Making physical estimates
Key idea: A physical estimate combines a simple model with sensible approximate values. Check its units and size, and quote only the precision those estimates support.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
Loading guide controls…
Guided lessons · coming soon
Guided lesson
Guided steps
- Choose a model and match its precision to your assumptions.
- Follow how assumed dimensions become a volume, then a mass.
- Change only the depth and calculate both masses and their ratio.
- Track weight and contact area separately before dividing to estimate pressure.
- Choose mass and height, then estimate useful power.
- Estimate the kettle’s energy and explain where it goes.
- Compare neighbouring powers of ten by ratio.
- Find the lifting energy’s nearest power of ten in joules.
- Find one sheet’s thickness in metres and its nearest power of ten.
- Explain why measuring a stack helps, and state your assumptions.
- Correct the water mass using density and a familiar size comparison.
On this page
You stopped at:
ContinueHow much does the water in a bath weigh? How much power do you develop running upstairs? You can answer questions like these without measuring anything, by building a simple physical model from values you already know. This lesson shows how to make a reasonable estimate, how precisely to state it, and how to tell whether it makes sense.
Making an estimate
An estimate follows the same steps every time:
- Choose a simple model: treat a bath as a box, a person as a 60 kg mass.
- State the values you assume, with units.
- Calculate, keeping the units.
- Give the answer to one or two significant figures. Your assumptions are only rough, so more digits would claim a precision you do not have.
- Check that the answer is sensible, by comparing it with something you know.
These values are worth knowing, because estimates are built from them:
| Quantity | Reasonable value |
|---|---|
| mass of an adult | 60–70 kg |
| height of one storey of a building | 3 m |
| walking speed | 1.5 m s⁻¹ |
| speed of a car on a highway | 25 m s⁻¹ |
| density of water | 1000 kg m⁻³ |
| density of air | 1.2 kg m⁻³ |
| acceleration of free fall | about 10 m s⁻² |
| atmospheric pressure | 1 × 10⁵ Pa |
| power of a kettle | 2 kW |
Worked example 1
Mass of water in a bath
Problem
Estimate the mass of water in a full domestic bath.
Show full solution
Choose a model
Method
Treat the water as a box with typical bath dimensions.Reason
A simple shape makes the assumptions clear and the volume easy to find.Working
Answer
Length 1.5 m, average width 0.55 m, depth of water 0.25 m.
Find the volume
Reason
Mass follows from volume and density.Working
Answer
V ≈ (1.5)(0.55)(0.25) = 0.21 m³
Use the density of water
Working
Answer
m = ρ V ≈ (1000)(0.21) ≈ 2 × 10² kg
Quick check
0.2 m³ is 200 litres, which matches the size of a bath. Writing 206.25 kg would claim far more precision than the guessed dimensions allow.
Check your understanding 1
Keep the same length and width, but try water depths of 0.15 m and 0.30 m. Find both masses to two significant figures. Use the unrounded values to find the mass multiplier.
Show the steps Optional working
Show answer
The unrounded model gives 123.75 kg and 247.5 kg, so the multiplier is 2.0. The reported masses are about 120 kg and 250 kg. Doubling the depth doubles the mass, but both are still a few hundred kilograms. A reasonable change in one assumption changes the estimate by a small factor, not by a factor of ten, so the estimate is still useful.
Worked example 2
Pressure under one foot
Problem
Estimate the pressure a person exerts on the floor when standing on one foot.
Show full solution
Find the force
Method
Use the person’s weight.Reason
A person standing still pushes down on the floor with a force equal to their weight.Working
Answer
m ≈ 60 kg and g ≈ 10 m s⁻², so F ≈ 600 N.
Estimate the contact area
Method
Model the sole as a rectangle 25 cm by 8 cm.Reason
A familiar size gives a sensible area.Working
Answer
A ≈ 200 cm² = 2 × 10⁻² m²
Calculate the pressure
Working
Answer
p = F/A ≈ 600/(2 × 10⁻²) = 3 × 10⁴ Pa
Quick check
This is an average: the pressure is higher under the heel and ball of the foot. It is about a third of atmospheric pressure, which is a sensible size.
