ExaminerPrep › Question bank › IAL Physics › 1.9 Stress, strain and the Young modulus
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Stress, strain and the Young modulus for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which row gives correct units for stress, strain and the Young modulus?
| Answer | Mark | Notes |
|---|---|---|
| Stress = force ÷ area, so N m−2 (Pa). Strain = length ÷ length, so it has no unit. The Young modulus = stress ÷ strain, so it has the unit of stress, N m−2. | ✓ 1 | Answer B |
Answer: B
Exam tip: Because strain has no unit, the Young modulus always has the same unit as stress — Pa, MPa or GPa.
A fishing line of diameter 0.40 mm supports a tension of 18 N.
What is the tensile stress in the line?
| Answer | Mark | Notes |
|---|---|---|
| A = πd2/4 = π × (0.40 × 10−3)2 ÷ 4 = 1.26 × 10−7 m2. σ = F/A = 18 ÷ 1.26 × 10−7 = 1.4 × 108 Pa. | ✓ 1 | Answer C |
Answer: C
Common error: Using the diameter as the radius in πr2. That makes the area four times too large and the stress four times too small.
The haul rope of a cable car is 1.20 km long. When the cars are loaded the rope extends by 0.84 m.
What is the strain in the rope?
| Answer | Mark | Notes |
|---|---|---|
| ε = Δx/x = 0.84 ÷ 1200 = 7.0 × 10−4. Both lengths must be in the same unit. | ✓ 1 | Answer B |
Answer: B
Exam tip: Strain is a ratio, so it has no unit — but only if the two lengths are in the same unit before you divide.
A steel guy wire helps to hold a radio mast upright. The wire has a diameter of 6.0 mm and the tension in it is 3.2 kN. The Young modulus of the steel is 2.1 × 1011 Pa.
State what is meant by tensile stress.
Calculate the cross-sectional area of the wire.
Calculate the tensile stress in the wire.
Calculate the strain in the wire.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the stretching force acting per unit cross-sectional area / F/A for forces pulling outwards | ✓ 1 | |
| Part (b) | ||
| π × (6.0 × 10−3)2 ÷ 4 = 2.8 × 10−5 m2 (2.83 × 10−5 m2) | ✓ 1 | |
| Part (c) | ||
| 3200 ÷ 2.83 × 10−5 = 1.1 × 108 Pa (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| ε = σ/E = 1.13 × 108 ÷ 2.1 × 1011 = 5.4 × 10−4 (allow error carried forward) | ✓ 1 | |
Exam tip: Keep the unrounded stress (1.132 × 108 Pa) in your calculator for the strain; round only the final answer of each part.
When a person stands on one leg, the shaft of the thigh bone (femur) carries a compressive force of about 1.4 kN. The shaft is 0.45 m long and its bone has a cross-sectional area of 3.5 × 10−4 m2. The Young modulus of bone is 1.7 × 1010 Pa.
State what is meant by compressive strain.
Calculate the compressive stress in the bone.
Calculate the amount by which the shaft shortens.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the reduction in length divided by the original length | ✓ 1 | |
| Part (b) | ||
| 1400 ÷ 3.5 × 10−4 = 4.0 × 106 Pa | ✓ 1 | |
| Part (c) | ||
| ε = σ/E = 4.0 × 106 ÷ 1.7 × 1010 = 2.35 × 10−4 | ✓ 1 | |
| Δx = εx = 2.35 × 10−4 × 0.45 = 1.1 × 10−4 m (0.11 mm) (allow error carried forward) | ✓ 1 | |
Common error: Calling a compressive change in length an “extension” and adding it. In compression the sample gets shorter; the equations are the same, but the length decreases.
An engineer chooses materials for parts of a garden tool.
State what is meant by a brittle material.
State what is meant by a ductile material.
The cutting edge of the tool must not be scratched or dented by stones. Name the property the edge material needs.
The handle must bend very little when a large force is applied. State the property the handle material needs and the feature of its stress–strain graph that shows it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a material that breaks with little or no plastic deformation | ✓ 1 | |
| Part (b) | ||
| a material that undergoes large plastic deformation under tension / can be drawn out into a wire | ✓ 1 | |
| Part (c) | ||
| hard / hardness | ✓ 1 | |
| Part (d) | ||
| stiff, shown by a steep gradient of the straight section / large Young modulus | ✓ 1 | |
Exam tip: When asked for a property, give the single word (stiff, strong, hard, tough, brittle, ductile, malleable) and, where asked, the graph feature that goes with it.
The graph shows the stress–strain graph for a sample of an aluminium alloy, up to its limit of proportionality.
What is the Young modulus of the alloy?
| Answer | Mark | Notes |
|---|---|---|
| The line passes through the gridline intersection (2.0 × 10−3, 140 MPa). E = gradient = 140 × 106 ÷ 2.0 × 10−3 = 7.0 × 1010 Pa. | ✓ 1 | Answer C |
Answer: C
Exam tip: Read the axis labels before taking a gradient: here the stress is in MPa (× 106) and the strain is in units of 10−3. Write both powers of ten into the division.
