/*! 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 56 In the earth's rest frame, two p... [FREE SOLUTION] | 91Ó°ÊÓ

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In the earth's rest frame, two protons are moving away from each other at equal speed. In the frame of each proton, the other proton has a speed of 0.700\(c\). What does an observer in the rest frame of the earth measure for the speed of each proton?

Short Answer

Expert verified
Speed of each proton in Earth's rest frame is approximately 0.94c.

Step by step solution

01

Understand the Problem

We need to find the speed of each proton in the Earth's rest frame, given that in the rest frame of one proton, the other proton's speed is 0.700c.
02

Use the Velocity Addition Formula

We can use the relativistic velocity addition formula to find the speed of each proton in the Earth's rest frame. This formula is: \[ u = \frac{v + w}{1 + \frac{vw}{c^2}} \] where \(u\) is the velocity in the Earth's frame, \(v\) is the velocity in one proton's frame, and \(w\) is the velocity of the Earth in proton's frame.
03

Set Variables and Apply Formula

Since both protons move with the same speed relative to the Earth, assign \(v = 0.7c\) as the speed of the other proton in one proton's rest frame. The frame of the proton treats the earth as moving the opposite direction at speed \(v\). We need to find \(w\), which should also be equal to \(v\) due to symmetry. Apply the formula \[ u = \frac{0.7c + 0.7c}{1 + \frac{(0.7c)(0.7c)}{c^2}} \]
04

Simplify and Calculate

Simplify the expression: \[ u = \frac{1.4c}{1 + 0.49} \] Calculate \(u\): \[ u = \frac{1.4c}{1.49} \approx 0.94c \] So, the speed of each proton in the Earth's rest frame is approximately 0.94c.

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Key Concepts

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

The Nature of Protons
Protons are one of the essential building blocks of matter. They are found in the nucleus of an atom and are positively charged. This charge has profound effects on how protons interact with other particles, both electromagnetically and through the strong nuclear force.
Due to their relatively large mass compared to electrons, protons play a significant role in determining the properties of atoms. They are composed of smaller particles known as quarks, held together by gluons. These interactions define many aspects of particle physics and influence how protons behave when they are accelerated to high velocities near the speed of light.
To understand exercises dealing with protons at relativistic speeds, it's vital to grasp these fundamental properties, which outline their behavior in various frames of reference.
Understanding Rest Frame
The concept of the rest frame is critical in physics, particularly in special relativity. The rest frame of an object is the perspective from which the object is observed to be at rest \( i.e., \, not \, moving \).
When analyzing relativistic problems, picking the right frame of reference simplifies calculations and enhances understanding.
  • The Earth's rest frame: where our standard measurements are taken, and the Earth is treated as stationary.
  • The proton's rest frame: where a proton is seen as at rest, and hence any other object, like a second proton, seems to be moving with respect to it.

Examining problems from different rest frames allows us to appreciate how speed and motion are relative concepts. This is essential in exercises that involve relativistic velocity addition.
Speed of Light in Relativistic Context
The speed of light, denoted as \( c \), is approximately \( 3 \times 10^8 \) meters per second. It represents the maximum speed at which information or matter can travel in the universe. This universal constant underpins many principles of special relativity.
In relativistic physics, objects approaching the speed of light exhibit time dilation and length contraction. These effects complicate classical notions of speed and velocity.
Relativistic velocity addition further addresses how speeds combine when moving close to \( c \). No matter how fast two objects move relative to each other, the speed of light remains constant in all frames of reference. This peculiar aspect of physics is a linchpin in solving problems involving relativistic speeds, such as those concerning colliding or receding protons.
Understanding Velocity
Velocity is the rate of change of an object's position with respect to time. It's a vector quantity, meaning it has both magnitude and direction. Clarifying these attributes is essential when exploring relativistic effects.
In special relativity, adding velocities doesn't follow the intuitions we have from everyday experiences. Instead, the relativistic velocity addition formula is needed. This ensures that the sum of velocities doesn't exceed the speed of light. The formula is:
\[ u = \frac{v + w}{1 + \frac{vw}{c^2}} \]
The case of two protons moving in opposite directions highlights the importance of using relativistic principles. They illustrate how traditional concepts of velocity are transformed when considering high speeds near \( c \.\)
Understanding these advanced concepts allows us to better grasp the nature of particles like protons moving at significant fractions of the speed of light.

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Most popular questions from this chapter

A spaceship moving at constant speed u relative to us broadcasts a radio signal at constant frequency \(f_0\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_0\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_0\)? (\(Hint\): In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1 /T. Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_0\) = 345 MHz as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f - f_0\)? (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift (\(f - f_0\)) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

A proton has momentum with magnitude \(p_0\) when its speed is 0.400c. In terms of \(p_0\) , what is the magnitude of the proton's momentum when its speed is doubled to 0.800c?

If a muon is traveling at 0.999c, what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

An electron is acted upon by a force of 5.00 \(\times\) 10\(^{-15}\) N due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is 1.00 km/s. (b) The electron's speed is 2.50 \(\times\) 10\(^8\) m/s and the force is parallel to the velocity.

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 \(\mu\)s. 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 - \(\mu\)s 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- \(\mu\)s 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 \(\mu\)s, 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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