Edexcel International A Level Physics · Unit 3 · Practical Skills in Physics I
The AS core practicals
28 free, high-quality past-paper-style questions on The AS core practicals 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 · 200 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U3-3-B-02Unit 3 · ASEasy6 marks
A student finds the terminal velocity of a small steel ball falling through engine oil in a tall measuring cylinder. Three rubber bands are placed round the cylinder: the first is 8 cm below the surface of the oil and the other two are each 0.150 m below the one above.
Figure: positions of the three bands on the measuring cylinder; not to scale
(a)
Explain why the first band is placed well below the surface of the oil.
(1)
(b)
Describe how the student can use the three bands to check that the ball is moving at terminal velocity.
(2)
(c)
The times for the ball to fall between the first and second bands in three trials are 4.21 s, 4.35 s and 4.28 s. Calculate the mean time and its uncertainty.
(2)
(d)
Calculate the terminal velocity of the ball.
(1)
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Answer
Mark
Notes
Part (a)
the ball is still accelerating near the surface, so timing starts only once it has reached terminal velocity
✓ 1
Part (b)
time the ball between the first and second bands and between the second and third bands
✓ 1
the two times are equal (within their uncertainty) if the velocity is constant
✓ 1
Part (c)
mean = 4.28 s
✓ 1
uncertainty = half the range = (4.35 − 4.21) ÷ 2 = ±0.07 s
✓ 1
Part (d)
0.150 ÷ 4.28 = 0.0350 m s−1
✓ 1
Common error: Quoting the mean to more decimal places than the readings. Times read to 0.01 s give a mean to 0.01 s.
Before measuring the Young modulus of copper, a student measures the diameter of a copper wire with a micrometer screw gauge at five different points and orientations. The readings, in mm, are 0.37, 0.38, 0.38, 0.39 and 0.37.
(a)
Explain why the diameter is measured at several points and in different orientations.
(1)
(b)
Calculate the mean diameter and its uncertainty.
(2)
(c)
Calculate the cross-sectional area of the wire in m2.
(2)
(d)
Calculate the percentage uncertainty in the diameter.
(1)
(e)
Explain why a long wire is used in this experiment.
(1)
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Answer
Mark
Notes
Part (a)
the wire may not be uniform along its length or exactly circular in cross-section, so averaging gives a representative value
✓ 1
Part (b)
mean = 0.378 mm, quoted as 0.38 mm
✓ 1
uncertainty = half the range = (0.39 − 0.37) ÷ 2 = ±0.01 mm
✓ 1
Part (c)
A = πd2/4 = π × (0.378 × 10−3)2 ÷ 4
✓ 1
= 1.1 × 10−7 m2 (1.12 × 10−7)
✓ 1
Part (d)
0.01 ÷ 0.378 × 100 = 2.6%
✓ 1
Part (e)
a longer wire gives a larger extension for the same stress, so the percentage uncertainty in the extension is smaller
✓ 1
Common error: Using the diameter in place of the radius in πr2, which makes the area four times too large. Either halve d first or use πd2/4.
A loudspeaker emits a steady note. Two microphones are connected to a two-beam oscilloscope with its timebase set to 0.2 ms per division. One complete cycle of each trace occupies 2.5 divisions.
Figure: the two traces; timebase 0.2 ms per division (one square = one division)
(a)
Calculate the period of the sound.
(1)
(b)
Calculate the frequency of the sound.
(1)
(c)
One microphone is moved directly away from the loudspeaker. The traces come back into phase 10 times while it moves 1.72 m. Calculate the wavelength.
(1)
(d)
Calculate the speed of sound.
(1)
(e)
The distance of 1.72 m has an uncertainty of ±0.01 m. Calculate its percentage uncertainty.
(1)
(f)
Explain why the student measures the distance for 10 wavelengths rather than for one.
(1)
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Answer
Mark
Notes
Part (a)
2.5 × 0.2 ms = 0.50 ms
✓ 1
Part (b)
1 ÷ 0.50 × 10−3 = 2.0 kHz (2000 Hz)
✓ 1
Part (c)
1.72 ÷ 10 = 0.172 m
✓ 1
Part (d)
2000 × 0.172 = 344 m s−1 (accept 340 m s−1)
✓ 1
Part (e)
0.01 ÷ 1.72 × 100 = 0.58%
✓ 1
Part (f)
the uncertainty in judging the in-phase positions is the same, but the distance is 10 times larger, so the percentage uncertainty is 10 times smaller
✓ 1
Common error: Reading the time for one division as the period. Count the divisions for one whole cycle, then multiply by the timebase setting.
A student determines the resistivity of nichrome. A nichrome wire of diameter 0.300 mm is connected in a circuit with a sliding contact, and the resistance R of different lengths l is found from voltmeter and ammeter readings. A graph of R (y-axis) against l (x-axis) has a gradient of 15.6 Ω m−1 and an intercept of 0.20 Ω on the R-axis.
Figure: the student's graph of R against l
(a)
Explain why a graph of R against l is expected to be a straight line.
