AQA Physics A

1. Materials AQA Physics A

2. An old riddle! • Which is heavier, a pound of lead or a pound of feathers? • People who are confused by this riddle do so because they do not understand the difference between mass and density,

3. Density. • Density depends upon the mass of the atoms inside a substance and how closely packed together they are. • We can calculate density using: • Density = mass volume ρ = m V • S.I. units are kgm-3 but density can also be given in gcm-3

4. Examples • The density of water is 1000kgm-3 • This means that a volume of 1m3 has a mass of 1000kg • The density of ice is 917kgm-3. Ice floats on water because its density is less. • The densities of solids and liquids overlap but gases have much lower values of density. This is due to the larger spacing of the molecules • e.g. Air has a density of just 1.2kgm-3

5. Calculations • Q Calculate the mass of air in a typical lab measuring 8 x 6 x 3m • A Volume = 144m3 • mass = density x volume • = 1.2 x 144 • = 172.8kg

6. Measuring density • A regular shaped object: Calculate its volume after measuring its dimensions using a ruler, micrometer or vernier callipers. • Measure its mass using an electric balance and then calculate its density using ρ = m V

7. An irregular shaped object: • Find the volume of the object using the displacement of water. • (This method was first used by Archimedes when asked to check whether the King’s crown was made of pure gold) • Measure the mass using an electric balance and then calculate the density

8. A liquid: Measure its volume using a measuring cylinder. • Find the mass of the liquid and the cylinder and then subtract the mass of the empty cylinder to find the mass of the liquid. • Calculate the density of the liquid using • ρ = m V

9. Further example • Q The crown that Archimedes had to investigate contained 12.5 x 10-5m3 of gold and 2.2 x 10-5m3 of silver. Calculate its mass and density. (Density of gold is 19300kgm-3 and density of silver is 10500kgm-3). • A Mass of gold= ρ x V = 19300 x12.5 x 10-5 • = 2.41kg • Mass of silver = ρ x V = 10500 x 2.2.x 10-5 • = 0.23kg • Total mass = 2.64kg • Total volume = 14.7 x 10-5m3 • Density of alloy = m = 2.64 = 18000kgm-3 • V 14.7 x 10-5

10. Stretching Springs • Measure the original length of the spring • Set up the apparatus and apply a force of 1N. This gives a tension of 1N in the spring. • Record the new length and find the extension by subtracting the original length. • Repeat until you have a total force of 8N • Plot a graph of Force (F) against extension (ΔL).

11. Hooke’s law • The graph should be a straight line through the origin showing direct proportion. • Hooke’s law states that the force applied to a spring is directly proportional to its extension provided it does not pass the elastic limit. • Force = constant x extension • F = k(ΔL) • (Springs can be used to measure forces using a Newton meter)

12. X F/ N Y Extension / m Spring Constant • k = spring constant or stiffness constant and is found from the gradient of the graph of F against ΔL • A typical spring used in school has k = 50Nm-1 • The stiffer the spring, the greater the value of k Spring X is stiffer than spring Y

13. Example • Q A spring has a length of 20mm and stretches to a length of 60mm with a load of 2N. Calculate its spring constant. • F = kΔL • k = F = 2 = 50Nm-1 • ΔL 40 x 10-3 • Q Calculate the length of the spring when a load of 7N is added to it • ΔL = F = 7 = 0.14m • k 50 • New length = 0.02 +0.14 = 0.16m

14. Elastic limit • If the spring is overstretched it does not return to its original length. It is permanently distorted. • The elastic limit is the point beyond which the material is no longer elastic and will not return to its original length when the load is removed. • F • x x = elastic limit • ΔL

15. Springs in series and parallel

16. Springs in parallel • Repeat the experiment for stretching a spring but use 2 springs (P and Q) in parallel. • Calculate the gradient of the graph to find the combined spring constant (k). • Theory (see P165) shows that: • k = kp + kQ where kp and kQ are the spring constants for springs P and Q ( the springs are stiffer and more difficult to stretch) • Find the percentage difference between your value of k and the theoretical value

17. Springs in series • Repeat the experiment for stretching a spring but use 2 springs (P and Q) in series. • Calculate the gradient of the graph to find the combined spring constant (k). • Theory (see P165) shows that: • 1 = 1 + 1 k kP kQ • where kp and kQ are the spring constants for springs P and Q (the springs are now easier to stretch) • Find the percentage difference between your value of k and the theoretical value

18. F/N ΔL Extension/m Energy stored • Work is done when stretching a spring and this equals the elastic potential energy stored in it. • Since work = force x distance • = area under graph • = 1 F ΔL = 1 kΔL2 • 2 2 Elastic potential energy stored in a stretched spring = 1 kΔL2 2

