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Collision/conservation of momentum & energy

  1. Nov 1, 2008 #1
    1. In Fig. 10-41, ball 1 with an initial speed of 12 m/s collides elastically with stationary balls 2 and 3 that are initially in contact with each other. The centers of balls 2 and 3 are on a line perpendicular to the initial velocity of ball 1. The three balls are identical. Ball 1 is aimed directly at the contact point, and all motion is frictionless.(Hint: With friction absent, each impulse is directed along the line connecting the centers of the colliding balls, normal to the colliding surfaces.)



    2. momentum = mv, KE = 1/2mv2



    3. after the collision, marbles two and three are going at the same velocity and angle; lets call them vcos(x). marble 1 is going in the opposite direction it had been traveling at initially. so since v2 = v3 , my final equations are 12 = 2v2cos(x) - v1 (mass cancels out). since there are two unknowns, i wrote a second equation: 1/2(12)^2 = v1^2 + 1/2(v2cos(x))^2, tried to solve using substitution, and didnt work. For angle x, i've tried both 30 degrees and 45 degrees. I did this using various methods about ten times. help :(
     
  2. jcsd
  3. Nov 2, 2008 #2

    Doc Al

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    Staff: Mentor

    OK.
    Redo this one. The KE must be calculated using the full velocity of the ball, not just one component. Set the total KE before equal to the total KE after.
    Figure out the angle using geometry. What angles do their centers make during the collision?
     
  4. Nov 2, 2008 #3
    Thank you, Doc Al! I got my new answers to be v1 = 3.4 m/s, v2=v3 = 8.3 m/s (they were correct).
     
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