/*! 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 77 In high-energy physics, new part... [FREE SOLUTION] | 91Ó°ÊÓ

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In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\mathrm{kaon}\left(\mathrm{K}^{+}\right) :\) $$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kimetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

Short Answer

Expert verified
The minimum kinetic energy for the initial moving proton colliding into a stationary target arises from the necessity of energy equalized with two kaon rest energies; colliding beams notably diminish the necessary energy input.

Step by step solution

01

Identify the Energy Requirements

The reaction requires enough kinetic energy to compensate for the rest mass energy of the new kaons created in the collision. This is the minimum energy required for the reaction to occur.
02

Calculate Rest Mass Energy of Created Particles

The rest mass energy of one kaon (either \(K^{-}\) or \(K^{+}\)) is 493.7 MeV. As two kaons are created, their total rest mass energy is \(2 \times 493.7 = 987.4\) MeV.
03

Use Conservation of Energy and Momentum

We work in the zero-total-momentum frame, where the total momentum before and after the collision is zero for calculations. The necessary kinetic energy of the colliding proton, when both protons are included, must be equal to or greater than the total rest mass energy of the two new kaons.
04

Initial Calculation of Minimum Kinetic Energy

In the center-of-mass frame, the energy of the system after the collision is the sum of the rest mass energy of the final particles plus the kinetic energy of the initial moving particles. The equivalent amount of energy must come from the incident proton's kinetic energy.The combined rest mass energy of the system after the reaction, transitioning into the rest mass energy of the two protons plus the two kaons:\(2M_p + 2M_K = 2 \times 938.3 + 2 \times 493.7 = 2863.4\) MeV.To reach zero total momentum where the system only presents the rest mass energy of the kaons, we calculate the kinetic energy in such momentum equality, specifically compensating for the kinetic energy of the moving proton.
05

Calculate Total Kinetic Energy

The proton's initial kinetic energy is needed to transform into the rest mass energy of the kaons. Solving via conservation frameworks and energy balance suggests a minimum combined energy existence when employing counter-active motion which is simplified by using colliding beams:For both moving protons: the combined kinetic energy is \(2 \times T_K = 287.9\) MeV, which is significantly lower than when just one proton moves against a stationary target.

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Key Concepts

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

Proton-Proton Collision
Proton-proton collisions are fundamental interactions in high-energy physics experiments. They provide a way to create new particles by colliding a fast-moving proton with another proton. During such collisions, some of the kinetic energy from the incoming proton is used to produce new particles with greater mass. These conditions mimic the energies found in the universe's earliest moments. Scientists use these collisions to study the building blocks of matter and forces that govern their interactions.
By accelerating protons to near-light speeds, the amount of energy available for these interactions is significant. This energy can then transform into particles like kaons, which are heavier than the original protons. The study of these interactions deepens our understanding of how matter behaves at the most fundamental level.
Proton-proton collisions are instrumental at facilities like the Large Hadron Collider (LHC), where researchers search for new physics beyond known theories.
Kinetic Energy Calculation
Calculating the kinetic energy in these reactions is crucial for determining the conditions under which certain reactions occur. Kinetic energy is the energy an object possesses due to its motion.
When particles collide, their kinetic energy can transform into other energy forms, such as rest mass energy of new particles. The minimum kinetic energy required for a reaction is dictated by the energy needed to produce new particles' rest mass. This involves ensuring that the energy after the collision matches the sum of the new particles' rest energy as calculated using \[ E = mc^2 \].
If creating two kaons, each with a rest mass energy of 493.7 MeV, the minimum total kinetic energy must at least equal the sum of these energies. This helps us understand energies required in high stakes experiments in laboratories.
Conservation of Energy and Momentum
The conservation of energy and momentum principles are key when analyzing particle collisions. Energy cannot be created or destroyed but can change forms. Similarly, momentum, a measure of motion, remains constant in isolated systems.
In particle collisions, the total energy before and after the collision remains constant. Pre-collision kinetic energy gets converted into rest mass energy and other forms. The **momentum** remains conserved, which also means the system's inertial properties remain unchanged. This conservation helps predict post-collision behavior of particles.
For the proton-proton collision, calculations ensure pre- and post-collision energies and momenta align. The result mirrors the equations of both principles, guiding us to predict the resulting conditions accurately upon the reaction's end.
Zero-Total-Momentum Frame
The zero-total-momentum frame simplifies collision calculations by analyzing events where the total momentum is zero. Also known as the center-of-mass frame, this simplifies the view of how energy distributes in high-speed collisions.
In this special frame, the collective motion of particles balances out, simplifying energy calculations. When figuring energies and outcomes straightforwardly, the zero-momentum frame aids in breaking down complex particle interactions into simpler parts.
The frame is useful in experiments as it allows physicists to focus purely on energy transformations undertaken, as momentum needs no adjustment for other involving variables. Using this frame, we calculate kinetic energies and resulting energies with greater accuracy.

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Most popular questions from this chapter

Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 \(\mathrm{m}\) from each other. Being very skilled players, they play with- out a net. The tennis ball has mass 0.0580 \(\mathrm{kg}\) . You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men. (a) Lorentz serves the ball at 80.0 \(\mathrm{m} / \mathrm{s}\) . What is the ball's kinetic energy? (b) Einstein slams a return at \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . What is the ball's kinetic energy? (c) During Einstein's return of the ball in part (a), a white rabbit runs beside the court in the direction from Einstein to Lorentz. The rabbit has a speed of \(220 \times 10^{8} \mathrm{m} / \mathrm{s}\) relative to the two men. What is the speed of the rabbit relative to the ball? (d) What does the rabbit measure as the distance from Einstein to Lorentz? (e) How much time does it take for the rabbit to run \(20.0 \mathrm{m},\) according to the players? (f) The white rabbit carries a pocket watch. He uses this watch to measure the time (as he sees a it) for the distance from Einstein to Lorentz to pass by under him. What time does he measure?

Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light-year is the distance that light travels in a year.) You plan to travel at constant speed in a \(1000-\mathrm{kg}\) rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{19} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }},\) \(t_{\text { earth }},\) \(\mathbf{t}_{\text { rouket }} \boldsymbol{E}\) \((\text { in } \mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) (c) \(0.9999 \mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.190 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 \(\mathrm{ms}\) . (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 \(c\) ?

Calculate the magnitude of the force required to give a \(0.145-\mathrm{kg}\) baseball an acceleration \(a=1.00 \mathrm{m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{m} / \mathrm{s} ;\) (b) 0.900 \(\mathrm{c}\) (c) \(0.990 c .(\mathrm{d})\) Repeat parts \((\mathrm{a}),\) and \((\mathrm{c})\) if the force and acceleration are perpendicular to the velocity.

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 \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) 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 \(f_{0}=345 \mathrm{MHz}\) 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, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) 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)?

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