Paper WPH11/01, worked through. Edexcel International A Level Physics, June 2025: worked solutions

Eighty marks in an hour and a half: a little over a minute a mark. Section A is ten one-mark multiple-choice questions, so aim to be through it in about twelve minutes. In Section B, write down the equation you are using before you put numbers in: the working earns marks even when the final answer goes wrong.

1. 1What the upthrust on a cube sitting in a liquid is equal to.[1]
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   1. Archimedes' principle: the upthrust on an object equals the weight of the fluid it displaces.
   2. The question tells you the weight of liquid displaced, so that is the upthrust. Nothing needs calculating.
   3. Check the other options by their units. Volume × density is a mass, not a force, and a volume on its own certainly isn't a force.

   AnswerD: the weight of liquid displaced.

   When a question gives you more quantities than you need, don't feel you have to use them all. Checking units is the quickest way to throw out wrong options.

   Revise this: Density and upthrust
2. 2The resultant force and resultant moment on a see-saw that is balanced.[1]
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   1. The beam is balanced and still, so it is in equilibrium. That needs two things: no resultant force and no resultant moment.
   2. Force: the weights pull down, and the support pushes up by exactly the same amount. Resultant force: zero.
   3. Moment: the two children weigh the same and sit the same distance from the support, so one turns the beam clockwise exactly as hard as the other turns it anticlockwise. Resultant moment: zero.

   AnswerA: both are zero.

   Don't forget the support. It is tempting to say the resultant force is downwards because the weights act downwards, but the support's upward push balances them.

   Revise this: Equilibrium and the principle of moments
3. 3How long a ball takes to fall from a table when its mass and its sideways speed are both doubled.[1]
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   1. A ball that rolls off a table is a projectile: its sideways motion and its downward motion have nothing to do with each other.
   2. Downwards, both balls start with no vertical velocity, fall the same height and have the same acceleration, g. So they take the same time.
   3. Mass makes no difference to g, and sideways speed makes no difference to the fall.

   AnswerB: the same time, t.

   The faster ball lands further from the table, not sooner. Time in the air is decided by the vertical motion alone.

   Revise this: Projectiles
4. 4Why a seated person's weight and the push of the chair on them are not a Newton's third law pair.[1]
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   1. A third law pair always acts on two different objects: A pushes on B, and B pushes back on A.
   2. Here both forces act on the same object, the person. That alone rules them out.
   3. The forces in a third law pair are equal and opposite too, so "opposite directions" and "same magnitude" can't be the reason these two fail.

   AnswerC: both forces act on the same object.

   The third law partner of the person's weight is the person's gravitational pull on the Earth. Two forces that balance on one object are never a third law pair.

   Revise this: Newton's laws
5. 5The vector equation linking the two tensions in a picture wire and the picture's weight.[1]
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   1. The picture hangs at rest, so it is in equilibrium and the resultant force on it is zero.
   2. With vectors, "resultant is zero" means all the forces added together come to zero: weight + first tension + second tension = 0.
   3. The direction of each force is already built into its arrow, so nothing is subtracted.

   AnswerB: the three vectors add up to zero.

   Minus signs are for when you are working with sizes along one line. When the forces are written as vectors, you simply add them all.

   Revise this: Equilibrium and the principle of moments
6. 6Choosing the displacement–time graph of something with a steady, non-zero acceleration.[1]
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   1. The gradient of a displacement–time graph is the velocity.
   2. A steady acceleration means the velocity changes steadily, so the gradient must change smoothly the whole way along. That is a smooth curve (a parabola, since s = ut + ½at²).
   3. Any straight section means a constant velocity, which is zero acceleration.

   AnswerB: the smoothly curving graph.

   A straight line on a displacement–time graph means no acceleration, however steep it is.

   Revise this: Graphs of motion
7. 7The expression for the resultant of three forces on a helicopter: two along the vertical and one horizontal.[1]
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   1. First deal with the two forces that lie along the same line. They are opposite, so they give 15 000 − 14 500 = 500 N.
   2. That 500 N and the 200 N sideways force are at right angles, so use Pythagoras: resultant = √(500² + 200²).
   3. Written with the original numbers: √((15 000 − 14 500)² + 200²).

   AnswerD: the square root of (15 000 − 14 500)² + 200².

   Subtract before you square. Squaring the two big forces separately and then subtracting gives a very different, wrong number.

