Notes for Edexcel Mechanics · Edexcel Mechanics Keywords: Mechanics,A Level, Edexcel Created Date: 10/26/2019 5:36:15 PM

Notes for Edexcel Mechanics

Dr Yu (郁　辰)

CasperYC.weebly.com/ALevelMechanics

October 26, 2019

http://CasperYC.weebly.com/ALevelMechanicshttp://CasperYC.weebly.com/ALevelMechanics

Note

This note is made for the purpose of revision for my students.

All materials are collated from the Internet and put together with some adaption. If you

see a content mistake, please let me know. If you see a grammar mistake, please let me know.

If you see a spelling mistake, please let me know. If you see any kind of mistake, please let me

know.

As we are currently migrating onto the new Edexcel Curriculum, a few of the topics need

to be re–mapped and chapters need to be re–written. Collaboration is warmly welcome and

contributors will be added and accredited accordingly.

i

Contents

Note i

I Year 12 1

1 Mathematical Models in Mechanics 2

2 Vectors in Mechanics 3

2.1 Magnitude and direction components . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Parallel vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.3 Adding vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.4 Resolving vectors in two perpendicular components . . . . . . . . . . . . . . . . 5

2.5 Vector algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.6 Vectors in mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.7 Velocity and displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.8 Relative displacement vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.9 Collision of moving particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.10 Closest distance between moving particles . . . . . . . . . . . . . . . . . . . . . 8

2.11 Relative velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Kinematics of a particle moving in a straight line 10

3.1 Constant acceleration formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.2 Vertical motion under gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.3 Speed–time graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Statics of a particle 13

4.1 Resultant forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.2 Resultant of three or more forces . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.3 Equilibrium of a particle under coplanar forces . . . . . . . . . . . . . . . . . . . 16

4.4 Types of force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.5 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.6 Coefficient of friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.7 Limiting equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 Dynamics of a particle moving in a straight line 21

5.1 Newton’s laws of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.2 Connected particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.3 Particles connected by pulleys . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.4 Force on pulley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.5 Impulse and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5.6 Internal and External Forces and Impulses . . . . . . . . . . . . . . . . . . . . . 29

5.7 Conservation of linear momentum – CLM . . . . . . . . . . . . . . . . . . . . . 29

ii

5.8 Impulse in string between two particles . . . . . . . . . . . . . . . . . . . . . . . 31

6 Moments 33

6.1 Moment of a Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.2 Sum of moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.3 Moments and Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.4 Non-uniform rods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.5 Nearly tilting rods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

II Year 13 36

7 Kinematics 37

7.1 Constant acceleration in a vertical plane . . . . . . . . . . . . . . . . . . . . . . 37

7.2 Variable acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.3 Using vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8 Centres of mass 42

8.1 Centre of mass of n particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

8.2 Centres of mass of simple laminas . . . . . . . . . . . . . . . . . . . . . . . . . . 42

8.3 Centres of mass of composite laminas . . . . . . . . . . . . . . . . . . . . . . . . 45

8.4 Laminas suspended freely under gravity . . . . . . . . . . . . . . . . . . . . . . . 47

8.4.1 Body with point mass hanging freely . . . . . . . . . . . . . . . . . . . . 48

8.5 Toppling on a slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

8.6 Centres of mass of wire frameworks . . . . . . . . . . . . . . . . . . . . . . . . . 49

9 Work, energy and power 51

9.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

9.2 Work done by a (constant) force . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

9.2.1 Forces parallel to the displacement . . . . . . . . . . . . . . . . . . . . . 51

9.2.2 Forces at an angle to the displacement . . . . . . . . . . . . . . . . . . . 51

9.2.3 Work done by gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

9.3 Work–energy equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

9.4 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

9.5 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

10 Collisions 59

10.1 Impulse and momentum – using vectors . . . . . . . . . . . . . . . . . . . . . . . 59

10.1.1 Impulse = change in momentum . . . . . . . . . . . . . . . . . . . . . . . 59

10.1.2 Conservation of linear momentum . . . . . . . . . . . . . . . . . . . . . . 59

10.2 Newton’s law of restitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

10.2.1 Coefficient of restitution . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

10.2.2 Collisions with a plane surface . . . . . . . . . . . . . . . . . . . . . . . . 60

iii

10.2.3 Multiple collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

10.3 Kinetic energy and impulses/collisions . . . . . . . . . . . . . . . . . . . . . . . 62

11 Statics of rigid bodies 64

11.1 Moment of a force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

11.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

11.3 Three non–parallel forces in equilibrium . . . . . . . . . . . . . . . . . . . . . . 67

11.4 Triangle of forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

A Conservation of linear momentum, C.L.M. 69

B Centre of mass of n particles 70

C Medians of a triangle 71

D Centre of mass of a triangle 72

E Centre of mass g 73

iv

Part I

Year 12

1

Chapter 1

Mathematical Models in Mechanics

Assumptions and approximations often used to simplify

the mathematics involved:

(a) a rigid body is a particle,

(b) no air resistance,

(c) no wind,

(d) force due to gravity remains constant,

(e) light pulleys and light strings etc. have no mass,

(f) the tension in a light string which remains taut will be constant throughout its length.

(g) if a pulley is light or smooth the tensions in the a string going round the pulley will be

equal on both sides; the same is true for a smooth peg,

(h) if a string is inextensible and remains taut, the accelerations of two particles, one fixed

at each end, will be equal,

(i) if a rod is uniform – constant mass per unit length – the centre of mass will be in the

middle,

(j) a lamina is a uniform flat object of negligible thickness,

(k) the earth’s surface, although spherical, is usually modelled by a plane,

(l) surface is smooth – no friction,

(m) forces behave like vectors.

2

Chapter 2

Vectors in Mechanics

A vector is a quantity which has both magnitude and direction.

A scalar is just a number – it has no direction – e.g. mass, time, etc.

Vectors should be underlined or with arrow above, a letter without the underlining

means the length of the vector. r is the length of the vector ~r.

2.1 Magnitude and direction components

Example 2.1.1: From component form, draw a sketch and use Pythagoras and

trigonometry to find the hypotenuse and angle.

x

5

3

180 -

r = (32 + 52)

From magnitude and direction form, draw a sketch and use trigonometry to find x and y

components.

