Collecting & Processing Data - IB Physics

1. The power $P$ dissipated in a resistor is calculated using $P = I^2R$. A current $I = (2.0 \pm 0.1)$ A is measured in a resistor of resistance $R = (10.0 \pm 0.2)$ $\Omega$. What is the absolute uncertainty in the calculated power?
   Answer: 4.8 W First, calculate the central value of the power: $P = (2.0)^2 \times 10.0 = 40.0$ W. Next, find the total percentage uncertainty by adding the percentage uncertainties of the components. The percentage uncertainty in $I$ is $(\Delta I/I) = (0.1/2.0) \times 100\% = 5\%$. The percentage uncertainty in $R$ is $(\Delta R/R) = (0.2/10.0) \times 100\% = 2\%$. The formula for power involves $I^2$, so the percentage uncertainty from the current term is $2 \times 5\% = 10\%$. The total percentage uncertainty in $P$ is $\Delta P/P = 2(\Delta I/I) + (\Delta R/R) = 10\% + 2\% = 12\%$. The absolute uncertainty is $\Delta P = 0.12 \times 40.0$ W $= 4.8$ W.
2. A student measures the angle of incidence $\theta_1 = (30 \pm 1)°$ and the angle of refraction $\theta_2 = (20 \pm 1)°$ for light entering a glass block. The refractive index $n$ is calculated using Snell's Law, $n = \frac{\sin\theta_1}{\sin\theta_2}$. What is the approximate percentage uncertainty in the calculated value of $n$?
   Answer: 8% To find the percentage uncertainty in $n$, we must add the percentage uncertainties of $\sin\theta_1$ and $\sin\theta_2$. A common approximation is to add the percentage uncertainties of the angles themselves. Percentage uncertainty in $\theta_1$ is $(1/30) \times 100\% \approx 3.3\%$. Percentage uncertainty in $\theta_2$ is $(1/20) \times 100\% = 5.0\%$. Summing these gives $3.3\% + 5.0\% = 8.3\%$, which is approximately 8%. A more precise method involves calculating $n_{max} = \sin(31°)/\sin(19°) \approx 1.582$ and $n_{min} = \sin(29°)/\sin(21°) \approx 1.353$ for a central value of $n = \sin(30°)/\sin(20°) \approx 1.462$. The percentage uncertainty is $((1.582 - 1.353)/2) / 1.462 \times 100\% \approx 7.8\%$, which also rounds to 8%.
3. A student investigates how the resistance $R$ of a wire depends on its temperature $T$ in Celsius. The relationship is modelled as $R = R_0(1 + \alpha T)$, where $R_0$ is the resistance at 0 °C and $\alpha$ is a constant. The student plots a graph of $R$ on the y-axis and $T$ on the x-axis. What does the y-intercept of this graph represent?
   Answer: The resistance of the wire at 0 °C, $R_0$. The equation $R = R_0(1 + \alpha T)$ can be expanded to $R = R_0 + R_0\alpha T$. When plotting $R$ on the y-axis against $T$ on the x-axis, this equation is in the linear form $y = c + mx$. The y-intercept $c$ is the value of $y$ when $x=0$. In this case, it is the value of $R$ when $T=0$ °C. According to the equation, when $T=0$, $R = R_0$. Therefore, the y-intercept represents the resistance of the wire at 0 °C.
4. A student uses a smartphone app to collect acceleration data during an oscillating motion experiment. The app samples at 100 Hz, but the motion has a frequency of approximately 2 Hz. The student notices that the collected data shows more scatter than expected from the measurement precision. Which processing approach would most effectively reduce this scatter while preserving the essential motion characteristics?
