Paper WPH16/01, worked through. Edexcel International A Level Physics, January 2025: worked solutions

Fifty marks in an hour and twenty minutes, so a little over a minute and a half a mark. Nothing is done at a bench: the paper tests whether you can plan an experiment, take a sensible reading, draw a graph and handle uncertainties. The same few ideas come up every sitting, and most of the marks here go to people who say exactly what they would do and why.

1. 1(a)Spotting one hazard when water is heated electrically in a glass beaker with a metal flask sitting in it, and saying what to do about it.[2]
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   1. A safety answer has two halves, a mark each: what could hurt someone, and the particular thing you would do about it.
   2. The hazard in plain sight is heat. The heater, the water, the beaker and the flask all get hot enough to burn skin.
   3. Now match the precaution to that hazard: switch the heater off before moving it, or handle the hot things with tongs or heat-resistant gloves.

   AnswerThe heater (or the beaker, the flask or the water) becomes hot and can burn, so switch off before touching it and move it with tongs or insulated gloves. The other accepted pair: the heater's leads could pull the beaker over, so clamp the heater in position.

   "Be careful" and "wear goggles" earn nothing. The precaution has to deal with the hazard you named, in this experiment.

   The gloves must be heat-resistant ones. Ordinary lab gloves do not stop a burn, and gloves are not how anyone avoids an electric shock.

2. 1(b)How to get one trustworthy temperature for the air sealed inside the flask, which cannot be measured directly.[3]
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   1. The air is shut in, so measure the water around the flask and make sure the water and the air are at the same temperature.
   2. Put a thermometer in the water, close to the flask and on the far side from the heater, so it reads the water the flask is actually sitting in rather than the hotter water beside the element.
   3. Stir the water, and give it a moment, so the water, the metal and the air inside all settle at one temperature before you read.
   4. Read the scale with your eye level with the top of the thread, square on to the scale.

   AnswerA thermometer in the water, placed next to the flask and away from the heater; stir so everything reaches the same temperature; read the scale at eye level.

   Three marks, three things: the instrument, where it goes, and what makes the reading one you can trust (stirring, or reading square on).

3. 1(c)(i)Averaging four repeated temperature readings.[1]
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   1. Add them: 42.5 + 41.0 + 42.0 + 43.5 = 169.0.
   2. Divide by four: 42.25.
   3. The readings were taken to one decimal place, so the mean is given to one decimal place.

   Answer42.3 °C

   Writing 42.25 loses the mark. A mean cannot be quoted more finely than the readings it came from.

   Revise this: Errors and uncertainties
4. 1(c)(ii)Turning the spread of those repeats into a percentage uncertainty.[2]
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   1. For repeated readings, the uncertainty is half the range: (43.5 − 41.0) ÷ 2 = 1.25 °C.
   2. As a percentage of the mean: 1.25 ÷ 42.3 × 100 = 2.96%.
   3. An uncertainty is quoted to one or two significant figures.

   Answer3%

   Half the range, not the whole range, and not the smallest division on the thermometer. The readings are scattered by far more than the scale could explain.

   Revise this: Errors and uncertainties
5. 1(c)(iii)Counting the molecules of air in a spherical flask from its width, its pressure and its temperature.[4]
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   1. Use pV = NkT, rearranged to N = pV ÷ kT. Everything has to be in SI units before it goes in.
   2. Temperature in kelvin: 42.3 + 273 = 315.3 K.
   3. The radius is half the diameter: 7.5 cm, which is 0.075 m. Volume of a sphere: V = 4/3 × π × r³ = 4/3 × π × 0.075³ = 1.77 × 10⁻³ m³.
   4. Pressure in pascals: 110 kPa = 1.10 × 10⁵ Pa.
   5. N = (1.10 × 10⁵ × 1.77 × 10⁻³) ÷ (1.38 × 10⁻²³ × 315.3) = 4.47 × 10²².

   Answer4.5 × 10²² molecules

   Three conversions are waiting for you: °C to kelvin, diameter to radius, and centimetres to metres. Each one missed puts the answer out by a large factor.

   It asks for a number of molecules, so reach for pV = NkT with the Boltzmann constant. Both are on the sheet at the back of the paper.

   Revise this: The ideal gas equation
6. 2(a)Finishing a circuit that lets the power going into a filament lamp be changed and measured.[2]
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   1. To change the power you need a supply and a way of varying the current: a variable power supply, or a fixed one with a variable resistor in series with the lamp.
   2. To find the power you need the current through the lamp and the potential difference across it. The ammeter goes in series with the lamp. The voltmeter goes in parallel, across the lamp alone.

   AnswerA supply and a variable resistor (or a variable supply) in series with the lamp and an ammeter, and a voltmeter connected across the lamp.

   The voltmeter goes across the lamp, not across the supply or the variable resistor. It is the lamp's power you are after.