(2)
(b)
Calculate the resistivity of nichrome.
(3)
(c)
Suggest a cause of the intercept and explain why it does not affect the value of resistivity.
(2)
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Answer
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Notes
Part (a)
R = ρl/A
✓ 1
ρ and A are constant, so R is proportional to l with gradient ρ/A
✓ 1
Part (b)
A = π × (0.150 × 10−3)2 = 7.07 × 10−8 m2
✓ 1
ρ = gradient × A = 15.6 × 7.07 × 10−8
✓ 1
= 1.10 × 10−6 Ω m
✓ 1
Part (c)
resistance of the contacts / connecting leads, which is in series with the wire
✓ 1
it adds the same amount to every reading, which moves the line up without changing its gradient
✓ 1
Common error: Using the diameter as the radius. The area of a wire is πd2/4; with d = 0.300 mm the radius is 0.150 mm.
A student finds the acceleration of free fall with a steel ball held by an electromagnet. Switching off the electromagnet releases the ball and starts an electronic timer; the ball stops the timer when it hits a trapdoor a height h below. The graph shows the student's values of h plotted against t2, where t is the fall time, with the best-fit line drawn.
(a)
Explain why the graph is expected to be a straight line through the origin, and state how g is related to the gradient.
(2)
(b)
Use the labelled point on the line to determine g.
(2)
(c)
Identify the anomalous result and suggest a cause that would produce a point on that side of the line.
(2)
(d)
State one safety precaution for this experiment.
(1)
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Answer
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Notes
Part (a)
the ball starts from rest, so s = ut + ½at2 gives h = ½gt2, so h ∝ t2
✓ 1
gradient = g/2, so g = 2 × gradient
✓ 1
Part (b)
gradient = 0.980 ÷ 0.200 = 4.90 m s−2
✓ 1
g = 2 × 4.90 = 9.80 m s−2 (allow 9.7–9.9)
✓ 1
Part (c)
the point for h = 0.60 m
✓ 1
its time is too long, e.g. the ball was released late because the electromagnet current was too large / residual magnetism, or the trapdoor was slow to open
✓ 1
Part (d)
place a box of sand or a soft mat under the trapdoor to catch the ball / keep hands and feet clear of the falling ball / switch off the electromagnet when not in use as it gets hot
✓ 1
Exam tip: When a labelled point lies on the best-fit line, use it for the gradient rather than a plotted cross: the line, not any single result, represents all the data.
A student investigates how the viscosity of an oil depends on temperature. The tube of oil stands in a water bath, and the same steel ball is used each time. The graph shows the terminal speed v of the ball at five temperatures θ.
(a)
Use the graph to find the terminal speed at 40 °C.
(1)
(b)
Describe how the terminal speed changes with temperature, and state what this shows about the viscosity of the oil.
(2)
(c)
For the same ball, the viscous drag at terminal speed is the same at every temperature, so the viscosity is inversely proportional to the terminal speed. Assuming the density of the oil does not change, calculate the ratio viscosity at 20 °C ÷ viscosity at 60 °C.
(1)
(d)
Suggest why the student waits several minutes after changing the water bath temperature before releasing the ball.
(1)
(e)
State one safety precaution when using a hot water bath.
(1)
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Answer
Mark
Notes
Part (a)
24 mm s−1 (allow 23–25 mm s−1)
✓ 1
Part (b)
the terminal speed increases with temperature, and increases more rapidly at higher temperatures / it doubles for each 20 °C rise
✓ 1
the viscosity of the oil decreases as its temperature increases
✓ 1
Part (c)
η ∝ 1/v, so ratio = 48 ÷ 12 = 4.0
✓ 1
Part (d)
so that all the oil reaches the temperature of the water bath (thermal equilibrium) and the temperature is uniform throughout the tube
✓ 1
Part (e)
keep the water below about 60 °C / handle the tube with tongs or heat-resistant gloves / keep the mains-powered heater away from spilt water
✓ 1
Common error: Thinking that a faster ball means the liquid is more viscous. A higher terminal speed means less drag at a given speed, so the viscosity is lower.
A loudspeaker connected to a signal generator emits a steady note. Two microphones, A and B, are placed in a straight line with the loudspeaker; B is 34 mm further from the loudspeaker than A, which is less than one wavelength. The microphones are connected to the two channels of an oscilloscope, and the screen is shown in the figure.
(a)
Use the figure to determine the period of the sound.
(1)
(b)
Calculate the frequency of the sound.
(1)
(c)
Determine the phase difference between the two traces, in degrees.
(1)
(d)
Use your answers to calculate the wavelength and the speed of sound.
(2)
(e)
Suggest a better way of using the two microphones to find the wavelength.
