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[Collision with 2 Clay Balls] [Inelastic Collisions] [Help Finding Theta]

  1. Oct 26, 2008 #1
    Here's my problem guys, I've been trying to puzzle this out for a while. Would really appreciate any help at all, thanks!

    1. The problem statement, all variables and given/known data

    A 50.0 g ball of clay traveling east at 5.50 m/s collides and sticks together with a 40.0 g ball of clay traveling north at 6.00 m/s.

    What is the speed of the resulting ball of clay?

    2. Relevant equations and 3. The attempt at a solution

    My work so far:
    Let Ball travelling East be 1
    Let ball travelling North be 2

    Known
    M1 = 0.05 kg
    M2 = 0.04
    M1 + M2 = 0.09

    (Vix)1 = 5.5 m/s; (Viy)1 = 0 m/s
    (Vix)2 = 0; (Viy)2 = 6 m/s

    (Vfx) = 5.5 + 0 = 5.5 m/s
    (Vfy) = 0 + 6 = 6 m/s


    So the X components are equal to Vf* (as in, the vector V)Cos(Theta)

    And the Y components are equal to Vf*Sin(Theta)

    => (M1 + M2)Vfx = (M1 + M2)VfCos(Theta) = M1(Vix)1 + M2(Vix)2 {which equals zero} = M1(Vix)1

    (M1 + M2)Vfy = (M1 + M2)VfSin(Theta) = M1(Viy)1{which equals zero} + M2(Viy)2 = M2(Viy)2
     
  2. jcsd
  3. Oct 26, 2008 #2
    Nevermind, I worked it out in the end. For anyone on the same problem, just get the momenta of the two balls, and then divide the momenta by the final mass (0.09kg) to get the velocities in the X and Y directions. Then simply use the baldy Greek's theorem (Pythagoras).


    x momentum = (0.05kg)(5.5 m/s) = 0.275 J
    y momentum = (0.04kg)(6 m/s) = 0.24 J
    (0.24 J) / (0.09 kg) = FInal y velocity = 2.667 m/s
    (0.275 J) / (0.09 kg) = Final x velocity = 3.056 m/s
    Final Speed = sqrt(Vfx^2 + Vfy^2) = 4.056 m/s
    Theta = arctan(2.667 / 3.056) = 41.1 deg
     
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