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\[ \begin{aligned} \delta A_{\mathrm{tot}} & =\sqrt{[\delta(a b)]^{2}+[\delta(c d)]^{2}} \\ & =\sqrt{\left(0.095028 \mathrm{~m}^{2}\right)^{2}+\left(0.060896 \mathrm{~m}^{2}\right)^{2}} \\ & =0.112866 \mathrm{~m}^{2} \approx 0.1 \mathrm{~m}^{2} \end{aligned} \] We round the total uncertainty to one significant figure and round the best estimate to the same decimal place as the uncertainty. The total area with uncertainty is \[ A_{\mathrm{tot}}=(6.4 \pm 0.1) \mathrm{m}^{2} \] 2.2 Exercises Complete these calculations of uncertainty propagation and submit them for the pre-lab assignment. You must show your work in your submission. Your video analysis of the motion of a marble gives it position in frame 23 as \( \left(x_{23}, y_{23}\right)= \) \( (0.134 \mathrm{~m}, 0.120 \mathrm{~m}) \) and its position in frame 24 as \( \left(x_{24}, y_{24}\right)=(0.122 \mathrm{~m}, 0.112 \mathrm{~m}) \). You estimate that you can measure the \( x \) and \( y \) positions with uncertainty \( \pm 0.003 \mathrm{~m} \). The frame rate of the video is 30 frames/s, which means the time interval between frames is \( \Delta t=0.033333 \mathrm{~s} \). The uncertainty of the frame rate of a video camera is VERY small. For the sake of this problem, use \( \delta(\Delta t)=1 \times 10^{-6} \mathrm{~s} \). The mass of the marble is \( (2.031 \pm 0.001) \times 10^{-2} \mathrm{~kg} \). Calculate the following quantities: 1. The value of the velocity component \( v_{x}=\frac{x_{i+1}-x_{i}}{\Delta t} \) and its uncertainty, 2. The value of the velocity component \( v_{y}=\frac{y_{i+1}-y_{i}}{\Delta t} \) and its uncertainty, 3. The value of the velocity magnitude \( v=\sqrt{v_{x}^{2}+v_{y}^{2}} \) and its uncertainty, 4. The value of the momentum component \( p_{x}=m v_{x} \) and its uncertainty, 5. The value of the momentum component \( p_{y}=m v_{y} \) and its uncertainty, and 6. The value of the kinetic energy \( K=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(v_{x}^{2}+v_{y}^{2}\right) \) and its uncertainty. Before you submit Write up your solutions to exercises 1-6 above clearly showing your work. Be sure to express your final values with uncertainties using the guidelines in Appendix A. Circle your answers. Make a PDF file with your work and submit it on Canvas in the Investigation 3 Pre-Lab assignment.

          \[
\begin{aligned}
\delta A_{\mathrm{tot}} & =\sqrt{[\delta(a b)]^{2}+[\delta(c d)]^{2}} \\
& =\sqrt{\left(0.095028 \mathrm{~m}^{2}\right)^{2}+\left(0.060896 \mathrm{~m}^{2}\right)^{2}} \\
& =0.112866 \mathrm{~m}^{2} \approx 0.1 \mathrm{~m}^{2}
\end{aligned}
\]
We round the total uncertainty to one significant figure and round the best estimate to the same decimal place as the uncertainty. The total area with uncertainty is
\[
A_{\mathrm{tot}}=(6.4 \pm 0.1) \mathrm{m}^{2}
\]
2.2 Exercises
Complete these calculations of uncertainty propagation and submit them for the pre-lab assignment. You must show your work in your submission.

Your video analysis of the motion of a marble gives it position in frame 23 as \( \left(x_{23}, y_{23}\right)= \) \( (0.134 \mathrm{~m}, 0.120 \mathrm{~m}) \) and its position in frame 24 as \( \left(x_{24}, y_{24}\right)=(0.122 \mathrm{~m}, 0.112 \mathrm{~m}) \). You estimate that you can measure the \( x \) and \( y \) positions with uncertainty \( \pm 0.003 \mathrm{~m} \). The frame rate of the video is 30 frames/s, which means the time interval between frames is \( \Delta t=0.033333 \mathrm{~s} \). The uncertainty of the frame rate of a video camera is VERY small. For the sake of this problem, use \( \delta(\Delta t)=1 \times 10^{-6} \mathrm{~s} \). The mass of the marble is \( (2.031 \pm 0.001) \times 10^{-2} \mathrm{~kg} \). Calculate the following quantities:
1. The value of the velocity component \( v_{x}=\frac{x_{i+1}-x_{i}}{\Delta t} \) and its uncertainty,
2. The value of the velocity component \( v_{y}=\frac{y_{i+1}-y_{i}}{\Delta t} \) and its uncertainty,
3. The value of the velocity magnitude \( v=\sqrt{v_{x}^{2}+v_{y}^{2}} \) and its uncertainty,
4. The value of the momentum component \( p_{x}=m v_{x} \) and its uncertainty,
5. The value of the momentum component \( p_{y}=m v_{y} \) and its uncertainty, and
6. The value of the kinetic energy \( K=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(v_{x}^{2}+v_{y}^{2}\right) \) and its uncertainty.
Before you submit
Write up your solutions to exercises 1-6 above clearly showing your work. Be sure to express your final values with uncertainties using the guidelines in Appendix A. Circle your answers. Make a PDF file with your work and submit it on Canvas in the Investigation 3 Pre-Lab assignment.
        