Figure 1 shows the cross-section of a hollow steel tube and a side view of the tube supporting a compressive load.
What is the compressive stress in the tube?
| Answer | Mark | Notes |
|---|---|---|
| Only the metal ring carries the load. A = π(D2 − d2)/4 = π × (0.0402 − 0.0302) ÷ 4 = 5.50 × 10−4 m2. σ = 55 × 103 ÷ 5.50 × 10−4 = 1.0 × 108 Pa. | ✓ 1 | Answer C |
Answer: C
Common error: Using the outer diameter alone. The hole carries no force, so subtract the area of the hole from the area of the full circle.
The graph shows stress–strain graphs for four materials, W, X, Y and Z, each tested until it breaks.
Which material is both stiff and ductile, compared with the others?
| Answer | Mark | Notes |
|---|---|---|
| Stiff means a steep initial gradient (large Young modulus): W and X. Ductile means a long plastic (curved) region before breaking: W and Y. Only W has both. | ✓ 1 | Answer A |
Answer: A
Exam tip: Judge stiffness only from the straight section at the start. Judge ductility from how far the graph continues after it stops being straight.
Figure 1 shows a crate hanging at rest from a lifting beam on four identical vertical steel cables. The Young modulus of the steel is 2.0 × 1011 Pa. g = 9.81 N kg−1.
Calculate the tension in each cable.
Calculate the tensile stress in each cable.
Calculate the extension of each cable.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1500 × 9.81 ÷ 4 = 3.7 × 103 N (3679 N) | ✓ 1 | |
| Part (b) | ||
| A = π × (10 × 10−3)2 ÷ 4 = 7.85 × 10−5 m2 | ✓ 1 | |
| σ = 3679 ÷ 7.85 × 10−5 = 4.7 × 107 Pa (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| ε = σ/E = 4.68 × 107 ÷ 2.0 × 1011 = 2.34 × 10−4 | ✓ 1 | |
| Δx = 2.34 × 10−4 × 6.0 = 1.4 × 10−3 m (1.4 mm) (allow error carried forward) | ✓ 1 | |
Common error: Using the whole weight of the crate for one cable. The four vertical cables share the weight equally, so each carries a quarter of it.
The graph shows the stress–strain graph for a sample of nylon, tested until it breaks at S.
State the name given to point P.
Use the graph to state the breaking stress (ultimate tensile stress) of the nylon.
P is at a strain of 0.020. Calculate the Young modulus of the nylon.
The load is removed when the sample reaches point Q. Describe what happens to the length of the sample.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| limit of proportionality | ✓ 1 | |
| Part (b) | ||
| 60 MPa / 6.0 × 107 Pa — the stress at R, the highest point | ✓ 1 | |
| Part (c) | ||
| E = 40 × 106 ÷ 0.020 = 2.0 × 109 Pa | ✓ 1 | |
| Part (d) | ||
| it does not return to its original length: the strain falls along a line parallel to OP and a permanent (plastic) extension remains | ✓ 1 | |
Exam tip: The breaking stress is the highest point of the graph, not the point where the line ends. Here the sample breaks at S, at a lower stress than R.
A concrete test cylinder is squeezed between two steel plates until it crushes. Figure 1 shows the cylinder and a graph of the compressive force against the decrease in length of the cylinder.
Calculate the compressive stress in the concrete when it crushes.
Calculate the compressive strain when the cylinder crushes.
Calculate the Young modulus of the concrete.
State the property of concrete shown by the shape of the graph, and the feature of the graph that shows it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = π × 0.1002 ÷ 4 = 7.85 × 10−3 m2 | ✓ 1 | |
| σ = 300 × 103 ÷ 7.85 × 10−3 = 3.8 × 107 Pa (allow error carried forward) | ✓ 1 | |
| Part (b) | ||
| 0.25 ÷ 200 = 1.3 × 10−3 (1.25 × 10−3) | ✓ 1 | |
| Part (c) | ||
| 3.82 × 107 ÷ 1.25 × 10−3 = 3.1 × 1010 Pa (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| brittle: the graph is a straight line right up to the point where it crushes, with no plastic region | ✓ 1 | |
Exam tip: Strain is a ratio of two lengths, so the decrease in length and the original length can both stay in millimetres — as long as they are in the same unit.
A strip of animal tendon is 0.12 m long and has a cross-sectional area of 3.0 × 10−5 m2. A tension of 600 N extends it by 2.4 mm, and it obeys Hooke's law.
What is the Young modulus of the tendon?
| Answer | Mark | Notes |
|---|---|---|
| σ = 600 ÷ 3.0 × 10−5 = 2.0 × 107 Pa. ε = 2.4 × 10−3 ÷ 0.12 = 0.020. E = σ/ε = 1.0 × 109 Pa. | ✓ 1 | Answer C |
Answer: C
Exam tip: Work out the stress and the strain separately, writing each with its power of ten, before dividing. It keeps the unit conversions visible.
