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The figure below shows a stress-strain graph for a copper wire. (a) Define tensile strain.
(1)
(b) State the breaking stress of this copper wire. answer = ______________________ Pa (1) (c) Mark on the figure above a point on the line where you consider plastic deformation may start. Label this point A. (1) (d) Use the graph to calculate the Young modulus of copper. State an appropriate unit for your answer. answer = ______________________ (3)
(e) The area under the line in a stress-strain graph represents the work done per unit volume to stretch the wire. (i) Use the graph to find the work done per unit volume in stretching the wire to a strain of 3.0 × 10–3. answer = ______________________J m– (2) (ii) Calculate the work done to stretch a 0.015 kg sample of this wire to a strain of 3.0 × 10–3. The density of copper = 8960 kg m–3. answer = ______________________J (2) (f) A certain material has a Young modulus greater than copper and undergoes brittle fracture at a stress of 176 MPa. On the figure above draw a line showing the possible variation of stress with strain for this material. (2) (Total 12 marks)
(iii) Describe and explain one feature of the crane that prevents it from toppling over when it is lifting a container.
(2) (b) Each cable has an area of cross–section of 3.8 × 10–4^ m^2. (i) Calculate the tensile stress in each cable, stating an appropriate unit. answer ____________________ (3) (ii) Just before the container shown in the diagram above was raised from the ship, the length of each lifting cable was 25 m. Show that each cable extended by 17 mm when the container was raised from the ship. Young modulus of steel = 2.1 × 10^11 Pa (2) (Total 12 marks) (a) (i) Explain what is meant by the spring constant k of a spring.
(ii) Give the unit of k.
(2)
(b) The figure below shows the stages in a bungee jump. In bungee jumping, the participant jumps from a high point attached to an elastic cord (step 1). After a period of free fall, the cord slows the fall of the jumper (step 2) with the system eventually undergoing oscillation (step 3). A bungee jump is to be set up from a suspension bridge with the jumper of weight 700 N falling towards the river below. The roadway of the bridge is 76 m above the river surface. The bungee cord is adjusted so that the jumper just reaches the river surface at the bottom of the first oscillation. The unstretched length of the elastic cord is to be 12 m. (i) Calculate the time taken before the cord begins to stretch. (ii) Show that, when jumping from the bridge to the river, the jumper loses about 53 kJ of gravitational potential energy. (iii) Calculate the extension of the cord when the jumper is at the bottom of the first oscillation. (iv) The gravitational potential energy is stored in the bungee cord. Calculate the spring constant of the cord. (v) Calculate the time period of oscillation of the jumper. (12) (c) (i) Calculate the tension in the cord when the jumper comes to rest for the first time. (ii) Forces on astronauts and ‘thrill seekers’ are often specified in terms of the g force acting on the participants. 1 g is equivalent to an acceleration of 9.8 m s−2. Calculate the maximum g force that acts on the jumper.
(b) A length of steel wire and a length of brass wire are joined together. This combination is suspended from a fixed support and a force of 80 N is applied at the bottom end, as shown in the figure below. Each wire has a cross-sectional area of 2.4 × 10–6^ m^2. length of the steel wire = 0.80 m length of the brass wire = 1.40 m the Young modulus for steel = 2.0 × 10^11 Pa the Young modulus for brass = 1.0 × 10^11 Pa (i) Calculate the total extension produced when the force of 80 N is applied.
(ii) Show that the mass of the combination wire = 4.4 × 10–2^ kg. density of steel = 7.9 × 10^3 kg m– density of brass = 8.5 × 10^3 kg m–
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(c) A single brass wire has the same mass and the same cross-sectional area as the combination wire described in part (b). Calculate its length.
(2) (Total 11 marks) The diagram below shows the tensile stress–tensile strain graphs for four materials, A , B , C and D , up to their breaking stress. (a) State what is meant by tensile stress and tensile strain. tensile stress ________________________________________________________
tensile strain ________________________________________________________
(2)
A student investigated how the extension of a rubber cord varied with the force used to extend it. She measured the extension for successive increases of the force and then for successive decreases. The diagram below shows a graph of her results. (a) (i) Give a reason why the graph shows the rubber cord does not obey Hooke’s law.
