Edexcel International A Level Physics · Unit 6 · Practical Skills in Physics II
The A2 core practicals
28 free, high-quality past-paper-style questions on The A2 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 · 195 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U6-3-B-01Unit 6 · A2Easy6 marks
A student tests the relationship between force and change of momentum. A trolley on a friction-compensated runway is pulled by a string that passes over a pulley to a hanging mass of 0.060 kg. The total mass of the trolley and the hanging mass is 0.800 kg. Two light gates record the speed of the trolley as 0.215 m s−1 at the first gate and 0.580 m s−1 at the second, and the time taken between the gates as 0.500 s.
Figure: trolley, light gates, pulley and hanging mass; not to scale
(a)
Calculate the accelerating force provided by the hanging mass.
(1)
(b)
Calculate the rate of change of momentum of the moving system.
(2)
(c)
Calculate the percentage difference between your answers to (a) and (b), using (a) as the reference.
(1)
(d)
Explain why the runway is tilted slightly before the experiment starts.
(1)
(e)
To change the force, the student moves slotted masses from the trolley to the hanger. Explain why.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
F = 0.060 × 9.81 = 0.589 N
✓ 1
Part (b)
Δp = 0.800 × (0.580 − 0.215) = 0.292 kg m s−1
✓ 1
Δp/Δt = 0.292 ÷ 0.500 = 0.584 N
✓ 1
Part (c)
(0.589 − 0.584) ÷ 0.589 × 100 = 0.8%
✓ 1
Part (d)
a component of the trolley's weight down the slope balances friction, so the resultant force is the pull of the hanging mass alone
✓ 1
Part (e)
the force accelerates the trolley and the hanging masses together, so moving masses keeps the total mass constant and only the force changes
✓ 1
Common error: Using the trolley's mass alone. The weight of the hanging mass accelerates the hanging mass too, so the momentum change is that of the whole moving system.
A 2.2 μF capacitor is connected in series with a 1.0 kΩ resistor to a signal generator producing a square wave. An oscilloscope connected across the capacitor shows it charging and discharging repeatedly.
Figure: p.d. across the capacitor during one discharge
(a)
Calculate the time constant of the circuit.
(1)
(b)
The capacitor is taken to be fully charged after five time constants. Calculate the highest square-wave frequency for which the capacitor fully charges and fully discharges in each cycle.
(2)
(c)
On the screen, the p.d. falls to half its initial value in 1.5 ms. Determine the capacitance from this measurement.
(2)
(d)
Suggest why an oscilloscope is used rather than a data logger that records a reading every 0.1 s.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
RC = 1.0 × 103 × 2.2 × 10−6 = 2.2 ms
✓ 1
Part (b)
each half-period must be at least 5 × 2.2 = 11 ms, so the period is at least 22 ms
✓ 1
f = 1 ÷ 0.022 = 45 Hz
✓ 1
Part (c)
t½ = RC ln 2, so RC = 1.5 ÷ 0.693 = 2.16 ms
✓ 1
C = 2.16 × 10−3 ÷ 1.0 × 103 = 2.2 μF (2.16 μF)
✓ 1
Part (d)
the whole discharge takes about 10 ms, far shorter than the logger's interval between readings, so the logger would miss it
✓ 1
Exam tip: Remember that the time to halve is RC ln 2 ≈ 0.69RC, not RC. In one time constant the p.d. falls to 37% of its starting value.
A student calibrates an NTC thermistor. The thermistor is connected in series with a 1.00 kΩ fixed resistor across a 5.00 V supply of negligible internal resistance. A voltmeter is connected across the thermistor. The thermistor and a thermometer are placed in a beaker of water that is heated slowly.
Figure: the calibration circuit; not to scale
(a)
Explain why the voltmeter reading decreases as the temperature of the water increases.
(2)
(b)
Explain why the water should be stirred and the reading taken a short time after the thermometer reading becomes steady.
(1)
(c)
At one temperature the voltmeter reads 1.85 V. Calculate the resistance of the thermistor.
(2)
(d)
State one safety precaution for this experiment.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the resistance of the thermistor decreases as its temperature rises
✓ 1
so it takes a smaller share of the 5.00 V supply / the p.d. across it falls
✓ 1
Part (b)
so that the thermistor and the thermometer are at the same temperature as the water
✓ 1
Part (c)
p.d. across the fixed resistor = 5.00 − 1.85 = 3.15 V, so I = 3.15 ÷ 1000 = 3.15 mA
✓ 1
R = 1.85 ÷ 3.15 × 10−3 = 587 Ω
✓ 1
Part (d)
stand the beaker on a heat-proof mat and do not move it while it is hot / keep connections and the supply away from the water
✓ 1
Exam tip: In a potential divider the current is the same in both resistors, so the p.d.s are in the same ratio as the resistances: Rth ÷ 1.00 kΩ = 1.85 ÷ 3.15.
A student investigates the absorption of gamma radiation by lead using a sealed gamma source and a Geiger–Müller tube connected to a counter. Before the source is taken out of its storage box, the counter records 31 counts in 60 s.
(a)
Calculate the background count rate.
(1)
(b)
With 6.0 mm of lead between the source and the tube, 1250 counts are recorded in 50 s. Calculate the corrected count rate.
(1)
(c)
State two precautions that reduce the student's exposure to radiation from the source.
(2)
(d)
Explain why the distance between the source and the tube must not change during the experiment.
