1. Year 12 physics summary
Use this as a reminder of content and a way of identifying weakness, NOT as a replacement for revision

2. Quantities and units
You will need to memorise the base units and be able to derive the other units
Derivation can be done simply by looking at the unit and finding a simple equation that uses it!
For example, the N is also one kgms-2 and this is derived from F=ma

3. Measurements and uncertainties
How to read vernier scales
Adding or subtracting?
ADD absolute uncertainties (e.g. the ones that have + in them)
Multiplying or Dividing?
ADD the percentage uncertainties (e.g. work out the % that the + is of the value)
Raising to a Power?
MULTIPLY the percentage uncertainty by the power
See where the vernier scale (bottom) begins at the main scale (this case=10cm)
Look for the only line on the vernier scale that is in-line with the main scale (this case the second one is!)
Main scale reading: 10.0 cm
Vernier scale reading: 0.02 cm
Measurement reading: cm
How to read micrometer scales
Main scale reading: 0.5mm
Microscale reading: 0.14mm
Measurement reading: 0.64mm

4. Decimal places and sig figs
When answering calculation questions, always give your answer first as is then to an appropriate number of sig.figs
Usually the appropriate number is the significant figures that are provided for values in the question

5. Scalar vs vector
You’ll be expected to be able to identify scalar values (no direction, only a magnitude) and vector (magnitude AND direction
Quick trick: If you draw the value on a diagram with an arrow it is usually a vector
Vectors can be combined to make a resultant

6. Resolving vectors: method 1
The simplest method for resolving a vector is to draw a scale diagram of it! Convert the data to an appropriate scale (e.g. 10ms-1 being 10 s) and draw the first vector From the end of that one draw the next (e.g. 5ms-1 at 20° from the vertical) and then join them up! (start to end)
In this, the question could be….A boat travels 40m east and then 30m south, what is the resultant displacement?
DON’T FORGET: give the angle! It’s a vector after all!
This can be worked out using trigonometry (SOHCAHTOA)

7. Resolving vectors: method 2
This one involves using more mathematics but can be completed quicker and is often more accurate and precise Either method ALWAYS draw a vector diagram before you calculate! A
Don’t forget, draw the vector components so that they encompass the angle! It doesn’t matter for example if the vertical component is acting at [ A], it can still be worked out by placing it over on the right hand side!
Then use simple trigonometry remembering you have the resultant and you’re calculating the components here

8. SUVAT Displacement Start Speed Final Speed Acceleration Time
You may need to know how to derive these equations so for that, look back at our old powerpoints! These can ONLY be used for UNIFORM ACCELERATION i.e. when acceleration is unchanging This tends (not always) relate to vertical motion as gravitational acceleration is constant
DO NOT FORGET the initial velocity has both horizontal and vertical components. If you’re dealing with the vertical motion USE THE VERTICAL COMPONENT of velocity, not that original.

9. Projectiles Things to remember:
1) Isolate the vertical/horizontal velocities
2) Use the vertical velocity to calculate the vertical motion (usually to find ‘time’)
3) Use the value of time with the horizontal velocity to calculate distance (s=d/t)
Projectiles
Things to remember:
You consider the components separately (use SUVAT for vertical and s=d/t for horizontal) and use the component velocities
You can consider the halfway point (with the vertical velocity here being zero) and then double the time if needed!
You can consider the whole journey and use s=ut+1/2at2 with s as zero as it hasn’t had any vertical displacement overall!

10. Moments
A moment has the unit of Nm and is the force multiplied by the PERPENDICULAR distance from the pivot Calculate all the CLOCKWISE moments and then calculate all the ANTICLOCKWISE moments. If they are equal, then the body is in equilibrium (NOT MOVING). Don’t forget, if the pivot is not directly at the centre of gravity of the board, the boards weight will contribute to the moments! For example, if the angle of the ladder was to increase, the weight of the man would act toward the left of the floor pivot, therefore contributing extra moment. This would make the moments unbalanced and therefore move the ladder/fall

11. Displacement-time graphs
The gradient represents the velocity, if its curved then you can take an instantaneous velocity by using a tangent If the graph shows a flat line, then the body is stationary as there is no displacement occurring If the graph returns back to the x axis, then the object has gone back to its original position!

12. Velocity-time graphs
The gradient of a velocity- time graph represents the acceleration If it’s a curve, the acceleration is not uniform (steadily increasing for e.g.) but you can use a tangent to calculate instantaneous acceleration The area underneath the graph represents the displacement. Divide up the graph into triangular and rectangular sections to do this! If a curve, use sections to estimate

13. Newton’s laws
1st law: A body will remain at rest or at a constant motion unless acted upon by a resultant force 2nd law: The acceleration of an object is directly proportional to the resultant force acting upon it (F=ma) 3rd law: If object A applies a force on object B, B will enact an equal and opposite force on object A
This plane can be described as “climbing at a constant speed”
It may be moving but the lift=weight as it’s a constant speed.
If all forces acting upon an object are balanced, then the object is in equilibrium and that means the resultant is 0 so no acceleration occurs!

