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A-Level Physics · guided topic map

Materials and deformation for Cambridge International AS & A Level Physics

Materials and deformation for A-Level Physics, organized into 1 syllabus topic and 5 mapped concept guides.

Syllabus topics
1
Mapped concept guides
5
Educational level
Cambridge International AS & A Level

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Work in order or jump to the concept named in your specification, course outline, or assignment.

6

Deformation of solids

AS Level foundations

5 guides
  1. 01Hooke's law and the limit of proportionalityMapped lesson
  2. 02Stress, strain, and Young modulusMapped lesson
  3. 03Elastic energy and force-extension graphsMapped lesson
  4. 04Stress-strain curvesMapped lesson
  5. 05Elastic and plastic behaviourMapped lesson

Diagrams

Materials and deformation as A-Level Physics draws it

The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 8.1Work, energy and power · Deformation of solidsA Level
A loaded steel wire hanging from a fixed supportfixed supportoriginal length2.50 msteel wirediameter 0.56 mm45 Nnot to scale

Figure comment

Fig. 8.1A steel wire hangs vertically from a rigid support drawn as a hatched ceiling, and a block is attached to its lower end. An arrow starting at the centre of that block points vertically downwards and is labelled 45 N. A dimension line to the left of the wire marks its original length as 2.50 m, and a leader line to the wire labels it as a steel wire of diameter 0.56 mm. The figure is not to scale.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

5 Work, energy and power6 Deformation of solidsOpen A Level question · AS Level foundations Q8
Guided questions 5 parts

Reading cue. The 45 N is the weight of the block, so it equals the tension only if the wire's own weight is ignored; 0.56 mm is a diameter, and 2.50 m is the length before loading.

  1. aCalculate Calculate the cross-sectional area of the wire labelled in Fig. 8.1.

    recall2 marks

    Check answer 2 marks
    1. A = πd²/4 with d = 0.56 × 10⁻³ m
    2. A = 2.5 × 10⁻⁷ m²

  2. bDetermine Steel of this type breaks at a tensile stress of 8.0 × 10⁸ Pa. Determine the greatest load this wire could support, and the factor by which the 45 N load could be increased before the wire breaks.

    routine3 marks

    Check answer 3 marks
    1. maximum load = stress × area = 8.0 × 10⁸ × 2.46 × 10⁻⁷
    2. = 2.0 × 10² N
    3. factor = 197 / 45 = 4.4

  3. cDeduce The wire is replaced by one of the same steel and the same original length but of diameter 1.12 mm, and the same 45 N load is hung from it. Deduce the factor by which the extension changes.

    demanding3 marks

    Check answer 3 marks
    1. doubling the diameter makes the cross-sectional area 4 times greater
    2. for the same load the stress, and hence the strain, is one quarter of its previous value
    3. the extension becomes one quarter of its previous value

  4. dSuggest The density of steel is 7800 kg m⁻³. Suggest, with a supporting calculation, whether taking the tension in the wire to be 45 N along its whole length is justified.

    top of the paper4 marks

    Check answer 4 marks
    1. volume of wire = 2.46 × 10⁻⁷ × 2.50 = 6.2 × 10⁻⁷ m³
    2. weight of wire = 7800 × 6.2 × 10⁻⁷ × 9.81 = 0.047 N
    3. this is about 0.1% of 45 N, and only the part of the wire below a given point adds to the tension there
    4. so treating the tension as 45 N throughout introduces no significant error

Transfer challenge

A climbing rope of unstretched length 12 m and cross-sectional area 1.1 × 10⁻⁴ m² stretches by 0.16 m when a climber of weight 750 N hangs at rest from it. Calculate the Young modulus of the rope material.

Check answer 3 marks
  1. stress = 750 / 1.1 × 10⁻⁴ = 6.8 × 10⁶ Pa
  2. strain = 0.16 / 12 = 0.013
  3. E = stress / strain = 5.1 × 10⁸ Pa