(1)
(e)
Explain why each count is taken over a long time.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
31 ÷ 60 = 0.52 s−1
✓ 1
Part (b)
1250 ÷ 50 − 0.52 = 24.5 s−1
✓ 1
Part (c)
handle the source with tongs, never with the hands
✓ 1
keep the source at arm's length and pointed away from people, and return it to its box as soon as possible
✓ 1
Part (d)
the count rate also changes with distance, so only the lead thickness should affect the reading
✓ 1
Part (e)
decay is random, so a larger number of counts gives a count rate that varies less between repeats / smaller percentage fluctuation
✓ 1
Exam tip: A safety precaution must say how it reduces exposure: tongs increase the distance from the hands; short handling times reduce the dose; the lead box absorbs the radiation during storage.
A student tests the link between force and rate of change of momentum. A trolley carrying a force sensor has a total mass of 0.920 kg. It runs on a friction-compensated runway, pulled by a string tied to the force sensor, which gives a steady reading of 0.455 N while the trolley speeds up. A motion sensor connected to a data logger records the velocity v of the trolley. Figure 1 shows v against time t with the best-fit line drawn.
(a)
Use the two labelled points on the line to determine the acceleration of the trolley.
(2)
(b)
Calculate the rate of change of momentum of the trolley.
(1)
(c)
Calculate the percentage difference between the force-sensor reading and your answer to (b), using the force-sensor reading as the reference.
(1)
(d)
Suggest why the rate of change of momentum is smaller than the force-sensor reading.
(1)
(e)
Explain how the graph shows that the resultant force on the trolley was constant.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient = (0.84 − 0.24) ÷ (1.50 − 0.25)
✓ 1
= 0.48 m s−2
✓ 1
Part (b)
Δp/Δt = ma = 0.920 × 0.48 = 0.442 N (allow 0.44 N)
✓ 1
Part (c)
(0.455 − 0.442) ÷ 0.455 × 100 = 2.9% (allow 3%)
✓ 1
Part (d)
the runway does not fully compensate for friction, so a small frictional force acts backwards and the resultant force is less than the pull of the string
✓ 1
Part (e)
the graph is a straight line, so the acceleration is constant; with constant mass the momentum changes at a constant rate, so the resultant force is constant
✓ 1
Exam tip: Take a gradient from points on the best-fit line that are far apart, not from two plotted results. The line averages out the scatter in the readings.
A student uses the circuit in Figure 1 to charge a capacitor C through a 100 kΩ resistor from a 6.0 V battery of negligible internal resistance. The capacitor is first fully discharged with switch S at b. At t = 0, S is moved to a and the data logger records the p.d. V across the capacitor, giving the graph in Figure 1.
(a)
Explain why V rises quickly at first and then more and more slowly.
(1)
(b)
Calculate the p.d. across the capacitor after one time constant.
(1)
(c)
Use the graph to determine the time constant.
(1)
(d)
Calculate the capacitance.
(1)
(e)
Calculate the current in the circuit at t = 0.
(1)
(f)
State the purpose of position b of the switch.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
as charge builds up, V increases, so the p.d. across the resistor and hence the current fall; charge arrives more slowly
✓ 1
Part (b)
(1 − e−1) × 6.0 = 0.63 × 6.0 = 3.8 V
✓ 1
Part (c)
time at which V = 3.8 V: 15 s (allow 14–16 s)
✓ 1
Part (d)
C = 15 ÷ (100 × 103) = 1.5 × 10−4 F = 150 μF (allow 140–160 μF)
✓ 1
Part (e)
6.0 ÷ (100 × 103) = 6.0 × 10−5 A = 60 μA
✓ 1
Part (f)
it connects the capacitor across the resistor without the battery, so the capacitor discharges and starts each run uncharged
✓ 1
Common error: Reading the time constant where a charging curve reaches 37% of the final p.d. That is the rule for discharge. When charging, after one time constant the p.d. has risen to 63% of its final value.
A student calibrates a thermistor by measuring its resistance R with an ohmmeter while it is in a water bath at different temperatures θ. The graph in Figure 1 shows the results. The thermistor is then used in the potential divider also shown in Figure 1, with a 1.00 kΩ fixed resistor and a 5.0 V supply of negligible internal resistance, to monitor the temperature of the bath.
(a)
State how the graph shows that the resistance of the thermistor has a negative temperature coefficient.
(1)
(b)
Use the graph to determine the resistance of the thermistor at 60 °C.
(1)
(c)
Calculate Vout when the bath is at 60 °C.
(2)
(d)
Use the labelled point to calculate Vout when the bath is at 40 °C.
(1)
(e)
During the calibration the thermistor is sealed inside a thin plastic sleeve. Suggest why.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the resistance decreases as the temperature increases
✓ 1
Part (b)
1.0 kΩ (allow 0.95–1.05 kΩ)
✓ 1
Part (c)
I = 5.0 ÷ (1.0 + 1.00) × 103 = 2.5 mA
✓ 1
Vout = 2.5 × 10−3 × 1.00 × 103 = 2.5 V
✓ 1
Part (d)
5.0 × 1.00 ÷ (2.00 + 1.00) = 1.7 V (1.67 V)
✓ 1
Part (e)
tap water conducts, so without the sleeve current could pass through the water in parallel with the thermistor and the ohmmeter would read too low a resistance
✓ 1
Exam tip: In a potential divider the output across one resistor is Vout = Vs × R1 ÷ (R1 + R2). Check which resistor the output is taken across before substituting.
A student finds the specific latent heat of fusion of ice. 0.0600 kg of crushed ice at −12 °C is placed in an insulated beaker with an immersion heater of power 40.0 W and a temperature probe. The heater is switched on at t = 0 and the contents are stirred continuously. Figure 1 shows the temperature θ recorded by a data logger.
