) A 6.50 kg rock starting from rest free-falls through a distance of 30.0 m. a. Assuming no air resistance, find the amount of momentum th

Question

) A 6.50 kg rock starting from rest free-falls through a distance of 30.0 m. a. Assuming no air resistance, find the amount of momentum that is transferred from the rock to earth as it collides with earth’s surface in a perfectly inelastic collision. b. What is the change in velocity of earth as a result of this momentum change? The earth’s mass is 5.972 x 10^24 kg. Show all your work, assuming the rock–earth system is closed.

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Kiệt Gia 4 years 2021-08-25T10:15:35+00:00 1 Answers 4 views 0

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  1. Euphemia
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    2021-08-25T10:16:35+00:00
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    Answer:

    Part a)

    Momentum transferred by the ball

    \Delta P = 157.95 kg m/s

    Part b)

    Change in the velocity of the ball is

    \Delta v = 2.64 \times 10^{-23} m/s

    Explanation:

    Velocity of the stone just before it will strike the Earth is given as

    v = \sqrt{2gh}

    so we will have

    v = \sqrt{2(9.81)(30)}

    v = 24.3 m/s

    Now by momentum conservation

    m_1v_i = (m_1 + m_2) v_f

    6.50(24.3) = (6.50 + 5.972 \times 10^{24})v

    v = 2.64 \times 10^{-23} m/s

    Part a)

    Momentum transferred by the stone is given as

    \Delta P = 6.50(24.3 - 2.64 \times 10^{-23})

    \Delta P = 157.95 kg m/s

    Part b)

    Change in velocity of Earth

    \Delta v = 2.64 \times 10^{-23} - 0

    \Delta v = 2.64 \times 10^{-23} m/s

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