   Revise this: Scalars and vectors
8. 8The size of the friction on a box sliding down a ramp at a steady speed.[1]
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   1. Steady speed means no acceleration, so the forces along the ramp balance.
   2. The part of the weight that acts down the ramp is W sin θ.
   3. Friction acts up the ramp and must be exactly as big: F = W sin θ.

   AnswerC: F = W sin θ.

   W cos θ is the part of the weight pressing into the ramp. If you can't remember which is which, picture a flat ramp (θ = 0): nothing pulls the box along, and sin 0 = 0.

   Revise this: Scalars and vectors
9. 9Finding a braking force from the average power, the stopping distance and the time.[1]
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   1. Energy transferred = power × time = 50 × 10³ × 2.3.
   2. That energy is the work done by the friction force: work = force × distance = F × 13.
   3. So F = (50 × 10³ × 2.3) ÷ 13.

   AnswerA: (50 × 10³ × 2.3) ÷ 13.

   Two equations joined together: P = W ÷ t and W = Fs. Power × time gives work; power × distance gives nothing useful.

   Revise this: Work, power and efficiency
10. 10The speed of a toy train and truck after they collide head-on and move off together.[1]
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    1. Take the train's direction as positive. Momentum before = (3m × v) + (m × −2v) = 3mv − 2mv = mv.
    2. Afterwards they move as one object of mass 4m. Momentum is conserved: 4m × new speed = mv.
    3. New speed = v ÷ 4.

    AnswerA: v ÷ 4.

    The truck is moving the other way, so its momentum is negative. Forget the sign and you get 5v ÷ 4, which is one of the options waiting for you.

    Revise this: Momentum and its conservation
11. 11Working out the weight of a boat from the resultant force on it and its acceleration.[3]
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    1. The resultant force and the acceleration give the mass: F = ma, so m = 4800 ÷ 0.31 = 15 500 kg.
    2. Weight is mass × g: W = 15 500 × 9.81.
    3. = 1.5 × 10⁵ N.

    Answer1.5 × 10⁵ N (150 000 N).

    Two steps, and the question only names the second. The 4800 N is the resultant force, not the weight: use it to find the mass first.

    Revise this: Newton's laws
12. 12(a)Why the displacement at the end of a winding ride is smaller than the distance cycled.[2]
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    1. Distance is the whole length of the path actually followed.
    2. Displacement is the straight line from the start to the finish.
    3. The path bends, so the straight line between its two ends is shorter than the path itself.

    AnswerDisplacement is the straight-line distance from start to end. Distance is the total length of the path, and the path is not straight, so it is longer.

    Saying only "displacement is a vector and distance is a scalar" earns one mark at most. Say what each one measures.

    Revise this: Scalars and vectors
13. 12(b)Finding the greatest acceleration from a velocity–time graph made of straight sections.[3]
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    1. Acceleration is the gradient of a velocity–time graph, so the greatest acceleration is the steepest section: between 4 s and 16 s.
    2. Read off both ends of that section: 1.8 m/s at 4 s, and 5.0 m/s at 16 s.
    3. Gradient = (5.0 − 1.8) ÷ (16 − 4) = 3.2 ÷ 12 = 0.27 m/s².

    Answer0.27 m/s².

    Use the whole of the steepest section, not a small piece of it: a bigger triangle gives a more reliable gradient. Don't divide a velocity by a time (5.0 ÷ 16): that isn't a gradient unless the line starts at the origin.

    Revise this: Graphs of motion
14. 13Checking the height of a basketball hoop from the ball's vertical speed when thrown and when it drops in.[3]
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    1. Only the vertical motion matters. Take upwards as positive: u = +5.1 m/s, v = −2.1 m/s (it is falling into the hoop) and a = −9.81 m/s².
    2. There is no time given, so use v² = u² + 2as: s = (2.1² − 5.1²) ÷ (2 × −9.81) = (−21.6) ÷ (−19.62) = 1.1 m.
    3. The ball enters the hoop 1.1 m above where it was thrown: 1.7 + 1.1 = 2.8 m above the floor. That is not 3.0 m.

    AnswerNo: the hoop was about 2.8 m above the floor, not 3.0 m.

    "Deduce" means finish with a comparison and a conclusion. A correct 2.8 m with no "so it was not 3.0 m" loses the last mark.