(a)
Explain why the temperature stays at 0 °C between 40 s and 520 s.
(1)
(b)
Use the graph to determine the time taken for the ice to melt.
(1)
(c)
Calculate the specific latent heat of fusion of ice.
(2)
(d)
The accepted value is 3.34 × 105 J kg−1. Calculate the percentage difference between the student's value and the accepted value.
(1)
(e)
Explain whether energy exchange with the surroundings could account for this difference.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the energy supplied breaks bonds between the molecules as the ice melts, increasing their potential energy but not their mean kinetic energy, so the temperature does not change
✓ 1
Part (b)
520 − 40 = 480 s (allow 470–490 s)
✓ 1
Part (c)
E = Pt = 40.0 × 480 = 1.92 × 104 J
✓ 1
L = 1.92 × 104 ÷ 0.0600 = 3.20 × 105 J kg−1
✓ 1
Part (d)
(3.34 − 3.20) ÷ 3.34 × 100 = 4.2%
✓ 1
Part (e)
yes — the ice is colder than the room, so it gains energy from the surroundings; less energy is needed from the heater to melt it, so the calculated L is too small
✓ 1
Common error: Using the whole heating time in E = Pt. Only the time spent at the melting point, where the graph is flat, goes into the latent heat.
A steel sphere of mass 0.064 kg rolls along a straight groove and strikes a stationary glass marble of mass 0.021 kg head-on. The collision is filmed at 120 frames per second and analysed frame by frame. Over 12 frames before the collision the steel sphere moves 5.0 cm. Over 12 frames after the collision the steel sphere moves 2.6 cm and the marble moves 7.3 cm, both in the original direction.
Figure: the head-on collision, before and after (both move in the original direction afterwards); not to scale
(a)
Calculate the time interval for 12 frames.
(1)
(b)
Calculate the velocity of each sphere after the collision.
(1)
(c)
Show that momentum is conserved in this collision.
(2)
(d)
Calculate the kinetic energy before and after the collision.
(2)
(e)
Each distance has an uncertainty of about 3%, so each kinetic energy has an uncertainty of about 6%. Deduce whether the collision can be described as elastic.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
12 ÷ 120 = 0.100 s
✓ 1
Part (b)
steel: 0.026 ÷ 0.100 = 0.26 m s−1; marble: 0.073 ÷ 0.100 = 0.73 m s−1
✓ 1
Part (c)
before: 0.064 × 0.50 = 0.0320 kg m s−1
✓ 1
after: 0.064 × 0.26 + 0.021 × 0.73 = 0.0166 + 0.0153 = 0.0320 kg m s−1, the same
the kinetic energy after is 3% less than before, which is within the 6% uncertainty, so the collision is elastic within the uncertainty of the experiment
✓ 1
Exam tip: Kinetic energy depends on v2, so its percentage uncertainty is twice that of the speed. Compare a percentage change with this before calling a collision inelastic.
A data logger records the p.d. V across a capacitor discharging through a 100 kΩ ± 5% resistor. t / s: 0, 10, 20, 30, 40 V / V: 9.00, 5.93, 3.91, 2.58, 1.70 The capacitor is labelled 220 μF ± 20%.
(a)
Explain why a graph of ln V against t should be a straight line, and state what its gradient represents.
(2)
(b)
Calculate ln (V / V) for t = 40 s.
(1)
(c)
The gradient of the best-fit line is −0.0417 s−1. Calculate the capacitance.
(2)
(d)
The steepest and shallowest acceptable lines give an uncertainty in the gradient of ±4%. Calculate the absolute uncertainty in the capacitance.
(2)
(e)
State whether the result is consistent with the label. Give a reason.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
V = V0e−t/RC, so ln V = ln V0 − t/RC, which has the form y = c + mx
✓ 1
gradient = −1/RC
✓ 1
Part (b)
ln 1.70 = 0.531
✓ 1
Part (c)
RC = 1 ÷ 0.0417 = 24.0 s
✓ 1
C = 24.0 ÷ 1.00 × 105 = 240 μF (2.40 × 10−4 F)
✓ 1
Part (d)
percentage uncertainty in C = 4% + 5% = 9%
✓ 1
0.09 × 240 = ±22 μF
✓ 1
Part (e)
yes — the label allows 176 μF to 264 μF and 240 μF lies within this range
✓ 1
Model sketch
Figure: ln (V / V) against t: a straight line
Common error: Taking the log of the unit as well as the number. Plot ln (V / V): the quantity inside the log is a pure number, and the log itself has no unit.
A thermostat for an aquarium heater uses a thermistor and a fixed resistor in series across a 9.00 V supply. The output p.d. is taken across the fixed resistor and fed to a switching circuit that turns the heater off when the output rises to 4.50 V. The calibrated thermistor has a resistance of 2.29 kΩ at 25.0 °C, 2.20 kΩ at 26.0 °C and 2.11 kΩ at 27.0 °C.
Figure: the thermostat potential divider (output taken across the fixed resistor R); not to scale
(a)
The water is to be kept at 26.0 °C. State the value of the fixed resistor needed, and explain your answer.
(2)
(b)
Calculate the output p.d. at 25.0 °C and at 27.0 °C, and hence the change in output per kelvin near 26.0 °C.
(3)
(c)
The switching circuit responds to changes of 0.01 V. Estimate the smallest temperature change the thermostat can detect.
