This document provides an introduction to dynamics and forces acting on particles moving in a straight line. It introduces Newton's second law of motion, which states that force is equal to mass times acceleration (F=ma). It defines key concepts like weight, normal reaction force, friction, tension, and thrust. Examples are provided on using F=ma to calculate acceleration given force and mass, or force given mass and acceleration. The document also discusses resolving forces into perpendicular components and using this to solve problems involving multiple forces acting on an object.

2. Introduction
• This chapter teaches you how to deal with
forces acting on an object
• You will learn how to use several formulae
(inspired by Isaac Newton)
• You will learn how to model situations
involving friction, particles moving on slopes
and when joined over pulleys
• You will also learn laws of momentum and
impulse

4. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
You need to understand all the forces at work in
various situations…
R
The Normal Reaction
The normal reaction acts
perpendicular to the
surface which an object is
resting on
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
mg (mass
x gravity)
It is equal and opposite to
the force exerted on the
surface by the object,
which is determined largely
by gravity and the mass of
the object
The table matches the force from the brick,
which is why the brick remains still on the
table (there of course would be a maximum
possible weight the table could take, but we
will not worry about this for now!
3A

5. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
You need to understand all the forces at work in
various situations…
Frictional
Force
Direction of
motion
Frictional Force
The frictional force opposes
motion between two ‘rough’
surfaces
Although it is a force, friction does not
cause movement in its own direction. It
just reduces the effect of another
force
Surfaces will have a maximum level of
friction where it is unable to completely
prevent movement
3A

6. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
You need to understand all the forces at work in
various situations…
Tension in
string
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
Tension
If an object is being pulled
along (for example by a string),
then the force acting on the
object is called the Tension
Tension = PULLING force
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

7. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
You need to understand all the forces at work in
various situations…
Thrust
Thrust
If an object is being pushed along (for
example by a rod), then the force acting on
the object is called the Thrust (or sometimes
compression)
Tension = PUSHING force
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

8. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
You need to understand all the forces at work in
various situations…
Resistance
Any object moving through air, fluid or a solid
will experience resistance caused by the
particles in the way
Gravity
F = ma
Gravity is the force between any object and
the earth.
 The Force caused by gravity acting on an
object is its weight
Force is measured in Newtons (N). A Newton is:
 Remember Newton’s formula…
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
𝐹 = 𝑚𝑎
𝑊 = 𝑚𝑔
 The Force is called the weight
 Mass is just mass!
 The acceleration due to gravity is
9.8ms-2 (or can be left as ‘g’
3A

9. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
You need to understand all the forces at work in
various situations…
Resolving
When there are multiple forces acting on an
object, we ‘resolve’ these forces in different
directions
 One direction will usually be the direction of
acceleration
 The other will be perpendicular to this
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

10. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Normal
R Reaction
𝐹 = 𝑚𝑎
𝑊 = 𝑚𝑔
Tension
mg (mass
x gravity)
Frictional
Force
Direction of
motion
Thrust
3A

11. 𝐹 = 𝑚𝑎
Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the weight in Newtons, of a particle
of mass 12kg
𝐹 = 𝑚𝑎
𝐹 = 12 × 9.8
The mass is already in kg, and use
acceleration due to gravity
Calculate
𝐹 = 117.6𝑁
𝐹 = 120𝑁
As the acceleration was given to
2sf, you should give you answer
to the same accuracy
 Ensure you use the exact
amount in any subsequent
calculations though!
3A

12. 𝐹 = 𝑚𝑎
Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the acceleration when a particle of
mass 1.5kg is acted on by a force of 6N
𝐹 = 𝑚𝑎
Sub in F and m
6 = 1.5𝑎
Divide by 1.5
4= 𝑎
𝑎 = 4𝑚𝑠 −2
3A

13. 𝐹 = 𝑚𝑎
Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the values of the missing forces
acting on the object in the diagram below
In this example you need to consider the horizontal
forces and vertical forces separately (This is called
resolving)
Resolving Horizontally
Take the direction of acceleration as the positive one
𝐹 = 𝑚𝑎
2ms-2
𝑋 − 4 = (2 × 2)
Y
𝑋−4=4
𝑋 = 8𝑁
4N
2kg
X
Sub in values. You must subtract any
forces acting in the opposite direction!
Calculate
Add 4
Resolving Vertically
Take the direction of the force Y as positive
𝐹 = 𝑚𝑎
2g N
𝑌 − 2𝑔 = (2 × 0)
𝑌 − 2𝑔 = 0
𝑌 = 2𝑔 (19.6𝑁)
Sub in values. Acceleration is 0 as
there is none in the vertical direction
Calculate
Add 2g
3A

14. 𝐹 = 𝑚𝑎
Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the values of the missing forces
acting on the object in the diagram below
In this example you need to consider the horizontal
forces and vertical forces separately (This is called
resolving)
Resolving Horizontally
Take the direction of acceleration as the positive one
𝐹 = 𝑚𝑎
2ms-2
80 − 𝑋 = (4 × 2)
Y
80 − 𝑋 = 8
20N
72𝑁 = 𝑋
80N
4kg
X
Sub in values. You must subtract any
forces acting in the opposite direction!
Calculate
Add X and
Subtract 8
Resolving Vertically
Take the direction of the force Y as positive
𝐹 = 𝑚𝑎
4g N
𝑌 − 20 − 4𝑔 = (4 × 0)
𝑌 − 20 − 4𝑔 = 0
𝑌 = 20 + 4𝑔 (59.2𝑁)
Sub in values. Acceleration
is 0 as there is none in the
vertical direction
Calculate
Add 20, add 4g
3A

16. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
a) The acceleration of the particle
b) The distance travelled by the particle in
the first 4 seconds
c) The magnitude of the normal reaction
between the particle and the table
Start by drawing
a diagram
R
4N
5kg
20N
5g N
a)
𝐹 = 𝑚𝑎
20 − 4 = (5 × 𝑎)
𝑎 = 3.2𝑚𝑠 −2
Resolve horizontally and sub in
values. Take the direction of
acceleration as positive
Calculate a
3B

17. Dynamics of a Particle moving in a
Straight Line
3.2ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
3.2ms-2
a) The acceleration of the particle –
b) The distance travelled by the particle in
the first 4 seconds
c) The magnitude of the normal reaction
between the particle and the table
Start by drawing
a diagram
R
4N
5kg
20N
Use SUVAT
5g N
b) 𝑠 = ?
𝑢=0
𝑠 = 𝑢𝑡 +
𝑣 =?
1 2
𝑎𝑡
2
1
𝑠 = (0 × 4) + (3.2)(42 )
2
𝑎 = 3.2
𝑡=4
Sub in values
Calculate
𝑠 = 25.6𝑚
3B

18. Dynamics of a Particle moving in a
Straight Line
3.2ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
Start by drawing
a diagram
R
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
4N
a) The acceleration of the particle – 3.2ms-2
b) The distance travelled by the particle in
the first 4 seconds – 25.6m
c) The magnitude of the normal reaction
between the particle and the table
c)
5kg
20N
5g N
𝐹 = 𝑚𝑎
𝑅 − 5𝑔 = (5 × 0)
𝑅 = 5𝑔 (49𝑁)
Resolve vertically, taking R
as the positive direction
Calculate
3B

19. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
T
2ms-2
A small pebble of mass 500g is attached
to the lower end of a light string. Find
the tension in the string when the pebble
is:
a)
Moving upwards with an acceleration
of 2ms-2
0.5kg
0.5g N
a)
𝐹 = 𝑚𝑎
𝑇 − 0.5𝑔 = (0.5 × 2)
b) Moving downwards with a
deceleration of 4ms-2
Start by drawing
a diagram
Resolve vertically, taking the
direction of the acceleration
as positive
Calculate T
𝑇 = 1 + 0.5𝑔
𝑇 = 5.9𝑁
3B

20. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
T
-4ms-2
A small pebble of mass 500g is attached
to the lower end of a light string. Find
the tension in the string when the pebble
is:
a)
Moving upwards with an acceleration
of 2ms-2 – 5.9N
0.5kg
0.5g N
b)
𝐹 = 𝑚𝑎
0.5𝑔 − 𝑇 = (0.5 × −4)
b) Moving downwards with a
deceleration of 4ms-2
Start by drawing
a diagram
In this case, the pebble is
moving downwards at a
decreasing rate, so you can
put the acceleration on as
negative
Resolve vertically, taking the
direction of movement as
positive
Calculate T
𝑇 = 2 + 0.5𝑔
𝑇 = 6.9𝑁
Even though the pebble is moving
downwards, there is more tension in the
string as the pebble is decelerating – the
string is working against gravity!
3B

21. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
A particle of mass 3kg is projected at an
initial speed of 10ms-1 in the horizontal
direction. As it travels, it meets a
constant resistance of magnitude 6N.
Calculate the deceleration of the particle
and the distance travelled by the time it
comes to rest.
a ms-2
R
6N
3kg
3g N
Start by drawing
a diagram
It is important to note that
the initial projection speed
is NOT a force, there are
actually no forces acting in
the positive direction
Deceleration
𝐹 = 𝑚𝑎
0 − 6 = (3 × 𝑎)
Take the direction of movement as
positive – remember to include 0
as the positive force!
Calculate a
−2 = 𝑎
So the deceleration is 2ms-2
3B

22. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
A particle of mass 3kg is projected at an
initial speed of 10ms-1 in the horizontal
direction. As it travels, it meets a
constant resistance of magnitude 6N.
Calculate the deceleration of the particle
and the distance travelled by the time it
comes to rest.
Deceleration = 2ms-2
a ms-2
Start by drawing
a diagram
R
6N
It is important to note that
the initial projection speed
is NOT a force, there are
actually no forces acting in
the positive direction
3kg
3g N
Distance travelled
𝑠 =?
𝑢 = 10
𝑣=0
𝑣 2 = 𝑢2 + 2𝑎𝑠
2
2
0 = 10 + 2(−2)𝑠
0 = 100 − 4𝑠
𝑎 = −2
𝑡 =?
Sub in values
Work through
to calculate s
𝑠 = 25𝑚
3B

24. Dynamics of a Particle moving in a
Straight Line
However, a force at an
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
A horizontal force has no effect on
the object in the vertical direction
angle will have some effect
in BOTH the horizontal and
vertical directions!
S
O
H
C
A
H
T
O
A
Opp = Sinθ x Hyp Cosθ x Hyp
Adj =
10N
Opp = Sin20Adj = Cos20 x 10
x 10
Hyp
10N
Opp
10sin20
20°
Adj
10N
A vertical force has
no effect on the
object in the
horizontal direction
10cos20
So a force can be split into its horizontal and
vertical components using Trigonometry!
3C

25. Dynamics of a Particle moving in a
Straight Line
y
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
9N
9Sin40
40°
Find the component of each force in the
x and y-directions
9Cos40
Force in the x-direction
x
Force in the y-direction
= 9Cos40
= 9Sin40
= 6.89N
= 5.79N
3C

26. Dynamics of a Particle moving in a
Straight Line
y
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
12N
12Sin23
Find the component of each force in the
x and y-directions
23°
12Cos23
Force in the x-direction
x
Force in the y-direction
= 12Cos23
= 12Sin23
= 11.05N
= 4.69N
= -11.05N
(This will be negative as it is
the opposite direction to x!)
3C

28. Dynamics of a Particle moving in a
Straight Line
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
It is dependent on two things:
1) The normal reaction between the two surfaces
2) The coefficient of friction between the two
surfaces
The maximum frictional force is calculated as
follows:
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
If a surface is described as ‘smooth’, the
implication is that the coefficient of friction is 0.
3D

29. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
Draw a diagram
R
F
5kg
10N
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
5g N
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a) P = 10N
b) P = 19.6N
c) P = 30N
We need to find the maximum possible frictional force
 To do this we need R, the normal reaction
𝐹 = 𝑚𝑎
𝑅 − 5𝑔 = (5 × 0)
𝑅 = 5𝑔 (49𝑁)
Resolve vertically
Calculate R
Now we can calculate the maximum possible frictional force
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.4 × 49)
𝐹 𝑀𝐴𝑋 = 19.6𝑁
Sub in values
Calculate FMAX
3D

30. Dynamics of a Particle moving in a
Straight Line
a ms-2
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
10N
F
Draw a diagram
𝐹 𝑀𝐴𝑋 = 19.6𝑁
5kg
10N
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
5g N
The maximum frictional force is 19.6 N
Any force will be opposed by friction up to
this value
For part a), the force is only 10N
 Therefore, the frictional force will match
this at 10N, preventing movement
 Hence, there is also no acceleration
a) P = 10N
b) P = 19.6N
c) P = 30N
3D

31. Dynamics of a Particle moving in a
Straight Line
a ms-2
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
19.6N
F
Draw a diagram
𝐹 𝑀𝐴𝑋 = 19.6𝑁
5kg
19.6N
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a) P = 10N
b) P = 19.6N
c) P = 30N
5g N
The maximum frictional force is 19.6 N
Any force will be opposed by friction up to
this value
For part b), the force is only 19.6N
 Therefore, the frictional force will match
this at 19.6N, preventing movement
 Hence, there is also no acceleration
 This situation is called ‘limiting
equilibrium’, as the object is on the point
of movement
3D

32. Dynamics of a Particle moving in a
Straight Line
a ms-2
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
19.6N
F
Draw a diagram
𝐹 𝑀𝐴𝑋 = 19.6𝑁
5kg
30N
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
5g N
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a) P = 10N
b) P = 19.6N
c) P = 30N
For part c), the force is 30N
The frictional force will oppose 19.6N of this,
but no more.
 Hence, the object will accelerate…
 Resolve horizontally!
𝐹 = 𝑚𝑎
30 − 19.6 = 5 × 𝑎
10.4 = 5𝑎
2.08 = 𝑎
Sub in values and resolve
horizontally
Calculate
Divide by 5
So the acceleration will be 2.08ms-2
3D

33. Dynamics of a Particle moving in a
Straight Line
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
F
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a) P is applied horizontally
b) P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
Draw a diagram
5kg
P
5g N
Resolve vertically to find the normal reaction
𝐹 = 𝑚𝑎
𝑅 − 5𝑔 = (5 × 0)
Sub in values and
resolve vertically
Calculate
𝑅 = 49𝑁
Now find the maximum frictional force
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.5 × 49)
Sub in values
Calculate
𝐹 𝑀𝐴𝑋 = 24.5𝑁
So P will have to exceed 24.5N to make the object move!
3D

34. Dynamics of a Particle moving in a
Straight Line
Draw a diagram
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
F
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a) P is applied horizontally – 24.5N
b) P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
0.6P
Psinθ
θ
5kg
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
P
Pcosθ
0.8P
5g N
We need to find the values of Cosθ and Sinθ. The
ratio for Tanθ can be used to find these!
S
O
We can find the
hypotenuse using
Pythagoras’
Theorem!
H
C
A
H
T
Hyp
5
O
Tanθ = 3/4
A
So Opp = 3
And Adj = 4
3 Opp
θ
4
Adj
Sinθ = 3/5
Cosθ = 4/5
Sinθ = 0.6
Cosθ = 0.8
3D

35. Dynamics of a Particle moving in a
Straight Line
Draw a diagram
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a) P is applied horizontally – 24.5N
b) P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
F
5kg
P
0.6P
θ
0.8P
5g N
Resolve vertically to find the normal reaction
𝐹 = 𝑚𝑎
𝑅 + 0.6𝑃 − 5𝑔 = (5 × 0)
Sub in values and
resolve vertically
𝑅 = 49 − 0.6𝑃
We find the normal
reaction in terms of P
Now find the maximum frictional force
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = 0.5(49 − 0.6𝑃)
𝐹 𝑀𝐴𝑋 = 24.5 − 0.3𝑃
Sub in
values
Find Fmax in
terms of P
So, 0.8P will have to exceed this if the box is to
move…
3D