   Answer: Apply a low-pass filter with a cutoff frequency of 10 Hz to remove high-frequency noise When analyzing oscillating motion data with unexpected scatter, you're dealing with a signal processing problem where noise interferes with your measurement of the actual physical motion. The key is understanding how different frequencies relate to your signal versus unwanted noise. Your motion oscillates at 2 Hz, which means any legitimate acceleration changes occur at this frequency and its harmonics (4 Hz, 6 Hz, etc.). However, measurement noise, vibrations, and electronic interference typically appear at much higher frequencies. A low-pass filter with a 10 Hz cutoff (option A) removes high-frequency noise above 10 Hz while preserving your 2 Hz motion and its important harmonics. This directly addresses the scatter problem without losing essential motion characteristics. Option B is problematic because reducing sampling to 10 Hz violates the Nyquist criterion—you need at least twice your signal frequency for proper reconstruction, so 2 Hz motion requires minimum 4 Hz sampling. Option C misunderstands the issue entirely; increasing sampling rate to 200 Hz would actually capture more high-frequency noise, worsening the scatter problem. Option D's band-pass filter centered at 2 Hz with 1 Hz bandwidth (1.5-2.5 Hz) is too restrictive—it would eliminate important harmonic information that characterizes the motion's shape, potentially distorting your acceleration measurements. For IB Physics data analysis questions, remember that signal processing aims to preserve your physical phenomenon while removing interference. Low-pass filtering is often the first approach when you have broadband noise contaminating a lower-frequency signal of interest.
5. The distance $s$ fallen by an object is given by $s = \frac{1}{2}gt^2$. A student measures time $t = (1.20 \pm 0.05)$ s. Assuming the value of $g$ is known precisely ($g = 9.81$ m s⁻²), what is the absolute uncertainty in the calculated distance $s$?
   Answer: 0.59 m First, calculate the central value of the distance: $s = \frac{1}{2}(9.81)(1.20)^2 = 7.0632$ m. The uncertainty in $s$ depends on the uncertainty in $t$. Since $s \propto t^2$, the percentage uncertainty in $s$ is twice the percentage uncertainty in $t$. The percentage uncertainty in $t$ is $(\Delta t/t) = (0.05 / 1.20) \approx 0.04167$ or 4.167%. The percentage uncertainty in $s$ is $\Delta s/s = 2 \times (\Delta t/t) = 2 \times 4.167\% = 8.333\%$. The absolute uncertainty in $s$ is $\Delta s = (0.08333) \times s = 0.08333 \times 7.0632 \approx 0.589$ m. This is closest to 0.59 m.
6. A student investigates the photoelectric effect by measuring the maximum kinetic energy of emitted electrons for different light frequencies. After collecting data, the student realizes that some measurements were taken under ambient lighting conditions that contributed additional low-energy photons. How should this contamination affect the data processing approach?
   Answer: The contamination only affects the number of emitted electrons, not their maximum kinetic energy, so no correction is needed When analyzing photoelectric effect data, you need to understand what determines the maximum kinetic energy of emitted electrons. Einstein's photoelectric equation shows that $E_{k,max} = hf - \phi$, where the maximum kinetic energy depends only on the photon frequency (f) and the material's work function (φ). The key insight is that each photon interacts individually with an electron. When ambient light adds low-energy photons to your experimental setup, these photons either lack sufficient energy to eject electrons (if below the threshold frequency) or they eject electrons with lower kinetic energies than your main light source. The highest-energy photons from your intended light source still produce electrons with the same maximum kinetic energy they would have produced alone. Answer A correctly recognizes that contamination affects the total number of emitted electrons but doesn't change the maximum kinetic energy, so no correction is needed for your energy measurements. Answer B is overly cautious—discarding usable data isn't necessary when the contamination doesn't affect your key measurement. Answer C misunderstands the physics by suggesting you subtract light intensity, but intensity affects photon quantity, not the maximum energy of individual photon-electron interactions. Answer D incorrectly assumes you need intensity ratio corrections, again confusing the number of photons with their individual energies. For IB Physics photoelectric effect questions, remember that maximum kinetic energy depends only on individual photon energy (frequency), not on light intensity or the presence of additional lower-energy photons.
7. During a collision experiment, a student records velocity data every 0.01 s using a motion sensor. The raw data shows significant noise due to vibrations. To calculate the momentum change most accurately, the student should:
   Answer: Apply a moving average filter to smooth the data, then use the filtered initial and final velocities A moving average filter reduces random noise while preserving the underlying trend, giving more accurate initial and final velocity values for momentum calculation. Option A ignores the benefit of multiple measurements and may use noisy endpoints. Option C introduces unnecessary complexity and doesn't address the noise issue directly. Option D is circular reasoning since the student doesn't know the theoretical prediction beforehand and may introduce bias.
