Original GioPhysics extended practice
MYP Physics · Extended Paper 1
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 175 minutes
- Questions
- 10
- Marks
- 126
- Answers
- Complete teacher key
Instructions
- Answer every task using evidence and scientific reasoning.
- Label each calculation, graph interpretation and evaluation clearly.
- Criteria A–D are represented across the two-paper pack.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Extract mass and friction from acceleration data
A trolley is pulled horizontally. The recorded pairs (applied force in N, acceleration in m/s²) are (0.8, 0), (1.2, 0), (1.6, 0.5), (2.0, 1.0), (2.4, 1.5). (a) Describe the pattern and identify the force needed before sustained motion appears. [3] (b) For the moving points, use F = ma + f to determine the trolley mass m and the approximately constant friction f. [5] (c) Assess whether all five readings support one straight-line model. [2] (d) Propose one improvement that would test the model more rigorously. [2]
Applied force and trolley acceleration
Acceleration stays at zero for the first two applied forces, then increases linearly across the three moving readings.
Read the exact data as a table
| Applied force / N | Acceleration / m s⁻² |
|---|---|
| 0.8 | 0 |
| 1.2 | 0 |
| 1.6 | 0.5 |
| 2 | 1 |
| 2.4 | 1.5 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Can the density identify an unknown alloy?
A cylindrical sample has mass (44.4 ± 0.2) g, diameter (2.40 ± 0.05) cm and length (5.00 ± 0.05) cm. Use π = 3.142. (a) Distinguish accuracy from precision. [2] (b) Calculate the cylinder volume and density. [4] (c) Estimate the percentage uncertainty in the density by adding percentage uncertainties; remember that diameter is squared. [4] (d) Reference densities are aluminium 2.70 g/cm³ and magnesium 1.74 g/cm³. Judge whether density alone securely identifies the sample. [2]
Open the interactive answer workspaceChoose a safer helmet liner using impulse evidence
A 4.0 kg instrumented headform moving at 6.0 m/s stops in a crash test. Liner P stops it in 0.012 s and is reusable after small impacts; liner Q stops it in 0.030 s but must be replaced after one severe impact. P costs $18, Q costs $29. (a) Calculate the momentum change and average force magnitude for each liner. [5] (b) Explain why the calculations do not prove either helmet prevents injury. [3] (c) Evaluate the physical, economic, environmental and user-access implications. [6] (d) Make a conditional recommendation and identify further evidence needed. [2]
Open the interactive answer workspaceDesign a fair material-stiffness comparison
A design team is choosing recycled sheet material for a reusable package. (a) Write a research question about sheet thickness and midpoint deflection under load. [2] (b) Give and justify a hypothesis. [3] (c) Design a method using five thicknesses, a fixed support span and small masses. Identify variables, repeats, measurements, safety and the graph to draw. [6] (d) Explain one limitation that repeats do not remove and a matching improvement. [3] Do not invent results.
Open the interactive answer workspacePredict waves entering two harbour gaps
Regular water waves have wavelength 12 m and frequency 0.40 Hz. They approach two harbour entrances of width 6 m and 30 m. (a) Calculate wave speed and period. [3] (b) Compare the expected diffraction at the two entrances. [3] (c) Explain how a two-dimensional wavefront sketch should differ behind each entrance. [2] (d) State two limits of applying this simple model to a real harbour. [2]
Open the interactive answer workspaceTest whether a transparent block has one refractive index
Light travels from air into a block. The recorded pairs (angle of incidence i in degrees, angle of refraction r in degrees) are (0, 0), (15, 10), (30, 19), (45, 28), (60, 35). (a) Describe the angle pattern. [2] (b) For each non-zero pair calculate n = sin i / sin r and find a representative value. [5] (c) Use that value to estimate the critical angle for light going from the block to air. [3] (d) Evaluate whether the evidence supports one constant refractive index and propose two improvements. [4]
Incident and refracted angles
The refracted angle rises with incident angle but remains smaller at each non-zero measurement.
