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On August \(2,1971,\) Astronaut David Scott, while standing on the surface of the Moon, dropped a 1.3 -kg hammer and a 0.030 -kg falcon feather from a height of \(1.6 \mathrm{~m} .\) Both objects hit the Moon's surface \(1.4 \mathrm{~s}\) after being released. What is the acceleration due to gravity on the surface of the Moon?

Short Answer

Expert verified
Answer: The acceleration due to gravity on the Moon's surface is approximately 1.63 m/s².

Step by step solution

01

Identify the relevant kinematic equation

We can use the following kinematic equation for uniformly accelerated motion: d = vi*t + 0.5*a*t^2 where d is the distance traveled, vi is the initial velocity (which is 0 in this case since the objects are dropped), a is the acceleration due to gravity on the Moon, and t is the time taken for the objects to fall.
02

Set up the equation with given values

Because the initial velocity (vi) is 0, the equation simplifies to: d = 0.5*a*t^2 Now, we can plug in the values given in the problem: 1.6 = 0.5*a*(1.4)^2
03

Solve for acceleration due to gravity (a)

We will now solve the equation for a, the acceleration due to gravity on the Moon: 1.6 = 0.5*a*(1.96) 1.6 = 0.98*a a = 1.6 / 0.98 a ≈ 1.63 m/s^2
04

Report the answer

The acceleration due to gravity on the surface of the Moon is approximately 1.63 m/s².

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Key Concepts

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

Kinematic Equations
Kinematic equations are the cornerstone of classical mechanics and essential for understanding motion. They allow us to predict the future position, velocity, and acceleration of an object when its motion is undergoing a constant acceleration. In our textbook exercise, the motion is vertical and the acceleration is due to gravity on the Moon. One of the fundamental kinematic equations used is \( d = v_i t + \frac{1}{2}a t^2 \), where \( d \) is the displacement, \( v_i \) the initial velocity, \(a\) the acceleration, and \(t\) the time. For objects dropped from rest, \( v_i = 0 \), simplifying our equation.

Understanding this equation is critical for solving problems involving uniformly accelerated motion and free fall, as it encapsulates the relationship between displacement, initial velocity, time, and acceleration. In our moon exercise, we rearranged this equation to solve for the acceleration \(a\), showing students the practical application of kinematic equations in real-world scenarios.
Uniformly Accelerated Motion
Uniformly accelerated motion describes any scenario where an object's velocity changes at a constant rate. This is the kind of motion that occurs in free fall, where gravity is the sole force acting on the object, causing it to accelerate uniformly. On Earth, this acceleration is approximately \(9.81 m/s^2\), but on the Moon, it's significantly less due to the Moon's smaller mass and gravitational force.

In the example from the textbook, the hammer and feather demonstrate uniformly accelerated motion under the Moon's gravity. By measuring the time it takes for them to fall a known distance, we can determine the acceleration due to Moon's gravity. This concept allows us to predict how any object will move under similar conditions, be it on the Moon, Earth, or any other celestial body with a gravitational field.
Free Fall
Free fall occurs when the only force acting on an object is gravity; this makes it a fascinating case study of uniformly accelerated motion. Objects in free fall, regardless of their mass, will have the same acceleration given that no other forces are acting upon them. This was famously demonstrated by Galileo and was again shown to be true during the Apollo 15 mission when an astronaut on the Moon dropped a hammer and a feather simultaneously, and they hit the ground together, just like in our textbook exercise.

In the absence of air resistance—such as on the Moon, where there's no atmosphere—objects will fall at the same rate. This notion is counter-intuitive when we observe objects falling on Earth since air resistance plays a significant role. The concept of free fall not only helps students to grasp the basics of gravitational acceleration but also underscores the universality of physical laws in different environments.

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Most popular questions from this chapter

On October 30 th, 2011 , in an NFL football game between the Detroit Lions and the Denver Broncos, Lions left cornerback Chris Johnson intercepted a pass 1 yard behind the goal line, ran down the entire length of the football field on a straight line, and scored a touchdown. Video analysis of this scoring play showed the following approximate times for him to cross the yard lines painted on the field. $$ \begin{aligned} &\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|} \hline \text { Yard } & -1 & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 \\ \hline \text { Time } & 0.00 & 0.23 & 1.16 & 1.80 & 2.33 & 2.87 & 3.37 & 3.87 & 4.33 & 4.80 & 5.27 \\ \hline \end{array}\\\ &\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|} \hline 50 & 45 & 40 & 35 & 30 & 25 & 20 & 15 & 10 & 5 & 0 & -1 \\ \hline 5.73 & 6.20 & 6.67 & 7.17 & 7.64 & 8.14 & 8.67 & 9.20 & 9.71 & 10.34 & 11.47 & 12.01 \\ \hline \end{array} \end{aligned} $$ a) What was his average speed from the time he caught the ball until he reached midfield? b) What was his average speed from the time he crossed midfield until he came to a stop 1 yard behind the opposite goal line? c) What was the average acceleration during his entire run?

Two cars are traveling at the same speed, and the drivers hit the brakes at the same time. The deceleration of one car is double that of the other. By what factor do the times required for the two cars to come to a stop differ?

A car travels at \(22.0 \mathrm{~m} / \mathrm{s}\) north for \(30.0 \mathrm{~min}\) and then reverses direction and travels at \(28.0 \mathrm{~m} / \mathrm{s}\) for \(15.0 \mathrm{~min} .\) What is the car's total displacement? a) \(1.44 \cdot 10^{4} \mathrm{~m}\) b) \(6.48 \cdot 10^{4} \mathrm{~m}\) c) \(3.96 \cdot 10^{4} \mathrm{~m}\) d) \(9.98 \cdot 10^{4} \mathrm{~m}\)

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How high above the projection point is it after \(1.00 \mathrm{~s}\) ?

An F-14 Tomcat fighter jet is taking off from the deck of the USS Nimitz aircraft carrier with the assistance of a steam-powered catapult. The jet's location along the flight deck is measured at intervals of \(0.20 \mathrm{~s}\). These measurements are tabulated as follows: $$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0.00 & 0.20 & 0.40 & 0.60 & 0.80 & 1.00 & 1.20 & 1.40 & 1.60 & 1.80 & 2.00 \\ \hline x(\mathrm{~m}) & 0.0 & 0.70 & 3.0 & 6.6 & 11.8 & 18.5 & 26.6 & 36.2 & 47.3 & 59.9 & 73.9 \\ \hline \end{array} $$ Use difference formulas to calculate the jet's average velocity and average acceleration for each time interval. After completing this analysis, can you say whether the F-14 Tomcat accelerated with approximately constant acceleration?

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