8. During a calorimetry experiment, a student records temperature versus time data while a heated metal sample cools in water. The data collection includes a 2-minute pre-heating period, the moment of sample insertion, and 8 minutes of cooling. Which data processing strategy would give the most accurate determination of the specific heat capacity?
   Answer: Extrapolate the pre-heating and post-cooling linear trends to find the theoretical temperature change at the moment of insertion Calorimetry experiments face a fundamental challenge: heat loss to the environment during data collection creates systematic errors. When you insert a hot metal sample into water, some thermal energy escapes to the surroundings rather than transferring between the metal and water, leading to inaccurate specific heat calculations. The most precise approach involves extrapolation to eliminate heat loss effects. By plotting temperature versus time, you can identify linear trends in both the pre-heating period (where water temperature may drift slightly) and the post-equilibrium cooling period (where the system loses heat to surroundings). Extrapolating these linear trends back to the moment of sample insertion gives you the theoretical temperatures that would have occurred without environmental heat loss. This method, shown in option D, provides the most accurate temperature change for your heat transfer calculations. Option A fails because the maximum recorded temperature and final equilibrium temperature both include heat loss effects, systematically underestimating the true temperature change. Option B incorrectly assumes that excluding early data improves accuracy, but exponential cooling behavior actually indicates ongoing heat loss to the environment. Option C compounds errors by averaging temperatures throughout the cooling period, which includes all the heat loss effects you want to eliminate. For IB Physics calorimetry questions, remember that the highest accuracy comes from correcting for systematic errors, not just improving measurement precision. Look for data processing methods that account for heat loss to the environment—extrapolation techniques are your most powerful tool for eliminating these systematic effects.
9. A student uses an analog voltmeter where the smallest division on the scale is 0.2 V. The needle is pointing exactly halfway between the 4.6 V and 4.8 V markings. What is the correct way to record this measurement and its uncertainty?
   Answer: $4.7 \pm 0.1$ V The reading is halfway between 4.6 V and 4.8 V, so the value is 4.7 V. For an analog instrument, the reading uncertainty is conventionally taken as half of the smallest division (or graduation) on the scale. The smallest division is 0.2 V, so the uncertainty is $0.2 / 2 = 0.1$ V. The value should be recorded to the same number of decimal places as the uncertainty. Thus, the correct reading is $4.7 \pm 0.1$ V. Recording it as 4.70 implies a higher precision than is possible with the instrument.
10. A student is investigating the factors affecting the period $T$ of a vertically oscillating mass-spring system. The student's hypothesis is that the period is proportional to the square root of the mass ($T \propto \sqrt{m}$). To test this hypothesis, the student measures the period for different masses. Which of the following is an essential control variable for this experiment?
    Answer: The spring constant of the spring. The experiment is designed to test the relationship between period $T$ (dependent variable) and mass $m$ (independent variable). A control variable is any other physical parameter that could affect the outcome and therefore must be kept constant. The theoretical period is given by $T = 2\pi\sqrt{m/k}$. This equation shows that the period depends on both mass $m$ and the spring constant $k$. To isolate the effect of $m$ on $T$, the spring constant $k$ must be kept constant. This means the same spring must be used for all trials.
11. A student measures the decay of a radioactive sample by counting particles detected in consecutive 1-minute intervals. The count rates are: 850, 823, 801, 774, 751, 730 counts per minute. To determine the half-life most accurately from this limited data set, which processing method should be used?