(1)
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Answer
Mark
Notes
Part (a)
one cycle = 4.0 divisions, so T = 4.0 × 0.1 ms = 0.40 ms
✓ 1
Part (b)
f = 1 ÷ 0.40 × 10−3 = 2500 Hz
✓ 1
Part (c)
B is 1.0 division behind A = ¼ cycle, so phase difference = 90°
✓ 1
Part (d)
34 mm is a quarter of a wavelength, so λ = 4 × 0.034 = 0.136 m
✓ 1
v = fλ = 2500 × 0.136 = 340 m s−1
✓ 1
Part (e)
start with the traces in phase, move B away until they have come back into phase many times (e.g. 10), measure the distance moved and divide by that number
✓ 1
Exam tip: Count divisions for a whole cycle, peak to peak, then multiply by the timebase. A shift of a quarter of that cycle between the traces is a phase difference of 90°.
A student sets up the circuit shown to find the resistivity of constantan. The wire has a diameter of 0.315 mm.
(a)
Identify the mistake in the circuit and explain why it would prevent the student from obtaining useful readings.
(2)
(b)
Use the figure to state the length of wire between A and B.
(1)
(c)
With the circuit corrected, the voltmeter reads 1.52 V and the ammeter reads 0.400 A. Calculate the resistance of this length of wire.
(1)
(d)
Calculate the resistivity of constantan.
(2)
(e)
Suggest why the switch should be opened between readings.
(1)
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Answer
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Notes
Part (a)
the ammeter and voltmeter have been swapped: the voltmeter is in series and the ammeter is in parallel with AB
✓ 1
the very high resistance of the voltmeter makes the current almost zero (and the ammeter would short-circuit the wire), so neither meter measures the current in or p.d. across AB
a current left on heats the wire, which increases its resistance
✓ 1
Common error: Measuring the wire length from the end of the metre rule rather than from contact A. Here A is at the 0 cm mark, so l equals the reading at B.
A student uses a single light gate to determine the acceleration of free fall. A steel ball is released from rest a height h above the light beam. The ball's diameter d is measured as 19.0 mm with vernier calipers. A timer records the time Δt for which the ball blocks the beam, and the speed of the ball is taken as v = d/Δt. The student repeats this for several values of h.
Figure: the ball is released from rest a height h above the light beam; not to scale
(a)
The calipers have a resolution of 0.1 mm. Taking the uncertainty as half the resolution, state the uncertainty in d and calculate its percentage uncertainty.
(2)
(b)
For h = 0.600 m, Δt = 5.56 ms. Calculate v.
(1)
(c)
Explain why a graph of v2 against h is expected to be a straight line through the origin, and state how g is found from it.
(2)
(d)
The gradient of the student's line is 19.4 m s−2. Calculate g.
(1)
(e)
Suggest why the light gate must be lined up so that the beam passes through the centre of the ball.
(1)
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Answer
Mark
Notes
Part (a)
half the resolution of the calipers = ±0.05 mm
✓ 1
0.05 ÷ 19.0 × 100 = 0.26%
✓ 1
Part (b)
19.0 × 10−3 ÷ 5.56 × 10−3 = 3.42 m s−1
✓ 1
Part (c)
from rest, v2 = u2 + 2as gives v2 = 2gh, so v2 ∝ h
✓ 1
g = gradient ÷ 2
✓ 1
Part (d)
19.4 ÷ 2 = 9.70 m s−2
✓ 1
Part (e)
off-centre, the ball blocks the beam along a chord shorter than d, so Δt is shorter and the calculated v too large
✓ 1
Exam tip: When the equation is v2 = 2gh, plot v2 on the y-axis and h on the x-axis, then compare with y = mx: the gradient is 2g, not g.
In a variation of the speed-of-sound experiment, two microphones are fixed 0.500 m apart in line with a loudspeaker and connected to the two channels of an oscilloscope. The frequency of the signal generator is slowly increased, and the student records each frequency at which the two traces are exactly in phase.
Figure: two microphones in line with the loudspeaker, connected to the two oscilloscope channels; not to scale
(a)
Explain why the traces are in phase when the separation of the microphones is a whole number of wavelengths.
(2)
(b)
Show that the difference between two successive in-phase frequencies is v/d, where v is the speed of sound and d the separation.
(2)
(c)
Three successive in-phase frequencies are 2.07 kHz, 2.76 kHz and 3.45 kHz. Calculate the speed of sound.
(2)
(d)
Suggest why the experiment should be done well away from walls and hard surfaces.
(1)
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Answer
Mark
Notes
Part (a)
the sound at the far microphone has travelled an extra whole number of wavelengths
✓ 1
so it is at the same point in its cycle as the sound at the near microphone / phase difference = 0 or a multiple of 360°
✓ 1
Part (b)
in phase when d = nλ = nv/f, so f = nv/d
✓ 1
successive values of n differ by 1, so Δf = v/d
✓ 1
Part (c)
mean Δf = (3.45 − 2.07) ÷ 2 = 0.690 kHz
✓ 1
v = 690 × 0.500 = 345 m s−1
✓ 1
Part (d)
sound reflected from hard surfaces reaches the microphones and superposes with the direct sound, distorting the traces
✓ 1
Exam tip: To find the average step, subtract the first reading from the last and divide by the number of gaps (here 2), not by the number of readings (3).