36. Dynamics of a Particle moving in a
Draw a diagram
Straight Line
𝐹
= 24.5 − 0.3𝑃
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑅 = 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a) P is applied horizontally – 24.5N
b) P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
24.5 – 0.3P
F
5kg
P
θ
𝑀𝐴𝑋
0.6P
0.8P
5g N
We need to find the value for P for which the
box is in ‘limiting equilibrium’ – that is, so the
horizontal forces cancel each other out…
Resolve horizontally…
𝐹 = 𝑚𝑎
0.8𝑃 − (24.5 − 0.3𝑃) = (5 × 0)
0.8𝑃 − 24.5 + 0.3𝑃 = 0
1.1𝑃 = 24.5
Sub in values and
resolve horizontally
Careful with the
bracket!
Rearrange and
solve
𝑃 = 22𝑁 (2𝑠𝑓)
P must exceed 22N, which is less than when P was horizontal
 The reason is because some of the force is upwards, this
alleviates some of the friction between the surfaces…
3D

38. Dynamics of a Particle moving in a
Straight Line
R
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
When an object is on an inclined plane,
we consider the forces acting parallel
to the plane and perpendicular to the
plane (instead of vertically and
horizontally)
This is because any movement will be
parallel to the plane (and we always
then consider the direction which is
perpendicular to any movement)
However, gravity will always work in a
vertical direction so must be split into
parallel and perpendicular directions…
60°
3g
30°
90°
30° 3gCos30
3gSin30
Above is a box resting on a plane inclined at an angle of 30° to
the horizontal
 Label gravity, which always acts vertically downwards
 Gravity must then be split into the parallel and
perpendicular components
 The angle in the triangle created is the same as the
angle the plane is inclined at (if you work out angles you
can see why!)
 Make sure you think carefully about which is Sine and
which is Cosine!
 Don’t forget the normal reaction, and any other forces
which are involved in the question!
3E

39. Dynamics of a Particle moving in a
Straight Line
R
2N
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
2gCos20
A box of mass 2kg is resting on a
smooth plane inclined at an angle of
20° to the horizontal. It meets
resistance of 2N as it travels down
the slope
a) Calculate the acceleration of the
box down the slope
b) If the box starts 10m up the plane,
calculate the velocity of the box at
the bottom of the plane
2g
20°
20°
2gSin20
As the plane is ‘smooth’, there is no need to consider friction or
the normal reaction
Resolve parallel to the plane
𝐹 = 𝑚𝑎
2𝑔𝑆𝑖𝑛20 − 2 = (2 × 𝑎)
19.6𝑆𝑖𝑛20 − 2 = 2𝑎
2.4 = 𝑎 (2𝑠𝑓)
Sub in values and
resolve parallel
Work out some parts (to keep
accuracy)
Round to 2sf as gravity is given
to this degree of accuracy
3E

40. Dynamics of a Particle moving in a
Straight Line
R
2N
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A box of mass 2kg is resting on a
smooth plane inclined at an angle of
20° to the horizontal. It meets
resistance of 2N as it travels down
the slope
a) Calculate the acceleration of the
box down the slope – 2.4ms-2
b) If the box starts 10m up the plane,
calculate the velocity of the box at
the bottom of the plane
2gCos20
2g
20°
20°
2gSin20
𝑠 = 10
𝑢=0
𝑣 =?
𝑣 2 = 𝑢2 + 2𝑎𝑠
2
2
𝑣 = 0 + 2(2.4)(10)
𝑣 2 = 02 + 2(2.4)(10)
2
𝑣 = 47.03 …
𝑣 = 6.6𝑚𝑠 −1 (2𝑠𝑓)
𝑎 = 2.4
𝑡 =?
Sub in values
Remember to use the exact
value for a, not the rounded one!
Calculate
Square root
3E

41. Dynamics of a Particle moving in a
Straight Line
0.8mg
R
a
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A particle is held at rest on a rough
plane inclined at an angle of θ to the
horizontal, where Tanθ is 0.75. If the
coefficient of friction between the
particle and the plane is 0.5, find the
acceleration of the particle.
You can find the below values (based
on tanθ = 0.75) by drawing a right
angled triangle and finding the
hypotenuse (as in section 3D)
Tanθ = 3/4
Sinθ = 3/5 (0.6)
Cosθ = 4/5 (0.8)
F
mg
θ
θ
mgCosθ
0.8mg
mgSinθ
We need the normal reaction in order to find the maximum
frictional force
 Resolve perpendicular to the plane
𝐹 = 𝑚𝑎
𝑅 − 𝑚𝑔𝐶𝑜𝑠θ = (𝑚 × 0)
𝑅 = 𝑚𝑔𝐶𝑜𝑠θ
𝑅 = 0.8𝑚𝑔
Resolve perpendicular
to the plane
We have to use ‘m’ for now as
we do not know the mass…
We know the value of Cosθ
3E

42. Dynamics of a Particle moving in a
Straight Line
0.8mg
a
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A particle is held at rest on a rough
plane inclined at an angle of θ to the
horizontal, where Tanθ is 0.75. If the
coefficient of friction between the
particle and the plane is 0.5, find the
acceleration of the particle.
You can find the below values (based
on tanθ = 0.75) by drawing a right
angled triangle and finding the
hypotenuse (as in section 3D)
0.4mg
F
mg
θ
θ
0.8mg
mgSinθ
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.5 × 0.8𝑚𝑔)
𝐹 𝑀𝐴𝑋 = 0.4𝑚𝑔
Sub in values
Calculate
Tanθ = 3/4
Sinθ = 3/5 (0.6)
Cosθ = 4/5 (0.8)
3E

43. Dynamics of a Particle moving in a
Straight Line
0.8mg
a
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A particle is held at rest on a rough
plane inclined at an angle of θ to the
horizontal, where Tanθ is 0.75. If the
coefficient of friction between the
particle and the plane is 0.5, find the
acceleration of the particle.
You can find the below values (based
on tanθ = 0.75) by drawing a right
angled triangle and finding the
hypotenuse (as in section 3D)
Tanθ = 3/4
Sinθ = 3/5 (0.6)
Cosθ = 4/5 (0.8)
0.4mg
mg
θ
0.8mg
mgSinθ
θ
Now we have all the forces involved acting in the required
directions, we can now calculate the acceleration of the particle…
 Resolve parallel to the plane
𝐹 = 𝑚𝑎
𝑚𝑔𝑆𝑖𝑛θ − 0.4𝑚𝑔 = 𝑚𝑎
𝑔𝑆𝑖𝑛θ − 0.4𝑔 = 𝑎
(9.8 × 0.6) − (0.4 × 9.8) = 𝑎
Sub in values and resolve
parallel
The m’s cancel (meaning the mass
does not affect the acceleration!)
Calculate a (remember, we
know Sinθ)
2.0𝑚𝑠 −2 = 𝑎 (2𝑠𝑓)
3E

44. Dynamics of a Particle moving in a
2.5ms
Straight Line
-2
43N
R
F
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A mass of 4kg is pushed up a plane by
a horizontal force of magnitude 25N.
The plane is inclined to the horizontal
at 10° and accelerates at 2.5ms-2.
Calculate the coefficient of friction
between the box and the plane.
 In this type of question you should
proceed as if you did have the
coefficient of friction. You will
end up with an equation where you
can solve for µ.
25N
10°
25Sin10
25Cos10
4g
10°
4gCos10
4gSin10
10°
We need to split the forces into parallel and perpendicular components…
Now we can resolve perpendicular to find the normal reaction, and
hence, the maximum frictional resistance created
Resolve perpendicular to the plane…
𝐹 = 𝑚𝑎
𝑅 − 25𝑆𝑖𝑛10 − 4𝑔𝐶𝑜𝑠10 = (4 × 0)
Sub in values and resolve
perpendicular
𝑅 = 25𝑆𝑖𝑛10 + 4𝑔𝐶𝑜𝑠10
Rearrange
Calculate
𝑅 = 43𝑁 (2𝑠𝑓)
3E