Wires X and Y are made of the same metal. Wire Y is three times as long as wire X and has half its diameter. The same tension is applied to each, and both obey Hooke's law.
What is the value of (extension of Y) ÷ (extension of X)?
| Answer | Mark | Notes |
|---|---|---|
| Δx = Fx/(EA), with A ∝ d2. Tripling x triples Δx; halving d quarters A, which multiplies Δx by 4. Ratio = 3 × 4 = 12. | ✓ 1 | Answer D |
Answer: D
Exam tip: Write the extension in terms of the quantities that change, Δx ∝ x/d2, then substitute the factors. E and F are the same and cancel.
Four materials are tested to breaking point and a stress–strain graph is plotted for each. The material with the greatest area under its graph, up to the breaking point, is best described as the most
| Answer | Mark | Notes |
|---|---|---|
| The area under a stress–strain graph up to breaking measures the energy absorbed per unit volume before the material breaks. A material that absorbs a lot of energy before breaking is tough. | ✓ 1 | Answer D |
Answer: D
Exam tip: Match each property to one feature of the graph: stiff — gradient; strong — highest stress; ductile — long plastic region; tough — area under the curve.
The stress–strain graph for a new ceramic is a steep straight line from the origin that ends suddenly at a very high stress, with no curved section.
Which description of the ceramic is correct?
| Answer | Mark | Notes |
|---|---|---|
| Steep gradient → large Young modulus → stiff. Breaks at a very high stress → strong. No plastic (curved) region before breaking → brittle. | ✓ 1 | Answer A |
Answer: A
Common error: Thinking that brittle means weak. A brittle material may need a very large stress to break; brittle describes how it breaks — suddenly, with no plastic deformation.
A titanium-alloy tie rod has a circular cross-section of diameter 6.0 mm. The breaking stress of the alloy is 9.0 × 108 Pa.
What is the greatest tension the rod can withstand?
| Answer | Mark | Notes |
|---|---|---|
| A = π × (6.0 × 10−3)2 ÷ 4 = 2.83 × 10−5 m2. F = σA = 9.0 × 108 × 2.83 × 10−5 = 2.5 × 104 N. | ✓ 1 | Answer A |
Answer: A
Exam tip: The breaking stress belongs to the material; the breaking force belongs to this rod. Multiply the breaking stress by the rod's own cross-sectional area.
A steel guitar string has a vibrating length of 0.65 m and a diameter of 0.25 mm. When it is tuned, the tension in it is 70 N. The Young modulus of the steel is 2.0 × 1011 Pa and its breaking stress is 2.2 × 109 Pa. Assume the string obeys Hooke's law up to this tension.
Calculate the tensile stress in the tuned string.
Calculate the extension of the 0.65 m length as it is tightened from zero tension.
Calculate the greatest tension the string can withstand.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = π × (0.25 × 10−3)2 ÷ 4 = 4.91 × 10−8 m2 | ✓ 1 | |
| σ = 70 ÷ 4.91 × 10−8 = 1.4 × 109 Pa (1.43 × 109 Pa) | ✓ 1 | |
| Part (b) | ||
| ε = σ/E = 1.43 × 109 ÷ 2.0 × 1011 = 7.13 × 10−3 | ✓ 1 | |
| Δx = 7.13 × 10−3 × 0.65 = 4.6 × 10−3 m (4.6 mm) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| 2.2 × 109 × 4.91 × 10−8 = 107.99 N, so 107 N (rounded down: 108 N or 110 N would exceed the breaking stress) | ✓ 1 | |
Exam tip: A diameter in millimetres must be converted before squaring: (0.25 × 10−3 m)2 = 6.25 × 10−8 m2, not 0.0625.
A polymer fibre is tested to breaking point. Its stress–strain graph is a straight line from the origin to a stress of 60 MPa at a strain of 0.020. It then curves, with a decreasing gradient, to a maximum stress of 85 MPa at a strain of 0.15. The stress falls slightly after this and the fibre breaks at a strain of 0.18 and a stress of 78 MPa.
Calculate the Young modulus of the polymer.
State the breaking stress (ultimate tensile stress) of the polymer.
A fibre of this polymer has a diameter of 0.20 mm. Calculate the greatest load it can support.
State and explain whether the polymer is brittle or ductile.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient of the straight section = 60 × 106 ÷ 0.020 | ✓ 1 | |
| = 3.0 × 109 Pa | ✓ 1 | |
| Part (b) | ||
| 85 MPa / 8.5 × 107 Pa — the maximum stress, not the stress at fracture | ✓ 1 | |
| Part (c) | ||
| A = π × (0.20 × 10−3)2 ÷ 4 = 3.14 × 10−8 m2 | ✓ 1 | |
| F = 85 × 106 × 3.14 × 10−8 = 2.670 N, so 2.67 N (rounded down: 2.7 N would exceed the breaking stress) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| ductile: most of its strain before breaking (from 0.020 to 0.18) is plastic deformation | ✓ 1 | |
Common error: Quoting the stress at the end of the line (78 MPa) as the breaking stress. The breaking stress, or ultimate tensile stress, is the greatest stress the material withstands.
