Physics: Measurement Uncertainty in Motion Labs
Estimating, calculating, and reporting uncertainty in motion measurements
Physics: Measurement Uncertainty in Motion Labs
Estimating, calculating, and reporting uncertainty in motion measurements
Physics - Grade 9-12
- 1
A student measures a cart's displacement with a meterstick marked every 1 mm. The cart moves from 12.4 cm to 86.7 cm. Estimate the uncertainty in each position reading and calculate the displacement with its uncertainty.
For an analog scale, the reading uncertainty is usually about half the smallest division.
Each position reading has an uncertainty of about ±0.05 cm, since the smallest marking is 0.1 cm. The displacement is 86.7 cm - 12.4 cm = 74.3 cm. For subtraction, absolute uncertainties add, so the displacement uncertainty is ±0.10 cm. The result is 74.3 ± 0.1 cm. - 2
A stopwatch records a time interval of 3.42 s. The stopwatch has a resolution of 0.01 s, but the student's reaction time adds an estimated uncertainty of ±0.20 s. Which uncertainty should dominate the measurement, and how should the time be reported?
The reaction time uncertainty dominates because ±0.20 s is much larger than the stopwatch resolution uncertainty. The time should be reported as 3.42 ± 0.20 s, or more simply 3.4 ± 0.2 s. - 3
A motion sensor measures a cart's position as 1.250 m with an uncertainty of ±0.005 m. Calculate the percent uncertainty in the position measurement.
Percent uncertainty equals absolute uncertainty divided by the measured value, multiplied by 100%.
The percent uncertainty is (0.005 m / 1.250 m) × 100% = 0.4%. The position measurement has a percent uncertainty of 0.4%. - 4
A cart travels 2.40 ± 0.02 m in 1.20 ± 0.10 s. Calculate the average speed and estimate its percent uncertainty.
For multiplication or division, add the percent uncertainties.
The average speed is 2.40 m / 1.20 s = 2.00 m/s. The percent uncertainty in distance is (0.02 / 2.40) × 100% = 0.83%. The percent uncertainty in time is (0.10 / 1.20) × 100% = 8.33%. For division, percent uncertainties add, so the total percent uncertainty is about 9.16%, or 9%. The speed is about 2.00 m/s with an uncertainty of 0.18 m/s, so it can be reported as 2.0 ± 0.2 m/s. - 5
A group measures the time for a toy car to travel 1.00 m in five trials: 1.26 s, 1.31 s, 1.29 s, 1.24 s, and 1.30 s. Calculate the mean time and estimate the uncertainty as half the range.
The mean time is (1.26 + 1.31 + 1.29 + 1.24 + 1.30) / 5 = 1.28 s. The range is 1.31 s - 1.24 s = 0.07 s, so half the range is 0.035 s, which rounds to about 0.04 s. The time can be reported as 1.28 ± 0.04 s. - 6
A student reports a measured acceleration as 0.847362 m/s² ± 0.08 m/s². Rewrite the result using an appropriate number of decimal places.
Round the measured value to the same decimal place as the uncertainty.
The uncertainty ±0.08 m/s² is given to the hundredths place, so the measured value should also be rounded to the hundredths place. The result should be reported as 0.85 ± 0.08 m/s². - 7
In a constant-velocity lab, a student makes a position-time graph. The best-fit line has a slope of 0.62 m/s. A steep reasonable line has a slope of 0.67 m/s, and a shallow reasonable line has a slope of 0.58 m/s. Estimate the uncertainty in the slope.
Use half the spread between the steepest and shallowest reasonable lines.
The uncertainty can be estimated as half the difference between the steepest and shallowest reasonable slopes. The difference is 0.67 m/s - 0.58 m/s = 0.09 m/s, so half is 0.045 m/s. The slope can be reported as about 0.62 ± 0.05 m/s. - 8
A student uses photogates to measure the time for a cart to pass between two gates. The distance between gates is 0.500 ± 0.002 m, and the time is 0.721 ± 0.001 s. Which measurement contributes more to the uncertainty in the calculated speed?
