Cambridge International A-Level Physics · topic 4
Forces, density and pressure
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.
Is a hollow object made from the claimed metal?
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.
The load that tips a mobile platform
A 20 kg platform has supports at x = 0 and x = 1.2 m. Its centre of mass is at x = 0.60 m. A 30 kg load is placed at variable x. Use g = 10 N kg⁻¹. The graph supplies the right-support reaction while both supports remain in contact. Supplied data (Load position x / m; Right-support reaction / N): (0, 100), (0.4, 200), (0.8, 300), (1.2, 400), (1.6, 500). (a) Find both support forces for x = 1.0 m and determine the load position at which tipping begins. [3 marks] (b) Find the combined centre of mass at the tipping threshold and assess lowering the platform's centre of mass on this ideal level-surface threshold. [3 marks]
Right-support reaction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Right-support reaction / N
Horizontal axis (x): Load position x / m
Read the exact data as a table
| Load position x / m | Right-support reaction / N |
|---|---|
| 0 | 100 |
| 0.4 | 200 |
| 0.8 | 300 |
| 1.2 | 400 |
| 1.6 | 500 |
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 both support forces for x = 1.0 m and determine the load position at which tipping begins.
Reveal the worked solution
- Taking moments about the left support: RR(1.2) = 200(0.60)+300x. At x = 1.0, RR = 350 N.
- Vertical force balance gives RL = 500−RR = 150 N.
- Tipping begins when the left reaction reaches zero, so RR = 500 N and 500(1.2) = 120+300x, giving x = 1.6 m. Beyond that point the two-contact static model fails.
- At xload = 1.6 m, xCM = [20(0.60)+30(1.6)]/(20+30) = 1.2 m, directly above the right support.
- At the threshold, the combined weight's line of action reaches the support boundary. A support cannot provide a downward reaction to maintain this contact model.
- Lowering the centre of mass can improve resistance to tipping during tilt or lateral acceleration, but it does not change the horizontal moment arms of vertical weights in this strictly level static setup.
Answer (a) At x = 1.0 m: right support = 350 N, left = 150 N; tipping threshold x = 1.6 m. (b) Combined centre of mass lies at x = 1.2 m; lowering it does not change this ideal horizontal overhang threshold.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A hydraulic lift below the reservoir surface
An ideal hydraulic system contains oil of density 800 kg m⁻³. Its small piston has area 2.0 cm² and its large piston has area 40 cm². The large piston is 0.50 m below the small piston. A 60 N force is applied downward to the small piston; atmospheric pressure acts on both external faces. Use g = 10 N kg⁻¹ and neglect piston weights and friction. Supplied data (Depth below small piston / m; Additional pressure / Pa): (0, 0), (0.25, 2000), (0.5, 4000). (a) Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance. [3 marks] (b) For a short motion, the small piston moves down 0.10 m. Find the large-piston rise and explain why force amplification does not create energy. [3 marks]
Hydrostatic gauge-pressure contribution below the small piston
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Additional pressure / Pa
Horizontal axis (x): Depth below small piston / m
Read the exact data as a table
| Depth below small piston / m | Additional pressure / Pa |
|---|---|
| 0 | 0 |
| 0.25 | 2000 |
| 0.5 | 4000 |
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 gauge pressure immediately below each piston and the greatest downward load the large piston can balance.
Reveal the worked solution
- Convert 2.0 cm² to 2.0×10⁻⁴ m². Applied gauge pressure is F/A = 60/(2.0×10⁻⁴) = 300000 Pa.
- The lower piston has an additional ρgh = 800(10)(0.50) = 4000 Pa, giving 304000 Pa gauge.
- Its area is 40 cm² = 0.0040 m², so the balanced load is pA = 304000(0.0040) = 1216 N. Atmospheric contributions cancel across each piston.
- Incompressibility gives equal displaced volumes: As xs = AL xL, so xL = (2/40)(0.10) = 0.0050 m.
- The large force acts through a smaller displacement. For equal piston heights the ideal input and output work are equal.
- Here the initial pressure difference also involves gravitational potential energy of the oil. For a finite stroke the height difference changes; a full energy audit must include the oil and the varying pressure, rather than multiplying the initial force by any arbitrary stroke.
Answer (a) Small-piston gauge pressure = 300 kPa; large = 304 kPa; load = 1216 N. (b) Large-piston rise = 0.0050 m; input work and gravitational-energy changes must both be included.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.