Edexcel International A Level Physics · Unit 3 · Practical Skills in Physics I
Measurements, uncertainties and graphs (AS)
28 free, high-quality past-paper-style questions on Measurements, uncertainties and graphs (AS) for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 174 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U3-2-B-01Unit 3 · ASEasy5 marks
A student releases a marble from rest at the top of a grooved ramp and uses a digital stopwatch, reading to 0.01 s, to time how long it takes to reach the bottom. Her four timings are:
t / s
1.62
1.58
1.71
1.60
(a)
Calculate the mean time.
(1)
(b)
Calculate the uncertainty in the mean time.
(1)
(c)
Calculate the percentage uncertainty in the mean time.
(1)
(d)
Explain why the uncertainty in the time is not ± 0.005 s, even though the stopwatch reads to 0.01 s.
the student's reaction time in starting and stopping the stopwatch varies from one timing to the next
✓ 1
this spread (about 0.1 s) is much larger than the resolution, so half the range of the repeats is the realistic uncertainty
✓ 1
Exam tip: For hand timings, the resolution of the stopwatch is almost never the limiting factor. Use the spread of repeated readings to estimate the uncertainty.
A student measures the current in a fixed resistor for different potential differences. The voltmeter reads to 0.01 V and the milliammeter to 0.1 mA. His table of results is:
V / V
I / mA
0.50
21.3
1.00
42.4
1.5
63.9
2.00
85.1
2.50
106.2
(a)
Identify the value that has been recorded incorrectly and explain your answer.
(2)
(b)
Calculate the resistance of the resistor using the readings for 2.00 V. Give your answer to an appropriate number of significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the potential difference “1.5”
✓ 1
all readings from the voltmeter should be given to two decimal places, matching its resolution of 0.01 V, so it should be written 1.50 V
✓ 1
Part (b)
R = V/I = 2.00 ÷ 0.0851 (current converted to amperes)
✓ 1
= 23.5 Ω (3 s.f., as the data are 3 s.f.)
✓ 1
Common error: Dividing by the current in milliamperes gives 0.0235, a resistance in kΩ. Convert mA to A before substituting into R = V/I.
To find the density of cooking oil, a student places an empty measuring cylinder on a digital balance and zeroes (tares) the balance. She pours in some oil and reads both instruments. Balance: reads to 0.1 g; reading 33.1 g. Measuring cylinder: scale divisions of 2 cm3; reading 36 cm3.
(a)
State the uncertainty in the mass reading.
(1)
(b)
Calculate the percentage uncertainty in the mass and in the volume.
(2)
(c)
Calculate the density of the oil. Give your answer to an appropriate number of significant figures.
(2)
(d)
Suggest one change to the method that would reduce the percentage uncertainty in the density.
= 0.92 g cm−3 (2 s.f., because the volume has 2 s.f.)
✓ 1
Part (d)
reduce the uncertainty in the volume, which is much the larger: use a narrower measuring cylinder with smaller scale divisions / use a larger volume of oil
✓ 1
Exam tip: Improve the measurement with the largest percentage uncertainty. Making the balance more precise here would change almost nothing.
The current in a small electric motor is measured with two different meters. Meter P: an analogue ammeter with scale divisions of 0.02 A; reading 0.46 A. Meter Q: a digital ammeter; reading 0.463 A.
(a)
State the uncertainty in the reading of meter P and calculate its percentage uncertainty.
(2)
(b)
State the uncertainty in the reading of meter Q and calculate its percentage uncertainty.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
half the resolution = ± 0.01 A
✓ 1
0.01 ÷ 0.46 × 100 = 2.2%
✓ 1
Part (b)
resolution 0.001 A, so ± 0.0005 A
✓ 1
0.0005 ÷ 0.463 × 100 = 0.11%
✓ 1
Exam tip: The resolution of a digital meter is one unit in its last displayed digit; the uncertainty is half of that.
A student measures the temperature of some warm water with a liquid-in-glass thermometer. Figure 1 shows part of the thermometer scale and three possible positions, A, B and C, for the student's eye.
(a)
State which eye position the student should use, and explain why.
(1)
(b)
Record the temperature shown, with its uncertainty.
(2)
(c)
Calculate the percentage uncertainty in the temperature.
