/*! 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 71 The creation and study of new el... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The energy available for creating the product particle in a perfectly inelastic collision of two protons is given by \(M c^{2} = 2 m_{p} c^{2} \sqrt{1 + K/(2 m_{p} c^{2})}\). When the kinetic energy \(K\) is much greater than the rest energy of the proton, the mass \(M\) is approximately proportional to the square root of \(K\), implying that a ninefold increase in \(K\) increases \(M\) only by a factor of three.

Step by step solution

01

Analyze the Collision

Analyze the perfectly inelastic collision between the moving proton and the stationary proton. The total momentum before the collision must equal the total momentum after the collision because momentum is conserved.
02

Derive the Energy Relation

To find the energy available for the creation of the product particle, use the conservation of energy and momentum. The total energy before the collision is the sum of the rest energy of both protons and the kinetic energy of the moving proton. The total energy after the collision is the energy of the product particle, which is at rest.
03

Conservation of Momentum

The initial momentum is given by the moving proton, and after the collision, the system still carries this momentum, which means there remains some kinetic energy apart from the rest energy transformed into the particle's mass.
04

Solve for Mass M

Using the conservation laws, solve for the mass M of the product particle in terms of the proton mass, the speed of light, and the kinetic energy of the incoming proton.
05

Relate K with the New Particle's Mass

From the relation obtained, analyze the case when K is much greater than the rest energy of the proton. Find how the mass M of the created particle relates to the kinetic energy of the incoming proton.

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Key Concepts

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

Collision of Particles
The collision between particles is a fundamental process in physics, allowing us to create and study new elementary particles. When two protons collide, the incident proton with a certain kinetic energy (\(K\)) and momentum (\(p\)), interacts with a target proton, resulting in the formation of a new particle. However, not all of the kinetic energy from the incident proton is available to create the new particle.

During this process, the conservation of momentum must be maintained. That is, the momentum before and after the collision must be the same. If the collision is perfectly inelastic, as in our exercise, the two protons merge into one heavy product particle, and because of the conservation of momentum, some energy must remain as kinetic energy and cannot be used for creating the massive particle. Understanding this intricate dance of energy and momentum is crucial for uncovering the mysteries of particle physics.
Conservation of Momentum
One of the core principles guiding the behavior of particles during collisions is the conservation of momentum. In physics, momentum is a measure of the motion of an object and is conserved in isolated systems, such as colliding protons in a vacuum. This means the total momentum of the system before the collision is equal to the total momentum after the collision.

For the problem we're examining, this principle implies that even after an incident proton collides with a stationary target proton to create a new particle, there's still momentum left over, corresponding to kinetic energy that cannot be used for particle creation. Thus, even if we increase the energy of the incoming proton, the mass of the particle created increases by a lesser factor than the increase in energy, as elegantly expressed by the provided equation.
Particle Accelerators
Particle accelerators are sophisticated machines that propel charged particles, such as protons, to high speeds, often close to the speed of light. These devices are vital in experimental physics for they enable particles to gain the energy necessary for creating new, massive particles upon collision.

In our exercise, traditional accelerators where a single beam of particles hits a stationary target aren't as efficient as colliding beam accelerators for massive particle creation. Colliding beam accelerators, like those at CERN or Fermilab, use two beams of identical particles accelerating towards each other. This setup has the advantage of potentially bringing the total system momentum to zero, allowing all of the kinetic energy to be converted into mass, according to the energy-mass equivalence principle, maximizing the ability to create massive particles.
Energy-Mass Equivalence
Albert Einstein's famous equation, E=mc^2, underpins the entire concept of particle collisions and creation. It postulates that energy (\(E\)) and mass (\(m\)) are interchangeable; mass can be seen as a dense form of energy. In the context of particle collisions, high levels of kinetic energy required to create a massive particle are a direct application of this energy-mass equivalence principle.

The more energy available in a collision, the more massive the particles that can be produced. However, as illustrated by our exercise, due to conservation of momentum, not all of the kinetic energy of the incident proton can be used to form the new particle's mass. Nevertheless, when we increase the kinetic energy significantly, we find that the mass of the new particle increases, yet it isn’t directly proportional to the energy increase unless we use colliding beams, which leverages the equation in its complete form, accounting for the total kinetic energy in the system.

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Most popular questions from this chapter

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?

Find the kinetic energy of a \(78.0-\mathrm{kg}\) spacecraft launched out of the solar system with speed 106 \(\mathrm{km} / \mathrm{s}\) by using (a) the classical equation \(K=\frac{1}{2} m u^{2} .\) (b) What If? Calculate its kinetic energy using the relativistic equation.

Police radar detects the speed of a car (Fig. P39.19) as follows. Microwaves of a precisely known frequency are broadcast toward the car. The moving car reflects the microwaves with a Doppler shift. The reflected waves are received and combined with an attenuated version of the transmitted wave. Beats occur between the two microwave signals. The beat frequency is measured. (a) For an electromagnetic wave reflected back to its source from a mirror approaching at speed \(v,\) show that the reflected wave has frequency $$ f=f_{\text { source }} \frac{c+v}{c-v} $$ where \(f_{\text { source }}\) is the source frequency. (b) When \(v\) is much less than \(c,\) the beat frequency is much smaller than the transmitted frequency. In this case use the approximation \(f+f_{\text { source }} \approx 2 f_{\text { source }}\) and show that the beat frequency can be written as \(f_{\text { beat }}\) \(=2 v / \lambda .\) (c) What beat frequency is measured for a car speed of 30.0 \(\mathrm{m} / \mathrm{s}\) if the microwaves have frequency 10.0 \(\mathrm{GHz}\) ? (d) If the beat frequency measurement is accurate to \(\pm 5 \mathrm{Hz},\) how accurate is the velocity measurement?

A \(2000-\mathrm{kg}\) car moving at 20.0 \(\mathrm{m} / \mathrm{s}\) collides and locks together with a \(1500-\mathrm{kg}\) car at rest at a stop sign. Show that momentum is conserved in a reference frame moving at 10.0 \(\mathrm{m} / \mathrm{s}\) in the direction of the moving car.

Find the momentum of a proton in MeV/c units assuming its total energy is twice its rest energy.

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