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Ch 37: Special Relativity
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 37, Problem 12

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?

Verified step by step guidance
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Step 1: For part (a), calculate the time it takes for the particle to travel 45.0 km as measured by the scientist on Earth. Use the formula for time: t = dv, where d is the distance (45.0 km) and v is the speed of the particle (0.99540c). Convert the distance to meters and use the speed of light c = 3.00 × 10⁸ m/s for calculations.
Step 2: For part (b), use the length-contraction formula to calculate the distance as measured in the particle’s frame. The formula is Lp = Le 1 - vc2, where Le is the distance measured by the scientist (45.0 km), v is the speed of the particle, and c is the speed of light.
Step 3: For part (c), calculate the time in the particle’s frame using the time dilation formula. The formula is tp = te1 - vc2, where te is the time measured by the scientist in part (a).
Step 4: Alternatively, calculate the time in the particle’s frame using the contracted distance from part (b) and the particle’s speed. Use the formula tp = Lpv, where Lp is the contracted distance and v is the speed of the particle.
Step 5: Compare the two results for the time in the particle’s frame from step 3 and step 4. Verify that they agree, demonstrating consistency between the time dilation formula and the contracted distance approach.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relativity of Time and Length

In the theory of relativity, time and space are not absolute but relative to the observer's frame of reference. This means that time can appear to pass at different rates for observers in different states of motion, and lengths can contract depending on the relative velocity between observers. This is crucial for understanding how measurements differ between the scientist on Earth and the particle moving at a significant fraction of the speed of light.
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Time Dilation

Time dilation is a phenomenon predicted by Einstein's theory of relativity, where a clock moving relative to an observer ticks slower than a clock at rest with respect to that observer. For the particle traveling at 0.99540c, the time experienced by the particle will be less than the time measured by the scientist on Earth. This concept is essential for calculating the time it takes for the particle to reach the Earth's surface from both frames of reference.
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Length Contraction

Length contraction is another relativistic effect where the length of an object in motion is measured to be shorter than its length at rest, as observed from a stationary frame. The formula for length contraction is L = L0√(1 - v²/c²), where L0 is the proper length, v is the relative velocity, and c is the speed of light. This concept is necessary for determining the distance the particle travels as measured in its own frame of reference.
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Related Practice
Textbook Question

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800c 0.800c relative to you. At the instant the spaceracer passes you, both of you start timers at zero.

(a) At the instant when you measure that the spaceracer has traveled 1.20×1081.20\(\times\)10^8 m past you, what does the race pilot read on her timer?

(b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her?

(c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

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Textbook Question

A rocket ship flies past the earth at 91.0% of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction in which the ship is moving. (a) If his height is measured to be 2.00 m by his doctor inside the ship, what height would a person watching this from the earth measure? (b) If the earth-based person had measured 2.00 m, what would the doctor in the spaceship have measured for the astronaut’s height? Is this a reasonable height?

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Textbook Question

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory?

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Textbook Question

Why Are We Bombarded by Muons? Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 μs. They are produced when cosmic rays bombard the upper atmosphere about 10 km above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2 μs lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2 μs lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 μs, so how does it make it to the ground? What is the thickness of the 10 km of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

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Textbook Question

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.1500.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.012.0 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 cc?

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Textbook Question

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

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