(1)
(d)
Another student records the reading as 46.00 °C. Explain why this is not an appropriate way to record it.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
B, level with the top of the liquid column, so that there is no parallax error in reading the scale
✓ 1
Part (b)
46 °C
✓ 1
± 0.5 °C (half of the 1 °C division)
✓ 1
Part (c)
0.5 ÷ 46 × 100 = 1.1%
✓ 1
Part (d)
46.00 claims a resolution of 0.01 °C, but the scale can only be read to about 1 °C (half a division), so the extra zeros are not justified
✓ 1
Exam tip: Read a scale with your eye level with the point being read, and quote the uncertainty as half of one scale division.
A student releases a paper cone from rest and uses a stopwatch, reading to 0.01 s, to time how long it takes to fall through a height of 1.50 m. The graph shows the time t for each of seven trials.
(a)
Identify the anomalous reading and suggest one possible cause.
(2)
(b)
Calculate the mean time, leaving out the anomalous reading.
(1)
(c)
Calculate the uncertainty in the mean time and its percentage uncertainty.
(2)
(d)
Calculate the average speed of the cone over the 1.50 m, giving your answer to an appropriate number of significant figures.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
trial 5, 2.30 s
✓ 1
e.g. the cone tilted or drifted sideways and fell more slowly / the stopwatch was stopped late
1.50 ÷ 1.83 = 0.820 m s−1 (3 s.f., as both data are 3 s.f.)
✓ 1
Common error: Leaving the anomaly in the half-range: (2.30 − 1.78) ÷ 2 = ± 0.26 s, five times too large. Remove it before finding the mean and the spread.
A student wants to find the mass of one steel ball bearing. Figure 1(a) shows a digital balance with nothing on its pan. Figure 1(b) shows the same balance with 25 identical ball bearings on the pan.
(a)
State the zero error of the balance.
(1)
(b)
Calculate the mass of one ball bearing.
(2)
(c)
State the type of error caused by the zero error, and how the student could remove it before taking the reading.
(1)
(d)
Explain why the student weighs 25 ball bearings rather than one.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
+0.8 g (the balance reads 0.8 g too high)
✓ 1
Part (b)
corrected mass of 25 bearings = 29.3 − 0.8 = 28.5 g
✓ 1
28.5 ÷ 25 = 1.14 g
✓ 1
Part (c)
a systematic error (every reading is 0.8 g too large); press the tare / zero button with the pan empty
✓ 1
Part (d)
the ± 0.05 g uncertainty of the reading is shared over 25 bearings, so the percentage uncertainty is 25 times smaller (about 0.2% instead of about 4%)
✓ 1
Exam tip: Check the zero of any instrument before you start. A zero error is systematic: repeating the reading never removes it, but subtracting it does.
To find the thickness of a metal washer, a student stacks 10 identical washers on a bench next to an upright millimetre ruler whose zero is at the bench surface. She lays a set square across the top of the stack to read its height, as shown in the figure.
(a)
State the height of the stack.
(1)
(b)
State the uncertainty in this reading and calculate its percentage uncertainty.
(2)
(c)
Calculate the thickness of one washer and its uncertainty.
(1)
(d)
Explain why this method is better than measuring one washer with the same ruler.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
32 mm
✓ 1
Part (b)
± 0.5 mm (half the 1 mm resolution)
✓ 1
0.5 ÷ 32 × 100 = 1.6%
✓ 1
Part (c)
32 ÷ 10 = 3.20 mm ± 0.05 mm (both the reading and its uncertainty are divided by 10)
✓ 1
Part (d)
one washer (3.2 mm ± 0.5 mm) would have a percentage uncertainty of about 16% (0.5 ÷ 3.2 × 100); the stack spreads the same reading uncertainty over 10 washers, making it 10 times smaller
✓ 1
Exam tip: To measure something small, measure many of them together and divide. The absolute uncertainty is divided by the number, so the percentage uncertainty falls.
A student drops a tennis ball from different heights h0 onto a hard floor and measures the height h of the first bounce against a vertical metre rule. She repeats each bounce three times.
h0 / cm
h1 / cm
h2 / cm
h3 / cm
40.0
28.5
29.0
28.0
60.0
43.0
42.5
43.5
80.0
57.0
56.5
64.5
100.0
71.5
72.0
70.5
120.0
86.0
85.0
86.5
(a)
Identify the anomalous reading and state what the student should do about it.
