/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A lunar lander is making its des... [FREE SOLUTION] | 91Ó°ÊÓ

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A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The speed of the lunar lander just before it touches the surface is \(4.08 \mathrm{~m/s}\).

Step by step solution

01

Identify the relevant parameters

The initial velocity (\(v_i\)) of the lander when the engine is cut is \(0.8 \mathrm{~m/s}\). The lander then falls a distance of \(5.0 \mathrm{~m}\) under the influence of moon's gravity of \(1.6 \mathrm{~m/s^{2}}\) (acceleration, \(a\)). The final velocity (\(v_f\)) at the end of this fall, i.e., just before it hits the ground, is what we are asked to find.
02

Apply the second equation of motion

The second equation of motion is \(v_f^2 = v_i^2 + 2ad\). Substituting the known values into this equation, we get \(v_f^2 = (0.8 \mathrm{~m/s})^2 + 2 * 1.6 \mathrm{~m/s^2} * 5.0 \mathrm{~m}\).
03

Solve for the final speed

Upon performing the computation, you’ll find \(v_f^2 = 0.64 \mathrm{~m^2/s^2} + 16 \mathrm{~m^2/s^2} = 16.64 \mathrm{~m^2/s^2}\). Now, we apply the square root on both sides, since velocity is not squared in normal practice. The square root of \(16.64\) will give us the final velocity \(v_f\) as \(4.08 \mathrm{~m/s}\).

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Key Concepts

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

Understanding Free Fall
Free fall is a fascinating concept in physics. Imagine an object that is falling solely under the influence of gravity, without any other forces acting on it. In the case of the lunar lander in the problem, once the engines stop, it enters this state of free fall. Here it is influenced only by the Moon's gravitational force which has an acceleration of \(1.6 \, \mathrm{m/s^2}\). This is different from Earth's gravity, which is much stronger at \(9.8 \, \mathrm{m/s^2}\).

When something is in free fall, its only acceleration is due to gravity. There is no interference from other forces like air resistance, as there would be on Earth. This makes calculations on the Moon both exciting and a tad bit simpler!

It's important to note that the velocity of an object in free fall increases steadily as it falls, as it's being pulled faster and faster towards the gravitational body. This increase in speed under lunar gravity is neatly explained using the equations of motion.
Equations of Motion Simplified
The equations of motion are handy tools that help us predict how objects move. They relate velocities, accelerations, distances, and times in a neat formulaic way.

In this problem, we used the equation: \[v_f^2 = v_i^2 + 2ad\] where:
  • \(v_f\) is the final velocity, which we want to find out.
  • \(v_i\) is the initial velocity (\(0.8 \, \mathrm{m/s}\) when the engines stop).
  • \(a\) is the acceleration due to moon's gravity (\(1.6 \, \mathrm{m/s^2}\)).
  • \(d\) is the distance the lander falls (\(5.0 \, \mathrm{m}\)).
This equation comes from combining other basic motion equations, distilling them into a form perfect for objects under constant acceleration, like our lunar lander's free fall. Using this specific equation allowed us to calculate the final speed effortlessly. It's a clear and systematic way to solve for the unknown while knowing initial conditions and acceleration.
Lunar Gravity: A Different World
The Moon's gravity is significantly weaker than Earth's. It is only about \(1/6\) of Earth's gravitational pull. This difference is essential not only for calculations but also for how we consider landing spacecraft or even imagining how walking would be on the Moon.

Because the Moon has less gravity, objects will fall slower than they would on Earth. This means a descending spacecraft like our lunar lander can take advantage of this weaker pull, requiring less thrust to land softly. The lower gravity affects all aspects of physics on the lunar surface, from how high you can jump to how objects settle.

For astronauts, this means both challenges and advantages, navigating an environment with different physical laws. Understanding lunar gravity is crucial for planning lunar missions, influencing engineering designs and mission protocols. Calculating movement under lunar gravity, as we saw, involves familiar formulas but applied in novel scenarios.

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Most popular questions from this chapter

The acceleration of a particle is given by \(a_{x}(t)=\) \(-2.00 \mathrm{~m} / \mathrm{s}^{2}+\left(3.00 \mathrm{~m} / \mathrm{s}^{3}\right) t\) Find the initial velocity \(v_{0 x}\) such that the particle will have the same \(x\) coordinate at \(t=4.00 \mathrm{~s}\) as it had at t=0 . \text { (b) What will be the velocity at } t=4.00 \mathrm{~s} ?

A car's velocity as a function of time is given by \(v_{x}(t)=\alpha+\beta t^{2}, \quad\) where \(\quad \alpha=3.00 \mathrm{~m} / \mathrm{s} \quad\) and \(\quad \beta=0.100 \mathrm{~m} / \mathrm{s}^{3}\) (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\). (b) Calculate the instantancous acceleration for \(t=0\) and \(t=5.00 \mathrm{~s}\). (c) Draw \(v_{x}-t\) and \(a_{x}-t\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{~s}\)

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for \(10.0 \mathrm{~s}\), Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back \(7.0 \mathrm{~s}\) after leaving the helicopter, and then he has a constant downward acceleration with magnitude \(2.0 \mathrm{~m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

The engineer of a passenger train traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) sights a freight train whose caboose is \(200 \mathrm{~m}\) ahead on the same track (Fig. \(\mathrm{P} 2.62\) ). The freight train is traveling at \(15.0 \mathrm{~m} / \mathrm{s}\) in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of \(0.100 \mathrm{~m} / \mathrm{s}^{2}\) in a direction opposite to the train's velocity, while the freight train continues with constant speed. Take \(x=0\) at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{~m}+\left(4.80 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}-\) \(\left(0.100 \mathrm{~m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{~s}\)

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