## Wednesday, 23 November 2011

### D1 Worked solutions

May 2010

January 2010

May 2009

January 2009

Mock paper

Practice paper A

Practice paper B

## Tuesday, 15 November 2011

### D1 Algorithms revision and linear programming 15/11/11

- Revision of.....
- Binary search algorithm
- Bubble sort, quick sort
- Bin packing

## Tuesday, 1 November 2011

### D1 Critical path analysis 01/11/11

- Precedence/dependence tables
- Drawing network from precedence table
- Early and late event times
- Slack on events
- Float on activities
- Critical activities
- Critical path
- Cascade (Gannt) chart

## Tuesday, 18 October 2011

### D1 Matchings 18/10/11

- Reminder of bipartite graphs
- Matchings - trivial, maximal, complete
- Alternating paths to improve matchings
- Exam questions involving matchings

## Wednesday, 12 October 2011

## Tuesday, 4 October 2011

### D1 Route inspection 04/10/11

- Reminder of traversable (Eulerian) and semi-traversable (semi-Eulerian) graphs
- The route inspection problem
- Examples of using the route inspection problem

## Tuesday, 27 September 2011

### D1 Spanning trees 27/09/11

- Spanning trees
- Minimum spanning trees
- Kruskal's algorithm
- Prim's algorithm
- Using a distance matrix with Prim's algorithm

## Tuesday, 20 September 2011

### D1 More graphs 20/09/11

- Digraphs (directed graphs)
- Modelling situations with graphs
- Moving around graphs
- Defining path, cycle, tree
- Traversing graphs - Eulerian & Semi-Eulerian graphs
- Odd/even vertices

## Tuesday, 13 September 2011

### D1 Introduction to graphs 13/09/11

- Defining terms relating to graphs
- - nodes, vertices, arcs, edges, isomorphic, connected, disconnected, complete
- - subgraph, loops, multiple edges, bipartite, complete bipartite, degrees
- Relationship between edges and sum of degrees
- Describing a graph using a vertex set and edge set
- Describing a graph using an adjacency matrix
- Network (weighted graph) and distance matrix

- Homework set - revision of Algorithms (past exam questions)

## Wednesday, 8 June 2011

### More Core 3&4 worked solutions

Core 3 Jan 2011

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzJmNDEyMmUtZWMyYS00MmJlLTkzMWItNTg5Yjc1N2UyYmYw&hl=en_US

Core 3 Jan 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYjRjYWY5MTgtN2ExYy00ZmY3LWIxODgtZmVjYmVlNGM1NWNj&hl=en_US

Core 3 June 2005

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzhkMDNkMGUtZmIzNS00OWJhLWIxNGUtNjU5Njc5MzEyZTkx&hl=en_US

Core 4 Jan 2011

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzA4NTJkMTQtZjc2NS00NGEyLTlmOTktYmNiZDM2MmNmY2U2&hl=en_US

Core 4 June 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTcyYjViMTgtN2NmMS00NzJmLWJiMjgtZDhhMDQwMzE1OWQ3&hl=en_US

Core 4 Jan 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOGQ2ZDE4ZTItMzMxNi00NmU3LTliYjYtZWJiMTg1Njc2MGUw&hl=en_US

Core 4 June 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpM2FlYjlhNzktZjUzZS00Mjk1LTkzYjAtNjFhOGQ3ODZlZmRh&hl=en_US

Core 4 Jan 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNWEyZWEwMTQtNDUyYi00NWUxLTlkZTUtOGM2N2IwNmI5Y2U3&hl=en_US

Core 4 June 2005

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZjRjYWUwYTgtMDBiOS00MzFiLTlhMGMtMmE4YTcwOWQwYzk3&hl=en_US

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzJmNDEyMmUtZWMyYS00MmJlLTkzMWItNTg5Yjc1N2UyYmYw&hl=en_US

Core 3 Jan 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYjRjYWY5MTgtN2ExYy00ZmY3LWIxODgtZmVjYmVlNGM1NWNj&hl=en_US

Core 3 June 2005

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzhkMDNkMGUtZmIzNS00OWJhLWIxNGUtNjU5Njc5MzEyZTkx&hl=en_US

Core 4 Jan 2011

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzA4NTJkMTQtZjc2NS00NGEyLTlmOTktYmNiZDM2MmNmY2U2&hl=en_US