Example 2.1.2: ~r is a vector of length 7 cm making an angle of −50◦ with the x-axis.

3

CHAPTER 2. VECTORS IN MECHANICS Page 4 of 73

Example 2.1.3: Find a vector p of length 15 in the direction of ~q =(−6

8

).

2.2 Parallel vectors

Two vectors are parallel ⇐⇒ one is a multiple of another:

6i− 8j = 2(3i− 4j) ⇐⇒ 6i− 8j and 3i− 4j are parallel,

or (6

−8

)= 2×

(3

−4

)⇐⇒

(6

−8

)and

(3

−4

)are parallel.

2.3 Adding vectors

p

p + q

q

p

p + q

q

1) In component form: (a

b

)+

(c

d

)=

(a+ c

b+ d

)2) To add two vectors which are given in magnitude and direction form:

c©Dr Yu 2019 Last updated: October 26, 2019


(a) convert to component form, add and convert back,

(b) sketch a vector triangle and use sine or cosine rule.

Example 2.3.1: Add together, 3 miles on a bearing of 60◦ and 4 miles on a bearing of

140◦.

2.4 Resolving vectors in two perpendicular components

x

y

FF

F



2.5 Vector algebra

Notation,−→OA = ~a,

−−→OB = ~b.

a

b

A

O B

To get from A to B, we use

−→AB =

−→AO +

−−→OB = −~a+~b or ~b− ~a.

2.6 Vectors in mechanics

Forces behave as vectors (the physicists tell us so) – modelling.

Velocity is a vector so must be given either in component form or as magnitude and direction.

Speed is the magnitude of the velocity so is a scalar and has no direction.

Acceleration is a vector so must be given either in component form or as magnitude and

direction.

2.7 Velocity and displacement

If a particle moves from the point (2,4) with a constant velocity ~v = 3i − 4j for 5 seconds,then its displacement vector will be velocity × time = (3i− 4j)× 5 = 15i− 20j and so its newposition will be given by ~r = (2i+ 4j) + (15i− 20j) = 17i− 16j.

Example 2.7.1: A particle is initially at the point (4,11) and moves with velocity(

3−7

)ms−1. Find its position vector after t seconds.



2.8 Relative displacement vectors

If you are standing at a point C and X is standing at a point D then the position vector of X

relative to you is the vector−−→CD

and−−→CD = ~d− ~c = ~rD − ~rC .

Thus if a particle A is at ~rA = 3i − 4j and B is at ~rB = 7i + 2j, then the position of Arelative to B is

−→BA = ~a−~b = (3i− 4j)− (7i+ 2j) = −4i− 6j.

2.9 Collision of moving particles

Example 2.9.1: Particle A is initially at the point (3,4) and travels with velocity

(9i− 2j) ms−1.Particle B is initially at the point (6,7) and travels with velocity (6i− 5j) ms−1

(a) Find the position vectors of A and B at time t.

(b) Show that the particles collide and find the time and position of collision.



2.10 Closest distance between moving particles

Example 2.10.1: Two particles, A and B, are moving so that their position vectors

at time t are

~rA =

(5− 3t2 + t

)and ~rB =

(4− t3 + 2t

).

(a) Find the vector−→AB at time t.

(b) Find the distance between A and B at time t in terms of t.

(c) Find the minimum distance between the particles and the time at which this occurs.

2.11 Relative velocity

This is similar to relative position in that if C and D are at positions ~rC and ~rD, then the

position of D relative to C is ~rD−~rC , which leads on, by differentiating with respect to time, ifC and D are moving with velocities ~vC and ~vD then the velocity of D relative to C is ~vD − ~vC .



Example 2.11.1: Particles A and B have velocities ~vA = (12t − 3)i + 4j and ~vB =(3t2−1)i+2tj. Find the velocity of A relative to B and show that this velocity is parallelto the x- axis for a particular value of t which is to be determined.


Chapter 3

Kinematics of a particle moving in a straight line

3.1 Constant acceleration formulae

v = u+ at, s = ut+1

2at2, s =

1

2(u+ v)t, v2 = u2 + 2as

Units must be consistent – e.g. change kmh−1 to ms−1 before using the formulae.

Example 3.1.1: A particle moves through a point O with speed 13 ms−1 with accel-

eration −6 ms−2. Find the time (s) at which the particle is 12m from O.

3.2 Vertical motion under gravity

1) The acceleration always acts downwards whatever direction the particle is moving.

2) We assume that there is no air resistance, that the object is not spinning or turning and

that the object can be treated as a particle.

3) We assume that the gravitational acceleration remains constant and is 9.8 ms−2.

4) Always state which direction (up or down) you are taking as positive.

Example 3.2.1: A ball is thrown vertically upwards from O with a speed of 14 ms−1.

(a) Find the greatest height reached.

(b) Find the total time before the ball returns to O.

(c) Find the velocity after 2 seconds.

10

CHAPTER 3. KINEMATICS OF A PARTICLE MOVING IN A STRAIGHT LINEPage 11 of 73

3.3 Speed–time graphs

(1) The area under a speed–time graph represents the distance travelled.

(2) The gradient of a speed–time graph is the acceleration or deceleration.

Example 3.3.1: A particle is initially travelling at a speed of 2 ms−1 and immediately

accelerates at 3 ms−2 for 10 seconds; it then travels at a constant speed before decelerating

at a 2 ms−2 until it stops.

(a) Find the maximum speed and the time spent decelerating.

(b) Sketch a speed–time graph.

(c) If the total distance travelled is 1130 metres, find the time spent travelling at a

constant speed.


CHAPTER 3. KINEMATICS OF A PARTICLE MOVING IN A STRAIGHT LINEPage 12 of 73


Chapter 4

Statics of a particle

4.1 Resultant forces

As forces behave like vectors you can add two forces geometrically using a triangle or a paral-

lelogram.

or

R

F1

F2 R F2

F1

To find R from a diagram either draw accurately or, preferably, use sine and/or cosine rules.

Example 4.1.1: ~F1 and ~F2 are two forces of magnitudes 9N and 5N and the angle

between their directions is 100◦. Find the resultant force.

80o 100o

R

9

5

xo

Example 4.1.2: Find the resultant of ~P = 5i− 7j and ~Q = −2i+ 13j.

