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VCE
PHYSICS
Unit 3
Topic 1
Motion in 1 & 2 Dimensions
Unit Outline
To achieve this outcome the student should demonstrate the knowledge and skills to:
• apply Newton’s laws of motion to situations involving two or more forces acting along a straight line and in
two dimensions;
•
analyse the uniform circular motion of an object moving in a horizontal plane (FNET = mv2/R) such as a
vehicle moving around a circular road; a vehicle moving around a banked track; an object on the end of a
string.
•
Apply Newton’s 2nd Law to circular motion in a vertical plane; consider forces at the highest and lowest
positions only;
•
investigate and analyse the motion of projectiles near the Earth’s surface including a qualitative description
of the effects of air resistance;
apply laws of energy and momentum conservation in isolated systems;
•
analyse impulse (momentum transfer) in an isolated system, for collisions between objects moving along a
straight line (FΔt = mΔt);
•
apply the concept of work done by a constant force
work done = constant force x distance moved in the direction of the force
work done = area under force distance graph
•
analyse relative velocity of objects along a straight line and in two dimensions;
•
analyse transformations of energy between: kinetic energy; strain potential energy; gravitational potential
energy; and energy dissipated to the environment considered as a combination of heat, sound and
deformation of material;
kinetic energy i.e. ½ mv2; elastic and inelastic collisions in terms of conservation of kinetic energy
strain potential energy i.e. area under force-distance graph including ideal springs obeying Hooke’s
Law ½ kx2
gravitational potential energy i.e. mgΔh or from area under force distance graph and area under field
distance graph multiplied by mass
•
apply gravitational field and gravitational force concepts g = GM/r2 and F = GM1M2/r2
•
apply the concepts of weight (W = mg), apparent weight (reaction force, N) , weightlessness (W = 0) and
apparent weightlessness (N = 0)
•
model satellite motion (artificial, moon, planet) as uniform circular orbital motion (a = v2/r = 4π2r/T2)
•
identify and apply safe and responsible practices when working with moving objects and equipment in
investigations of motion.
Chapter 1
Topics covered:
• The S.I. System.
• Position.
• Scalars &Vectors.
• Vector Addition & Components.
1.0 The S. I. System
The system of units used in
Physics is the “Systeme
Internationale d’Units” or
more simply the S. I.
System.
The system has two
important characteristics;
Different units for the same
physical quantity are related
by factors of 10.(eg. mm;
cm; km)
Length:
Unit
Metre
(m)
Mass:
Unit
Kilogram
(kg)
Velocity:
Force:
Unit
N = ms
kg -1ms-2
Time:
Unit
Second
(s)
The system is based on 7
Fundamental Units, each
of which is strictly
defined.
S.I. System:
7 Fundamental
Units
Electric
Current:
Unit
Ampere (A)
Velocity
Force - Derived Unit
All other units, so called
DERIVED UNITS, are
simply combinations of 2
or more of the
Fundamental Units.
Temperature:
Unit
Kelvin
(K)
Luminous
Intensity:
Unit
Candela (cd)
Amount of
Substance:
Unit
Mole (M)
1.1 Position
To specify the POSITION
of an object, a point of
ORIGIN needs to be
defined. It is from this
point all measurements
can be taken.
-30
-25
-20
-15
For example on the number line
below the point labelled 0 is the
origin and all measurements
are related to that point.
The Number Line
0
-10 -5
5
10
- 15 Units
Thus a number called -15 is 15
units to the left of 0 on the
number line.
15
20
25
30
+ 30 Units
A number called +30 is 30 units to
the right of 0.
1.2 Scalars & Vectors
• Before proceeding, it is
important to define two general
classes of quantities.
• 1. SCALAR QUANTITIES:
• These are COMPLETELY
specified by:
• A MAGNITUDE (ie a NUMBER)
• and A UNIT
• Examples of Scalar Quantities
would be: Temperature (17oC),
Age (16 years), Mass (2.5 kg),
Distance (150 m).
•
2. VECTOR QUANTITIES:
•
•
•
•
These are COMPLETELY specified by:
A MAGNITUDE (ie. A NUMBER)
and A UNIT
and A DIRECTION
•
Examples of Vector Quantities would
be: Displacement (2.7 km, West),
Force (15 N, Downward), Acceleration
(1.5 ms-2, S.E.)
VECTORS ARE GENERALLY REPRESENTED BY ARROWS:
The length of the arrow represents the magnitude of the vector.
The orientation of the arrow represents the direction of the vector.
N
15
A vector of magnitude 15 units
directed East
1.3 Vector Addition
VECTOR ADDITION
Two Forces act at the Centre of Mass of a body.
The first of 4N East and the second of 3N South
Which way will
the body move ?
SINGLE VECTOR
DIAGRAM
N
Centre of Mass

450
4
5
55
3
Magnitude =
=
A Vector of:
Magnitude; 5 units
Direction; NE or N45E or 45T
42
+
25
32
3
In a direction, and
with a force,
that is the sum
of the 2 vectors
Direction:
Sin  = 3/5
  = Sin-1 3/5
= 36.90
THE RESULTANT FORCE HAS A MAGNITUDE OF
5 N DIRECTED AT E 36.90 S
1.4 Vector Subtraction
An object moving East at 8.0 ms-1 changes its velocity to 8.0 ms-1 South
The velocity change (v) is given by vf - vi
vi
(- vI ) is a negative vector. It can be
vf
8.0 ms-1 E
converted to a positive one by reversing
its direction.
What is the object’s
8.0 ms-1 S
Then, by performing a vector addition, the
change in velocity ?
velocity change v can be obtained.

v
Magnitude v =
=
vf
8.0 ms-1 S
82 + 82
128
= 11.3 ms-1
- vi
8.0 ms-1 W
Direction:
Tan  = 8/8 = 1.0
  = Tan-1 1.0
= 450
THE CHANGE IN VELOCITY = 11.3 ms-1 AT S 450W
1.5 Vector Components
A “Jump Jet” is launched from a 150 ramp at a velocity of 40 ms-1
What are the vertical and horizontal components of its velocity ?
40 ms-1
Vertical Component:
V VERTICAL = 40 Sin 150
= 10.4 ms-1
V VERTICAL
150
VHORIZONTAL
V VERTICAL and VHORIZONTAL are the
COMPONENTS of the plane’s velocity.
Horizontal Component:
VHORIZONTAL = 40 Cos 150
= 38.6 ms-1
Motion - Revision Questions
Question type: Vectors
Adam is testing a trampoline. The diagrams show Adam at successive stages of his
downward motion.
Figure C shows Adam at a time when he is travelling DOWNWARDS and SLOWING
DOWN.
A
B
Q1: What is the direction of
Adam’s acceleration at the time
shown in Figure C ? Explain your
answer.
C
D
A: Acc is UPWARD.
In order to meet the requirements set travelling downward BUT slowing
down, he must be decelerating ie.
Accelerating in a direction opposite to
his velocity. Thus acc is upward.
Chapter 2
Topics covered:
• Distance versus Displacement.
• Speed versus Velocity.
• Acceleration.
• Graphical Representations.
2.0 Distance vs Displacement
•
•
•
•
•
Distance is a Scalar Quantity
having a magnitude and a
unit. The S.I. unit for Distance
is the metre (m)
Distance is best thought of as:
“How far you have travelled in
your journey”.
Displacement is a Vector
Quantity having a magnitude,
a unit and a direction. The S.I.
unit for Displacement is the
metre (m), plus a direction
Displacement is best thought
of as: “How far from your
starting point you are at the
end of your journey”.
Distance and Displacement
may or may not be
numerically equal, depending
on the nature of the journey.
JOURNEY No 1.