   Revise this: Potential difference and power
7. 2(b)Planning an experiment to test a prediction that the power supplied to a lamp is proportional to the fourth power of the reading on a light sensor.[6]
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   1. Keep it a fair test. The sensor stays the same distance from the lamp, pointing the same way, for every reading.
   2. Keep other light out. Work in a darkened room or shield the sensor, or take the background reading and subtract it.
   3. At each setting record three things: the current through the lamp, the potential difference across it, and the sensor reading X.
   4. Work out the power each time from P = IV.
   5. Change the current and repeat, until you have at least five pairs of P and X spread over a wide range.
   6. Plot P against X⁴. If the prediction is right, the points lie on a straight line through the origin. Or plot log P against log X, which should be a straight line of gradient 4.

   AnswerSensor at a fixed distance and angle; no stray light; measure I and V and calculate P = IV; at least five values of P with their X; then P against X⁴ (straight, through the origin) or log P against log X (straight, gradient 4).

   "Plot a graph" earns nothing by itself. Say what goes on each axis and what the graph would look like if the prediction held.

   The log graph is the stronger test. Its gradient is the power, so you find out whether it is 4 instead of assuming so.

   Revise this: Potential difference and power
8. 3(a)Why the amplitude of a swinging pendulum cannot sensibly be read to the nearest millimetre.[2]
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   1. One reason is about time. The bob is at its furthest point only for an instant before it turns back, so you cannot be sure you read the rule at exactly that moment.
   2. The other is about position. The cone hangs above the rule, not against it, so unless your eye is directly in line the reading shifts. The swing may also wander out of line with the rule.

   AnswerIt is hard to judge the instant of greatest displacement, because the pendulum reverses quickly; and the cone is not close to the rule, so it is hard to line the eye up at right angles to the scale.

   Two reasons means two different ideas. "It is moving" said twice in different words is one mark.

   Revise this: Uncertainty in a measurement
9. 3(b)(i)Showing how a logarithmic graph turns an exponential fall in amplitude into a straight line, and where the decay constant is on it.[2]
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   1. Take natural logs of both sides of the equation given: ln A = ln A₀ − λn.
   2. Set that beside y = mx + c. Here y is ln A and x is n, so the graph is a straight line with gradient −λ and intercept ln A₀.

   Answerln A = ln A₀ − λn has the form y = c + mx, so a graph of ln A against n is a straight line whose gradient is −λ. The value of λ is the gradient with its sign changed.

   The second mark depends on the first. Write the log equation out before you compare it with y = mx + c.

   Revise this: Damping and resonance
10. 3(b)(ii)Working out ln A for six amplitude readings and plotting them against the number of swings.[5]
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    1. Fill in the spare column, every value to the same number of decimal places: ln 8.5 = 2.140, ln 7.0 = 1.946, ln 5.5 = 1.705, ln 5.0 = 1.609, ln 4.0 = 1.386, ln 3.5 = 1.253.
    2. Label the axes: ln (A / cm) up the side, n along the bottom. A logarithm has no unit, which is why the unit sits inside the bracket.
    3. Pick scales that are easy to read and spread the points over more than half the grid. The grid is 8 large squares wide and 12 tall, so n from 0 to 40 at 5 a square, and ln A from 1.2 to 2.4 at 0.1 a square, fits exactly.
    4. Plot each point to within half a small square, then rule one thin straight line of best fit with the points scattered evenly either side of it.

    AnswerA column of ln A values (2.140, 1.946, 1.705, 1.609, 1.386, 1.253), both axes labelled, sensible scales, six points plotted accurately and one straight best-fit line sloping downwards.

    The points will not sit exactly on a line, because the amplitudes were only read to the nearest 5 mm. Do not join the dots.

    Starting the n axis at zero costs nothing here and lets you read the intercept straight off the graph two parts later.

    Scales that go up in threes or sevens lose the scale mark and invite plotting slips.

    Revise this: Damping and resonance
11. 3(b)(iii)Finding the decay constant from the gradient of the log graph.[3]
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    1. Draw a large triangle on the best-fit line, using two points on the line itself that are far apart.
    2. For example, the line passes close to (5.5, 2.10) and (30, 1.23). Gradient = (1.23 − 2.10) ÷ (30 − 5.5) = −0.87 ÷ 24.5 = −0.0355.
    3. λ is the gradient with its sign changed, so it is positive. It has no unit, because n is only a count.

    Answerλ ≈ 0.036. Anything from 0.0335 to 0.0384 is accepted, given to two or three significant figures.

    A small triangle loses the first mark. So does one drawn between two plotted points that are not on your line.

    Give λ as a positive number with no unit.

    Revise this: Damping and resonance
12. 3(b)(iv)Finding the starting amplitude from where the log graph meets the vertical axis.[3]
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    1. Where the line crosses the ln A axis, at n = 0, the value is ln A₀. Read it off: about 2.30.
    2. If your axis does not start at n = 0, use the gradient and a point on the line instead: ln A₀ = 2.10 + 0.0355 × 5.5 = 2.295.
    3. Undo the logarithm: A₀ = e^2.295 = 9.9.
    4. The amplitudes were in centimetres, so this is too, to the nearest millimetre.