A student investigates how the frequency of the first harmonic depends on the mass per unit length μ of a string. Several strings are tested in turn, each with a vibrating length of 0.600 m and a tension of 19.6 N.
(a)
Describe how the student could find μ for one of the strings using a balance and a metre rule.
(2)
(b)
Show that the frequency of the first harmonic is f = (1/2l)√(T/μ).
(2)
(c)
One string has μ = 4.5 × 10−3 kg m−1. Calculate the expected frequency of its first harmonic.
(2)
(d)
State the graph the student should plot to obtain a straight line through the origin.
(1)
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Answer
Mark
Notes
Part (a)
measure the length of a long sample (e.g. 2 m or more) with the metre rule and its mass on the balance
✓ 1
μ = mass ÷ length
✓ 1
Part (b)
for the first harmonic the length is half a wavelength, so λ = 2l
✓ 1
f = v/λ with v = √(T/μ) gives the expression
✓ 1
Part (c)
v = √(19.6 ÷ 4.5 × 10−3) = 66.0 m s−1
✓ 1
f = 66.0 ÷ (2 × 0.600) = 55 Hz (55.0 Hz)
✓ 1
Part (d)
f against 1/√μ (or f2 against 1/μ)
✓ 1
Exam tip: Measure the mass of a long piece of string. A short piece has a tiny mass whose percentage uncertainty on the balance is large.
A student uses a red laser of wavelength 633 nm to find the number of lines per millimetre on a diffraction grating. The grating is 1.500 m from a screen, and the distance between the two second-order spots on the screen is 1.232 m.
Figure: the central spot and the two second-order spots on the screen (seen from above); not to scale
(a)
Explain why the student measures the distance between the two second-order spots rather than from the central spot to one second-order spot.
(1)
(b)
Calculate the angle between the second-order beam and the central beam.
(2)
(c)
Calculate the number of lines per millimetre on the grating.
(3)
(d)
State one safety precaution when using the laser.
(1)
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Answer
Mark
Notes
Part (a)
the measured distance is twice as large, so its percentage uncertainty is halved (and the centre of the pattern need not be located)
✓ 1
Part (b)
x = 1.232 ÷ 2 = 0.616 m
✓ 1
θ = tan−1(0.616 ÷ 1.500) = 22.3°
✓ 1
Part (c)
d = 2λ ÷ sin θ = 2 × 633 × 10−9 ÷ sin 22.3°
✓ 1
= 3.33 × 10−6 m
✓ 1
1 ÷ 3.33 × 10−6 = 3.0 × 105 m−1 = 300 lines per mm
✓ 1
Part (d)
never look along the beam or at its reflections / keep shiny objects out of the beam path
✓ 1
Common error: Forgetting the order number. For the second-order spots nλ = d sin θ uses n = 2, so d = 2λ/sin θ.
A student adapts the resistivity experiment to find the resistivity of the graphite “lead” from a mechanical pencil. The lead is a rod of nominal diameter 0.70 mm. Crocodile clips are attached 0.100 m apart along the rod, and the student measures the p.d. across this length and the current through it.
(a)
Name a suitable instrument for measuring the diameter of the rod. State its resolution and calculate the percentage uncertainty in a single diameter reading of 0.70 mm.
(2)
(b)
The p.d. is 1.42 V and the current is 0.273 A. Calculate the resistance of the 0.100 m length.
(1)
(c)
Calculate the resistivity of the graphite.
(2)
(d)
The student notices that the readings drift while the current flows. Explain why, and suggest how to reduce this effect.
the current heats the rod, so its resistance changes as its temperature changes
✓ 1
use a small current / switch off between readings so the rod stays at room temperature
✓ 1
Exam tip: A single reading gives a resistivity, but a graph of R against l for several clip separations is better: contact resistance then appears as an intercept and does not affect the gradient.
A student makes a cell by pushing a copper strip and a zinc strip into a lemon, and investigates its e.m.f. and internal resistance. The currents are very small, so a milliammeter and a variable resistor of range 0–10 kΩ are used. A graph of terminal p.d. V (y-axis) against current I (x-axis) is a straight line that meets the V-axis at 0.94 V and passes through the point (0.40 mA, 0.58 V).
Figure: terminal p.d. against current for the lemon cell
(a)
Describe the circuit the student should use.
(2)
(b)
Explain why the terminal p.d. decreases as the current increases.
(2)
(c)
Determine the e.m.f. and the internal resistance of the lemon cell.
(3)
(d)
Explain why a voltmeter with a very high resistance is needed in this experiment.
(1)
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Answer
Mark
Notes
Part (a)
lemon cell in series with the variable resistor and the milliammeter
✓ 1
voltmeter connected directly across the terminals (strips) of the lemon cell
✓ 1
Part (b)
there is a p.d. across the internal resistance (“lost volts”) equal to Ir
✓ 1
this increases with current, and V = ε − Ir with ε constant
the cell's currents are tiny (below a milliampere), so the voltmeter must take a negligible current compared with them or it would change the readings
✓ 1
Model sketch
Figure: model circuit: variable resistor and milliammeter in series, voltmeter across the cell; not to scale
Common error: Quoting the gradient of the V–I graph as the internal resistance with its minus sign. The gradient is −r, so r = 900 Ω, a positive value.