    Answer: Plot the natural logarithm of count rate versus time and use the slope to find the decay constant When analyzing radioactive decay data, you're dealing with an exponential process that follows the equation $N(t) = N_0 e^{-\lambda t}$, where λ is the decay constant. The key insight is recognizing which mathematical approach gives the most accurate results with limited data points. Option D is correct because taking the natural logarithm of both sides gives $\ln N(t) = \ln N_0 - \lambda t$. This transforms the exponential relationship into a linear one, where plotting ln(count rate) versus time produces a straight line with slope -λ. Linear regression on this transformed data is statistically robust and uses all data points equally, minimizing the impact of random measurement uncertainties. Option A attempts to fit the original exponential curve, but non-linear curve fitting is mathematically complex and more sensitive to measurement errors, especially with only six data points. Option B uses only two data points, completely ignoring four measurements and making the result highly sensitive to random errors in those specific measurements. Option C calculates ratios between consecutive measurements, but averaging these ratios doesn't properly account for the time dependence and can amplify measurement uncertainties. The linearization technique in option D is a standard approach in experimental physics because it converts complex exponential fitting into simple linear regression, which is more reliable and gives better uncertainty estimates. Remember: when you encounter exponential decay problems with real data, always consider whether taking logarithms can linearize the relationship for more accurate analysis.
12. During a projectile motion experiment, a student tracks the trajectory of a ball using video analysis. The analysis software provides position coordinates at regular time intervals, but some coordinate pairs show obvious tracking errors where the software briefly lost the ball. Which data processing strategy would give the most reliable kinematic analysis?
    Answer: Exclude erroneous data points and perform kinematic analysis using only the remaining valid coordinates Projectile motion analysis relies on fitting parabolic trajectories to determine initial conditions and acceleration. Using only valid data points for this fitting process avoids introducing artificial correlations or systematic errors from interpolated points. Option A risks introducing bias if the quadratic assumption doesn't perfectly match the data. Option B oversimplifies the trajectory shape. Option D may oversmooth the data and mask real physical effects or measurement systematic errors.
13. A student investigates the relationship between the period $T$ and the length $L$ of a simple pendulum. The theoretical relationship is $T = 2\pi\sqrt{\frac{L}{g}}$. To determine the acceleration due to gravity, $g$, from a graph, which quantities should be plotted to produce a straight line that passes through the origin?
    Answer: $T^2$ on the y-axis and $L$ on the x-axis. To obtain a linear relationship of the form $y = mx + c$, the equation $T = 2\pi\sqrt{\frac{L}{g}}$ must be rearranged. Squaring both sides gives $T^2 = (2\pi)^2 \frac{L}{g}$, which can be written as $T^2 = \left(\frac{4\pi^2}{g}\right)L$. This equation matches the form $y = mx$, where $y = T^2$, $x = L$, the gradient $m = \frac{4\pi^2}{g}$, and the y-intercept is zero. Therefore, plotting $T^2$ versus $L$ will produce a straight line through the origin.
14. A student investigates conservation of momentum in a collision. The total momentum before the collision is calculated as $p_{initial} = (0.50 \pm 0.02)$ kg m s⁻¹ and the total momentum after the collision is $p_{final} = (0.47 \pm 0.02)$ kg m s⁻¹. Which conclusion is justified by these results?
    Answer: Momentum was conserved because the difference between the values is within the combined experimental uncertainty. To check for agreement between two experimental values, we must consider their uncertainties. The difference between the central values is $\Delta p = 0.50 - 0.47 = 0.03$ kg m s⁻¹. The uncertainty in this difference is the sum of the absolute uncertainties: $\Delta(\Delta p) = 0.02 + 0.02 = 0.04$ kg m s⁻¹. The result for the difference is $0.03 \pm 0.04$ kg m s⁻¹. Since this range includes zero, the experimental results are consistent with the hypothesis that the true difference is zero. Therefore, the data supports the conclusion that momentum was conserved within the experimental uncertainty.
15. A student performs an experiment to measure the specific heat capacity of a metal block. They conduct five trials and obtain the following values: 450, 452, 449, 510, and 451 J kg⁻¹ K⁻¹. The accepted value is 450 J kg⁻¹ K⁻¹. What is the most appropriate next step for the student to take?
    Answer: Discard the 510 value as an outlier and average the other four values. The data set {449, 450, 451, 452, 510} shows four values clustered closely together and one value (510) that is significantly different. This value is a statistical outlier, likely resulting from a mistake in that particular trial. Including the outlier would skew the average away from the true value. The scientifically appropriate procedure is to identify and discard clear outliers before calculating the mean. The average of the four consistent values (449, 450, 451, 452) is 450.5 J kg⁻¹ K⁻¹, which is a reliable estimate.