Core 4 June 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTcyYjViMTgtN2NmMS00NzJmLWJiMjgtZDhhMDQwMzE1OWQ3&hl=en_US

Core 4 Jan 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOGQ2ZDE4ZTItMzMxNi00NmU3LTliYjYtZWJiMTg1Njc2MGUw&hl=en_US

Core 4 June 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpM2FlYjlhNzktZjUzZS00Mjk1LTkzYjAtNjFhOGQ3ODZlZmRh&hl=en_US

Core 4 Jan 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNWEyZWEwMTQtNDUyYi00NWUxLTlkZTUtOGM2N2IwNmI5Y2U3&hl=en_US

Core 4 June 2005

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZjRjYWUwYTgtMDBiOS00MzFiLTlhMGMtMmE4YTcwOWQwYzk3&hl=en_US

## Monday, 16 May 2011

### Core4 Worked solutions 2008,2009,2010

June 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGJhODk2ODQtMjY4Yi00MTIyLWFlODgtZDI2Zjc0YjE4YWMz&hl=en

January 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYWFjZmQ4NzYtZDY5Zi00YzEzLWI2OTUtOTk4MDI4OTUzNmMy&hl=en

June 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMTA4MzRlZjAtYWM3Zi00OGRiLWJhMDUtMzkzZWQ3MDhlZTJl&hl=en

January 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpODBkNmQ4MjMtOTg0ZC00OWUzLWIwMmItYjIwNTJjOTc2MTdk&hl=en

June 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGNiNTJiN2YtOThjZi00MDdlLTk5N2YtNDBiNmMwYmQ0Yjdm&hl=en

January 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMGVlOTFlY2UtOTRkYi00NzhiLTg1MjYtY2QxZWQyMWRhZjUw&hl=en

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGJhODk2ODQtMjY4Yi00MTIyLWFlODgtZDI2Zjc0YjE4YWMz&hl=en

January 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYWFjZmQ4NzYtZDY5Zi00YzEzLWI2OTUtOTk4MDI4OTUzNmMy&hl=en

June 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMTA4MzRlZjAtYWM3Zi00OGRiLWJhMDUtMzkzZWQ3MDhlZTJl&hl=en

January 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpODBkNmQ4MjMtOTg0ZC00OWUzLWIwMmItYjIwNTJjOTc2MTdk&hl=en

June 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGNiNTJiN2YtOThjZi00MDdlLTk5N2YtNDBiNmMwYmQ0Yjdm&hl=en

January 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMGVlOTFlY2UtOTRkYi00NzhiLTg1MjYtY2QxZWQyMWRhZjUw&hl=en

## Wednesday, 4 May 2011

## Tuesday, 3 May 2011

## Wednesday, 27 April 2011

## Monday, 4 April 2011

## Thursday, 31 March 2011

### Core3 Worked solutions

June 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzhkYzc0MGItNDBkOS00MTI3LThhNjYtOTlkM2VmYzhmMzUw&hl=en

Jan 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYjRhNWY3NzEtYTM2Ni00YmVmLTgzYjctYWYwMzYyMjZiZmUx&hl=en

June 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMWFjYmFhMmEtZjQ0Yy00ZGE2LThjNGYtMDBjMjdhNmMwNGE2&hl=en

Jan 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOTkxY2MyYTctYzNmNC00MzZjLTk1OTQtZDBmZTc4MDA3YTk0&hl=en

June 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYTUwNzAzZTQtYjBkYi00ODlhLTgzMDEtMWQ3ZmExZDEyMmE1&hl=en

Jan 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDBmODNkYmItNzI5Yi00Njg1LThjODktMDRmZTAxZWZlNGI3&hl=en

June 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzA3OWY5YjEtNWJiZi00NjkzLTk5YjctY2RiYTY5OTcxNzBl&hl=en

Jan 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWRmZDI0ODktNzJjZi00NWYwLWExNmMtNzJkNjNkNDFiYmIz&hl=en

June 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzQ2YTU5YzktZjMxNC00N2ZjLWIzMDctYjI3ZjcyZjIxYzJk&hl=en

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzhkYzc0MGItNDBkOS00MTI3LThhNjYtOTlkM2VmYzhmMzUw&hl=en

Jan 2010

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYjRhNWY3NzEtYTM2Ni00YmVmLTgzYjctYWYwMzYyMjZiZmUx&hl=en

June 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMWFjYmFhMmEtZjQ0Yy00ZGE2LThjNGYtMDBjMjdhNmMwNGE2&hl=en