45. Dynamics of a Particle moving in a
2.5ms
Straight Line
-2
43N
43µ
F
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A mass of 4kg is pushed up a plane by
a horizontal force of magnitude 25N.
The plane is inclined to the horizontal
at 10° and accelerates at 2.5ms-2.
Calculate the coefficient of friction
between the box and the plane.
 In this type of question you should
proceed as if you did have the
coefficient of friction. You will
end up with an equation where you
can solve for µ.
25N
10°
25Sin10
25Cos10
4g
10°
4gCos10
4gSin10
10°
Now we can find the maximum frictional force created between the
surfaces…
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = 𝜇 × 43
𝐹 𝑀𝐴𝑋 = 43𝜇
Sub in values (remember to use exact
values, not rounded ones)
Leave in terms of µ
3E

46. Dynamics of a Particle moving in a
2.5ms
Straight Line
-2
43N
43µ
You can extend this process to
particles on an inclined plane, by
considering forces parallel and
perpendicular to the plane
A mass of 4kg is pushed up a plane by
a horizontal force of magnitude 25N.
The plane is inclined to the horizontal
at 10° and accelerates at 2.5ms-2.
Calculate the coefficient of friction
between the box and the plane.
25N
10°
25Sin10
25Cos10
4g
10°
4gCos10
4gSin10
10°
Now we have all the forces acting perpendicular to the plane, we can
find the value of µ
Resolve parallel to the plane…
 In this type of question you should
𝐹 = 𝑚𝑎
proceed as if you did have the
coefficient of friction. You will
25𝐶𝑜𝑠10 − 43𝜇 − 4𝑔𝑆𝑖𝑛10 = (4 × 2.5)
end up with an equation where you
can solve for µ.
25𝐶𝑜𝑠10 − 4𝑔𝑆𝑖𝑛10 − (4 × 2.5) = 43𝜇
7.8131 … = 43𝜇
0.18 = 𝜇
Sub in values and resolve
parallel
Rearrange to find µ
Solve exactly
Remember to use
exact values!!
3E

48. Dynamics of a Particle moving in a
Straight Line
a
You can solve problems involving connected
particles by considering the particles
separately
If the system involves the motion of more than
one particle, you can consider them separately
If all parts of the system are moving in the
same straight line, you can treat the whole
system as a single particle
Two particles, P and Q, of masses 5kg and 3kg
respectively, are connected by a light
inextensible string. Particle P is pulled along by
a horizontal force of magnitude 40N along a
rough horizontal plane. The coefficient of
friction between the blocks and the plane is
0.2.
a)
c)
Find the acceleration of each particle
b) Find the tension in the string
Explain how the assumption that the string
is light and inextensible has been used in
the question
Q
5g
R1
3g
R2
T
0.6g
F2
3kg
3g
P
5kg
40N
T
Fg
1
5g
The only force acting vertically is the weight, so the normal
reaction for each particle will be equal to this…
We can now work out the maximum friction for each particle…
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.2 × 5𝑔)
𝐹 𝑀𝐴𝑋 = 𝑔 𝑁
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.2 × 3𝑔)
𝐹 𝑀𝐴𝑋 = 0.6𝑔 𝑁
Sub in values for particle P
Calculate (in terms of g)
Sub in values for particle Q
Calculate (in terms of g)
3F

49. Dynamics of a Particle moving in a
Straight Line
3.04ms-2
a
You can solve problems involving connected
particles by considering the particles
separately
If the system involves the motion of more than
one particle, you can consider them separately
If all parts of the system are moving in the
same straight line, you can treat the whole
system as a single particle
Two particles, P and Q, of masses 5kg and 3kg
respectively, are connected by a light
inextensible string. Particle P is pulled along by
a horizontal force of magnitude 40N along a
rough horizontal plane. The coefficient of
friction between the blocks and the plane is
0.2.
a)
c)
Find the acceleration of each particle
b) Find the tension in the string
Explain how the assumption that the string
is light and inextensible has been used in
the question
Q
5g
3g
T
0.6g
3kg
P
5kg
40N
T
g
3g
5g
Now we can calculate the acceleration for the system as a
whole, since it is all travelling in the same direction
 You will need to consider all forces except the tension in the
string. As it is connecting the particles it is effectively
cancelling itself out!
𝐹 = 𝑚𝑎
40 − 𝑔 − 0.6𝑔 = (8 × 𝑎)
24.32 = 8𝑎
Resolve parallel to the plane. Use
the total mass of the particles
Calculate
Divide by 8
3.04 = 𝑎
As the particles are connected, they will both accelerate
together (the string is ‘inextensible’, meaning a fixed length)
3F

50. Dynamics of a Particle moving in a
Straight Line
3.04ms-2
You can solve problems involving connected
particles by considering the particles
separately
If the system involves the motion of more than
one particle, you can consider them separately
If all parts of the system are moving in the
same straight line, you can treat the whole
system as a single particle
Two particles, P and Q, of masses 5kg and 3kg
respectively, are connected by a light
inextensible string. Particle P is pulled along by
a horizontal force of magnitude 40N along a
rough horizontal plane. The coefficient of
friction between the blocks and the plane is
0.2.
a)
c)
Find the acceleration of each particle –
3.04ms-2
b) Find the tension in the string
Explain how the assumption that the string
is light and inextensible has been used in
the question
Q
5g
3g
T
0.6g
3kg
P
5kg
40N
T
g
3g
5g
Calculating the tension for particle P
Only include forces acting on P, and resolve parallel
𝐹 = 𝑚𝑎
40 − 𝑔 − 𝑇 = (5 × 3.04)
40 − 𝑔 − (5 × 3.04) = 𝑇
15 = 𝑇
Resolve parallel to the
plane. Only particle P
should be considered
Rearrange for T
Calculate
3F

51. Dynamics of a Particle moving in a
Straight Line
3.04ms-2
You can solve problems involving connected
particles by considering the particles
separately
If the system involves the motion of more than
one particle, you can consider them separately
If all parts of the system are moving in the
same straight line, you can treat the whole
system as a single particle
Two particles, P and Q, of masses 5kg and 3kg
respectively, are connected by a light
inextensible string. Particle P is pulled along by
a horizontal force of magnitude 40N along a
rough horizontal plane. The coefficient of
friction between the blocks and the plane is
0.2.
a)
Find the acceleration of each particle –
3.04ms-2
b) Find the tension in the string – 15N
c) Explain how the assumption that the string
is light and inextensible has been used in
the question
Q
5g
T
0.6g
3kg
3g
P
5kg
3g
40N
T
g
5g
Imagine we calculated the tension from particle Q instead…
Only include forces acting on Q, and resolve parallel
𝐹 = 𝑚𝑎
𝑇 − 0.6𝑔 = (3 × 3.04)
𝑇 = 3 × 3.04 + 0.6𝑔
𝑇 = 15
Resolve parallel to the plane.
Only particle Q should be
considered
Rearrange for T
Calculate
So it does not matter which particle we choose –
we get the same value for the tension either way!
3F

52. Dynamics of a Particle moving in a
Straight Line
3.04ms-2
You can solve problems involving connected
particles by considering the particles
separately
If the system involves the motion of more than
one particle, you can consider them separately
If all parts of the system are moving in the
same straight line, you can treat the whole
system as a single particle
Two particles, P and Q, of masses 5kg and 3kg
respectively, are connected by a light
inextensible string. Particle P is pulled along by
a horizontal force of magnitude 40N along a
rough horizontal plane. The coefficient of
friction between the blocks and the plane is
0.2.
Find the acceleration of each particle –
3.04ms-2
b) Find the tension in the string – 15N
c) Explain how the assumption that the string
is light and inextensible has been used in
the question
Q
5g
3g
T
0.6g
3kg
3g
P
5kg
40N
T
g
5g
How have we used the fact that the string is light and
inextensible?
Light – the string has no mass and the tension will be
consistent
Inextensible – Acceleration is the same across both masses
a)
3F

