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To what angle does each ball rebound?

  1. Dec 1, 2004 #1
    need help ASAP [solved] :)

    I've been trying for hours..... and couldn't get it right though

    A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
    Question: To what angle does each ball rebound.?
    Answer: m =100 g >>> 79.3 degree
    m = 200 g>>>>>14.7 degree
    -----------
    1) find (y_0)A when the 100 ball is pulled to the left w/ 45 degree angle
    (y_1)A = L(1-cos(theta))
    =1(1-cos(45)
    = 0.293
    2) From that I find (V_1)A = sqr(2*g*(y_0)A)
    (V_1)A = 2.40 m/s

    3) Now I find the (V_2)A = (m1-m2)/(m1+m2)*(V_1)A
    V_2)A = - 0.8 m/s

    4) Now I find (Y_2)A = (V_2)A ^2/2*g

    4) Now I need to find the angle that rebound for Ma=100 g

    (theta) = arcosin(1-(Y_2)A /L) = 37 degree....

    but the answer is 79.3 degree, please help me what i've done wrong?
     
    Last edited: Dec 1, 2004
  2. jcsd
  3. Dec 1, 2004 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Think of this in terms of an elastic collision between the two balls at the bottom. You can determine their speeds at the bottom using:
    mgh = mv2/2

    Since the collision speed is independent of m, v1 = -v2
    Using conservation of momentum you can determine their respective velocities after collision. From that, determine their respective kinetic energies immediately after the collision. Using mgh = mv2/2 you can determine the maximum height the respective masses will reach.

    AM
     
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