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Getting Ready to Teach – Online Course Further Mechanics 1 and 2
The A level reforms
• All new AS and A levels will be assessed at the same standard as they are currently
• All new AS and A levels will be fully linear
• AS levels will be stand-alone qualifications
• The content of the AS level can be a sub-set of the A level content to allow co-teachability, but marks achieved in the AS will not count towards the A level
A few questions for you!
The A level reforms
A level Further Mathematics
• 50% core (all pure mathematics)
• 50% optional and can include
– pure mathematics
– mechanics
– statistics
– decision mathematics
– any other
The A level reforms - content
AS level Further Mathematics
•20% core (all pure mathematics)
•10% compulsory (selected from the A level core)
•70% optional (same options as A level)
•We have made an additional 20% of the content compulsory (taken from A level core)
The A level reforms
Requirement for the assessment of problem solving, communication, proof, modelling, application of techniques
Requirement that candidates have a calculator with
• the ability to compute summary statistics and access probabilities from standard statistical distributions
• an iterative function
• the ability to perform calculations with matrices up to at least order 3 × 3 (further mathematics only)
Overview of the specification
A Level Further mathematics
Paper 1: Core Pure Mathematics 125%1 hour 30 minutes75 marks Compulsory content – any content
on either paperPaper 2: Core Pure Mathematics 225%1 hour 30 minutes75 marks
Paper 3: Further Mathematics Option 125%1 hour 30 minutes75 marks
Students take two optional papers with options available in
• Further Pure Mathematics
• Further Statistics
• Further Mechanics
• Decision Mathematics
Paper 4: Further Mathematics Option 225%1 hour 30 minutes75 marks
A level Further Mathematics options
For papers 3 and 4 students choose a pair of options, either•any two from column A, or•a matching pair from columns A and B
Overview of the specification
AS Further Mathematics
Interactive Scheme of Work
Schemes of Work
Exam Papers
AS Level
A Level
Existing Resources AS Further Mechanics 1
Unit TitleOld
Module
Chapter
Reference
1 Momentum and impulse M1 3
a Momentum and impulse; derivation of units and formulae
Impulse-momentum principle. Conservation of momentum
applied to collisions and ‘jerking’ string problems
M1 3
2 Work, energy and power M2 3
a Work, kinetic energy; derivation of units and formulae M2 3
b Potential energy, work-energy principle, conservation of
mechanical energy, problem solvingM2 3
c Power; derivation of units and formula M2
3 Elastic collisions in one dimension M2 3
a Direct impact of elastic spheres. Newton’s law of restitution.
Loss of kinetic energy due to impact M2 4
b Problem solving (including ‘successive’ impacts) M2 4
Existing Resources AL Further Mechanics 1
UnitTitle Old Module
Chapter
Reference
4 Momentum and impulse (part 2) M2 4
a Momentum as a vector (i, j problems)
Impulse-momentum principle in vector formM2 4
5 Elastic strings and springs and elastic energy M3 2
a Hooke’s law and definition of modulus of elasticity. Derivation of elastic potential
energy formula.M3 2
bProblem solving: equilibrium and using the work-energy principle M3 2
6 Elastic collisions in two dimensions M4 2
aOblique impact of a smooth sphere with a fixed surface Successive oblique impacts
of a sphere with smooth plane surfaces M4 2
b Oblique impact of two smooth spheres of equal radius M4 2
Existing Resources AS Further Mechanics 2
UnitTitle Old Module
Chapter
Reference
1 Motion in a circle M3 4
aAngular speed, central force, radial acceleration M3 4
b Uniform motion in a horizontal circle M3 4
2 Centres of mass of plane figures M2 2
aMoment of a force
Centre of mass of a discrete mass distribution in one and two dimensions,
framework and uniform lamina (rectilinear shapes)
M2 2
bCentre of mass of triangular, circular-based and composite laminas and
centre of mass of a uniform circular arcM2 2
c Modelling equilibrium: hanging bodies and systems free to rotate (about a
fixed horizontal axis)M2 2
3 Further kinematics M3 1