Distance = Displacement
+ve direction
100 m
Start
At the end of the run:
Distance = 100 m.
Displacement = +100 m
Finish
JOURNEY No 2.
Distance  Displacement
400 m track
Start/Finish
At the end of the one lap run:
Distance = 400 m.
Displacement = 0 m
2.1 Speed vsINSTANTANEOUS
Velocity
vs AVERAGE
•
•
•
•
•
•
•
•
Speed is defined as the Time Rate
•
of Change of Distance.
Speed is a Scalar Quantity.
•
Mathematically:
Speed = Distance/Time
The S.I. unit for Speed is
metres/sec (ms-1)
•
Velocity is defined as the Time
Rate of Change of Displacement. •
Velocity is a Vector Quantity.
•
Mathematically:
•
Velocity = Displacement/Time
The S.I. unit for Velocity is
metres/sec (ms-1), plus a direction
IN ALL CALCULATIONS AND
EQUATIONS USED IN THE COURSE,
ASSUME INSTANTANEOUS VALUES
ARE REQUIRED UNLESS OTHERWISE
STATED.
VELOCITY
The term velocity can be misleading
unless a specific label is attached.
The label indicates whether the
velocity is an Average value calculated
over a long period of time OR an
Instantaneous value calculated at any
instant of time.
A simple example illustrates:
A journey of 40 km across the suburbs
takes 1 hour;
VAV = 40/1 = 40 kmh-1
BUT VINST could be anything from 0
kmh-1 (stopped at traffic lights) to VINST
= 100 kmh-1 (travelling along the
freeway).
2.2 Some Common Speeds
Event
Speed (ms-1)
Speed (kmh-1)
1. Grass Growing
5.0 x 10-8
1.8 x 10-7
1-2
4-8
3. Marathon
Runner
5
18
4. 100 m
Sprinter
10
36
5. Suburban
Speed Limit
16.7
60
6. Freeway Speed
Limit
30.6
110
7. Boeing 737
Cruising
246
886
2.99 x 108
1.1 x 109
2. Walking Pace
8. Speed of Light
2.3 Acceleration
•
•
•
•
•
•
Acceleration is defined as the Time
Rate of Change of Velocity.
Acceleration is a Vector Quantity.
Mathematically:
Acceleration = Velocity/Time
The S.I. unit for acceleration is
metres/sec/sec (ms-2)
Since acceleration is a vector
quantity, a body travelling with a
constant speed but in a constantly
changing direction must be
accelerating.
So a cyclist travelling around a
corner at constant speed is, in fact,
accelerating ! (More of this later).
ACCELERATING VEHICLE
v
a
The velocity and acceleration
are in the same direction
DECELERATING VEHICLE
v
a
The velocity and acceleration
are in opposite directions.
•
2.4 Graphical Representations
Much of the information delivered in this Physics course is presented
graphically.
• Generally, graphs “tell a story” and you need to develop the ability to “read”
the story the graph is telling.
• There are two basic families of graphs you should be familiar with:
(a) Sketch Graphs, paint a broad brush, general picture of the relationship
between the quantities graphed.
(b) Numerical Graphs from which exact relationships may be deduced and/or
exact values may be calculated.
Velocity
Displacement
Distance
SKETCH
GRAPHS
Time
The Story:
The
The
Story:
AsStory:
time
passes, the distance of the object
The
Story:
As
As
time
time
passes,
passes,
the
the
velocity
displacement
of of
thethe
of
object
the
is
from
itspasses,
starting
point
is increasing
in aobject
As
time
the
distance
object
increasing
isfrom
increasing
in amore
uniform
quickly
manner
(the
slope
(the
slope
is This
is
uniform
manner
(the
slope
is
constant).
This
its starting
point
does
not
change.
constant).
increasing
This
at a
aan
is
constant
the graph
rate).
a constantly
isthe
thegraph
graph
object
moving
at constant
is
stationary
object.
accelerating
object
speed.
This
is the graph a constantly accelerating
object
Motion - Revision Questions
Question type: Sketch Graphs
In a road test, a car was uniformally accelerated from rest over a
distance of 400 m in 19 sec. The driver then applied the brakes,
stopping in 5.1 sec with constant deceleration.
The graphs A to F below should be used to answer the questions below. The
horizontal axis represents time and the vertical axis could be velocity or
distance.
A
B
C
D
E
F
Q2: Which of the graphs, A to F, represents the velocity time graph for the
entire journey ?
A: Graph B
Q3: Which of the graphs, A to F, best represents the distance time graph of
the car for the entire journey ?
A: Graph E
2.5 Exact Graphical
Relationships
•
•
•
•
•
You are required to be familiar with graphs of:
Distance or Displacement Versus Time;
Speed or Velocity Versus Time and
Acceleration Versus Time
These graph types and the exact information obtainable from them can be
summarised in the table given below. Put this table on your cheat sheet.
Graph Type
Read directly
from Graph
Slope of Graph
gives:
Area under
Graph gives:
Distance or
Displacement
versus Time
Distance
or
Displacement
Speed
or
Velocity
No Useful
Information
Speed or Velocity
versus Time
Speed
or
Velocity
Acceleration
Distance
or
Displacement
Acceleration
versus Time
Acceleration
No Useful
Information
Velocity
2.6 The Equations of Motion
•
•
•
•
•
•
•
•
These are a series of equations
linking velocity, acceleration,
displacement and time.
THE EQUATIONS CAN ONLY BE
USED IN SITUATIONS WHERE THE
ACCELERATION IS CONSTANT.
The 3 most important of these
equations are:
1.
v = u + at
2.
v2 = u2 +2ax
3.
x = ut + ½at2
where u = Initial Velocity (ms-1)
v = Final Velocity (ms-1)
a = Acceleration (ms-2)
x = Displacement (m)
t = Time (s)
u=
v=
a=
x=
t=
HINTS FOR EQUATIONS’ USE
When using these equations
always list out the information
supplied in the question and what
you are required to calculate; then
choose the appropriate equation to
use.
Make a list like this.
+ve
Sometimes it is necessary to
choose a positive direction ie. up
or down for vertical motion
questions or left or right for
horizontal motion questions.
Questions are often asked which
require a 2 step process to get to
the answer, ie. A value for
acceleration may be needed before
the final velocity can be found.
Motion - Revision Questions
Question type: Equations of Motion
In a road test, a car was uniformally accelerated from rest over a distance of
400 m in 19 sec.
The driver then applied the brakes, stopping in 5.1 sec with constant
deceleration.
Q4: Calculate the acceleration of the car for the first 400 m.
A: Firstly, list information:
u=0
v=?
a=?
x = 400 m
t = 19 s
Choose the appropriate equation:
x = ut + ½at2
400 = 0 + ½a(19)2
a = 2.22 ms-2
Motion - Revision Questions
Question type: Average Speed
In a road test, a car was uniformally accelerated from rest over a distance of 400 m in
19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant
deceleration.
Q5: Calculate the average speed for the entire journey, covering both the
accelerating and braking sections.
A: Average Speed = Total Distance Need to know u, the initial speed for the
braking section which equals the final
Total Time
For the accelerated part of journey: speed for the accelerating section.
For accelerating section:
u=0
v = u + at
v=?