    AnswerA₀ ≈ 9.9 cm

    The intercept is ln A₀, not A₀. Forgetting to raise e to that power is the usual slip.

    Revise this: Damping and resonance
13. 3(b)(v)Why the starting amplitude read from the graph may not be the real one.[2]
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    1. The plotted points are scattered about the line, so the best-fit line could fairly be drawn a little differently, and the intercept would move with it.
    2. Another route: every amplitude might be misread the same way (by looking from one side, say), which shifts the whole line up or down.
    3. A third: the first swings may have been too wide for the simple theory to hold, so running the straight line back to n = 0 is not justified.

    AnswerOne linked pair. For example: the data are scattered, so the position of the best-fit line, and with it the intercept, is uncertain.

    Two marks for a reason and what follows from it. What follows only counts if the reason is there.

    Revise this: Errors and uncertainties
14. 4(a)Why the difference between two heights measured with a metre rule is uncertain by a whole millimetre.[3]
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    1. A metre rule is marked in millimetres. That is its resolution: 1 mm.
    2. The uncertainty in a single reading is half the resolution: 0.5 mm.
    3. The height difference comes from two readings, one taken from the other. Uncertainties add even when the readings are subtracted: 0.5 + 0.5 = 1 mm.

    AnswerThe rule's resolution is 1 mm, so each height is uncertain by 0.5 mm. Two heights are used, and their uncertainties add to give 1 mm.

    Say "resolution". "Precision" and "accuracy" are not accepted for it.

    Subtracting two readings never makes the uncertainty smaller.

    Revise this: Uncertainty in a measurement
15. 4(b)How timing the sphere over a longer stretch of ramp changes the percentage uncertainty in the time.[3]
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    1. A longer distance takes the sphere longer to cover, so the time measured is larger.
    2. The uncertainty in each timing comes from the person and the stopwatch, so it stays about the same.
    3. Percentage uncertainty is the uncertainty divided by the value. The same uncertainty on a bigger value is a smaller percentage.

    AnswerThe time increases, the absolute uncertainty in it stays the same, so the percentage uncertainty in the time decreases.

    Keep the two kinds of uncertainty apart in your answer. One stays the same and the other falls.

    Revise this: Uncertainty in a measurement
16. 4(c)(i)Calculating g from the time a sphere takes to roll down the ramp, using the formula the question gives.[2]
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    1. Rearrange the formula to make g the subject: g = 14s² ÷ (5t²Δh).
    2. Put the lengths into metres: s = 0.900 m and Δh = 0.021 m.
    3. g = 14 × 0.900² ÷ (5 × 3.36² × 0.021) = 11.34 ÷ 1.185 = 9.57.

    Answerg = 9.6 m/s² (9.57 m/s² is accepted too)

    Both lengths need converting, and they start in different units: one in centimetres, the other in millimetres.

17. 4(c)(ii)Combining the percentage uncertainties in a distance, a height and a time to get the one in g.[3]
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    1. Find each percentage. Distance: 0.1 ÷ 90.0 × 100 = 0.11%. Height difference: 1 ÷ 21 × 100 = 4.76%. Time: 0.03 ÷ 3.36 × 100 = 0.89%.
    2. In g = 14s² ÷ (5t²Δh) the distance and the time are squared, so their percentages are doubled. Then everything is added, whether a quantity is on the top or the bottom.
    3. 2 × 0.11 + 4.76 + 2 × 0.89 = 0.22 + 4.76 + 1.78 = 6.76%.

    Answer6.8%, which is about 7%

    In a "show that", finish with one more significant figure than the number you were asked for: 6.8%, not just 7%.

    Nearly all of it comes from the height difference, a small length measured with a metre rule. That is the measurement worth improving.

    Revise this: Errors and uncertainties
18. 4(c)(iii)Deciding whether the measured g agrees with the accepted value once its uncertainty is taken into account.[2]
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    1. Work out how far the result could stretch. 6.8% of 9.57 is 0.65, so g lies between about 8.9 and 10.2 m/s².
    2. The accepted value, 9.81 m/s², is inside that range.
    3. The same thing by percentages: the difference from 9.81 is (9.81 − 9.57) ÷ 9.81 × 100 = 2.4%, which is less than the 6.8% uncertainty.

    AnswerYes, it is accurate: 9.81 m/s² lies within the range of the student's value, whose upper limit is about 10.2 m/s².

    "Deduce" wants a number and then a conclusion. "Yes, it is close" with nothing to back it scores nothing.

    Revise this: Errors and uncertainties

These explanations are Papermunch’s own, written to teach the method. The answers have been checked against the exam board’s mark scheme, which has the final say: open it above. The question paper and mark scheme belong toPearson Education: each question shown here is drawn from the paper itself as you read, and is not kept on this site.