aMotion in a straight line when the acceleration is a function of time (t);
Setting up and solving differential equationsM3 1
bMotion in a straight line when the acceleration is a function of the velocity
(v); Setting up and solving differential equationsNew
Existing Resources AL Further Mechanics 2
UnitTitle Old Module
Chapter
Reference
4 Motion in a circle (part 2)
aMotion in a vertical circle: radial and tangential acceleration; Conservation
of energy in this contextM3 4
5 Further centres of mass M3 5
aCentre of mass of uniform/non-uniform rod, lamina, 3-D rigid body using
integration (and symmetry); Deriving formulae in formula bookM3 5
bCentre of mass of composite bodies; Simple cases of equilibrium of rigid
bodies.M3 5
c Conditions for toppling/sliding M3 5
6 Further kinematics (part 2)
aMotion in a straight line when the acceleration is a function of time (t);
Setting up and solving differential equationsM3 1
bMotion in a straight line when the acceleration is a function of the velocity
(v); Setting up and solving differential equationsNew
7 Further dynamics
aParticle moving in straight line with variable applied force; Using F = ma
to set up differential equations and solvingM3 1
b Newton’s law of gravitation M3 3
c Simple harmonic motion (SHM) M3 3
Key PrerequisitesAS Further Mechanics 1
Unit Title Prerequisite Knowledge
1 Momentum and impulse SUVAT Formulae (introduction)
a Momentum and impulse; derivation of units and formulae
Impulse-momentum principle. Conservation of momentum applied
to collisions and ‘jerking’ string problems
2Work, energy and power
Resolving Forces
Frictional Forces
a Work, kinetic energy; derivation of units and formulae
b Potential energy, work-energy principle, conservation of
mechanical energy, problem solving
c Power; derivation of units and formula
3Elastic collisions in one dimension
SUVAT Formulae
Geometric Progression Summation
a Direct impact of elastic spheres. Newton’s law of restitution. Loss of
kinetic energy due to impact
b Problem solving (including ‘successive’ impacts)
Key Prerequisites AL Further Mechanics 1
UnitTitle Prerequisite Knowledge
4 Momentum and impulse (part 2) SUVAT in 2D
a Momentum as a vector (i, j problems)
Impulse-momentum principle in vector form
5Elastic strings and springs and elastic energy
Resolving Forces, Basic
Integration, Friction
a Hooke’s law and definition of modulus of elasticity. Derivation of elastic potential
energy formula.
b Problem solving: equilibrium and using the work-energy principle
6Elastic collisions in two dimensions
AS Maths Trig Identities, AS
FM Scalar product
a Oblique impact of a smooth sphere with a fixed surface Successive oblique impacts
of a sphere with smooth plane surfaces
b Oblique impact of two smooth spheres of equal radius
Key Prerequisites AS Further Mechanics 2
Unit Title Prerequisite Knowledge
1Motion in a circle
A Level Maths Radians
A Level Maths Resolving Components
aAngular speed, central force, radial acceleration
b Uniform motion in a horizontal circle
2 Centres of mass of plane figures A Level Maths Moments
aMoment of a force
Centre of mass of a discrete mass distribution in one and two
dimensions, framework and uniform lamina (rectilinear shapes)
bCentre of mass of triangular, circular-based and composite laminas and
centre of mass of a uniform circular arc
c Modelling equilibrium: hanging bodies and systems free to rotate
(about a fixed horizontal axis)
3Further kinematics
A Level Maths Integration separation of
variables
aMotion in a straight line when the acceleration is a function of time
(t); Setting up and solving differential equations
bMotion in a straight line when the acceleration is a function of the
velocity (v); Setting up and solving differential equations
Key Prerequisites AL Further Mechanics 2
Unit Title
4 Motion in a circle (part 2) Radians, Projectile motion, resolving forces
aMotion in a vertical circle: radial and tangential
acceleration; Conservation of energy in this context
5 Further centres of mass Integration
a
Centre of mass of uniform/non-uniform rod, lamina, 3-D
rigid body using integration (and symmetry); Deriving
formulae in formula book
bCentre of mass of composite bodies; Simple cases of
equilibrium of rigid bodies.
