AQA Physics A

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,

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

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

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

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

• 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

• 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

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

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).

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)

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

F/

N

Extension / m

X

Y

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

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

Springs in series and parallel

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

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

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

F/N

ΔLExtension/m

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

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

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

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.

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

Ductile

Brittle

St

re ss

Strain

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

• 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

Typical graph

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

• 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 is a ductile material

• e.g copper

• D is plastic material

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

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

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

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.

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.

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.

Explaining stiffness and elasticity

Metals

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a metal is an array ofpositive ions bondedby negative electron'glue'

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stretching has to pull bonds apart

Elastic extensibility ~ 0.1% Young modulus

~1011 — 1012 Pa

Stretching a metal stretches bonds — but not much.

gaps open up a little

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.

Shape.

B

C

A

Strain

Stress/ N

• 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.

Shape.

B

C

A

Strain

Stress/ N

Rubber

Rubber stretches and contracts by chains uncoiling and coiling up again. Rubber is elastic, not plastic.

In stretched rubber, the chain bonds rotate, and chainsfollow straighter paths between cross-links. When let go,the chains fold up again and the rubber contracts.

In unstretched rubber, chains meander randomlybetween sulphur cross-links.

sulphur cross-linkssulphur cross-links

Elastic extensibility > 100%

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.