Jan 2009

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOTkxY2MyYTctYzNmNC00MzZjLTk1OTQtZDBmZTc4MDA3YTk0&hl=en

June 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYTUwNzAzZTQtYjBkYi00ODlhLTgzMDEtMWQ3ZmExZDEyMmE1&hl=en

Jan 2008

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDBmODNkYmItNzI5Yi00Njg1LThjODktMDRmZTAxZWZlNGI3&hl=en

June 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzA3OWY5YjEtNWJiZi00NjkzLTk5YjctY2RiYTY5OTcxNzBl&hl=en

Jan 2007

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWRmZDI0ODktNzJjZi00NWYwLWExNmMtNzJkNjNkNDFiYmIz&hl=en

June 2006

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzQ2YTU5YzktZjMxNC00N2ZjLWIzMDctYjI3ZjcyZjIxYzJk&hl=en

## Wednesday, 30 March 2011

### Core4 Rates of change 28/03/11

- Going through homework question about implicit differentiation
- Introducing connected rates of change
- Problems involving rates of change
- Past exam questions involving rates of change

Flipchart:

Past exam questions worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpZTk2M2EwNjAtZTY4OS00NDgxLWIxNmMtOTE3NmM5MjFkOTk4&hl=en

## Wednesday, 23 March 2011

### Core4 Implicit differentiation 23/03/11

- Intro to implicit differentiation
- Using chain rule to differentiate terms in y with respect to x
- Including use of product rule for terms in both x and y
- Using to solve problem involing tangents and normals to curve defined implicitly
- HOMEWORK SET on problem involving stationary points on a curve defined implicitly

## Wednesday, 9 March 2011

## Monday, 7 March 2011

### Core3 a cos + b sin 07/03/11

- Recalling sin and cos addition formulae
- Using form a sin theta + b cos theta = R cos (theta - alpha) etc.

## Thursday, 3 March 2011

### Thursday 3rd March 2011, Period 5

Just did more differentiation practice.

Differentiate:

y = sec x

y = cosec x

y = cot x

using

Exercise 5F

Q8 (all parts)

due Wednesday 9th March 2011

Differentiate:

y = sec x

y = cosec x

y = cot x

using

__either__Quotient Rule or Chain Rule.**HOMEWORK**Exercise 5F

Q8 (all parts)

due Wednesday 9th March 2011

## Wednesday, 2 March 2011

### Core4 Using parametric equations 02/03/11

- Reminder of differentiating parametric equations
- Finding gradient at given point of curve defined by parametric equations
- Finding equations of tangent to curve at given point

## Monday, 28 February 2011

### Core4 Parametric equations 28/02/11

- Sketching curves where x and y are defined in terms of a third variable
- Defining parametric equations
- Converting from parametric equations to Cartesian form
- Differentiation from parametric equations

## Wednesday, 16 February 2011

### Core4 Vector equations 16/02/11

- Reminder identifying perpendicular vectors
- Introduce concept of vector equation of line
- Vector equation of line from given point and direction
- Vector equation of line from two given points
- HOMEWORK SET

## Monday, 14 February 2011

### Core4 More vectors 14/02/11

- Using vectors geometrically
- Scalar (dot) products
- Scalar products with i-j-k form
- Perpendicular and parallel
- Solving problems involving scalar products
- HOMEWORK SET

## Wednesday, 9 February 2011

### Core4 Vectors - moving from M1 09/02/11

- Vectors revision from M1 (magnitude, direction and unit vectors)
- Introducing column vectors
- Introducing 3 dimensional vectors (column and i-j-k)
- Magnitude of 3 dimensional vectors
- Homework set (from worksheet given in lesson)

## Monday, 7 February 2011

### Core4 Binonial series / Partial fractions exam questions 07/02/11

- Core 4 exam questions
- Binomial series including use of partial fractions
- HOMEWORK SET

Flipchart:

Classwork sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpODVjMzg0MTMtMDk4Ni00ZDdkLTljMGItZWI3NjMzYmExYzRi&hl=en

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpMTJmZDFkODktNzg4My00ZDAxLWEwNDYtMmU1NTUwOWJlMDlj&hl=en

### Core3 Trig exam questions 07/02/11

- Reminder of double angle formulae
- Some Core 3 trig exam questions
- HOMEWORK SET