14. Newton pairs
Newton Pairs of forces are ones that obey the following rules:
They act on different objects
They have the same magnitude
They are the same type (e.g. gravitational) of force
They act in opposite directions
In this example, the earth pulls the book down (gravitational pull/weight) so the book pulls the earth up (gravitational pull) in the opposite direction
In this example, there’s also another at play. That weight also acts down on the table (contact) so the table acts upon the book (contact)

15. Resolving forces
When resolving forces don’t forget what component is actually relevant. In this example they want to know the weight of the painting. This acts vertically downward This MUST be being balanced by all the upward vertical forces. I.e. vertical component of tension in each cable! Therefore you could use 30Sin(45) to work out one, then double for both tensions.
vertical

16. Object on a slope
40°
40°
If an object is at equilibrium (i.e. moving at a constant speed or stationary) on a slope of a known angle, it’s weight will be acting as a vector (at an angle-just look, turn the image so the slope is horizontal!) This is the ONLY downward force and is being matched by the friction and reaction force. To work out either of those, just calculate the components of the weight Hint: the angle of the slope is also the angle at the top of the component triangle (usually…slope friction=sine, contact force=cosine)

17. momentum
The law of conservation of momentum states that ‘the total momentum before a situation = the total momentum after”
Momentum (p) = mass x velocity
BEFORE
20ms-1
Divide up the situation and in the before, calculate the momentum of each object individually then add them together to get the total momentum
0ms-1
500kg
400kg
AFTER
? ms-1
e.g. in this instance:
Total= (20 x 500) + ( 0 x 400)
=10,000
After=10,000 so then divide by 900kg (total mass) = 11.1m/s! Kg
As the total momentum is the same before and after, to calculate the velocity they go off at together, just divide the total momentum by the total mass!

18. Task: What is the velocity of the gun?
Momentum- Explosions
0ms-1
BEFORE
Gun: 1.2kg
Bullet: 0.005kg
The momentum of both objects together must equal 0 as both are stationary!kgms-1
? ms-1
400ms-1
AFTER
1.7ms-1 (bullet has p of 2kgms-1)
Momentum before (P(b)) = Momentum after(P(a))
The total momentum must still be 0. This is because the momentum of the bullet will be positive and that of the gun will be negative as it’s moving backward!
e.g. in this instance:
Total= 0. The momentum of the bullet=400 x 0.005= 2
So: the backward momentum must= -2, so -2 divided by the mass of the gun gives (-1/1.2)= 0.83ms-1
Task: What is the velocity of the gun?

19. 𝑭𝒐𝒓𝒄𝒆= 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒕𝒊𝒎𝒆
Momentum- Change
Momentum is a VECTOR quantity – it has a size and direction.
Momentum=
mass x velocity
200kg
10m/s
So this car= 2000kgm/s!
When a car crashes (for example into a wall), all its momentum will be transferred to the it!
As such it experiences a change in momentum. This is always a set value that is dependent on the momentum just prior to colliding
This quick change in momentum will produce a huge force (impact) on both the car and passengers. Linked by this equation:
𝑭𝒐𝒓𝒄𝒆= 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒕𝒊𝒎𝒆

20. So as the man is pushing the block 4m with 15N, he has used 15x4=60J!
Work done
Work Done = Force x Distance (in direction of force)
When a force pushes an object it makes it move by transferring energy to it.
We say the force “does work”.
Work done is the same as energy transferred so is measured in Joules (J).
15N
So as the man is pushing the block 4m with 15N, he has used 15x4=60J! 4m

21. Gravitational Potential Energy
GPE= mass x gravitational field strength x height
Gravitational Potential Energy (GPE) is exactly the same for lifting an object as you are moving the weight (force) a vertical distance!
This is exactly the same as the previous equation! Weight is a Force (mass x g). And height is a distance. So GPE is the same as calculating work! Which is the same as energy!
60kg
So as the man is lifting the 60kg dumbbells 1.5m with he has used
60 x 10 x 1.5=900J!
1.5m

22. Kinetic energy
GPE= mass x gravitational field strength x height
This is exactly the same as the previous equation! Weight is a Force (mass x g). And height is a distance. So GPE is the same as calculating work! Which is the same as energy!
60kg
So as the man is lifting the 60kg dumbbells 1.5m with he has used
60 x 10 x 1.5=900J!
1.5m

23. Kinetic Energy = ½ x mass x velocity2
Note – only the velocity is squared!
The amount of kinetic energy an object has depends on its mass and how fast it is moving…
So if both of these were travelling at 10m/s, the car would have more kinetic energy as its mass is greater!