c Conditions for toppling/sliding
6Further kinematics (part 2)
Integration
A Level Maths and Further Maths
a Motion in a straight line when the acceleration is a
function of time (t); Setting up and solving differential
equations
b Motion in a straight line when the acceleration is a
function of the velocity (v); Setting up and solving
differential equations
7
Further dynamics
AL Maths Trigonometry, Differentiation
AL FM1 Hooke’s Law
AL FM Integration, Differentiation, Hyperbolics
a
Particle moving in straight line with variable applied
force; Using F = ma to set up differential equations
and solving
b Newton’s law of gravitation
c Simple harmonic motion (SHM)
Getting Ready to Teach – Online Course Elastic Collisions–Further Mechanics 1 (AS)
Objectives
•
Objectives
• be able to solve problems of the following types involving elastic impacts:
– successive collisions between pairs of spheres (horizontal motion);
– bouncing ball (off a horizontal elastic plane);
– successive collisions including two spheres and sphere against a wall;
– determination of number of collisions or deriving the possible range of values of e.
Newton’s (Experimental) Law of Restitution
•
Key Formulae
•
Example
•
151
AS FM 1 Qu 1
Getting Ready to Teach – Online Course Elastic Collisions in 2D–Further Mechanics 1 (A Level)
Elastic collisions in two dimensions: Prior Knowledge
Covered so far
•Momentum and impulse (Unit 1)
•Direct collisions, Newton’s (experimental) law of restitution (Unit 3)
•Change in kinetic energy due to impact (Unit 3)
•Momentum and Impulse-momentum principle in vector form (Unit 4)
•Change in kinetic energy due to impact (Unit 5)
•AS Mathematics – Pure content
– Work with i, j vectors, magnitude & direction (See SoW Year 1 Unit 5)
– Trigonometric identities (See SoW Year 1 Unit 7)
•A level Mathematics – Pure content
– Trigonometric identities (See SoW Year 2 Unit 8)
•AS Further Mathematics – Core Pure content
– Scalar product of vectors in 2-D (See SoW Unit 7b)*
•* Note that, although an understanding of scalar product is useful in problem solving, only vectors in two dimensions are being considered and so other techniques are possible
Elastic collisions in two dimensions
• 5.1 Oblique impact of smooth elastic spheres and a smooth sphere with a fixed surface. Loss of kinetic energy due to impact.
• 5.2 Successive oblique impacts of a sphere with smooth plane surfaces.
Elastic collisions in two dimensions: Objectives
By the end of the sub-unit, students should:
•understand that during an impact the impulse acts perpendicularly to the surface through the centre of the sphere;
•be able to apply Newton’s (experimental) law of restitution in the direction of the impulse;
•appreciate that perpendicular to the impulse, the velocity component does not change;
•understand and be able to calculate an angle of deflection;
•be able to calculate the kinetic energy ‘lost’ in an impact;
•be able to work in speeds and angles or in velocity vectors (i, j).
Impulse in 2 DimensionsA Level
Teaching Points
•
Example
•
Example
•
Situations
• Oblique impact of a smooth sphere with a fixed surface
• Successive oblique impacts of a sphere with smooth plane surfaces
• Oblique impact of two smooth spheres of equal radius (Spin is ignored)
Two Spheres
Two Spheres
Angle of Impact?
Two Spheres
Angle of Impact?