16. A student measures the period of a pendulum for different lengths and records the data in a table. To determine the relationship between period $T$ and length $L$, which data processing approach would most effectively linearize the expected theoretical relationship $T = 2\pi\sqrt{\frac{L}{g}}$?
    Answer: Plot $T^2$ versus $L$ and determine the slope of the linear relationship To linearize $T = 2\pi\sqrt{\frac{L}{g}}$, square both sides to get $T^2 = \frac{4\pi^2}{g}L$. This shows $T^2$ is directly proportional to $L$, making option B correct. Option A doesn't linearize the relationship. Option C would work but is unnecessarily complex compared to the direct $T^2$ vs $L$ approach. Option D creates an inverse relationship that doesn't match the theoretical form.
17. In a standing wave experiment, a student measures the positions of nodes along a string for different frequencies. The raw position data contains measurement uncertainty of ±0.5 cm. To determine the wavelength most accurately, which data processing method should be prioritized?
    Answer: Measure the distance spanning several wavelengths and divide by the number of wavelengths counted Measuring over multiple wavelengths reduces the relative impact of measurement uncertainty. If each position has ±0.5 cm uncertainty, measuring one wavelength gives high relative error, but measuring across many wavelengths allows the uncertainties to partially cancel when averaged. Option A uses minimal data and maximizes relative uncertainty. Option B improves on A but still doesn't optimize the measurement span. Option D reduces the data set unnecessarily and may introduce bias toward easier-to-measure positions.
18. A student performs an experiment to find the internal resistance $r$ of a cell with emf $\varepsilon$. The student measures the terminal potential difference $V$ across the cell for different values of current $I$ drawn from it. The relationship is $\varepsilon = V + Ir$. To find $r$, the student plots $V$ on the y-axis against $I$ on the x-axis. The student's voltmeter has a zero error, causing it to read 0.5 V higher than the true value. How will this affect the graph and the calculated value of $r$?
    Answer: The y-intercept of the graph will be incorrect, but the gradient will be correct. The equation rearranged for the graph is $V = -rI + \varepsilon$. This is of the form $y = mx + c$, where the gradient $m = -r$ and the y-intercept $c = \varepsilon$. The voltmeter's zero error is a systematic error, so every measured voltage is $V_{meas} = V_{true} + 0.5$. The plotted equation becomes $V_{meas} = -rI + (\varepsilon + 0.5)$. The gradient of this new line is still $-r$, so the calculated internal resistance $r$ will be correct. However, the y-intercept is now $\varepsilon + 0.5$, which is an incorrect value for the emf $\varepsilon$.
19. A student investigates Hooke's law by plotting applied force $F$ (y-axis) against extension $x$ (x-axis) for a spring. The graph shows a straight line with equation $F = kx + F_0$, where $F_0$ is a positive constant. What does this result indicate about the spring?
    Answer: The spring has a natural tension when unstretched. Hooke's law states that $F = kx$ for an ideal spring, where the graph should pass through the origin. The positive y-intercept $F_0$ indicates that a force is required even when there is no additional extension ($x = 0$). This suggests the spring has some natural tension or pre-stress when in its 'unstretched' position. The spring still obeys Hooke's law, but with an offset due to this initial tension.
20. A quantity $N$ is measured as a function of time $t$. The theoretical model predicts an exponential relationship $N = N_0 e^{-\lambda t}$, where $N_0$ and $\lambda$ are constants. To find the constant $\lambda$ from a linear graph, which quantities should a student plot?
    Answer: $\ln(N)$ versus $t$ To linearize the exponential equation $N = N_0 e^{-\lambda t}$, we take the natural logarithm of both sides: $\ln(N) = \ln(N_0 e^{-\lambda t})$. Using logarithm properties, this becomes $\ln(N) = \ln(N_0) + \ln(e^{-\lambda t})$, which simplifies to $\ln(N) = \ln(N_0) - \lambda t$. This equation is in the form of a straight line, $y = c + mx$. By plotting $y = \ln(N)$ on the vertical axis and $x = t$ on the horizontal axis, the resulting graph will be a straight line with a gradient $m = -\lambda$ and a y-intercept $c = \ln(N_0)$.

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