(2)
(b)
Calculate the mean bounce height for h0 = 80.0 cm.
(1)
(c)
For h0 = 100.0 cm, calculate the uncertainty in the mean bounce height and its percentage uncertainty.
(2)
(d)
Suggest one improvement to the way the bounce height is measured.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
64.5 cm (for h0 = 80.0 cm)
✓ 1
repeat the bounce for that drop height / if it cannot be repeated, leave it out of the mean
✓ 1
Part (b)
(57.0 + 56.5) ÷ 2 = 56.8 cm (56.75 cm)
✓ 1
Part (c)
half the range = (72.0 − 70.5) ÷ 2 = ± 0.75 cm
✓ 1
mean = 71.3 cm; 0.75 ÷ 71.3 × 100 = 1.1%
✓ 1
Part (d)
film the bounce against the rule (slow motion / frame by frame) so that the highest point can be read, with the camera level with the top of the bounce to avoid parallax
✓ 1
Exam tip: An anomaly is a reading clearly outside the spread of its repeats. Name it, then say what you would do: repeat it, or exclude it from the mean.
A student hangs weights from a spring and measures its extension x for each applied force F.
F / N
x / mm
0.0
0.0
1.0
8.4
2.0
16.6
3.0
25.3
4.0
33.4
5.0
41.9
6.0
50.1
She plots x on the y-axis against F on the x-axis, on a grid that is 8 large squares wide and 12 large squares high.
(a)
Suggest a suitable scale for each axis. Explain your choice.
(2)
(b)
Her best-fit line passes through (0.0 N, 0.0 mm) and (6.0 N, 50.2 mm). Calculate the gradient of the line.
(1)
(c)
Use your answer to (b) to calculate the stiffness of the spring in N m−1.
(2)
(d)
Explain why the gradient should be found from a triangle that covers most of the line rather than from two neighbouring points.
(1)
(e)
State what the shape of the graph shows about the spring.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
F: 1 N per large square (6 of 8 squares used); x: 5 mm per large square (just over 10 of 12 squares used; 50.2 mm still fits)
✓ 1
both scales are easy to read (steps of 1 or 5) and the points occupy more than half the grid in each direction
✓ 1
Part (b)
50.2 ÷ 6.0 = 8.37 mm N−1
✓ 1
Part (c)
stiffness = F/x = 1 ÷ gradient, with the extension in metres: 6.0 ÷ 0.0502
✓ 1
= 120 N m−1 (119.5 N m−1)
✓ 1
Part (d)
a larger triangle makes Δx and ΔF large, so the reading errors are a smaller fraction of them / neighbouring data points carry random error that a small triangle magnifies
✓ 1
Part (e)
a straight line through the origin: the extension is proportional to the force (the spring obeys Hooke's law over this range)
✓ 1
Model sketch
Figure: model graph: 1 N and 5 mm per large square; points and best-fit line
Exam tip: Choose scales of 1, 2 or 5 units (× a power of ten) per large square, and check that the points fill more than half the grid both across and up.
A mass m hanging from a spring oscillates vertically. A student times 10 oscillations for each mass. The period T is expected to obey T = 2π√(m/k), where k is the stiffness of the spring.
m / kg
10T / s
0.100
3.98
0.200
5.61
0.300
6.90
0.400
7.93
0.500
8.90
0.600
9.72
(a)
Calculate T2 for m = 0.300 kg, to an appropriate number of significant figures.
(1)
(b)
Explain why a graph of T2 against m should be a straight line through the origin, and state what its gradient represents.
(2)
(c)
The best-fit line passes through (0.100 kg, 0.158 s2) and (0.600 kg, 0.947 s2). Calculate the gradient.
(1)
(d)
Determine the stiffness of the spring.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
T = 0.690 s, T2 = 0.476 s2
✓ 1
Part (b)
T2 = (4π2/k) m, which has the form y = mx with no constant term
A glider on a tilted air track passes a first light gate with speed u and accelerates uniformly. A second light gate, a distance s further down the track, measures the speed v. The student moves the second gate and records:
s / m
v / m s−1
0.20
0.51
0.40
0.65
0.60
0.77
0.80
0.87
1.00
0.97
1.20
1.05
(a)
Calculate v2 for s = 0.60 m.
(1)
(b)
Explain how a graph of v2 against s can be used to find the acceleration a and the speed u.