SAM Questions
Getting Ready to Teach – Online Course Elastic Springs and EnergyFurther Mechanics 1 (A Level)
Elastic Springs and Elastic Energy
• Covered so far– KE, GPE and the work-energy principle (Unit 2)
• AS Mathematics – Pure content
– 8.3 Basic integration (See SoW Year 1 Unit 4)
• AS Mathematics – Mechanics content – 8.1, 8.2, 8.4 Types of forces (in particular tension) and Newton’s
laws
– (See SoW Year 1 Unit 9)
• A level Mathematics – Mechanics content – 8.2, 8.4, 8.5, 8.6 Resolving forces, friction, equilibrium and dynamics
– (See SoW Year 2 Units 4, 7 and 8)
Elastic Springs and Elastic Energy: Objectives
•
Hooke’s Law
•
Example
•
Further Problems
•
Further Problems
•
Work Energy PrincipleObjectives
By the end of the sub-unit, students should:
•be able to calculate the tension in a string or spring when a system is held in equilibrium;
•be able to include EPE when using the work-energy principle;
•know the conditions for conservation of mechanical energy;
•be able to solve string/spring problems involving work and energy (i.e. KE, GPE and EPE).
Energy
•
Example
•
Example
•
Getting Ready to Teach Online course
Hookes Law A Level
Getting Ready to Teach – Online Course Horizontal Circular Motion– Further Mechanics 2 (AS)
Horizontal Circular MotionPRIOR KNOWLEDGE
AS level Mathematics – Pure content
Calculus: Rates of Change, Integration (unit 6 & 7)
Vectors (Unit 5)
A level Mathematics – Pure content
Trigonometry: Radians (Unit 6)
Differentiation: sin & cos, function of a function (Unit 8)
AS Mathematics – Mechanics content
Kinematics 2 : Variable acceleration (Unit 4)
A level Mathematics – Mechanics content
Resolving using Newton’s Second Law: F = ma (Units 4 & 8b)
Friction (Unit 7)
Modelling assumptions made throughout this course
(e.g. particle, light, rod, inextensible, smooth)
Horizontal Circular Motion
Horizontal Circular Motion
OBJECTIVES 2
• understand that a particle moving in a horizontal circle with constant speed has an acceleration of magnitude rω2 directed towards the centre of the circle
• know that the resultant force acting on the particle has magnitude mrω2 and is directed towards the centre of the circle
• be able to model a variety of physical situations which give rise to uniform circular motion, including use of the period, T
• understand that a banked surface is a plane that is at an angle to the horizontal
• be able to consider how conditions affect the described circular motion, for example the greatest and least possible angular speeds
Horizontal Circular MotionAS SAMs Question
Horizontal Circular MotionDERIVATION
If the particle P moves in the circle shown then the angular speed
is defined as rate of change of angle θ w.r.t. time t or dθ/dt .
We can formalise this relationship by writing down the arc length
formula: s = rθ (where r is the radius)
Differentiating both sides with respect to t gives: ds/dt = r dθ/dtas r is constant for a circle.
Denoting angular speed dθ/dt by ω we have the important
relationship
The angular velocity, ω will be measured an rad s–1 and rev min–1 v = rω
When a particle moves in a circle with a constant speed the direction of its linear
velocity (which is tangential to the circle) is constantly changing; this means there
is an acceleration (even though the speed is constant). This acceleration is always
directed towards the centre of the circle (radial acceleration)
The next approach shows why the force is acting in this direction:
If dθ/dt = ω, then if we integrate both sides we get θ = ∫ ω dt
So, θ = ωt + C (Taking θ = 0 when t = 0 gives C = 0) Therefore, θ = ωt
Horizontal Circular Motion
Horizontal Circular Motion
Horizontal Circular Motion
Horizontal Circular MotionMATHEMATICAL MODELLING
For each of the following examples it is a good idea to draw a
diagram and label the forces on the particle. Discuss where the
centre of the circle is and which force, or combination of forces,
provides the force towards the centre.