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpODM3ZTJiM2ItMDQ2Yy00YTU0LWEwMmEtMzg2MGY2MjM2ZWRi&hl=enClasswork sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpM2U2ZDY2YzgtNTQxMS00OTI1LTg5NmYtZjljZDQzZjU0ZmJh&hl=en

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpNWNkNjk5OGEtMGJkOS00OTIxLWIyNGMtZDZiMGI4NDU1ZGZm&hl=en

## Wednesday, 2 February 2011

### Core2and3 Trigonometry revision 02/02/11

- Solving simple trig equations
- Analysing trig graphs
- Multiple solutions to trig equations
- Core 3 inverse trig functions
- Core 3 trig identities
- Using the above to solve Core 3 trig equations

## Monday, 31 January 2011

### Core4 Binomial series 31/01/11

- Recap of Core 2 binomial series
- Including using for approximation
- Introduce Core 4 binomial series - negative and fractional indices
- Expanding binomial series with negative and fractional indices
- Including using for approximation
- HOMEWORK set - Core 4 exam questions on binomial series

Flipchart:

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpNDMwZTFhNjktYTZmNC00YjcyLWE3YzEtYTQ2ZmM5OGEzZDgx&hl=en

## Wednesday, 26 January 2011

### 13B Cover Work - C2 Differentiation

http://dl.dropbox.com/u/15976318/13B%20Cover%20Work%20C2%20Differentiation%20Exam%20Q.doc

As I will not be here on Thrusday 27th January 2011, I have set some cover work. Questions are from past C2 exams (including one question from the exam I sat when I did A-Level Maths!) and are based on Differentiation in preparation for the next topic, which is... Differentiation!

Mr San.

As I will not be here on Thrusday 27th January 2011, I have set some cover work. Questions are from past C2 exams (including one question from the exam I sat when I did A-Level Maths!) and are based on Differentiation in preparation for the next topic, which is... Differentiation!

Mr San.

### Core4 More partial fractions 26/01/11

- Recap of splitting into partial fractions (C4 exam question)
- Dealing with improper fractions (degree of numerator greater than or equal to that of denominator)
- Dividing the polynomials and then splitting the remainder into partial fractions
- Homework set

## Monday, 24 January 2011

### Core4 Partial fractions 24/01/11

- Revision of Core 3 simplifying algebraic fractions and combining to single fraction
- Revision of basic principles of improper fractions and mixed numbers
- Revision of Core 2 dividing polynomials including finding remainders
- Introduce concept of partial fractions as inverse process of Core 3 level combining fractions into a single fraction
- Examples of partial fractions using substitution method, equating coefficient method, cover up and combining substitution and equating coefficients
- Examples with two or three distinct linear terms in denominator
- Examples with repeated linear terms in denominator
- Homework questions set to consolidate

## Wednesday, 12 January 2011

### Mechanics1 Worked solutions Jan 2009 paper

Link to worked solutions for Jan 2009 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGJmZDA2NjgtZTA1Zi00Y2VjLThmZTYtNGZjMDg3ZTVjZDUx&hl=en

Further supporting notes to follow....

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGJmZDA2NjgtZTA1Zi00Y2VjLThmZTYtNGZjMDg3ZTVjZDUx&hl=en

Further supporting notes to follow....

### Mechanics1 Worked solutions May 2008 paper

Link to worked solutions for May 2008 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDA0MDA2YWEtMzkzYS00ZGVhLWI5NTAtZDE3YWVkNTgxNTUy&hl=en

Further supporting notes to follow...

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDA0MDA2YWEtMzkzYS00ZGVhLWI5NTAtZDE3YWVkNTgxNTUy&hl=en

Further supporting notes to follow...

## Monday, 10 January 2011

### Mechanics1 Numerical solutions - all papers

Links below to numerical solutions for all papers

Jan 2006

https://docs.google.com/document/d/1l-PB6Tl_Feo__8HNDV-EGQn228Es2Q3cnWSKiuiJiKk/edit?hl=en

June 2006

https://docs.google.com/document/d/1K7w-cpfNncITKjEsCIKZMAVawV6Z_KLXhulttZMBdcM/edit?hl=en

Jan 2007

https://docs.google.com/document/d/16xd2Aa9ra_nA3m3fPUGVhkyAU4e3rVBse2Ss3RmMTyQ/edit?hl=en

June 2007

https://docs.google.com/document/d/1PdknJVpREC3c7D2L1gGRrItd6CqZdBUOiMARUsUlxZc/edit?hl=en