Exam-style question 1
Power climbing stairs
Examination question
Estimate the useful power a student develops when running up one flight of stairs, from one floor of a building to the next, in about 5 s. State the values you assume. [3 marks]
Write your answer before viewing the mark scheme
Show solution step by step
State the assumptions
1 markMethod
Choose a mass and a height.Reason
The useful work is the gain in gravitational potential energy.Working
Answer
Mass about 60 kg; height of one storey about 3 m; g ≈ 10 m s⁻².
Calculate the power
2 marksWorking
Answer
P = mgh/t ≈ (60)(10)(3)/5 ≈ 4 × 10² W
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark each point. Any sensible mass and height earn the first mark.
Check your understanding 2
Estimate the electrical energy a kettle transfers when it boils water for 3 minutes. Is all of it used to heat the water?
Show answer
E = Pt ≈ (2 × 10³)(180) ≈ 4 × 10⁵ J. Not all of it heats the water: some heats the kettle itself and escapes to the surroundings.
Order of magnitude
Sometimes only the order of magnitude of a quantity is wanted: the power of ten closest to it. The midpoint between two powers of ten, such as 10 and 100, is not 55 but square root of 10 ≈ 3.16 times the lower one, about 31.6, because powers of ten go up by multiplying. So a value with a standard-form coefficient below square root of 10 rounds down to the power below, and one above it rounds up. At exactly square root of 10, the two powers are equally near by ratio.
Spot the mistake 3
Nearest power of ten
Learner claim
Lifting 2 kg through 2 m gives about 40 J. A learner says the order of magnitude is 10 J, because 40 is closer to 10 than to 100. Correct the claim.
Find the order of magnitude
Show solution step by step
Estimate the energy
Reason
Near the Earth, Δ Eₚ = mgh.Working
Answer
Δ Eₚ ≈ (2)(10)(2) = 40 J = 4 × 10¹ J
Compare by ratio
Working
Answer
40 is 4 times 10 but only 2.5 times smaller than 100, so it is nearer 100 by ratio. Equivalently, 4 > 3.16. The order of magnitude is 10² J.
Common mistake
Powers of ten are spaced by multiplying, so “nearest” means nearest by ratio, not by difference.
Try it yourself 4
Thickness of one sheet of paper
Problem
A stack of 200 identical sheets of paper is 18 mm thick. Find the thickness of one sheet in metres and the nearest power of ten in metres. Enter each as a number using decimal or e notation. Then explain why you would measure a stack rather than one sheet.
Find the thickness and its order of magnitude
Show the steps Optional working
Hints
Hint 1
average first
Divide by 200, convert to metres, then write the answer in standard form.
Show solution step by step
Find one sheet
Reason
Assuming identical sheets with no gaps, the stack is 200 equal thicknesses.Working
Answer
t = (18 mm)/200 = 0.09 mm = 9 × 10⁻⁵ m
Order of magnitude and method
Working
Answer
9 > 3.16, so the order of magnitude is 10⁻⁴ m. One sheet is too thin to measure with a ruler, but the stack is 18 mm: the same reading uncertainty is a much smaller fraction of the stack.
Check your understanding 3
Explain why measuring the stack is better, and identify an assumption about the sheets. You may write your reasoning on paper.
Show answer
The same ruler uncertainty is a smaller fraction of 18 mm than of one sheet’s thickness. Dividing by 200 reduces the absolute uncertainty by 200 while preserving the stack’s fractional uncertainty. The sheets must be alike and pressed together without gaps; air gaps would make the estimate too large.
Check that the answer makes sense
Check your understanding 4
A learner estimates that 0.20 m³ of water has a mass of 0.20 kg. How can you tell this is wrong, and what is a better estimate?
Show answer
0.20 kg is the mass of a cup of water, but 0.20 m³ is a bathful. The learner used a density of 1 kg m⁻³; water is 1000 kg m⁻³, so the mass is about 200 kg.
Spotted an error? Report a correction
Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
Last reviewed: