1. “Teach A Level Maths” Yr1/AS Mechanics
Sample 1
© Christine Crisp

2. Whilst these presentations are relevant to any introductory course in Mechanics, particular attention has been paid to the material in the Year1/AS specifications offered by the three English Awarding Bodies: AQA, Edexcel and OCR.
OCR has two specifications so, to distinguish between them, the 2nd is referred to as OCR/MEI.

3. They need to be viewed as a slide show.
The slides that follow are extracts from 4 of the 13 presentations for Mechanics.
They need to be viewed as a slide show.
Equations of Motion for Constant Acceleration
The Resultant of Two Forces
Newton’s 2nd Law
Connected Particles and Newton’s 3rd Law

4. 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.

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

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

7. v - u a = t a t = v - u  u + a t = v  v = u + a t
We usually learn the formula with v as the “subject”.
Multiplying by t:
a t = v - u
 u + a t = v 
v = u + a t
The velocity, u, at the start of the time is often called the initial velocity.

8. The Resultant of Two Forces
In this presentation students are introduced to force as a vector.
In Year 1 they only study perpendicular forces so they have no need to find components of forces. Here they are shown how to collect up forces in two directions and find the resultant.

9. Suppose we have several forces trying to pull a body in different directions . . .3 15 7 2
We model the body as a particle . . .
so the “lines of action” of the forces all act at a point.
We will only consider sets of perpendicular forces whose lines of action lie in one plane.
The standard unit of force is a newton. There is no need to show standard units on diagrams.

10. The direction of a force clearly makes a difference to its effect . . .
so force is a vector. Q
To see the effects of forces on a body, we need to add all the forces – and that means adding vectors. R
The single force representing the others is called the resultant.

11. e.g.1 Find the magnitude and direction of the resultant of the forces shown in the diagram.3 15 7 2
Solution: First collect the forces in two perpendicular directions.
Across the page:
P = 15 – 3
= 12
The force of magnitude 3 is in the opposite direction so we must subtract.
Up the page:
Q = 7 – 2
= 5

12. We have reduced the 4 forces to 2, as shown12 5

13. 5 F 5 12 12 We have reduced the 4 forces to 2, as shown
We can draw the forces head-to-tail . . .
and the resultant completes the triangle.
The lines are not drawn to scale but if we draw them looking roughly in proportion to the magnitudes of the forces, the angle of the resultant will look about the right size.

14. 5
22·6 13 F 5 q 12 12
Using Pythagoras’ theorem:
F 2 =
 F = 13
Force is a vector so, unless we are asked for the magnitude only, we must also give the direction. 5 12
tanq =
 q = 22·6 ( 3 s.f. )
The resultant has magnitude 13 newtons at an angle of 22·6 to the 12 newton force.

15. Newton’s 2nd Law
This is the first of a variety of examples used to illustrate Newton’s 2nd Law.
Students are always encouraged to draw careful diagrams and state methods.

16. e. g. 1. A pebble of mass 0·2 kg is dropped from the top of a cliff
e.g.1 A 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.
Solution:
particle
Tip: When we are given mass rather than weight, it’s a good idea to use W = mg to replace W on the diagram. W
0·2g

17. e. g. 1. A pebble of mass 0·2 kg is dropped from the top of a cliff
e.g.1 A 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.
Solution:
particle
0·06
We don’t substitute for g at this stage since we need to see both mass and weight clearly in the diagram. a
0·2g

18. On the right-hand side we have mass.
e.g.1 A 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.
Solution:
particle
0·06
N2L: a
Resultant force = mass  acceleration
0·2g
We always resolve in the direction shown for the acceleration.
Beware !
On the left-hand side of the equation we have force, so we need weight.
On the right-hand side we have mass.

19. 0·06 a 0·2g - 0·06 = 0·2  a 0·2g  0·2  9·8 - 0·06 = 0·2a 1·9 0·2 
e.g.1 A 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.
Solution:
particle
0·06
N2L: a
Resultant force = mass  acceleration
0·2g
- 0·06 =
0·2  a
0·2g 
0·2  9·8 - 0·06 = 0·2a
1·9
0·2 
= a 
a = 9·5 ms-2
Our model will now assume that the acceleration due to gravity has magnitude 98 ms-2

20. Connected Particles and Newton’s 3rd Law
The following example shows a pulley problem. It covers the methods used to find the forces and acceleration of the system.
Further examples illustrate Newton’s 3rd Law.
To give the students confidence, they are sometimes asked to share ideas with a partner.

21. e.g.1. The diagram shows 2 particles, A and B, each of mass 3 kg, connected by a string which runs over a fixed pulley. A B 3g
The particles are held at rest.
Decide with your partner what will happen if the particles are released.
Ans: The particles remain at rest.
( The extra weight due to the longer string on the side of B is very small and our models ignore it. There is also likely to be friction from the pulley and again this is ignored. )

22. e.g.1. The diagram shows 2 particles, A and B, each of mass 3 kg, connected by a string which runs over a fixed pulley. A B 3g
We make the following modelling assumptions when we solve problems: T
The string is light (no weight), and inextensible (it doesn’t stretch). T
The pulley is smooth so the
tension in the string is the same throughout.
Since the forces on A and B are the same, it is not possible for B to accelerate downwards and A upwards.

23. A has the greater mass so it will accelerate downwards.
Suppose now that A has mass 4 kg whilst B still has mass 3 kg. A B 3g T
A has the greater mass so it will accelerate downwards.
The particles are joined together so the magnitude of the acceleration is the same for B. a a 3g 4g
To find the values of a and T, we use Newton’s 2nd Law on each particle separately.

24. A B 3g T 4g a A: 4g - T = 4a - - - - (1) B: T - 3g = 3a - - - - (2)
N2L:
Resultant force = mass  acc. A: 4g - T = 4a
(1) B: T -
We always resolve in the direction shown for the acceleration. 3g =
There are 2 unknowns so we need another equation before we can solve. 3a
(2)
Adding the equations will eliminate T and we can then solve for a.
Tip: It’s usually easier to substitute for g at the end of the question. I shall assume g = 9.8 ms-2

25. A B 3g T 4g a A: 4g - T = 4a - - - - (1) B: T - 3g = 3a - - - - (2)
N2L:
Resultant force = mass  acc. A: 4g - T = 4a
(1) B: T - 3g = 3a
(2)
(1) + (2):

26. A B 3g T 4g a A: 4g - T = 4a - - - - (1) B: T - 3g = 3a - - - - (2)
N2L:
Resultant force = mass  acc. A: 4g - T = 4a
(1) B: T - 3g = 3a
(2)
(1) + (2): g

27. A B 3g T 4g a A: 4g - T = 4a - - - - (1) B: T - 3g = 3a - - - - (2)
N2L:
Resultant force = mass  acc. A: 4g - T = 4a
(1) B: T - 3g = 3a
(2)
(1) + (2): g = 7a 
a = g 7 
g = 9·8
m s -2
a = 1·4
Substitute in (2): T = + 3g 3a
 T
= 3  1·4 +
3  9·8
 T = 33·6 newtons
Equations (1) and (2) are sometimes called the equations of motion of the particles.

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