Questions on springs
1. (a) The extension of a spring is directly proportional to the applied force M1
as long as the elastic limit is not exceeded) A1
(b) (i) Correct pair of values read from the graph
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
(ii) extension, x = × 80 (= 133.33) (mm) C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J) A1
(iii) The spring has not exceeded its elastic limit B1
(iv) (elastic potential energy = kinetic energy)
M1
m and k are constant, therefore x v. M1
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2. (a) Young modulus = stress/strain
(As long as elastic limit is not exceeded) B1
(b) Strain has no units because it is the ratio of two lengths. B1
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3. (a) A brittle material does not have a plastic region / it breaks at its elastic limit. B1
(b) Ultimate tensile strength is breaking stress for a material B1
Materials can be chosen / tested to prevent collapse of the bridge B1
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4. (i) 1. Measure the time t for N oscillations. M1
frequency f = N/t A1
2. Measure the amplitude A of the oscillations using the ruler. M1
maximum speed is calculated using: vmax = (2πf )A A1
(ii) The maximum speed is doubled B1
because the frequency is the same and vmax A B1
(iii) F = (–) kx and F = ma
Therefore ma = (–) kx M2
ω2 = k / m
T = M1
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5. (a) One reading from the graph e.g. 1.0 N causes 7 mm C1
Hence 5.0 (N) causes 35 0.5 (mm) A1
(allow one mark for 35 1 (mm)
(b) (i) Force on each spring is 2.5 (N) C1
extension = 17.5 (mm) allow 18 (mm) or reading from graph A1
[allow ecf from (a)]
(ii) strain energy = area under graph / ½ F e C1
= 2 0.5 2.5 17.5 10–3
= 0.044 (J) A1
[allow ecf from (b)(i)]
(c) E = stress / strain C1
Stress = force / area and strain = extension / length C1
extension = (F L) / (A E)
= (5 0.4) / (2 10–7 2 1011)
= 5.(0) 10–5 (m) A1
(d) strain energy is larger in the spring B1
extension is (very much larger) (for the same force) for the spring B1
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6. (i) 1 Elastic as returns to original length (when load is removed) B1
2 Hooke’s law is obeyed as force is proportional to the extension B1
Example of values given in support from table B1
(ii) Measure (original) length with a (metre) rule / tape B1
Suitable method for measuring the extension e.g.
levelling micrometer and comparison wire or fixed
scale plus vernier or travelling microscope and marker / pointer B1
(iii) E = stress / strain C1
= (25 1.72) / (1.8 10–7 1.20 10–3) C1
= 1.99 1011 (Pa) A1
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7. (i) Density = mass / volume B1
Area length = mass / density
Area = (2.0 10–3) / (7800 0.5) or 2.56 10–7 / 0.5 B1
= 5.1(3) 10–7 m2 A0
(ii) E = (F l) / (A e) / stress = F / A (1.6 108 and strain
= e / l (8 10–4) C1
F = (E A e) / l
= (2 1011 5.1 10–7 4.0 10–4) / 0.5 C1
=82 (N) (81.6) A1
(iii) Diameter for D is half G hence area is ¼ of G
Extension is 4x greater
Tension required is the same = 82 (N) A1
(iv) The extension is proportional to the force / Hooke’s B1
law (OWTE)
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8. (i) Graph through origin with (short) linear section then reducing gradient. 1
(ii) Straight section - elastic; (1)
Curved section - plastic. (1) 2
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9. (a) P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3
(b) 24 N m–1 × 100 m = 2400 N 1
(c) elastic P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1) 2
(d) loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2
(e) idea that a given (unit) extension for a shorter rope requires a greater force 1
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10. (a) (i) Stress = force / area C1
force = stress area
= 180 106 1.5 10–4
= 27000 (N) A1
(ii) Y M = stress / strain C1
= 180 106 / 1.2 10–3 or using the gradient C1
= 1.5 1011 N m–2 A1
(b) brittle
elastic/ graph shown up to elastic limit
obeys Hooke’s law / force α extension / stress α strain
no plastic region B3
MAX 3
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