Jan 2008

https://docs0.google.com/document/d/16T4AYaF4aiICP6Hrr24MPhgNyANX7682cNF-JxnAhPk/edit?hl=en

June 2008

https://docs1.google.com/document/d/1x80p9Zt-WMXo3D7Jf9j4bOjhbr8yYa65q50gvYVNgLI/edit?hl=en

Jan 2009

https://docs2.google.com/document/d/1H9TFASQJyZi7Z-I9bknAFWEAi1QrazHIFYq7Welz3cc/edit?hl=en

June 2009

https://docs3.google.com/document/d/1RTMlnasQtPU3t1kru-ptsKquMgJUq89XOplfsfQ8meQ/edit?hl=en

Jan 2010

https://docs2.google.com/document/d/1N5Nl7yzpcVK4nq9LSh1oJbAxRAWoFIxGAe7WQTLYjP8/edit?hl=en

June 2010

https://docs3.google.com/document/d/138cr5TBF2a8QIP9fQDPJqyrZ6u9oEi8fuQyweZVMIEE/edit?hl=en

Jan 2006

https://docs.google.com/document/d/1l-PB6Tl_Feo__8HNDV-EGQn228Es2Q3cnWSKiuiJiKk/edit?hl=en

June 2006

https://docs.google.com/document/d/1K7w-cpfNncITKjEsCIKZMAVawV6Z_KLXhulttZMBdcM/edit?hl=en

Jan 2007

https://docs.google.com/document/d/16xd2Aa9ra_nA3m3fPUGVhkyAU4e3rVBse2Ss3RmMTyQ/edit?hl=en

June 2007

https://docs.google.com/document/d/1PdknJVpREC3c7D2L1gGRrItd6CqZdBUOiMARUsUlxZc/edit?hl=en

Jan 2008

https://docs0.google.com/document/d/16T4AYaF4aiICP6Hrr24MPhgNyANX7682cNF-JxnAhPk/edit?hl=en

June 2008

https://docs1.google.com/document/d/1x80p9Zt-WMXo3D7Jf9j4bOjhbr8yYa65q50gvYVNgLI/edit?hl=en

Jan 2009

https://docs2.google.com/document/d/1H9TFASQJyZi7Z-I9bknAFWEAi1QrazHIFYq7Welz3cc/edit?hl=en

June 2009

https://docs3.google.com/document/d/1RTMlnasQtPU3t1kru-ptsKquMgJUq89XOplfsfQ8meQ/edit?hl=en

Jan 2010

https://docs2.google.com/document/d/1N5Nl7yzpcVK4nq9LSh1oJbAxRAWoFIxGAe7WQTLYjP8/edit?hl=en

June 2010

https://docs3.google.com/document/d/138cr5TBF2a8QIP9fQDPJqyrZ6u9oEi8fuQyweZVMIEE/edit?hl=en

## Friday, 7 January 2011

### "EVERYTHING YOU NEED TO KNOW FOR MECHANICS 1"

http://dl.dropbox.com/u/15976318/Everything_You_Need_to_Know_for_Mechanics_1.doc

Something I made for a past A-Level revision seesion I ran at another school.

Make sure you use it as a guide, not a cheat sheet. That is, you should be attempting exams papers independently only refering to it when you are stuck. As you do more exam papers, you should not have to refer to the sheet as much.

Something I made for a past A-Level revision seesion I ran at another school.

Make sure you use it as a guide, not a cheat sheet. That is, you should be attempting exams papers independently only refering to it when you are stuck. As you do more exam papers, you should not have to refer to the sheet as much.

## Thursday, 6 January 2011

### Mechanics1 Worked solutions May 2010 paper

Link to worked solutions for May 2010 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTE2YTNjODMtOTI0Zi00MDJhLWFlYTEtYzk1NzcyYzllN2U5&hl=en

Further supporting notes for the worked solutions:

1. position at time t=6s subtract 4s x velocity to find position at t=2s then magnitude of the position vector to find the distance (using pythagoras)

2. diagams shows speeds and directions before and after collision

(a) use principle of conservation of momentum - considering whole system - initial momentum (mass x velocity) = final momentum - when equation for initial=final is set up, m and u can be eliminated so k can be found

(b) considering only particle P find the change in momentum (final momemtum - initial momentum) and this is equal to the impulse