    Revise this: Projectiles
15. 14The efficiency of charging a battery from the brakes of an electric car that stops while going uphill.[4]
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    1. Kinetic energy at the start = ½mv² = ½ × 1800 × 14² = 176 000 J.
    2. As it stops, the car climbs 0.76 m, so some of that energy becomes gravitational potential energy: mgΔh = 1800 × 9.81 × 0.76 = 13 400 J.
    3. Energy the brakes had to take away = 176 000 − 13 400 = 163 000 J. That is the total input to the braking.
    4. Efficiency = useful output ÷ total input = 45 000 ÷ 163 000 = 0.28.

    Answer0.28 (28%).

    The hill is the catch. Not all the kinetic energy goes through the brakes: part of it lifts the car. Also change 45 kJ to 45 000 J before dividing.

    Revise this: Work, power and efficiency
16. 15(a)The extra elastic strain energy stored in a brick as the force on it rises, read from a force–compression graph.[3]
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    1. The energy stored is the area under the force–compression graph. The line is straight through the origin, so up to any point the area is a triangle: ½ × force × compression.
    2. Read the compressions from the graph: about 60 nm at 85 N, and about 97.5 nm at 140 N.
    3. At 140 N: ½ × 140 × 97.5 × 10⁻⁹ = 6.83 × 10⁻⁶ J. At 85 N: ½ × 85 × 60 × 10⁻⁹ = 2.55 × 10⁻⁶ J.
    4. Increase = 6.83 × 10⁻⁶ − 2.55 × 10⁻⁶ = 4.3 × 10⁻⁶ J.

    Answer4.3 × 10⁻⁶ J.

    It is the difference between two triangles, not one small triangle: ½ × (140 − 85) × (97.5 − 60) leaves out the rectangle underneath. And nm means 10⁻⁹ m.

    Revise this: Elastic and plastic behaviour
17. 15(b)What elastic deformation means.[1]
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    1. "Elastic" is about what happens when the force is taken away again.

    AnswerThe brick returns to its original shape and size when the force is removed.

    Mention the force being removed. "It stretches and goes back" without saying when is too vague.

    Revise this: Elastic and plastic behaviour
18. 16(a)What else has to be measured to find the viscosity of an oil with a falling ball.[1]
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    1. At terminal velocity the forces on the ball balance: weight = upthrust + viscous drag.
    2. The drag is where the viscosity comes in (Stokes' law), so to find it you need the weight of the ball, and for that you need its mass.

    AnswerThe mass (or weight) of the ball-bearing. The density of the oil is also accepted.

    The diameter doesn't count as the "other" measurement here: the question goes on to give it, so it has clearly been measured already.

    Revise this: Viscous drag and Stokes' law
19. 16(b)Calculating viscosity from the drag on a ball falling at terminal velocity.[3]
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    1. At terminal velocity the speed is steady, so speed = distance ÷ time = 0.200 ÷ 1.3 = 0.154 m/s.
    2. Stokes' law needs the radius, not the diameter: r = 1.24 mm ÷ 2 = 0.62 × 10⁻³ m.
    3. Rearrange F = 6πηrv: η = F ÷ (6πrv) = 5.6 × 10⁻⁵ ÷ (6π × 0.62 × 10⁻³ × 0.154).
    4. = 0.031 Pa s.

    Answer0.031 Pa s.

    Three easy slips in one question: forgetting to halve the diameter, leaving 20.0 cm in centimetres, and leaving the radius in millimetres.

    Revise this: Viscous drag and Stokes' law
20. 16(c)How colder oil changes the time the ball takes to fall.[3]
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    1. Colder oil is more viscous: viscosity goes up as the temperature goes down.
    2. With a higher viscosity, the drag needed to balance the ball's weight is reached at a lower speed, so the terminal velocity is smaller.
    3. A smaller speed over the same 20.0 cm means a longer time.

    AnswerThe time is longer: the oil's viscosity is greater at the lower temperature, so the terminal velocity is smaller.

    Three marks, three links in the chain: viscosity, then velocity, then time. Jumping straight to "it takes longer" earns very little.

    Revise this: Viscous drag and Stokes' law
21. 17(a)Working out which violin string a force–extension graph belongs to, from its stiffness.[4]
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    1. Stiffness is force ÷ extension, the gradient of the straight line.
    2. Read a point as far up the line as you can: 80 N at an extension of 29 mm, which is 0.029 m.
    3. k = 80 ÷ 0.029 = 2760 N/m.
    4. That is closest to 2700 N/m, so it is string Y.

    AnswerString Y: the graph gives a stiffness of about 2760 N/m, which matches 2700 N/m.

    Change millimetres to metres, or your stiffness is a thousand times too small. Finish by saying which value yours matches: the comparison is a mark of its own.