24. Power= current x voltage
Power= energy ÷ time
Power= work done ÷ time
Power= current x voltage
Power is the rate of doing work so it’s the energy transferred from one form into another per second. Often, questions will ask you to first work out the energy transferred (perhaps by calculating GPE) and then you just divide by the time!
The unit is the WATT (W) and this is equivalent to Js-1

25. Electrical quantities 1
We defined current as the rate of flow of charge and this can be applied to anything that is carrying a charge (charge carriers) such as conventionally, the electron.
This can be expressed simply mathematically with the formula:
𝐶𝑢𝑟𝑟𝑒𝑛𝑡= 𝐶ℎ𝑎𝑟𝑔𝑒 𝑇𝑖𝑚𝑒 aka 𝐼= ∆𝑄 ∆𝑡
So if 1C worth of charge flows by in 1s, then 1A of current is flowing! The ampere is the base unit of current but as 1A is actually a large current, often values are given in mA, so don’t forget to convert to amperes.

26. Electrical quantities 2
Potential Difference
Potential difference is the more specific term for voltage, as it highlights the transformation of energy from electrical into another type
It is the amount of electrical energy transferred by a component into a different form per unit charge
In order to transfer the energy, the component is said to have done work and this is why the symbol has changed.
𝑷.𝒅(𝑽) = 𝑬𝒏𝒆𝒓𝒈𝒚 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 (𝑱) 𝑪𝒉𝒂𝒓𝒈𝒆 𝑷𝒂𝒔𝒔𝒊𝒏𝒈(𝑪)
𝑽= 𝑾 𝑸

27. Electrical quantities 3
Electromotive Force
This is basically the supply voltage, i.e. if you were to put a voltmeter around the power-pack or battery etc, this would be the reading
It is the energy transferred to each coulomb of charge from a supply. It is given symbol of epsilon.
𝑬𝑴𝑭(𝑽)= 𝑬𝒏𝒆𝒓𝒈𝒚 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 (𝑱) 𝑪𝒉𝒂𝒓𝒈𝒆 𝑷𝒂𝒔𝒔𝒊𝒏𝒈(𝑪)
ε = 𝑾 𝑸

28. The charge on a single electron is 1.6 x 10-19C
Quantities and units
Some particles are said to have an electric charge such as the electron which has a negative charge.
The charge on a single electron is 1.6 x 10-19C
6.25 x electrons to one C!
This fundamental charge on a particle is unchangeable, remember we can’t destroy nor create charge (conservation of charge)

29. Kirchoff’s Laws: conservation of charge
This rule states that the algebraic sum of all the currents entering a junction is equal to zero
ΣI=0
In order to conserve electrical charge (law of conservation of charge), the sum of all charge arriving at a point (a junction), must be equal to those leaving it!
So for this circuit:
If a current is arriving we give it a positive value and if it is leaving we give it a negative value
Q(T) = Q1 + Q2 + Q3 + Q4
And as I= Q/t, and t=1
I(T) = I1 + I2 + I3 + I4
And as I=V/R, and v=1
1 𝐼𝑡𝑜𝑡𝑎𝑙 = 1 𝐼 𝐼 𝐼 𝐼4 Q1 Q2 Q3 Q4

30. Σε= ΣV Kirchoff’s Laws: emf
In order for energy to be conserved in any loop of a circuit (not necessarily the whole circuit), the sum of EMFs must equal the sum of the potential differences around a loop.
Σε= ΣV
Note: The potential differences are a result of ohms law calculations based upon the current and the resistance. For this reason, the P.d. may have a negative or positive value dependent on the direction of the current

31. Combining Resistance 1 𝑅𝑡𝑜𝑡𝑎𝑙 = 1 𝑅1 + 1 𝑅2 + 1 𝑅3 + 1 𝑅4
We can combine resistors in series, and in parallel.
Resistors in series add values:   
Rtotal = R1 + R2  Rn
Resistors in parallel work a little differently. The general formula for computing resistance in parallel is:
Don’t forget! When you’re calculating the total resistance, you end up with the value for 1/Rtotal. So remember to finish the step by finding the inverse (1/your answer).
1 𝑅𝑡𝑜𝑡𝑎𝑙 = 1 𝑅 𝑅 𝑅 𝑅4
If you have a combination of the two, you should work out the parallel total first then use that in series with the rest!

32. Resistance
A wire is made up of metal ions surrounded by a sea of free electrons which flow (a current)
If the voltage increases, the temperature of a wire will increase. This will cause the ions to vibrate with higher amplitude and therefore increase the frequency of electrons colliding with the ions. This would impede flow, thus RESISTANCE INCREASES.

33. Resistance graphs
The gradient of a voltage-current graph tells us the resistance of that component. The flatter it is, the greater the resistance because it means we’re getting very little increase in current!
In this case, the gradient isn’t changing which means the resistance is fixed. It’s a fixed resistor!