= 0 + (2.2)(19)
-2
a = 2.2 ms
= 41.8 ms-1
x = 400 m
Now can calc s
t = 19 s
x = ut + ½at2
Need
to
calc
acc
To get braking distance, Braking list
= (41.8)(5.1) + ½(-8.2)(5.1)2
v = u + at
use Eqns of Motion
becomes
0 = 41.8 + a(5.1) = 213.2 - 106.6
u = ? List does not
u = 41.8 ms-1
= 106.6 m
a = - 8.2 ms-2
v = 0 contain
v = 0 Still not
a = ? enough info
a = ? enough
Total Distance = 400 + 106.6 = 506.6 m
x = ? the calculate s
x = ? info
Total Time = 19 + 5.1 = 24.1 s
t = 5.1s
t = 5.1s
So, Average Speed = 506.6/24.1 = 21 ms-1
Distance = 400 m
Time = 19 sec
For the braking part of journey:
Distance = needs to be calculated
Time = 5.1 sec
Chapter 3
Topics covered:
• Newton’s Laws.
• Force in Two Dimensions.
• Momentum and Impulse.
• Conservation of Momentum.
3.0 Newtonian Motion
•
•
•
Sir Isaac Newton (1642 - 1727) was
unique for a number of reasons, but
mostly because he developed a set
of laws describing the motion of
objects in the universe.
Prior to Newton, scientists believed
that a set of laws existed which
explained motion on Earth and
these laws had to be modified to
describe motions in all other parts
of the universe.
Newton was the first scientist to
realise that all motion anywhere in
the universe could be described by
a single set of laws which then had
to be modified for use in the friction
riddled confines of the Earth.
Isaac Newton
Aged 26
3.1 Newton’s Laws
INERTIA – That property
possessed by all bodies with
mass whereby they tend to
resist changes to their
motion.
It is associated with an
object’s mass – more mass,
more inertia.
Inertia is NOT a force.
Newton developed
3 laws covering
motion in the
universe, they are:
LAW 1.
THE LAW OF INERTIA.
A body will remain at rest, or
in a state of uniform motion,
unless acted upon by a net
external force.
LAW 2.
The acceleration of a body is
directly proportional to the
net force applied and
inversely proportional to its
mass. (a = F/m)
LARGE MASS – LARGE INERTIA
LAW 3.
For every action there is an
equal and opposite reaction
small mass – small inertia
Motion - Revision Questions
Question type: Newton’s Laws
A seaplane of mass 2200 kg takes off from a smooth lake as shown. It starts from rest,
and is driven by a CONSTANT force generated by the propeller. After travelling a
distance of 500 m, the seaplane is travelling at a constant speed, and then it lifts off
after travelling a further 100 m.
Total Opposing
Force (N)
10000
8000
6000
4000
2000
0
The total force opposing the motion
of the seaplane is not constant. The
graph shows the TOTAL FORCE
OPPOSING THE MOTION of the
seaplane as a function of the
distance travelled.
Distance (m)
100 200 300 400 500 600
Q6: What is the magnitude of the net force
acting on the seaplane after it has travelled a
distance of 500 m from the start ?
A: At d = 500 m the plane is travelling at
CONSTANT VELOCITY,
So ΣF = 0
Motion - Revision Questions
Question type: Newton’s 2nd Law
Q7: What is the magnitude of the seaplane’s acceleration at the 200 m mark ?
A: At d = 500 m the seaplane is subject to 0 net force (see previous question).
Thus, Driving Force = Opposing Force = 10,000 N (read from graph).
At d = 200 m the total opposing force = 2000 N (read from graph)
So ΣF = 10,000 - 2000 = 8000 N
Now, we know that ΣF = ma
So, a = ΣF/m = 8000/2200 = 3.64 ms-2
Total Opposing
Force (N)
10000
8000
6000
4000
2000
0
Distance (m)
100 200 300 400 500 600
Motion - Revision Questions
Question type: Work
Q8: Estimate the work done by the seaplane against the opposing forces in
travelling for a distance of 500 m.
A: Work = Force x Distance = Area under F vs d graph.
Area needs to be calculated by “counting squares”.
Each square has area = 2000 x 100 = 2 x 105 J
Total number of squares (up to d = 500 m) = 9 whole squares (x)
+ 6 part squares (p) = 12 whole squares.
So Work done = (12) x (2 x 105) = 2.4 x 106 J
Total Opposing
Force (N)
10000
p
8000
p
x
x
p
x
x
x
x
x
x
6000
4000
p
2000
p
0
p
x
100 200 300 400 500 600
Distance (m)
Motion - Revision Questions
Question type: Newtons 2nd Law
In Figure 2, a car of mass 1000 kg is being towed on a level road by a van of
mass 2000 kg. There is a constant retarding force, due to air resistance and
friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at
a constant speed.
Figure 2
Q9: What is the magnitude of the force driving the van?
As the vehicles are travelling at constant speed, a = 0 and thus ΣF = 0
Thus driving force = total retarding force
Thus driving Force = 800 N
Q10: What is the value of the tension, T, in the towbar?
Looking at the towed car alone the forces acting are tension in the
towbar and the retarding force.
Since the car is travelling at constant velocity, ΣF = 0, so tension =
retarding force = 300 N
Motion - Revision Questions
Question type: Newton’s 3rd Law
a
The figure shows a cyclist with the bicycle
wheels in CONTACT with the road surface.
The cyclist is about to start accelerating
forward.
FTR
FRT
Q11: Explain, with the aid of a clear force diagram, how the rotation of the wheels
result in the cyclist accelerating forwards.
A: The wheels rotate in the direction shown.
The force labelled FTR is the force the tyre exerts on the road. This force is directed
in the opposite direction to the acceleration and thus cannot be the force producing
that acceleration.
The force labelled FRT is the Newton 3 reaction force arising from the action of FTR.
It is this force (directed in the same direction as the acceleration) that actually
produces the acceleration of the bike and rider.
3.2 Newton’s Laws
Restrictions & Consequences
RESTRICTIONS
CONSEQUENCES
The laws only apply at speeds
st law is concerned
As
far
as
Newton’s
1
much, much less than the
“rest” and “uniform motion” are the same
speed of light.
The laws apply equally in ALL state.
You cannot perform any test which can
inertial frames of reference.
show whether you are stationary or moving
FRAME OF REFERENCE ?
at constant velocity.
A frame of reference is best
The action and reaction law requires there to
described as “your point of
be TWO bodies interacting, the ACTION force
observation.”
acting on one body and the REACTION acting
An inertial frame of reference
on the other.
is one that is either
Deciding when an Action / Reaction situation
stationary or moving with
exists can be done by answering the
constant velocity.
question:
A non inertial frame of
Does the second (reaction) force disappear
reference is accelerating.
immediately the first (action) force
Humans in non inertial
disappears ?
frames tend to invent
If the answer is yes, you have an action
“fictitious forces” to explain
reaction pair.
their experiences
Motion - Revision Questions
Question type: Relative Motion
A train is travelling at a constant velocity on a level track. Lee is standing
in the train, facing the front, and throws a ball vertically up in the air, and
observes its motion.
Q12: Describe the motion of the ball as seen by Lee.
Lee sees the ball move straight up and down.
Sam, who is standing at a level crossing, sees Lee throw the ball into the air.
Q13: Describe and explain the motion of the ball as seen by Sam.
From Sam’s point of view, the ball follows a parabolic path
made up of the vertical motion imparted by Lee and the
horizontal motion due to the train.
3.3 Force in Two Dimensions
AN OBJECT UNDER THE
ACTION OF 4 FORCES
F1
F4
F3
FRES
F2
Perform a vector addition
In which direction will
of the forces. The
the object accelerate ?
Resultant Force (FRES ) will
give the direction
F of the acc.
1
F2
F3
FRES
F4
The object will be subject
to FRES and accelerate in
that direction
• A Force is either a Push or a Pull.
• A Force is either a CONTACT type force
or a FIELD type force.
• Force is a Vector quantity having both
a magnitude, a unit and a direction.
• Force is NOT one of the 7 fundamental
units of the S.I. System and thus it is a
Derived quantity.
