13 Solid materials Exam practice questions

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Magnus Perry
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1 Pages Exam practice questions 1 a) The toughest material has the largest area beneath the curve the answer is C. b) The strongest material has the greatest breaking stress the answer is B. c) A polymer stretches easily as the chain molecules untangle, and then stiffens (the gradient rises) when they are aligned the answer is D. d) A brittle material has little or no plastic extension before it breaks the answer is B. 2 a) Hooke s law states that, up to a certain limit, the extension is directly proportional to the load. b) F 8.0 N k = = = 64 N m x m [Total 3 Marks] 3 Ductile materials can be drawn into wires Malleable metals can be readily hammered into thin sheets Copper can be drawn into wires for electrical work and gold can be hammered into very thin leaves for decoration. [Total 2 Marks] 4 If the plastic kettle is dropped, or receives a sharp blow, it is more likely to crack or shatter than the steel one. This is because the plastic is a brittle material. The steel is much tougher, and is capable of absorbing much more energy, and may only suffer from minor indentations due to plastic deformation. The surface of the steel is harder than the plastic and so is less likely to be scratched. The plastic kettle will retain its heat better than the steel one. It consists of long-chain molecules called polymers, which give it a low thermal conductivity. The plastic is also less dense than steel and so the kettle is lighter. (Six correct [3] 4/5 correct, 2/3 correct ) 5 a) i) C ii) iii) O A C D [Total 3 Marks] b) Calculate the area under the line. c) i) The wire returns to its original length. ii) The wire will be permanently extended. d) The wire regains its original stiffness it will follow an identical loading curve. [Total 9 Marks]

2 6 a) i) Stress (σ) = F A ii) Strain (ε) = l l iii) Young modulus E = σ ε b) σ = 50 N π ( m) = Pa ε = m 3 = m Pa 11 E = 3 = Pa [Total 5 Marks] 7 a) Hysteresis occurs when the unloading curve lags behind the loading curve. b) The blue (upper) line is the loading curve. c) The area enclosed by the loop represents the energy per unit volume (energy density) transferred to internal energy of the rubber during each cycle. d) The band is initially slightly stiff until the weak cross-links between the tangled chains are broken. The chains are then uncoiled, giving a large increase in strain for a little extra stress until the molecules become aligned. The band now becomes stiff as the strong covalent bonds between the atoms are stretched. On releasing the stress, the chains recoil until the initial amorphous state is regained. [Total 9 Marks] 8 a) Tension T = St St St E A l l T E A Cu Cu Cu = as l and l are the same for both wires T E A T T T T E A Cu Cu Pa π ( m) = = = 0.89 Cu St St St E A St St Pa π ( m) 4.00 kg 9.8 N kg = T Cu + T St = 0.89 T St + T St = 1.89 T St T St = 4.00 kg 9.8 N kg 1.89 = 20.8 N T Cu = 39.2 N 20.8 N = 18.4 N

3 b) l = Fl = 18.4 N 2.20 m = 1.4 EA Pa π ( m) [Total 8 Marks] 80 GPa 9) a) E = gradient of linear section = = 14 GPa b) Max. stress = 150 MPa c) v 2 = u 2 + 2as = m s m v = 6.3 m s mv 70 kg 6.3 m s F = = t 0.30 s = 1500 N 1460 N d) Compressive stress = π ( m) = Pa (Assuming that the impact is shared equally by both tibia.) This is much less than the maximum compressive stress that the bone can withstand (150 MPa). e) 10 a) Consider one half of the band: length = 0.75 m, extension = 0.40 m, cross-sectional area = 0.15 m m = m 2, tension in each half of the band = 100 N [Total 11 Marks] Fl Average E = A l 100 N 0.75 m = m m = Pa 3 MPa b) Initially the band extends easily because the long chain molecules are untangling, but, as the force increases the chains become aligned and require greater forces to stretch the intermolecular bonds. The band gets stiffer as the force increases, so its Young modulus gets bigger.

4 c) If Hooke s law were obeyed, F = kx would apply over the whole range of extension. Two bands in parallel would have twice the stiffness of a single band, and would therefore extend by half if the same force were applied. So we would expect an extension of 20 cm in this situation. With two bands, the tension in each section will be half that in the single band, but, because of the untangling of the molecules, the extension will be more than half of that of the single band with twice the tension. d) Work done in stretching = average force extension. Here, the average force is the same but the extension is smaller. So less work is done. e) Each time the band is stretched and released, hysteresis occurs. More work is done on the band on stretching than is done by the band when it contracts. The difference in work done is transferred as internal energy in the band. This increase in internal energy over many cycles results in a rise in temperature in the band. [Total 15 Marks] 11 a) i) Change in length = length after adding extra force length before C 4 = B 4 B 3 ( m = m m) ii) Work done = force change in length D 6 = A 6 C 6 (1.76 J = 16 N 0.11 m) iii) Total work done = work done before load added + work done after load added E 8 = E 7 + D 8 (2.72 J = 2.53 J J) b) The error is in column D. The work done should equal the average force times the extension, but here the maximum force is used. The model would have been better if an average was found by adding the initial force to the final force and dividing by two. e.g. D 6 = (A 5 + A 6)/2 C 6 Note: this assumes that the force varies in a linear fashion for this extension, which is not always true. c) Estimate the area under the curve by counting the number of squares. Calculate the value of work done for each square e.g. if 1 division on the force axis 2.0 N and 1 division on the extension axis m, one square J. A graph using these scale values will cover an area of about 150 squares, representing 3.0 J of work done.

5 d) The area under the curve is the sum of the products F δx, where δx represents a tiny increase in length for a given constant force (A = F dx). Drawing a smooth curve gives a better representation of the work done than using the model. The value will be smaller than that in cell E 12 because the values calculated in column D will always give a value that is too high. [Total 12 Marks]

Revision Guide for Chapter 4

Revision Guide for Chapter 4 Contents Student s Checklist Revision Notes Materials: properties and uses... 5 Materials selection charts... 5 Refraction... 8 Total internal reflection... 9 Conductors and