(2)
(c)
The best-fit line passes through (0.000 m, 0.086 m2 s−2) and (1.200 m, 1.103 m2 s−2). Calculate the acceleration.
(2)
(d)
Calculate u.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.772 = 0.593 m2 s−2 (allow 0.59)
✓ 1
Part (b)
v2 = u2 + 2as compared with y = mx + c: gradient = 2a, so a = gradient ÷ 2
✓ 1
intercept on the v2 axis = u2, so u = √intercept
✓ 1
Part (c)
gradient = (1.103 − 0.086) ÷ 1.200 = 0.8475 m s−2
✓ 1
a = 0.8475 ÷ 2 = 0.42 m s−2 (0.424)
✓ 1
Part (d)
u = √0.086 = 0.29 m s−1
✓ 1
Model sketch
Figure: v² against s: gradient = 2a, intercept = u²
Exam tip: Write the equation under y = mx + c and match the terms one by one; then the meaning of the gradient and the intercept is in your working.
A student shades a uniform strip on paper with a soft pencil. He presses the two probes of an ohmmeter onto the strip a distance l apart and records the resistance R.
l / cm
R / kΩ
2.0
4.6
4.0
8.1
6.0
11.9
8.0
15.2
10.0
18.9
12.0
22.4
A graph of R against l is a straight line. The best-fit line passes through (0.0 cm, 1.0 kΩ) and (12.0 cm, 22.4 kΩ).Figure: the student's results with the best-fit line
(a)
Calculate the gradient of the line.
(2)
(b)
The line does not pass through the origin. State the type of error this shows and suggest its cause.
(2)
(c)
Explain whether this error affects the value of the gradient.
(1)
(d)
Suggest one way of reducing the random error in the readings of R.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
(22.4 − 1.0) ÷ (12.0 − 0.0)
✓ 1
= 1.78 kΩ cm−1
✓ 1
Part (b)
a systematic error — the same extra resistance (about 1.0 kΩ) is added to every reading
✓ 1
resistance where the probes touch the graphite (contact resistance) / zero error of the ohmmeter
✓ 1
Part (c)
no — adding the same amount to every R moves the line up without changing its slope
✓ 1
Part (d)
press the probes with the same force each time / mark the positions for the probes and repeat each reading, using the mean
✓ 1
Exam tip: A constant systematic error changes the intercept, not the gradient. That is why a value from a gradient is often more reliable than one from a single reading.
A student measures her reaction time. A friend holds a metre rule vertically with its zero mark level with the top of her open fingers and lets go without warning; she catches it as quickly as possible. The distance d the rule falls is read at the top of her fingers. g = 9.81 N kg−1.
d / cm
14.2
17.8
15.1
16.4
15.7
(a)
Calculate the mean distance and its uncertainty.
(2)
(b)
Use d = ½gt2 to calculate her mean reaction time.
(2)
(c)
In a second set of trials the friend holds the rule with its 1.0 cm mark, instead of its zero mark, level with the top of her fingers, and the readings are not corrected. State the type of error this introduces.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
mean = 79.2 ÷ 5 = 15.8 cm (15.84 cm)
✓ 1
uncertainty = (17.8 − 14.2) ÷ 2 = ± 1.8 cm
✓ 1
Part (b)
t = √(2d/g) = √(2 × 0.1584 ÷ 9.81) (distance in metres)
✓ 1
= 0.18 s (0.180 s)
✓ 1
Part (c)
systematic error — every distance read is 1.0 cm too large
✓ 1
Common error: Leaving d in centimetres in t = √(2d/g) gives a time ten times too large. Convert to metres to match g in N kg−1 (m s−2).
A pressure sensor is lowered into a tall cylinder of salt solution. The graph shows the total pressure p at depth h below the surface, with a best-fit line. The pressure is expected to obey p = p0 + ρgh, where p0 is the atmospheric pressure and ρ is the density of the solution. g = 9.81 N kg−1.
(a)
Explain how the graph can be used to find p0 and ρ.
(2)
(b)
State the value of p0.
(1)
(c)
Determine the gradient of the line.
