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) 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.

Respuesta :

Answer:

Part a)

Momentum transferred by the ball

[tex]\Delta P = 157.95 kg m/s[/tex]

Part b)

Change in the velocity of the ball is

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

Explanation:

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

[tex]v = \sqrt{2gh}[/tex]

so we will have

[tex]v = \sqrt{2(9.81)(30)}[/tex]

[tex]v = 24.3 m/s[/tex]

Now by momentum conservation

[tex]m_1v_i = (m_1 + m_2) v_f[/tex]

[tex]6.50(24.3) = (6.50 + 5.972 \times 10^{24})v[/tex]

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

Part a)

Momentum transferred by the stone is given as

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

[tex]\Delta P = 157.95 kg m/s[/tex]

Part b)

Change in velocity of Earth

[tex]\Delta v = 2.64 \times 10^{-23} - 0[/tex]

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

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