Read the exact data as a table
| Angle of incidence / ° | Angle of refraction / ° |
|---|---|
| 0 | 0 |
| 15 | 10 |
| 30 | 19 |
| 45 | 28 |
| 60 | 35 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Account for energy inside a real battery
A 12.0 V battery with internal resistance 0.50 Ω supplies a 5.50 Ω load. (a) Calculate total resistance and current. [3] (b) Calculate terminal potential difference and show that it matches the load voltage. [3] (c) Calculate power in the load, power dissipated inside the battery and efficiency of transfer to the load. [4] (d) Explain why the battery terminal voltage falls when current increases. [2]
Open the interactive answer workspaceTest an inverse-square gravitational field claim
A simulation reports the pairs (distance r in arbitrary units, gravitational field g in arbitrary units) (1.0, 9.00), (1.5, 4.00), (2.0, 2.25), (2.5, 1.44), (3.0, 1.00). (a) Describe the pattern without calling it linear. [2] (b) Calculate gr² for every pair and use the values to test g ∝ 1/r². [4] (c) Predict g at r = 4.0 and explain the interpolation or extrapolation risk. [3] (d) State what this evidence can and cannot establish about real gravitational fields. [3]
Gravitational field against distance
Field strength falls steeply at first and then more gradually as distance increases.
Read the exact data as a table
| Distance / arbitrary units | Field / arbitrary units |
|---|---|
| 1 | 9 |
| 1.5 | 4 |
| 2 | 2.25 |
| 2.5 | 1.44 |
| 3 | 1 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Recover half-life from count-rate evidence
A sealed source gives 920 counts/min initially and 220 counts/min after 18 years. Background is 120 counts/min and is assumed constant. (a) Explain why background must be subtracted. [2] (b) Calculate both net count rates and the number of half-lives elapsed. [4] (c) Determine the half-life. [2] (d) Predict the total detector reading after a further 12 years and state two assumptions. [4]
Open the interactive answer workspaceEvaluate an ultrasound frequency–penetration claim
A training phantom gives the pairs (ultrasound frequency in MHz, maximum useful image depth in cm) (2, 14.0), (3, 10.0), (5, 6.5), (8, 4.0), (12, 2.8). A trainee claims useful depth is inversely proportional to frequency. (a) Describe the trend. [2] (b) Calculate frequency × depth for every pair and assess the claim using a ±10% consistency rule around the first product. [5] (c) Explain the physical trade-off between frequency, attenuation and image detail. [3] (d) Identify one limitation of transferring this phantom result to patients and one improvement. [2]
Ultrasound frequency and useful image depth
Useful depth decreases non-linearly as ultrasound frequency increases.
Read the exact data as a table
| Frequency / MHz | Maximum useful depth / cm |
|---|---|
| 2 | 14 |
| 3 | 10 |
| 5 | 6.5 |
| 8 | 4 |
| 12 | 2.8 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Teacher copyDetailed answer and marking guideShow 10 answers
Teacher copy
Detailed answer and marking guide
Accept equivalent physics expressed clearly. Award method credit where the reasoning is valid.
Extract mass and friction from acceleration data
Answer: The stationary readings show a threshold up to about 1.2 N. For the moving data Δa/ΔF = 1.25 kg⁻¹, so m = 0.80 kg and f = 1.20 N. Static and moving behaviour should not be forced onto one line.
- (a) Acceleration remains zero at 0.8 N and 1.2 N, then rises by 0.5 m/s² for each further 0.4 N. Sustained motion begins only above an applied force of about 1.2 N.
- (b) For moving points, a = (F − f)/m, so the gradient of acceleration against force is 1/m. Gradient = (1.5 − 0.5)/(2.4 − 1.6) = 1.25 kg⁻¹.