(2)
(d)
Calculate the density of the salt solution.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
compared with y = mx + c, the intercept on the p axis (at h = 0) is p0
✓ 1
the gradient is ρg, so ρ = gradient ÷ g
✓ 1
Part (b)
101.0 kPa (allow 100.9–101.1 kPa)
✓ 1
Part (c)
large triangle, e.g. from (0, 101.0) to (0.50, 107.0): (107.0 − 101.0) ÷ 0.50
A ball is released from rest at the top of a long straight ramp. A student uses a stopwatch to measure the time t taken for the ball to roll distances d from 0.40 m to 1.40 m down the ramp. The ball has a constant acceleration a, so d = ½at2. The graph shows d plotted against t2, with a best-fit line.
(a)
Explain why the graph is expected to be a straight line through the origin, and state what its gradient represents.
(2)
(b)
For d = 0.80 m the time recorded was 1.58 s. Calculate t2 to an appropriate number of significant figures.
(1)
(c)
Determine the gradient of the line.
(1)
(d)
Calculate the acceleration of the ball.
(1)
(e)
Explain why the uncertainty in the timing matters most for the shortest distances, and suggest an improvement.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
d = (a/2) t2 has the form y = mx with no constant term
✓ 1
gradient = a/2
✓ 1
Part (b)
1.582 = 2.50 s2 (2.4964)
✓ 1
Part (c)
e.g. 1.60 ÷ 5.0 = 0.32 m s−2 (allow 0.31–0.33)
✓ 1
Part (d)
a = 2 × 0.32 = 0.64 m s−2 (allow 0.62–0.66)
✓ 1
Part (e)
the reaction-time uncertainty is about the same for every timing, so it is a larger percentage of the shorter times
✓ 1
use light gates connected to a timer / data logger to remove reaction time
✓ 1
Exam tip: Processed values such as t2 keep the significant figures of the raw data (here 3 s.f.) and get their own unit, s2.
A card is fixed upright on a trolley. Figure 1(a) shows the card held against a millimetre ruler, with its left edge exactly at the zero mark. The trolley then passes through a light gate, and the timer in Figure 1(b) shows the time for which the card blocks the beam.
(a)
State the width of the card and its uncertainty.
(2)
(b)
Calculate the speed of the trolley. Give your answer to an appropriate number of significant figures.
(2)
(c)
Calculate the percentage uncertainty in the width and in the time.
(2)
(d)
Suggest one improvement to the measurements, giving a reason.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
48 mm
✓ 1
± 0.5 mm (half the 1 mm resolution)
✓ 1
Part (b)
v = 0.048 m ÷ 0.0627 s (width in metres)
✓ 1
= 0.77 m s−1 (0.766; 2 s.f., as the width has 2 s.f.)
Water in a beaker is heated steadily. Every 2 minutes a student reads its temperature θ on two thermometers held side by side in the water: A, a calibrated digital thermometer, and B, a liquid-in-glass thermometer. The graph shows both sets of readings, each with a best-fit line.
(a)
State the type of error in the readings of thermometer B and explain how the graph shows this.
(2)
(b)
State the size of the error in thermometer B.
(1)
(c)
Use thermometer A's line to determine the rate at which the temperature rises.
(2)
(d)
Explain why thermometer B gives the same rate of rise as thermometer A.
(1)
(e)
Suggest how the student could have detected this error before the experiment.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
a systematic error
✓ 1
B's line is parallel to A's but higher by the same amount at every time: every reading is shifted by the same amount
✓ 1
Part (b)
B reads 2.0 °C too high (allow 1.8–2.2 °C)
✓ 1
Part (c)
large triangle, e.g. (35.0 − 20.0) ÷ (10 − 0)
✓ 1
= 1.5 °C min−1 (allow 1.4–1.6)
✓ 1
Part (d)
the constant 2.0 °C offset cancels when a temperature change is found, so the gradient is unchanged
✓ 1
Part (e)
place B in melting ice (it should read 0 °C) / compare B with A at room temperature, then subtract the difference from every reading
✓ 1
Exam tip: On a graph, random error scatters points about the line; a systematic error moves the whole line. A shift with the same gradient means a constant offset.
Two students investigate how the current I in a 6 V filament lamp depends on the potential difference V across it. Student X takes four readings between 0.20 V and 0.80 V and concludes that the lamp has a constant resistance. Student Y takes seven readings from 0.5 V to 6.0 V. The graph shows both sets of results.
(a)
Comment on the number and range of student X's readings.
(2)
(b)
Use student Y's results to calculate the resistance of the lamp at 2.0 V and at 6.0 V.
