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6 October, 00:20

A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball's final speed is 0.94 m/s at an angle of θ with respect to its original line of motion? Find the eight ball's speed after the col - lision. Assume an elastic collision (ignoring friction and rotational motion).

Answer in units of m/s.

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Answers (1)
  1. 6 October, 00:40
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    speed of eight ball speed after the collision is 3.27 m/s

    Explanation:

    given data

    initially moving v1i = 3.4 m/s

    final speed is v1f = 0.94 m/s

    angle = θ w. r. t. original line of motion

    solution

    we assume elastic collision

    so here using conservation of energy

    initial kinetic energy = final kinetic energy ... 1

    before collision kinetic energy = 0.5 * m * (v1i) ²

    and

    after collision kinetic energy = 0.5 * m * (v1f) ² + 0.5 * m * (v2f) ²

    put in equation 1

    0.5 * m * (v1i) ² = 0.5 * m * (v1f) ² + 0.5 * m * (v2f) ²

    (v2f) ² = (v1i) ² - (v1f) ²

    (v2f) ² = 3.4² - 0.94²

    (v2f) ² = 10.68

    taking the square root both

    v2f = 3.27 m/s

    speed of eight ball speed after the collision is 3.27 m/s
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