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Mobile App Chapter 3: Problem 36
Based on Figure \(3,\) if \(Y_{0}\) were \(5 \mathrm{cm},\) the fall time on the Moon would be closest to: F. 0.4 sec. G. 0.8 sec. H. 1.9 sec. J. 2.3 sec.
Short Answer
Expert verified
Answer: F. 0.4 seconds
Step by step solution
01
Write down the general equation for the distance fallen in free fall.
We can start by writing the equation for free fall:
\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
In this equation, \(y\) is the final position, \(y_0\) is the initial position, \(v_0\) is the initial velocity, \(t\) is the time, and \(g\) is the acceleration due to gravity.
02
Fill in the given values and simplify the equation
We are given that the initial height \(y_0\) is 5 cm. Since the object is released from rest, its initial velocity \(v_0 = 0\). The final position \(y\) will be at ground level, so \(y = 0\).
On the Moon, \(g\) is approximately 1/6th of the Earth's acceleration due to gravity, which is approximately \(9.81 m/s^2\). So, \(g_{moon} = \frac{1}{6} \times 9.81 m/s^2 = 1.635 m/s^2\). Remember to convert centimeters to meters: \(y_0 = 0.05 m\). Thus, the equation becomes:
\[0 = 0.05 + 0 + \frac{1}{2}(1.635)t^2\]
03
Solve for time (t)
Now we need to solve the equation above for \(t\). First, we can simplify the equation:
\[-0.05 = 0.8175t^2\]
Divide by 0.8175:
\[-0.305 = t^2\]
Ignoring the negative sign, we get:
\[0.305 = t^2\]
Taking the square root of both sides, we get the time \(t\):
\[t = \sqrt{0.305} = 0.55 s\]
04
Determine which answer choice is closest
Reviewing the given answer choices, we see that 0.55 seconds is closest to 0.4 seconds (choice F). Thus, the fall time on the Moon would be closest to 0.4 seconds. The answer is F.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gravitational Acceleration
Gravitational acceleration is the acceleration that an object undergoes due to the pull of gravitational force when it is in free fall. On Earth, this acceleration is approximately equal to \(9.81 m/s^2\), a value that we often round off to \(10 m/s^2\) for ease of calculation in many physics problems.
In a different celestial environment such as the Moon, this value changes due to the Moon's weaker gravitational force. As seen in the exercise, the gravitational acceleration on the Moon is approximately \(1.635 m/s^2\), which is one sixth of the Earth's gravitational acceleration. Understanding this concept is vital, as it allows us to predict how the motion of a falling object will differ on other planets or moons.
In a different celestial environment such as the Moon, this value changes due to the Moon's weaker gravitational force. As seen in the exercise, the gravitational acceleration on the Moon is approximately \(1.635 m/s^2\), which is one sixth of the Earth's gravitational acceleration. Understanding this concept is vital, as it allows us to predict how the motion of a falling object will differ on other planets or moons.
Motion Equations
Motion equations, also known as equations of motion, are used to describe the relationship between an object's position, velocity, acceleration, and time. The key equation for free fall motion, when an object starts from rest, is given by:
\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
where \(y\) is the final position, \(y_0\) is the initial position, \(v_0\) is the initial velocity (which is zero for an object starting from rest), \(t\) represents time, and \(g\) denotes the acceleration due to gravity. By plugging in the known values, this equation can be manipulated algebraically to solve for the unknown variable, such as the time of fall in our exercise.
\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
where \(y\) is the final position, \(y_0\) is the initial position, \(v_0\) is the initial velocity (which is zero for an object starting from rest), \(t\) represents time, and \(g\) denotes the acceleration due to gravity. By plugging in the known values, this equation can be manipulated algebraically to solve for the unknown variable, such as the time of fall in our exercise.
Physics Problem-Solving
Physics problem-solving involves a structured approach to finding solutions to physical scenarios using mathematical formulas and scientific concepts. In the context of free fall motion, problem-solving often requires recognizing the known and unknown variables, selecting the appropriate motion equations, and performing the necessary algebraic manipulations to solve for the desired unknown.
The exercise provided demonstrates these steps clearly, starting from writing down the free fall equation, substituting known values for gravitational acceleration and initial positions, then solving for the unknown time. It's crucial to perform unit conversions when necessary (like converting from centimeters to meters) and to apply the correct physical principles, such as using the appropriate value of gravitational acceleration for the Moon.
The exercise provided demonstrates these steps clearly, starting from writing down the free fall equation, substituting known values for gravitational acceleration and initial positions, then solving for the unknown time. It's crucial to perform unit conversions when necessary (like converting from centimeters to meters) and to apply the correct physical principles, such as using the appropriate value of gravitational acceleration for the Moon.
ACT Prep
ACT prep often includes brushing up on science concepts, particularly physics, as they are commonly featured in the Science Reasoning section of the ACT test. Grasping the foundational physics concepts, such as free fall motion and gravitational acceleration, and knowing how to apply motion equations are essential skills for excelling in this section.
Practice problems similar to the one provided can help students to improve their problem-solving speed and accuracy, both of which are crucial for the timed nature of the ACT. In addition to mastering the content, students should also familiarize themselves with the format of the test and develop test-taking strategies, such as process of elimination and educated guessing, which was exemplified in the step-by-step solution process by selecting the closest answer to the calculated time of fall.
Practice problems similar to the one provided can help students to improve their problem-solving speed and accuracy, both of which are crucial for the timed nature of the ACT. In addition to mastering the content, students should also familiarize themselves with the format of the test and develop test-taking strategies, such as process of elimination and educated guessing, which was exemplified in the step-by-step solution process by selecting the closest answer to the calculated time of fall.