(2)
(c)
Explain why student X's conclusion is not valid.
(1)
(d)
Suggest the range of V in which student Y should take extra readings, and why.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
only four readings: at least six values are needed to establish a trend
✓ 1
0.20 V to 0.80 V is a narrow range at the bottom of the lamp's 0 to 6 V range; readings should cover the whole range
✓ 1
Part (b)
2.0 ÷ 0.200 = 10 Ω
✓ 1
6.0 ÷ 0.300 = 20 Ω
✓ 1
Part (c)
over the full range the graph curves and V/I doubles from 10 Ω to 20 Ω; X's narrow, low range only shows the nearly straight start of the curve
✓ 1
Part (d)
between 0 V and about 3 V (e.g. at 0.25, 1.5 and 2.5 V), where the gradient of the graph changes most quickly and Y's readings are widely spaced, so the shape of the curve is well defined
✓ 1
Exam tip: A narrow range can make a curve look straight. Spread at least six readings over the full range the apparatus allows before drawing a conclusion.
A student measures the current I in a resistor for several potential differences V. After taking her readings she notices that the analogue ammeter reads 0.04 A when the switch is open.
V / V
I / A (as read)
1.0
0.13
2.0
0.22
3.0
0.31
4.0
0.40
5.0
0.50
6.0
0.59
(a)
State the type of error and describe its effect on a graph of I against V.
(2)
(b)
A friend calculates the resistance as 6.0 ÷ 0.59 = 10.2 Ω. Calculate the resistance from the same row after correcting for the zero error.
(2)
(c)
The best-fit line through the uncorrected points passes through (1.0 V, 0.128 A) and (6.0 V, 0.589 A). Determine the resistance from the gradient.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
a systematic error (a zero error)
✓ 1
every point is 0.04 A too high, so the line is shifted up with an intercept of about 0.04 A, but its gradient is unchanged
A wire is stretched between two bridges with a fixed tension. A student changes the vibrating length l and finds the frequency f of the fundamental mode for each length. Theory gives f = v/2l, where v is the speed of the transverse waves on the wire.
l / m
f / Hz
0.300
107
0.400
79
0.500
65
0.600
53
0.700
46
0.800
40
(a)
Explain why a graph of f against 1/l is plotted, and state its expected gradient.
(2)
(b)
Calculate 1/l for l = 0.700 m. Give your answer to an appropriate number of significant figures and with its unit.
(1)
(c)
The best-fit line passes through the origin and through (3.50 m−1, 112 Hz). Determine the speed of the waves on the wire.
(3)
(d)
State what the graph passing through the origin shows.
(1)
(e)
Suggest why the range of lengths used is a good choice.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
f = (v/2) × (1/l) has the form y = mx, so the graph should be a straight line through the origin
✓ 1
gradient = v/2
✓ 1
Part (b)
1.43 m−1
✓ 1
Part (c)
gradient = 112 ÷ 3.50 = 32.0 Hz m (or m s−1)
✓ 1
v = 2 × gradient
✓ 1
= 64.0 m s−1
✓ 1
Part (d)
the frequency is proportional to 1/l (inversely proportional to l) — the data support f = v/2l with no systematic offset
✓ 1
Part (e)
six values spread from 0.300 m to 0.800 m (nearly a factor of 3), giving 1/l from 1.25 m−1 to 3.33 m−1, so the trend is clear and the gradient is well defined
✓ 1
Model sketch
Figure: f against 1/l: straight line through the origin, gradient v/2
Exam tip: Processed values such as 1/l need a unit (m−1) and the same number of significant figures as the raw data. Write them in an extra column.
A student investigates how the time taken for a toy car to roll down a ramp depends on the angle of the ramp. He releases the car by hand and times it with a stopwatch over the full 0.80 m length of the ramp. His results are:
angle
time
10°
0.81 s
12
0.7
14°
0.68 s
(a)
Evaluate the student's method and the way he has recorded his results. Suggest improvements.