- (b) Therefore m = 1/1.25 = 0.80 kg. Substituting F = 1.6 N and a = 0.5 m/s² gives f = F − ma = 1.6 − 0.80 × 0.5 = 1.20 N.
- (c) The three moving readings fit F = 0.80a + 1.20 exactly, but the two zero-acceleration readings represent adjustable static friction. They are not evidence that the same kinetic straight line extends to rest.
- (d) Add closely spaced force values around 1.2–1.6 N and repeat each measurement. A force sensor and motion sensor would show both the breakaway peak and the steadier force while moving.
Can the density identify an unknown alloy?
Answer: The volume is 22.62 cm³ and the density is 1.96 g/cm³. The estimated percentage uncertainty is 5.62%, or about ±0.11 g/cm³, so the interval does not match either reference closely enough to identify the alloy securely.
- (a) Precision describes how closely repeated readings agree; accuracy describes closeness to an accepted value. A set can be precise yet inaccurate if every reading has the same calibration offset.
- (b) Radius = 2.40/2 = 1.20 cm. Volume V = πr²L = 3.142 × 1.20² × 5.00 = 22.62 cm³, with units carried through the calculation.
- (b) Density ρ = m/V = 44.4/22.62 = 1.96 g/cm³ to three significant figures, matching the precision of the supplied measurements.
- (c) Mass uncertainty = 0.2/44.4 × 100 = 0.45%. Volume uncertainty ≈ 2(0.05/2.40 × 100) + 0.05/5.00 × 100 = 5.17%.
- (c) Density is a quotient, so the estimated percentage uncertainty is 0.45% + 5.17% = 5.62%. Its absolute uncertainty is 0.0562 × 1.96 ≈ 0.11 g/cm³.
- (d) The plausible range is approximately 1.85–2.07 g/cm³. It excludes the two pure reference values, suggesting an alloy or a systematic measurement problem; density alone cannot determine composition.
Choose a safer helmet liner using impulse evidence
Answer: The momentum change magnitude is 24 kg m/s. Average force is 2000 N for P and 800 N for Q. Q reduces average force in this test, but a recommendation also needs peak-force, fit, certification, replacement and user-context evidence.
- (a) Initial momentum is mv = 4.0 × 6.0 = 24 kg m/s and final momentum is zero, so the change has magnitude 24 kg m/s opposite the original motion.
- (a) Average force magnitude for P is Δp/Δt = 24/0.012 = 2000 N. For Q it is 24/0.030 = 800 N, so the longer stopping time lowers the average force in this test.
- (b) Injury can depend on peak rather than average force, rotational acceleration, impact direction, fit and biological variation. One headform speed is not a complete safety test or a guarantee for a rider.
- (c) Q offers the lower calculated average force in the stated severe impact, while P's reuse after minor impacts may reduce replacement cost and material waste. A damaged liner that looks unchanged also creates a behaviour and inspection risk.
- (c) Higher purchase and replacement costs can reduce access, especially for families or shared-cycle schemes. Manufacturers, users, emergency clinicians, waste services and regulators value different outcomes, so price alone is not the decision rule.
- (d) Prefer a certified design that keeps both peak linear and rotational acceleration within accepted limits and fits the user. If Q meets those tests materially better, recommend Q for severe-impact protection while making replacement guidance and support affordable.
- (d) Seek repeated impacts at several speeds and angles, peak-force and rotational data, fit testing, certification results, material life-cycle evidence and real replacement compliance before a final procurement decision.
Design a fair material-stiffness comparison
Answer: Vary sheet thickness and measure midpoint deflection under the same force, span and specimen geometry. Predict that thicker sheets deflect less, then use repeated loading measurements and a deflection–thickness graph to test the relationship.
- (a) A focused question is: How does the measured thickness of an equal-width recycled sheet affect its midpoint deflection under a fixed load across a fixed support span?
- (b) Predict that deflection decreases as thickness increases because distributing material farther from the neutral layer increases bending stiffness. This predicts a direction without pretending the exact exponent is already known.