34. V-I relationships FIXED RESISTOR DIODE FILAMENT BULB
Resistance Unchanged
Directly Proportional
Flatter=Higher Resistance
Resistance HIGH in one direction but LOW in the other
Forces current to only flow in one direction
Used to turn AC current into DC current
Resistance gradually increases as the PD goes up.
This is because the wire is heating up (see resistance in a wire)

35. This would reduce resistance and increase current!
semiconductors
Thermistors (left) and LDRs (right) are different to normal wire components as they are semiconductors. Semiconductors increase then number of charge carriers available when exposed to energy (heat or light).
This would reduce resistance and increase current!
The MORE LIGHT/MORE HEAT  RESISTANCE DECREASES  therefore INCREASING CURRENT

36. This ‘crawling’ motion towards the positive terminal is known as the
Drift velocity
A metal consists of positive metal ions with free electrons that move with random motion as they collide with the ions.
If a source of emf is connected across a metal, the electric field set up have a tendency to push the electrons toward the positive end of the metal.
This ‘crawling’ motion towards the positive terminal is known as the
drift velocity

37. Calculating drift velocity
We consider the shape of a wire to be a cylinder, the current that is flowing by can be calculated with the formula:
With ‘v’ being the drift velocity and ‘n’ the number density of electrons. ‘A’ is the cross-sectional area and the ‘q’ is the charge on a single electron!
𝐼=𝑛𝐴𝑣𝑞

38. resistivity
Resistivity is similar to resistance with the exception that it’s an inherent characteristic of the material itself, i.e.
“ a quantification of a material’s ability to resist the flow of electric current”
So this means that it is a feature of the type of metal for example, rather than anything to do with the dimensions of this metal in a wire.
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒(Ω)= 𝑹𝒆𝒔𝒊𝒔𝒕𝒊𝒗𝒊𝒕𝒚 Ω𝒎 ×𝑆𝑎𝑚𝑝𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ(𝑚) 𝐶𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 (𝑚2
𝑅= 𝜌𝑙 𝐴

39. Internal resistance ε =IR + Irinternal
When a current is flowing through a circuit, the electrons still have to travel through the cell.
A small portion of the voltage/energy will be lost here because the electrons have to overcome the internal resistance of the cell!
Because the cell has this internal resistance, the convention is to show the cell adjacent to a resistor.
This resistor would have a very low value.
ε =IR + Irinternal
(note, the r is the internal resistance)
IR = ε - Irinternal
V= ε – Irinternal
So this shows that the voltage produced at the terminals of the cell (i.e. outside of the cell) is effectively the EMF – the voltage overcoming internal resistance

40. Dividing potential You can calculate the ratio of the resistors
This allows you to work out the individual contribution of each resistor to the total resistance
E.g. R2’s contribution to the total resistance (R1+R2)

41. rheostats
Instead of using two different resistors we can also just use a single rheostat/potentiometer (variable resistor)
So by moving the contact, we can alter the p.d across the terminals (Vout)
Just think of the remaining part of the variable resistor as R1.
So just as before, work out the individual contribution by calculating the ratio!
Note: this can be done in terms of length along the rheostat!

42. As an object is submerged in a fluid, it will begin to displace it.
𝝆 = 𝒎 𝒗
upthrust
When an object is floating, the value of W will equal the U
To calculate U, you first use the density equation to calculate mass from the volume of fluid displaced. This will be equal to the volume of the object in the fluid! USE THE DENSITY OF THE FLUID not the object
You then just muliply by gravitational field strength to get the weight of that water!
As an object is submerged in a fluid, it will begin to displace it.
The volume of the object that is submerged is equal to the volume of the water displaced
So a submerged object will be subjected to a force equal to the WEIGHT of the fluid displaced (UPTHRUST)

43. Flow lines
Fluids can be said to move in either laminar flow or turbulent flow.
The movement velocity of a fluid (gas or liquid) can be represented by streamlines, which are arrowed lines showing the path taken by small regions of fluid.
a) In laminar flow, like slow moving water, adjacent layers do not cross over one another as there is no abrupt changes in speed or direction
b) However in turbulent flow the fluid swirls around in vortices or eddy currents so streamlines cross over

44. Flight
As the air moves faster over the top, this means that there is a difference in pressure on the wing. This means that there is more force applied to the bottom of the wing, resulting in uplift (and drag)

45. viscoscity
Viscosity is defined as ‘the magnitude of internal friction within a fluid’ a.k.a ‘how sticky it is’ and as such how resistant it is to flow
The higher the temperature of a liquid, the lower its viscosity which lets it flow at a faster velocity
The higher the temperature of a gas, the higher its viscosity which lets it flow at a slower velocity
Greek symbol: Eta
Coefficient of viscosity
(Nsm-2) η

46. Dropping a sphere F
When a sphere moves slowly through a liquid, the relative movement of the liquid around the sphere is laminar. U
As the molecules its passing through will stick to the surface as it travels, a viscous drag (F) is created
This force was shown to be related to the radius of the sphere, the velocity of the sphere and the coefficient of viscosity ( η ) W

