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“Teach A Level Maths”
Vol. 4: Mechanics 1
Sample 1
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spacebar to move on when the presentation pauses.
© Christine Crisp
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 2nd is referred to as MEI.
Where a topic relates to some specifications only, this
is indicated in a contents file and also at the start of
the presentation.
The slides that follow are samples from 7 of the 29
presentations.
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 i and j.
Instead of drawing diagrams to show vectors we can
use unit vectors. A unit vector has magnitude 1.
The unit vectors i and j
are parallel to the x- and
y-axes respectively.
e.g. A velocity v is given by
v = (3 i + 4 j ) m s -1
y
v
4
j
i
3
x
In text-books single letters for vectors are printed
in bold but we must underline them.
The magnitude of velocity
is speed, so, using
Pythagoras’ theorem,
v = 32 + 42
 v=5
So,We
if can
we have
writethe
v orunit
v
vector form,
we use the
for speed.
numbers in front of i and j
No i or j in
magnitude
j
y
i
v
4
3
x
v = 3i + 4j
v = 32 + 42
Tip: Squares of real numbers
are always positive so we
never need any minus signs.
v = 3i + 4 j
The direction of the vector
is found by using trig.
j
y
i
v
4
tan q =

q = 53·1 ( 3 s.f. )
3
BUT beware !
x
For v = 3 i + 4 j
we have q = 53·1 ( 3 s.f. )
Suppose v = -3 i - 4 j
Without a diagram we get
tan q = -4
-3
So again q = 53·1 ( 3 s.f. )
But, the vectors are not the same !
If we need the direction of a 3vector when unit
vectors are used, we must sketch qthe vector to
4 j the angle we have found.
v = 3 i + show
4
q
3
4
v = -3 i - 4 j
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.
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 0
velocity (ms-1)
. . . the velocity is u.
v
Constant acceleration means
the graph is a straight line.
u
0
t time (s)
At any time, t, we let the
velocity be v.
Remind your partner how to find acceleration from a
velocity-time graph.
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 0
velocity (ms-1)
. . . the velocity is u.
v
Constant acceleration means
v-u
the graph is a straight line.
u
0
t
t time (s)
So,
At any time, t, we let the
velocity be v.
a= v-u
t
From this equation we can find the value of any of the
4 quantities if we know the other 3.
a= v-u
t
We usually learn the formula with v as the “subject”.
Multiplying by t:
at = v - u
 u + at = v

v = u + at
The velocity, u, at the start of the time is often
called the initial velocity.
9: Displacement and Velocity
using Unit Vectors
Throughout the course students are encouraged to
draw diagrams to help them solve problems.
e.g.2 A ship is at a point A given by the position
vector
r A = (-4 i + 3 j ) km
The ship has a constant velocity of 6 i km h -1.
After half-an-hour the ship is at a point B.
Find
(a) the displacement of B from A, and
(b) the position vector of B.
Solution:
We can solve this problem without a diagram,
but a diagram can help us to see the method.
Velocity v = 6 i km h -1
After half-an-hour the ship is at a point B.
Solution:
(a) Find the displacement of B from A.
A: r A = (-4 i + 3 j ) km
Constant velocity 
displacement = velocity  time

s = 6 i  0·5 = 3 i km
(b) Find the position vector
of B.
rB = rA + s

r B = (-4 i + 3 j ) + 3 i
= (- i + 3 j ) km
Ax 6 i s Bx
rA
3
y
rB
x
4
O
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.
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:
• Draw before and after diagrams giving mass and
velocity ( showing the direction with an arrow ).
• Write the momentum statement (can abbreviate ).
+ to show the positive direction.
• Draw




• Write the equation remembering to add the
momentums but watching for negative velocities.
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.
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
6
60
4
x
The 6 and 10 newton
forces lie along the
axes so we only need
to resolve the 4
newton force.
You may be able to see directly what the components
are but don’t take chances.
If you need a triangle to help you, draw it
to one side.
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
6
60
x
4
4
We can always find 1 angle in the triangle from
the original diagram.
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
4cos 60
6
60
60
x
4
4sin 60
4
Tell your partner what the expression
is for
10
each component.
Ans: Across the 60 gives 4cos 60. . .
the other component is 4sin 60
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
4cos 60
6
60
60
x
4
4sin 60
4
Now we collect all the components in the x and y
directions.
X = -10 + 4cos 60
and then
check
I must
the
arrow the
on
remember