Also note how the vertical forces balance.
• Particle on the rough surface of a rotating disc – the friction
force acts towards centre. (e.g. Roundabout in a playground)
• Conical pendulum (including elastic string) – a light string has
one end attached to a fixed point and the other attached to a
particle which is moving in a horizontal circle. The horizontal
component of tension provides the central force.
• Smooth bead on a rotating wire – component of reaction acts
towards centre of circular path.
Horizontal Circular MotionMATHEMATICAL MODELLING 1 (Friction)
The person is standing 5m from the axis of the roundabout and is rotating at 1 rads-1
Solution
Resolving vertically: R – 40g = 0 (Equilibrium) So, R = 40gN
For circular motion: X = mrω2 (These two forces identical) So, X = 40 x 5 x (12) = 200N
Now, Friction force X = µR (at point of sliding) Therefore, 40g µ = 200N
Giving µ = 200/40g = 0.51
So, to prevent sliding at this speed µ ≥ 0.51
Horizontal Circular MotionMATHEMATICAL MODELLING 2 (Conical Pendulum)
The chord is 1.2m long and the 0.2kg stone is spun in a horizontal circle at 5 rads-1
Solution
Resolving vertically: T cos θ – 0.2g = 0 (Equilibrium) So, T cos θ = 0.2gN (1)
For circular motion: T sin θ = mrω2 (2)
By Trigonometry the radius r of the circle is 1.2 sin θ (since the hypotenuse is 1.2m)
Substitute r and given ω into equation (2) gives T sin θ = 0.2 x 1.2 sin θ x 52
Cancel sin θ to give T = 0.2 x 1.2 x 52 = 6N The tension in the chord is 6N
Subst. T into equation (1) gives cos θ = g/30, so θ = 70.90 (1 d.p.)
Horizontal Circular MotionMATHEMATICAL MODELLING 3 (Banked Tracks)
The coefficient of friction between the tyres and the road is 0.6, find the max speed of the car in km h-1.
Source:Mr Barton Maths
Horizontal Circular MotionSUMMARY
Source:MEI
Horizontal Circular Motion
PROBLEM SOLVING 1
Particle supported by two light strings (See diagram) – Discuss why the
tensions in this case cannot be equal and which one is greater. What is
a condition for circular motion to be possible in this configuration with
both strings taut? The fact that the tension in the lower string must be
greater than zero leads to a minimum possible value for the angular
speed (and consequently a maximum value for the period of revolution).
T
This is like the SAMs question we saw earlier
You could also discuss what happens
to the tension in AP as the angular
speed increases leading to the
possibility that the string breaks so that
there is a minimum value for the period
of revolution.
Horizontal Circular MotionPROBLEM SOLVING 2
Particle moving on a smooth fixed surface – for example a particle
moving on the inside of a smooth fixed hollow cone.
Source: Mr Barton Maths
Horizontal Circular MotionPROBLEM SOLVING 3
Particle moving on a smooth fixed surface – for example a particle
moving on the inside of a smooth fixed hemispherical shell as shown
in the diagram below: Here it is important to note that:
the reaction is directed towards the
centre of the hemisphere and not
towards the centre of the circular path of
the particle.
The radius of the circular path is not the
same as the radius of the hemisphere
(called r here, so care must be taken
when quoting standard formula ‘rω2’)
The radius of circle can be found by
using Pythagoras theorem and the angle
of the reaction with horizontal is found by
trigonometry.
Horizontal Circular MotionWHAT THE EXAMINER SAID
3. A rough disc rotates about its centre in a horizontal plane with constant angular speed 80 revolutions per minute. A particle P lies on the disc at a distance 8 cm from the centre of the disc. The coefficient of friction between P and the disc is μ. Given that P remains at rest relative to the disc, find the least possible value of μ.
