Measurements and their Errors
5 questions, 48 marks, every one with its mark scheme.
Or open them one at a time, as you finish each question.
Question 01
8 marksWhich of these is an SI base unit?
Tick (✓) one box.
- joule
- kelvin
- newton
- volt
Mark scheme
- kelvin1
The stiffness k of a spring is defined by F = k ΔL, where F is the force applied and ΔL is the extension.
Determine the SI base units of k.
Mark scheme
- k = F / ΔL, so the unit is N m−11
- kg s−21
A technician records the temperature of a water bath as 18 °C. All other temperatures in her results table are in kelvin.
Write the temperature of the water bath as it should appear in her table, with its unit.
Mark scheme
- 291 K1
Do not accept: 291 °K; 291 k
A proton leaving a small accelerator carries 2.4 MeV of kinetic energy.
Calculate the kinetic energy of the proton in J.
Mark scheme
- 2.4 × 1.60 × 10−131
- 3.8 × 10−13 J1
Estimate the gravitational potential energy gained by an adult who walks up the stairs from the ground floor to the first floor of a house.
Show your working.
Mark scheme
- sensible estimates: mass of adult 50 kg to 100 kg and height gained 2.5 m to 4 m1
- use of ΔEp = m g Δh to give a value between 1000 J and 4000 J, to 1 or 2 significant figures1
Question 02
11 marksA student investigates how the depth of water in a plastic tank changes as the water drains out through a small hole in the base. A vertical ruler is fixed inside the tank with its end resting on the base. The student reads the depth from the ruler every 20 s, using a stopwatch.
The student reads the ruler while standing, looking down at the water surface at an angle.
Name the cause of random error this introduces into the depth readings and describe how the student can avoid it.
Mark scheme
- parallax error1
- placing eye level at the level of the water surface1
The zero mark of the ruler is 3 mm from its end.
Name the type of error this causes and state its effect on the depth readings.
Mark scheme
- systematic error: zero error on ruler (zero mark not at the end)1
- every depth reading is 3 mm less than the true depth1
Do not accept: zero error on its own
The student's teacher suggests filming the ruler and a digital clock together with a video camera instead.
Explain one advantage of this method.
Mark scheme
- freeze frame1
- allows the observer to take simultaneous readings of depth and time1
Two students, P and Q, each time how long the full tank takes to empty. Each repeats the measurement five times. Their results are shown in the table.
| Student | Time to empty / s | ||||
|---|---|---|---|---|---|
| P | 182.4 | 185.1 | 179.6 | 184.0 | 181.9 |
| Q | 183.3 | 182.7 | 183.5 | 190.8 | 182.9 |
Determine the mean time to empty and its absolute uncertainty from the results of student Q.
Give the uncertainty to an appropriate number of significant figures.
Mark scheme
- 190.8 s identified as anomalous and removed1
- mean = 183.1 s1
- uncertainty = half the range = 0.4 s1
Do not accept: an uncertainty to 3 or more significant figures
Compare the precision of the results of the two students. Use data from the table in your answer.
Mark scheme
- the results of Q (without the anomaly) are more precise1
- range of Q is 0.8 s, range of P is 5.5 s: small variation in repeated measurements1
Question 03
9 marksA student determines the kinetic energy of a glider on an air track. A card fixed to the glider passes through a light gate, which records the time t for the card to pass. The length L of the card is measured with a ruler that has a smallest division of 1 mm, by taking a reading at each end of the card. The mass m of the glider is measured on a balance.
| Quantity | Value | Absolute uncertainty |
|---|---|---|
| m | 0.452 kg | 0.001 kg |
| L | 10.0 cm | 0.1 cm |
| t | 0.125 s | 0.001 s |
Explain why the absolute uncertainty in L is 0.1 cm.
Mark scheme
- uncertainty in each reading is half the smallest division = 0.05 cm1
- subtraction of data leads to addition of uncertainties: 2 × 0.05 cm1
Calculate the kinetic energy of the glider.
Mark scheme
- v = 0.100 / 0.125 = 0.800 m s−11
- Ek = ½ × 0.452 × 0.8002 = 0.145 J1
Calculate the absolute uncertainty in the kinetic energy of the glider.