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(ii) Give a reason why the graph shows the rubber cord does not exhibit plastic behaviour.
(1) (iii) What physical quantity is represented by the area shaded on the graph between the loading curve and the extension axis?
(1)
(b) Describe, with the aid of a diagram, the procedure and the measurements you would make to carry out this investigation. The quality of your written answer will be assessed in this question.
(6) (Total 9 marks)
(c) The spring is attached at a distance of 8.0 cm from the pivot and the spring has a stiffness of 100 N m−1. Calculate the extension of the spring when the rod is horizontal and the spring is vertical. You may assume the mass of the pen and the mass of the rod are negligible. extension = ____________________ m (3) (d) Before an earthquake occurs, the line being drawn on the graph paper is horizontal. Explain what happens to the line on the graph paper when an earthquake is detected and the frame of the seismometer accelerates rapidly downwards.
(2) (Total 11 marks)
A wire of length L and cross-sectional area A is stretched a distance e by a tensile force. The
Young modulus of the material of the wire is E.
Which expression gives the elastic energy stored in the stretched wire? A B C D (Total 1 mark)
A load of 3.0 N is attached to a spring of negligible mass and spring constant 15 N m–1. What is the energy stored in the spring? A 0.3 J B 0.6 J C 0.9 J D 1.2 J (Total 1 mark)
Two vertical copper wires X and Y of equal length are joined as shown. Y has a greater diameter than X. A weight W is hung from the lower end of Y. Which of the following is correct? A The strain in X is the same as that in Y. B The stress in Y is greater than that in X. C The tension in Y is the same as that in X. D The elastic energy stored in X is less than that stored in Y. (Total 1 mark)
Two masses hang at rest from a spring, as shown in the diagram. The string separating the masses is burned through. Which of the following gives the accelerations of the two masses as the string breaks?
acceleration of free fall = g
acceleration of 1 kg mass upwards in m s– acceleration of 2 kg mass downwards in m s–
A 3 g 1 g
B 2 g 2 g
C 2 g 1 g
D 1 g 1 g
(Total 1 mark)
(iii) the counterweight (1) provides a (sufficiently large) anticlockwise moment (about Q) or moment in opposite direction ( to that of the container to prevent the crane toppling clockwise) (1) or left hand pillar pulls (down) (1) and provides anticlockwise moment or the centre of mass of the crane(‘s frame and the counterweight) is between the two pillars (1) which prevents the crane toppling clockwise /to right (1) 7 (b) (i) (tensile) stress ecf (a) (i) (1) = 1.4(2) × 10^8 (1) Pa (or N m–2) (1) (ii) extension = (1) = and (= 1.7 × 10–2^ m) = 17 (mm) (1) 5 [12] (a) (i) k = force/extension (1) (ii) N m-1^ (1) 2
(b) (i) s = ut + ½ at^2 or alt used (1) t^2 – 12/4.9 (1) 1.6 s (1) (ii) weight × height change seen (1) 53200 J (1) (iii) 76 – 12 = 64 m (1) 12
(iv) ½ kx^2 = energy stored seen (1) k = 2 × 53200 /(64)^2 (1) 25.9 N (1) (v) T = 2π√( k/m ) seen (1) subst (1) 1.2 s (1) (c) (i) F = kx seen (1) = 25.9 × 64 = 1660 N (1) (ii) 1660/700 seen (1) 2.4 g (1) (iii) stiffer cord (1) less elongation so longer natural length (1) 6 [20] (a) Hooke’s law: the extension is proportional to the force applied (1) up to the limit of proportionality or elastic limit [or for small extensions] (1) 2
(b) (i) (use of E = gives) ΔL s = (1) = 1.3 × 10–4^ (m) (1) (1.33 × 10–4^ (m)) ΔL b = = 4.7 × 10–4^ (m) (1) (4.66 × 10–4^ (m)) total extension = 6.0 × 10–4^ m (1) (ii) m = ρ × V (1) m s = 7.9 × 10^3 × 2.4 × 10–6^ × 0.8 = 15.2 × 10–3^ (kg) (1) m b = 8.5 × 10^3 × 2.4 × 10–6^ × 1.4 = 28.6 × 10–3^ (kg) (1) (to give total mass of 44 or 43.8 × 10–3^ kg) 7
