IB Physics HL B 3 1 Pressure Questions

Practise IB Physics HL B.3.1 by applying pressure differences, hydrostatic depth and gas laws to forces in fluid systems.

Exam points

Apply p=F/A and pressure differences to calculate force magnitude and direction on surfaces, doors, stoppers and fluid-exposed objects.
Derive and use hydrostatic pressure ρgh with fluid density, depth and area to calculate pressure or lifting force.
Use gas-pressure data and the ideal-gas equation to calculate pressure in enclosed systems and the force or load they can support.
Evaluate whether a pressure-derived force is physically plausible by commenting on its magnitude and the assumptions in the model.

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Study B.3.1—Pressure conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 5]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

Figure for Question IB Physics HL B 3 1 Pressure Questions question 1 — IB Physics HL

(a)

When fully inflated, an unloaded airbag has a diameter of 1.80 m and a length of 24.0 m . At a temperature of $15^{\circ} \mathrm{C}$ , an airbag can hold 4200 mol of gas.

[ 5 ]

(i)

Show that the pressure in an airbag is about 0.2 MPa .

When the boat is placed on the airbags, the airbags are compressed so that the effective contact area is as shown in the diagram.

In this arrangement, the maximum safe pressure before an airbag bursts is four times the value calculated in (a)(i). The boat is supported by airbags on a slope that is at an angle of $4.0^{\circ}$ to the horizontal. There are fifteen airbags supporting the boat at all times.

[ 2 ]

$\begin{aligned} V=\pi r^{2} h=\pi 0.9^{2} \times 24=61.07 \mathrm{~m}^{3} P=\frac{(4200)(8.31)(273+15)}{61.07}=0.16 \mathrm{MPa} \end{aligned}$

Allow MP1 to be seen in MP2.

(ii)

Estimate the maximum safe mass that this arrangement can hold.

[ 3 ]

$\begin{aligned} \mathrm{F}=4 P A=4\left(1.6 \times 10^{5}\right)(15 \times 24 \times 1.8)=4.14 \times 10^{8} \mathrm{~N} \mathrm{F}=m g \cos \theta=\mathrm{m}(9.8)(\cos 4)=9.78 \mathrm{~m} \mathrm{~N} m=\frac{4 P A}{g \cos \theta}=\frac{4.14 \times 10^{8}}{9.78}=4.3 \times 10^{7} \mathrm{~kg} \end{aligned}$

Do not award MP2 if the cos θterm is omitted.

Allow ECF for MP3.

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IB Physics HL B 3 1 Pressure Questions question 1

Question (a)

Question (i)

Question (ii)