13

CHAPTER 4. STATICS OF A PARTICLE Page 14 of 73

4.2 Resultant of three or more forces

Note. We can resolve vectors in two perpendicular components as shown. ~F has components

F cos(θ) and F sin(θ).

x

y

F F sin

F cos

To find the resultant of three forces

1) convert into component form (i and j) , add and convert back

2) sketch a vector polygon and use sine/cosine rule to find the resultant of two, then combine

this resultant with the third force to find final resultant.

For more than three forces continue with either of the above methods.

Example 4.2.1: Find the resultant of the four forces shown in the diagram.

9 N

8 N 7 N

4 N

60o

25o



Example 4.2.2: Use a vector polygon to find the resultant of the three forces shown

in the diagram.

To sketch the vector polygon, draw the forces end to end. I have started with the 3 N,

then the 4 N and finally the 2 N force.

48o

o 4 N

3 N

R2 N

R1 N

2 N

48o

3 N

2 N

4 N

Combine the 3 N and 4 N forces to find the resultant ~R1 = 5N with θ = 36.9◦, and now

combine ~R1 with the 2N force to find the final resultant ~R2 using the cosine and/or sine

rule.



4.3 Equilibrium of a particle under coplanar forces

If the sum of all the forces acting on a particle is zero (or if the resultant force is 0 N) then the

particle is said to be in equilibrium.

Example 4.3.1: Three forces ~P =(

7−2

)N, ~Q =

(−34

)N and ~R =

(ab

)N are acting on a

particle which is in equilibrium. Find the values of a and b.

Example 4.3.2: A particle is in equilibrium atO under the forces shown in the diagram.

Find the magnitudes of ~P and ~Q.

P

Q

60o

12



4.4 Types of force

Definition 4.4.1.

Contact forces: tension, thrust, friction, normal (i.e. perpendicular to the surface) reaction.

Non-contact forces: weight / gravity, magnetism, force of electric charges.

N.B. ALWAYS DRAW A DIAGRAM SHOWING ALL FORCES, but never

mark a force on a diagram without knowing what is providing it.

4.5 Friction

If we try to pull a box across the floor there is a friction force between the box and the floor.

F

R

P

mg

If the box does not move the friction force will be equal to the force ~P and as ~P increases

from 0 N, the friction force will also increase from 0 N until it reaches its maximum value~Fmax, after which the box will no longer be in equilibrium.

When friction force is at its maximum and the box is on the point of moving the box is said

to be in limiting equilibrium.



N.B. The direction of the friction force is always opposite to the direction of motion

(or the direction in which the particle would move if there was no friction).

Example 4.5.1: A particle of mass 2 kg rests in equilibrium on a plane which makes

an angle of 25◦ with the horizontal. Find the magnitude of the friction force and the

magnitude of the normal reaction.

R

2g

F

25o 25o

2g cos 25

2g sin 25

R

F

DRAW A DIAGRAM SHOWING ALL FORCES – the weight 2g N, the friction ~F N

and the normal reaction ~R N. Remember that the particle would move down the slope

without friction so friction must act up the slope. Then draw a second diagram showing

forces resolved along and perpendicular to the slope.

4.6 Coefficient of friction

There is a maximum value, or limiting value, of the friction force between two surfaces. The

ratio of this maximum friction force to the normal reaction between the surfaces is constant for

the two surfaces and is called the coefficient of friction.

~Fmax = µ~R,

where µ is the coefficient of friction and ~R is the normal reaction.



Example 4.6.1: A particle of mass 3 kg lies in equilibrium on a slope of angle 25◦. If

the coefficient of friction is 0.6, show that the particle is in equilibrium and find the value

of the friction force.

Solution:

25o

R F

3g

R F

3g sin 25 3g cos 25

4.7 Limiting equilibrium

When a particle is in equilibrium but the friction force has reached its maximum or limit-

ing value and the particle is on the point of moving, the particle is said to be in limiting

equilibrium.

Example 4.7.1: A particle of mass 6 kg on a slope of angle 30◦ is being pushed by a

horizontal force of ~P N. If the particle is in limiting equilibrium and is on the point of

moving up the slope find the value of ~P , given that µ = 0.3.

DRAW A DIAGRAM SHOWING ALL FORCES


Chapter 5

Dynamics of a particle moving in a straight line

5.1 Newton’s laws of motion

1) A particle will remain at rest or will continue to move with constant velocity in a straight

line unless acted on by a resultant force.

2) For a particle with constant mass, m kg, the resultant force ~F N acting on the particle and

its acceleration ~a ms−2 satisfy the equation ~F = m~a.

3) If a body A exerts a force on a body B, then body B exerts an equal force on body A but

in the opposite direction.

Example 5.1.1: A box of mass 30 kg is being pulled along the ground by a horizontal

force of 95 N. If the acceleration of the trolley is 1.5ms−2, find the coefficient of friction.

Example 5.1.2: A ball of mass 2 kg tied to the end of a string. The tension in the

string is 30 N. Find the acceleration of the ball and state in which direction it is acting.

21

CHAPTER 5. DYNAMICS OF A PARTICLE MOVING IN A STRAIGHT LINEPage 22 of 73

Example 5.1.3: A particle of mass 25 kg is being pulled up a slope at angle of 25◦

above the horizontal by a rope which makes an angle of 15◦ with the slope. If the tension

in the rope is 300 N and if the coefficient of friction between the particle and the slope is14, find the acceleration of the particle.

Solution:

R + 300 sin 15

a

F + 25g sin 25

300 cos 15

25g cos 25 25o

15o

25g

300 R

F

a

25o



5.2 Connected particles

In problems with two or more connected particles, draw a large diagram in which the particles

are clearly separate. Then put in all forces on each particle: don’t forget Newton’s third law

there will be some ‘equal and opposite’ pairs of forces.

Example 5.2.1:

A lift of mass 600 kg is accelerating upwards carrying a man of mass 70 kg. If the tension

in the lift cables is 7000 N find the acceleration of the lift and the force between the floor

and the man’s feet.

Note. Draw a clear diagram with all forces on lift AND all forces on man.

Draw a second diagram showing the combined system, that is, the man and the lift as

‘one particle’.