(1)
(d)
Suggest why the supply p.d. should not be made much larger to increase the sensitivity.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
2.20 kΩ
✓ 1
the output is half the supply (4.50 V) when the two resistances are equal, which happens at 26.0 °C
✓ 1
Part (b)
25.0 °C: 9.00 × 2.20 ÷ (2.20 + 2.29) = 4.41 V
✓ 1
27.0 °C: 9.00 × 2.20 ÷ (2.20 + 2.11) = 4.59 V
✓ 1
(4.594 − 4.410) ÷ 2.0 = 0.092 V K−1 (92 mV K−1)
✓ 1
Part (c)
0.01 ÷ 0.092 = 0.11 K (about 0.1 K)
✓ 1
Part (d)
a larger current would heat the thermistor itself, so it would be warmer than the water and the thermostat would switch off too early
✓ 1
Exam tip: A potential divider is most sensitive to temperature where the thermistor's resistance is close to that of the fixed resistor. That is why the fixed resistor is matched to the set temperature.
A student measures the specific latent heat of vaporisation of water. An immersion heater keeps water boiling in a beaker standing on a balance. With the heater supplying 60.0 W, the mass decreases at 0.0222 g s−1. With the heater supplying 40.0 W, the mass decreases at 0.0134 g s−1.
(a)
Explain why using two different powers allows the energy lost to the surroundings to be eliminated.
(2)
(b)
Calculate the specific latent heat of vaporisation of water.
(2)
(c)
Calculate the rate at which energy is lost to the surroundings.
(2)
(d)
Explain why the student waits until the water is boiling steadily before starting to time.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the water is at 100 °C in both runs, so the rate of energy loss to the surroundings, h, is the same
✓ 1
P1 = Lr1 + h and P2 = Lr2 + h; subtracting removes h, giving L = (P1 − P2) ÷ (r1 − r2)
✓ 1
Part (b)
L = (60.0 − 40.0) ÷ ((0.0222 − 0.0134) × 10−3)
✓ 1
= 2.27 × 106 J kg−1
✓ 1
Part (c)
h = P1 − Lr1 = 60.0 − 2.273 × 106 × 2.22 × 10−5
✓ 1
= 60.0 − 50.5 = 9.5 W
✓ 1
Part (d)
until then some of the energy supplied is raising the temperature of the water rather than turning it to steam
✓ 1
Exam tip: Convert rates of mass loss to kg s−1 before dividing: 0.0088 g s−1 is 8.8 × 10−6 kg s−1.
A sealed plastic syringe containing air is connected to a pressure sensor. The plunger is pushed in slowly and the volume and pressure are recorded at a temperature of 21 °C. V / cm3: 50, 40, 30, 25 p / kPa: 101, 126, 168, 202
(a)
Show that these results are consistent with pressure being inversely proportional to volume.
(2)
(b)
Calculate the number of air molecules in the syringe.
(2)
(c)
State the graph that should be plotted to give a straight line through the origin.
(1)
(d)
Explain why the plunger is pushed in slowly.
(1)
(e)
Suggest why the student should avoid holding the barrel of the syringe in their hand.
A stiff spring carrying a mass m hangs from a vibration generator. For each of several known masses, the student finds the driving frequency f at which the mass oscillates with the largest amplitude, and plots T2 = 1/f2 against m. The best-fit line has a gradient of 0.162 s2 kg−1 and an intercept of 0.004 s2.
Figure: the student's best-fit line of T² against m
(a)
Show that T2 is expected to be proportional to m, and state what the gradient represents.
(2)
(b)
Calculate the spring constant.
(1)
(c)
An unknown mass resonates at 3.10 Hz. Determine the unknown mass.
(2)
(d)
Describe how the student can find the resonant frequency as precisely as possible.
(1)
(e)
Suggest a reason why the line does not pass through the origin.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
at resonance the driving frequency equals the natural frequency, with T = 2π√(m/k)
✓ 1
so T2 = (4π2/k)m; gradient = 4π2/k
✓ 1
Part (b)
k = 4π2 ÷ 0.162 = 244 N m−1
✓ 1
Part (c)
T2 = 1 ÷ 3.102 = 0.1041 s2
✓ 1
m = (0.1041 − 0.004) ÷ 0.162 = 0.618 kg
✓ 1
Part (d)
approach the frequency of maximum amplitude slowly from below and from above, record both and take the mean (or check the frequency on an oscilloscope)
✓ 1
Part (e)
the spring itself has mass, part of which oscillates, adding an effective mass to every load
✓ 1
Exam tip: To read an unknown from a calibration line, use the equation of the line including its intercept: m = (y − c) ÷ gradient. Leaving the intercept out gives a mass that is too large.
Sphere A, of mass 0.050 kg, rolls along a straight groove and hits sphere B, of mass 0.030 kg, which is at rest. A video of the collision is analysed with software that records the position x of the centre of each sphere every 0.10 s. Figure 1 shows the results.
(a)
Use the graph to determine the velocity of A before the collision.
(1)
(b)
Determine the velocities of A and B after the collision.
(2)
(c)
Calculate the total momentum before and after the collision, and the percentage change.
(2)
(d)
Suggest a physical reason why the momentum after the collision is less than before.