- (c) Measure each specimen's thickness at several positions with a micrometer. Cut equal widths and lengths, place each across the same two supports, add the same small central load and measure vertical deflection against a fixed scale or from calibrated photographs.
- (c) Thickness is independent and deflection is dependent. Control material batch, width, span, grain direction, load, load position, loading time, temperature and whether a specimen has previously been bent.
- (c) Test at least five thicknesses and three fresh specimens per thickness. Record raw thickness and deflection with units, calculate mean and range, and plot mean deflection against measured thickness with uncertainty visible.
- (c) Keep fingers clear of loads, use a tray for falling masses and stay well below sudden-failure loads. Stop and replace any creased specimen rather than quietly reusing it.
- (d) Lamination adhesive or fibre orientation may change with nominal thickness, producing a material-structure difference rather than thickness alone. Build specimens from the same layer stock, standardise adhesive mass and alternate test order.
Predict waves entering two harbour gaps
Answer: Wave speed is 4.8 m/s and period is 2.5 s. The 6 m gap is narrower than one wavelength and produces strong spreading; the 30 m gap is 2.5 wavelengths wide and produces less spreading.
- (a) Wave speed v = fλ = 0.40 × 12 = 4.8 m/s. The period is T = 1/f = 1/0.40 = 2.5 s.
- (b) For the 6 m entrance, width/wavelength = 6/12 = 0.50, so diffraction is strong and energy spreads through a broad range of directions.
- (b) For the 30 m entrance, width/wavelength = 30/12 = 2.5, so most wavefronts continue roughly forward with only edge spreading. Diffraction is not literally zero.
- (c) Behind 6 m, draw widely curved, nearly semicircular fronts with 12 m separation. Behind 30 m, draw a broad central region of straighter fronts and curvature mainly near both edges.
- (d) Real seas contain many frequencies and directions, while water depth changes wave speed. Reflection from walls, irregular entrances, wind and breaking also violate the simple uniform-wave model.
Test whether a transparent block has one refractive index
Answer: The refracted angle increases more slowly than the incident angle. The non-zero n values are about 1.50, 1.54, 1.51 and 1.51, giving n ≈ 1.51 and a critical angle sin⁻¹(1/1.51) ≈ 41.5°.
- (a) Refraction is towards the normal because r < i for every non-zero reading. Both angles increase, but equal angle increments do not produce equal refracted-angle increments.
- (b) The ratios sin i/sin r are approximately 1.49, 1.54, 1.51 and 1.51 for 15°, 30°, 45° and 60°. The 0° pair cannot be used because it gives 0/0.
- (b) A representative mean is about 1.51. The small scatter is plausible for degree-scale angle readings and supports, but does not prove, a constant index for this wavelength and material.
- (c) At the block–air critical angle, sin c = n_air/n_block ≈ 1/1.51. Therefore c ≈ sin⁻¹(0.662) = 41.5°.
- (d) Repeat independently, use a narrow laser line, draw a thin normal and measure from the normal rather than the surface. Plot sin i against sin r; a straight line through the origin tests the model more directly.
- (d) Wavelength and block homogeneity are uncontrolled in the claim of one universal value, so repeat with stated colours and several entry positions before generalising.
Account for energy inside a real battery
Answer: Total resistance is 6.00 Ω and current is 2.00 A. Terminal/load voltage is 11.0 V. Load power is 22.0 W, internal dissipation is 2.00 W and transfer efficiency is 91.7%.
- (a) The internal and external resistances are in series for the complete circuit, so R_total = 0.50 + 5.50 = 6.00 Ω.
- (a) Current is I = emf/R_total = 12.0/6.00 = 2.00 A.
- (b) Lost voltage inside the battery is Ir = 2.00 × 0.50 = 1.00 V, so terminal potential difference is 12.0 − 1.00 = 11.0 V.