If the normal reaction on the man is R newtons then this means that the lift floor is

pushing up on the man with a force of R newtons, therefore the man must be pushing

down on the lift floor with an equal sized force of R newtons.

Draw a LARGE lift and put the man in the middle (not touching the floor).

Lift and man separate Lift and man combined

7000

a

670g

600g + R

R

70g

a

7000



Example 5.2.2: A truck of mass 1300 kg is pulling a trailer of mass 700 kg. The

driving force exerted by the truck is 1500 N and there is no resistance to motion.

Find the acceleration of the truck and trailer, and the force in the tow bar between the

truck and the trailer.

Trailer and truck separate

R1

T

700g

T

1300g

R2 a

1500

a

Trailer and truck combined

2000g

R a

1500

Note that the truck and trailer both have the same acceleration, assuming a rigid tow

bar.



5.3 Particles connected by pulleys

The string will always be inextensible and light and the pulley (or peg) will always be smooth

or light.

1) As the string is inextensible the accelerations of the two particles at its ends will have the

same magnitude.

2) As the string is light, the tension in the string will be constant along its length.

3) As the pulley (or peg) is smooth or light, the tensions in the string on either side of the

pulley (or peg) will be equal.

Example 5.3.1: Particles of mass 3 kg and 5 kg are attached to the ends of a light

inextensible string which passes over a fixed smooth pulley. The 5 kg particle is initially

2 m above the floor.

The system is released from rest; find the greatest height of the lighter mass above its

initial position in the subsequent motion. Assume that the lighter mass does not reach

the pulley.



5.4 Force on pulley

Example 5.4.1: A block of mass 4 kg is on a slope which makes an angle of 40◦ with

the horizontal. The block is attached to an inextensible, light string which passes over a

light, smooth pulley. The other end of the string is attached to a ball of mass 5 kg. The

coefficient of friction between the block and the slope is 14

. The block is accelerating up

the slope.

(a) What can you assume because the string is light and inextensible?

(b) What can you assume because the pulley is light and smooth?



(c) Find the magnitude of the force exerted on the pulley by the string.



5.5 Impulse and Momentum

(a) We know that the velocity v ms−1 of a body of mass m kg moving with a constant accel-

eration a ms−2 for time t seconds is given by

v = u+ at

where u is the initial velocity.

mv = mu+mat and mat = mv −mu

Newton’s Second Law states that F = ma.

Ft = mat =⇒ Ft = mv −mu

Note that F must be constant since a is constant.

(b) We define the impulse, I, of a constant force F N acting for a time tseconds to be I = Ft

Newton–seconds (Ns).

(c) We define the momentum of a body of mass m kg moving with velocity v ms−1 to be mv

kgms−1.

(d) The equation I = Ft = mv −mu of paragraph (a) can now be thought of as

Impulse = Change in Momentum

N.B. Impulse and Momentum are vectors

Example 5.5.1: A ball of mass 2 kg travelling in a straight line at 4 m/s is acted on by

a constant force of 3 N acting in the direction of motion. Find the speed after 5 seconds.



Example 5.5.2: A ball of mass 1.5 kg is struck by a bat in the opposite direction to

the motion of the ball. Before the impulse the ball is travelling at 16m/s and the impulse

of the bat on the ball is 50 Ns. Find the velocity of the ball immediately after impact.

Before During After

+

50 Ns

x m s 1 16 m s 1

5.6 Internal and External Forces and Impulses

1) If a hockey ball is hit by a hockey stick then the impulse on the ball is an external impulse

on the ball.

2) If two cricket balls collide then the impulses between the balls at the moment of collision

are internal when considering the two balls together. If we were considering just one ball

then the impulse of collision would be external to that ball.

3) If an explosion separates a satellite from a rocket then the impulse of the explosion is

internal when considering the rocket and the satellite together. If we were considering

the satellite alone then the impulse of the explosion would be external to the satellite.

5.7 Conservation of linear momentum – CLM

If there are no external impulses acting on a system then the total momentum of that system

is conserved (i.e. remains the same at different times).

or total momentum before impact equals total momentum after impact.

Note that if there is an external impulse acting on the system then the momentum perpen-

dicular to that impulse is conserved.



before during after

m2 m1 m2 m1 m2 m1

u1 v1 u2 v2 I I

Note that the total impulse on the system is −I + I = 0, as the impulses when consideringboth balls together are internal.

CLM ⇐⇒ m1u1 +m2mu = m1v1 +m2v2

Example 5.7.1: A railway truck of mass 1500 kg is travelling in a straight line at

3m/s. A second truck of mass 1000 kg is travelling in the opposite direction at 5 m/s.

They collide (without breaking up) and couple together. With what speed and in what

direction are they moving after the impact?



Example 5.7.2: Two balls A and B are travelling towards each other with speeds

uA = 5 m/s and uB = 6 m/s. After impact A is now travelling in the opposite direction

at 3 m/s, and B continues to travel in its original direction but with speed 2 m/s. The

mass of A is 2 kg. Find the mass of B.

5.8 Impulse in string between two particles

If a string links two particles which are moving apart then the string will become taut and, at

that time, there will be an impulse in the string. In this case the impulses on the two particles

will be equal in magnitude but opposite in direction. Thus when considering the two particles

as one system there is no external impulse and the problem can be treated in a similar way

to collisions. The assumptions involved are that the string is light (mass can be ignored) and

inextensible (does not stretch).

Example 5.8.1: Two particles P and Q of masses 2 kg and 5 kg are connected by a

light inextensible string. They are moving away from each other with speeds uP = 3 m/s

and uQ = 4 m/s. After the string becomes taut the particles move on with the same

velocity.

(a) Find this common velocity.

(b) Find the impulse in the string.


Chapter 6

Moments

6.1 Moment of a Force

Definition 6.1.1. The moment of a force F about a point P is the product of the magnitude

of F and the perpendicular distance from P to the line of action of the force. Moments are

measured in newton-metres, Nm and the sense – clockwise or anti–clockwise should always be

given.

moment = F d clockwise moment = F PM = F d sin clockwise

or

P

d

F

M

P

F

d

6.2 Sum of moments

Example 6.2.1: The force (7i + 4j) N acts at the point (5, 3), find its moment about

the point (2, 1).