(1)
(e)
Explain why the camera should be placed some distance away, with its line of sight perpendicular to the groove.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
40 cm in 1.00 s = 0.40 m s−1
✓ 1
Part (b)
A: (50 − 40) cm ÷ 1.00 s = 0.10 m s−1
✓ 1
B: (90 − 42) cm ÷ 1.00 s = 0.48 m s−1
✓ 1
Part (c)
before: 0.050 × 0.40 = 0.0200 kg m s−1; after: 0.050 × 0.10 + 0.030 × 0.48 = 0.0050 + 0.0144 = 0.0194 kg m s−1
✓ 1
change = 0.0006 ÷ 0.0200 × 100 = 3% decrease
✓ 1
Part (d)
friction / rolling resistance from the groove is an external force on the spheres, so momentum is not exactly conserved over the time measured
✓ 1
Part (e)
so that distances along the groove appear in the image without parallax or perspective distortion, and a scale in the plane of the groove gives true distances throughout the motion
✓ 1
Exam tip: On a position–time graph the velocity is the gradient. Read positions at whole numbers of seconds where the points sit on gridlines, and use a long time interval.
A capacitor discharges through a 680 Ω resistor. A square-wave signal generator charges and discharges it repeatedly, and an oscilloscope connected across the capacitor shows the trace in Figure 1. The Y-gain is 1.0 V div−1 and the time-base is 0.20 ms div−1. The discharge starts 1.0 division from the left-hand edge of the screen.
(a)
State the p.d. across the capacitor at the start of the discharge.
(1)
(b)
Determine the time taken for the p.d. to fall to 1.0 V.
(1)
(c)
Show that the time constant RC is about 0.43 ms.
(2)
(d)
Calculate the capacitance.
(1)
(e)
The time can be read from the screen to ±0.1 division. Calculate the percentage uncertainty in the time in (b).
(1)
(f)
Suggest a change to the oscilloscope settings that would reduce this percentage uncertainty. Explain your answer.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
4.0 div × 1.0 V div−1 = 4.0 V
✓ 1
Part (b)
3.0 div × 0.20 ms div−1 = 0.60 ms (allow 0.58–0.62 ms)
✓ 1
Part (c)
1.0 = 4.0 e−t/RC, so t/RC = ln 4
✓ 1
RC = 0.60 ÷ 1.386 = 0.433 ms
✓ 1
Part (d)
C = 0.433 × 10−3 ÷ 680 = 6.4 × 10−7 F = 0.64 μF
✓ 1
Part (e)
0.1 ÷ 3.0 × 100 = 3.3%
✓ 1
Part (f)
use a time-base of 0.10 ms div−1, so the same time covers 6.0 divisions and ±0.1 division is a smaller fraction of it
✓ 1
Exam tip: Measure times on an oscilloscope from the start of the decay, not from the edge of the screen, and convert divisions using the time-base before using any formula.
A thermistor and a 3.30 kΩ resistor are connected in series to a 6.00 V supply of negligible internal resistance, and a voltmeter measures the p.d. V across the thermistor (inset in Figure 1). The graph in Figure 1 shows V against the temperature θ of the thermistor. A tangent to the curve has been drawn at 50 °C.
(a)
Use the tangent to determine the rate of change of V with temperature at 50 °C.
(2)
(b)
At 30 °C, V = 3.00 V. State the resistance of the thermistor at 30 °C and explain your answer.
(1)
(c)
Use the labelled point at 70 °C to calculate the resistance of the thermistor at this temperature.
(2)
(d)
The circuit is to be used to control an incubator kept at 70 °C. Suggest a change to the circuit and explain why it gives better control.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient = (0.46 − 3.30) ÷ (80 − 20)
✓ 1
= −0.0473 V K−1, so V falls by 47 mV K−1 (allow 46–49 mV K−1)
✓ 1
Part (b)
3.30 kΩ — the thermistor has half of the 6.00 V supply, so it has the same resistance as the fixed resistor
✓ 1
Part (c)
p.d. across the fixed resistor = 6.00 − 1.12 = 4.88 V, so I = 4.88 ÷ 3300 = 1.48 mA
✓ 1
R = 1.12 ÷ 1.48 × 10−3 = 757 Ω
✓ 1
Part (d)
replace the 3.30 kΩ resistor with one of about 760 Ω, equal to the thermistor's resistance at 70 °C
✓ 1
V changes most rapidly with temperature when the two resistances are about equal, so a small temperature change near 70 °C gives a larger, easily detected change in V
✓ 1
Exam tip: A tangent gives the gradient at one point on a curve. Draw it long, and read two points on it that are far apart and on gridlines.
Crushed ice at 0 °C is packed round an immersion heater in a funnel, and the water that drips out is collected in a beaker on a balance. The heater is switched on at t = 200 s; it is connected to a 12.0 V supply and the current in it is 2.50 A. Figure 1 shows the mass m of water collected against time t, with lines of best fit drawn.
(a)
Explain why water is collected before the heater is switched on.
(1)
(b)
Use the graph to determine the rate at which water is collected before and after the heater is switched on.
(2)
(c)
Assume that energy from the surroundings melts ice at the same rate throughout. Calculate the specific latent heat of fusion of ice.
(2)
(d)
The voltmeter reading has an uncertainty of ±0.1 V and the ammeter reading ±0.05 A. Calculate the percentage uncertainty in the power.
(1)
(e)
Suggest why crushed ice is used rather than large lumps.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the ice is colder than the room, so energy flows in from the surroundings and melts some ice
✓ 1
Part (b)
before: 2.0 ÷ 200 = 0.010 g s−1
✓ 1
after: (62.0 − 2.0) ÷ (800 − 200) = 0.100 g s−1
✓ 1
Part (c)
rate due to the heater = 0.100 − 0.010 = 0.090 g s−1 = 9.0 × 10−5 kg s−1, and P = 12.0 × 2.50 = 30.0 W
crushed ice packs closely round the heater with few air gaps, so the energy from the heater goes into melting ice rather than warming air or escaping
✓ 1
Exam tip: A gradient taken after a change in conditions must use points from the new straight section only. Here both labelled points lie on the steeper line, from 200 s onwards.