• The unit for Force is kgms-2. This was
assigned the name the NEWTON (N), in
honour of Sir Isaac.
• Forces can act in any direction and the
TOTAL, NET or RESULTANT force is
the vector sum of all forces acting on a
body.
• The body will then ACCELERATE in the
direction of the RESULTANT FORCE,
according to Newton 2.
Motion - Revision Questions
Question type: Force in 1 & 2 Dimensions
A cyclist is towing a small trailer along a
level bike track (Figure 1). The cyclist
and bike have a mass of
90 kg, and the trailer has a mass of 40
kg.
There are opposing constant forces of
190 N on the rider and bike, and 70 N on
the trailer.
These opposing forces do not depend
on the speed of the bike.
The bike and trailer are initially
travelling at a constant speed
of 6.0 m s-1.
Q14. What driving force is being exerted on the road by the rear tyre of the bicycle?
A: Constant speed implies ΣF = 0; So driving force = total retarding force
Total retarding force = 190 + 70 = 260 N = driving force
3.4 Momentum & Impulse
• Newton called Momentum the
“quality of motion” and it is a
measure of a body’s
translational motion - its
tendency to continue moving in
a particular direction.
• Roughly speaking, a body’s
momentum indicates which way
the body is heading and just
how difficult it was to get the
body moving with its current
velocity.
• Momentum is a Vector Quantity.
• Mathematically:
Momentum (p) = m.v
where, m = mass (kg),
v = velocity (ms-1)
p = momentum (kgms-1)
• Impulse is the “transfer
mechanism” for momentum.
• In order to change the momentum
of a body you need to apply a
force for a certain length of time to
produce the change.
• Impulse is a Vector Quantity.
• Mathematically:
Impulse (I) = F.t
where
F = Force (N)
t = time (s)
I = Impulse (Ns)
• From Newton 2 (F = ma) and the
definition of acceleration, (a = v/t),
we get:
F = mv/t
 Ft = mv
•
Thus: Impulse = Momentum
Motion - Revision Questions
Question type: Momentum
A small truck of mass 3.0 tonne collides
with a stationary car of mass 1.0 tonne.
They remain locked together as they move
off.
The speed immediately after the collision
was known to be 7.0 ms-1 from the jammed
reading on the car speedometer.
Robin, one of the police investigating the
crash, uses conservation of momentum to
estimate the speed of the truck before the
collision.
Q15 : What value did Robin obtain?
A: PBEFORE = PAFTER
PBEFORE = (3000)(x)
PAFTER = (3000 + 1000)(7.0)
So, 3000x = 28,000
x = 9.3 ms-1
The calculated value is questioned by the other
investigator, Chris, who believes that
conservation of momentum only applies in
elastic collisions.
A: Momentum is conserved in ALL
Q16: Explain why Chris’s comment is wrong.
types of collisions whether they be
elastic or inelastic.
KE is not conserved in this type
(inelastic) collision.
Motion - Revision Questions
Question type: Momentum
A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides
with a truck that is stationary at a set of traffic lights. After the collision
they are stuck together and move off with a speed of 2.0 ms–1
Q17 : How much momentum did the car transfer to the truck?
A: Mom is ALWAYS conserved.
Mom of car before = (1000)(5) = 5000 kgms-1
Mom of car after = (1000)(2) = 2000 kgms-1
Since mom is conserved
Mom loss by car = Mom gain by truck
So, Mom transferred to truck = 3000 kgms-1
Q18 : What is the mass of the truck?
A: p = mv
so, m = p/v
= 3000/2
= 1500 kg
Motion - Revision Questions
Question type: Impulse
A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a
truck that is stationary at a set of traffic lights. After the collision they are stuck
together and move off with a speed of 2.0 ms–1
Q19: If the collision took place over a period of 0.3 s, what was the
average force exerted by the car on the truck?
For the truck
Impulse = Change in momentum
Ft = mv
F(0.3) = 3000
F = 10,000 N
Motion - Revision Questions
Question type: Momentum
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary
railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and
move off as one.
10 tonnes
5 tonnes
-1
6.0 ms
Before Collision
X
Y
stationary
v ms-1
After Collision
X
Y
Q20: Calculate the final speed of the joined railway trucks after collision.
A: In ALL collisions Momentum is conserved.
So Mom before collision = Mom after collision
Mom before = (10 x 103)(6.0) + (5 x 103)(0)
Mom after = (15x 103)(v)
So v = 60,000/15,000
= 4.0 ms-1
Motion - Revision Questions
Question type: Impulse
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary
railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together
and move off as one at a speed of 4.0 ms-1.
10 tonnes
Before
Collision
6.0
5 tonnes
ms-1
X
Y
stationary
After Collision
X
4.0 ms-1
Y
Q21: Calculate the magnitude of the total impulse that truck Y exerts on truck X
A: Impulse = Change in Momentum
Truck X’s change in momentum = Final momentum – Initial Momentum
= (10 x 103)(4.0) – (10 x 103)(6.0)
= -2.0 x 104 kgms-1
The mechanism for this change in momentum is the impulse supplied by Truck Y
So, I = 2.0 x 104 Ns
3.5 Elastic & Inelastic Collisions
All collisions (eg, cars with trees, cyclists with the footpath, neutrons with
uranium atoms, bowling balls with pins etc.) fall into one of 2 categories:
(a) ELASTIC COLLISIONS,
where BOTH Momentum
AND Kinetic Energy are
conserved. (Very few
collisions are of this type).
If anywhere, these will most
likely occur on the atomic
or subatomic level.
(b) INELASTIC COLLISIONS,
where Momentum is conserved
BUT Kinetic Energy is NOT
conserved. (Most collisions are of
this type).
The “lost” Kinetic Energy has
been converted to other forms of
energy eg, heat, sound, light.
U
n
Neutron
INELASTIC COLLISION
ELASTIC COLLISION
Motion - Revision Questions
Question type: Elastic/Inelastic Collisions
A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0
tonne. They remain locked together as they move off.
The speed immediately after the collision was known to be 7.0 ms-1 from the
jammed reading on the car speedometer.
Robin, one of the police investigating the crash, uses conservation of
momentum to estimate the speed of the truck before the collision at 9.3 ms-1
Q22: Use a calculation to show whether the collision was elastic or inelastic.
A: Total KE before collision = ½ mv2 = ½ (3000)(9.3)2
= 129735 J
Total KE after collision = ½ mv2 = ½ (4000)(7.0)2
= 98,000 J
KE is NOT conserved. So collision is INELASTIC
Motion - Revision Questions
Question type: Inelastic Collisions
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary
railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and
move off as one at 4.0 ms-1.
10 tonnes
5 tonnes
-1
6.0
ms
Before Collision
X
Y
stationary
After Collision
X
4.0 ms-1
Y
Q23: Explain why this collision is an example of an inelastic collision. Calculate
specific numerical values to justify your answer.
A: In Inelastic collisions Momentum is conserved BUT Kinetic Energy is not.
For this collision
pBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1
pAFTER = (15 x 103)(4.0)
= 6 x 104 kgms-1
KEBEFORE = ½mv2 = ½(10 x 103)(6.0)2 = 1.8 x 105 J
KEAFTER = ½mv2 = ½(15 x 103)(4.0)2 = 1.2 x 105 J
Thus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic
collision
3.5 Conservation of Momentum
• The Law of Conservation of
Momentum states that in an
isolated system, (one subject
to no outside influences), the
total momentum is conserved.
• So in a collision (say a car
hitting a tree and coming to a
stop), the total momentum
before the collision = total
momentum after the collision.
