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Sample 1 Teach A Level Maths Vol. 4: Mechanics 1 © Christine Crisp Use the mouse or the keyboard forward arrow or spacebar to move on when the presentation pauses.

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Where a topic relates to some specifications only, this is indicated in a contents file and also at the start of the presentation. Whilst these presentations are relevant to any introductory course in Mechanics, particular attention has been paid to the material in the M1 specifications offered at GCE A/AS level by the three English Awarding Bodies: AQA, Edexcel and OCR. OCR has two specifications so, to distinguish between them, the 2 nd is referred to as MEI.

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The slides that follow are samples from 7 of the 29 presentations.

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5: Vectors for Mechanics This presentation about vectors has been structured so that teachers can cover, or revise, the vector theory that is required in mechanics. The other presentations that use vectors are hyperlinked to this one to make it easy to dip in and out. The next 4 slides come from the section on the unit vectors and. i j

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j i e.g. A velocity v is given by v 3 4 ) m s -1 i j x y j i 3 4 v Instead of drawing diagrams to show vectors we can use unit vectors. A unit vector has magnitude 1. The unit vectors and are parallel to the x- and y- axes respectively. ij In text-books single letters for vectors are printed in bold but we must underline them.

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v 3 4 i j The magnitude of velocity is speed, so, using Pythagoras theorem, v No or in magnitude i j v So, if we have the unit vector form, we use the numbers in front of and ij x y 3 4 v j i Tip: Squares of real numbers are always positive so we never need any minus signs. v 5 We can write v or v for speed.

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v i j The direction of the vector is found by using trig. tan 53·1 ( 3 s.f. ) BUT beware ! 3443 x y 3 4 v j i

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If we need the direction of a vector when unit vectors are used, we must sketch the vector to show the angle we have found. v 3 4 i j Suppose Without a diagram we get tan ·1 ( 3 s.f. ) So again But, the vectors are not the same ! v i j 53·1 ( 3 s.f. ) 3443 For we have 3 4 i j 3443 v 3 4 v 3 4 i j

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6: Equations of Motion for Constant Acceleration In this extract the students are shown how to use a velocity-time graph to develop the equations of motion for constant acceleration. To give the students confidence, they are sometimes asked to share ideas with a partner.

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We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t0 Suppose when the time is 0... At any time, t, we let the velocity be v. the velocity is u. Ans: The gradient gives the acceleration. Remind your partner how to find acceleration from a velocity-time graph. Constant acceleration means the graph is a straight line.

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We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t a v u t So, v u 0 Suppose when the time is 0... the velocity is u. t From this equation we can find the value of any of the 4 quantities if we know the other 3. At any time, t, we let the velocity be v. Constant acceleration means the graph is a straight line.

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a v u t a t v u v u a t Multiplying by t : u a t v We usually learn the formula with v as the subject. The velocity, u, at the start of the time is often called the initial velocity.

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9: Displacement and Velocity using Unit Vectors Throughout the course students are encouraged to draw diagrams to help them solve problems.

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e.g.2A ship is at a point A given by the position vector r A 4 3 ) km i j Solution: Find (a) the displacement of B from A, and After half-an-hour the ship is at a point B. We can solve this problem without a diagram, but a diagram can help us to see the method. i The ship has a constant velocity of 6 km h -1. (b) the position vector of B.

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O y x 4 3 A x 6 i After half-an-hour the ship is at a point B. B x displacement velocity time 3 km i s 6 i 4 3 ) i j 3 i 3 ) km i j r A 4 3 ) km i j A:A: Velocity v 6 km h -1 i r B r A s r B r Br B r Ar A Solution: (a)Find the displacement of B from A. 0·5 (b)Find the position vector of B. Constant velocity s

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12: Momentum and Collisions Each presentation has a simple summary page that teachers can ask students to copy into their notes. A version of the summary is given, without colour, at the end of each presentation so that photocopies can be made if preferred.

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SUMMARY Momentum mass velocity Momentum is a vector. The units of momentum are kg m s -1. In a collision, the conservation of momentum gives: total momentum before the total momentum after Solving collision problems: Write the momentum statement (can abbreviate ). Write the equation remembering to add the momentums but watching for negative velocities. Draw before and after diagrams giving mass and velocity ( showing the direction with an arrow ). Draw to show the positive direction.

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17: The Resultant of Forces by Resolving In the presentation before this one students have practised resolving forces. Here they see how to collect components and then to find the resultant. At this stage some students may still be having difficulties finding components, so they are reminded of the process. The calculator symbol appears where students may benefit from doing the calculation themselves.

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The 6 and 10 newton forces lie along the axes so we only need to resolve the 4 newton force. e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. If you need a triangle to help you, draw it to one side x y You may be able to see directly what the components are but dont take chances.

