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The creation and study of new and very massive elementary particles is an important part of contemporary physics. 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 proton to collide with a similar target particle. 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 $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by $$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{\mu} c^{2}}}$$ This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2}$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. 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 the initial kinetic energy can be used for particle creation. Show that $$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$ where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2}$, we have $M$ directly proportional to $K$ as we would desire.

          The creation and study of new and very massive elementary particles is an important part of contemporary physics. 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 proton to collide with a similar target particle. 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 $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle.
(a) Show that the energy available to create a product particle is given by $$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{\mu} c^{2}}}$$
This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2}$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. 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 the initial kinetic energy can be used for particle creation. Show that
$$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$
where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2}$, we have $M$ directly proportional to $K$ as we would desire.

Added by Encarnacion W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert David Morabito

12/04/2021

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Step 1: Start with the conservation of momentum for the perfectly inelastic collision between two protons: \[p + 0 = p' \] \[p = p' \] Show more…

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The creation and study of new and very massive elementary particles is an important part of contemporary physics. 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 proton to collide with a similar target particle. 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 $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by $$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{\mu} c^{2}}}$$ This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2}$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. 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 the initial kinetic energy can be used for particle creation. Show that $$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$ where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2}$, we have $M$ directly proportional to $K$ as we would desire.

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00:01 In this problem, we're going to look at the way of creating a particle of mass m and two types of collisions, one with the incoming proton, collides with a stationary proton, and one where protons are moving in opposite directions and collide.

00:14 So we'll look at the energetics of each one of those.

00:17 Before we get into the analysis of this first case, let's remind ourselves or maybe discuss something that you've not heard before, but when you first were introduced to momentum conservation, they talk about the conservation of the momentum of the center of mass, but you also have to have conservation of the energy of the center of mass.

00:39 So to have both the momentum and energy of the center of mass conserved, some of this energy has to be reserved for that purpose.

00:48 So not all of it can go to creating of the mass m.

00:53 So that's important.

00:55 That's important that you have to conserve both energy and momentum of the center of mass.

01:01 And that's why we have different energetics depending on case a and case b.

01:07 We're going to be dealing.

01:09 This is a.

01:12 So now let's look at our conservation laws for this problem.

01:15 Momentum.

01:17 We have p is equal to p prime, which gives me that p and p prime are the same in magnitude, square of both sides and multiplied by c squared.

01:28 P squared, c squared, is equal to p prime squared, c squared, c squared, and now using the relationships that we know from relativity, this is e prime squared minus capital m squared c to the fourth.

01:47 Now let me rewrite the left hand side now and using the bottom expression here too for the energy.

01:55 So this will be k plus mc squared, squared, minus m squared, c to the fourth is equal to e prime, minus m squared c to the fourth this is equation one now we have energy to deal with that will be kinetic energy plus the two rest energies k plus two mc squared is equal to e prime now if i stick e prime into equation one so i'll call this equation two if i stick e prime into equation one i'm going to have it in terms of variables i want notice my m is the massive proton the m with the feet on it, that is the mass of the end product at the end of this resulting product.

02:54 So let's put e prime into one.

02:57 And it's just a matter of algebra now.

02:59 So let me write this out.

03:00 So i got k plus mc squared squared minus m squared c to the fourth is equal to k plus 2 mc squared mc squared, m squared, minus capital m squared, c to the fourth.

03:20 And now let me solve for this on the, put this on the left, m squared, c to the fourth, is equal.

03:27 Now i'm going to have to expand out these squares.

03:31 So let me expand this one out.

03:32 I could write it over again.

03:34 I'm just going to give you the end result of it, k squared plus 4k, mc squared, plus 4k, m squared c to the fourth that's the expansion of this now minus k squared minus 2k mc squared minus m squared c to the fourth and then we still got to bring the last term on the left over to the right so plus m squared c to the fourth so what do we have k squared minus k square that's gone minus m squared c to fourth plus m squared these two terms are gone so i'm left with k 2 m c squared plus and that's four minus two that's where this comes from four minus two and then we got the four m squared c to the fourth now i'm trying to get into the form they want so let me give myself uh common factor.

04:46 So 4m squared, c to the fourth, plus.

04:51 Now, to have the same factor here, i've got to multiply by 2mc squared over 2mc squared.

04:59 So k, give me 4m squared, c to the fourth, over 2mc squared.

05:10 And now i can rewrite this as 4m squared, c to the fourth, 1 plus k over 2 mc squared.

05:26 And to get then capital m c squared, taking the square root, i get 2 mc squared, square root, i get 2 mc squared, square root 1 plus k over 2 mc squared.

05:44 And that's the relationship they wanted us to prove.

05:46 Notice, as they mentioned, that when k is much larger than the rest of energy, that with this square root factor, increase k by 9, you're going to get a factor of 3 in the capital m.

06:01 That's obviously not a good thing.

06:04 Now, case b, now we have both moving...

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