Mass
Mass of
of car
car plus
plus
passengers == m
m
passengers
v=0
Velocity = v
Momentum
Momentum
0
pp==mv
THE CONSERVATION LAW DOES NOT ALLOW
MOMENTUM TO DISAPPEAR.
THE APPARENTLY “LOST” MOMENTUM HAS,
IF FACT, BEEN TRANSFERRED THROUGH THE
TREE TO THE EARTH.
Since the Earth has a huge mass (6.0 x 1024 kg)
the change in its velocity is negligible.
3.6 The Physics of Crumple
Zones & Air Bags
A car crashes into a concrete barrier.
The change in momentum suffered by the car
(and passengers) is a fixed quantity.
So, Impulse, (the product of F and t), is also
fixed.
However, individual values of F and t can vary as
long as their product is always the same.
So if t is made longer, consequently F must be
smaller.
Crumple Zones increase the time (t) of the
collision.
So, F is reduced and the passengers are less
likely to be injured.
The same logic can also be applied to Air Bags
The air bag increases the time it takes
for the person to stop.
So the force they must absorb is
lessened.
So they are less likely to be seriously
injured.
A further benefit
is this lesser
force is
distributed over
a larger area
Motion - Revision Questions
Question type: Momentum/Impulse
In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an
air bag as shown. The time taken for the driver’s head to come to a complete
stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal
collision between the head of mass 7.0 kg and the air bag.
Q24: Calculate the magnitude of the average contact force that the air bag exerts on the
driver’s head during this collision.
A:
Impulse = Change in Momentum
FΔt = Δ(mv)
So F = Δ(mv)/Δt
= (7.0)(8.0)/(1.6 x 10-1)
= 350 N
Motion - Revision Questions
Question type: Air Bags/Crumple Zones
In a car the driver’s head is moving horizontally at 8.0 ms-1 and
collides with an air bag as shown. The time taken for the driver’s
head to come to a complete stop is 1.6 x 10-1 s. This collision
may be modelled as a simple horizontal collision between the
head of mass 7.0 kg and the air bag.
Q25: Explain why the driver is less likely to suffer a head injury in a collision
with the air bag than if his head collided with the car dashboard, or other
hard surface.
A: The change in momentum suffered by the driver’s head is a FIXED quantity no
matter how his head is brought to rest.
• Therefore the product of F and t (ie Impulse) is also a fixed quantity.
• However the individual values of F and t may be varied as long as their
product always remains the same.
• The air bag increases the time over which the collision occurs, therefore
reducing the size of the force the head must absorb so reducing the risk of
injury.
• The air bag also spreads the force over a larger area, reducing injury risk.
• Without the air bag the driver’s head may hit a hard surface decreasing the
time to stop his head and necessarily increasing the force experienced and
thus the likelihood of injury.
• In addition the force will be applied over a much smaller area increasing the
likelihood of severe injury
Chapter 4
Topics covered:
• Centre of Mass.
• Weight.
• Reaction Force.
• Bouncing Balls.
• Friction.
• Various Force Applications
4.0 Centre of Mass
In dealing with large objects it is
useful to think of all the object’s
mass being concentrated at one
point, called the Centre of Mass of
the object.
The C of M of the object is the point
around which it will spin if a torque
or turning force is applied to the
object.
Centre of Mass
C of M
For regularly shaped objects eg. squares
or rectangles, cubes or spheres the
Centre of Mass of the object is in the
geometric centre of the object
For oddly shaped objects
eg. a boomerang, the C of M
may fall outside the
perimeter of the object.
4.1 Centre of Mass - Systems
For a system of 2 or more bodies, the
position of the C of M may be determined
from the formula:
XCofM = (m1x1 + m2x2 + m3x3 + …)
(m1 + m2 + m3 +…)
WHAT IS THE CENTRE OF
MASS OF THIS SYSTEM?
XC of M is the position of the Centre of
Mass of the System as measured
from a CHOSEN REFERENCE POINT.
m1,m2,m3, etc are the masses of the
individual components of the System
x1, x2, x3 etc are the distances
measured from the CHOSEN
REFERENCE POINT to the centres of
mass of the system’s components.
5.0 m
1.0 m
Centre of Mass
of each mass
2.875 m
A
2.5 m
3.5 m
1.0 m
30 kg
mass
CHOSEN REFERENCE POINT IS A
X C of M = [(50 x 2.5) + (30 x 3.5)]
50 kg beam
Centre of Mass
of System
(50 + 30)
= 2.875 m from A
4.2 Weight
Object of Mass (m)
Centre of Mass
•
•
•
Weight = mg
Weight Force acts through
the Centre of Mass and is
directed toward the Centre
of the Earth
•
•
The effect of a Gravitational Field on a
Mass is called its WEIGHT.
Weight is a FORCE and therefore a
Vector quantity.
Mathematically:
W = mg
where W = Weight (N)
m = mass (kg)
g = Gravitational Field
Strength (Nkg-1)
Weight acts through the Centre of Mass
of the body and is directed along the line
joining the centres of the the two bodies
between which the Gravitational Field is
generated.
On Earth, the Gravitational Field of
Strength 9.8 Nkg-1 gives any mass under
its influence alone an acceleration of 9.8
ms-2
4.3 Reaction Force
•
•
•
Any stationary object which is
under the influence of the Earth’s
Gravitational Field must be
subject to a force equal in size,
but opposite in direction to, its
weight.
This equal but opposite force is
called the NORMAL REACTION
FORCE.
The Normal Reaction Force only
arises from the action of the
weight force and DOES NOT
EXIST AS AN ISOLATED FORCE
IN ITS OWN RIGHT.
R - Normal Reaction Force
Object
Mass = m
Table
W - Weight
R acts upwards from the
boundary between the table and
the mass through the Centre of
Mass
W acts downwards from the
Centre of Mass toward the centre
of the Earth
R and W are NOT an
ACTION - REACTION PAIR
WHY ?
4.4 Bouncing Balls.
When a ball, falling
under the action of
gravity, arrives at a
hard surface, it is the
Normal Reaction
Force (R), which
provides;
(a) The Force required
to decelerate the ball
to a stop and then,
(b) The Force needed
to accelerate the ball
away from the
surface.
vvv==00
R
v
v
W
v
v
W
R
W
W
v
v
W
W
v=0
W
W
W
The ball is dropped:
a = g Velocity = 0
Acceleration = g
a=g
a=g
a=g
aa==gg
a = g Ball strikes surface, R
aa arises and begins to
decelerate ball.
R > W (acc now up)
4.5 Friction
•
Friction is the most unusual of all
forces as it cannot start an object
moving.
• Friction can, however, slow or stop
an object once it is moving or
prevent it from starting to move.
• Slowing a moving object is the result
of DYNAMIC FRICTION, while
stopping an object from starting to
move is the result of STATIC
FRICTION.
• Generally speaking, for the same
pair of surfaces, Static Friction 
Dynamic Friction.
• The size of the frictional force
depends upon:
(a) The Roughness of the surfaces
measured by the Coefficient of
Friction ()
(b) The Separation of the surfaces.
• Friction does NOT depend on the
AREA in contact.
OBJECT HELD STATIONARY
BY STATIC FRICTION
R
Fr
F
(Static Friction)
(Pulling Force)
W
The size of the Frictional
Force Fr is calculated from:
Fr = R
4.6 Friction Applications
•
Friction is NOT always a
hindrance to living on Earth,
often it is vital for movement
over the Earth’s surface.
Consider the following:
A car is accelerating along a
Rough (meaning having
Friction) Road Surface.
Acceleration
•
•
FTR
Force of
TYRE on
ROAD.
ACTION
FORCE
FRT
Direction of rotation
Force of Driven Wheel
ROAD on
TYRE.