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We can always find 1 angle in the triangle from the original diagram. e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. 4 6 x y

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60 4 cos sin 60 Tell your partner what the expression is for each component. Ans: Across the 60 gives 4 cos the other component is 4 sin 60 e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant x y 10 60

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X 10 Now we collect all the components in the x and y directions. I must remember the arrow 4 4cos 60 4 sin 60 and then check the arrow on each force Y 6 4sin 60 X 8 Y 2·5358… Store this in a memory on your calculator. e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant x y Y 2·54 ( 3 s.f. )

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X Now we have just 2 components, we can find the resultant. 8 Y 2·54 e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. 4 6 x y 10 60

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X Now we have just 2 components, we can find the resultant. 8 Y 2·54 8 Reversing the arrow takes care of the minus sign. e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. 4 6 x y 10 60

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X Now we have just 2 components, we can find the resultant. 8 Y 2·54 8 e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. 4 6 x y 10 60

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X Now we have just 2 components, we can find the resultant. 8 Y 2·54 8 F F ·54 2 F 8·39 newtons ( 3 s.f. ) tan 2· ·6 ( 3 s.f. ) We can write 3 s.f. but we use the number stored on the calculator. e.g.1The forces shown in the diagram are in newtons. Find the magnitude and direction of the resultant. ( as shown in the sketch ) 4 6 x y 10 60

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21: Newtons 2 nd Law This is the first of a variety of examples used to illustrate Newtons 2 nd Law.

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e.g.1A pebble of mass 0·2 kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of 0·06 newtons. particle W 0·2g Tip: When we are given mass rather than weight, its a good idea to use W = mg to replace W on the diagram. Solution:

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e.g.1A pebble of mass 0·2 kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of 0·06 newtons. particle a 0·06 0·2g Solution: We dont substitute for g at this stage since we need to see both mass and weight clearly in the diagram.

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e.g.1A pebble of mass 0·2 kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of 0·06 newtons. Resultant force mass acceleration particle a 0·06 N2L: 0·2g Beware ! On the left-hand side of the equation we have force, so we need weight. On the right-hand side we have mass. Solution: We always resolve in the direction of the acceleration.

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e.g.1A pebble of mass 0·2 kg is dropped from the top of a cliff. Find the acceleration of the pebble as it falls, assuming that the pebble falls in a straight line and there is a constant resistance of 0·06 newtons. Resultant force mass acceleration particle a 0·06 N2L: 0·2g 0·06 0·2 a Solution: a = 9·5 ms -2 0·2 9·8 0·06 0·2a a 1·9 0·2

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24: Connected Particles and Newtons 3 rd Law The following example shows a pulley problem. It covers the methods used to find the forces and acceleration of the system, including the force on the pulley.

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e.g.2.A particle, A, of mass 0·6 kg, is held at rest on a smooth table. A is connected by a light, inextensible string, which passes over a smooth fixed pulley at the edge of the table, to another particle, B, of mass 0·4 kg hanging freely. The string is horizontal and at right angles to the edge of the table. A B A is released. Find (a) the magnitude of the reaction of the table on A, (b) the acceleration of the system (c) the tension in the string, and (d) the force on the pulley due to the particles.

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A B A : mass 0·6 kg B : mass 0·4 kg T R 0·4g 0·6g T Solution: A:A: B:B: 0·6g (1) (2) R 0 a a R 0·6g T 0·6a T 0·4g 0·4a (3) (a) Find R. (b) Find a. A:A: N2L: Resultant force mass acceleration A accelerates horizontally, so there is no vertical component of acceleration. R 5·88 newtons ( 3 s.f. )

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A B A : mass 0·6 kg B : m a ss 0·4 kg T R 0·4g 0·6g T (2) a a R 5·88 N T 0·6a T 0·4g 0·4a (3) (2) (3): a 0·4g a 0·4 9·8 a 3·92 m s -2 (b) Find a. Substitute in (2): T 0·6 3·92 T 2·35 newtons ( 3 s.f. ) (c) Find T.

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A B A : mass 0·6 kg B : mass 0·4 kg T R 0·4g 0·6g T a a a 3·92 m s -2 2·35 N T T T T T What can you say about the direction of the resultant ? Since the 2 forces are equal, the resultant lies between them, at an equal angle to each. 45 2T cos 45 F F 3·33 newtons ( 3 s.f. ) F T cos 45 T cos 45 (d) Find the force on the pulley due to the particles. Reminder: Use the value of T from your calculator so that your answer will be correct to 3 s.f.

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The next 2 slides contain the names of all 29 presentations. Number 20, Bodies in Equilibrium is provided in full as Sample 2. The title is hyperlinked to the presentation.

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2.Distance and Speed 3.Displacement and Velocity 5.Vectors for Mechanics 7.Vertical Motion under Gravity 8.Bearings 1.Introducing Mechanics 4.Acceleration and Graphs 6.Constant Acceleration Equations 9.Displacement and Velocity using Unit Vectors 10.Relative Velocity in 1 Dimension (MEI) 11.Resultant Velocity (AQA) Teach A Level Maths – Mechanics D Collisions (AQA) 14.Impulse in Collisions (Edexcel) 16.Components of a Force 18.Equilibrium 19.Force Diagrams and Newtons 1 st Law 12.Momentum and Collisions 15.The Resultant of 2 Forces 17.The Resultant of Forces by Resolving 20.Bodies in Equilibrium 21.Newtons 2 nd Law continued

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23.Coefficient of Friction, 2 (AQA, Edexcel and OCR) 24.Connected Particles and Newtons 3 rd Law 26.Projectiles, 1 (AQA and MEI) 28.Moment of a Force (Edexcel) 29.Motion Using Calculus (OCR and MEI) 22.Coefficient of Friction, 1 (AQA, Edexcel and OCR) 25.Connected Particles, 2 27.Projectiles, 2 (AQA and MEI) Teach A Level Maths – Mechanics 1

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