3. considering vertical forces allows expression for normal reaction to be found which then allows expression for the frictional force to be found by considering max friction = coefficient of friction x reaction force

- then considering horizontal forces allows an equation to be set up which can be solved for m

4. the unknown distance is set up as x being the distance from Tom to the pivot (to make calculations relatively simple) which so as the pivot is 1m away from the end of the beam Sophie is x+1 from Tom (double distance from end of beam)

- using the given info about the reactions and considering up=down allows the reaction forces to be calculated

then moments about D has x as the only unknown

- not forgetting that the distance from Tom to the end of the beam is x+1 when stating the final answer

5. (a) both P and Q are horizontal (for the period of constant speed) and then a straight line diagonally down to meet at the same point (for the period of deceleration) - the lines must cross so that there is a period of time where Q is faster than P so that it catches up

(b) area under a speed-time graph represents distance - both have travelled 800m - use area of trapezium on Q to find X and then same for P to find T

6. (a) suvat on journey from starting point to max height to find distance - then add the 49m to find height above ground

(b)&(c) suvat on whole journey until hits floor to find speed at that point and time taken

7. because P is horizontal it is most efficient to work vertically first to eliminate P to find R and then horizontally to find P

8. direction of motion indicated is because A is heavier than B

(a)&(b) F=ma on each particle finds two equations in a and T which can then be solved simultaneously

(c) suvat on motion for 0.5 seconds before string breaks to find distance travelled in that time - add this distance to 1m for height above ground when string breaks

then suvat for motion after string breaks (travelling freely under gravity)

solve quadratic for t using quadratic formula - the negative solution for t does not apply to the situation

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTE2YTNjODMtOTI0Zi00MDJhLWFlYTEtYzk1NzcyYzllN2U5&hl=en

Further supporting notes for the worked solutions:

1. position at time t=6s subtract 4s x velocity to find position at t=2s then magnitude of the position vector to find the distance (using pythagoras)

2. diagams shows speeds and directions before and after collision

(a) use principle of conservation of momentum - considering whole system - initial momentum (mass x velocity) = final momentum - when equation for initial=final is set up, m and u can be eliminated so k can be found

(b) considering only particle P find the change in momentum (final momemtum - initial momentum) and this is equal to the impulse

3. considering vertical forces allows expression for normal reaction to be found which then allows expression for the frictional force to be found by considering max friction = coefficient of friction x reaction force

- then considering horizontal forces allows an equation to be set up which can be solved for m

4. the unknown distance is set up as x being the distance from Tom to the pivot (to make calculations relatively simple) which so as the pivot is 1m away from the end of the beam Sophie is x+1 from Tom (double distance from end of beam)

- using the given info about the reactions and considering up=down allows the reaction forces to be calculated

then moments about D has x as the only unknown

- not forgetting that the distance from Tom to the end of the beam is x+1 when stating the final answer

5. (a) both P and Q are horizontal (for the period of constant speed) and then a straight line diagonally down to meet at the same point (for the period of deceleration) - the lines must cross so that there is a period of time where Q is faster than P so that it catches up

(b) area under a speed-time graph represents distance - both have travelled 800m - use area of trapezium on Q to find X and then same for P to find T

6. (a) suvat on journey from starting point to max height to find distance - then add the 49m to find height above ground

(b)&(c) suvat on whole journey until hits floor to find speed at that point and time taken

7. because P is horizontal it is most efficient to work vertically first to eliminate P to find R and then horizontally to find P

8. direction of motion indicated is because A is heavier than B

(a)&(b) F=ma on each particle finds two equations in a and T which can then be solved simultaneously

(c) suvat on motion for 0.5 seconds before string breaks to find distance travelled in that time - add this distance to 1m for height above ground when string breaks

then suvat for motion after string breaks (travelling freely under gravity)

solve quadratic for t using quadratic formula - the negative solution for t does not apply to the situation

## Wednesday, 5 January 2011

### Mechanics1 Worked solutions Jan 2010 paper

Link to worked solutions for Jan 2010 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGI1MTc1ZmUtNWQ2OC00NDlkLTkyMTItZDQ4ZTc4MzQ2MjNl&hl=en

Supporting notes for worked solutions:

1. The diagram shows the system before and after the collision.

(a) considering just particle A calculate the change in momentum by subtracting the intial momentum from the final momentum - the impulse is the change in momentum

(b) considering the whole system and using the principle of conservation of momentum, the total initial momentum = the total final momentum - and solve for m