each
force
arrow
Y = +6 - 4sin 60
X = -8
Y = 2·5358…
Store this in a memory on
Y = your
2·54 calculator.
( 3 s.f. )
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
6
X = -8
60
x
4
Now we have just 2 components,
we can find the resultant.
Y = 2·54
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
6
X = -8
60
x
Y = 2·54
4
Now we have just 2 components,
we can find the resultant.
Reversing the arrow takes
care of the minus sign.
8
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
10
6
X = -8
60
x
Y = 2·54
4
Now we have just 2 components,
we can find the resultant.
2·54
8
e.g.1 The forces shown in the diagram are in newtons.
Find the magnitude and direction of the resultant.
y
6
X = -8
60
10
x
Y = 2·54
4
Now we have just 2 components,
we can find the resultant.
F
2·54
= +
q
8
 F = 8·39 newtons ( 3 s.f. )
317·6
s.f. but we use the
2·54canwrite
tanq = We
q
=
( 3 s.f. )
8
number stored on the calculator.
F2
82
2·542
( as shown in the sketch )
21: Newton’s 2nd Law
This is the first of a variety of examples used to
illustrate Newton’s 2nd Law.
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:
Tip: When we are given mass
rather than weight, it’s a good
idea to use W = mg to replace
W on the diagram.
particle
0·2g
W
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:
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
particle
0·2g
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:
N2L:
Resultant force = mass  acceleration
0·06
a
particle
0·2g
Beware
We always resolve
in the!
On the left-hand
side of the equation
direction
of the acceleration.
we have force, so we need weight.
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:
N2L:
Resultant force = mass  acceleration
0·2g - 0·06 = 0·2  a
 0·2  9·8 - 0·06 = 0·2a
1·9 = a  a = 9·5 ms-2

0·2
0·06
a
particle
0·2g
24: 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, including the force on
the pulley.
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
A is released. Find
(a) the magnitude of the
reaction of the table on A,
B (b) the acceleration of the
system
(c) the tension in the string, and
(d) the force on the pulley due to the particles.
A: mass 0·6 kg
B: mass 0·4 kg
R
A
0·6g
a
T
T
B
0·4g
a
Solution:
N2L: Resultant force = mass  acceleration
(a) Find R.
R - 0·6g = 0  R = 0·6g - - - - (1)
A:
 R = 5·88 newtons ( 3 s.f. )
(b) Find a.
A accelerates horizontally,
so there is
A:
T = 0·6a - - - - (2)
no vertical component of acceleration.
0·4g - T = 0·4a - - - - (3)
B:
A: mass 0·6 kg
B: mass 0·4 kg
R
A
0·6g
a
T
T
a
R = 5·88N
B
0·4g
(b) Find a.
T = 0·6a - - - - (2)
0·4g - T = 0·4a - - - - (3)
(2) + (3):
0·4g = a  a = 0·4  9·8
 a = 3·92 m s -2
(c) Find T.
Substitute in (2):
T = 0·6  3·92
 T = 2·35 newtons ( 3 s.f. )
A: mass 0·6 kg
B: mass 0·4 kg
a = 3·92 m s -2
T = 2·35 N
R
A
0·6g
T
a
T
T
T
B
0·4g
a
(d) Find the force on the pulley due to the particles.
T
Since
the
2
forces
are
equal,
the
45
What can you say about the
resultant
liesofbetween
them, ?at
direction
the resultant
F
T
an equal angle to each.
F = Tcos45 + Tcos45
= 2Tcos45  F = 3·33 newtons ( 3 s.f. )
Reminder: Use the value of T from your calculator
so that your answer will be correct to 3 s.f.
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.
Teach A Level Maths – Mechanics 1
1.
Introducing Mechanics
11. Resultant Velocity (AQA)
2.
Distance and Speed
12. Momentum and Collisions
3.
Displacement and Velocity
13. 2-D Collisions (AQA)
4.
Acceleration and Graphs
14. Impulse in Collisions (Edexcel)
5.
Vectors for Mechanics
15. The Resultant of 2 Forces
6.
Constant Acceleration
Equations
Vertical Motion under
Gravity
16. Components of a Force
7.
8.
Bearings
9.
Displacement and Velocity
using Unit Vectors
10. Relative Velocity in 1
Dimension (MEI)
17. The Resultant of Forces
by Resolving
18. Equilibrium
19. Force Diagrams and
Newton’s 1st Law
20. Bodies in Equilibrium
21. Newton’s 2nd Law
continued
Teach A Level Maths – Mechanics 1
22. Coefficient of Friction, 1 (AQA, Edexcel and OCR)
23. Coefficient of Friction, 2 (AQA, Edexcel and OCR)
24. Connected Particles and Newton’s 3rd Law
25. Connected Particles, 2
26. Projectiles, 1 (AQA and MEI)
27. Projectiles, 2 (AQA and MEI)
28. Moment of a Force (Edexcel)
29. Motion Using Calculus (OCR and MEI)
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