3.6 x 106 m/s. Find the magnitude and direction of the velocity of the third particle.

    Magnitude: ______ m/s
    Direction: ______ ° counterclockwise from the +x-axis

    This is what I did....

    1) M = m1 + m2 + m3

    (2.7 x 10-26) = (1.0 x 10-26) + (1.2 x 10-26) + m3
    (2.7 x 10-26) - (1.0 x 10-26) - (1.2 x 10-26) = m3
    5.0 x 10-27 = m3

    2) Determine your systems

    1st particle:
    m1 = 1.0 x 10-25 kg
    vinitial = 0 m/s
    v1final = 4.8 x 106 m/s

    2nd particle:
    m2 = 1.2 x 10-26 kg
    vinitial = 0 m/s
    v2final = 3.6 x 106 m/s

    3rd particle:
    m3 = 5.0 x 10-27 kg
    Vinitial = 0 m/s
    V3final = ? m/s

    3) Conservation of momentum equation

    pinitial = pfinal

    m1vinitial + m2vinitial + m3vinitial = m1v1final + m2v2final + m3v3final
    (m1)(0) + (m2)(0) + (m3)(0) = (1.0 x 10-25)(4.8 x 106) + (1.2 x 10-26)(3.6 x 106) + (5.0 x 10-27)(v3final)
    0 = (4.8 x 10-20) + (4.32 x 10-20) + (5.0 x 10-27)(v3final)
    -(9.12 x 10-20) = (5.0 x 10-27)(v3final)
    -18240000 = v3final
    -1.82e+07 = v3final

    "INCORRECT. CORRECT ANSWER IS: 1.27e+07"

    I don't understand!!! How did they get 1.27e+07? Help!!!
     
  2. jcsd
  3. Oct 24, 2012 #2

    SammyS

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    Staff Emeritus
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    Re: Conservation of Momentum???

    Velocity and momentum are vector quantities.

    You need to keep track of vector components.
     
  4. Oct 24, 2012 #3
    Re: Conservation of Momentum???

    I STILL get the same answer though...

    0 = (4.8 x 10-20) + (4.32 x 10-20) + (5.0 x 10-27)(v3final)

    x component


    0 = [STRIKE](4.8 x 10-20)[/STRIKE] + (4.32 x 10-20) + (5.0 x 10-27)(v3final)
    0 = (4.32 x 10-20) + (5.0 x 10-27)(v3final)
    -(4.32 x 10-20) = (5.0 x 10-27)(v3final)
    -(4.32 x 10-20)/(5.0 x 10-27) = v3final
    -8.64 x 106= v3final in the x direction

    y component


    0 = (4.8 x 10-20) + [STRIKE](4.32 x 10-20)[/STRIKE] + (5.0 x 10-27)(v3final)
    0 = (4.8 x 10-20) + (5.0 x 10-27)(v3final)
    -(4.8 x 10-20) = (5.0 x 10-27)(v3final)
    -(4.8 x 10-20)/(5.0 x 10-27) = v3final
    -9.6 x 106 = v3final in the y direction

    Add vectors

    -8.64 x 106 - 9.6 x 106 = -18240000 = -1.82 x 10^7 m/s

    -____-
     
  5. Oct 24, 2012 #4

    Doc Al

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    Staff: Mentor

    Re: Conservation of Momentum???

    How do you find the magnitude of a vector given its components? (You don't just add the components!)
     
  6. Oct 25, 2012 #5
    Re: Conservation of Momentum???

    But I just did!!!
     
  7. Oct 25, 2012 #6
    Re: Conservation of Momentum???

    Yeah, he's saying it's wrong to just add the components.

    Consider the vector and it's components as a triangle (I assume you're familiar with this representation). You're trying to the find the length of the hypotenuse.
     
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