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Edexcel International GCSE Physics · Section 1 Forces and motion
Hooke's Law and Elastic Behaviour — Required Practical
In this lesson you'll learn to: describe the practical to find how the extension of a
spring, a wire or an elastic band depends on the force; state Hooke's law and spot it on
a graph; explain what elastic means; and explain the loading and unloading curves of an
elastic band.
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Prefer to watch? Scroll to the bottom for the video on this practical
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A video walkthrough is coming soon — it will appear at the bottom of this page
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Start here — run the practical
Don't read anything yet. Hang 50 g masses on the spring one at a time and watch the table and the graph fill in. What happens to the extension each time? Keep going past 350 g, then take them all off — does the spring go back? Then try the elastic band, loading and unloading.
| mass / g | force / N | original length / cm | stretched length / cm | extension / cm |
|---|
Force on the spring = weight of the masses = m × g. A 50 g mass: 0.050 kg × 9.8 N/kg = 0.49 N. Extension = stretched length − original length. Read the ruler at eye level, level with the pointer.
right first time 0 / 0
1 · Does it obey Hooke's law?
2 · Did it behave elastically?
Look for a straight line through the origin. For elastic: does the unloading line come back to zero extension?
Click the steps in the right order
Your method
Start with the step you would do first.
Physics With Kate · physicswithkate.com — Edexcel International GCSE Physics
▲ The graph is extension on the y-axis, force on the x-axis — the way round Edexcel draws it.
Straight line through the origin? It obeys Hooke's law.
Back to its original length when the force is removed? It's elastic. The notes give you the practical, the two definitions and the elastic band's loop.
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Revision notes
🎯 Aim
Investigate how the extension of a spring (or a metal wire, or a rubber band) varies with the force applied to it.
Apparatus
- clamp stand, boss and clamp
- a spring (or an elastic band)
- a metre ruler, clamped vertically beside it — with a set square or pointer to read it level
- a mass hanger and 50 g masses
Method (workbook page 11)
- Measure the original length of the spring with the ruler.
- Add a 50 g mass to the spring.
- Calculate the force: F = m × g = 0.050 kg × 9.8 N/kg = 0.49 N.
- Measure the stretched length of the spring.
- Calculate the extension: extension = stretched length − original length.
- Add another 50 g and repeat, five more times.
- Plot a graph of extension (y-axis) against force (x-axis).
| Variable | In this practical |
|---|---|
| Independent — what you change | the force (the number of 50 g masses) |
| Dependent — what you measure | the extension |
| Control — what you keep the same | the same spring, the same temperature |
⚠ Trap 1 — extension is not length
Extension = stretched length − original length. Plotting the length itself gives a line that doesn't go through the origin, and you can't then say it obeys Hooke's law.
Example results
| mass / g | force / N | original length / cm | stretched length / cm | extension / cm |
|---|---|---|---|---|
| 50 | 0.49 | 5.0 | 8.0 | 3.0 |
| 100 | 0.98 | 5.0 | 11.0 | 6.0 |
| 150 | 1.47 | 5.0 | 14.0 | 9.0 |
| 200 | 1.96 | 5.0 | 17.0 | 12.0 |
| 250 | 2.45 | 5.0 | 20.0 | 15.0 |
| 300 | 2.94 | 5.0 | 23.0 | 18.0 |
Hooke's law
⭐ Learn this word for word
The extension of a spring is directly proportional to the force applied to it — as long as the limit of proportionality is not exceeded.
On a graph of extension against force, an object that obeys Hooke's law gives a straight line through the origin. You need both: straight, and through the origin. Double the force and the extension doubles.
You don't need an equation for Hooke's law in Edexcel International GCSE — you need the statement, and you need to recognise it on a graph.
Elastic behaviour
⭐ Definition
A material is elastic if it returns to its original shape and length once the forces causing the deformation are removed.
Stretch a spring too far and it goes past its elastic limit: when you take the masses off it stays longer than it was. It has been permanently deformed, so it did not behave elastically.
Springs, wires and elastic bands
| Object | Extension–force graph | Hooke's law? |
|---|---|---|
| helical spring | straight through the origin, then curves past the limit of proportionality | yes, up to the limit |
| metal wire | straight through the origin for a small extension, then stretches a lot (permanently) | yes, at first |
| elastic (rubber) band | curved from the start, an S shape | no |
The elastic band: loading and unloading
Load an elastic band with masses, then take them off one at a time. The unloading curve follows a different path from the loading curve, but still comes back to zero extension:
- back to zero → it returns to its original length → the band is elastic;
- not a straight line → it does not obey Hooke's law;
- the area between the two curves is energy that has been transferred to the surroundings as heat — the band warms up.
Evaluation
| What to do | Why | |
|---|---|---|
| Precaution | Read the ruler at eye level, using a pointer or set square | avoids parallax error |
| Precaution | Wait for the spring to stop bouncing before reading | a steady reading |
| Improvement | Repeat with the masses coming off, and take a mean | reduces the effect of random error |
| Safety | Wear eye protection; keep feet clear; clamp the stand to the bench | springs can snap back; masses can fall |
How to write it in the exam
"Does it obey Hooke's law? Explain."
Yes / No — the graph is / is not a straight line through the
origin, so the extension is / is not directly proportional to the force.
answer the question describe the graph — both parts say "directly proportional"
A spring stretches by 12.5 mm with a 250 g mass on it. A second 250 g mass is added. The spring obeys Hooke's law. State the new extension and explain.
25 mm. The force doubles, and because the spring obeys Hooke's law the extension is directly proportional to the force, so the extension doubles too.
Calculate the force on a spring from a 250 g mass. (g = 10 N/kg)
250 g = 0.25 kg. W = m × g = 0.25 × 10 = 2.5 N
Explain how the loading and unloading graph shows that an elastic band behaves elastically.
The unloading curve returns to the origin, so when the force is removed the band goes back to its original length.
What does the area between the loading and unloading curves represent?
Energy that has been transferred to the surroundings as heat — not all the energy stored while loading is given back while unloading.
✅ Check your answer
For Hooke's law, say both "straight line" and "through the origin", then "directly proportional". For elastic, say original length and force removed. Extension, not length.
📘 Now do it in your workbook
Page 11: draw the set-up, fill the gaps in the method, complete the results, plot the graph and write the conclusion. Page 12: state Hooke's law, define elastic, and read the spring and elastic band graphs.
Edexcel iGCSE Forces and Motion Workbook · pages 11–12✅ Can you do it? Tick as you go
0% complete🎉 Nice work! You've ticked off every objective for this practical.
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Prefer to watch? Here's the whole thing
Video walkthrough — coming soon
The whole practical, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
The whole practical, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it. The video is on its way.
🎬
Video coming soon
The walkthrough for this practical is being filmed and will appear right here. Until then, the notes and the free workbook pages above cover every mark.
Now put it into practice
Pages 11 and 12 of the Forces and Motion workbook are free: the practical write-up, Hooke's law, elastic behaviour and two exam-style questions. The full book covers the whole topic with worked answers.
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