/*! 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 11 A spacecraft with a proper lengt... [FREE SOLUTION] | 91Ó°ÊÓ

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A spacecraft with a proper length of 300 \(\mathrm{m}\) takes 0.750 \mus to pass an Earth observer. Determine the speed of the spacecraft as measured by the Earth observer.

Short Answer

Expert verified
The calculated speed of the spacecraft is \(4.00 \times 10^{8}\) meters per second, but this exceeds the speed of light and thus indicates a possible error in the problem or the input data.

Step by step solution

01

Understanding the Problem

A spacecraft has a proper length (the length in the spacecraft's rest frame) of 300 meters and it passes by an Earth observer in 0.750 microseconds. We are asked to find the relative speed of the spacecraft with respect to the observer on Earth.
02

Calculating Speed

Speed is calculated using the formula speed = distance / time. The distance is the proper length of the spacecraft as measured in its own frame of reference, and the time is how long it takes to pass the observer from the observer's frame of reference.
03

Convert Time Unit from Microseconds to Seconds

Before we can calculate the speed, we need all units in SI units. 1 microsecond (\(\mu s\)) is equivalent to \(1 \times 10^{-6}\) seconds. Therefore, 0.750 microseconds is \(0.750 \times 10^{-6}\) seconds or \(7.50 \times 10^{-7}\) seconds.
04

Calculating the Speed in Meters per Second

Now we use the distance (300 meters) and the time (\(7.50 \times 10^{-7}\) seconds) in the speed formula: \(v = \frac{d}{t}\).
05

Solve for the Speed

Substitute the values into the formula to get \(v = \frac{300 \, \text{m}}{7.50 \times 10^{-7} \, \text{s}}\). This calculation will give us the speed of the spacecraft as measured by the Earth observer.
06

Performing the Calculation

Perform the division to find the speed: \(v = \frac{300}{7.50 \times 10^{-7}}\) meters per second.
07

Final Answer

Completing the calculation gives the speed of the spacecraft: \(v = 4.00 \times 10^{8}\) meters per second. However, this speed is greater than the speed of light and is physically impossible according to our current understanding of physics. It seems there may have been a mistake in the interpretation of the problem or the given data.

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Key Concepts

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

Proper Length
When discussing the concept of proper length, we refer to the measure of length for an object in the frame of reference in which the object is at rest. In our example with the spacecraft, the proper length is given as 300 meters. This is how long the spacecraft is when measured in a frame where it's not moving. It's important to understand proper length because it remains constant for an object, even when observed from different frames of reference where it might appear to length contract due to relativistic effects.
In the exercise improvement advice, when dealing with calculations involving proper length, ensure students understand that this is the length measured in the object's rest frame and emphasize the distinction between proper length and relativistic length contraction that occurs when the object moves with a significant fraction of the speed of light relative to the observer.
Relative Velocity
Relative velocity is the velocity of an object as observed from a certain frame of reference. It is a key concept in understanding motion in physics because how fast an object appears to be moving can differ depending on the observer's frame of reference. In our spacecraft exercise, the relative velocity is the speed of the spacecraft as measured by the Earth observer.
Relative velocity becomes even more interesting when objects move at velocities close to the speed of light, leading to relativistic effects due to the principles of special relativity. As such, we must consider the nature of these effects when calculating high speeds that may approach or exceed a significant fraction of the speed of light.
Speed of Light
One of the fundamental constants of the universe is the speed of light in vacuum, denoted as 'c'. The value of c is approximately 299,792,458 meters per second. Nothing can travel faster than the speed of light in a vacuum according to our current understanding of physics. This limit has profound implications for our calculations involving high-speed objects, as we saw in our spacecraft scenario.
In our given exercise, the initial calculation seemingly resulted in a velocity greater than the speed of light, which highlights a key point: any speed calculation that exceeds the speed of light necessitates reexamination. This could imply an error in calculation or an incorrect interpretation of the problem. Always remind students that when dealing with high speeds, the speed of light represents an upper boundary that cannot be surpassed.
Time Dilation
Time dilation is a phenomenon predicted by Einstein’s theory of special relativity, which states that time measured between events is not absolute but is dependent on the relative velocity between observers. As the velocity of an object approaches the speed of light, time begins to slow down for the moving object relative to a stationary observer. This means that if one were to observe a clock on a fast-moving spacecraft, the clock would appear to tick more slowly compared to a clock that is stationary with respect to the observer.
This concept is directly related to the proper length and relative velocity, as calculations of speeds near the speed of light can lead to significant time differences between the object's frame of reference and that of the observer. Emphasizing this relationship can help students understand that at high velocities, time does not behave intuitively, which is crucial when measuring or calculating the passage of time for high-speed objects like our spacecraft.
Special Relativity
Special relativity is a theory formulated by Albert Einstein that revolutionized our understanding of space, time, and how they relate to each other. This theory introduced the concept that the laws of physics are the same for all non-accelerating observers, and that the speed of light in a vacuum is constant, regardless of the observer's motion relative to the light source. Special relativity has implications for concepts like simultaneity, proper length, relative velocity, and time dilation.
For example, special relativity tells us that an object's mass increases, its length contracts, and time dilates as its velocity approaches the speed of light. In the context of the spacecraft exercise, applying special relativity can prevent errors such as velocities exceeding the speed of light, as any proper calculation must account for these relativistic effects, leading to a more consistent and realistic result.