33

CHAPTER 6. MOMENTS Page 34 of 73

6.3 Moments and Equilibrium

If several forces are in equilibrium then

(i) The resultant force must be zero ⇐⇒ the resultant force in any direction must be zero.

(ii) The sum of the moments of all the forces about any point must be zero.

Example 6.3.1: If the forces in the diagram are in equilibrium find the force F and

the distance x m.

C

4 N F N

13 N 7 N

A B D

3 m 2 m x m

6.4 Non-uniform rods

A uniform rod has its centre of mass at its mid–point. A non-uniform rod (e.g. a tree trunk)

would not necessarily have its centre of mass at its mid-point

Example 6.4.1: A non-uniform rod AB of mass 25 kg is of length 8 metres. Its centre

of mass is 3 metres from A. The rod is pivoted about M , its mid point. A mass of 20

kg is placed at P so that the system is in equilibrium. How far should this mass be from

the end A? What is now the reaction at the pivot?

A G

25g

M B

20g

1m x m 3 m

R

P


CHAPTER 6. MOMENTS Page 35 of 73

6.5 Nearly tilting rods

If a rod is supported at two points A and B, then when the rod is about to tilt about B the

normal reaction at A will be 0.

Example 6.5.1: A uniform plank PQ of mass 8 kg rests on two supports at A and B.

PQ = 2 m, PA = 0.6 m and AB = 0.7 m. A mass M kg is placed at X on the rod

between B and Q at a distance of 0.5m from B. The rod is on the point of tilting about

B, find the value of M .


Part II

Year 13

36

Chapter 7

Kinematics

7.1 Constant acceleration in a vertical plane

We can think of the horizontal and vertical motion separately and use the formulae:

v = u+ at, s = ut+1

2, v2 = u2 + 2as, s =

1

2(u+ v)t

Example 7.1.1: A stone is thrown at a speed of 20m/s at an angle of 35◦ to the

horizontal. Find

(a) the greatest height reached,

(b) the direction in which it is moving after 1 second,

(c) the height of the stone after it has travelled a horizontal distance of 25m.

37

CHAPTER 7. KINEMATICS Page 38 of 73

Example 7.1.2: A ball is projected from a point on horizontal ground with a speed of

U , making an angle of α with the horizontal.

(a) Show that the ball moves along a parabola.

(b) Find the range.

(c) If α is allowed to vary, find the maximum range.



7.2 Variable acceleration

When a is given as a function of t (not constant),

v(t) =

∫a(t) dt,

and

s(t) =

∫v(t) dt.

When s (or v) is given as a function of t,

v(t) =ds

dt

and

a(t) =dv

dtor a(t) =

d2s

dt2

Note that s is the displacement (the distance from the origin), which is not necessarily

the same as the distance travelled (the particle may have moved forwards and backwards).

Example 7.2.1: A particle is moving along the x–axis with an acceleration (5 − 2t)m/s2. At time t = 0, the particle moves through the origin with speed 6 m/s in the

direction of the positive x–axis.

(a) Find the displacement of the particle after 9 seconds.

(b) Find the distance travelled in the first 9 seconds.



7.3 Using vectors

This is just combining horizontal and vertical motion into one expression.

Example 7.3.1: A particle moves with velocity v =(3t2

−4t

)m/s. It is initially at the

point (6, 3). Find

(a) its acceleration after 2 seconds, and

(b) its displacement at time t.



Example 7.3.2: A particle A is at (1, 2) when t = 0, moving with velocity(4t−3

)m/s.

A second particle, B, is at the origin when t = 0, moving with velocity(

2t−2t

)m/s.

Investigate whether A and B collide.


Chapter 8

Centres of mass

8.1 Centre of mass of n particles

The centre of mass, (x̄, ȳ), of n particles, which have masses m1,m2, . . . ,mn

at points (x1, y1, ), (x2, y2), . . . , (xn, yn) is given by

M

(x̄

ȳ

)=∑

mi

(xiyi

)if and only if

Mx̄ =∑

mixi and Mȳ =∑

miyi

where M is the total mass, M =∑mi, or

Mr =∑

miri.

8.2 Centres of mass of simple laminas

1) The centre of mass of a uniform rod is at its mid point.

2) The centre of mass of a uniform rectangular lamina (sheet) is at its point of symmetry.

G

3) The centre of mass of a uniform triangular lamina.

A

B CD

EF

G

(a) G is at the point where the three medians meet.

G divides each median in the ratio 2:1.

AG

GD=BG

GE=CG

GF=

2

1

42

CHAPTER 8. CENTRES OF MASS Page 43 of 73

(b) If A,B and C are the points (a1, a2), (b1, b2, ) and (c1, c2), then the centre of mass, G, is

at the point

G

(1

3(a1 + b1 + c1),

1

3(a2 + b2 + c2)

)or

~r =1

3(~a+~b+ ~c)

4) The centre of mass of a uniform lamina in the shape of a sector of a circle, with radius r

and angle 2α.

GO

r

r

Circle centre O with radius r.

Angle of sector is 2α.

G lies on the axis of symmetry and

OG =2r sin(α)

3α



Example 8.2.1: A uniform triangular lamina has mass 3 kg, and its vertices are

O(0, 0), A(5, 0) and B(4, 3). Masses of 2, 4 and 5 kg are attached at O,A and B re-

spectively. Find the centre of mass of the system.



8.3 Centres of mass of composite laminas

In the following examples ρ is the surface density, or mass per unit area.

Example 8.3.1: Find the centre of mass of a uniform rectangular lamina attached to a

uniform semi circular lamina, as shown in the diagram. O is the centre of the semi–circle.



Example 8.3.2: Find the centre of mass of the uniform lamina ABCDE. FC is an

axis of symmetry. CB = CD. AE = 10, AB = 20, FC = 14.



8.4 Laminas suspended freely under gravity

Example 8.4.1: The lamina in Example 8.3 is suspended from A, and hangs in equi-

librium. Find the angle made by AB with the downward vertical.



8.4.1 Body with point mass hanging freely

The best technique will probably be to take moments about the point of suspension.

Example 8.4.2: A uniform rectangular lamina ABCD, where AB = 6 m and BC = 4

m, has mass 2M . A particle of mass M is attached to the edge of the rectangle at B.

The compound body is freely suspended under gravity from O, the mid–point of AB.

