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Two-Dimensional Collisions

  1. Nov 28, 2004 #1

    Bri

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    The mass of the blue puck in the figure is 20.0% greater than the mass of the green one. Before colliding the, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds of the pucks after the collision if half the kinetic energy is lost.

    Code (Text):

    [COLOR=Green]
              /
             /
            /   A
    O-->---   [/COLOR][COLOR=Blue]---<--O
         B   /
            /
           /
    [/COLOR]
     
    Angles A and B are 30 degrees.

    I calculated the initial velocity of the blue puck to be 25/3 m/s. I've tried setting the initial Kinetic energy in the x and y directions equal to two times the final Kinetic energy in the x and y directions and solving the system of equations, but it doesn't work out.
    Could someone please tell me how I should set this up so I can solve it?
    Thanks.
     
  2. jcsd
  3. Nov 28, 2004 #2

    Doc Al

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    Staff: Mentor

    You need to apply:
    (1) conservation of momentum
    (2) the fact that the total KE after the collision is half what it was before the collision.
     
  4. Nov 28, 2004 #3

    Bri

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    Do I need conservation of momentum to solve for the final speeds?
    Should I be separating it into two equations for KE and Momentum, one for in the x direction and one for in the y direction? And wouldn't I end up with 4 equations then? What would I do with them?

    I used momentum to find the initial velocity of the blue puck. So far I've been using these equations to try to solve this...

    KE(Blue, initial, x direction) + KE(Green, initial, x direction) = 2*KE(Blue, final, x direction) + 2*KE(Green, final, x direction)

    KE(Blue, initial, y direction) + KE(Green, initial, y direction) = 2*KE(Blue, final, y direction) + 2*KE(Green, final, y direction)

    I simplified those and solved the system.
     
    Last edited: Nov 28, 2004
  5. Nov 29, 2004 #4

    Doc Al

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    Staff: Mentor

    It's much simpler than all that. You have two unknowns (the final speeds of the pucks), but you also have two equations:
    (1) conservation of momentum: What's the total momentum of the system?
    (2) the fact that the total KE after the collision is half what it was before the collision

    That's all you need. (Here's a hint: Let the mass of the green puck = m; then the mass of the blue puck = 1.2m. You know the initial speeds of the pucks so you should be able to figure out the initial KE, at least in terms of m.)
     
  6. Nov 29, 2004 #5

    Bri

    User Avatar

    Ok, so I used:

    m(b) = 1.2m(g)
    v(bi) = 25/3 m/s
    v(gi) = 10 m/s

    m(b)v(bi) + m(g)v(gi) = m(b)v(bf) + m(g)v(gf)
    1.2m(g)v(bi) + m(g)v(gi) = 1.2m(g)v(bf) + m(g)v(gf)
    1.2v(bi) + v(gi) = 1.2v(bf) + v(gf)
    20 = 1.2v(bf) + v(gf)

    .5m(b)v(bi)^2 + .5m(g)v(gi)^2 = m(b)v(bf)^2 + m(g)v(gf)^2
    275/3 = 1.2v(bf)^2 + v(gf)^2

    I solved for v(gf) in the momentum equation and put it in the kinetic energy equation and set the equation equal to zero to get

    0 = 2.64v(bf)^2 - 48v(bf) + 925/3

    There's no solution.
     
  7. Nov 29, 2004 #6
    If I didn't make any math errors, the KE before the collision is 65*m(g)

    So the KE after the collision is 32.5*m(g) which will be equal to

    .5*m(g)*v(g2)^2 + .5*(1.2*m(g))*v(b2)^2

    Note that the m(g) terms cancel out leaving
    32.5 = .5*v(g2)^2 + .5*1.2*v(b2)^2

    constrained by m(g)*v(g2) + 1.2*m(g)*v(b2) = 0
    (post momentum = pre momentum = 0)

    this is the same as saying v(g2) = -1.2*v(b2) ---- so it you stick the value of v(g2) in terms of v(b2) into the Post-KE expression, don't you get v(b2) and thus v(g2) or have I missed something?
     