(7 marks)
Candidates seemed to have great difficulty changing from revolutions perminute to radians per second. This seemed to be wrong at least as often as itwas correct. A surprising number of solutions involved inequalities, not alwaysthe correct way round. Unfortunately, some gave their final answer as aninequality and so failed to answer the question as set. Many did not notice thatthe distance was given in centimetres and so used 8 instead of 0.08 in theircalculation. Those who knew how to tackle this question produced succinctsolutions.
Horizontal Circular MotionReturn to AS SAMs Question
Horizontal Circular MotionAS SAMs Question
Horizontal Circular MotionAS SAMs Question
Horizontal Circular MotionMark Scheme
Horizontal Circular MotionMark Scheme
Horizontal Circular Motion
This topic is extended in
AL Further Mechanics 2:
Vertical Circular Motion
Thank you for your attention
Getting Ready to Teach – Online Course Centres of Mass – Further Mechanics 2 (AS/AL)
Centres of Mass - Moments
PRIOR KNOWLEDGE
A level Mathematics – Pure content
Sigma notation (Unit 4b)
AS Mathematics – Mechanics content
Modelling assumptions made throughout this course
(e.g. particle, rigid, light, lamina, etc.)
Types of forces and force diagrams (Unit 8a)
A level Mathematics – Mechanics content
Equilibrium and statics (Unit 7a)
Moments (Unit 4)
Centres of Mass - MomentsOBJECTIVES 1
• be able to extend the principle of moments to develop an
approach and formula for finding the centre of mass of a
one dimensional discrete mass distribution;
• be able to find the centre of mass of a two dimensional
discrete mass distribution and a framework;
• be able to find the centre of mass of a plane uniform
lamina made up of rectilinear shapes e.g. L shape.
• know and be able to quote the position of the centre of
mass for a uniform triangular lamina;
• know and be able to quote the position of the centre of
mass for a uniform circular arc or sector of a circle;
• be able to adapt the given formulae for semi-circular
laminas and arcs;
Centres of Mass - Moments
OBJECTIVES 2
• be able to find the centre of mass of composite 2-D shapes.
• be able to use the position of the centre of mass for a 2-D lamina or framework to determine the orientation of the shape relative to either the vertical or horizontal when suspended freely from a point;
• be able to use the position of the centres of mass for a 2-D lamina or framework to establish a condition for equilibrium to exist when the body is free to rotate about a fixed horizontal axis.
Centres of Mass - MomentsAS SAMs Question
Centres of Mass - MomentsAS SAMs Question
Centres of Mass - Moments
TEACHING POINTS 1
Centres of Mass - Moments Centre of mass of a discrete mass distribution in two dimensions, framework and lamina. (rectilinear shapes)
1) 2)
3a) 3b)
Centres of Mass - MomentsCentre of mass of uniform bodies
Triangular lamina, arc & sector of a circle (Special case:- Semi-circular lamina)
Centres of Mass - Moments
MATHEMATICAL MODELLING (Hanging Body)
In the example below the lamina is freely
suspended from A. It is required to find
the size of the angle that the side AB
makes with the vertical.
The diagram illustrates that the creation
of a right-angled triangle is the most
effective way to find the size of this angle,
reading off the distances to the centre of
mass to find the lengths and using ‘arc
tan’ to find the size of the required angle
(usually in degrees to 1 decimal place).
Note that, although when suspended the
centre of mass lies vertically below A, it is
not necessary to draw a new diagram.
Centres of Mass - MomentsPROBLEM SOLVING
The mass of the lamina is M. A particle of mass kM is attached to
the lamina at D to form the system S. The system S is freely
suspended from A and hangs in equilibrium with AO horizontal.
A horizontal force of magnitude P is applied at C in the direction CD.
The loaded plate L remains suspended from A and rests in
equilibrium with AB horizontal and C vertically below B.
This topic is extended in
AL Further Mechanics 2:
Further Centres of Mass
Thank you for your attention
Centres of Mass - Moments
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