REACTION
FORCE
The frictional force between the
tyre and the road (FTR) is directed
backwards. This force CANNOT
provide the forward propulsion.
In order for the Car to accelerate in the
direction shown a force must exist in that
direction.
Thus it is the Force of the Road on the
Tyre (FRT) which gives rise to the acceleration
IT IS THE REACTION FORCE
WHICH PRODUCES THE
ACCELERATION
4.7 Various Force Situations:
Stationary and Falling Bodies
MASS - STATIONARY
ON A TABLE
+ ve direction
R
Normal Reaction Force
Acts from the point of
contact b/w Mass &Table
MASS - FALLING WITH
AIR RESISTANCE
FR
+ve
a
Centre of Mass
a=0
W = mg
From Newton 2
 F = ma
F = R - W = ma
but a = 0
So, R - W = 0
and R = W
Weight acts directly down
from the Centre of Mass
W = mg
 F = ma
 F = W - FR
So, W - FR = ma
and FR = ma - mg
= m(a - g)
4.8 Various Force Situations:
Lifts
LIFT - ACCELERATING UPWARDS
LIFT - ACCELERATING DOWNWARDS
+ve
R
+ve
a
W = mg
a
R is a measure of the
Mass (m) on Floor
weight” of the
of Lift
W = mg
mass. When accelerating
For the Mass (m)
upward the mass is
F = ma
“heavier” than normal, by an
F = W - R
amount (ma), and when
so ,W - R = ma
accelerating downward it is
and R = W - ma
“lighter” than normal, by
= mg - ma
(ma).
Thus R = m(g - a)
Mass (m) on Floor
“apparent
of Lift
For the Mass (m)
F = ma
F = R – W
so, R - W = ma
and R = W + ma
= mg + ma
Thus R = m(g + a)
R
4.9 Various Force Situations:
R
Inclined Planes
a

RR
RR

mg Cos 
W = mg
mg Sin 
Body of mass ( m) on an inclined plane
Weight
and
Normal
Reaction
act triangle
on mass
Force
triangle
and
plane
Weight
broken
upinclined
into
2 components
Parallel
(to
the
plane)
component
= mgSin 
Parallel
Component
can be transferred
are
similar,
’s are equal.
IT IS THIS
THAT
to act through
the CCOMPONENT
of M.
ACCELERATES
THE MASS (mg
DOWN
Perpendicular Component
CosTHE
) =PLANE
R
a
a
a
a
W
W
W
W
Notice the Component of the Weight
parallel to the plane (mg Sin) increases
as the plane gets steeper thus increasing
the acceleration of the mass down the
plane.
4.10 Various Force Situations:
Connected Masses
Frictionless
Pulley
Masses connected by
Light Inextensible
String. (Tension (T)
the same everywhere)
a
+ve
Frictionless Table
Frictionless
Pulley
m1
a
T
m2
T
m2
W = m1g
+ve
For m1: T - m1g = m1a
F = ma
For m2 : W - T = m2a
T
m1
T
For m1: T = m1a
Light inextensible
String. (Tension (T)
the same everywhere)
For m2: m2g - T = m2a
W = m2 g
W = m2g
Solve simultaneous equations to get acceleration
Chapter 5
Topics covered:
• Work.
• Work & Energy.
• Power.
• Energy Types.
• Conservation of Energy.
5.0 Work
•
•
•
•
•
•
•
The term WORK has a very strict
definition in the physics world.
WORK DONE BY A VARIABLE FORCE
If a FORCE moves an object
through a DISTANCE, WORK has
Force
been done on that object.
F
Mathematically:
W = F.d
Area = Work done
by the Force
where W = Work (Joules)
 Work = ½F x d
F = Force (N)
d = Distance (m)
This formula can only be used if the
d
Force remains constant through the
distance
course of doing the work.
Work is a SCALAR quantity.
If the Force varies in the course of
No work has been done if the force
doing the Work (as in stretching a
has not caused the object to move.
spring), the work done can only be
No work has been done if the object
calculated from the area under the
begins and ends its movement in
Force versus distance graph.
the same place. ie travels in a circle.
5.1 Work & Energy
In the Physics world, Work and
Energy are intimately related.
Energy is very difficult to
define. It is easy to say what
energy can do but not so easy
to say what it is.
Thus energy is defined in
terms of work.
An object is said to possess
energy if it has the ability or
capacity to do work.
Work is the “Transfer Mechanism”
for Energy meaning that if some
work has been done on an object
the amount of energy it possesses
has been changed.
WORK DONE = ENERGY TRANSFERRED
Motion - Revision Questions
Question type: Work and Energy
A model rocket of mass 0.20 kg is launched by means of a spring, as shown
in Figure 1. The spring is initially compressed by 20 cm, and the rocket
leaves the spring as it reaches its natural length. The force-compression
characteristic of the spring is shown in Figure 2.
Q26 : How much energy is stored in the spring when it is compressed?
A: Compressing the spring requires work to be done on it. (this work is stored
as elastic potential energy in the spring)
Work done = area under graph up to a compression of 0.2 m
= ½ (0.2)(1000)
= 100 J
Motion - Revision Questions
Question type: Energy Conversion
Q27: What is the speed of the rocket as it leaves the spring?
A: All the elastic potential energy stored at max compression
(100 J) will be converted to Kinetic Energy at release. So
100 = ½ mv2
v = √1000
= 31.6 ms-1
Motion - Revision Questions
Question type: Equations of Motion
Q28: What is the maximum height, above the
spring, reached by the rocket? You should ignore
air resistance on the way up since the rocket is
very narrow.
+ve
u = 31.6 ms-1
v=0
a = -10 ms-2
x=?
t=?
v2 = u2 +2ax
0 = (31.6)2 + 2(-10)x
x = 1000/20
= 50 m
Motion - Revision Questions
Question type: Newtons 2nd Law
Retarding
Force (R)
(Newtons)
time (s)
Q29: What is the acceleration of the
rocket at a time 5 s after the parachute
+ve
opens?
A: At t = 5.0 s, R = 1.8 N
Weight of Rocket W = mg
= (0.2)(10)
= 2.0 N
R
W
When the rocket reaches
its maximum height, the
parachute opens and the
system begins to fall. In the
following questions you
should still ignore the
effects of air resistance on
the rocket, but of course it
is critical to the force on the
parachute. This retarding
force due to the parachute
is shown as R in Figure 3,
and its variation as a
function of time after the
parachute opened is shown
in Figure 4.
ΣF = ma
a = ΣF/m = (2.0 – 1.8)/ 0.2
= 1.0ms-2
Motion - Revision Questions
Question type: Work and Energy
In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second
floor, and slides from rest to the ground floor below, as shown in Figure 4. The slide
has a linear length of 6.0 m, and is designed to provide a constant friction force
of 50 N on the box. The box reaches the end of the slide with a speed of 8.0 m s–1
Figure 4
Q30: What is the height, h, between the floors?
At the second floor the box has Potential Energy = mgh
On reaching the ground floor this has been converted to Kinetic Energy
(½mv2) plus the work done against friction in moving down the slope
thus,
mgh = ½ mv2 + Fd
(30.0)(10)h = ½ (30)(8.0)2 + 50(6.0)
h = 4.2 m
5.2 Energy Types
•
•
KINETIC ENERGY
The Energy of Motion. It is the
energy possessed by moving
objects
Mathematically:
K.E. = ½mv2
where, K.E. = Kinetic Energy (J )
m = mass (kg)
v = velocity (ms-1)
ELASTIC POTENTIAL ENERGY
•The energy stored in elastic
materials (eg. Springs & rubber bands)
•Mathematically:
Es = ½ kx2
where, Es = Elastic P. E. (J )
k = Spring Constant (N kg-1)
x = Extension or
Compression (m)
GRAVITATIONAL POTENTIAL ENERGY
• The Energy of Position. The energy
possessed by an object due to its
position.
