MYP Physics · topic MYP-08.04
Buoyancy and Archimedes' principle
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
Work through the linked stages and explain what each result means.
A floating instrument that reaches its loading limit
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An instrument housing has external volume 5.0×10⁻³ m³ and mass 3.0 kg. It floats upright in fresh water of density 1000 kg m⁻³. Use g = 10 N kg⁻¹. Assume its sealed walls do not deform and any added ballast is inside. The supplied graph gives buoyant force against displaced volume. Supplied data (Displaced volume / litre; Upthrust / N): (0, 0), (1, 10), (3, 30), (5, 50). (a) Find the fraction initially submerged and the maximum ballast mass before complete submersion. [3 marks] (b) It is moved with 1.0 kg ballast into a liquid of density 800 kg m⁻³. Predict its behaviour and explain why density alone does not establish stability. [3 marks]
Fresh-water upthrust
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upthrust / N
Horizontal axis (x): Displaced volume / litre
Read the exact data as a table
| Displaced volume / litre | Upthrust / N |
|---|---|
| 0 | 0 |
| 1 | 10 |
| 3 | 30 |
| 5 | 50 |
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 fraction initially submerged and the maximum ballast mass before complete submersion.
Reveal the worked solution
- Floating equilibrium requires upthrust equal to the 30 N weight. Archimedes' principle gives Vdisplaced = m/ρ = 0.0030 m³.
- The submerged fraction is 0.0030/0.0050 = 0.60. The maximum upthrust before full submersion is 1000(10)(0.0050) = 50 N.
- The largest supported total mass is 5.0 kg, leaving 2.0 kg ballast. At exactly that limit there is no freeboard, so a safe operating load must be lower.
- The total mass is 3.0+1.0 = 4.0 kg; complete displacement in the new liquid yields 800(0.0050) = 4.0 kg of displaced liquid.
- It therefore has no positive freeboard in the ideal model, and waves or added water would make its operation unreliable.
- Whether a tilted housing returns upright depends on the locations of its centre of mass and the changing centre of buoyancy. Average density predicts floating capacity, not rotational stability.
Answer (a) Initially 60% submerged; limiting ballast mass = 2.0 kg. (b) Its total 4.0 kg mass exactly equals the displaced-liquid mass at full submersion; it is at the neutral-buoyancy limit.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.