Two fibres, A and B, are considered for the core of a climbing rope. To a good approximation, the stress–strain graph of each is a straight line from the origin to its breaking point.
Fibre A breaks at a stress of 0.90 GPa and a strain of 0.20.
Fibre B breaks at a stress of 3.0 GPa and a strain of 0.025.
Calculate the Young modulus of each fibre.
State which fibre is stronger.
The area under a stress–strain graph up to breaking is the energy absorbed per unit volume. Show that fibre A absorbs more than twice as much energy per unit volume as fibre B before breaking.
The rope must stop a falling climber without breaking. Suggest which fibre is more suitable, giving a reason.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A: 0.90 × 109 ÷ 0.20 = 4.5 × 109 Pa | ✓ 1 | |
| B: 3.0 × 109 ÷ 0.025 = 1.2 × 1011 Pa | ✓ 1 | |
| Part (b) | ||
| B, because its breaking stress is greater | ✓ 1 | |
| Part (c) | ||
| A: ½ × 0.90 × 109 × 0.20 = 9.0 × 107 J m−3; B: ½ × 3.0 × 109 × 0.025 = 3.75 × 107 J m−3 | ✓ 1 | |
| ratio = 9.0 ÷ 3.75 = 2.4, which is more than 2 | ✓ 1 | |
| Part (d) | ||
| A, because it is tougher: it absorbs more energy per unit volume before breaking (and its lower stiffness gives a more gradual stop) | ✓ 1 | |
Exam tip: Stronger and tougher are different questions. Answer “stronger” from the breaking stress and “tougher” from the area under the graph.
A student is asked to determine the Young modulus of copper using a length of copper wire.
Describe a method the student could use. Include the measurements, the instruments used and how the results are processed to give the Young modulus.
State one safety precaution for this experiment.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Indicative content — one mark for each point made: | — | |
| use a long wire (2 m or more), clamped at one end and passing over a pulley, with a small load to straighten it | ✓ 1 | |
| measure the original length from the clamp to a pointer on the wire with a metre rule or tape measure | ✓ 1 | |
| measure the diameter with a micrometer at several places along the wire, in two directions at right angles, and take the mean | ✓ 1 | |
| add masses in equal steps and record the extension from the movement of the pointer against a fixed scale or vernier | ✓ 1 | |
| calculate stress = mg/A with A = πd2/4, and strain = Δx/x, for each load | ✓ 1 | |
| plot stress against strain; the Young modulus is the gradient of the straight section | ✓ 1 | |
| Part (b) | ||
| wear eye protection in case the wire snaps / put a box of padding under the masses and keep feet clear | ✓ 1 | |
Exam tip: In a method description, link every quantity in the final equation to an instrument and say how it is used. Finish by saying exactly what is plotted and what the gradient gives.
Figure 1 shows a micrometer screw gauge closed on a test wire. The micrometer has no zero error.
What is the cross-sectional area of the wire?
| Answer | Mark | Notes |
|---|---|---|
| Sleeve: the half-millimetre mark is showing, so 0.50 mm. Thimble: 38 × 0.01 mm = 0.38 mm. d = 0.88 mm. A = πd2/4 = π × (0.88 × 10−3)2 ÷ 4 = 6.1 × 10−7 m2. | ✓ 1 | Answer B |
Answer: B
Common error: Missing the half-millimetre mark below the datum line. If a half-millimetre mark is showing past the last whole millimetre, add 0.50 mm to the thimble reading.
The graph shows how the stress varies with strain for two materials, M and N, up to the point where each breaks.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| Strength is the greatest stress before breaking: M reaches 400 MPa, N only 250 MPa, so M is stronger. Toughness is shown by the area under the graph up to breaking (energy absorbed per unit volume): M gives ½ × 400 × 106 × 0.010 = 2.0 × 106 J m−3; N gives about 9.1 × 106 J m−3, so N is tougher. | ✓ 1 | Answer B |
Answer: B
Exam tip: Strong and tough are answered from different features: strong from the highest point, tough from the area under the whole graph up to breaking.
Wires of a nickel alloy with different cross-sectional areas are pulled until they break. The graph shows the breaking force plotted against the cross-sectional area, with a best-fit line through the origin.
What is the breaking stress (ultimate tensile stress) of the alloy?
| Answer | Mark | Notes |
|---|---|---|
| Breaking stress = breaking force ÷ area = gradient of the line. Using (2.0 mm2, 1.2 kN): 1.2 × 103 N ÷ 2.0 × 10−6 m2 = 6.0 × 108 Pa. | ✓ 1 | Answer D |
Answer: D
Common error: Converting mm2 to m2 with 10−3 or 10−4. Since 1 mm = 10−3 m, 1 mm2 = (10−3)2 m2 = 10−6 m2.