Figure 1 shows two views of an apparatus used to test Boyle's law. Air is trapped above oil in a sealed glass tube of uniform cross-sectional area 1.20 cm2. A pump pushes oil from the reservoir into the tube, and the gauge reads the pressure of the trapped air. The scale reads zero at the closed top of the tube. Between the two views the oil was pushed in slowly and the temperature did not change.
(a)
Read the pressure of the air and the length of the air column in each view.
(2)
(b)
Calculate the volume of the air in each view.
(1)
(c)
Show that these results are consistent with Boyle's law.
(1)
(d)
Explain why the length of the air column can be used in place of its volume when a graph is plotted to test Boyle's law.
(1)
(e)
The length can be read to ±0.2 cm. Calculate the percentage uncertainty in the length in view 2.
(1)
(f)
State how the student should read the scale to avoid parallax error.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
view 1: 110 kPa and 36.0 cm
✓ 1
view 2: 180 kPa and 22.0 cm
✓ 1
Part (b)
1.20 × 36.0 = 43.2 cm3 and 1.20 × 22.0 = 26.4 cm3
✓ 1
Part (c)
pV = 110 × 43.2 = 4750 kPa cm3 and 180 × 26.4 = 4750 kPa cm3: the product is the same (4752 in each case), so p ∝ 1/V
✓ 1
Part (d)
the tube has a uniform cross-section, so V = AL with A constant and V ∝ L; a graph of p against 1/L is still a straight line through the origin
✓ 1
Part (e)
0.2 ÷ 22.0 × 100 = 0.91%
✓ 1
Part (f)
with the eye level with the oil surface, looking at right angles to the scale
✓ 1
Exam tip: Before reading a dial, work out what one small division is worth. On this gauge the labels are 50 kPa apart with five divisions between them, so each division is 10 kPa.
A student measures the count rate from a sealed gamma source as sheets of lead are placed between the source and a Geiger–Müller tube. The background count rate is 0.45 s−1. The graph in Figure 1 shows the corrected count rate C against the total thickness x of lead, with a smooth curve drawn through the points.
(a)
Use the curve to determine the half-value thickness, the thickness of lead that halves the corrected count rate.
(2)
(b)
Use the curve to show that a further 12 mm of lead halves the count rate again, and state what this shows about the absorption.
(1)
(c)
Calculate the absorption coefficient μ in C = C0e−μx.
(1)
(d)
With 40 mm of lead, 839 counts were recorded in 100 s. Taking the uncertainty in a count N as √N, calculate the percentage uncertainty in this count.
(1)
(e)
Show that the corrected count rate from this reading agrees with the curve.
(1)
(f)
The thickness of each lead sheet is measured with a micrometer screw gauge. Suggest why it is measured at several places on the sheet.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
C = 80 s−1 at x = 0, so find where C = 40 s−1
✓ 1
x = 12 mm (allow 11–13 mm)
✓ 1
Part (b)
at 24 mm C = 20 s−1, half of 40 s−1; equal thicknesses reduce C by the same factor, so the decrease is exponential
✓ 1
Part (c)
μ = ln 2 ÷ 12 = 0.058 mm−1
✓ 1
Part (d)
√839 = 29, and 29 ÷ 839 × 100 = 3.5%
✓ 1
Part (e)
839 ÷ 100 − 0.45 = 7.94 s−1, which matches the curve's value of about 8 s−1 at 40 mm
✓ 1
Part (f)
the sheet may not be of uniform thickness, so a mean of several readings gives a better value and reduces random error
✓ 1
Exam tip: Counting longer reduces the percentage uncertainty only as 1/√N: four times as many counts halves it. This matters most for the thickest absorbers, where the count rate is lowest.
A tennis ball of mass 0.057 kg is dropped from rest from a height of 1.08 m onto a force plate connected to a data logger. The force–time graph for the impact is approximately a triangle with a peak force of 46 N and a total contact time of 0.020 s. Air resistance is negligible.
Figure: approximate force–time graph for the impact; not to scale
(a)
Calculate the speed of the ball just before it hits the plate.
(2)
(b)
Use the graph to determine the impulse exerted by the plate on the ball.
(2)
(c)
Taking the impulse to be equal to the change of momentum of the ball, calculate the speed at which it rebounds.
(2)
(d)
The weight of the ball acts throughout the contact. Show that ignoring it makes the rebound speed in (c) about 0.2 m s−1 too large.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
v2 = 2gh = 2 × 9.81 × 1.08
✓ 1
v = 4.60 m s−1
✓ 1
Part (b)
impulse = area under the force–time graph = ½ × 46 × 0.020
✓ 1
= 0.46 N s
✓ 1
Part (c)
velocity reverses, so 0.46 = 0.057(v + 4.60)
✓ 1
v = 8.07 − 4.60 = 3.47 m s−1
✓ 1
Part (d)
impulse of the weight = 0.057 × 9.81 × 0.020 = 0.011 N s, which acts against the plate's push
✓ 1
resultant impulse 0.449 N s gives v = 0.449 ÷ 0.057 − 4.60 = 3.27 m s−1, about 0.2 m s−1 smaller
✓ 1
Exam tip: When a body bounces, the change in velocity is the sum of the two speeds. Write the equation with signs — taking upwards as positive, Δv = v − (−u) — before substituting.