47. Calculating terminal velocity of a sphere in a liquid
This relationship is called
Stoke’s law
Radius of sphere (m)
Rho (kgm-3), 2100kgm-3, temperature
Drag force (N)
Velocity of sphere (ms-1) W
It is only obeyed when the sphere is not near the wall of the cylinder (Bernoulli’s principle at play) and when the sphere is at terminal velocity (when F+U=W)

48. Extension
Whenever a force acts on a material, the materials will be deformed into a different size or shape.
If a material is made longer, then the force is referred to as tension and the extra length is known as extension
If a material is shortened, then the force is referred to as compression and the reduced size is known as negative extension

49. Force applied (N) = Stiffness constant (Nm-1) x Extension (m)
Hooke’s Law
Hooke’s law states that the force needed to extend a spring is proportional to the extension of the spring UNTIL the limit of proportionality has been reached
Force applied (N) = Stiffness constant (Nm-1) x Extension (m)
∆𝐹=𝑘 ∆𝑥

50. Hooke’s Law: In series 1 𝑘𝑡𝑜𝑡𝑎𝑙 = 1 𝑘1 + 1 𝑘2
If you combine springs, you must combine their spring constants.
For springs in series, if a force is applied to them, they will have an overall greater extension (more stretchy)
To do this you treat them like parallel resistors….
1 𝑘𝑡𝑜𝑡𝑎𝑙 = 1 𝑘 𝑘2

51. Hooke’s Law: In parallel
For springs in parallel, if a force is applied to them, they share the force and as such are harder to extend (less stretchy)
To combine them, simply add their stiffness constants
K(total) = K1 + K2

52. definitions
Yield Point
The point at which the material will elongate without any extra added load (acts plastically)
Elastic Limit
The maximum extension that a material can undergo and still return to its original dimensions when the load is removed
Strength (Ultimate tensile)
The highest stress a material can take before breaking (the peak of the graph)

53. elastic strain energy (Eel)
Elastic energy
The work done in deforming a material before it reaches the elastic limit will be stored internally as
elastic strain energy (Eel)
The force varies with different extensions so to calculate the work done we need to use the average force over the distance of extension
A.K.A : The area under the graph!
E(el)= ½ FΔ𝑥

54. Stress/Strain
In physics, when stress is applied to a material, the material will be put under strain (i.e. strain is the result of stress)
This is a little like force and extension except this time, it takes into account the area of the material; so that it’s specific to the type of the material that is used and is not specific to a certain dimension

55. Brittle Vs Ductile
A brittle material can often be strong (high UTS) but will break soon after the elastic limit
A ductile material can be stretched into wires because it will have a large plastic region (After elastic limit)

56. Different metals
HCS is stronger as it has a higher UTS, it is however more brittle as it breaks under a lower strain limit and little plastic deformation, copper is weaker as it has a low UTS but has high plasticity therefore also making it more ductile. Copper- electricity wires and high-carbon steel--? Drill bits (high force-UTS)
In this case: HC steel is strong but brittle as it has a high UTS but breaks soon after elastic limit. Copper is malleable and ductile as it undergoes significant plastic deformation

57. Young modulus (E) = 𝑺𝒕𝒓𝒆𝒔𝒔 (𝑷𝒂) 𝑺𝒕𝒓𝒂𝒊𝒏 (𝒏𝒐 𝒖𝒏𝒊𝒕)
Young’s modulus
Named after Thomas Young, Young’s modulus is the measure of the stiffness of a material but taking into account the dimensions.
I.e. it’s like the stiffness constant but for a stress-strain graph! Just work out the gradient of the linear part of the graph
Young’s Modulus:
Young modulus (E) = 𝑺𝒕𝒓𝒆𝒔𝒔 (𝑷𝒂) 𝑺𝒕𝒓𝒂𝒊𝒏 (𝒏𝒐 𝒖𝒏𝒊𝒕)

58. stress/strain graphs
As Young’s modulus is the gradient of the stress-strain graph line, what is the area underneath?
Well as stress = 𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 and
strain = 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
The area underneath would be stress x strain aka:
𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 x 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
Which is: 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
Makes sense as it’s energy per unit volume!

59. Nature of waves Frequency= 1/period
HCS is stronger as it has a higher UTS, it is however more brittle as it breaks under a lower strain limit and little plastic deformation, copper is weaker as it has a low UTS but has high plasticity therefore also making it more ductile. Copper- electricity wires and high-carbon steel--? Drill bits (high force-UTS)
Frequency= 1/period

60. Transverse waves
In transverse waves, the oscillations are perpendicular to the direction the wave is travelling.
All electromagnetic waves are transverse waves! Maximum displacement occurs at a crest or trough with no displacement along the rest line.