Samples of the same nylon cord, of different original lengths, each carry a tension of 50 N. The cord has a diameter of 1.6 mm. The graph shows the extension of each sample plotted against its original length.
What is the Young modulus of the nylon?
| Answer | Mark | Notes |
|---|---|---|
| The gradient is the strain: 30 × 10−3 m ÷ 2.0 m = 0.015. A = π × (1.6 × 10−3)2 ÷ 4 = 2.01 × 10−6 m2, so σ = 50 ÷ 2.01 × 10−6 = 2.49 × 107 Pa. E = σ/ε = 1.7 × 109 Pa. | ✓ 1 | Answer D |
Answer: D
Exam tip: Ask what the gradient of a graph means before using it. Extension ÷ original length is the strain, which is the same for every sample because each one carries the same stress.
In an experiment to find the Young modulus of a metal, a vernier scale reads the position of a pointer fixed to a test wire of original length 2.00 m. Figure 1 shows the vernier readings with no load and with the full load.
What is the strain in the wire with the full load?
| Answer | Mark | Notes |
|---|---|---|
| No load: 23 mm + 4 × 0.1 mm = 23.4 mm. Full load: 25 mm + 1 × 0.1 mm = 25.1 mm. Extension = 1.7 mm = 1.7 × 10−3 m. ε = 1.7 × 10−3 ÷ 2.00 = 8.5 × 10−4. | ✓ 1 | Answer A |
Answer: A
Exam tip: On a vernier, take the whole millimetres from the main-scale mark just before the vernier zero, then add 0.1 mm for each vernier division up to the line that matches a main-scale mark.
A student determines the Young modulus of copper using a wire of original length 2.40 m. Micrometer readings of the diameter, taken at different places along the wire, are: 0.42 mm, 0.41 mm, 0.42 mm, 0.43 mm, 0.42 mm. The graph shows the student's results with a best-fit line.
Calculate the mean diameter of the wire.
Determine the gradient of the best-fit line, in N m−1.
Calculate the Young modulus of copper.
Explain why the diameter was measured at several places along the wire.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.420 mm | ✓ 1 | |
| Part (b) | ||
| uses a large triangle on the line, e.g. 28 N ÷ 4.0 × 10−3 m | ✓ 1 | |
| = 7.0 × 103 N m−1 (allow 6.8–7.2 × 103) | ✓ 1 | |
| Part (c) | ||
| A = π × (0.420 × 10−3)2 ÷ 4 = 1.39 × 10−7 m2 | ✓ 1 | |
| E = gradient × x/A = 7.0 × 103 × 2.40 ÷ 1.39 × 10−7 = 1.2 × 1011 Pa (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the wire may not be uniform (or perfectly circular), so a mean of several readings gives a more representative value of the diameter / reduces the effect of random error | ✓ 1 | |
Exam tip: With load on the y-axis and extension on the x-axis, the gradient is F/Δx = EA/x, so E = gradient × x/A. Convert the extension to metres before taking the gradient.
Figure 1 shows a student's arrangement for determining the Young modulus of a metal wire. The pointer moves against a millimetre scale fixed to the bench. The student records the original length as 2.50 m and reads each extension from the movement of the pointer.
State the original length that should be used with the extensions read at the pointer. Give its value from Figure 1.
Calculate the factor by which the student's value of the Young modulus is wrong, and state whether it is too large or too small.
At the largest load the pointer moves 1.0 mm. Using an uncertainty of half the scale resolution, calculate the percentage uncertainty in this extension.
The pointer is moved to 2.40 m from the clamp and the same load is applied. Calculate the new extension at the pointer and its percentage uncertainty.
Before any readings are taken, a small load is hung on the wire. Explain why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the length from the clamp to the pointer, 0.60 m | ✓ 1 | |
| Part (b) | ||
| E = Fx/(AΔx) is proportional to x; factor = 2.50 ÷ 0.60 = 4.2 (4.17) | ✓ 1 | |
| too large | ✓ 1 | |
| Part (c) | ||
| 0.5 ÷ 1.0 × 100 = 50% | ✓ 1 | |
| Part (d) | ||
| extension = 1.0 × 2.40 ÷ 0.60 = 4.0 mm | ✓ 1 | |
| 0.5 ÷ 4.0 × 100 = 13% (12.5%; allow 12–13%) (allow error carried forward) | ✓ 1 | |
| Part (e) | ||
| to pull the wire straight / remove kinks, so that the measured extensions are due to stretching only | ✓ 1 | |
Exam tip: The original length must be measured to the point where the extension is measured. A long length from clamp to pointer gives a larger extension and a smaller percentage uncertainty.
The graphs show the stress–strain graph for spider dragline silk, drawn on two different strain scales. P is the limit of proportionality. A single thread of silk has a diameter of 4.0 μm.
Use panel A to determine the Young modulus of the silk.
Calculate the tension in a single thread at the limit of proportionality.
Use panel B to state the breaking stress of the silk.