(6)
Show mark scheme
Answer
Mark
Notes
Indicative content:
—
only three angles — take at least six different values
✓ 1
the range 10° to 14° is too narrow — extend it, e.g. 5° to 40°
✓ 1
no repeats — repeat each timing (3 or more times), calculate a mean and check for anomalies
✓ 1
times of under a second are dominated by reaction time (± about 0.1 s, over 10% of the reading) — use light gates and a data logger, or a longer ramp
✓ 1
the table needs units in the column headings (angle / °, t / s) and the times recorded to a consistent number of decimal places (0.7 should be e.g. 0.70)
✓ 1
release the car from the same marked point each time without pushing it, so the starting speed is always zero
✓ 1
Exam tip: Make each improvement specific and give its reason: “use light gates to remove reaction time” earns credit; “use better equipment” does not.
A hacksaw blade is clamped horizontally to the edge of a bench, with a length L overhanging. A small mass is fixed to its free end, which is pulled down and released. A student times 5 oscillations for each overhang length, using a stopwatch. She estimates that each timing has an uncertainty of ± 0.2 s due to her reaction time.
L / m
5T / s
0.20
1.9
0.22
2.2
0.24
2.5
0.26
2.8
0.28
3.2
Figure: the clamped hacksaw blade with overhang L; not to scale
(a)
Calculate the period for L = 0.20 m and its percentage uncertainty.
(2)
(b)
Calculate the smallest number of oscillations she should time, with the same timing uncertainty, to make the percentage uncertainty in T less than 2% for L = 0.20 m.
(2)
(c)
Comment on the number and range of overhang lengths she used.
(2)
(d)
Suggest one technique that would help her count the oscillations accurately.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
T = 1.9 ÷ 5 = 0.38 s
✓ 1
0.2 ÷ 1.9 × 100 = 11% (the same percentage as for 5T)
✓ 1
Part (b)
the total time must exceed 0.2 ÷ 0.02 = 10 s
✓ 1
10 ÷ 0.38 = 26.3, so at least 27 oscillations (allow “about 30”)
✓ 1
Part (c)
only five lengths — at least six are needed for a reliable graph
✓ 1
range 0.20 m to 0.28 m is narrow — extend it, e.g. 0.10 m to 0.40 m, as far as the blade allows
✓ 1
Part (d)
use a fiducial mark (e.g. a pointer) at the equilibrium position and start and stop timing as the blade passes it in the same direction
✓ 1
Exam tip: Timing N oscillations divides the percentage uncertainty in T by N, because the reaction-time uncertainty is shared over all the cycles.
A vibrating tuning fork of frequency f is held over the open end of a tube that is closed at the bottom. For each of six tuning forks, from 256 Hz to 512 Hz, a student adjusts the length L of the air column until the first resonance is heard. Theory gives L + e = v/4f, where v is the speed of sound and e is a constant called the end correction. The graph shows L against 1/f, with a best-fit line.
(a)
Explain why a graph of L against 1/f should be a straight line, and state its gradient and its intercept on the L axis.
(2)
(b)
Calculate 1/f for the 384 Hz fork. Give your answer to an appropriate number of significant figures, with its unit.
(1)
(c)
Use the graph to determine the speed of sound.
(2)
(d)
Determine the end correction e.
(1)
(e)
Explain why the value of v found from the gradient is not affected by the end correction.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
L = (v/4)(1/f) − e has the form y = mx + c, so the graph is a straight line with gradient v/4
✓ 1
intercept = −e
✓ 1
Part (b)
1 ÷ 384 = 2.60 × 10−3 s
✓ 1
Part (c)
gradient from a large triangle, e.g. (0.330 − (−0.010)) ÷ (4.0 × 10−3) = 85 m s−1 (allow 84–86)
✓ 1
v = 4 × gradient = 340 m s−1 (allow 336–344)
✓ 1
Part (d)
intercept = −0.010 m, so e = 0.010 m (allow 0.008–0.012 m)
✓ 1
Part (e)
e shortens every measured length by the same amount (a systematic offset), which moves the line down without changing its gradient
✓ 1
Exam tip: A line that crosses an axis below the origin is telling you something: here the intercept measures the end correction, while the gradient is free of it.
The resistance R of a coil of copper wire is measured at temperatures θ from 25 °C to 85 °C, with the coil in a water bath. The resistance is expected to obey R = R0(1 + αθ), where R0 is the resistance at 0 °C and α is a constant. The graph shows the results with a best-fit line. Look carefully at the scales on both axes.
(a)
Explain why R0 is not the value at which the line meets the vertical axis of this graph.
(1)
(b)
Determine the gradient of the line.
(2)
(c)
Calculate R0.