2. (a) on a speed-time graph, constant acceleration is a straight line sloping up, constant speed is a horizontal line and constant deceleration is a straight line sloping down

(b) simplest method is to consider the distance as the area under the graph and use area of trapezium to set up an equation with T as the only unknown

- alternatively the journey can be split into three parts and suvat used on the final part of the journey with T as the unknown

- finally solve for T

3. (a) considering horiztonal the tension in B is the only unknown

(b) so then use vertical to find m

4. (a) moments about A eliminates the tension in A to find an expression for tension in C

(b) use up = down and the answer to (a) to find expression for tension in A

(c) answer to (a) equals 8 x answer to (b) and solve for W

5. (a) the given information about the particle's motion enables a suvat equation to be set up to find the accelearation

(b) use F=ma to find the resultant force down the slope

** the remainder of this question is missing and will be updated***

6. (a) use F=ma on only particle A to find the tension

(b) use F=ma on only particle B using tension from (a) to set up an equation from which the value of k can be found

(c) the pulley being smooth means there is no force (eg friction) resisting motion

(d) suvat on particle A falling can find the distance it falls beforce reaching the plane, particle B rises the same amount so ends up double that distance above the plane

7. (a) the displacement of S is the position after 4s minus the initial position (when doing this make sure the initial position is in brackets), then the velocity is the displacement/time and finally the speed is the magnitude of the velocity (use pythagoras on the i and j part of the velocity)

(b) the diagram shows the velocity vector - the angle needed for the bearing is the same as the angle indicated inside the triangle

(c) the position at any given time is the initial position + (time x velocity)

(d) displacement of S from L is (position of S - position of L) so then distance is magnitude of this vector (using pythagoras) subst t=T and distance = 10 and then solve for T

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGI1MTc1ZmUtNWQ2OC00NDlkLTkyMTItZDQ4ZTc4MzQ2MjNl&hl=en

Supporting notes for worked solutions:

1. The diagram shows the system before and after the collision.

(a) considering just particle A calculate the change in momentum by subtracting the intial momentum from the final momentum - the impulse is the change in momentum

(b) considering the whole system and using the principle of conservation of momentum, the total initial momentum = the total final momentum - and solve for m

2. (a) on a speed-time graph, constant acceleration is a straight line sloping up, constant speed is a horizontal line and constant deceleration is a straight line sloping down

(b) simplest method is to consider the distance as the area under the graph and use area of trapezium to set up an equation with T as the only unknown

- alternatively the journey can be split into three parts and suvat used on the final part of the journey with T as the unknown

- finally solve for T

3. (a) considering horiztonal the tension in B is the only unknown

(b) so then use vertical to find m

4. (a) moments about A eliminates the tension in A to find an expression for tension in C

(b) use up = down and the answer to (a) to find expression for tension in A

(c) answer to (a) equals 8 x answer to (b) and solve for W

5. (a) the given information about the particle's motion enables a suvat equation to be set up to find the accelearation

(b) use F=ma to find the resultant force down the slope

** the remainder of this question is missing and will be updated***

6. (a) use F=ma on only particle A to find the tension

(b) use F=ma on only particle B using tension from (a) to set up an equation from which the value of k can be found

(c) the pulley being smooth means there is no force (eg friction) resisting motion

(d) suvat on particle A falling can find the distance it falls beforce reaching the plane, particle B rises the same amount so ends up double that distance above the plane

7. (a) the displacement of S is the position after 4s minus the initial position (when doing this make sure the initial position is in brackets), then the velocity is the displacement/time and finally the speed is the magnitude of the velocity (use pythagoras on the i and j part of the velocity)

(b) the diagram shows the velocity vector - the angle needed for the bearing is the same as the angle indicated inside the triangle

(c) the position at any given time is the initial position + (time x velocity)

(d) displacement of S from L is (position of S - position of L) so then distance is magnitude of this vector (using pythagoras) subst t=T and distance = 10 and then solve for T

### Mechanics1 Worked solutions May 2009 paper

Link to worked solutions for May 2009 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNWI3NGU2ZjQtZTk4OC00ZDVlLWIyZWUtZDkyYzFhM2Y4Y2Yy&hl=en

Supporting notes for worked solutions:

1. For journeys P to Q and P to R, distance and time are known so for each journey an equation in terms and u & a can be set up - then solve the equations simultaneously