A steel wire has a similar breaking stress but breaks at a strain of about 0.01. Explain which material is tougher.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient of the straight section, e.g. 200 × 106 ÷ 0.020 | ✓ 1 | |
| = 1.0 × 1010 Pa | ✓ 1 | |
| Part (b) | ||
| A = π × (4.0 × 10−6)2 ÷ 4 = 1.26 × 10−11 m2 | ✓ 1 | |
| F = 200 × 106 × 1.26 × 10−11 = 2.5 × 10−3 N (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| 1100 MPa / 1.1 × 109 Pa (allow 1080–1120 MPa) | ✓ 1 | |
| Part (d) | ||
| silk: it reaches a similar stress but breaks at a much larger strain (0.27), so the area under its graph — the energy absorbed per unit volume before breaking — is much greater | ✓ 1 | |
Exam tip: When a graph is drawn on two scales, read each quantity from the panel where it is clearest: the gradient from the enlarged start, the breaking point from the whole curve.
The graph shows force against extension for two wires, A and B. The wires are made of the same metal and have the same original length, 1.50 m. Wire A has a diameter of 0.30 mm.
Use line A to determine the Young modulus of the metal.
Deduce the diameter of wire B.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| reads a point on line A, e.g. 10 N at 2.0 mm | ✓ 1 | |
| A = π × (0.30 × 10−3)2 ÷ 4 = 7.07 × 10−8 m2 | ✓ 1 | |
| E = Fx/(AΔx) = 10 × 1.50 ÷ (7.07 × 10−8 × 2.0 × 10−3) = 1.1 × 1011 Pa | ✓ 1 | |
| Part (b) | ||
| for the same extension B needs 4 times the force, so its cross-sectional area is 4 times larger (same E and x) | ✓ 1 | |
| d ∝ √A, so d = 2 × 0.30 = 0.60 mm | ✓ 1 | |
Common error: Doubling the diameter when the area is doubled. Area depends on d2, so 4 times the area needs only twice the diameter.
In an experiment on a metal wire of original length 2.20 m and diameter 0.36 mm, a student plots extension on the y-axis against load on the x-axis. The straight line through the origin has a gradient of 1.8 × 10−4 m N−1.
What is the Young modulus of the metal?
| Answer | Mark | Notes |
|---|---|---|
| E = Fx/(AΔx) = x ÷ (A × gradient), since gradient = Δx/F. A = π × (0.36 × 10−3)2 ÷ 4 = 1.02 × 10−7 m2. E = 2.20 ÷ (1.02 × 10−7 × 1.8 × 10−4) = 1.2 × 1011 Pa. | ✓ 1 | Answer C |
Answer: C
Exam tip: Rearrange E = Fx/(AΔx) so that the gradient appears as one group. Here Δx/F is the gradient, so E = x/(A × gradient).
A uniform steel wire hangs vertically from a fixed point. The greatest stress is at the top, where the wire supports the weight of all the wire below it. The breaking stress of the steel is 1.2 × 109 Pa and its density is 7800 kg m−3. g = 9.81 N kg−1.
What is the greatest length of wire that can hang without breaking under its own weight?
| Answer | Mark | Notes |
|---|---|---|
| Stress at the top = weight ÷ area = ρALg/A = ρgL; the area cancels. Breaking when ρgL = σb: L = 1.2 × 109 ÷ (7800 × 9.81) = 1.568 × 104 m, so 1.56 × 104 m (rounded down: a longer wire would break). | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the weight as density × volume × g = ρALg. Dividing by A shows that the answer does not depend on how thick the wire is.
In an experiment to measure the Young modulus of copper, the original length of the wire is 1.80 m. A micrometer gives these readings of the diameter: 0.46 mm, 0.45 mm, 0.47 mm, 0.46 mm, 0.46 mm. The student plots extension against load. For loads up to 30 N the points lie on a straight line through the origin, and this line passes through the point (20.0 N, 1.73 mm). For loads above 30 N the points lie above the line.
Calculate the mean diameter and its percentage uncertainty, using half the range of the readings.
Calculate the cross-sectional area of the wire.
Use the point on the line to determine the Young modulus of copper.
Explain why the points for loads above 30 N lie above the line, and why they are not used to find the Young modulus.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mean = 0.46 mm (0.460 mm) | ✓ 1 | |
| uncertainty = (0.47 − 0.45) ÷ 2 = 0.01 mm; percentage = 0.01 ÷ 0.46 × 100 = 2.2% | ✓ 1 | |
| Part (b) | ||
| π × (0.460 × 10−3)2 ÷ 4 = 1.66 × 10−7 m2 | ✓ 1 | |
| Part (c) | ||
| E = Fx/(AΔx) = 20.0 × 1.80 ÷ (1.66 × 10−7 × 1.73 × 10−3) | ✓ 1 | |
| = 1.3 × 1011 Pa (1.25 × 1011 Pa) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| above 30 N the wire has passed its limit of proportionality, so the extension increases more than in proportion to the load | ✓ 1 | |
| the Young modulus is defined only where stress is proportional to strain, so only the straight section can be used | ✓ 1 | |
Exam tip: With extension on the y-axis, points above the line mean “more extension than Hooke's law predicts” — the first sign that the wire is starting to yield.