53. Dynamics of a Particle moving in a
Straight Line
10.3N
T
You can solve problems involving connected
particles by considering the particles
separately
0.5ms-2
A light scale-pan is attached to a vertical
light inextensible string. The scale pan
carries two masses, A and B. The mass of A
is 400g and the mass of B is 600g. A rests
on top of B.
The scale pan is raised vertically with an
acceleration of 0.5ms-2.
a) Find the Tension in the string
b) Find the force exerted on mass B by
mass A
c) Find the force exerted on mass B by
the scale pan
A
0.4g
To find the tension in
the string you should
consider the system as
a whole, as all the
forces will affect it!
B
0.6g
Resolving vertically
𝐹 = 𝑚𝑎
𝑇 − 0.4𝑔 − 0.6𝑔 = (1 × 0.5)
𝑇 = 1 × 0.5 + 1𝑔
Resolve vertically. There is no
normal reaction as the pan is
not on a surface
Rearrange to find T
Calculate
𝑇 = 10.3𝑁
3F

54. Dynamics of a Particle moving in a
Straight Line
0.5ms-2
A light scale-pan is attached to a vertical
light inextensible string. The scale pan
carries two masses, A and B. The mass of A
is 400g and the mass of B is 600g. A rests
on top of B.
A
0.4g
B
0.6g
The scale pan is raised vertically with an
acceleration of 0.5ms-2.
a) Find the Tension in the string – 10.3N
b) Find the force exerted on mass B by
mass A
c) Find the force exerted on mass B by
the scale pan
We cannot consider mass B on its own
at this point.
10.3N
You can solve problems involving connected
particles by considering the particles
separately
However, the force exerted on mass
B by mass A, will be the same as the
force exerted on mass A by mass B
 So we can consider mass A instead
(the scale pan is not acting on it)
Resolving forces on A
 R is the normal reaction, the force
of B acting on A
R
0.4kg A
0.4g
The reason is that the scale pan is
also acting on mass B, and we do not
know the magnitude of this force
0.5ms-2
B
𝐹 = 𝑚𝑎
𝑅 − 0.4𝑔 = (0.4 × 0.5)
Sub in forces
Calculate
𝑅 = 4.1𝑁 (2𝑠𝑓)
The magnitude of the force from B acting on A is 4.1N.
Therefore, the force from A acting on B must be equal to this!
(since the two masses are staying together)
3F

55. Dynamics of a Particle moving in a
Straight Line
Now as we have to involve the scale
pan, we will consider the forces acting
on Mass B
10.3N
You can solve problems involving connected
particles by considering the particles
separately
0.5ms-2
A light scale-pan is attached to a vertical
light inextensible string. The scale pan
carries two masses, A and B. The mass of A
is 400g and the mass of B is 600g. A rests
on top of B.
Draw a diagram for B, remember to
include the force exerted by A which
pushes down, and the force from the
scale pan which pushes up, from
beneath…
A
0.4g
B
0.6g
The scale pan is raised vertically with an
acceleration of 0.5ms-2.
a) Find the Tension in the string – 10.3N
b) Find the force exerted on mass B by
mass A – 4.1N
c) Find the force exerted on mass B by
the scale pan
Resolving forces on B
 S is the force exerted by the
scale pan on mass B
4.1N
0.6kg
B
0.6g
0.5ms-2
𝐹 = 𝑚𝑎
𝑆 − 4.1 − 0.6𝑔 = (0.6 × 0.5)
S
Sub in
forces
Calculate
𝑆 = 10.3𝑁
This type of question can be very tricky
to get the hang of – make sure you get
lots of practice!
3F

56. Dynamics of a Particle moving in a
Straight Line
Draw a diagram with all the
forces on…
You can solve problems involving connected
particles by considering the particles
separately
Particles P and Q, of masses 2m and 3m, are
attached to the ends of a light inextensible
string. The string passes over a small,
smooth, fixed pulley and the masses hang
with the string taut.
a) Find the acceleration of each mass
b) Find the tension in the string, in terms
of m
c) Find the force exerted on the pulley by
the string
d) Find the distance travelled by Q in the
first 4 seconds, assuming that P does
not reach the pulley
e) Comment on any modelling assumptions
used
T
T
T
The heavier particle will move
downwards, pulling the lighter
one upwards
T
3m
2m
a
P
2mg
Q
Sometimes you have to set up
two equations with the
information given, and combine
them…
a
3mg
Equation using P
𝐹 = 𝑚𝑎
𝑇 − 2𝑚𝑔 = 2𝑚𝑎
Equation using Q
Sub in
values
𝐹 = 𝑚𝑎
3𝑚𝑔 − 𝑇 = 3𝑚𝑎
Sub in
values
𝑇 − 2𝑚𝑔 = 2𝑚𝑎
3𝑚𝑔 − 𝑇 = 3𝑚𝑎
Add the equations
together
𝑚𝑔 = 5𝑚𝑎
1.96 = 𝑎
Cancel m’s and
divide g by 5
3F

57. Dynamics of a Particle moving in a
Straight Line
Draw a diagram with all the
forces on…
You can solve problems involving connected
particles by considering the particles
separately
Particles P and Q, of masses 2m and 3m, are
attached to the ends of a light inextensible
string. The string passes over a small,
smooth, fixed pulley and the masses hang
with the string taut.
a)
Find the acceleration of each mass –
1.96ms-2
b) Find the tension in the string, in terms
of m
c) Find the force exerted on the pulley by
the string
d) Find the distance travelled by Q in the
first 4 seconds, assuming that P does
not reach the pulley
e) Comment on any modelling assumptions
used
T
T
T
The heavier particle will move
downwards, pulling the lighter
one upwards
T
3m
2m
a
P
2mg
Q
Sometimes you have to set up
two equations with the
information given, and combine
them…
a
3mg
Equation using P
𝐹 = 𝑚𝑎
𝑇 − 2𝑚𝑔 = 2𝑚𝑎
Equation using Q
Sub in
values
𝑇 − 2𝑚𝑔 = 2𝑚𝑎
𝑇 = 2𝑚𝑎 + 2𝑚𝑔
𝐹 = 𝑚𝑎
3𝑚𝑔 − 𝑇 = 3𝑚𝑎
Rearrange
to find T
𝑇 = 2 × 1.96 𝑚 + 2 × 9.8 𝑚
𝑇 = 23.52𝑚
Sub in g
and a
Group up
for m
Sub in
values
3F

58. Dynamics of a Particle moving in a
Straight Line
The force on the pulley is the
tension on both sides – these
must be added together
You can solve problems involving connected
particles by considering the particles
separately
Particles P and Q, of masses 2m and 3m, are
attached to the ends of a light inextensible
string. The string passes over a small,
smooth, fixed pulley and the masses hang
with the string taut.
a)
Find the acceleration of each mass –
1.96ms-2
b) Find the tension in the string, in terms
of m – 23.52m
c) Find the force exerted on the pulley by
the string
d) Find the distance travelled by Q in the
first 4 seconds, assuming that P does
not reach the pulley
e) Comment on any modelling assumptions
used
T
23.52m
23.52m
T
T
2m
a
P
2mg
T
3m
Q
23.52m + 23.52m
= 47.04m
a
3mg
3F

59. Dynamics of a Particle moving in a
Straight Line
As P does not meet the
pulley, we assume Q moves
consistently
You can solve problems involving connected
particles by considering the particles
separately
Particles P and Q, of masses 2m and 3m, are
attached to the ends of a light inextensible
string. The string passes over a small,
smooth, fixed pulley and the masses hang
with the string taut.
Find the acceleration of each mass –
1.96ms-2
b) Find the tension in the string, in terms
of m – 23.52m
c) Find the force exerted on the pulley by
the string – 47.04m
d) Find the distance travelled by Q in the
first 4 seconds, assuming that P does
not reach the pulley
e) Comment on any modelling assumptions
used
23.52m
23.52m
T
T
3m
2m
a
P
a)
Q
2mg
𝑠 =?
3mg
𝑢=0
𝑠 = 𝑢𝑡 +
a
𝑣 =?
1 2
𝑎𝑡
2
1
𝑠 = 0 (4) + (1.96)(4)2
2
𝑎 = 1.96
𝑡=4
Sub in
values
Calculate
𝑠 = 15.7𝑚
3F

60. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving connected
particles by considering the particles
separately
Particles P and Q, of masses 2m and 3m, are
attached to the ends of a light inextensible
string. The string passes over a small,
smooth, fixed pulley and the masses hang
with the string taut.
a)
Find the acceleration of each mass –
1.96ms-2
b) Find the tension in the string, in terms
of m – 23.52m
c) Find the force exerted on the pulley by
the string – 47.04m
d) Find the distance travelled by Q in the
first 4 seconds, assuming that P does
not reach the pulley – 15.7metres
e) Comment on any modelling assumptions
used
23.52m
23.52m
T
2m
a
P
2mg
T
3m
Q
a
3mg
Comment on the modelling assumptions used:
Light string  The string has no mass
Inextensible string  The particles move with
the same acceleration
Smooth pulley – No Frictional force, tension
equal on both sides
3F

61. Dynamics of a Particle moving in a
Draw a diagram and
Straight Line0.4g
label all the forces
R
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
T
0.08g
F
T
A
T
0.4g
T
a
As we do not know the tension, we will have
to set up and solve 2 equations for a
 Remember particle A will also be
affected by friction – we need to know this
first…
a
B
0.8g
0.5m
Find the normal reaction for A (so we can then find the
frictional force)
 Resolve vertically for A
a) The acceleration of the system
b) The velocity at which B hits the ground
𝐹 = 𝑚𝑎
c) The total distance travelled by A
before it comes to rest
d) The force the string exerts on the 𝑅 − 0.4𝑔 = (0.4 × 0)
pulley
𝑅 = 0.4𝑔
Resolve
vertically
Rearrange…
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.2 × 0.4𝑔)
Sub
in
Solve
𝐹 𝑀𝐴𝑋 = 0.08𝑔
3F

62. Dynamics of a Particle moving in a
Draw a diagram and
Straight Line0.4g
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
T
0.08g
T
A
T
0.4g
T
0.6g
a
 Now we have the frictional force, we can
set up two equations for A and B
 As the particles are connected, resolving
horizontally for A and vertically for B are
equivalent…
Resolving horizontally for A
0.5m
0.8g
Resolving vertically for B
a) The acceleration of the system
𝐹 = 𝑚𝑎
𝐹 = 𝑚𝑎
Resolve
b) The velocity at which B hits the ground
horizontally
𝑇 − 0.08𝑔 = 0.4𝑎
0.8𝑔 − 𝑇 = 0.8𝑎
c) The total distance travelled by A
before it comes to rest
Add the two
d) The force the string exerts on the equations together 𝑇 − 0.08𝑔 = 0.4𝑎
pulley
 The T’s cancel out 0.8𝑔 − 𝑇 = 0.8𝑎
0.72𝑔 = 1.2𝑎
0.6𝑔 = 𝑎
0.6g
a
B
Divide by 1.2
Resolve
vertically
3F

63. Dynamics of a Particle moving in a
Draw a diagram and
Straight Line0.4g
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
a) The acceleration of the system – 0.6g
b) The velocity at which B hits the ground
c) The total distance travelled by A
before it comes to rest
d) The force the string exerts on the
pulley
T
T
A
0.08g
T
0.4g
T
0.6g
 We can use SUVAT to calculate the velocity
of B as it hits the ground
𝑠 = 0.5
𝑢=0
𝑣 =? 𝑎 = 0.6𝑔
𝑣 2 = 𝑢2 + 2𝑎𝑠
2
2
𝑣 = (0) +2(0.6𝑔)(0.5)
𝑣 2 = 0.6𝑔
𝑣 = 2.42𝑚𝑠
−1
𝑡 =?
B
0.8g
0.6g
0.5m
Sub in values
Calculate
Square
root
3F

64. Dynamics of a Particle moving in a
Draw a diagram and
Straight Line0.4g
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
a) The acceleration of the system – 0.6g
b) The velocity at which B hits the ground
– 2.42ms-1
c) The total distance travelled by A
before it comes to rest
d) The force the string exerts on the
pulley
T
0.08g
T
A
T
0.4g
T
0.2g
0.6g
Particle A will travel 0.5m by the time B
hits the floor
When B hits the floor, A will be moving at
speed (the same as B as it hit the floor…) and
will decelerate due to the frictional force…
 We need to know the deceleration of a…
𝐹 = 𝑚𝑎
𝑇 − 0.08𝑔 = (0.4 × 𝑎)
0 − 0.08𝑔 = 0.4𝑎
B
0.8g
0.6g
0.5m
Resolve
horizontally for A
T = 0 now as the
string will be slack
Divide by 0.4
−0.2𝑔 = 𝑎
3F

65. Dynamics of a Particle moving in a
Draw a diagram and
Straight Line0.4g
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
a) The acceleration of the system – 0.6g
b) The velocity at which B hits the ground
– 2.42ms-1
c) The total distance travelled by A
before it comes to rest
d) The force the string exerts on the
pulley
T
T
A
0.08g
T
0.4g
T
0.2g
Now we can use SUVAT again to find the
distance A travels before coming to rest…
𝑠 =?
𝑢 = 2.42
𝑣 = 0 𝑎 = −0.2𝑔
𝑣 2 = 𝑢2 + 2𝑎𝑠
02 = 2.422 + 2(−0.2𝑔)(𝑠)
0 = 5.88 − 3.92𝑠
𝑠 = 1.5𝑚
𝑠 = 2𝑚
𝑡 =?
0.6g
B
0.8g
0.5m
Sub in values (remember the
initial velocity of A)
Calculate
Rearrange to
find s
Remember to add on the 0.5m A
has already travelled!
3F

66. Dynamics of a Particle moving in a
Straight Line0.4g
Draw a diagram and
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
T
A
0.08g
3.136N
T
0.4g
T
0.6g
The two tensions are the forces acting on the
pulley
 We first need to know the value of T, and we
can use an equation from earlier to find it…
𝑇 − 0.08𝑔 = 0.4𝑎
a) The acceleration of the system – 0.6g
b) The velocity at which B hits the ground
– 2.42ms-1
c) The total distance travelled by A
before it comes to rest – 2m
d) The force the string exerts on the
pulley
3.136N
T
𝑇 − 0.08𝑔 = 0.4(0.6g)
B
0.8g
0.6g
0.5m
We worked out acceleration
earlier
Calculate T
𝑇 = 3.136
3F

67. Dynamics of a Particle moving in a
Straight Line0.4g
Draw a diagram and
label all the forces
You can solve problems involving connected
particles by considering the particles
separately
Two particles A and B of masses 0.4kg and
0.8kg respectively are connected by a light
inextensible string. Particle A lies on a
rough horizontal table 4.5m from a small
smooth fixed pulley which is attached to
the end of the table. The string passes over
the pulley and B hangs freely, with the
string taut, 0.5m above the ground. The
coefficient of friction between A and the
table is 0.2. The system is released from
rest. Find:
a) The acceleration of the system – 0.6g
b) The velocity at which B hits the ground
– 2.42ms-1
c) The total distance travelled by A
before it comes to rest – 2m
d) The force the string exerts on the
pulley
T
0.08g
3.136N
A
3.136N
0.4g
T
0.6g
The overall force on the pulley is the resultant
of the two tensions
 Since they are acting at 90° to each other,
the resultant will be at a 45° angle between
them (effectively an angle bisector)
3.136N
45°
45° 3.136N
Hyp
Adj
F
Opp
Draw a diagram and show
the resultant force
 Then you use GCSE Trig!
𝐻𝑦𝑝 =
𝐻𝑦𝑝 =
𝐴𝑑𝑗
𝐶𝑜𝑠𝜃
3.136
𝐶𝑜𝑠45
𝐻𝑦𝑝 = 4.43𝑁
𝐹 = 8.87𝑁
B
0.8g
0.6g
0.5m
Sub in values
This will be the
force from one
part of the
string
Double for the
total!
3F