2. (a) the vector

(b) To find the resultant of F1 & F2 just add them - then for R to be parallel to

3. Because the impulse is given work with each particle separately - define right as positive - particle A's initial momentum is positive and the impulse acts left so negative - particle B's initial momentum is negative and the impulse acts right so positive - then as momentum = mass x velocity, velocity = momentum/mass

4. Aside about the angle - given info can be used with a 3,4,5 triangle to find the sin and cos (could work out the angle with inverse tan but this will be a long decimal)

- perpendicular to the plane to find normal reaction which is then used with coefficient of friction to find the friction force - then work out the resultant force down the slope and use F=ma to find the acceleration

5. both vertical and horizontal can set up an expression for R in terms of P which are then equated to find P

6. (a) consider the whole system (eliminating the tension forces) using F=ma to find the acceleration

(b) consider just the car (could equally have considered just the trailer) using F=ma to find the tension in the towbar

(c) thrust forces in the towbar act towards the car and trailer (opposite to the tension in (a) and (b)) - considering the trailer only using F=ma enable the acceleration (actually deceleration) to be found which the braking force to be calculated using F=ma on the car only

7. (a) moments about Q enables the unknown tension at Q to eliminated to the the tension in P can be found in terms of x

(b) considering up=down enables the tension in Q to be found in terms of x

(c) the question states x is between 0 and 1.4 so the minimum tension in Q & maximum tension in P will be x=0 (ie at P) and the maximum tension in Q & minimum tension in P will be x=1.4 (ie at Q)

(d) use the information given to state that Q=3P and substitue the answers to (a) and (b) and solve for x

8. (a) velocity of H is given so speed is jsut pythagoras on the

(b) the position vector for H is the inital position (100

(c) the position vector for K is the inital position plus time x velocity then the displacement of K from H is the position vector for K subtract the position vector for H

(d) for H & K to meet both

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNWI3NGU2ZjQtZTk4OC00ZDVlLWIyZWUtZDkyYzFhM2Y4Y2Yy&hl=en

Supporting notes for worked solutions:

1. For journeys P to Q and P to R, distance and time are known so for each journey an equation in terms and u & a can be set up - then solve the equations simultaneously

2. (a) the vector

**i**+ 2**j**is parallel to F2 so can be used to find the angle - it is the angle with the vertical that is needed(b) To find the resultant of F1 & F2 just add them - then for R to be parallel to

**i**the**j**part is zero - then solve for p3. Because the impulse is given work with each particle separately - define right as positive - particle A's initial momentum is positive and the impulse acts left so negative - particle B's initial momentum is negative and the impulse acts right so positive - then as momentum = mass x velocity, velocity = momentum/mass

4. Aside about the angle - given info can be used with a 3,4,5 triangle to find the sin and cos (could work out the angle with inverse tan but this will be a long decimal)

- perpendicular to the plane to find normal reaction which is then used with coefficient of friction to find the friction force - then work out the resultant force down the slope and use F=ma to find the acceleration

5. both vertical and horizontal can set up an expression for R in terms of P which are then equated to find P

6. (a) consider the whole system (eliminating the tension forces) using F=ma to find the acceleration

(b) consider just the car (could equally have considered just the trailer) using F=ma to find the tension in the towbar

(c) thrust forces in the towbar act towards the car and trailer (opposite to the tension in (a) and (b)) - considering the trailer only using F=ma enable the acceleration (actually deceleration) to be found which the braking force to be calculated using F=ma on the car only

7. (a) moments about Q enables the unknown tension at Q to eliminated to the the tension in P can be found in terms of x

(b) considering up=down enables the tension in Q to be found in terms of x

(c) the question states x is between 0 and 1.4 so the minimum tension in Q & maximum tension in P will be x=0 (ie at P) and the maximum tension in Q & minimum tension in P will be x=1.4 (ie at Q)

(d) use the information given to state that Q=3P and substitue the answers to (a) and (b) and solve for x

8. (a) velocity of H is given so speed is jsut pythagoras on the

**i**and**j**parts of the velocity (ie finding the magnitude of the velocity)(b) the position vector for H is the inital position (100

**j**) plus time x velocity(c) the position vector for K is the inital position plus time x velocity then the displacement of K from H is the position vector for K subtract the position vector for H

(d) for H & K to meet both

**i**and**j**parts of the displacement found in (c) must be zero at the same value of t - then substitute the found value of t into one of the position vectors
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