A student measures the terminal velocity v of steel balls of different radii r falling through an oil in a wide tube. Density of steel = 7800 kg m−3; density of the oil = 960 kg m−3. The graph shows v plotted against r2 with the best-fit line drawn for the first four points.
(a)
Use Stokes' law to explain why v is expected to be proportional to r2.
(2)
(b)
Use the labelled point on the line to determine the viscosity of the oil.
(3)
(c)
The point for the largest ball lies below the line. Suggest a reason.
(1)
(d)
Describe how the student should measure the radius of a ball.
(1)
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Answer
Mark
Notes
Part (a)
at terminal velocity weight = upthrust + viscous drag, so ⁴⁄₃πr3(ρs − ρf)g = 6πηrv
✓ 1
v = 2g(ρs − ρf)r2 ÷ 9η; everything except r2 is constant, so v ∝ r2
the largest ball is close to the walls of the tube, which increase the drag / it falls fast enough for the flow to stop being laminar, so Stokes' law no longer applies
✓ 1
Part (d)
measure the diameter with a micrometer screw gauge in several orientations (or for several balls of the same size), take the mean and halve it
✓ 1
Exam tip: Read the powers of ten in the axis labels before finding a gradient. Here 10.0 on the v-axis means 0.100 m s−1 and 8.0 on the r2-axis means 8.0 × 10−6 m2.
Figure 1 shows a student's apparatus for finding the Young modulus of steel. The wire has a diameter of 0.30 mm. A 0.50 kg mass keeps the wire taut while the first reading is taken; a further 1.50 kg is then added. The magnified views show the pointer against the millimetre scale for the two loads.
(a)
Read the two pointer positions and calculate the extension of the wire.
(2)
(b)
Calculate the cross-sectional area of the wire.
(1)
(c)
Calculate the stress produced by the extra 1.50 kg.
(1)
(d)
Calculate the Young modulus of the steel.
(2)
(e)
Explain why the percentage uncertainty in the extension is large in this experiment, and suggest one way to reduce it.
the extension is only a few millimetres, while each reading on a millimetre scale is uncertain by about ±0.5 mm, so the uncertainty is a large fraction of the extension
✓ 1
use a longer wire / add more load (staying within the limit of proportionality) / read the pointer with a vernier scale or travelling microscope
✓ 1
Exam tip: The extension is the change in the pointer reading caused by the added load. The first reading with the small taut-wire load is the zero; never use a scale reading itself as the extension.
A student measures the wavelength λ of sound from a loudspeaker at six frequencies f between 1.43 kHz and 5.00 kHz, using two microphones and an oscilloscope. The graph shows λ plotted against 1/f, with the best-fit line drawn.
(a)
Explain why the graph is a straight line through the origin, and state what the gradient represents.
(2)
(b)
Use the labelled point to determine the speed of sound.
(2)
(c)
Identify the anomalous point and suggest how it could have been caused.
(2)
(d)
Suggest why the student records the air temperature during the experiment.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
v = fλ gives λ = v × (1/f), and the speed of sound does not depend on frequency, so λ ∝ 1/f
✓ 1
the gradient is the speed of sound
✓ 1
Part (b)
gradient = 0.240 ÷ 0.700 × 10−3
✓ 1
= 343 m s−1 (allow 340–346 m s−1)
✓ 1
Part (c)
the point at 1/f = 0.40 ms (f = 2.5 kHz)
✓ 1
its wavelength is too large: the student miscounted the number of in-phase positions (counted one too few) or misread the distance moved
✓ 1
Part (d)
the speed of sound depends on the temperature of the air, so the value found applies to that temperature
✓ 1
Common error: Using 0.700 instead of 0.700 × 10−3 for 1/f. The axis is in milliseconds, so the raw gradient of 0.343 must be multiplied by 1000 to give m s−1.
Figure 1 shows the apparatus a student uses to investigate stationary waves on a string. With the signal generator set to 96.0 Hz, the string between the two bridges vibrates as shown.
(a)
Use the figure to determine the wavelength of the stationary wave.
(2)
(b)
Calculate the speed of the waves on the string.
(1)
(c)
Calculate the mass per unit length of the string.
(2)
(d)
Predict the lowest frequency at which the string would form a stationary wave between the bridges.
(1)
(e)
State one safety precaution for this experiment.
(1)
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Answer
Mark
Notes
Part (a)
there are three loops between the bridges, each half a wavelength long
✓ 1
λ = 2 × 0.900 ÷ 3 = 0.600 m
✓ 1
Part (b)
v = fλ = 96.0 × 0.600 = 57.6 m s−1
✓ 1
Part (c)
T = 0.800 × 9.81 = 7.85 N
✓ 1
μ = T/v2 = 7.85 ÷ 57.62 = 2.37 × 10−3 kg m−1
✓ 1
Part (d)
the first harmonic has one loop, so f = 96.0 ÷ 3 = 32.0 Hz
✓ 1
Part (e)
keep feet clear of the hanging mass in case the string snaps / place a soft mat or box under the masses
✓ 1
Exam tip: Count loops, not nodes. Each loop between neighbouring nodes is half a wavelength, so n loops on a length l give λ = 2l/n.