Show more…

    δ Atot    =√([δ(a b)]^2+[δ(c d)]^2)
        =√((0.095028  m^2)^2+(0.060896  m^2)^2)
        =0.112866  m^2≈ 0.1  m^2

We round the total uncertainty to one significant figure and round the best estimate to the same decimal place as the uncertainty. The total area with uncertainty is

    Atot=(6.4 ± 0.1) m^2

2.2 Exercises
Complete these calculations of uncertainty propagation and submit them for the pre-lab assignment. You must show your work in your submission.

Your video analysis of the motion of a marble gives it position in frame 23 as (x23, y23)= (0.134  m, 0.120  m) and its position in frame 24 as (x24, y24)=(0.122  m, 0.112  m). You estimate that you can measure the x and y positions with uncertainty ± 0.003  m. The frame rate of the video is 30 frames/s, which means the time interval between frames is Δ t=0.033333  s. The uncertainty of the frame rate of a video camera is VERY small. For the sake of this problem, use δ(Δ t)=1 × 10^-6 s. The mass of the marble is (2.031 ± 0.001) × 10^-2 kg. Calculate the following quantities:
1. The value of the velocity component vx=(xi+1-xi)/(Δ t) and its uncertainty,
2. The value of the velocity component vy=(yi+1-yi)/(Δ t) and its uncertainty,
3. The value of the velocity magnitude v=√(vx^2+vy^2) and its uncertainty,
4. The value of the momentum component px=m vx and its uncertainty,
5. The value of the momentum component py=m vy and its uncertainty, and
6. The value of the kinetic energy K=(1)/(2) m v^2=(1)/(2) m(vx^2+vy^2) and its uncertainty.
Before you submit
Write up your solutions to exercises 1-6 above clearly showing your work. Be sure to express your final values with uncertainties using the guidelines in Appendix A. Circle your answers. Make a PDF file with your work and submit it on Canvas in the Investigation 3 Pre-Lab assignment.

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After doing a number of the exercises with carts and fans on ramps, it is easy to draw the conclusion that everything that moves is moving at cither a constant velocity or a constant acceleration. Let's cxamine the horizontal motion of a triangular frame with a pendulum at its center that has been given a push. It undergoes an unusual motion. You should determine whether or not it is moving at either a constant velocity or constant acceleration. (Note: You may want to look at the motion of the triangular frame by viewing the digital movie entitled PASCO070. This movie is included on the VideoPoint compact disk. If you are not using VideoPoint, your instructor may make the movie available to you some other way.) The images in Fig. $2-41$ are taken from the 7 th, 16 th, and 25 th frames of that movie. Data for the position of the center of the horizontal bar of the triangle were taken every tenth of a second during its first second of motion. The origin was placed at the zero centimeter mark of a fixed meter stick. These data are in the table below. (a) Examine the position vs. time graph of the data shown above. Does the triangle appear to have a constant velocity throughout the first second? A constant acceleration? Why or why not? (b) Discuss the nature of the motion based on the shape of the graph. At approximately what time, if any, is the triangle changing direction? At approximately what time does it have the greatest negative velocity? The greatest positive velocity? Explain the reasons for your answers. (c) Use the data table and the definition of average velocity to calculate the average velocity of the triangle at each of the times between $0.100 \mathrm{~s}$ and $0.900 \mathrm{~s}$. In this case you should use the position just before the indicated time and the position just after the indicated time in your calculation. For example, to calculate the average velocity at $t_{2}=0.100$ seconds, use $x_{3}=44.5 \mathrm{~cm}$ and $x_{1}=52.1$ $\mathrm{cm}$ along with the differences of the times at $t_{3}$ and $t_{1} .$ Hint: Use only times and positions in the gray boxes to get a velocity in a gray box and use only times and positions in the white boxes to get a velocity in a white box. (d) Since people usually refer to velocity as distance divided by time. maybe we can calculate the average velocities as simply $x_{1} / t_{1}, x_{2} / l_{2}$, $x_{y} / l_{3}$, and so on. This would be easier. Is this an equivalent method for finding the velocities at the different times? Try using this method of calculation if you are not sure. Give reasons for your answer. (e) Often, when an oddly shaped but smooth graph is obtained from data it is possible to fit a polynomial to it. For example, a fourth-order polynomial that fits the data is $$\left.x=\mid\left(-376 \mathrm{~cm} / \mathrm{s}^{4}\right) t^{4}+\left(719 \mathrm{~cm} / \mathrm{s}^{3}\right) t^{3}-\left(347 \mathrm{~cm} / \mathrm{s}^{2}\right) t^{2}+(5.63 \mathrm{~cm} / \mathrm{s}) t+52.1 \mathrm{~cm}\right\}$$ Using this polynomial approximation, find the instantaneous velocity at $t=0.700 \mathrm{~s}$. Comment on how your answer compares to the average velocity you calculated at $0.700 \mathrm{~s}$. Are the two values close? Is that what you expect?