• Mathematically:
P.E. = mgh
where, P.E. = Grav. P. E. (J )
m = mass (kg)
g = gravitational field
strength (N kg-1)
h = height above zero
point (m)
• The zero point for measuring height
is usually, but not always, the
surface of the Earth.
Energy is not a directional
quantity so all these forms of
energy are Scalar Quantities.
Motion - Revision Questions
Question type: Elastic Potential Energy
Figure 4
The box then slides along the frictionless floor, and is momentarily
stopped by a spring of stiffness 30 000 N m–1
Q31: How far has the spring compressed when the box has come to rest?
KE of box is converted to Elastic Potential Energy in the spring
Thus ½ mv2 = ½ kx2
½ (30.0)(8.0)2 = ½ (30,000)(x)2
x = 0.25 m
•
5.3 Power
Power is defined as the Time rate of
doing Work.
• Mathematically:
P = W/t
where,
• P = Power (W - Watts)
W = Work (J )
t = time (s)
• Power is a Scalar Quantity
• SINCE WORK DONE
= ENERGY TRANSFERRED
• Power can also be defined as the Time
rate of Energy Transfer.
• Mathematically:
P = E/t
where,
P = Power (W - Watts)
E = Energy (J )
t = time (s)
• It is useful to note that since W = F.d
and v = d/t the original power formula
P = W/t = (F.d)/t =F.v
• This allows the calculation of the
power of a body moving at constant
velocity
5.4 Conservation of Energy
•
•
The law of conservation of
energy states:
ENERGY CAN BE NEITHER
CREATED NOR DESTROYED
BUT ONLY TRANSFORMED
FROM ONE FORM TO ANOTHER.
Energy conversion processes
convert useful, ORDERED energy
(energy capable of doing useful
work) into useless,
DISORDERED energy. This
disordered energy is sometimes
called Thermal Energy, or Low
Grade Unrecoverable Heat.
The percentage efficiency of energy
transfers is calculated from:
%Eff = EOUT/EIN x 100/1
No energy transfer process can be
100% efficient in our friction ridden
world.
If 100% efficiency could be attained
perpetual motion machines would then
be possible.
5.5 “Seeing” Energy (1)
With a little practice you can
“watch” energy flow through a
system just as an accountant
can “watch” money flow
through an economy.
The most obvious form of
energy is Kinetic Energy, the
energy of motion. It is easy
to see when kinetic energy is
transferred to or from an
object.
As KE leaves the object it
slows down: as in a car
slowing down when you lift off
the accelerator pedal.
A water polo ball speeds up when
you do work on it during the action
of throwing, you are transferring
energy from your body into the ball,
where the energy shows up in the
motion of the ball.
5.6 Seeing Energy (2)
Potential Energy is more difficult to
“see”.
It can take many different forms, as
shown in the table below.
In each case nothing is moving; but
because the objects still have a great
potential to do work, they possess
Potential Energy.
At certain points all a body's energy
may be potential whilst at another
all its energy may be kinetic.
Recognizing these points will
enable the solutions to most
problems in this area.
Form of Potential Energy
Example
1. Gravitational Potential Energy
A person standing at the top of a
building
2. Elastic Potential Energy
A wound clock spring
3. Electrostatic Potential Energy
A cloud in a thunderstorm
4. Chemical Potential Energy
A firecracker
5. Nuclear Potential Energy
Uranium
Chapter 6
Topics covered:
• The Experience of Acceleration.
• Circular Motion.
• Centripetal Force.
• Vertical Circles.
• Projectile Motion.
• Projection Angles
• Projectile Graphs
• Real Life Projectiles
6.0 The Experience of Acceleration
Nothing is more central to the
understanding of the laws of
motion than understanding the
relationship between force and
acceleration.
Up to now we have looked at
forces and noticed they can
produce accelerations (Newtons
2nd Law).
Now we will reverse the process –
looking at accelerations and
noticing that they require a force.
For you to accelerate, something
must push or pull on you.
Just where and how that force is
exerted on you determines how
you “feel” when you accelerate.
The “backward” force you experience
when a car accelerates is caused by
your body’s “inertia”, its tendency not to
want to accelerate.
The car and your seat are accelerating, and
since the seat back acts to keep you from
falling “through” its surface, it provides a
forward support force which causes you to
accelerate forward.
But the seat can’t exert a force uniformly
throughout your body. Instead, it pushes
only on your back and your back then
pushes on your bones, internal organs and
tissues to make them accelerate forward.
The experience of the accelerating car
seat is very similar to the experience of
“gravity” when you stand still on the
Earth’s surface. In this situation you feel
“heavy” ie. you experience your weight
6.1 Fictitious Forces
When the car seat is causing you to
accelerate forward, you also feel
“heavy”; your body senses all the
internal forces needed to accelerate its
pieces forward, and you interpret these
sensations as “weight”.
This time you experience the weight
directed toward the back of the car.
The gravity-like “force” that you
experience as you accelerate is truly
indistinguishable from the force of
gravity.
No laboratory instrument can
determine directly whether you are
experiencing gravity or simply
accelerating.
However despite the convincing
sensations, the backward, heavy
feeling in your gut as you accelerate
is not due to a real force.
This experience of acceleration is
explained by the supposed action of a
Fictitious Force.
This Fictitious Force always points in
the direction opposite to the
acceleration that causes it and its
strength is proportional to the
acceleration.
a
FFict
6.2 Apparent Weight
With the car stationary the
only force you are subject to
is your Weight
As the car accelerates you
are “pushed back” in your
seat by the fictitious force
The vector sum of the Weight
and Fictitious force produces
the Apparent Weight you
experience whilst accelerating.
Fictitious
Force
Apparent
Weight
a
Fictitious
Fictitious
Force
Force
aa
Apparent
Apparent Weight
Weight
Weight
Weight
The larger the Acceleration, the
larger the Fictitious force, the more
“backward” your Apparent Weight
becomes.
Weight
6.3 Circular Motion
•
•
•
•
•
•
Newton’s 1st law says, in part, an
object experiencing no net force
will travel in a straight line at
constant speed.
Thus, in order to make an object
travel in a circle ie. constantly
change the direction of its velocity,
a force must be constantly applied
to it. This means the object is
constantly accelerating.
The number of times an object
spins around per second is called
its Frequency (f).
An object travelling around a circle
completes one full spin in a time
interval called a Period (T). Period
and frequency (f) are the inverse of
one another. Thus:
T = 1/f
T =Period (sec)
f = Frequency (Hz or sec-1)
CIRCLE OF RADIUS = R
R
Circumference of circle = 2  R
Time for one revolution = Period = T
Speed = Distance , So v = 2 R = 2 Rf
Time
T
V
6.4 Centripetal Acceleration
Since the direction of the velocity at any
point is the tangent to the circle at that
point, the direction of the velocity is
constantly changing, thus an object
travelling in a circular path is subject to
a constant acceleration.
The size of that acceleration (called the
CENTRIPETAL (or centre seeking)
ACCELERATION) is given by:
aC = v2/R
where
aC =centripetal acc (ms-2)
v = linear velocity (ms-1)
R = radius of circle (m)
Substituting for v = 2R/T
gives aC = 4 2 R/T2
CIRCLE VIEWED FROM ABOVE
aC = vf - vi
aC
- vi
vi
vf
THE CENTRIPETAL ACCELERATION, aC ,IS ALWAYS
DIRECTED TOWARD THE CENTRE OF THE CIRCLE.
6.5 Centripetal Force
The fact that an acceleration exists, directed toward the centre of the circle
requires there to be a force acting in the
aFCC
same direction.
