OCR A Physics · Module 3: Forces and motion
Materials: practice questions with answers
6 exam-style questions on materials. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Unit prefixes2 marks
A spring has a spring constant of 250 N m⁻¹ and obeys Hooke's law. Calculate the elastic energy stored when it is extended by 4.5 cm.
E = J
Show the answer and working
Answer E = 0.253 J
Convert x = 4.5 cm = 0.045 m. E = ½kx² = ½ × 250 × 0.045² = 0.253 J.
- Got 2530 J? You used 4.5 cm without converting to metres. x = 0.045 m, so E = ½kx² = ½ × 250 × 0.045² = 0.253 J.
- Got 5.63 J? You did not square the extension. E = ½kx² = ½ × 250 × 0.045² = 0.253 J.
- Using the right equation3 marks
A steel wire of length 2.50 m and diameter 0.40 mm supports a load of 60 N. The Young modulus of steel is 2.0 × 10¹¹ Pa. Calculate the extension of the wire, in mm.
x = mm
Show the answer and working
Answer x = 5.97 mm
A = π(d/2)² = 1.26 × 10⁻⁷ m². E = stress ÷ strain, so x = FL ÷ (AE) = (60 × 2.5) ÷ (1.26 × 10⁻⁷ × 2.0 × 10¹¹) = 5.97 mm.
- Got 1.49 mm? You used the diameter as the radius. A = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m², giving x = FL ÷ AE = 5.97 mm.
- Got 4.69 mm? You used d² for the area. The area is πd²/4 = 1.26 × 10⁻⁷ m², so x = 5.97 mm.
- Unit prefixes2 marks
A rod of cross-sectional area 1.2 cm² carries a tensile force of 18 kN. Calculate the stress in the rod, in MPa.
σ = MPa
Show the answer and working
Answer σ = 150 MPa
A = 1.2 × 10⁻⁴ m². σ = F ÷ A = 18 000 ÷ 1.2 × 10⁻⁴ = 1.5 × 10⁸ Pa = 150 MPa.
- Got 1.5 MPa? You converted cm² to m² using 10⁻². 1 cm² = 10⁻⁴ m², so A = 1.2 × 10⁻⁴ m² and σ = F ÷ A = 150 MPa.
- Got 15000 MPa? You converted cm² to m² using 10⁻⁶ (that is mm²). The factor is 10⁻⁴, so σ = 150 MPa.
- Reading graphs2 marks
A force-extension graph for a wire is a straight line through the origin up to a force of 12 N at an extension of 3.0 mm. Calculate the work done stretching the wire to this point, in mJ.
W = mJ
Show the answer and working
Answer W = 18 mJ
Work = area under the graph = ½ × 12 N × 3.0 × 10⁻³ m = 0.018 J = 18 mJ.
- Got 36 mJ? You used the area of a rectangle. The graph is a triangle, so W = ½Fx = ½ × 12 × 0.0030 = 18 mJ.
- Got 0.018 mJ? That is the answer in joules (0.018 J). Multiply by 1000 to give mJ: 18 mJ.
- Using the right equation1 mark
A metal wire is stretched beyond its elastic limit and the load is then removed. Which describes the unloading graph of force against extension?
- A A line parallel to the loading line that ends at a non-zero extension.
- B The same curve as the loading, returning to the origin.
- C A horizontal line at the maximum force.
Show the answer and working
Answer A: A line parallel to the loading line that ends at a non-zero extension.
Plastic deformation leaves a permanent extension and the unloading line is roughly parallel to the elastic part.
- Why not B? Beyond the elastic limit the deformation is plastic, so the wire does not return to its original length.
- Why not C? The force falls as the load is removed, so it cannot stay constant.
- Using the right equation1 mark
What is the energy stored per unit volume in a material obeying Hooke's law?
- A stress × strain
- B ½ × stress × strain
- C ½ × stress ÷ strain
Show the answer and working
Answer B: ½ × stress × strain
Energy per volume = area under the stress-strain graph = ½ × stress × strain.
- Why not A? That is twice the energy stored per unit volume; the average force is half the final force.
- Why not C? Dividing stress by strain gives the Young modulus, not an energy density.