    Revise this: Hooke's law and springs
22. 17(b)Finding the Young modulus of a string from its tension, its change in length and its radius.[5]
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    1. Cross-sectional area = πr² = π × (0.85 × 10⁻³)² = 2.27 × 10⁻⁶ m².
    2. Stress = force ÷ area = 36 ÷ 2.27 × 10⁻⁶ = 1.59 × 10⁷ Pa.
    3. Extension = 0.752 − 0.750 = 0.002 m, so strain = 0.002 ÷ 0.750 = 2.67 × 10⁻³.
    4. Young modulus = stress ÷ strain = 1.59 × 10⁷ ÷ 2.67 × 10⁻³ = 6.0 × 10⁹ Pa.

    Answer6.0 × 10⁹ Pa (6.0 GPa).

    The extension is the difference between the two lengths, not the new length. And check whether you've been given the radius or the diameter: here it is the radius.

    Revise this: Stress, strain and the Young modulus
23. 18(a)The mass of a steel sphere from its radius and the density of steel.[3]
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    1. Volume of a sphere = (4/3)πr³ = (4/3) × π × (7.0 × 10⁻³)³ = 1.44 × 10⁻⁶ m³.
    2. Mass = density × volume = 7.8 × 10³ × 1.44 × 10⁻⁶.
    3. = 1.1 × 10⁻² kg.

    Answer1.1 × 10⁻² kg (about 11 g).

    The formula for the volume of a sphere isn't on the formula sheet: learn it. Put brackets round the radius on your calculator so the whole thing, power of ten included, is cubed.

    Revise this: Density and upthrust
24. 18(b)The acceleration of a hanging sphere at the moment it is let go with its string at an angle.[2]
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    1. Two forces act on the sphere: its weight, and the tension in the string. The sphere can only move at right angles to the string, and the tension has no part in that direction.
    2. The part of the weight in that direction is mg sin 38°, since the string is at 38° to the vertical.
    3. a = F ÷ m = g sin 38° = 9.81 × sin 38° = 6.0 m/s².

    Answer6.0 m/s².

    The mass cancels, so you don't need your answer to part (a). To choose between sin and cos, try a string hanging straight down (0°): the sphere wouldn't accelerate at all, and sin 0 = 0.

    Revise this: Scalars and vectors
25. 18(c)Describing how the energy and momentum of the first and last spheres of a Newton's cradle change through one swing.[6]
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    1. As sphere A swings down, it loses gravitational potential energy and gains kinetic energy.
    2. It is getting faster, so its momentum increases as it falls.
    3. In the collision A stops, so its momentum drops to zero. E moves off with that momentum: momentum is conserved.
    4. A's kinetic energy is passed on to E as well.
    5. As E swings up, its kinetic energy and its momentum both fall to zero at the top, while its gravitational potential energy rises to a maximum.
    6. With no air resistance the total energy stays the same throughout, so E ends with the same gravitational potential energy that A started with.

    AnswerA turns potential energy into kinetic energy and gains momentum; the collision passes its momentum and kinetic energy to E; E then turns the kinetic energy back into the same amount of potential energy, its momentum falling to zero.

    In a starred question some of the marks are for a logical order. Tell it as a story in time order (falling, colliding, rising) and cover both energy and momentum, for both spheres, at each stage.

    Revise this: Momentum and its conservation
26. 19(a)Stating the principle of moments.[1]
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    1. It is a statement about an object that is balanced: the turning effects one way equal the turning effects the other way.

    AnswerFor an object in equilibrium, the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about that point.

    "Clockwise = anticlockwise" needs the word moments in it, and it needs the word sum (or total).

    Revise this: Equilibrium and the principle of moments
27. 19(b)(i)The smallest push on the end of a hinged rail that will lift it off the ground, with a piston helping.[5]
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    1. Take moments about the pivot. Only the part of each force at right angles to the rail turns it.
    2. Weight: it acts at the middle of the uniform rail, 0.70 m from the pivot. The rail is at 42° to the ground, so the part of the weight at right angles to the rail is 95 cos 42°. Moment = 95 × cos 42° × 0.70 = 49.4 N m, turning the rail down.
    3. Piston: its force is at 18° to the rail, so the part at right angles is 160 sin 18°. Moment = 160 × sin 18° × 0.37 = 18.3 N m, helping to lift.
    4. F is already at right angles to the rail, 1.40 m from the pivot. When the end is just lifting: (F × 1.40) + 18.3 = 49.4.
    5. F = (49.4 − 18.3) ÷ 1.40 = 22 N.

