1.74PH1 · Section 1: Forces and motion

Forces and elasticity

Key learnings

Hooke's Law: F = ke
Hooke's Law: extension is directly proportional to the applied force, up to the limit of proportionality. $F = ke$ ( $F$ in N, $k$ = spring constant in N/m, $e$ = extension in m). Extension $e$ = stretched length − natural length. On a force–extension graph the gradient = $k$ (not $1/k$ ). The limit of proportionality is where the line stops being straight — not the same as the elastic limit (permanent deformation), which lies slightly beyond it.
Elastic vs plastic deformation
Elastic deformation: the object returns to its original shape and size when the force is removed — no permanent change (elastic limit not exceeded). Plastic (inelastic) deformation: beyond the elastic limit the object does NOT return — a permanent change. The mark-scheme definition of elastic deformation needs both elements: (1) it returns to its original shape, (2) when the deforming force is removed.
Elastic PE: E = ½ke²
Stretching or compressing a spring elastically stores elastic potential energy $E = \frac{1}{2}ke^2$ ( $E$ in J, $k$ in N/m, $e$ in m). This equals the triangular area under a force–extension graph: $\frac{1}{2} \times F \times e = \frac{1}{2}ke^2$ . Units: N/m × m² = N·m = J. The extension must be in metres before substituting, and $e$ is squared. E.g. $k = 200$ N/m, $e = 0.08$ m: $E = \frac{1}{2} \times 200 \times 0.08^2 = 0.64$ J.

Common mistakes

Drawn from real examiner reports.

Total length used instead of extension

Extension is the CHANGE in length: $e = \text{stretched} - \text{natural length}$ . A spring with natural length 12 cm stretched to 20 cm has $e = 8$ cm = 0.08 m, NOT 20 cm. Substituting the total length into $F = ke$ gives an answer 2–3 times too large. Ask "change in length or total length?", and convert cm to m ( $\div 100$ ) so $k$ comes out in N/m.

Recurring pattern in Paper 1P calculations — candidates using total stretched length rather than extension, compounded by failure to convert cm to m. November 2024 Paper 1P general comment: "know the SI units and be able to convert from non-SI units."

"Elastic limit" written for the graph kink

Two distinct points, treated as non-interchangeable. Limit of proportionality: where the force–extension line stops being straight ( $F$ no longer ∝ $e$ ); the spring may still return to its length. Elastic limit: slightly beyond, where permanent (plastic) deformation begins. Label the kink "limit of proportionality" — writing "elastic limit" there is not credited.

June 2024 Paper 1P Q5(a): candidates who labelled the start of curvature on their graph as "elastic limit" were not credited — the examiner required "limit of proportionality" for that point on the graph.

Curve added to a Hooke's Law graph

A graph of a spring "obeying Hooke's Law" is a straight line through the origin only. Adding a curve after the straight section (to show behaviour beyond the limit of proportionality) is wrong for that question and costs the mark. Draw a curve only if the question also asks for behaviour beyond the limit — then include a labelled straight section AND the curve.

June 2024 Paper 1P Q5(a): the examiner reported this was more challenging than anticipated. Many candidates scored 2/3 marks because they included a curved section, which was incorrect for a graph showing a material obeying Hooke's Law.

F–e gradient inverted (gives 1/k)

The spring constant is the gradient of the F–e line: $k = \Delta F/\Delta e$ (N/m). Computing $\Delta e/\Delta F$ gives $1/k$ (units m/N), a wrong and much smaller number. Always divide the y-axis quantity (force) by the x-axis quantity (extension), and check the unit reads N/m.

Not squaring e in E = ½ke²

In $E = \frac{1}{2}ke^2$ the extension is squared, and it must be in metres. For $k = 80$ N/m and $e = 5$ cm: convert to 0.05 m, then $E = \frac{1}{2} \times 80 \times 0.05^2 = 0.10$ J. Forgetting the square, or leaving $e$ in cm, gives wildly wrong values (e.g. 200 J or more). Convert, then square.

Larger k misread as "stretches more"

A larger spring constant means a stiffer spring, not one that stretches more easily. $k = 200$ N/m needs 200 N for a 1 m extension, so it is harder to stretch than a $k = 50$ N/m spring. For a given force, the larger- $k$ spring extends LESS ( $e = F/k$ ).

Exam tips

F–e gradient = k, not 1/k
The gradient of a straight F–e line is the spring constant $k = \Delta F/\Delta e$ (N/m), not $\Delta e/\Delta F$ . Use two widely-spaced points: rise = change in force (N), run = change in extension (m). The gradient unit N/m confirms it is $k$ .
Convert cm to m before substituting
Convert every extension from cm to m ( $\div 100$ ) before substituting, so $F = ke$ yields $k$ in N/m and $E = \frac{1}{2}ke^2$ yields joules. Report $k$ in N/m unless the question specifically asks for N/cm.
Read the graph question precisely
"A spring obeying Hooke's Law" means draw a straight line through the origin ONLY — no curve. A curve is required only when the question also asks for behaviour beyond the limit of proportionality. Match what you draw to exactly what is asked.

Revision notes