MYP Physics · unit 06
Matter, density, and materials
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Is a hollow object made from the claimed metal?
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A sealed metal object has mass 156 g. It raises a measuring-cylinder reading from 40 to 100 cm³. Each reading has uncertainty ±1 cm³, and the mass has uncertainty ±1 g. Solid samples of the claimed alloy give the mass–volume data below. Ignore the mass of air in the hollow object. Supplied data (Volume / cm³; Mass / g): (10, 78), (20, 156), (30, 234). (a) Use the reference data to find the alloy density and calculate the sealed object's cavity volume. [3 marks] (b) Estimate a conservative uncertainty in the external volume and in the average density; suggest a method improvement. [3 marks]
Solid alloy samples
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Mass / g
Horizontal axis (x): Volume / cm³
Read the exact data as a table
| Volume / cm³ | Mass / g |
|---|---|
| 10 | 78 |
| 20 | 156 |
| 30 | 234 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Use the reference data to find the alloy density and calculate the sealed object's cavity volume.
Reveal the worked solution
- The mass–volume gradient is 78/10 = 7.8 g cm⁻³, equivalent to 7800 kg m⁻³.
- Metal volume = mass/density = 156/7.8 = 20 cm³. The external volume is 100−40 = 60 cm³.
- Cavity volume = external volume − metal volume = 40 cm³. An average density of 2.6 g cm⁻³ does not by itself disprove the alloy claim.
- For a conservative bound, add the two reading uncertainties: ΔV = 1+1 = 2 cm³.
- The approximate fractional density uncertainty is 1/156 + 2/60 = 0.0397; absolute uncertainty is about 0.10 g cm⁻³.
- Use a narrower calibrated cylinder and read the meniscus at eye level. Ensure complete submersion without trapped external bubbles; repeated readings do not correct a trapped bubble.
Answer (a) Alloy density = 7.8 g cm⁻³; cavity volume = 40 cm³. (b) External volume = 60 ±2 cm³; average density ≈2.6 ±0.10 g cm⁻³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Heating ice while heat escapes
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 0.10 kg sample of ice begins at 0°C. A heater supplies 100 W, while a separate calibration shows a constant 20 W loss to the surroundings during this experiment. The sample-temperature observations are supplied below. Use latent heat of fusion 3.36×10⁵ J kg⁻¹ and water specific heat capacity 4200 J kg⁻¹ K⁻¹. Ignore vessel heat capacity. Supplied data (Time / s; Temperature / °C): (0, 0), (210, 0), (420, 0), (525, 20). (a) Predict the melting duration and then the time to reach 20°C after melting. Check both against the data. [3 marks] (b) Explain the constant-temperature interval with a particle model and evaluate a calculation that ignores losses. [3 marks]
Observed heating curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Temperature / °C
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Temperature / °C |
|---|---|
| 0 | 0 |
| 210 | 0 |
| 420 | 0 |
| 525 | 20 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Predict the melting duration and then the time to reach 20°C after melting. Check both against the data.
Reveal the worked solution
- Net heating power is 100−20 = 80 W, since the sample does not receive the full heater input.
- Melting needs mL = 0.10(3.36×10⁵) = 33600 J; duration is 33600/80 = 420 s.
- Warming liquid by 20 K needs mcΔT = 0.10(4200)(20) = 8400 J; duration is 8400/80 = 105 s. Both match the supplied curve.
- During melting, transferred energy changes intermolecular potential energy rather than increasing mean kinetic energy, so temperature need not rise.
- The internal energy still increases through the phase change; temperature is not a measure of the total internal energy.
- Ignoring losses gives 33600/100 = 336 s, 84 s too short. Insulation and a lid can reduce unwanted transfers; repeats alone do not remove systematic heat loss.
Answer (a) Melting takes 420 s; subsequent warming takes 105 s. (b) Temperature stays constant while internal energy increases through a change in particle arrangement; ignoring losses underestimates the required time.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The spring that stops obeying the first slope
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A wire of original length 1.0 m and cross-sectional area 1.0 mm² is loaded gradually. Its force–extension data are supplied below; segments between points are straight. The first 4 mm is reversible and proportional. After the 6 mm reading, unloading leaves 1 mm permanent extension. Supplied model: Use F = kx, stress = F/A, strain = ΔL/L and Young modulus = stress/strain in the proportional region. Supplied data (Extension / mm; Force / N): (0, 0), (2, 40), (4, 80), (6, 100). (a) Find the initial spring constant and Young modulus, converting the units explicitly. [3 marks] (b) Calculate the work done loading to 6 mm. Explain why ½Fmax xmax is invalid and whether all the work is recoverable. [3 marks]
Wire loading curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force / N
Horizontal axis (x): Extension / mm
Read the exact data as a table
| Extension / mm | Force / N |
|---|---|
| 0 | 0 |
| 2 | 40 |
| 4 | 80 |
| 6 | 100 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the initial spring constant and Young modulus, converting the units explicitly.
Reveal the worked solution
- Initial stiffness is 40/(0.002) = 2.0×10⁴ N m⁻¹. A millimetre is 10⁻³ m, while 1 mm² is 10⁻⁶ m².
- At 40 N, stress is F/A = 4.0×10⁷ Pa and strain is ΔL/L = 0.002.
- Young modulus = stress/strain = 2.0×10¹⁰ Pa. This material property requires the original geometry; k alone describes this particular wire.
- From 0 to 4 mm the area is ½(80)(0.004) = 0.16 J.
- From 4 to 6 mm it is ½(80+100)(0.002) = 0.18 J, so total loading work is 0.34 J.
- The full graph is not one straight proportional line, so ½Fmax xmax would give an incorrect 0.30 J. Permanent extension shows plastic deformation; without an unloading curve the recoverable energy cannot be quantified.
Answer (a) k = 2.0×10⁴ N m⁻¹; Young modulus = 2.0×10¹⁰ Pa. (b) Loading work = 0.34 J; not all is recoverable after permanent extension.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.