    Answer22 N.

    The two angles are measured from different things. The 18° is between the piston's force and the rail, so use sin for the part at right angles. The 42° is between the rail and the ground, which makes the weight 42° away from the perpendicular to the rail, so use cos.

    Revise this: Moments and centre of gravity
28. 19(b)(ii)How the force needed to lift the rail changes if it is applied straight up, not at right angles to the rail.[3]
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    1. Either way, the student's force has to give the same moment about the pivot.
    2. A vertical force is not at right angles to the rail, so only part of it turns the rail. (Put another way: the perpendicular distance from the pivot to the line of the force is shorter.)
    3. To give the same moment with less turning effect per newton, the force has to be bigger.

    AnswerA larger force is needed. The moment required is the same, but a vertical force has a smaller perpendicular distance from the pivot (only its component at right angles to the rail turns it).

    A force turns a lever best when it is at right angles to it. Any other direction wastes part of the force.

    Revise this: Moments and centre of gravity
29. 20(a)Deciding whether air resistance mattered in the first second and a half of a skydive.[4]
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    1. The skydiver starts with no vertical velocity, so s = ½at², which gives a = 2s ÷ t² = 2 × 11 ÷ 1.5² = 9.78 m/s².
    2. Compare with the test value: 0.95g = 0.95 × 9.81 = 9.32 m/s². (Or as a fraction of g: 9.78 ÷ 9.81 = 0.997.)
    3. 9.78 m/s² is greater than 9.32 m/s², so the mean acceleration was more than 0.95g.

    AnswerYes, air resistance was negligible: the mean acceleration was 9.78 m/s², which is 0.997g and so greater than 0.95g.

    Another "deduce": calculate, compare, conclude. Work out 0.95g as a number (or your acceleration as a fraction of g) so that the two things you compare are the same kind of quantity.

    Revise this: Equations of motion
30. 20(b)Sketching a velocity–time graph from the displacement–time graph of a skydiver in free fall.[2]
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    1. Velocity is the gradient of the displacement–time graph. The displacement is negative (downwards) and getting more negative, so the velocity is negative.
    2. The displacement graph starts flat and gets steeper until about 12 s. So the velocity starts at zero and becomes more and more negative, quickly at first and then more slowly: a curve that flattens out.
    3. After about 12 s the displacement graph is a straight line, so the velocity is constant: a horizontal line.

    AnswerA curve starting at zero and going downwards, getting less steep, then a horizontal line below the time axis from about 12 s onwards.

    The flat part must not lie along the time axis. Constant velocity is not zero velocity: he is falling at terminal velocity.

    Revise this: Graphs of motion
31. 20(c)Following the skydiver's energy from the jump until just before the parachute opens.[4]
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    1. All the way down, his gravitational potential energy is decreasing.
    2. While he is speeding up, his kinetic energy increases. But not all the potential energy lost becomes kinetic energy: some is used doing work against air resistance and is dissipated to the surroundings.
    3. At terminal velocity his kinetic energy is no longer changing, so all of the potential energy he loses is being dissipated by the work done against air resistance.

    AnswerGravitational potential energy falls throughout. While he accelerates it becomes kinetic energy plus energy dissipated by work against air resistance; at terminal velocity kinetic energy is constant, so all the potential energy lost is dissipated.

    At terminal velocity energy is still being transferred: he is still losing height. "No energy change" is the common mistake.

    Revise this: Kinetic and potential energy
32. 20(d)How the forces on the skydiver make his acceleration change after the parachute opens.[5]
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    1. When the parachute opens, the air resistance suddenly becomes much greater than his weight.
    2. So the resultant force is upwards, and his acceleration is upwards: he slows down.
    3. As his speed falls, the air resistance falls, so the resultant force gets smaller and so does the deceleration.
    4. This goes on until the air resistance equals his weight.
    5. Then the resultant force is zero and the acceleration is zero: he has reached a new, lower terminal velocity.

    AnswerAir resistance exceeds weight, giving an upward resultant force and a deceleration; as he slows, air resistance falls, so the deceleration decreases until air resistance equals weight and the acceleration is zero.

    An upward acceleration doesn't mean he moves upwards: he is still falling, just more slowly. Keep "force", "acceleration" and "speed" as three separate ideas and link them in that order.

    Revise this: Falling with air resistance

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.