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Physics

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A 14.6-g marble is dropped from rest onto the floor 1.23 m below. If the marble bounces straight upward to a height of 0.665 m, what is the magnitude of the impulse delivered to the marble by the floor.

If the marble had bounced to a height of 0.884 m, what would the magnitude of the impulse delivered to the marble by the floor be?

  • Physics -

    m1v1 = (0.0146)(v1) (vertically downward is neg direction)
    v1 = √2gh (where h = 1.23)
    v1 = 4.91 m/s
    m1v1 = (0.0146)(4.91) = - 0.0717 kg-m/s (Momentum Before Floor)

    m2v2 = (0.0146)v2 (vertically upward)
    v2 = √2gh (where h=.665)
    v2 = 3.61 m/s
    m2v2 = (0.0146)(3.61) = 0.0527 kg-m/s (Momentum After Floor)

    Change in Momentum is:
    m2v2 - m1v1 = 0.0527 - (- 0.0717) = 0.124 kg-m/s (upward direction)
    Impulse = 0.124 N s (upward direction)

    Do the same thing with .884 instead of .665 for the second question.

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