- (b) Across the load, V = IR = 2.00 × 5.50 = 11.0 V, which independently checks the terminal-voltage calculation.
- (c) Load power is I²R = 2.00² × 5.50 = 22.0 W; internal power is I²r = 2.00² × 0.50 = 2.00 W. Total 24.0 W also equals emf × current.
- (c) Efficiency to the load is 22.0/24.0 × 100 = 91.7%. The remainder heats the battery internally.
- (d) A larger current produces a larger internal drop Ir, so terminal voltage emf − Ir falls even if the emf is approximately unchanged.
Test an inverse-square gravitational field claim
Answer: The field decreases non-linearly. Every gr² value is 9.00, so the dataset exactly follows g = 9/r²; at r = 4.0 the model predicts g = 0.5625. Simulation agreement does not independently verify nature.
- (a) As distance increases, field strength decreases by progressively smaller absolute amounts. A straight line through the raw g–r points would not describe the curved pattern.
- (b) The products gr² are 9.00×1.0² = 9.00, 4.00×1.5² = 9.00, 2.25×2.0² = 9.00, 1.44×2.5² = 9.00 and 1.00×3.0² = 9.00.
- (b) A constant gr² means g = constant/r² for these values, so the supplied record is exactly consistent with an inverse-square model.
- (c) At r = 4.0, g = 9.00/4.0² = 0.5625 arbitrary units. This is extrapolation beyond the measured 1.0–3.0 range and depends on the same model continuing to apply.
- (d) The evidence checks that the simulation implements an inverse-square rule over its sampled range. It does not measure a real source mass, establish the units, reveal uncertainty or independently test Newtonian gravity in nature.
Recover half-life from count-rate evidence
Answer: Net activity falls from 800 to 100 counts/min, a factor of eight or three half-lives in 18 years, so the half-life is 6 years. After 12 more years the predicted total is 25 + 120 = 145 counts/min.
- (a) The detector records environmental and instrumental background as well as the source. Radioactive decay applies to the source contribution, not to the unchanged background count.
- (b) Initial net count = 920 − 120 = 800 counts/min; later net count = 220 − 120 = 100 counts/min.
- (b) The net count falls 800 → 400 → 200 → 100, which is three halvings, so three half-lives have elapsed in 18 years.
- (c) Half-life = 18/3 = 6 years.
- (d) A further 12 years is two half-lives, so net count becomes 100/4 = 25 counts/min. Add background back for the detector prediction: 25 + 120 = 145 counts/min.
- (d) Assumptions include constant detector geometry and efficiency, constant mean background, no contamination or source loss, and a single radionuclide with unchanged decay probability.
Evaluate an ultrasound frequency–penetration claim
Answer: Useful depth falls as frequency rises. Products are 28.0, 30.0, 32.5, 32.0 and 33.6 MHz·cm; the last exceeds ±10% of 28.0, so the complete dataset fails the stated consistency rule even though the inverse trend is approximate.
- (a) Maximum useful image depth decreases as frequency increases, with large losses at first and smaller absolute losses at the highest frequencies. The raw relationship is curved rather than linear.
- (b) The products frequency × depth are 2×14.0 = 28.0, 3×10.0 = 30.0, 5×6.5 = 32.5, 8×4.0 = 32.0 and 12×2.8 = 33.6 MHz·cm.
- (b) Ten percent of the first product is 2.8, so the allowed interval is 25.2–30.8. The 5, 8 and 12 MHz products lie above that interval; the full record does not meet the stated inverse-proportion rule.
- (c) Higher-frequency waves generally have shorter wavelength and can resolve smaller structures, but tissue attenuates them more strongly, reducing useful depth. The operator chooses a compromise for the target anatomy.
- (d) A uniform phantom cannot reproduce varied tissue, body size, bone, gas or operator technique. Repeat with tissue-equivalent phantoms of several compositions and blinded image-quality criteria before informing clinical practice.