19. Example • Q How much energy is stored in a spring which is 32cm long and stretches to 54cm when a force of 6.0N is applied to it. • A ΔL = 22cm = 0.22m • Ep = 1 FΔL 2 • = 1 x 6.0 x 0.22 2 • = 0.66J

20. Tensile Stress • Tensile stress = tension in wire x-sectional area of wire • σ = T A Units:Nm-2 or Pa • Stress is a property of the material rather than the individual object being stretched • Q Calculate the stress in a wire of x-sectional area 4 x 10-8m2 when it is stretched with a force of 200N • A 5 x 109 Pa

21. Tensile strain • Tensile strain = extension of wire original length • ε = ΔL L • Since it is a ratio it has no units and is once again a property of the material. • Q Calculate the strain in a wire of length 2.3m which stretches by 1.5mm • A 6.5 x 10-4

22. Stretching different materials • A rubber band and a polythene strip can be stretched using the same apparatus as used with the spring. • A copper wire can be stretched along a bench. • In all cases plot a graph of stress against strain. • The shape of a stress-strain graph is the same as the shape of a force-extension graph but is common to the material used and not the individual wires.

23. Brittle Stress Ductile  Strain Examples • Copper is an example of a ductile material. • Initially it obeys Hooke’s law and returns to its original length if the load is removed. • There is then a large increase in its extension and if the load is removed it does not return to its original length. • Glass is an example of a brittle material

24. More definitions • Limit of proportionality: Point beyond which Hooke’s law is no longer obeyed. • Elastic limit: Point beyond which the material is no longer elastic and will not return to its original length when the load is removed. • Yield point: Point at which there is marked increase in extension

25. Elastic deformation: The material will return to its original shape when the load is removed. • Plastic deformation: The material is permanently stretched when the load is removed. • Ductile materials like copper can be drawn into wires and undergo large plastic deformation. • Brittle materials fracture suddenly without any noticeable yield e.g. glass • Strong materials require a high stress to break them • Ultimate tensile stress: maximum stress that can be applied. This is sometimes called the Breaking stress

26. Typical graph • E = elastic limit • Y = yield point • P = limit of proportionality • UTS = ultimate tensile stress • S = wire snaps

27. A is a brittle material which is also strong e.g glass • B is a strong material which is not ductile e.g steel • C isa ductile material • e.g copper • D is plastic material

28. Young modulus of elasticity • Young modulus = stress strain • E = σ ε • = T/ΔL A L • TL AΔL • E is measured in Pa, the same as stress. • It is a measure of how difficult it is to change the shape of the material and can be found from the gradient of a stress-strain graph

29. Examples • Q1 The Young modulus for cast iron is 2.1 x 1011Pa and is snaps when the strain reaches 0.0005. What is the stress needed to break it? • A 1.1 x 108Pa • Q2 A wire made of a particular material is loaded with a load of 500 N.  The diameter of the wire is 1.0 mm.  The length of the wire is 2.5 m, and it stretches 8 mm when under load.  What is the Young Modulus of this material? • A 2.0 x 1011Pa

30. Experiment to measure Young modulus • Measure the original length of the wire • Measure the diameter of the wire at different points along its length using a micrometer and find is average diameter. Calculate the x=sectional area of the wire. • Load the wire as before and measure its extension. • Plot a stress-stain graph and measure its gradient = Young modulus

31. Area under stress-strain graph • Area under graph = average stress x strain • = F x ΔL A L • = work done volume • The area gives us the energy stored per unit volume and is a measure of the toughness of the material. A tough material needs a large amount of energy to break it.

32. Loading and unloading materials • A material can be stretched as before by increasing the load on it and the extension measured. • It can the unloaded gradually and the extension measured again. • A graph can then be plotted showing both the loading and unloading. • The shape of this graph will depend upon the type of material used and the size of the load.

33. Metal wire • The loading and unloading give the same straight line provided the elastic limit has not been passed. • The atoms in the metal are pulled slightly further apart and then return to their original positions when the load is removed. • If taken beyond this point the unloading will give a line parallel to the first but leaving a permanent extension.

35. B Shape.  Stress/ N A C Strain Rubber band • The rubber band returns to its original length when the load is removed but the unloading curve is below the loading curve. • The molecules are like long strands of tangled spaghetti that become untangled when the rubber is stretched and the become tangled again when the load is removed.

36. B Shape.  Stress/ N A C Strain • The shaded area represents the energy stored in the form of internal (heat) energy by the rubber band when it is unloaded. • The rubber used in car types should give a small area to reduce the energy lost by heat.

38. Polythene • Polythene does not return to its original length when the load is removed. • Like rubber, polythene is a polymer, but when it is stretched new cross links form which prevent the molecules from returning to their original state.