A 12.4-g marble is dropped from rest onto the floor 1.56 m below. If the marble bounces straight upward to a height of 0.614 m, what is the

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A 12.4-g marble is dropped from rest onto the floor 1.56 m below. If the marble bounces straight upward to a height of 0.614 m, what is the magnitude of the impulse delivered to the marble by the floor

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Dâu 4 years 2021-07-28T20:31:07+00:00 1 Answers 21 views 0

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  1. Nho
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    2021-07-28T20:32:57+00:00
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    Answer:

    The impulse is    I= 0.1116\ kg \cdot m/s

    Explanation:

    Generally Impulse which the change in momentum  is mathematically represented as

                              I = m\Delta v

    Where m is the mass  with a value 12.4g = \frac{12.4}{1000} = 12.4*10^{-3}kg

               \Delta v is the change in velocity which is mathematically represented as

                      \Delta v = v_2 -v_1

    Where v_1 s the velocity of the marble drooping and v_2 is the velocity of the marble bouncing back

    setting up in a coordinate y-axis would show that v_1 is moving in the negative y-direction so the  value would be v_1 = -v_1 and  v_2 is moving in the positive  y-direction so the  value would be v_2 = +v_2

    So the formula for \Delta v would be

               \Delta v = v_2 -( -v_1)

               \Delta v = v_1 +v_2

    General v i mathematically represented as

                      v = \sqrt{2gh}

     Substituting this into the formula for \Delta v

                     \Delta v = \sqrt{2gh_2} + \sqrt{2gh_1}

    Now substituting values as given in the question

                     \Delta v = \sqrt{2 * 9.8 *0.614 } + \sqrt{2 * 9.8 *1.56 }

                           = 8.9986 m/s

    The impulse is

                     I = 12.4*10^{-3} * 8.9986

                       I= 0.1116\ kg \cdot m/s

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