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figure-44-14-shows-part-of-the-experimental-arrangement-in-which-antiprotons-were-discovered-in-th-2

Figure 44-14 shows part of the experimental arrangement in which antiprotons were discovered in the 1950 s. A beam of $6.2 \mathrm{GeV}$ 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 and $$ \begin{aligned} &\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\mathrm{p}+\overline{\mathrm{p}} \\ &\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\mathrm{p}+\overline{\mathrm{p}} \end{aligned} $$ However, even if these reactions did occur, they would be rare compared to the reactions and $$ \begin{aligned} &\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\pi^{+}+\pi^{-} \\ &\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\pi^{+}+\pi^{-} \end{aligned} $$ Thus, most of the particles produced by the collisions between the $6.2 \mathrm{GeV}$ 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 the second magnetic field (Q1) had to be negatively charged (either a por a $\pi^{-}$ ) and have a momentum of $1.19 \mathrm{GeV} / \mathrm{c}$. Field $\mathrm{Q} 1$ 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 $\mathrm{S} 1 .$ The passage of a charged particle through the counter triggered a signal, with each signal indicating the passage of either a $1.19 \mathrm{GeV} / \mathrm{c}$ $\pi^{-}$ or (presumably) a $1.19 \mathrm{GeV} / c \overline{\mathrm{p}}$. After being refocused by magnetic field $\mathrm{Q} 2$, the particles were directed by magnetic field M2 through a second scintillation counter $\mathrm{S} 2$ and then through two Cerenkov counters $\mathrm{C} 1$ and $\mathrm{C} 2$. 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 $0.79 c$ would trigger $\mathrm{C} 1$ and a particle with a speed between $0.75 c$ and $0.78 c$ would trigger $\mathrm{C} 2$. 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 $1.19 \mathrm{GeV} / c \overline{\mathrm{p}}$ differs from that of a $1.19$ GeV/c $\pi^{-}$ : (1) According to calculations, a p would trigger one of the Cerenkov counters and a $\pi^{-}$ would trigger the other. (2) The time interval $\Delta t$ between signals from $S 1$ and $\mathrm{S} 2$, which were separated by $12 \mathrm{~m}$, would have one value for a $\overline{\mathrm{p}}$ 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 (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. Segrè, C. Wiegand, and T. Ypsilantis, "Observation of Antiprotons," Physical Review, Vol. 100, pp. 947-950 (1955).]

Fundamentals of Physics

figure-44-14-shows-part-of-the-experimental-arrangement-in-which-antiprotons-were-discovered-in-the-

Figure $44-14$ shows part of the experimental arrangement in which antiprotons were discovered in the 1950 s. A beam of $6.2 \mathrm{GeV}$ 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 and $p+p \rightarrow p+p+p+\bar{p}$ $p+n \rightarrow p+n+p+\bar{p}$. However, even if these reactions did occur, they would be rare compared to the reactions and $\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\pi^{+}+\pi^{-}$ $\mathrm{p}+\mathrm{n} \rightarrow \mathrm{p}+\mathrm{n}+\pi^{+}+\pi^{-}$. Thus, most of the particles produced by the collisions between the $6.2 \mathrm{GeV}$ 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 the second magnetic field (Q1) had to be negatively charged (either a $\overline{\mathrm{p}}$ or a $\pi^{-}$ ) and have a momentum of $1.19 \mathrm{GeV} / \mathrm{c}$. Field $\mathrm{Q} 1$ 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 $\mathrm{S} 1 .$ The passage of a charged particle through the counter triggered a signal, with each signal indicating the passage of either a $1.19 \mathrm{GeV} / \mathrm{c} \pi$ or (presumably) a $1.19 \mathrm{GeV} / \mathrm{c} \overline{\mathrm{p}}$. After being refocused by magnetic field $\mathrm{Q} 2,$ the particles were directed by magnetic field $\mathrm{M} 2$ through a second scintillation counter $\mathrm{S} 2$ and then through two Cerenkov counters $\mathrm{C} 1$ and $\mathrm{C} 2$. 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 $0.79 c$ would trigger $\mathrm{C} 1$ and a particle with a speed between $0.75 c$ and $0.78 c$ 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 $1.19 \mathrm{GeV} / \mathrm{c} \overline{\mathrm{p}}$ differs from that of a $1.19 \mathrm{GeV} / \mathrm{c} \pi:$ (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 $\overline{\mathrm{p}}$ 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. Segrè, C. Wiegand, and T. Ypsilantis, "Observation of Antiprotons," Physical Review, Vol. $100,$ pp. $947-950(1955)$.]

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