A green laser beam is directed at right angles onto a diffraction grating with 300 lines per mm. A metre rule is fixed to a screen 1.200 m from the grating. The figure shows the positions of the spots on the rule.
(a)
Use the positions of the two first-order spots to calculate the angle between a first-order beam and the central beam.
(2)
(b)
Calculate the wavelength of the laser light.
(2)
(c)
Calculate the wavelength again, using the second-order spots.
(2)
(d)
Explain why the value from the second-order spots is likely to be more precise.
(1)
(e)
State one precaution for using the laser safely.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
distance from the centre = (69.4 − 30.6) ÷ 2 = 19.4 cm
✓ 1
θ = tan−1(0.194 ÷ 1.200) = 9.18°
✓ 1
Part (b)
d = 1 ÷ (300 × 103) = 3.33 × 10−6 m
✓ 1
λ = d sin θ = 3.33 × 10−6 × sin 9.18° = 5.32 × 10−7 m
the spot positions have the same uncertainty, but the second-order distance (40.4 cm) is about twice as large, so its percentage uncertainty is smaller
✓ 1
Part (e)
never look along the beam or at its reflection / remove shiny objects such as watches from the beam path / do not point the laser at anyone
✓ 1
Exam tip: The spots lie on a flat screen, so the angle comes from tan θ = distance ÷ D, not sin. Then use sin θ in nλ = d sin θ.
A student uses the circuit shown to find the e.m.f. ε and internal resistance r of a used cell. The graph shows the terminal p.d. V against the current I for six settings of the variable resistor, with the best-fit line drawn.
(a)
Explain how ε and r can be found from this graph.
(2)
(b)
Use the two labelled points to determine r and ε.
(3)
(c)
Explain why the switch is closed only while each pair of readings is taken.
(1)
(d)
Explain why the student must not set the variable resistor to zero.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
V = ε − Ir, which has the form y = mx + c
✓ 1
the intercept on the V-axis at I = 0 is ε, and the gradient is −r
a steady current would warm the cell and run it down, changing ε and r during the experiment
✓ 1
Part (d)
the cell would be short-circuited: the current would be about ε/r = 1.52 ÷ 0.40 = 3.8 A, which would heat the cell and wires and could damage the ammeter
✓ 1
Exam tip: Look at where each axis starts. Here the V-axis starts at 0.90 V, but the I-axis starts at zero, so the line still meets the V-axis at I = 0 and that intercept is the e.m.f.
A steel ball of diameter 3.00 mm falls through glycerol in a wide tube and reaches a terminal velocity of 0.0230 m s−1. Density of steel = 7850 kg m−3; density of glycerol = 1260 kg m−3.
(a)
Show that, at terminal velocity, the viscosity of the liquid is given by η = 2r2g(ρs − ρf) ÷ 9v where r is the radius of the ball and ρs and ρf are the densities of steel and liquid.
(3)
(b)
Calculate the viscosity of the glycerol.
(2)
(c)
The tube is placed in a warm water bath and the experiment repeated. The ball now falls faster. State what this shows about the viscosity of glycerol.
(1)
(d)
Explain why a wide tube is used.
(1)
(e)
State one safety precaution for this experiment, other than wearing goggles.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
weight = ⁴⁄₃πr3ρsg and upthrust = ⁴⁄₃πr3ρfg
✓ 1
at terminal velocity weight = upthrust + 6πηrv
✓ 1
6πηrv = ⁴⁄₃πr3g(ρs − ρf), which rearranges to the expression given
near the walls the drag on the ball is larger than Stokes' law predicts, so a wide tube keeps the ball far from the walls
✓ 1
Part (e)
wipe up spilt glycerol at once because it makes the floor slippery / clamp the tall tube so that it cannot topple
✓ 1
Model sketch
Figure: forces on the ball at terminal velocity (arrow lengths in proportion)
Exam tip: In a “show that” derivation, write each force as an expression before equating them. A single line that jumps to the final equation loses the method marks.
A steel wire of original length 1.80 m and diameter 0.45 mm hangs vertically beside an identical reference wire. Loads are added and the extension is measured with a micrometer and spirit level fitted between the two wires. The results are: load / N: 0, 10, 20, 30, 40, 50 extension / mm: 0, 0.57, 1.13, 1.70, 2.26, 3.10
(a)
Identify the reading that suggests the wire has passed its limit of proportionality. Explain your answer.
(2)
(b)
Show that the Young modulus E = kl/A, where k is the gradient of a graph of load against extension, l is the original length and A is the cross-sectional area.
(2)
(c)
For the straight part of the graph, k = 1.77 × 104 N m−1. Calculate E.