The distance percent uncertainty is (0.002 / 0.500) × 100% = 0.4%. The time percent uncertainty is (0.001 / 0.721) × 100% = about 0.14%. The distance measurement contributes more to the uncertainty in the calculated speed because it has the larger percent uncertainty. - 9
A motion lab result gives an experimental acceleration of 9.6 ± 0.4 m/s² for a falling object. The accepted value is 9.8 m/s². Does the accepted value fall within the experimental uncertainty range?
Compare the accepted value with the lowest and highest possible values from the uncertainty interval.
The experimental range is 9.6 - 0.4 = 9.2 m/s² to 9.6 + 0.4 = 10.0 m/s². The accepted value 9.8 m/s² is inside this range, so the result is consistent with the accepted value within uncertainty. - 10
Two groups measure the same cart speed. Group A reports 1.20 ± 0.03 m/s. Group B reports 1.28 ± 0.04 m/s. Do the two results agree within uncertainty? Explain.
Group A's range is 1.17 m/s to 1.23 m/s. Group B's range is 1.24 m/s to 1.32 m/s. The ranges do not overlap, so the two results do not agree within their stated uncertainties. - 11
A student calculates velocity from two position readings: x1 = 0.15 ± 0.01 m and x2 = 1.05 ± 0.01 m. The time interval is 2.0 ± 0.1 s. Calculate the velocity and estimate its uncertainty.
First find the displacement uncertainty, then use percent uncertainty for the division by time.
The displacement is 1.05 m - 0.15 m = 0.90 m. The displacement uncertainty is ±0.02 m because the two position uncertainties add. The velocity is 0.90 m / 2.0 s = 0.45 m/s. The displacement percent uncertainty is (0.02 / 0.90) × 100% = 2.2%, and the time percent uncertainty is (0.1 / 2.0) × 100% = 5.0%. The total percent uncertainty is about 7.2%, so the velocity uncertainty is 0.072 × 0.45 m/s = about 0.03 m/s. The result is 0.45 ± 0.03 m/s. - 12
A ticker tape timer makes dots every 0.020 s. A student measures the distance between two dots as 3.6 cm with an uncertainty of ±0.1 cm. What is the speed for that interval, and what is the percent uncertainty due to the distance measurement?
The speed is 3.6 cm / 0.020 s = 180 cm/s, or 1.8 m/s. The percent uncertainty due to distance is (0.1 cm / 3.6 cm) × 100% = about 2.8%. If the timing uncertainty is ignored, the speed can be reported with about 2.8% uncertainty. - 13
A student says, "Our motion sensor gives digital readings, so there is no measurement uncertainty." Explain why this statement is incorrect.
Think about the limits of the measuring device and the way it is used.
The statement is incorrect because all measurements have uncertainty. A digital sensor may reduce reading uncertainty, but it still has limits such as resolution, calibration error, sampling rate, alignment error, and random noise. The uncertainty may be small, but it is not zero. - 14
A cart rolls down a ramp. The class collects this position-time data: at 0.0 s, x = 0.00 m; at 0.5 s, x = 0.12 m; at 1.0 s, x = 0.48 m; at 1.5 s, x = 1.08 m; at 2.0 s, x = 1.92 m. The position uncertainty is ±0.02 m. Describe whether the uncertainty is likely to change the conclusion that the cart is accelerating.
The uncertainty is not likely to change the conclusion that the cart is accelerating. The position increases by larger amounts during each equal 0.5 s time interval: 0.12 m, 0.36 m, 0.60 m, and 0.84 m. These increases are much larger than the ±0.02 m position uncertainty, so the pattern clearly shows increasing speed. - 15
In a lab report, a student writes the final result as acceleration = 1.3 m/s². List two pieces of information missing from this result that would make it scientifically stronger.
A strong measurement result should tell the reader how reliable the number is.
The result is missing an uncertainty value, such as ±0.1 m/s², and it is missing context about how the acceleration was found, such as from the slope of a velocity-time graph or from repeated trials. It would also be stronger if the student included units consistently, a comparison to an accepted or predicted value, and a brief discussion of major sources of uncertainty.