(2)
(d)
Calculate α.
(1)
(e)
Suggest one precaution to make sure the temperature of the coil is the same as the temperature read on the thermometer.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the θ axis starts at 20 °C, not 0 °C (a false origin), so the line meets the R axis at θ = 20 °C, where R = 13.5 Ω
✓ 1
Part (b)
large triangle, e.g. (17.0 − 13.5) ÷ (90 − 20)
✓ 1
= 0.050 Ω °C−1 (allow 0.048–0.052)
✓ 1
Part (c)
use a point on the line in R = R0 + (gradient)θ: R0 = 13.5 − 0.050 × 20
stir the water bath and wait for the reading to become steady before measuring R / place the thermometer next to the coil
✓ 1
Common error: Reading the “intercept” where the line meets a vertical axis drawn at 20 °C. With a false origin, calculate the intercept from c = y − mx using a point on the line.
A cell of e.m.f. ε and internal resistance r is connected in series with a resistance box and an ammeter of negligible resistance (Figure 1(a)). A student records the current I for six values of the resistance R, then plots 1/I against R (Figure 1(b)).
(a)
Show that 1/I = R/ε + r/ε.
(2)
(b)
For R = 6.0 Ω the current was 0.228 A. Calculate 1/I, with its unit, to an appropriate number of significant figures.
(1)
(c)
Use the graph to determine ε.
(2)
(d)
Use the graph to determine r.
(2)
(e)
Explain why the student should open the switch between readings.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
ε = I(R + r)
✓ 1
dividing both sides by εI: 1/I = (R + r)/ε = R/ε + r/ε
✓ 1
Part (b)
1 ÷ 0.228 = 4.39 A−1
✓ 1
Part (c)
gradient from a large triangle, e.g. (8.40 − 0.40) ÷ 12 = 0.667 A−1 Ω−1 (allow 0.65–0.68)
✓ 1
ε = 1 ÷ gradient = 1.50 V (allow 1.47–1.54 V)
✓ 1
Part (d)
intercept = 0.40 A−1 = r/ε (allow 0.38–0.42)
✓ 1
r = 0.40 × 1.50 = 0.60 Ω (allow 0.56–0.64 Ω)
✓ 1
Part (e)
so the cell does not warm up or run down: ε and r must stay constant for the graph to be a straight line
✓ 1
Exam tip: Write the rearranged equation directly under y = mx + c. Here y = 1/I, x = R, m = 1/ε and c = r/ε.
Steel ball bearings of radius r are dropped one at a time into a tall tube of glycerol, and the terminal speed v of each is found by timing it between two marks on the tube. Stokes' law predicts v = 2(ρs − ρg)gr2/9η, where η is the viscosity of glycerol, ρs = 7850 kg m−3 (steel) and ρg = 1260 kg m−3 (glycerol). g = 9.81 N kg−1. The graph shows v against r2 with a best-fit line.
(a)
Explain why the graph is expected to be a straight line through the origin, and state its gradient.
(2)
(b)
Identify the anomalous result and suggest a cause for it.
(2)
(c)
Determine the gradient of the line in SI units.
(2)
(d)
Calculate the viscosity of glycerol.
(1)
(e)
Describe how the student could check that each ball has reached its terminal speed before the timing starts.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
v = [2(ρs − ρg)g/9η] r2 has the form y = mx, everything in the bracket being constant
✓ 1
gradient = 2(ρs − ρg)g/9η
✓ 1
Part (b)
the point at r2 = 1.00 mm2, v = 7.0 mm s−1, well below the line
✓ 1
e.g. timing started before the ball reached terminal speed / the ball fell close to the wall of the tube, which slowed it
✓ 1
Part (c)
e.g. 35.0 ÷ 3.5 = 10.0 mm s−1 per mm2 (allow 9.8–10.2)
✓ 1
= 10.0 × 10−3 ÷ 10−6 = 1.00 × 104 m−1 s−1
✓ 1
Part (d)
η = 2 × 6590 × 9.81 ÷ (9 × 1.00 × 104) = 1.44 Pa s (allow 1.41–1.47)
✓ 1
Part (e)
time the ball over two equal distances, one below the other: equal times show the speed is constant
✓ 1
Common error: Using the gradient in mm s−1 per mm2 without converting: the factor is 10−3 ÷ 10−6 = 103, not 10−3.
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