Welling Maths Year 13: January 2011

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:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZmM5ZjNlMDctNmMwMS00ZDE4LTk4MzMtN2RiOGNiZWFjN2Fi&hl=en

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.

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

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOWIwMzVlZGUtYjA4MS00NGUzLWFmZjUtZTNmMmUwZDZiODc2&hl=en

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

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzk2NGU2ZGMtZWVmNi00MDU1LWIxMjktM2Q2MTRlYmM1NGI3&hl=en

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....

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...

Monday 10 January 2011

Mechanics1 Numerical solutions - all papers

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.

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

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

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 i + 2j 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 p

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

Welling Maths Year 13