61. longitudinal waves
In longitudinal waves, the oscillations move parallel to the direction of the wave’s travel
Sound waves are longitudinal and they still have wavelengths which can be measured from compression to compression

62. Polarisation of light
“restricting the vibrations of a transverse wave wholly or partially to a specific direction”
If you polarise light, it is restricted to a specific direction and if you use a second filter and slowly turn it, the intensity of light getting through will reduce. At 90°, no light will get through

63. Echo-pulse technique
Just like sonar, an emitter sets out a pulse of sound and this reflects off the object and then is detected. The time taken is then used along with the speed of the sound to calculate the distance travelled. DON’T forget that’s there AND back.

64. Nature of waves
All points in a wave will be at some point through their cycle.
One full cycle (360°) is one full wavelength (λ) which is 2π rads
Each peak would be ¼ of a wavelength (90 °/ λ 𝟒 / π 𝟐 )
HCS is stronger as it has a higher UTS, it is however more brittle as it breaks under a lower strain limit and little plastic deformation, copper is weaker as it has a low UTS but has high plasticity therefore also making it more ductile. Copper- electricity wires and high-carbon steel--? Drill bits (high force-UTS)

65. So in this example, the top wave is π/2 ahead of the bottom one!
Phase Difference
Phase Difference is the difference in phase(measured in radians) between one point of a wave and the same point on another wave of identical frequency (coherent) that started at the same time
So in this example, the top wave is π/2 ahead of the bottom one!

66. Path Difference
Path difference is the difference in distance that two waves have travelled when they reach the same point (P). Measured in wavelengths P
Constructive=
Whole number wavelengths path difference
Delta
Destructive=
Half wavelengths path difference

67. superpositioning
If two waves interfere in phase the waves will make a resultant with a larger amplitude
This is known as constructive interference and only works when they have phase differences of an EVEN number (rads) e.g. (0, 2π, 4π)
If their phase difference is an ODD number (rads)
Then they interact destructively (e.g. π, 3π, 5π)

68. One full cycle (360°) is one full wavelength (λ) which is 2π rads
interference
REMEMBER
One full cycle (360°) is one full wavelength (λ) which is 2π rads
Coherent = have the same wavelength and frequency and a fixed phase difference between them

69. Standing waves
A standing wave is the super-positioning of two progressive waves that are coherent in opposite directions
From node to node or antinode to antinode will be half a wavelength. So for example the third harmonic will be three half wavelengths (1 ½ lambda)
Maximum Kinetic energy is absorbed by the wave at an antinode but none is used at a node. This is because at an antinode, there is maximum amplitude

70. lenses
Lenses can both magnify and focus. There are two types of lenses you need to deal with in A-level.
Converging lenses are convex in shape
Diverging lenses are concave in shape
A real image is formed when the focal point is actually formed where the light converges. A virtual image is when the focal point forms where the light appears to focus and appears behind the lens on a diverging lens

71. The unit of power is the dioptre(D) which is equivalent to (m-1)
Lens power
The power of a lens relates to the ability of the lens to refract the light that passes through it.
“The stronger the power of a lens, the more it refracts light, so it produces a focal point closer to the lens (short focal length)”
The unit of power is the dioptre(D) which is equivalent to (m-1)
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑎 𝑙𝑒𝑛𝑠= 1 𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
The power of a converging lens is always a positive value, and that of a diverging lens is a negative value. If lots of lenses are in combination, you just ADD all the powers!
𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 (𝑃𝑐)=𝑃1+𝑃2+𝑃3

72. Drawing lens diagrams- convergent
Draw the principle ray (horizontal to the lens then down through the focal length of the lens)
Draw the optical focus ray (through the centre of the lens and through the other side)
Where they cross is the point that the image is produced. If you compare the image distance to the object distance, this will tell you the magnification

73. Drawing lens diagrams- divergent
Draw the principle ray (horizontal to the lens then up through at an angle that would make the line APPEAR to hit the focal length to the left of the lend
Draw the optical focus ray (through the centre of the lens and through the other side)
Where they APPEAR to cross is the point that the image is produced. If you compare the image distance to the object distance, this will tell you the magnification

74. Lens equation
There is a relationship between the object distance (u), the image distance (v) and the focal length of the lens (f)
Image distance (v)
Object Distance (u)
Focal length(f)
Delta
Don’t forget, if you are using a diverging lens, use a negative value for focal length or power!
1 𝑢 𝑣 = 1 𝑓

75. 𝑆𝑖𝑛∅1 𝑆𝑖𝑛∅2 = 𝑛1 𝑛2 refraction
Refraction is the bending of light due to a change in speed which is a result of the light travelling through a change in density
REMEMBER: the angle of incidence and angle of reflection are the angles from the NORMAL to the ray.
‘n’ is the refractive index (how much the material refracts radiation)
They have the following relationship (like equivalent ratios)
𝑆𝑖𝑛∅1 𝑆𝑖𝑛∅2 = 𝑛1 𝑛2

76. They have the following relationship (like equivalent ratios)
Critical angle
The critical angle is the angle of incidence that results in the ray refracting at 90°. Any angle greater than this will result ins total internal reflection (TIR)
As ∅2 would be 90°, and Sin90 is equal to ‘1’. At critical angle, this can be omitted from the equation!
They have the following relationship (like equivalent ratios)
𝑆𝑖𝑛∅𝑐 = 𝑛1 𝑛2

77. Nature of waves
Diffraction is the spreading out of waves as they pass through an aperture
Diffraction is the spreading out of waves as they pass through an aperture

78. Interference Peak Trough
Where the peaks cross, we know the waves must be in phase (green dots)
Wave 1
Where it is a peak for one wave but the trough for another then the waves at that point are out of phase (red dots)
Easy: if a point that has both line cross = constructive and if only on one line=destructive
Wave 2

79. Interference Pattern
So if a screen is placed away from the interfering waves you’ll get a pattern like this…
There is a central maximum and fringes of brightness (maxima) where light interferes constructively and fringes of darkness (minima) where light interferes destructively
If the source was sound, you’d hear the source as loud in the maxima and quiet in the minima

80. Double Slit Experiment
λ = 𝑆𝑥 𝐷
Fringe Separation (x)
Note: when taking the measurement of fringe separation you need to take it from the centre of the bright fringe to the centre of the next.
To reduce uncertainty you should take the measurement of a number of fringes and then divide by the number (e.g. 6 fringes, then divide by 6).
Slit separation (S)
mms-1 or 9.67x10-4 ms-1
Screen Distance (D)

81. Diffraction gratings
White light is a mixture of colours and if you put it through a diffraction grating, the different colours (different wavelengths) of light are diffracted by different amounts
Each order in the pattern becomes a spectrum with red on the outskirts and violet more central (longer wavelengths diffract more)
The central maxima will be white as all light will pass straight through

82. Evidence for wave particle duality

83. photons
A ‘photon’ is a discrete ‘packet’ of energy. The energy of that photon is directly proportional to its frequency. Frequency if not provided can be worked out by c/λ.
Electrons in atoms exist in discrete energy levels given by whole numbers (e.g. n=1). If electrons move down an energy level, they emit a packet of energy known as a photon
There are set values for the energy of these photons due to the set energy levels. Each photon would therefore have a set frequency dependent on the transition from one energy level to another.

84. Emission spectra
If you heat a gas to a high temperature, the electrons of the atom may jump to a higher energy level
They will then quickly drop back down to their energy level and release a photon with equivalent energy (and as such, frequency)
If the light from a hot gas is split through a prism, it will produce an emission spectrum of coloured lines on a black background
These correspond to the different wavelengths of photons emitted. As there are only discrete energy levels to drop from/to, there are specific wavelengths produced.

85. absorption spectra
If you heat a gas to a high temperature, the electrons of the atom may jump to a higher energy level
They will then quickly drop back down to their energy level and release a photon with equivalent energy (and as such, frequency)
If the light from a hot gas is split through a prism, it will produce an emission spectrum of coloured lines on a black background
These correspond to the different wavelengths of photons emitted. As there are only discrete energy levels to drop from/to, there are specific wavelengths produced.

86. Reducing uncertainty:
Core Practical 1
Determine the value of g for free fall of an object
Method:
Drop a dowel of measured length (l) through a guiding pipe through light gates
Record the time taken for the dowel to go through the light gate
Repeat for different heights of release
Use the value of time for the dowel to pass through to calculate the speed it was travelling through (s=d/t)
Graph:
Plot a graph of v2 against height
Work out the gradient (m) of the line of best fit
‘g’ = gradient/2
Reducing uncertainty:
Taking measurement of height on the level (avoid parallax error)
Half of range of readings for t as uncertainty of t
Use a high resolution ruler (30cm ruler) for measurement of dowel length
Dowel needs to pass straight through the beam

87. Reducing uncertainty:
Core Practical 2
Determine the resistivity of a material
Method:
Get a meter rule with a metal wire attached and attach a crocodile clip at the zero end
Use a 4mm plug at the free end of the second lead to make a circuit with the zero end and a digital multimeter set to ohms
Every 10cm take a reading of resistance
Use a micrometer to find the diameter of the wire used to calculate the cross sectional area
Graph:
Plot a graph of R against I and measure the gradient ‘m’
Resistivity= m x A (A being the cross sectional area of the wire)
Reducing uncertainty:
Diameter readings should be taken 5 times along the wire, in both a vertical and horizontal direction and then an average taken
Look at the connection from above to reduce parallax error

88. Core Practical 3 Determine the emf and internal resistance of a cell
Method:
Set up the circuit as shown with a 100ohm resistor in series with a cell and a variable resistor
Measure the potential difference (V) across the variable resistor and the current going through it.
Graph:
Plot a graph of V against I.
This gives a gradient= -r (the internal resistance)
This is because V=-Ir + EMF