    Last edited: Nov 29, 2004
  8. Nov 29, 2004 #7

    Doc Al

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    Staff: Mentor

    Realize that the pucks move in opposite directions.

    Since "the pucks approach each other with momenta of equal magnitudes and opposite directions", what must be the total momentum?
     
  9. Nov 29, 2004 #8

    Bri

    User Avatar

    Woohoo!! I got it now...
    Had to set the momentum to 0, not 20. Setting one of their velocities to negative because of their opposite direction slipped past me.
    Thanks so much for all the help!
     
  10. Feb 2, 2011 #9


    PSE 5E PRO 9.30
    PSE 6E PRO 9.31

    The mass of the blue puck in Figure P9.30 is 20.0% greater than the
    mass of the green one. Before colliding, the pucks approach each other
    with equal and opposite momenta, and the green puck has an initial
    speed of 10.0 m/s. Find the speeds of the pucks after the collision if
    half the kinetic energy is lost during the collision.

    ----------------------------------------------------------------------

    The mass of the blue puck in Figure P9.30 is 20.0% greater
    than the mass of the green one.

    (1) m2 = m1 + 0.2 m1

    m2 = m1 ( 1 + 0.2 )

    m2 = m1 ( 1.2 )

    m2 = 1.2 m1

    Before colliding, the pucks approach each other with equal and
    opposite momenta,

    (2) pi1 = - pi2

    pi1 = m1 vi1

    pi2 = m2 vi2

    the green puck has an initial speed of 10.0 m/s

    (3) vi1 = 10

    (4) pi = pf

    pi = pi1 + pi2

    pf = pf1 + pf2

    pf1 = m1 vf1

    pf2 = m2 vf2

    (5) Ef = 1/2 Ei

    Ei = Ki1 + Ki2

    Ef = Kf1 + Kf2

    Ki1 = 1/2 m1 vi1^2

    Ki2 = 1/2 m2 vi2^2

    Kf1 = 1/2 m1 vf1^2

    Kf2 = 1/2 m2 vf2^2

    ----------------------------------------------------------------------

    (2): pi1 = - pi2

    m1 vi1 = - m2 vi2

    m1 vi1 = - (1.2 m1) vi2

    vi1 = - 1.2 vi2

    vi2 = - vi1 / 1.2

    vi2 = - 8.33

    (4): pi = pf

    pi1 + pi2 = pf1 + pf2

    (- pi2) + pi2 = pf1 + pf2

    0 = pf1 + pf2

    0 = m1 vf1 + m2 vf2

    0 = m1 vf1 + (1.2 m1) vf2

    0 = vf1 + (1.2) vf2

    vf2 = - vf1 / 1.2

    ----------------------------------------------------------------------

    (5): Ef = 1/2 Ei

    Kf1 + Kf2 = 1/2 (Ki1 + Ki2)

    1/2 m1 vf1^2 + 1/2 m2 vf2^2 = 1/2 (1/2 m1 vi1^2 + 1/2 m2 vi2^2)

    m1 vf1^2 + m2 vf2^2 = 1/2 m1 vi1^2 + 1/2 m2 vi2^2

    m1 vf1^2 + (1.2 m1) vf2^2 = 1/2 m1 vi1^2 + 1/2 (1.2 m1) vi2^2

    vf1^2 + (1.2) vf2^2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

    vf1^2 + (1.2) (- vf1 / 1.2)^2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

    vf1^2 + vf1^2 / 1.2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

    vf1^2 ( 1 + 1 / 1.2 ) = 1/2 vi1^2 + 1/2 (1.2) vi2^2

    vf1^2 = ( 1/2 vi1^2 + 1/2 (1.2) vi2^2 ) / ( 1 + 1 / 1.2 )

    vf1 = sqrt[ ( 1/2 vi1^2 + 1/2 (1.2) vi2^2 ) / ( 1 + 1 / 1.2 ) ]

    vf1 = 7.07

    ----------------------------------------------------------------------

    vf2 = - 5.89

    ----------------------------------------------------------------------


    Also available at: https://gist.github.com/808567
     
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