Find the angle made by AB with the horizontal.



8.5 Toppling on a slope

Example 8.5.1: What angle of slope would cause a 4× 6 uniform rectangular laminato topple (assuming that the friction is large enough to prevent sliding).

8.6 Centres of mass of wire frameworks

In the following examples ρ is mass per unit length.

(1) The centre of mass of a uniform straight wire is at its centre.

(2) The centre of mass of a uniform circular arc, of radius r and angle at the centre 2α, lies on

the axis of symmetry and OG = r sin(α)α

.

r

G O



Example 8.6.1: A uniform wire framework is shown in the diagram. OABC is a

rectangle and the arc is a semi–circle. OA = 10 and OC = 4. Find the position of the

centre of mass.


Chapter 9

Work, energy and power

v2 = u2 + 2as

mas =1

2mv2 − 1

2mu2

Fs =1

2mv2 − 1

2mu2

9.1 Definitions

We define

• The kinetic energy, K.E., of a body of mass m moving with speed v is 12mv2.

• Work done by a (constant) force F is the magnitude of the force × distance moved in thedirection of the force.

• The units of kinetic energy and work are Joules, J .

9.2 Work done by a (constant) force

9.2.1 Forces parallel to the displacement

(a) Work done by a force of 6 N when a particle is moving from A to B (AB = 3 m) in the

direction of the force, is 6× (+3) = 18J .

A B

6 N

3 m

(b) Work done by a force of 6 N when a particle is moving from B to A in the opposite direction

to the force, is 6× (−3) = −18J .

A B

6 N

3 m

9.2.2 Forces at an angle to the displacement

Work done can be calculated in two ways

(a) As a body moves from A to B,

51

CHAPTER 9. WORK, ENERGY AND POWER Page 52 of 73

A B

F N

s m

N

we can think that it has moved through the distance A N in the direction of the force

AN = s cos(θ),

work done is

F × s cos(θ) = Fs cos(θ).

(b) Or, we can resolve the force F parallel and perpendicular to the displacement.

A B

F sin N

s m F cos N

The component F sin(θ) is perpendicular to the displacement, and so does no work.

The component F cos(θ) is parallel to the displacement, work done is

F × s cos(θ) = Fs cos(θ).

as in part (a).

9.2.3 Work done by gravity

• If a particle of mass m falls a vertical distance h, then the work done by gravity is mgh,force and displacement are in the same direction.

• If a particle of mass m rises a vertical distance h, then the work done by gravity is −mgh,force and displacement are in opposite directions.

When a particle is moving on a slope, it is usual to consider the vertical distance moved

and multiply by mg to calculate the work done by gravity.

mg

A

B

d

h



From A to B the partial moves a distance d, but its vertical movement is h = d sin(θ).

Work done by gravity

mgh = mgd sin(θ).

Work done by gravity is also known as loss or gain in Gravitational Potential Energy,

G.P.E.

9.3 Work–energy equation

The equation

Fs =1

2mv2 − 1

2mu2,

can be rearranged as1

2mv2 =

1

2mu2 + Fs,

which can be thought of as

Final K.E = Initial K.E±Work done

Notice the ±,

• If a force increases the K.E. then the work done is positive.

• If a force decreases the K.E. then the work done is negative.

Example 9.3.1: A particle of mass 5 kg is being pulled up a rough slope by a force

of 50 N parallel to the slope. The coefficient of friction is 0.2, and the slope makes an

angle of α = tan−1(3/4) with the horizontal.

The particle is observed to be moving up the slope with a speed of 3 m/s. Find its speed

when it has moved 12 m up the slope.

B

A

h

3

v

5g

R

F

50



Example 9.3.2: Tarzan swings on a rope, lets go and falls to the ground. If Tarzan

was initially 7 m above the ground and not moving, and if he lets go when he is 3 m

above the ground after the rope has passed the vertical, with what speed does he hit the

ground?

mg

T

7 m 3 m



Example 9.3.3: Tarzan now goes skiing in Switzerland. It is much colder than Africa

so he is wearing lots of warm clothes and his mass is 90 kg. He starts with a speed of 3

m/s and skis along a path as shown in the diagram when he comes to a cliff. The total

length of his path is 150 m, and he experiences a constant resistance of 95 N. Find his

speed as he launches himself into thin air.

30 m

R 95 N

90g 13 m

3 ms 1

v ms 1



9.4 Potential energy

When a body falls it gains K.E. The higher its starting point the greater the gain in K.E. We

say that the Gravitational Potential Energy, G.P.E., of a body depends on its height above

some fixed point.

When a body falls a distance h, the loss in G.P.E. is the work done by gravity, mgh.

1

2mv2 =

1

2mu2 +mgh =⇒ mgh = 1

2mv2 − 1

2mu2

or Loss in G.P.E. = Gain in K.E.

Similarly when a body rises a distance h, the gain in G.P.E. = mgh.

1

2mv2 =

1

2mu2 −mgh =⇒ mgh = 1

2mu2 − 1

2mv2

or Gain in G.P.E. = Loss in K.E.

You will do better using the energy equation than “gain in GPE = loss in KE”,

etc.

You are expected to understand the terms loss in G.P.E. and gain in G.P.E., but all you

can always use work done by gravity if you wish.

Example 9.4.1: A block, of mass 20 kg, is pulled up a rough slope of angle

α, tan−1(3/4), by a rope parallel to the slope. The block starts from rest and is moving

at 6 m/s when it has moved 10 m. If the coefficient of friction is µ = 0.3, find the tension

in the rope.

20g F

R T 10 m 0

6



9.5 Power

Power is the rate of doing work. The units are Watts = Joules/second. For a constant force,

F , moving a distance s, the work done is W = Fs.

The power

P =d

dtFs = F

ds

dt= Fv

since F is constant.

The power developed by a constant force F moving its point of application at a speed v is

P = Fv.

Example 9.5.1: A car of mass 900 kg is travelling up a slope of 5◦ at a constant speed.

Assume that there is no resistance to motion – other than gravity.

(a) If the engine of the car is working at a rate of 20 kW, find the speed of the car.

(b) The car later travels up a slope of 8◦ at the same constant speed. Find the power

developed by the engine.