A helicopter lifts a cargo net of total mass 900 kg on a steel cable of cross-sectional area 1.5 × 10−4 m2. The helicopter and the net accelerate vertically upwards at 1.2 m s−2. The length of cable between the helicopter and the net is 40 m. The Young modulus of the steel is 2.0 × 1011 Pa and its breaking stress is 1.5 × 109 Pa. Ignore the mass of the cable. g = 9.81 N kg−1.
Calculate the tension in the cable.
Calculate the stress in the cable.
Calculate the extension of the 40 m length of cable.
Calculate the ratio breaking stress ÷ working stress for the cable at this moment.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T − mg = ma, so T = 900 × (9.81 + 1.2) | ✓ 1 | |
| = 9.9 × 103 N (9909 N) | ✓ 1 | |
| Part (b) | ||
| 9909 ÷ 1.5 × 10−4 = 6.6 × 107 Pa (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| ε = 6.61 × 107 ÷ 2.0 × 1011 = 3.30 × 10−4 | ✓ 1 | |
| Δx = 3.30 × 10−4 × 40 = 0.013 m (13 mm) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| 1.5 × 109 ÷ 6.61 × 107 = 23 | ✓ 1 | |
Common error: Using the weight alone as the tension. When the load accelerates upwards the cable must also supply the resultant force ma, so the tension is greater than mg.
A hanging sculpture is supported by a composite wire: a steel section 1.20 m long joined end to end to a brass section 0.80 m long. Both sections have a diameter of 0.80 mm. The wire carries a load of 60 N. The Young modulus of steel is 2.0 × 1011 Pa and that of brass is 1.0 × 1011 Pa. Both sections obey Hooke's law, and the weight of the wire is negligible.
Explain why the stress is the same in the two sections.
Calculate the stress in the wire.
Calculate the total extension of the composite wire.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the same tension acts all the way along the wire (the sections are in series) and the cross-sectional areas are equal | ✓ 1 | |
| Part (b) | ||
| A = π × (0.80 × 10−3)2 ÷ 4 = 5.03 × 10−7 m2 | ✓ 1 | |
| σ = 60 ÷ 5.03 × 10−7 = 1.2 × 108 Pa (1.19 × 108 Pa) | ✓ 1 | |
| Part (c) | ||
| steel: Δx = σx/E = 1.19 × 108 × 1.20 ÷ 2.0 × 1011 = 7.2 × 10−4 m | ✓ 1 | |
| brass: 1.19 × 108 × 0.80 ÷ 1.0 × 1011 = 9.5 × 10−4 m | ✓ 1 | |
| total = 1.7 × 10−3 m (1.67 mm) (allow error carried forward) | ✓ 1 | |
Exam tip: For sections joined end to end, find each extension separately with its own length and Young modulus, then add. Do not average the two Young moduli.
Figure 1 shows a brass rod made of two sections, P and Q, joined end to end. Forces of 6.0 kN pull on the ends of the rod. The Young modulus of brass is 1.0 × 1011 Pa.
What is the total extension of the rod?
| Answer | Mark | Notes |
|---|---|---|
| The same 6.0 kN tension acts in both sections. Δx = Fx/(EA) for each. P: A = π × 0.0122 ÷ 4 = 1.13 × 10−4 m2, Δx = 6000 × 0.40 ÷ (1.0 × 1011 × 1.13 × 10−4) = 2.12 × 10−4 m. Q: A = 5.03 × 10−5 m2, Δx = 6000 × 0.25 ÷ (1.0 × 1011 × 5.03 × 10−5) = 2.98 × 10−4 m. Total = 5.1 × 10−4 m = 0.51 mm. | ✓ 1 | Answer B |
Answer: B
Exam tip: Sections in series carry the same force but have different stresses when their areas differ. Find each extension with its own length and area, then add.
The graph shows the stress–strain graph for a metal wire. P is the limit of proportionality. The wire is loaded until it reaches point L and the load is then removed completely.
What is the strain that remains in the wire?
| Answer | Mark | Notes |
|---|---|---|
| Beyond the elastic limit, unloading follows a straight line parallel to the original line, of gradient E = 200 × 106 ÷ 2.0 × 10−3 = 1.0 × 1011 Pa. The strain recovered is 260 × 106 ÷ 1.0 × 1011 = 2.6 × 10−3. Permanent strain = 4.0 × 10−3 − 2.6 × 10−3 = 1.4 × 10−3. | ✓ 1 | Answer A |
Answer: A
Exam tip: When a ductile material is unloaded from beyond its elastic limit, sketch a line from the point parallel to the first straight section. Where it meets the strain axis is the permanent strain.
