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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 \(0.650 c,\) and the speed of each particle relative to the other is \(0.950 c .\) What is the speed of the second particle, as measured in the laboratory?

Short Answer

Expert verified
The speed of the second particle with respect to the laboratory is approximately \(0.989c\).

Step by step solution

01

Understand the Problem

We have two particles moving in opposite directions. The speed of one particle with respect to the laboratory is given as \( 0.650c \). The relative speed of the two particles with respect to each other is \( 0.950c \). We are to find the speed of the second particle with respect to the laboratory.
02

Recall the Relativistic Velocity Addition Formula

In the context of relativity, when two objects move relative to each other, their velocities add up according to the formula: \[ v = \frac{u + v'}{1 + \frac{uv'}{c^2}} \] where \( u \) is the speed of the first particle with respect to the lab, \( v \) is the speed of the second particle with respect to the lab, and \( v' \) is their relative speed. Here, \( c \) is the speed of light.
03

Set Up the Equation

Let \( u = 0.650c \) (speed of the first particle with respect to the lab), \( v' = 0.950c \) (relative speed of the particles). We substitute these values into the velocity addition formula to find \( v \), the speed of the second particle with respect to the lab: \[ v = \frac{0.950c + 0.650c}{1 + \frac{(0.950c)(0.650c)}{c^2}} \]
04

Simplify the Numerator

Evaluate the numerator: \( 0.950c + 0.650c = 1.600c \).
05

Simplify the Denominator

Calculate the term in the denominator: \( \frac{(0.950)(0.650)c^2}{c^2} = 0.6175 \). Then, add 1 to this value: \( 1 + 0.6175 = 1.6175 \).
06

Compute the Final Speed

Substitute the simplified numerator and denominator back into the equation: \[ v = \frac{1.600c}{1.6175} \]. Calculate \( v \) to find the speed of the second particle with respect to the laboratory: \[ v \approx 0.989c \].

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Key Concepts

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

Relativity
Relativity is a cornerstone of modern physics. It provides the framework needed to understand high-speed particles and light. Developed by Albert Einstein, the theory of relativity consists of two parts: special relativity and general relativity. Special relativity is most applicable to this exercise.

Special relativity deals with objects moving close to the speed of light. It challenges common notions of time and space. One key principle is that the laws of physics are the same for all observers, regardless of their constant motion relative to each other.

This leads to concepts like time dilation and length contraction. In our context, relativistic velocity addition is based on these principles. It shows how the velocities of objects in different reference frames are added together. This addition is not as simple as in classical physics, where you might just sum the speeds. Instead, the relativistic formula considers the effects of moving close to light speed.
High-Energy Particles
High-energy particles are fundamental in understanding high-speed processes and cosmic phenomena. Often generated in particle accelerators like the Large Hadron Collider, these particles travel at speeds approaching that of light.

In practical terms, high-energy particles are used to explore the fundamental nature of matter. They smash into each other at high speeds to reveal the building blocks of the universe. Scientists use these interactions to test theories of physics under extreme conditions.

In our exercise, the two particles moving in opposite directions in a high-energy accelerator illustrate these principles. Knowing their speeds and how they interact helps scientists make predictions about similar high-energy environments found naturally, like those in stars or cosmic rays.

Understanding how to calculate their velocity using relativistic principles helps scientists pinpoint the exact conditions during these high-energy collisions.
Speed of Light
The speed of light, denoted by \( c \), is a fundamental constant in physics. It has a value of roughly \( 299,792,458 \) meters per second. This speed remains constant in all reference frames, a key feature of Einstein's relativity.

The constancy of light speed leads to some fascinating consequences, such as the fact it represents a cosmic speed limit. Nothing can travel faster than the speed of light in a vacuum.

In the exercise we looked at, the calculated speeds of particles were given as a fraction of \( c \), such as \( 0.650c \) or \( 0.950c \). These numbers indicate how close the particles are traveling relative to the speed of light.

The processes that happen at these speeds called relativistic effects, make calculations different from those we would perform at everyday speeds. That’s why understanding the speed of light's role in high-speed contexts becomes crucial. The relativistic velocity addition formula mathematically captures these effects.

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Most popular questions from this chapter

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earth-based observers? Do present-day jet planes approach such speeds?

A particle of mass \(m\) is moving at a speed \(v .\) (a) At what fraction of the speed of light is the momentum of the particle twice the nonrelativistic Newtonian momentum \(p=m v ?\) (b) Can the magnitude of the relativistic momentum ever be less than that of the Newtonian momentum?

A muon is created \(55.0 \mathrm{~km}\) above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is \(2.20 \mu \mathrm{s}\), and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of \(0.9860 \mathrm{c}\). (a) In the muon's frame, what is its initial height above the surface of the earth? (b) In the muon's frame, how much closer does the earth get during the lifetime of the muon? What fraction is this of the muon's original height, as measured in the muon's frame? (c) In the earth's frame, what is the lifetime of the muon? In the earth's frame, how far does the muon travel during its lifetime? What fraction is this of the muon's original height in the earth's frame?

Neutron stars are the remains of exploded stars, and they rotate at very high rates of speed. Suppose a certain neutron star has a radius of \(10.0 \mathrm{~km}\) and rotates with a period of \(1.80 \mathrm{~ms}\). (a) Calculate the surface rotational speed at the equator of the star as a fraction of c. (b) Assuming the star's surface is an inertial frame of reference (which it isn't, because of its rotation), use the Lorentz velocity transformation to calculate the speed of a point on the equator with respect to a point directly opposite it on the star's surface.

A particle is moving with a speed of \(0.80 c\). Calculate the ratio of its kinetic energy to its rest energy.

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