A 2.00-GeV proton hits another 2.00-GeV proton in a headon collision. (a) Calculate v, β, p, K, and E for each of the initial protons. (b) What happens to the kinetic energy? Strategy (a) By the convention just discussed, a 2.00-GeV proton has a kinetic energy of 2.00 GeV. We use Equation (2.65) to

Question

A 2.00-GeV proton hits another 2.00-GeV proton in a headon collision. (a) Calculate v, β, p, K, and E for each of the initial protons. (b) What happens to the kinetic energy?

Strategy (a) By the convention just discussed, a 2.00-GeV proton has a kinetic energy of 2.00 GeV. We use Equation (2.65) to determine the total energy and Equation (2.70) to determine momentum if we know the total energy. To determine β and v, it helps to first determine the relativistic factor γ, which we can use Equation (2.65) to find. Then we use Equation (2.62) to find β and v. These are all typical calculations that are performed when doing relativistic computations.

[latex]\beta^{2}=\frac{\gamma^{2}-1}{\gamma^{2}}=\frac{(1.049)^{2}-1}{(1.049)^{2}}=0.091[/latex] (2.62)

[latex]E=\gamma m c^{2}=\frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}}[/latex] (2.65)

[latex]E^{2}=p^{2} c^{2}+E_{0}^{2}[/latex] (2.70)

Accepted Answer

(a) We use K = 2.00 GeV and the proton rest energy, 938 MeV, to find the total energy from Equation (2.65),

[latex]E=K+E_{0}=2.00 GeV +938 MeV =2.938 GeV[/latex]

The momentum is determined from Equation (2.70).

[latex]\begin{gathered}p^{2} c^{2}=E^{2}-E_{0}^{2}=(2.938 GeV )^{2}-(0.938 GeV )^{2} \=7.75 GeV ^{2}\end{gathered}[/latex]

The momentum is calculated to be

[latex]p=\sqrt{7.75( GeV / c)^{2}}=2.78 GeV / c[/latex]

Notice how naturally the unit of GeV/c arises in our calculation.

In order to find β we first find the relativistic factor γ. There are several ways to determine γ; one is to compare the rest energy with the total energy. From Equation (2.65) we have

[latex]E=\gamma E_{0}=\frac{E_{0}}{\sqrt{1-u^{2} / c^{2}}}[/latex]

[latex]\gamma=\frac{E}{E_{0}}=\frac{2.938 GeV }{0.938 GeV }=3.13[/latex]

We use Equation (2.62) to determine β.

[latex]\beta=\sqrt{\frac{\gamma^{2}-1}{\gamma^{2}}}=\sqrt{\frac{3.13^{2}-1}{3.13^{2}}}=0.948[/latex]

The speed of a 2.00-GeV proton is 0.95c or [latex]2.8 imes 10^{8} m / s[/latex].

(b) When the two protons collide head-on, the situation is similar to the case when the two blocks of wood collided head-on with one important exception. The time for the two protons to interact is less than [latex]10^{-20}[/latex] s. If the two protons did momentarily stop at rest, then the two-proton system would have its mass increased by an amount given by Equation (2.68), [latex]2 K / c^{2} ext { or } 4.00 GeV / c^{2}[/latex]. The result would be a highly excited system. In fact, the collision between the protons happens very quickly, and there are several possible outcomes. The two protons may either remain or disappear, and new additional particles may be created. Two of the possibilities are

[latex]\Delta M=M-2 m=\frac{2 K}{c^{2}}[/latex] (2.68)

[latex]p+p ightarrow p+p+p+\bar{p}[/latex] (2.78)

[latex]p+p ightarrow \pi^{+}+d[/latex] (2.79)

where the symbols are [latex]p ext { (proton), } \bar{p} ext { (antiproton), } \pi ext { (pion) }[/latex], and d (deuteron). We will learn more about the possibilities later when we study nuclear and particle physics. Whatever happens must be consistent with the conservation laws of charge, energy, and momentum, as well as with other conservation laws to be learned. Such experiments are routinely done in particle physics. In the analysis of these experiments, the equivalence of mass and energy is taken for granted.

Question 2.13: A 2.00-GeV proton hits another 2.00-GeV proton in a headon c...

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