A 14 kg rock starting from rest free falls through a distance of 5.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of earth’s velocity. Earth’s mass is 6.0 × 1024 kg. Show all your work, assuming the rock-earth system is closed.

v = sqrt (2 g h) = sqrt (2*9.8*5)

= 9.9 m/s

total momentum before = 0
total momentum after
= 14*9.9 - 6*10^24 v = 0

v = (14 * 9.9/6)*10^-24

So, let's go. Vf^2{squared final velocity}= Vi^2{squared initial velocity}+2a{acceleration}d{distance}. Acceleration is 9.8 for gravity, good luck my dudes.

To find the momentum change of the rock caused by its fall, we can use the formula:

Δp = m * Δv

Where Δp is the change in momentum, m is the mass of the rock, and Δv is the change in velocity.

Given:
mass of the rock (m) = 14 kg
change in velocity (Δv) = ?

To find the change in velocity, we need to use the equation for gravitational potential energy and equate it to the change in kinetic energy:

m * g * h = 0.5 * m * Δv^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height (5.0 m in this case).

By rearranging the equation, we can solve for Δv:

Δv = √(2 * g * h)

Substituting the given values:

Δv = √(2 * 9.8 m/s^2 * 5.0 m)
= √(98 m^2/s^2)
= 9.9 m/s

Now, we can calculate the momentum change:

Δp = m * Δv
= 14 kg * 9.9 m/s
= 138.6 kg·m/s

Therefore, the momentum change of the rock caused by its fall is 138.6 kg·m/s.

To find the resulting change in the magnitude of Earth's velocity, we divide the momentum change by the mass of Earth:

change in Earth's velocity = Δp / mass of Earth
= 138.6 kg·m/s / (6.0 × 10^24 kg)

This calculation gives a very small change in Earth's velocity, which is negligible. Therefore, we can ignore the change in the magnitude of Earth's velocity.

To find the momentum change of the rock caused by its fall, we can use the equation

Δp = m * Δv

where Δp represents the change in momentum, m is the mass of the rock, and Δv is the change in velocity of the rock.

First, let's find the final velocity of the rock. We can use the equation for free fall:

v² = u² + 2gh

where v is the final velocity, u is the initial velocity (which is 0 m/s since the rock starts from rest), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height the rock falls through (5.0 m in this case).

Plugging in the values, we get:

v² = 0 + 2 * 9.8 * 5.0
v² = 98
v ≈ √98
v ≈ 9.90 m/s

Now that we have the final velocity, we can calculate the change in velocity:

Δv = v - u
Δv = 9.90 - 0
Δv = 9.90 m/s

Next, we can calculate the momentum change:

Δp = m * Δv
Δp = 14 kg * 9.90 m/s
Δp = 138.6 kg·m/s

So, the momentum change of the rock caused by its fall is 138.6 kg·m/s.

To find the resulting change in the magnitude of Earth's velocity, we can use the law of conservation of momentum. Since the rock-Earth system is closed, the total momentum before the rock falls is zero, and after the rock falls, the total momentum remains zero.

Thus, the change in the magnitude of Earth's velocity is equal to the change in the magnitude of the rock's velocity, but in the opposite direction.

Therefore, the change in the magnitude of Earth's velocity is also 9.90 m/s, but in the opposite direction.