Two identical snooker balls are placed on a smooth, level table. Ball P is rolled into stationary ball Q so that they collide off-centre and move off at different angles. A video camera and analysis software are available.
(a)
Describe how the student should record and analyse the collision to test whether momentum is conserved. Your answer should include the measurements to be made and how the results are used.
(6)
Show mark scheme
Answer
Mark
Notes
Indicative content — one mark for each point made:
—
camera mounted on a stand directly above the table, looking vertically down, with the whole collision in view
✓ 1
a ruler or scale placed on the table in the plane of the balls to calibrate distances
✓ 1
step through the video frame by frame, recording the position of the centre of each ball; time between frames = 1 ÷ frame rate
✓ 1
velocity of each ball = displacement over several frames ÷ time, measured just before and just after the collision
✓ 1
resolve each velocity into components along and perpendicular to P's original direction
✓ 1
since the masses are equal, check that the sum of the velocity components is the same before and after in each direction
✓ 1
Exam tip: In a two-dimensional collision, momentum must be checked in two perpendicular directions separately. One direction alone does not test conservation.
A student determines the specific latent heat of fusion of ice. Crushed ice is packed round an immersion heater in a funnel; an identical funnel of ice without a heater stands beside it. A joulemeter shows that the heater supplies 15.2 kJ ± 0.1 kJ. In the same time, 55.6 g of water collects below the heated funnel and 9.3 g below the other. The balance reads to ±0.1 g. The accepted value is 3.34 × 105 J kg−1.
(a)
State the purpose of the funnel without a heater.
(1)
(b)
Calculate the specific latent heat of fusion of ice.
(2)
(c)
Calculate the absolute uncertainty in your value of L.
(3)
(d)
Deduce whether the result agrees with the accepted value and suggest a reason for any difference.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
it measures the mass of ice melted by energy from the surroundings, which is subtracted from the mass melted in the heated funnel
✓ 1
Part (b)
Δm = 55.6 − 9.3 = 46.3 g
✓ 1
L = 15 200 ÷ 0.0463 = 3.28 × 105 J kg−1
✓ 1
Part (c)
uncertainty in Δm = 0.1 + 0.1 = ±0.2 g, which is 0.43%
✓ 1
uncertainty in energy = 0.1 ÷ 15.2 = 0.66%; total = 0.43 + 0.66 = 1.1%
percentage difference = (3.34 − 3.28) ÷ 3.34 × 100 = 1.7%, larger than the 1.1% uncertainty, so it does not agree
✓ 1
the measured value is too small because extra water reached the beaker — e.g. water already on the surface of the ice drained through, or the heated funnel gained more energy from the room than the control
✓ 1
Exam tip: When a mass is found by subtracting two readings, add the absolute uncertainties first (±0.1 g + ±0.1 g), then turn the total into a percentage of the difference.
The corrected count rate C of gamma radiation passing through lead of thickness x is expected to follow C = C0e−μx, where μ is a constant. A student plots ln (C / s−1) against x. The best-fit line passes through (4.0 mm, 4.12) and (24.0 mm, 2.86).
Figure: the student's best-fit line of ln (C / s⁻¹) against x
(a)
Explain why the graph should be a straight line.
(2)
(b)
Determine μ.
(2)
(c)
Calculate the half-value thickness of lead for this radiation.
(1)
(d)
Calculate the thickness of lead that would reduce the corrected count rate to 5.0% of its value with no lead.
(2)
(e)
Explain why the background count rate must be subtracted before the logarithms are taken.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
taking natural logs: ln C = ln C0 − μx
✓ 1
this has the form y = c + mx with ln C0 and μ constant, so the graph is straight with gradient −μ
✓ 1
Part (b)
gradient = (2.86 − 4.12) ÷ (24.0 − 4.0)
✓ 1
= −0.063 mm−1, so μ = 0.063 mm−1 (63 m−1)
✓ 1
Part (c)
ln 2 ÷ 0.063 = 11 mm (11.0 mm)
✓ 1
Part (d)
0.050 = e−μx, so x = ln 20 ÷ μ
✓ 1
= 3.00 ÷ 0.063 = 48 mm
✓ 1
Part (e)
the background does not fall exponentially with x; if it is included, the large-thickness points lie above the line and the graph curves
✓ 1
Common error: Reading the gradient of a log graph and forgetting the sign. The line falls, so the gradient is negative; μ is minus the gradient and is positive.
A trolley of mass 0.600 kg carries a card 5.00 cm long. It runs along a friction-compensated track, passes through a light gate and hits a force sensor fitted with a spring buffer. It rebounds and passes back through the same light gate. The card blocks the light gate for 0.0800 s before the collision and for 0.2000 s after it. Figure 1 shows the force-sensor readings, taken every 5 ms by a data logger, during the collision.
(a)
Calculate the speed of the trolley before and after the collision.
(2)
(b)
Calculate the magnitude of the change in momentum of the trolley.
(2)
(c)
Use the graph to determine the impulse of the force on the trolley.
(2)
(d)
Calculate the mean force on the trolley during the collision.
(1)
(e)
The spring buffer is replaced by a rigid block, and the contact now lasts about 4 ms. Explain why the impulse found with the same data-logger settings would be unreliable.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
before: 0.0500 ÷ 0.0800 = 0.625 m s−1
✓ 1
after: 0.0500 ÷ 0.2000 = 0.250 m s−1
✓ 1
Part (b)
the velocity reverses, so Δv = 0.250 − (−0.625) = 0.875 m s−1
✓ 1
Δp = 0.600 × 0.875 = 0.525 kg m s−1
✓ 1
Part (c)
impulse = area under the graph = ½ × (45 + 25) × 10−3 × 15.0
✓ 1
= 0.525 N s, equal to the change in momentum
✓ 1
Part (d)
0.525 ÷ 0.045 = 11.7 N
✓ 1
Part (e)
readings are taken only every 5 ms, so there would be at most one reading during the contact and the shape and area of the force–time graph could not be found; a much higher sampling rate is needed
✓ 1
Common error: Subtracting the speeds when a trolley bounces back. Velocity is a vector: with the original direction positive, the change is +0.250 − (−0.625), so the speeds add.