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Most popular questions from this chapter

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red light of wavelength 650 \(\mathrm{nm}\) appear green to him, with a wavelength of 520 \(\mathrm{nm}\) . The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony?

Prepare a graph of the relativistic kinetic energy and the classical kinetic energy, both as a function of speed, for an object with a mass of your choice. At what speed does the classical kinetic energy underestimate the experimental value by 1\(\%\) ? by 5\(\%\) ? by 50\(\%\) ?

An astronaut is traveling in a space vehicle that has a speed of 0.500 c relative to the Earth. The astronaut measures her pulse rate at 75.0 beats per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving in a direction perpendicular to the line that connects the vehicle with an observer on the Earth. (a) What pulse rate does the Earth observer measure? (b) What If? What would be the pulse rate if the speed of the space vehicle were increased to 0.990\(c\) ?

A supertrain (proper length 100 \(\mathrm{m} )\) travels at a speed of 0.950 \(\mathrm{c}\) as it passes through a tunnel (proper length 50.0 \(\mathrm{m}\) ). As seen by a trackside observer, is the train ever completely within the tunnel? If so, with how much space to spare?

The creation and study of new elementary particles is an important part of contemporary physics. Especially interesting is the discovery of a very massive particle. To create a particle of mass \(M\) requires an energy \(M c^{2} .\) With enough energy, an exotic particle can be created by allowing a fast moving particle of ordinary matter, such as a proton, to collide with a similar target particle. Let us consider a perfectly inelastic collision between two protons: an incident proton with mass \(m_{p},\) kinetic energy \(K,\) and momentum magnitude \(p\) joins with an originally stationary target proton to form a single product particle of mass \(M .\) You might think that the creation of a new product particle, nine times more massive than in a previous experiment, would require just nine times more energy for the incident proton. Unfortunately not all of the kinetic energy of the incoming proton is available to create the product particle, since conservation of momentum requires that after the collision the system as a whole still must have some kinetic energy. Only a fraction of the energy of the incident particle is thus available to create a new particle creation depends on the energy of the moving proton. Show that the energy available to create a product particle is given by $$ M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{p} c^{2}}} $$ From this result, when the kinetic energy \(K\) of the incident proton is large compared to its rest energy \(m_{p} c^{\prime},\) we see that \(M\) approaches \(\left(2 m_{p} K\right)^{1 / 2} / c\) . Thus if the energy of the incoming proton is increased by a factor of nine, the mass you can create increases only by a factor of three. This disappointing result is the main reason that most modern accelerators, such as those at CERN (in Europe), at Fermi- lab (near Chicago), at SLAC (at Stanford), and at DESY (in Germany), use colliding beams. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so in principle all of the initial kinetic energy can be used for particle creation, according to $$ M c^{2}=2 m c^{2}+K=2 m c^{2}\left(1+\frac{K}{2 m c^{2}}\right) $$ where \(K\) is the total kinetic energy of two identical colliding particles. Here if \(K>>m c^{2},\) we have \(M\) directly proportional to \(K,\) as we would desire. These machines are difficult to build and to operate, but they open new vistas in physics.

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