(3)
(d)
Give a reason why the student should wear safety goggles.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the 50 N reading (3.10 mm)
✓ 1
the extension increases by about 0.56–0.57 mm per 10 N up to 40 N, but by 0.84 mm from 40 N to 50 N, so extension is no longer proportional to load
✓ 1
Part (b)
E = stress ÷ strain = (F/A) ÷ (Δx/l)
✓ 1
= (F/Δx) × l/A, and F/Δx = k
✓ 1
Part (c)
A = π × (0.225 × 10−3)2 = 1.59 × 10−7 m2
✓ 1
E = 1.77 × 104 × 1.80 ÷ 1.59 × 10−7
✓ 1
= 2.0 × 1011 Pa (2.00 × 1011)
✓ 1
Part (d)
the wire may snap under load and whip back towards the face
✓ 1
Model sketch
Figure: the 50 N point lies well off (to the right of) the straight line through the other points
Exam tip: Use only the straight section of a load–extension or stress–strain graph for the gradient. A point beyond the limit of proportionality is not an anomaly to be repeated — it is the material behaving differently.
A student is asked to investigate how the frequency of the first harmonic of a stretched nylon string depends on the tension in the string. A vibration generator, a signal generator, a pulley, two bridges and a set of slotted masses are available.
(a)
Describe how the student should carry out the investigation and use the results to show the relationship f ∝ √T.
(6)
Show mark scheme
Answer
Mark
Notes
Indicative content — one mark for each point made:
—
string runs over two bridges and a pulley to a mass hanger; tension T = mg
✓ 1
vibration generator near one bridge, driven by the signal generator
✓ 1
increase the frequency from a low value until the string vibrates in a single loop with maximum amplitude; read f from the generator
✓ 1
keep the length between the bridges and the string (mass per unit length) the same throughout
✓ 1
repeat for at least five tensions over a wide range, finding each frequency more than once and averaging
✓ 1
plot f2 against T (or f against √T): a straight line through the origin confirms f ∝ √T
✓ 1
Exam tip: A 6-mark method answer should be a sequence a reader could follow: set-up, how each quantity is measured, what is kept constant, range and repeats, and the graph that tests the relationship.
A grating with 500 lines per mm is used with a green laser of unknown wavelength. The angle θ of each order n is found, and a graph of sin θ (y-axis) against n (x-axis) is plotted for n = 1, 2 and 3.
(a)
Explain why the graph should be a straight line through the origin, and state what its gradient represents.
(2)
(b)
The gradient of the line is 0.266. Calculate the wavelength.
(2)
(c)
Determine the highest order that can be observed with this grating.
(2)
(d)
Suggest one advantage of finding λ from a graph rather than from a single order.
(1)
(e)
Explain why the screen is placed as far from the grating as is practical when the angles are found from spot positions.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
nλ = d sin θ gives sin θ = (λ/d)n, with λ and d constant
✓ 1
gradient = λ/d
✓ 1
Part (b)
d = 1 ÷ 500 × 103 = 2.00 × 10−6 m
✓ 1
λ = 0.266 × 2.00 × 10−6 = 5.32 × 10−7 m (532 nm)
✓ 1
Part (c)
sin θ cannot exceed 1, so n ≤ d/λ = 3.76
✓ 1
highest order = 3
✓ 1
Part (d)
the gradient uses all the readings, reducing the effect of random error in any one, and an anomalous order shows up off the line
✓ 1
Part (e)
the spots are further apart, so the same uncertainty in measuring their positions is a smaller percentage
✓ 1
Model sketch
Figure: sin θ against n: a straight line through the origin
Exam tip: The highest order is the whole number just below d/λ. Round down, even when the decimal part is large.
In a free-fall experiment the timer starts when the electromagnet is switched off, but the ball is released a short, constant time δ later. The measured time is therefore t = t0 + δ, where t0 is the true fall time. The student plots √h against t for six drop heights h from 0.20 m to 1.20 m, as shown in the graph.
(a)
Show that √h = √(g/2) × (t − δ).
(2)
(b)
Use the two labelled points to determine the gradient of the line, and hence calculate g.
(3)
(c)
Determine δ.
(1)
(d)
For h = 1.20 m the measured time is 0.513 s. A second student uses only this result and g = 2h/t2. Calculate the value obtained and explain why it differs from your answer to (b).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
from rest, h = ½gt02 and t0 = t − δ, so h = ½g(t − δ)2
✓ 1
taking the square root of both sides gives √h = √(g/2) × (t − δ)
δ is the intercept on the t-axis: 0.115 − 0.220 ÷ 2.20 = 0.015 s (allow 0.014–0.016 s)
✓ 1
Part (d)
g = 2 × 1.20 ÷ 0.5132 = 9.1 m s−2 (9.12)
✓ 1
the measured time includes the delay δ, so t is too long and g too small; the graph method puts the delay into the intercept, so the gradient is unaffected
✓ 1
Common error: Reading the gradient as g or as g/2. Here the gradient is √(g/2), so it must be squared and doubled: g = 2 × gradient2.
