/*! 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 12 An unstable particle is created ... [FREE SOLUTION] | 91Ó°ÊÓ

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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 km. (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Short Answer

Expert verified
(a) 1.508 × 10^-4 s; (b) 4307 m; (c) 1.574 × 10^-3 s, results agree.

Step by step solution

01

Calculate time as measured by the scientist

To find the time taken for the particle to travel 45.0 km at speed 0.99540c, use the formula \( t = \frac{d}{v} \), where \( d = 45000 \text{ m} \) and \( v = 0.99540c \). First calculate the velocity: \[ v = 0.99540 \times 3 \times 10^8 \text{ m/s} = 2.9862 \times 10^8 \text{ m/s} \]Next, use the formula to calculate time:\[ t = \frac{45000}{2.9862 \times 10^8} = 1.508 \times 10^{-4} \text{ s} \]
02

Calculate contracted distance in particle's frame

Using the length-contraction formula \( L = L_0 \sqrt{1 - v^2/c^2} \), where \( L_0 = 45000 \text{ m} \) and \( v = 0.99540c \), find \( L \):\[ L = 45000 \times \sqrt{1 - (0.99540)^2} \]Calculate the contraction factor:\[ \sqrt{1 - (0.99540)^2} = \sqrt{1 - 0.99082} = \sqrt{0.00918} = 0.0958 \]So, contracted distance:\[ L = 45000 \times 0.0958 = 4307 \text{ m} \]
03

Calculate time in particle's frame using time dilation

Using the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \), substitute \( t = 1.508 \times 10^{-4} \text{ s} \) and solve for \( t' \):\[ t' = \frac{1.508 \times 10^{-4}}{0.0958} = 1.574 \times 10^{-3} \text{ s} \]
04

Calculate time in particle's frame using distance

For the particle's frame, use \( t' = \frac{L}{v} \) with \( L = 4307 \text{ m} \) and \( v = 0.99540c = 2.9862 \times 10^8 \text{ m/s} \):\[ t' = \frac{4307}{2.9862 \times 10^8} = 1.574 \times 10^{-3} \text{ s} \]
05

Compare results

Both methods to find time in the particle's frame give \( t' = 1.574 \times 10^{-3} \text{ s} \). The results agree, confirming the calculations.

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Key Concepts

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

Time Dilation
Time dilation is a fascinating concept in physics that comes from Einstein's theory of special relativity. It tells us that time passes differently for two observers who are moving relative to each other. To understand this, imagine a scientist and a fast-moving particle.
For the scientist, who is stationary on Earth, time ticks at a normal pace. However, for the particle whizzing toward Earth at nearly the speed of light (0.99540c, where c is the speed of light), time slows down. This slowing of time means that the particle, in its own "frame of reference," experiences less time passing compared to the scientist who is observing from a stationary position.
To calculate the time experienced by the particle, we use the time dilation formula \( t' = \frac{t}{\sqrt{1 - v^2/c^2}} \). Here, \( t \) is the time measured by the observer (the scientist), and \( t' \) is the dilated time experienced by the particle. Hence, even though the scientist measures the journey to take \( 1.508 \times 10^{-4} \) seconds, the particle experiences a longer time: \( 1.574 \times 10^{-3} \) seconds.
This example shows how high velocities can significantly affect time perception, which is only observable at speeds close to the speed of light. It’s exciting to think about how flexible time becomes in such extreme conditions.
Length Contraction
Length contraction is another intriguing result of special relativity. It explains how the length of an object is perceived to be shorter when it is moving at high speeds relative to an observer. Imagine the path the particle takes from the atmosphere to the Earth's surface.
From the scientist's perspective on Earth, the distance is a standard \( 45.0 \) kilometers or \( 45000 \) meters. But for the particle moving at the speed of \( 0.99540c \), the distance appears shorter. According to the formula \( L = L_0 \sqrt{1 - v^2/c^2} \), the length \( L \) becomes \( 4307 \) meters in the particle's frame.
This occurs because, at such high speeds, the dimensions along the direction of motion contract. This effect doesn’t affect motions or objects we encounter in our daily lives but becomes significant when dealing with particles or objects moving at speeds close to that of light.
Length contraction is a key piece of understanding how interstellar travel or high-speed journeys would alter human perceptions and measurements, a captivating point for future explorations.
Special Relativity
Special relativity is a theory introduced by Albert Einstein in 1905. It reshaped our understanding of space and time. One of its core ideas is that the laws of physics are the same for all non-accelerated observers, meaning there are no "special" or "preferred" reference frames.
One of the most famous outcomes of special relativity is the equation \( E=mc^2 \), linking mass (m) and energy (E). However, the theory also leads to phenomena like time dilation and length contraction, as discussed earlier. These phenomena reveal how different observers, moving relative to one another, can perceive time and space in contrasting ways.
Special relativity becomes especially crucial when understanding cosmic rays or particles traveling at light-like speeds because it gives us the tools to calculate how time and space change in such extraordinary conditions. Its implications for high-speed travel, particle physics, and even the fabric of the universe continue to be areas of active research and wonder.
Whether it's calculating times or distances in unique scenarios, special relativity helps us comprehend the universe's intricate dance as particles flit across it, governed by these timeless laws.
Cosmic Rays
Cosmic rays are high-energy particles that travel through space and sometimes enter Earth's atmosphere. They can be electrons, protons, and other atomic nuclei. When these cosmic rays strike the Earth's atmosphere, they often produce a shower of secondary particles.
These particles collide with atmospheric atoms and create more particles, a phenomenon scientists observe at high altitudes. In the provided problem, the creation of an unstable particle from a cosmic ray in the upper atmosphere is being studied.
The particle's high speed (0.99540c) as it descends toward Earth brings special relativity into play. Time dilation and length contraction reflect the incredible speeds involved. Studying cosmic rays isn't just an exercise in theory but an exploration of nature’s powerful, energetic phenomena that might give us clues about the universe's origins.
Cosmic rays offer a fascinating avenue for both experimental and theoretical physics, allowing scientists to test theories like special relativity under extreme conditions and providing insights into high-energy processes far beyond Earth’s boundaries.

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Most popular questions from this chapter

Many of the stars in the sky are actually \(binary\space stars\), in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called \(spectroscopic\space binary\space stars. \textbf{Figure P37.68}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m\), orbiting their center of mass in a circle of radius \(R\). The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of 4.568110 \(\times\) 10\(^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) Hz. Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (\(Hint\): The speeds involved are much less than \(c\), so you may use the approximate result \(\Delta f/f = u/c\) given in Section 37.6.) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, 1.99 \(\times\) 10\(^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, 1.50 \(\times\) 10\(^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

The negative pion (\(\pi^-\)) is an unstable particle with an average lifetime of 2.60 \(\times\) 10\(^{-8}\)s (measured in the rest frame of the pion). (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be 4.20 \(\times\) 10\(^{-7}\) s. Calculate the speed of the pion expressed as a fraction of c. (b) What distance, measured in the laboratory, does the pion travel during its average lifetime?

As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 m. (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of 4.00 \(\times\) 10\(^7\) m/s relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light \(c\)?

A spaceship flies past Mars with a speed of 0.985c relative to the surface of the planet. When the spaceship is directly overhead, a signal light on the Martian surface blinks on and then off. An observer on Mars measures that the signal light was on for 75.0 ms. (a) Does the observer on Mars or the pilot on the spaceship measure the proper time? (b) What is the duration of the light pulse measured by the pilot of the spaceship?

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