89. Reducing uncertainty:
Core Practical 4
Using the falling-ball method to determine the viscosity of a liquid
Method:
Get a variety of different radius balls and by finding their mass, use density=mass/volume to measure their density
Place three rubber bands around a tall tube, with the first far down enough to ensure the ball is moving at terminal velocity
Calculate the time taken for the ball to pass from one band to another and measure the distance between these bands
Use v=d/t to find v and then do this for all balls of this diameter and then again for the other balls
Use the value of ‘v’ and ‘r’ etc in the Stoke’s law equation to calculate the viscosity of the liquid.
Reducing uncertainty:
The ball must not go near the sides of the tube as this will introduce a new source of drag as explained by bernoulli’s principle.
When considering the ball moving past a point, this must be viewed straight on as to remove parallax error

90. Reducing uncertainty:
Core Practical 5
Determine the Young modulus of a material
Method:
1. Fix the bench pulley at the end of the bench. Trap one end of the wire between the two wood blocks and secure these to the bench approximately 3 m from the pulley. Lay out the wire so that it passes over the pulley and attach the slotted mass hanger
to the end. Measure the diameter d of the wire.
2. Lay the metre ruler under the wire near the pulley and attach the sticky label to act as a length marker. You judge the length by looking vertically down over the edge of the paper onto the scale of the metre ruler.
3. Measure the length of wire L from the wood blocks to the edge of the paper.
4. Add masses to the hanger and record the position of the marker against the metre ruler. Calculate the extension x for each mass added.
5. You might notice significant creep occurring at higher loads. This indicates the elastic limit has been exceeded
Use the value of mass against extension, the gradient is the young modulus
Reducing uncertainty:
Diameter readings should be taken 5 times along the wire, in both a vertical and horizontal direction and then an average taken
Look at the connection from above to reduce parallax error

91. Core Practical 6
Determine the speed of sound in air using an oscilloscope, a generator etc..
1. The oscilloscope will display on two traces the signal fed to the loudspeaker and the signal
received by the microphone. As the distance between the microphone and the speaker is
increased, the phase of the signals varies and the traces on the screen move past each other.
2. Place the microphone next to the oscilloscope and place the speaker about 50 cm away, facing
the microphone. Turn on the signal generator and set it to about 4 kHz. Adjust the oscilloscope
to show the microphone signal with about three cycles on the screen.
3. Connect the signal generator output to the second oscilloscope input (as well as the speaker)
and adjust the controls to display three cycles of this signal.
4. Adjust the spacing on the screen and the distance between the speaker and microphone so
that the bottom of one trace is just level with the top of the other.
5. Adjust the separation so that a trough on the top trace exactly coincides with a peak on the
lower trace. Place the metre ruler alongside the microphone and speaker and record the
distance between the microphone and speaker.
6. Move the speaker away from the microphone and observe one trace sliding over the other.
Move the speaker so that the trace has moved exactly one cycle. The troughs and peaks
should just touch again. Record the new distance between the microphone and speaker. The
difference between the two distances is one wavelength.
7. Continue to move the speaker away from the microphone and record each successive distance
where the peaks of one trace coincide with the troughs of the other.
8. Calculate a mean value for the wavelength of the sound and estimate the uncertainty in this
measurement.
9. Use one of the traces to determine the frequency of the sound. You will achieve a greater
resolution this way than using the scale on the signal generator. Estimate the uncertainty in this
measurement. You should be able to measure the frequency to three significant figures and the
wavelength to at least two.
10. Using the scale on the signal generator, halve the frequency and repeat the measurements for
frequency and wavelength. You might need to increase the separation beyond 1 m.
11. You might try this experiment at much higher and lower frequencies to observe the effect.

92. Core Practical 7
Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string
Method:
1. Attach one end of the ‘string’ to the vibration transducer. Pass the other end over the bench pulley and attach the mass hanger.
2. Add masses until the total mass is 100 g.
3. Turn on the signal generator to set the rubber oscillating. Vary the oscillating length by moving
the vibration generator until resonance is observed.
4. In this investigation, you are observing standing
waves. These can occur at a variety of resonant frequencies.
5. Can change either the length of the string or the mass added (therefore changing tension)
So for example: Keeping the Tension the same (1.96N) you could plot a graph of wavelength against (1/f) to get a value of μ (mass per unit length)

93. Core Practical 8
Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string
Method:
1. Place the laser approximately 4 m away from a large wall and place the diffraction grating in front of it. Arrange for the beam to pass through the grating at normal incidence and meet the wall perpendicularly.
2. Measure the distance D between the grating and the wall.
3. Turn on the laser and identify the zero order maximum (straight through). Measure the distance ‘s’ to the nearest two first order maxima. Calculate the mean of these two
values. The first order is the maximum produced
according to n = 1 in the equation
nλ = d sin θ
4. Measure s for increasing orders.
5. Repeat for a diffraction grating with a different number of slits/mm.
You can find the angle by using trigonometry (the distance to the wall and then the distance between two fringes)
We use monochromatic light as white light consists of multiple frequencies and wavelengths of light. These will diffract by different amounts, causing the fringes to be a coloured blur with red on the outside and violet towards the central maxima