Puck A, of mass 0.060 kg, glides across an air table and hits puck B, of mass 0.040 kg, which is at rest. A camera fixed directly above the table films the collision. Figure 1 shows the positions of the centres of the pucks, on a 1 cm grid, in images taken every 0.10 s. The collision happens at image 4.
(a)
Determine the velocity of A before the collision.
(1)
(b)
Determine the x- and y-components of the velocities of A and B after the collision.
(2)
(c)
Show that momentum is conserved in both the x- and y-directions.
(2)
(d)
Calculate the percentage of the kinetic energy that is lost in the collision.
(2)
(e)
Calculate the angle between the paths of A and B after the collision.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
8 cm every 0.10 s along the x-axis = 0.80 m s−1 in the +x direction
✓ 1
Part (b)
A: 4 cm and 2 cm every 0.10 s, so vx = 0.40 m s−1, vy = 0.20 m s−1
✓ 1
B: 6 cm and −3 cm every 0.10 s, so vx = 0.60 m s−1, vy = −0.30 m s−1
✓ 1
Part (c)
x: before 0.060 × 0.80 = 0.048 kg m s−1; after 0.060 × 0.40 + 0.040 × 0.60 = 0.024 + 0.024 = 0.048 kg m s−1
✓ 1
y: before 0; after 0.060 × 0.20 + 0.040 × (−0.30) = 0.012 − 0.012 = 0
Exam tip: Kinetic energy is a scalar: square the full speed, v2 = vx2 + vy2, and add the energies. Momentum is a vector and must be added component by component.
Air is trapped in a cylinder by a piston. A gauge connected to the cylinder reads the amount by which the pressure of the air exceeds atmospheric pressure (the gauge reading). The volume V of the air is reduced in steps at a constant temperature of 20 °C. Figure 1 shows the gauge reading plotted against 1/V, with the best-fit line extended back to 1/V = 0.
(a)
Show that, if the air obeys Boyle's law, the graph should be a straight line that meets the vertical axis at minus atmospheric pressure.
(2)
(b)
Use the graph to determine atmospheric pressure.
(1)
(c)
Use the labelled points to determine the gradient of the line, and express it in joules.
(2)
(d)
Calculate the amount of air, in mol, in the cylinder.
(1)
(e)
Explain why the student waits for about a minute after each change of volume before taking the reading.
(1)
(f)
Suggest one safety precaution for this experiment.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the true pressure p = gauge reading + patm, and Boyle's law gives p = k/V
✓ 1
so gauge reading = k(1/V) − patm: a straight line of gradient k with intercept −patm
compressing the air does work on it and warms it; waiting lets it return to room temperature so the temperature is the same for every reading
✓ 1
Part (f)
wear eye protection / use a safety screen, because the cylinder is under pressure and could burst; do not exceed the rated pressure of the apparatus
✓ 1
Common error: Treating a gauge reading as the pressure of the gas. A gauge that reads zero when open to the air shows pressure above atmospheric, so p = reading + patm must be used in pV = nRT.
A spring hangs from a vibration generator driven by a signal generator. A load is hung on the spring, and the student varies the driving frequency f and measures the amplitude of oscillation of the load. This is done for loads of 0.200 kg and 0.500 kg, and for an object of unknown mass X. Figure 1 shows the results.
(a)
Use the graph to state the resonant frequency for each load.
(1)
(b)
Part of the spring, of effective mass ms, oscillates with the load. Show that 1/f2 = (4π2/k)(m + ms), where m is the load, k the spring constant and f the resonant frequency.
(1)
(c)
Use the results for the two known loads to calculate k.
(2)
(d)
Calculate ms.
(1)
(e)
Determine the mass of X.
(2)
(f)
Suggest how the experiment could be improved to give a more reliable value for the mass of X.
at resonance f equals the natural frequency, f = (1/2π)√(k/(m + ms)); squaring and inverting gives the result
✓ 1
Part (c)
1/f2 = 0.2066 s2 and 0.4756 s2
✓ 1
4π2/k = (0.4756 − 0.2066) ÷ 0.300 = 0.8967 s2 kg−1, so k = 4π2 ÷ 0.8967 = 44.0 N m−1
✓ 1
Part (d)
ms = 0.2066 ÷ 0.8967 − 0.200 = 0.0304 = 0.030 kg
✓ 1
Part (e)
1/1.702 = 0.3460 s2
✓ 1
m = 0.3460 ÷ 0.8967 − 0.0304 = 0.355 kg (unrounded value 0.35547 kg; 0.356 kg from intermediate values rounded to 0.897 and 0.030 is also correct; allow 0.345–0.365 kg, the range from reading the peak of X to ±0.02 Hz)
✓ 1
Part (f)
use five or more known loads and plot 1/f2 against m; read the mass of X from the best-fit line, so a random error in any one frequency has less effect
✓ 1
Common error: Taking f ∝ 1/m or ignoring the mass of the spring. The resonant frequency depends on 1/√(m + ms), so use 1/f2, which is linear in m.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.