Chapter 10

Collisions

10.1 Impulse and momentum – using vectors

10.1.1 Impulse = change in momentum

I = mv −mu

where v is the final velocity and u is the initial velocity.

10.1.2 Conservation of linear momentum

If there is no external impulse during a collision, then the total momentum before impact equals

the total momentum after impact.

m1u1 +m2u2 = m1v1 +m2v2

Example 10.1.1: A ball, A, of mass 3kg is moving with velocity(

3−2

)m/s when it

collides with another ball, B, of mass 2kg moving with velocity(−3−1

)m/s. After the

collision A moves with velocity(

1−3

)m/s. Find the velocity of B after the collision, and

the impulse on A during the collision.

59

CHAPTER 10. COLLISIONS Page 60 of 73

10.2 Newton’s law of restitution

This is also known as Newton’s Experimental Law, NEL.

10.2.1 Coefficient of restitution

The coefficient of restitution in a collision, e, is defined as

e =speed of separation

speed of approach0 ≤ e ≤ 1

If e = 1, then the collision is perfectly elastic and K.E. is conserved during the collision.

As usual it is essential to draw good diagrams and to take care over positive and negative

values.

Example 10.2.1: Particles P and Q with masses 2kg and 3kg are moving towards each

other with velocities of 7m/s and 5 m/s respectively. If the coefficient of restitution is34, find the velocities of P and Q after the collision.

10.2.2 Collisions with a plane surface

The velocity of the surface before and after is 0.



m kg u m kg v

e =speed of separation

speed of approach=v − 0u− 0

=v

u

10.2.3 Multiple collisions

Treat each collision as a new problem – the final velocities from one collision become the initial

velocities for the next collision.

Example 10.2.2: Two particles, A and B, are of equal mass and are moving towards

each other with speed of 3m/s and 2m/s respectively and collide. Particle B then

strikes a plane surface which is perpendicular to it direction of motion and rebounds.

The coefficient or restitution between the two particles is 12, and between B and the

plane surface is 57.

Show that B collides with A a second time, and find the velocities of both particles after

this collision.



10.3 Kinetic energy and impulses/collisions

K.E. will be generated if there is an external impulse, but in collisions K.E. will be lost (unless

the collision is perfectly elastic, e = 1).

Example 10.3.1: A rifle of mass 5kg fires a bullet of mass 25 grams with a muzzle

velocity of 800m/s. The rifle is pointing in a horizontal direction and is free to move.

Find the K.E. generated in firing the rifle.



Example 10.3.2: Particle A, mass 3kg, and particle B, mass 4kg, are moving towards

each other with speeds of 5m/s and 2m/s respectively. If e = 12, find the K.E. lost in the

collision.


Chapter 11

Statics of rigid bodies

11.1 Moment of a force

Example 11.1.1: The moment of a force F about a point P is the magnitude of the

force multiplied by the perpendicular distance from P to the line of action of F .

PN PN

Nm Nm

P

N

mP

N

m

64

CHAPTER 11. STATICS OF RIGID BODIES Page 65 of 73

11.2 Equilibrium

A system of forces will be in equilibrium if

(i) The sum of the resolved forces in any direction is zero.

(ii) The moment about any point is zero.

Example 11.2.1:

A uniform ladder of mass 20kg and length 8m is leaning against a smooth vertical wall

on rough ground. The ladder makes an angle of 60◦ with the ground, and the coefficient

of friction between the ladder and the ground is 0.5.

What is the maximum height that a man of mass 80kg can climb up the ladder before it

starts to slip?



Example 11.2.2:

A non-uniform rod AB, of length 4m, is freely hinged to a vertical wall at A. It is held

in equilibrium by a string which makes an angle of 40◦ with the rod, and is attached to

the wall above A. The tension in the string is 65N , the mass of the rod is 6kg and the

rod makes an angle of 70◦ with the upwards vertical. Find the position of the centre of

mass of the rod, and the magnitude and direction of the reaction at the hinge.



11.3 Three non–parallel forces in equilibrium

If three forces are not concurrent, as shown in the diagram, then the moment about A,

intersection of F1 and F2, can never be zero, and the forces cannot be in equilibrium.

A

F

F

F

Thus, if three forces are in equilibrium, their lines of action must pass through

one point.

F

F

F

Note that for the three forces to be in equilibrium, the sum of the resolved forces in any

direction must also be zero.

Example 11.3.1:

A non-uniform rod of length 6m and mass m kg is supported at its ends by two strings,

which make angles of 50◦ and 35◦ with the horizontal, as shown.

TT

mg

If the rod is horizontal and in equilibrium, find the position of its centre of mass.


CHAPTER 11. STATICS OF RIGID BODIES 68

11.4 Triangle of forces

If three forces P,Q and R are in equilibrium, then their (vector) sum must be zero. Thus the

three forces must form a triangle.

Example 11.4.1:

The three forces shown are in equilibrium. Find the magnitudes of P and Q.

70o 65o

45o P

Q

50

115o

135o 110o

P Q

50

Appendix A

Conservation of linear momentum, C.L.M.

Two balls, masses m1 and m2, are moving with speeds u1 and u2. They collide and after impact

are moving with speeds v1 and v2, as shown in the diagram. There are no external impulses

acting on the system. Let the internal impulse between the balls be I, acting in opposite di-

rections on each ball.

before during after

m2 m1 m2 m1 m2 m1

u1 v1 u2 v2

I I

Using I = mv −mu, for m1.−I = m1v1 −m1u1,

for m2,

I = m2v2 −m2u2.

Which can be shown,

m1u1 +m2u2 = m1v1 +m2v2.

or total momentum before collision = total momentum after collision. Thus linear

momentum is conserved, giving the Conservation of Linear Momentum, C.L.M.

Note. Note that if there is an external impulse acting on the system then the momentum

perpendicular to that impulse is conserved.

69

Appendix B

Centre of mass of n particles

M2 OCTOBER 2016 SDB 27

Appendix

Centre of mass of n particles

Consider three particles with masses m1, m2 and m3 at

points with position vectors r1, r2 and r3.

Let the force of m1 on m2 be Q12, of m2 on m1 be

Q21. Then Q12 and Q21 are internal forces and

Q12 + Q21 = 0.

The other internal forces are defined in a similar way.

Let P1, P2 and P3 be external forces on m1, m2 and m3.

