Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
69 original questions across 13 selected topic areas. No account or payment required to practise.
5 questions in this topic. Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Uncertainty and dimensions · Projectile components · Upthrust and moments · Young modulus · Sound Doppler effect
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Question 1A cylindrical specimen has mass (8.0 +/- 0.1) g, radius (1.00 +/- 0.02) cm and height (5.00 +/- 0.05) cm. Calculate its density and the worst-case first-order percentage uncertainty. State the SI base dimensions of density.4 indicative marks
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A cylindrical specimen has mass (8.0 +/- 0.1) g, radius (1.00 +/- 0.02) cm and height (5.00 +/- 0.05) cm. Calculate its density and the worst-case first-order percentage uncertainty. State the SI base dimensions of density.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked cylindrical volume, factor-two radius sensitivity, g/cm3 to kg/m3 conversion, dimension M L^-3 versus units kg m^-3, and worst-case addition; 1.3.3 supports propagation, not laboratory performance. This is conceptual placement, not completion of the entire referenced subsection.
The volume is pi x 1.00^2 x 5.00 = 15.708 cm^3; 8.0/15.708 = 0.5093 g cm^-3. Multiplying by 1000 converts this to kg m^-3. Since volume has dimension L^3, density has dimension M/L^3 = M L^-3, distinct from its kg m^-3 units.
The mass, radius and height percentage contributions are 1.25%, 4.00% and 1.00%. Their first-order sum is 6.25%; this is not quadrature or a statistical confidence interval.
Self-marking points
rho = m/(pi r^2 h) = 0.509 g cm^-3, approximately 509 kg m^-3.
Percentage uncertainty is 100(0.1/8.0 + 2 x 0.02/1.00 + 0.05/5.00) = 6.25%, about 6.3%.
Density has dimensions M L^-3 and SI base units kg m^-3 (mass per volume).
Worst-case contributions add in magnitude; the squared radius doubles its fractional contribution.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2A particle is launched at 20.0 m s^-1 at 30.0 degrees above horizontal and returns to its launch height. Ignore drag; use g = 10.0 m s^-2. Calculate flight time, horizontal range and maximum height. Explain why the horizontal velocity is constant.4 indicative marks
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A particle is launched at 20.0 m s^-1 at 30.0 degrees above horizontal and returns to its launch height. Ignore drag; use g = 10.0 m s^-2. Calculate flight time, horizontal range and maximum height. Explain why the horizontal velocity is constant.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Independently resolved launch components and solved equal-height vertical quadratic; 2.1.9 explicitly combines perpendicular uniform and accelerated motion. This is conceptual placement, not completion of the entire referenced subsection.
Resolve the initial velocity: 20cos30 = 17.3205 and 20sin30 = 10.0 m s^-1. Vertical displacement is zero on landing, so 0 = 10t - 5t^2 gives the nonzero root t = 2.00 s.
Use unrounded horizontal speed to get 34.641 m. At the top vertical speed is zero, so 0 = 10^2 - 2(10)h gives 5.00 m. Gravity supplies only vertical acceleration.
Self-marking points
Initial components are ux = 17.3 m s^-1 and uy = 10.0 m s^-1.
Flight time = 2uy/g = 2.00 s.
Range = ux t = 34.6 m; maximum height = uy^2/(2g) = 5.00 m.
There is no horizontal resultant force in the no-drag model.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3In a hypothetical static setup, an object of mass 3.00 kg and volume 2.00 x 10^-3 m^3 is completely immersed in water of density 1000 kg m^-3. A vertical string supports it without contact with the container. Use g = 9.80 m s^-2. Find upthrust and string tension. The tension pulls perpendicular to a lever at 0.400 m from its pivot; find its moment magnitude.4 indicative marks
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In a hypothetical static setup, an object of mass 3.00 kg and volume 2.00 x 10^-3 m^3 is completely immersed in water of density 1000 kg m^-3. A vertical string supports it without contact with the container. Use g = 9.80 m s^-2. Find upthrust and string tension. The tension pulls perpendicular to a lever at 0.400 m from its pivot; find its moment magnitude.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked submerged-volume assumption, vertical force balance and perpendicular moment arm; combines explicit 4.3.5-6 with 4.1.2 without prescribing a real apparatus. This is conceptual placement, not completion of the entire referenced subsection.
The displaced volume equals the entire object volume because it is fully immersed. Archimedes' relation gives 1000 x 9.80 x 0.00200 = 19.6 N.
Balance vertical forces: T + 19.6 = 3.00 x 9.80. Multiply the resulting 9.80 N by the perpendicular pivot distance; no angle or compass sense is supplied for a signed moment.
Self-marking points
Upthrust = rho gV = 19.6 N upward.
Weight = 29.4 N, so equilibrium tension = 29.4 - 19.6 = 9.80 N.
Moment magnitude = 9.80 x 0.400 = 3.92 N m.
The buoyant force is due to a hydrostatic pressure difference, not disappearance of weight.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 4An idealised wire of length 2.00 m and cross-sectional area 1.00 x 10^-6 m^2 extends by 1.00 mm under tensile force 100 N, within its proportional limit. Find stress, strain, Young modulus and stored elastic energy. State why the proportional-limit condition matters.4 indicative marks
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An idealised wire of length 2.00 m and cross-sectional area 1.00 x 10^-6 m^2 extends by 1.00 mm under tensile force 100 N, within its proportional limit. Find stress, strain, Young modulus and stored elastic energy. State why the proportional-limit condition matters.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified stress/strain units, millimetre conversion, modulus 200 GPa and triangular work; leaves actual wire measurement as a separate practical gap. This is conceptual placement, not completion of the entire referenced subsection.
Convert 1.00 mm to 0.00100 m before calculating the dimensionless strain. Divide 100/(1.00e-6) by 0.00100/2.00 to obtain the modulus.
The force rises linearly from zero to 100 N, giving area 0.5 x 100 x 0.00100 = 0.0500 J. A non-linear loading graph would require its actual area instead.
Self-marking points
Stress = F/A = 1.00 x 10^8 Pa; strain = extension/length = 5.00 x 10^-4.
Young modulus = stress/strain = 2.00 x 10^11 Pa.
Stored energy = Fx/2 = 0.0500 J.
Within the proportional limit force is proportional to extension, so the force-extension area is triangular.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 5A source emits sound at 1000 Hz and travels at 30.0 m s^-1 directly towards a stationary observer. Sound speed is 330 m s^-1. Calculate the observed frequency while approaching and while receding at the same speed. Explain the wavelength change in front of the source.3 indicative marks
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A source emits sound at 1000 Hz and travels at 30.0 m s^-1 directly towards a stationary observer. Sound speed is 330 m s^-1. Calculate the observed frequency while approaching and while receding at the same speed. Explain the wavelength change in front of the source.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked correct moving-source denominators and independent forward wavelength 0.300 m; no unsupported moving-observer formula is required. This is conceptual placement, not completion of the entire referenced subsection.
This is a moving-source/stationary-observer case, the scope of 7.3. Substitute 1000 x 330/300 = 1100 Hz, then 1000 x 330/360 = 916.667 Hz.
One emission period is 0.001 s. In that interval the source advances 0.0300 m; the forward wavefront separation is 0.330 - 0.030 = 0.300 m.
Self-marking points
Approach: fo = fs v/(v-vs) = 1100 Hz.
Recession: fo = fs v/(v+vs) = 916.7 Hz, about 917 Hz.
The approaching source emits successive wavefronts closer together in front, reducing wavelength while the medium's sound speed stays fixed.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