68. Dynamics of a Particle moving in a
5gCos25 a
R
Straight Line
T
You can solve problems involving connected
particles by considering the particles
separately
Two particles, P and Q, of masses 5kg and
10kg are connected by a light inextensible
string. The string passes over a small
smooth pulley which is fixed at the top of a
plane inclined at an angle of 25° to the
horizontal. P is resting on the plane and Q
hangs freely with the string vertical and
taut. The coefficient of friction between P
and the plane is 0.2.
a)
Find the acceleration of the system
b) Find the tension in the string
Draw a diagram
 Remember to split the
forces into parallel and
perpendicular (where
appropriate!)
P
T
F
5g
25˚
25˚
5gCos25
5gSin25
Q
a
10g
We will need to form 2 equations, one for each particle
 For particle P, we need to calculate the frictional
force first
 Resolve perpendicular to find the normal reaction
𝐹 = 𝑚𝑎
𝑅 − 5𝑔𝐶𝑜𝑠25 = (5 × 0)
Resolve
perpendicular for P
Rearrange
𝑅 = 5𝑔𝐶𝑜𝑠25
3F

69. Dynamics of a Particle moving in a
5gCos25 a
Straight Line
T
You can solve problems involving connected
particles by considering the particles
separately
Two particles, P and Q, of masses 5kg and
10kg are connected by a light inextensible
string. The string passes over a small
smooth pulley which is fixed at the top of a
plane inclined at an angle of 25° to the
horizontal. P is resting on the plane and Q
hangs freely with the string vertical and
taut. The coefficient of friction between P
and the plane is 0.2.
a)
Find the acceleration of the system
Draw a diagram
 Remember to split the
forces into parallel and
perpendicular (where
appropriate!)
P
T
F
gCos25
5g
25˚
25˚
𝐹 𝑀𝐴𝑋 = 𝜇𝑅
𝐹 𝑀𝐴𝑋 = (0.2 × 5𝑔𝐶𝑜𝑠25)
𝐹 𝑀𝐴𝑋 = 𝑔𝐶𝑜𝑠25
5gCos25
5gSin25
Q
a
10g
Sub in values for particle P
Calculate
b) Find the tension in the string
3F

70. Dynamics of a Particle moving in a
5gCos25 a
Straight Line
T
You can solve problems involving connected
particles by considering the particles
separately
Two particles, P and Q, of masses 5kg and
10kg are connected by a light inextensible
string. The string passes over a small
smooth pulley which is fixed at the top of a
plane inclined at an angle of 25° to the
horizontal. P is resting on the plane and Q
hangs freely with the string vertical and
taut. The coefficient of friction between P
and the plane is 0.2.
a)
Find the acceleration of the system
b) Find the tension in the string
Draw a diagram
 Remember to split the
forces into parallel and
perpendicular (where
appropriate!)
P
T
gCos25
5g
25˚
25˚
5gCos25
Q
5gSin25
a
10g
Now we can form 2 equations using P and Q
Equation for P
 Resolve Parallel
𝐹 = 𝑚𝑎
Equation for Q
 Resolve Vertically
Resolve
Parallel for P
𝐹 = 𝑚𝑎
𝑇 − 𝑔𝐶𝑜𝑠25 − 5𝑔𝑆𝑖𝑛25 = 5𝑎
𝑇 − 𝑔𝐶𝑜𝑠25 − 5𝑔𝑆𝑖𝑛25 = 5𝑎
10𝑔 − 𝑇 = 10𝑎
10𝑔 − 𝑇 = 10𝑎
Add the equations together
 The T’s cancel out!
10𝑔 − 𝑔𝐶𝑜𝑠25 − 5𝑔𝑆𝑖𝑛25 = 15𝑎
4.56 = 𝑎
Resolve
Vertically
for Q
Solve for a
3F

71. Dynamics of a Particle moving in a
5gCos25 a
Straight Line
T
You can solve problems involving connected
particles by considering the particles
separately
Two particles, P and Q, of masses 5kg and
10kg are connected by a light inextensible
string. The string passes over a small
smooth pulley which is fixed at the top of a
plane inclined at an angle of 25° to the
horizontal. P is resting on the plane and Q
hangs freely with the string vertical and
taut. The coefficient of friction between P
and the plane is 0.2.
a)
Find the acceleration of the system –
4.56ms-2
Draw a diagram
 Remember to split the
forces into parallel and
perpendicular (where
appropriate!)
P
T
gCos25
5g
25˚
25˚
5gCos25
5gSin25
Q
a
10g
Use one of the previous equations to find the tension
10𝑔 − 𝑇 = 10𝑎
10𝑔 − 𝑇 = 10(4.56)
We now know
the acceleration
Solve for T
𝑇 = 52.4𝑁
b) Find the tension in the string
3F

73. Dynamics of a Particle moving in a
Straight Line
You can calculate the momentum of
a particle and the impulse of a force
The momentum of a body of mass m
which is moving with velocity v is given
by mv
If the mass is in kg and the velocity is
in ms-1 then the momentum will be in
kgms-1
kgms-1 can be written as (kgms-2)s
As kgms-2 is Newtons…
Find the magnitude of the momentum of:
a) A cricket ball of mass 400g moving at 18ms-1
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 0.4 × 18
Sub in units (remember
to use kg)
Calculate
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 7.2𝑁𝑠
b) A lorry of mass 5 tonnes moving at 12ms-1
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
Sub in units (remember
to use kg)
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 5000 × 12
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 60,000𝑘𝑔𝑚𝑠
−1
Calculate
kgms-1 can be written as Ns
These are both acceptable units for
momentum
Either Ns or kgms-1 are acceptable units (make
sure you read the question in case you’re asked
for one specifically!)
3G

74. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
Dynamics of a Particle moving in a
Straight Line
You can calculate the momentum of
a particle and the impulse of a force
If a constant force F acts for time t
we define the impulse of the force to
be Ft
If force is measured in N and time in
seconds, then the units of impulse are
Ns
An example of impulse would be a
cricket bat hitting a ball
 In this case, the time the force is
exerted over is small, but if the force
is big enough it will transfer noticeable
impulse to the ball
Remember that the acceleration of an object is given by:
𝑣− 𝑢
𝑎=
𝑡
𝐹 = 𝑚𝑎
𝐹= 𝑚
𝑣− 𝑢
𝑡
𝐹𝑡 = 𝑚 𝑣 − 𝑢
𝐹𝑡 = 𝑚𝑣 − 𝑚𝑢
𝐼 = 𝑚𝑣 − 𝑚𝑢
Replace
acceleration
with the above
Multiply by t
Multiply the
bracket out
Force x time =
Impulse!
So Impulse = Final momentum – Initial momentum
So Impulse = Change in momentum
This is the Impulse-Momentum Principle!
3G

75. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
Straight Line
You can calculate the momentum of
a particle and the impulse of a force
a)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 6 × 4.5
A body of mass 2kg is initially at rest
on a smooth horizontal plane. A
horizontal force of magnitude 4.5N
acts on the body for 6s. Find:
a)
The magnitude of the impulse
given to the body by the force
27Ns
b) The final speed of the body
13.5ms-1
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
Sub in values
Calculate
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 27𝑁𝑠
b)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
27 = (2 × 𝑣) − (2 × 0)
Sub in impulse, the
mass and the initial
velocity
Calculate
13.5 = 𝑣
3G

76. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
Straight Line
You can calculate the momentum of
a particle and the impulse of a force
A ball of mass 0.2kg hits a vertical wall
at right angles with a speed of 3.5ms-1.
The ball rebounds from the wall with
speed 2.5ms-1. Find the magnitude of
the impulse the ball exerts on the wall.
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
3.5ms-1
0.2kg
I
As always, draw a diagram!
2.5ms-1
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = (0.2 × 3.5) − (0.2 × −2.5)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 0.7 − −0.5
Sub in values, remembering
the final velocity is in the
opposite direction
Careful with negatives!
Calculate
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 1.2𝑁𝑠
3G

78. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
Straight Line
You can solve problems involving
collisions using the principle of
Conservation of Momentum
By Newton’s third law, when two
bodies collide they exert equal and
opposite forces on each other.
The objects will also be in contact for
the same length of time, so the
impulse exerted by each will be equal
but opposite in direction
Therefore, these changes in
momentum cancel each other out, and
the overall momentum is unchanged
This is the principle of Conservation
of Momentum
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
Total momentum before impact = Total momentum after impact
u1
u2
Before collision
I
After collision
m1
m2
v1
v2
I
𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2
Total
momentum of
the two
particles
before impact
Total
momentum of
the two
particles after
impact
When solving problems involving this principle:
a) Draw a diagram and label velocities before and after impact
with their relevant directions
b) Draw impulses on where necessary
c) Choose a positive direction and apply the rules you know
3H

79. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑚1 𝑢1 + 𝑚2 𝑢2
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
= 𝑚 𝑣 + 𝑚 𝑣
Straight Line
1 1
2 2
You can solve problems involving
collisions using the principle of
Conservation of Momentum
A particle of mass 2kg is moving with
speed 3ms-1 on a smooth horizontal
plane. Particle Q of mass 3kg is at
rest on the plane. Particle P collides
with Q and after the collision Q
moves away with a speed of 21/3ms-1.
Find:
Before collision
I
After collision
3ms-1
0ms-1
P
Q
2kg
3kg
v ms-1
21/3ms-1
The speed and direction of the
motion of P after the collision
0.5ms-1 in the opposite direction
b) The magnitude of the impulse
received by P and by Q in the
collision
I
We aren’t sure which direction P goes after the collision – just
choose one for now…
 If the answer is negative, the direction is the other way!
𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2
a)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
(2 × 3) + (3 × 0) = (2 × 𝑣) + (3 × 2 1 3)
6 = 2𝑣 + 7
−0.5 = 𝑣
Sub in the values
from the diagram
Work out each side
Calculate v
So the direction of motion of P is
reversed by the collision and it moves
off at 0.5ms-1
3H

80. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑚1 𝑢1 + 𝑚2 𝑢2
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
= 𝑚 𝑣 + 𝑚 𝑣
Straight Line
1 1
2 2
You can solve problems involving
collisions using the principle of
Conservation of Momentum
A particle of mass 2kg is moving with
speed 3ms-1 on a smooth horizontal
plane. Particle Q of mass 3kg is at
rest on the plane. Particle P collides
with Q and after the collision Q
moves away with a speed of 21/3ms-1.
Find:
a)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
The speed and direction of the
motion of P after the collision
0.5ms-1 in the opposite direction
b) The magnitude of the impulse
received by P and by Q in the
collision
Before collision
3ms-1
P
Q
2kg
3kg
0.5ms-1
21/3ms-1
I
After collision
0ms-1
I
Impulse on P
𝐼 = 𝑚𝑣 − 𝑚𝑢
𝐼 = (2 × 0.5) − (2 × −3)
Take the direction of impulse
on P as the positive direction
Work out the brackets
𝐼 = 1 − −6
Calculate
𝐼 = 7𝑁𝑠
Impulse on Q
𝐼 = 𝑚𝑣 − 𝑚𝑢
𝐼 = (3 × 2 1 3) − (3 × 0)
𝐼 = 7𝑁𝑠
You can see the impulse
received by each is equal
and opposite!
Take the direction of impulse
on Q as the positive direction
Calculate
3H

81. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑚1 𝑢1 + 𝑚2 𝑢2
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
= 𝑚 𝑣 + 𝑚 𝑣
Straight Line
1 1
2 2
You can solve problems involving
collisions using the principle of
Conservation of Momentum
Two particles, A and B, of masses 8kg
and 2kg respectively, are connected
by a light inextensible string. The
particles are at rest on a smooth
horizontal plane with the string slack.
Particle P is projected directly away
from Q with speed 4ms-1.
a)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
Find the speed of the particles
when the string goes taut
v = 3.2ms-1
b) Find the magnitude of the impulse
transmitted through the string
when it goes taut
Before motion
0ms-1
B
After motion
2kg
4ms-1
I
I
v ms-1
8kg
v ms-1
A
The particles do
not collide – in this
case the impulse is
transmitted
through the
string…
When the string is taut, the particles will move together and
hence have the same final velocity
𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2
(2 × 0) + (8 × 4) = (2 × 𝑣) + (8 × 𝑣)
Sub in values from the
diagram, leaving v in both
cases
Work out brackets
32 = 10𝑣
Divide by 10
3.2 = 𝑣
3H

82. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑚1 𝑢1 + 𝑚2 𝑢2
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
= 𝑚 𝑣 + 𝑚 𝑣
Straight Line
1 1
2 2
You can solve problems involving
collisions using the principle of
Conservation of Momentum
Two particles, A and B, of masses 8kg
and 2kg respectively, are connected
by a light inextensible string. The
particles are at rest on a smooth
horizontal plane with the string slack.
Particle P is projected directly away
from Q with speed 4ms-1.
Before motion
After motion
0ms-1
B
2kg
4ms-1
I
3.2ms-1
Find the speed of the particles
when the string goes taut
v = 3.2ms-1
b) Find the magnitude of the impulse
transmitted through the string
when it goes taut
I = 6.4Ns
I
8kg
A
3.2ms-1
The particles do
not collide – in this
case the impulse is
transmitted
through the
string…
Calculating the impulse for B
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 2 × 3.2 − (2 × 0)
a)
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
Sub in values for particle
B, taking the impulse from
it as the positive direction
Calculate
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 6.4𝑁𝑠
This is all we need to do. The
impulse in the opposite direction
will be the same!
3H

83. 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑣
𝑚1 𝑢1 + 𝑚2 𝑢2
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
Dynamics of a Particle moving in a
= 𝑚 𝑣 + 𝑚 𝑣
Straight Line
1 1
2 2
You can solve problems involving
collisions using the principle of
Conservation of Momentum
Two particles, P and Q of mass 2kg
and 4kg respectively are moving
towards each other along the same
straight line on a smooth horizontal
plane. The particles collide. Before
the collision, the speeds of P and Q
are 3ms-1 and 2ms-1. Given that the
magnitude of the impulse due to the
collision is 7Ns, find:
Before collision
I
After collision
3ms-1
P
2ms-1
4kg
v1 ms-1-1
0.5ms
0.25ms-1
v2 ms-1
The speed and direction of P
after the collision
b) The speed and direction of Q
after the collision
I
For particle P
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
7 = (2 × 𝑣1 ) − (2 × −3)
7 = 2𝑣1 + 6
Sub in values from the diagram, using
impulse as the positive direction
Work out
brackets
Calculate
0.5 = 𝑣1
a)
If you do not know a
velocity’s direction, set it
the same as the direction
of the impulse (this will
keep it positive while you
work it out!)
Q
2kg
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑡
As this answer is
positive, it means the
direction we put on the
diagram is correct!
For particle Q
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑚𝑣 − 𝑚𝑢
7 = (4 × 𝑣2 ) − (4 × −2)
7 = 4𝑣2 + 8
−0.25 = 𝑣2
Calculate
Sub in values from the diagram, using
impulse as the positive direction
Work out
brackets
As this answer is
negative, it means the
direction we put on the
diagram is incorrect!
3H

84. Summary
• You have learnt a huge amount about
forces in this chapter
• It is very important that you practice
questions like these – it is very easy to
forget some of the forces involved in more
complicated questions
• Remember that the key to most questions
is the correct use of F = ma!