How would you solve this ?

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  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    a) begin by conserving "vertical" momentum, which is zero. Make "to the right" a negative angle:

    0 = M*v*sin53º + M*V*sin-37º

    where v is the velocity of the second ball.

    v = V*sin37º / sin53º = 0.7536V

    conserve momentum in horizontal direction, substituting for v:

    M*4m/s = M*0.7536V*cos53º + M*V*cos-37º

    4 m/s = 1.252V

    V = 3.2 ms ◄ originally stationary ball

    v = 0.7536*3.2m/s = 2.4 m/s ◄ originally moving ball

    b) this is basically just using this same process iteratively.

    However, we must make an assumption in order to find a solution -- elastic collisions. We know that for bodies of equal mass, an elastic collision results in a scatter angle of 90º (as in part a). Otherwise we don't have enough data to solve for the velocities and the angles.

    conserve "vertical" momentum for the cue-9 collision:

    0 = M*v*sin20º + M*V*sin-70º

    where v is the velocity of the cue ball

    v = V*sin70º / 20º = 2.747V

    and so horizontally

    M*6m/s = M*(2.747V)*cos20º + M*V*cos-70º

    6 m/s = 2.924V

    V = 2.05 m/s ≈ 2.1 m/s ◄ 9-ball

    v = 2.747V = 5.6 m/s ◄ cue ball after 1st collision

    second collision:

    conserve "vertical" momentum

    0 = M*v*sin40º + M*V*sin-50º

    where v is new post-collision velocity of the cue ball

    v = V*sin50º/sin40º = 1.192V

    and horizontally, substituting for v:

    M*5.6m/s = M*(1.192V)*cos40º + M*V*cos50º

    5.6 m/s = 1.556V

    V = 3.6 m/s ◄ 8-ball

    v = 1.192V = 4.3 m/s ◄ cue ball, final

    But check the math!

    If you find this helpful, please award Best Answer!

    • jennifer2 months agoReport

      How do you go from 
      0 = M*v*sin53º + M*V*sin-37º to v = V*sin37º / sin53º = 0.7536V

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  • Anonymous
    2 months ago

    Open up photoshop and go to image > rotate canvas

    Then save the new image as a pdf or a jpg. 

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