A copper wire of original length 2.00 m and diameter 0.40 mm is loaded until its extension is 4.0 mm, and the load is then removed in steps. The graph shows the load F against the extension x for loading and unloading.
(a)
Use the labelled point to determine the Young modulus of copper.
(3)
(b)
State the load at the limit of proportionality.
(1)
(c)
State the permanent extension of the wire after the load is removed.
(1)
(d)
Explain why the unloading line is parallel to the straight part of the loading line.
(1)
(e)
A student says the wire returned to its original length because the load returned to zero. Use the graph to explain why this is wrong.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient of the straight section = 8.0 ÷ 1.00 × 10−3 = 8.0 × 103 N m−1
✓ 1
A = π × (0.20 × 10−3)2 = 1.26 × 10−7 m2
✓ 1
E = gradient × l ÷ A = 8.0 × 103 × 2.00 ÷ 1.26 × 10−7 = 1.3 × 1011 Pa (1.27 × 1011)
✓ 1
Part (b)
10.0 N (allow 9.5–10.5 N): the graph stops being a straight line through the origin there
✓ 1
Part (c)
2.25 mm (allow 2.2–2.3 mm)
✓ 1
Part (d)
the bonds between the atoms are unchanged, so the elastic part of the extension is recovered with the same stiffness (Young modulus); only the plastic part remains
✓ 1
Part (e)
at zero load the extension is still 2.25 mm, not zero: the wire was loaded beyond its elastic limit and has deformed plastically
✓ 1
Common error: Calculating E from a point on the curved part of the graph, such as (4.0 mm, 14.0 N). The Young modulus is defined only for the straight, proportional section.
A student keeps the tension in a string constant, using a hanging mass of 0.500 kg, and moves a bridge to change the vibrating length L. For each length the frequency f of the first harmonic is found. The graph shows f plotted against 1/L.
(a)
Explain why the graph is a straight line through the origin.
(2)
(b)
Determine the gradient of the line.
(1)
(c)
Calculate the mass per unit length of the string.
(3)
(d)
Describe how the student can find the frequency of the first harmonic precisely for one length.
(1)
(e)
The experiment is repeated with a string that has four times the mass per unit length, under the same tension. State the gradient of the new line.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
for the first harmonic λ = 2L, so f = v/2L = (1/2L)√(T/μ)
✓ 1
T and μ are constant, so f ∝ 1/L with gradient ½√(T/μ)
✓ 1
Part (b)
80.0 ÷ 2.50 = 32.0 m s−1 (allow 31.5–32.5)
✓ 1
Part (c)
T = 0.500 × 9.81 = 4.91 N
✓ 1
√(T/μ) = 2 × 32.0 = 64.0 m s−1
✓ 1
μ = 4.91 ÷ 64.02 = 1.20 × 10−3 kg m−1
✓ 1
Part (d)
increase the frequency slowly until the amplitude of the single loop is greatest, repeat approaching from a higher frequency, and take the mean of the two readings
✓ 1
Part (e)
the gradient ∝ 1/√μ, so it halves to 16.0 m s−1
✓ 1
Common error: Setting the gradient equal to √(T/μ). The first harmonic has λ = 2L, which puts a factor of ½ into the gradient: the wave speed is twice the gradient.
Instead of changing the length, a student finds the resistivity of an alloy using six wires of the same alloy, each 0.800 m long, with diameters from 0.25 mm to 0.56 mm. For each wire the resistance R and the cross-sectional area A are found. The graph shows R plotted against 1/A.
(a)
Explain why the graph is expected to be a straight line through the origin, and state what the gradient represents.
(2)
(b)
Use the labelled point to determine the resistivity of the alloy.
(3)
(c)
The diameter of the thinnest wire, 0.25 mm, is measured with a micrometer of resolution 0.01 mm, and the uncertainty in this reading is taken as half the resolution. Calculate the percentage uncertainty in this diameter.
(1)
(d)
The point for the thinnest wire lies above the line. Suggest a reason, other than a random error in reading the meters.
(1)
(e)
Explain why the student must use the same length for every wire.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
R = ρl/A = ρl × (1/A), with ρ and l the same for every wire, so R ∝ 1/A
✓ 1
gradient = ρl
✓ 1
Part (b)
gradient = 6.00 ÷ 15.0 × 106 = 4.00 × 10−7 Ω m2
✓ 1
ρ = gradient ÷ l = 4.00 × 10−7 ÷ 0.800
✓ 1
= 5.0 × 10−7 Ω m (allow 4.9–5.1 × 10−7)
✓ 1
Part (c)
±0.005 mm ÷ 0.25 mm × 100 = 2.0%
✓ 1
Part (d)
its diameter was measured too large (e.g. at one point only, or the wire is thinner where it has been stretched), so the true 1/A is larger than plotted; a small error in d matters more for a thin wire
✓ 1
Part (e)
the gradient equals ρl, which is constant only if l is the same for all wires; otherwise the points would not lie on one line
✓ 1
Exam tip: Write the equation in the form y = mx before reading a gradient. With R against 1/A the gradient is ρl, so divide by the length to find ρ.
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