Newton’s second law for each particle gives

P1 + Q21 + Q31 = m1

P2 + Q32 + Q12 = m2

P3 + Q13 + Q23 = m3

Adding P1 + P2 + P3 = m1 + m2 + m3 I all internal forces, Q, cancel

Define =

=

= ( II

From I and II P1 + P2 + P3 = (

the point moves as if all the mass was concentrated at that point, and all the external

forces acted at that point. This point, , is called the centre of mass.

This can be generalised for n particles to give

M =

M =

M = and M =

where M is the total mass, M =

m1

Q12

m2

m3

Q21

Q32

Q23

Q13 Q31

P1

P3

P2

70

Appendix C

Medians of a triangle



A median of a triangle is a line joining one vertex to the mid-point of the opposite side.

Let BE and CF be medians of the triangle ABC.

F and E are the mid-points of the sides AC and AB

ABC is an enlargement of AFE, scale factor 2

FE =

BC and

FE is parallel to BC.

XFE = XCB and XEF = XBC

triangles XFE and XCB are similar in the ratio 2 : 1

FX and XC are corresponding sides FX =

XC

also EX =

XB.

Thus X lies on the two medians, dividing each one in the ratio 2 : 1.

Similarly, X lies on the third median, AD, in the ratio 2 : 1

the medians meet in a point, X, which divides each median in the ratio 2 : 1.

Centre of mass of a triangle

Divide the triangle into narrow strips parallel to one

side.

The centre of mass of each strip will be at the centre of

each strip

the centre of mass of the triangle must lie on the line

joining these centres of mass

the centre of mass lies on the median of the triangle.

Similarly the centre of mass of the triangle lies on the other two medians, and therefore lies at

the intersection of the medians.

G is the centre of mass of the triangle, and

AG : GD = BG : GE = CG : GF = 2 : 1

G

C

B

E

A

F

D

C

A

B

E F

X

71

Appendix D




A median of a triangle is a line joining one vertex to the mid-point of the opposite side.

Let BE and CF be medians of the triangle ABC.

F and E are the mid-points of the sides AC and AB

ABC is an enlargement of AFE, scale factor 2

FE =

BC and

FE is parallel to BC.

XFE = XCB and XEF = XBC

triangles XFE and XCB are similar in the ratio 2 : 1

FX and XC are corresponding sides FX =

XC

also EX =

XB.

Thus X lies on the two medians, dividing each one in the ratio 2 : 1.

Similarly, X lies on the third median, AD, in the ratio 2 : 1

the medians meet in a point, X, which divides each median in the ratio 2 : 1.


Divide the triangle into narrow strips parallel to one

side.

The centre of mass of each strip will be at the centre of

each strip

the centre of mass of the triangle must lie on the line

joining these centres of mass

the centre of mass lies on the median of the triangle.

Similarly the centre of mass of the triangle lies on the other two medians, and therefore lies at

the intersection of the medians.

G is the centre of mass of the triangle, and

AG : GD = BG : GE = CG : GF = 2 : 1

G

C

B

E

A

F

D

C

A

B

E F

X

72

Appendix E

Centre of mass g


Centre of mass a + b + c)

We know that the centre of mass of the triangle ABC lies on

the median AD in the ratio AG : GD = 2 : 1.

A, B, C, D and G have position vectors a, b, c, d and .

D is the mid-point of BC

d =

(b + c)

= d a =

(b + c) a

=

= = + = a +

=

G

C

B

A

D

73

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CoverNoteContentsI Year 121 Mathematical Models in Mechanics2 Vectors in Mechanics2.1 Magnitude and direction components2.2 Parallel vectors2.3 Adding vectors2.4 Resolving vectors in two perpendicular components2.5 Vector algebra2.6 Vectors in mechanics2.7 Velocity and displacement2.8 Relative displacement vectors2.9 Collision of moving particles2.10 Closest distance between moving particles2.11 Relative velocity

3 Kinematics of a particle moving in a straight line3.1 Constant acceleration formulae3.2 Vertical motion under gravity3.3 Speed–time graphs

4 Statics of a particle4.1 Resultant forces4.2 Resultant of three or more forces4.3 Equilibrium of a particle under coplanar forces4.4 Types of force4.5 Friction4.6 Coefficient of friction4.7 Limiting equilibrium

5 Dynamics of a particle moving in a straight line5.1 Newton's laws of motion5.2 Connected particles5.3 Particles connected by pulleys5.4 Force on pulley5.5 Impulse and Momentum5.6 Internal and External Forces and Impulses5.7 Conservation of linear momentum – CLM5.8 Impulse in string between two particles

6 Moments6.1 Moment of a Force6.2 Sum of moments6.3 Moments and Equilibrium6.4 Non-uniform rods6.5 Nearly tilting rods

II Year 137 Kinematics7.1 Constant acceleration in a vertical plane7.2 Variable acceleration7.3 Using vectors

8 Centres of mass8.1 Centre of mass of n particles8.2 Centres of mass of simple laminas8.3 Centres of mass of composite laminas8.4 Laminas suspended freely under gravity8.4.1 Body with point mass hanging freely

8.5 Toppling on a slope8.6 Centres of mass of wire frameworks

9 Work, energy and power9.1 Definitions9.2 Work done by a (constant) force9.2.1 Forces parallel to the displacement 9.2.2 Forces at an angle to the displacement9.2.3 Work done by gravity

9.3 Work–energy equation9.4 Potential energy9.5 Power

10 Collisions10.1 Impulse and momentum – using vectors10.1.1 Impulse = change in momentum10.1.2 Conservation of linear momentum

10.2 Newton's law of restitution10.2.1 Coefficient of restitution10.2.2 Collisions with a plane surface10.2.3 Multiple collisions

10.3 Kinetic energy and impulses/collisions

11 Statics of rigid bodies11.1 Moment of a force11.2 Equilibrium11.3 Three non–parallel forces in equilibrium11.4 Triangle of forces

AppendicesA Conservation of linear momentum, C.L.M.B Centre of mass of n particlesC Medians of a triangleD Centre of mass of a triangleE Centre of mass g

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Notes for Edexcel Mechanics · Edexcel Mechanics Keywords: Mechanics,A Level, Edexcel Created Date: 10/26/2019 5:36:15 PM