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Chapter 4: Problem 43

Free Fall On the moon, the acceleration due to gravity is 1.6 \(\mathrm{m} / \mathrm{sec}^{2} .\) (a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later? (b) How far below the point of release is the bottom of the crevasse? (c) If instead of being released from rest, the rock is thrown into the crevasse from the same point with a downward velocity of \(4 \mathrm{m} / \mathrm{sec},\) when will it hit the bottom and how fast will it be going when it does?

Short Answer

Expert verified
The answers are (a) the rock will be moving at 48 m/s when it hits the bottom (b) the bottom of the crevasse is 720 m below the point of release and (c) the rock will hit the bottom after 27.5 seconds and at a velocity of 48 m/s.

Step by step solution

01

- Calculate Final Velocity for Part (a)

To solve this, use the formula for velocity under constant acceleration without an initial velocity: Velocity = Acceleration * Time. Given acceleration on moon as 1.6 m/s^2 and time as 30 seconds, Velocity = 1.6 m/s^2 * 30 s = 48 m/s.
02

- Calculate Distance for Part (b)

To solve this, use formula for distance in case of motion under constant acceleration: Distance = 0.5 * Acceleration * Time^2. Substituting the values given, Distance = 0.5 * 1.6 m/s^2 * (30 s)^2 = 720 m.
03

- Calculate Time and final Velocity for Part (c)

Firstly, calculate the time it would take for the rock to hit the bottom if thrown at an initial velocity, using the equation of motion: Time = (Final Velocity - Initial Velocity) / Acceleration. However, in this case, we already calculated the final velocity when the rock is dropped (48 m/s), we know the initial velocity is 4 m/s, and the acceleration is 1.6 m/s^2 . So, Time = (48 m/s - 4 m/s) / 1.6 m/s^2 = 27.5 s. Secondly, calculate the final velocity when the rock hits the bottom with the same equation from Step 1. But this time, it has an initial velocity: Final Velocity = Initial Velocity + Acceleration * Time. Substituting the values, Final Velocity = 4 m/s + 1.6 m/s^2 * 27.5 s = 48 m/s. Note that the final velocity is same as that in part (a) as the distance of fall remains the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravity on the Moon
The concept of gravity varies from one celestial body to another. On Earth, we often use the value of approximately 9.8 m/s虏. However, the moon's gravity is much weaker, approximately 1.6 m/s虏. This difference impacts how objects behave in motion. If you drop a rock on the moon, it will fall slower than it would on Earth, but it will still accelerate over time. Understand that the weaker the gravitational pull, the slower the acceleration towards the surface.
Velocity Calculation
When calculating velocity under constant acceleration, it's helpful to use the formula:
  • Final Velocity = Initial Velocity + (Acceleration * Time).
In free fall problems, if an object is dropped, the initial velocity is typically zero. For a rock dropped on the moon, the final velocity after 30 seconds would be: \[0 \text{ m/s} + (1.6 \text{ m/s}^2 \times 30 \text{ s}) = 48 \text{ m/s}.\]This means that the rock will be moving at 48 m/s just before impact.
Distance Calculation
Distance calculation in free fall requires knowing the equation:
  • Distance = \(0.5 \times \text{Acceleration} \times \text{Time}^2\).
For an object dropped from rest on the moon:\[0.5 \times 1.6 \text{ m/s}^2 \times (30 \text{ s})^2 = 720 \text{ m}.\]This tells us the rock will fall to a depth of 720 meters in 30 seconds. This method calculates the distance by considering the constant acceleration due to gravity.
Equations of Motion
Equations of motion are useful tools in physics, especially for predicting how objects move. When an object is thrown with an initial velocity, the equation becomes:
  • Final Velocity = Initial Velocity + (Acceleration * Time)
For a rock thrown downward, adjust the equation:
  • Initial velocity is now included.
  • If thrown at 4 m/s, the final velocity needs the adjustment for that initial push.
  • Use the calculated final velocities and distances to solve for time as needed.
The integrity of equations remains, they utilize constant values like acceleration to evaluate where and how quickly an object will move.

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Most popular questions from this chapter

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that \(X\) -ray photographs confirm that the trachea contracts about this much during a cough. (b) Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

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