Figure 1 shows a reinforced-concrete column. The steel bars run the full height of the column and are bonded to the concrete, so the steel and the concrete shorten together. The Young modulus of the concrete is 3.0 × 1010 Pa and that of the steel is 2.0 × 1011 Pa.
Explain why the strain is the same in the steel and the concrete.
Calculate the total cross-sectional area of the steel bars and the cross-sectional area of the concrete.
Show that the strain in the column is about 5.9 × 10−4.
Calculate the stress in the steel and the fraction of the load carried by the steel bars.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| they have the same original length and, being bonded, shorten by the same amount, so Δx/x is the same | ✓ 1 | |
| Part (b) | ||
| steel: 4 × π × 0.0252 ÷ 4 = 1.96 × 10−3 m2 | ✓ 1 | |
| concrete: 0.3002 − 1.96 × 10−3 = 8.80 × 10−2 m2 | ✓ 1 | |
| Part (c) | ||
| load = σcAc + σsAs = ε(EcAc + EsAs) | ✓ 1 | |
| ε = 1.8 × 106 ÷ (3.0 × 1010 × 8.80 × 10−2 + 2.0 × 1011 × 1.96 × 10−3) = 1.8 × 106 ÷ 3.03 × 109 = 5.93 × 10−4 | ✓ 1 | |
| Part (d) | ||
| σs = 2.0 × 1011 × 5.93 × 10−4 = 1.2 × 108 Pa (1.19 × 108 Pa) | ✓ 1 | |
| fraction = 1.19 × 108 × 1.96 × 10−3 ÷ 1.8 × 106 = 0.13 (13%) (allow error carried forward) | ✓ 1 | |
Exam tip: When two materials are bonded side by side, they share the strain, not the stress. The stiffer material carries the larger stress.
Figure 1 shows a light rigid bar hanging from two vertical wires attached to its ends. A load hangs from the bar. Before the load is added, the bar is horizontal. The Young modulus of steel is 2.0 × 1011 Pa and that of brass is 1.0 × 1011 Pa.
Calculate the tension in each wire.
Calculate the stress in the steel wire.
Calculate the extension of each wire and state which end of the bar is lower.
Calculate the distance from the left end at which the load should hang so that the bar stays horizontal.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moments about the left end: Tbrass × 1.20 = 300 × 0.40, so Tbrass = 100 N | ✓ 1 | |
| Tsteel = 300 − 100 = 200 N | ✓ 1 | |
| Part (b) | ||
| A = π × (0.60 × 10−3)2 ÷ 4 = 2.83 × 10−7 m2; σ = 200 ÷ 2.83 × 10−7 = 7.1 × 108 Pa (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| steel: 200 × 2.00 ÷ (2.0 × 1011 × 2.83 × 10−7) = 7.1 × 10−3 m; brass: A = 5.03 × 10−7 m2, 100 × 2.00 ÷ (1.0 × 1011 × 5.03 × 10−7) = 4.0 × 10−3 m | ✓ 1 | |
| the left (steel) end is lower (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| equal extensions need Ts/(EsAs) = Tb/(EbAb), so Ts/Tb = 56.5 ÷ 50.3 = 1.125 and Tb = 300 ÷ 2.125 = 141 N | ✓ 1 | |
| distance = 141 × 1.20 ÷ 300 = 0.56 m | ✓ 1 | |
Exam tip: For a rigid bar on two wires, use moments to find the tensions first; only then use E = Fx/(AΔx) for each wire separately.
In an experiment on a steel wire, a student calculates the stress and strain for each load and plots the graph shown. The best-fit line does not pass through the origin.
Use the graph to determine the Young modulus of the steel.
Suggest why the best-fit line does not pass through the origin.
Explain why the gradient still gives a valid value of the Young modulus.
A straight 3.0 m length of this wire, of diameter 0.50 mm, is used to hang a picture. Calculate the load that extends it by 2.0 mm.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient from a large triangle, e.g. (4.0 − 2.0) × 108 ÷ (2.2 − 1.2) × 10−3 | ✓ 1 | |
| = 2.0 × 1011 Pa (allow 1.9–2.1 × 1011 Pa) | ✓ 1 | |
| Part (b) | ||
| the wire was not straight / had kinks at the start, so the first loads straightened it and each measured extension includes the same extra amount | ✓ 1 | |
| Part (c) | ||
| the extra extension is the same for every reading (a systematic shift), so it moves the line sideways without changing its gradient | ✓ 1 | |
| Part (d) | ||
| ε = 2.0 × 10−3 ÷ 3.0 = 6.67 × 10−4; σ = Eε = 2.0 × 1011 × 6.67 × 10−4 = 1.33 × 108 Pa | ✓ 1 | |
| A = π × (0.50 × 10−3)2 ÷ 4 = 1.96 × 10−7 m2 | ✓ 1 | |
| F = σA = 26 N (allow error carried forward) | ✓ 1 | |
Exam tip: A line that is straight but shifted along an axis still has a meaningful gradient. Use it, and say what caused the shift.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy