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Answers

Additional Problems
Figure 44-14 shows part of the experimental arrangement in which antiprotons were discovered in the 1950s. A beam of 6.2 GeV protons emerged from a particle accelerator and collided with nuclei in a copper target. According to theoretical predictions at the time, collisions between protons in the beam and the protons and neutrons in those nuclei should produce antiprotons via the reactions
$$\begin{array}{l}{\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\mathrm{p}+\overline{\mathrm{p}}} \\ {\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\mathrm{p}+\overline{\mathrm{p}}}\end{array}$$
and
However, even if these reactions did occur, they would be rare compared to the reactions
$$\begin{array}{l}{\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\pi^{+}+\pi^{-}} \\ {\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\pi^{+}+\pi^{-}}\end{array}$$
Thus, most of the particles produced by the collisions between the 6.2 GeV protons and the copper target were pions.
To prove that antiprotons exist and were produced by some limited number of the collisions, particles leaving the target were sent into a series of magnetic fields and detectors as shown in Fig. $44-14 .$ The first magnetic field (M1) curved the path of any charged particle passing through it; moreover, the field was arranged so that the only particles that emerged from it to reach arranged so that the only particles thad to be negatively charged the second magnetic field (Q1) had to be negatively charged (either a $\overline{\text { p}}$ or a $\pi^{-} )$ and have a momentum of 1.19 GeV/c. Field Q1 was a special type of magnetic field (a quadrapole field) that focused the particles reaching it into a beam, allowing them to pass through a hole in thick shielding to a scintillation counter S1. The passage of a charged particle through the counter triggered a signal, with each signal indicating the passage of either a 1.19 GeV/c $\pi^{-}$ or (presumably) a 1.19 GeV/c $\overline{\mathrm{p}} .$
After being refocused by magnetic field $\mathrm{Q} 2,$ the particles were directed by magnetic field $\mathrm{M} 2$ through a second scintillation counter S2 and then through two Cerenkov counters C1 and C2. These latter detectors can be manufactured so that they send a signal only when the particle passing through them is moving with a speed that falls within a certain range. In the experiment, a particle with a speed greater than 0.79c would trigger C1 and a particle with a speed between 0.75c and 0.78c would trigger C2.
There were then two ways to distinguish the predicted rare antiprotons from the abundant negative pions. Both ways involved the fact that the speed of a 1.19 GeV/c $\overline{p}$ differs from that of a 1.19 GeV/c $\pi^{-} :(1)$ According to calculations, a $\overline{\mathrm{p}}$ would trigger one of the Cerenkov counters and a $\pi^{-}$ would trigger the other. (2) The time interval $\Delta t$ between signals from $\mathrm{S} 1$ and $\mathrm{S} 2,$ which were separated by $12 \mathrm{m},$ would have one value for a $\mathrm{p}$ and another value for a $\pi^{-} .$ Thus, if the correct Cerenkov counter were triggered and the time interval $\Delta t$ had the correct value, the experiment would prove the existence of antiprotons.
What is the speed of (a) an antiproton with a momentum of 1.19 $\mathrm{GeV} / \mathrm{c}$ and $(\mathrm{b})$ a negative pion with that same momentum? (The speed of an antiproton through the Cerenkov detectors would actually be slightly less than calculated here because the antiproton would lose a little energy within the detectors. Which Cerenkov detector was triggered by (c) an antiproton and (d) a negative pion? What time interval $\Delta t$ indicated the passage of (e) an antiproton and (f) a negative pion? [Problem adapted from O. Chamberlain, E. Segre, C. Wiegand, and T. Ypsilantis, "Observation of Antiprotons," Physical Review, Vol. $100,$ pp. $947-950(1955) . ]$

Okay. So after reading that paragraph, what is the speed of an anti proton? With a momentum of 1.19 Giga electron volts per seat. So they have a relativistic relationships. So we know that the momentum is gamma and the so we have The oversee is equal to two square root of 1 -1 over pc over EMC squared squared plus one. So for an anti proton EMC squared is 938 0.3 mega electron volts. PC is 1.19 Giga electron volts Which is 1190. My collection bolts. So, plugging that in and we get a velocity 0.785 C. Part B a negative pi on with the same momentum. So for a negative pion we have EMC squared is a 193.6 mega electron volts. It has the same pc. So it gives us a velocity of 0.993. See So for part C, an anti proton. So this beat of an anti proton is about .78 C. But not Over .79 seed. So an anti proton will trigger C. Two a negative pi on. So the speed of the negative pi on is over .79 c. So a negative pylon will trigger see one what time interval adult ET indicated the passage of an electron or for part of a negative pi on. So we have that delta T. Is the over V. We know the distance is 12 m. We know those velocities. So for part E. We have adult it T. of 51 nanoseconds and for part F. Yes, We have adult itty of 40 nanoseconds.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.


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