Mark scheme
- percentage uncertainty in v = 1.0 % + 0.8 % = 1.8 %1
- percentage uncertainty in Ek = 0.2 % + 2 × 1.8 % = 3.8 %1
- absolute uncertainty = 0.145 × 3.8 / 100 = 0.0055 J1
Do not accept: addition of absolute uncertainties for a product
Suggest and explain one change to the card that would reduce the percentage uncertainty in L.
Mark scheme
- use a longer card1
- the absolute uncertainties are unchanged and L is larger, so the percentage uncertainty is reduced1
Question 04
14 marksA marble is released from rest at different distances d up a straight ramp. For each distance, a student measures the mean time t for the marble to roll to the bottom. The absolute uncertainty in each mean value of t is 0.03 s.
The student plots a graph of t2 on the vertical axis against d on the horizontal axis, with error bars.
Complete the table.
| d / m | mean t / s | t2 / s2 |
|---|---|---|
| 0.20 | 1.06 | 1.12 |
| 0.40 | 1.50 | 2.25 |
| 0.60 | 1.82 | |
| 0.80 | 2.08 | |
| 1.00 | 2.35 | 5.52 |
Mark scheme
- 3.311
- 4.331
Calculate the absolute uncertainty in t2 for d = 0.40 m.
Mark scheme
- percentage uncertainty in t2 = 2 × (0.03 / 1.50 × 100 %) = 4.0 %1
- absolute uncertainty = 2.25 × 4.0 / 100 = 0.09 s21
State how the student should draw the line of best fit, the line of maximum gradient and the line of minimum gradient on the graph.
Mark scheme
- the line of best fit should pass through all error bars1
- maximum gradient line from the bottom of the first error bar to the top of the last1
- minimum gradient line from the top of the first error bar to the bottom of the last1
The maximum gradient is 5.86 s2 m−1 and the minimum gradient is 5.10 s2 m−1.
Determine the best value of the gradient and its percentage uncertainty.
Mark scheme
- best gradient = (5.86 + 5.10) / 2 = 5.48 s2 m−11
- uncertainty in gradient = (5.86 − 5.10) / 2 = 0.381
- percentage uncertainty of gradient = 0.38 / 5.48 × 100 = 6.9 %1
The acceleration a of the marble is given by
a = 2 / gradient
State the percentage uncertainty in a. Give a reason for your answer.
Mark scheme
- 6.9 %1
- for a reciprocal the percentage uncertainty is the same as that in the gradient1
Another student measures every value of d from the wrong mark on the ramp, so each value of d is 2.0 cm too large.
Explain the effect of this on the gradient of the graph.
Mark scheme
- the gradient is not affected by zero error: the systematic error only affects the intercept1
- because the same error affects every measurement1
Question 05
6 marksA student is given a bag of identical glass marbles, each with a diameter of about 1.6 cm. The student has a micrometer screw gauge, vernier calipers and a top-pan balance with a resolution of 0.01 g.
Describe how the student should determine the density of the glass and the percentage uncertainty in the density.
Your answer should include how the student reduces the uncertainties in the measurements.
Mark scheme
| Level | Marks | Description |
|---|---|---|
| 3 | 5-6 | A coherent method covering measurement of the diameter and the mass, reduction of uncertainties in both, and the calculation of the density and its percentage uncertainty, including the factor of 3 for the diameter. |
| 2 | 3-4 | A method covering most of measurement, reduction of uncertainties and calculation, with some omissions; the density is calculated, but the combination of uncertainties may be incomplete. |
| 1 | 1-2 | Some relevant measurements or techniques are given, but the method is incomplete and the density or its uncertainty is not clearly determined. |
| 0 | No relevant content |
Indicative content
- Diameter: use a micrometer screw gauge; the resolution of the micrometer is better than the calipers, so it leads to a smaller percentage uncertainty in the diameter
- close the micrometer with nothing between the jaws and subtract any zero error from every reading
- measure the diameter in more than one direction and on more than one marble
- repeats and averages: calculate the mean diameter after removing anomalies
- uncertainty in the diameter = half the range
- Mass: measure the mass of a large number of marbles together and divide by the number of marbles
- the absolute uncertainty in the balance reading is unchanged and the mass is larger, so the percentage uncertainty is reduced
- Calculation: volume of one marble V = 4/3 π r3 with r = d / 2
- density = mass / volume
- percentage uncertainty in V = 3 × percentage uncertainty in d
- percentage uncertainty in density = percentage uncertainty in mass + 3 × percentage uncertainty in d