This force is called the CENTRIPETAL
FORCE (FC) and is defined in terms of
Newton’s 2nd law.
Mathematically:
Girl Riding a
FC = maC = mv2/R = m42R/T2
Playground Ride
T
Centripetal Force is NOT a REAL
FORCE in its own right, but is
Fictitious
supplied by other real, measurable
Force
FC
forces.
Centre of Mass In the case opposite, the girl
requires a Centripetal Force to
Direction of
travel her circular path
Rotation
The Tension (T) of her muscular
grip on the pivot pole of the ride
provides the Centripetal Force.
She, of course, will use the idea of a
Fictitious Force, centrifugal force, to
Linear Velocity
explain the “outward” pull she feels while
on the constantly accelerating ride.
Motion - Revision Questions
Question type: Centripetal Force
Mark Webber and his Formula 1 racing car are taking a corner at the
Australian Grand Prix. A camera views the racing car head on at point X on
the bend where it is travelling at constant speed. At this point the radius of
curvature is 36.0 m.
The total mass of the car and driver is 800 kg.
FC
Q32: On the diagram showing the
camera’s view of the racing car, draw
an arrow to represent the direction of
the NET force acting on the racing
car at this instant.
Camera's head on view
of racing car at point X
The magnitude of the horizontal force
on the car is 6400 N.
36.0 m
Q33: Calculate the speed of the car.
X
A:
FC = mv2/R
So v =√FCR/m
= √(6400)(36.0)/(800)
= 17 ms-1
Camera
Motion - Revision Questions
Question type: Centripetal Force
Q34: Referring to the racing car from the previous slide, explain:
(a) Why the car needs a horizontal force to turn the corner.
(b) Where this force comes from.
A: (a) Newton 1 states all objects will continue in a straight line unless acted upon by a
net force.
In order for the car to travel in a circular path a constant force directed at right angles
to the direction of motion (toward the centre of the circle) must exist.
(b) The horizontal force arises from the frictional force exerted BY THE ROAD ON
THE TYRES and is directed toward the centre of the circle the car is travelling in.
Motion - Revision Questions
Question type: Centripetal Force
Fc
The safe speed for a train taking a curve on
level ground is determined by the force that
the rails can take before they move sideways
relative to the ground. From time to time
trains derail because they take curves at
speeds greater than that recommended for
safe travel.
Figure 5 shows a train at position P taking a
curve on horizontal ground, at a constant
speed, in the direction shown by the arrow.
Q35: At point P shown on the figure,
draw an arrow that shows the direction
of the force exerted by the rails on the
wheels of the train.
Motion - Revision Questions
Question type: Centripetal Force
The radius of curvature of a track that is safe
at 60 km/h is approximately 200 m.
A: Centripetal Force
Fc = mv2
R
Q36: What is the radius of curvature of a track
that would be safe at a speed of 120 km/h, This can be solved as a ratio question
assuming that the track is constructed to the
same strength as for a 60 km/h curve?
m(60)2 = m(120)2
200
x
x = 800 m
Q37: At point Q the driver applies the
brakes to slow down the train on the curve.
Which of the arrows (A to D) indicates the
direction of the net force exerted on the
wheels by the rails?
A: B
6.6 Centrifugal Force
Despite its fictitious
nature, centrifugal
force creates a
compelling sensation
of gravity like force.
The girl on the ride
feels as though
gravity is pulling her
outward as well as
down and must hold
the handle tight in
order not to fall off.
Linear Velocity
Fictitious forces,
such as Centrifugal
Force, do NOT
contribute to the net
force experienced by
an object.
Thus if the girl lets go
she will fly off the
ride in the direction
of the linear velocity,
NOT in the direction
of the fictitious force.
Anyone who talks about
centrifugal force as being a
real, measurable force is
talking rubbish and does not
understand physics, be wary
of ALL they say !!!!!
A stationary object
on Earth experiences
a weight force of 1g.
The fictitious
centrifugal force
experienced by 5 kg
of clothes during the
spin cycle in a
washing machine,
travelling in a 0.25 m
radius circle at 20
ms-1 is about 163 g’s
and they have an
apparent weight of
815 kg.
6.7
Banked
Corners
Race track and road designers often build their tracks or roads with “banking” on
the corners. This design feature is used to enhance the safety of the track or road
allowing users to round corners at higher speeds (and with a greater margin of
safety) than they could if the corner was not banked.
Any vehicle travelling around a corner with velocity v needs a
centripetal force (FC) acting toward the centre of the corner
In the case of the flat, non banked, corner this centripetal force
(FC) is supplied by the friction of the road against the tyres (FRT)
With the banked corner the centripetal force has an extra
component - that being a component of the car’s weight force
acting toward the centre of the corner.
This gives a larger overall Centripetal Force (FC) larger than in the non
R
banked case.
FC
v
R
F LAT
CORNER
Component of W acting
toward centre of corner
BANKED
CORNER
FC
FC
FRT
FRT
W
FRT
W
FRT
The larger FC allows the car either a greater margin for safety or a faster speed around
a banked corner compared to a flat corner of the same radius
6.8 Vertical Circles
Objects travelling in
vertical circles are
subject to the
acceleration due to
gravity, thus their speed
will vary depending
where in their motion
observations are made.
Analysis of this type of
motion is based on
energy considerations
and the fact that the
motion takes place in a
uniform gravitational
field.
In theoretical situations,
the TOTAL ENERGY
REMAINS CONSTANT,
but varies between
Kinetic and Potential,
depending on where in
the circle you choose to
look.
A soccer ball of mass 0.5 kg moves along a friction
free track at 20.0 ms-1 toward a vertical circle 10.0
m in diameter.
v = 20
ms-1
PE = 49 J
KE = 51 J
PE = 24.5
J
10 m
KE =100J
KE = 75.5 J
PE = 0
Its initial K.E. = ½mv2 = ½(0.5)(20)2 = 100 J
PE = mgh =(0.5)(9.8)(5.0) = 24.5 J
KE = 100 J KE = 100 - 24.5 = 75.5 J
PE = 0 J
5.0 m
h=0
6.9 Projectile Motion
Projectile Motion is the motion
which objects, launched at
some angle to the Earth’s
gravitational field, will undergo.
Projectile motion is a combination of
TWO INDEPENDENT MOTIONS.
(a) HORIZONTAL MOTION which is
CONSTANT VELOCITY motion.
(b) VERTICAL MOTION which is
CONSTANT ACCELERATION motion.
vHORIZ
vHORIZ
vVERT
The horizontal motion, being
constant velocity motion, is
covered by only one equation:
v = d/t.
The vertical motion, being
constantly accelerated motion, is
covered by the Equations of
Motion
The only common
factor between the the
two motions is THE
TIME OF FLIGHT (t).
vHORIZ
vVERT
vHORIZ
vHORIZ
vVERT
vVERT
vHORIZ
vHORIZ remains constant throughout
vVERT increases under the action of gravity
vVERT
Motion - Revision Questions
Question type: Projectile Motion
A car takes off from a ramp and the path of its centre of mass through the air is
shown below.
First model the motion of the car assuming that air resistance is small enough to
neglect.
Q38: Which of the directions (A - H), best shows the VELOCITY of the car at X ?
A: Direction C.
Q39: Which of the directions (A - H), best shows the VELOCITY of the car at Y ?
A: Direction D
A
B
A
B
H
G
C
D
Direction of Motion
H
G
C
X
F
D
E
Y
F
E
Motion - Revision Questions
Question type: Projectile Motion
Q40: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?
A: Direction E.
Q: Which of the directions (A - H), best shows the ACCELERATION of the car at Y ?
A: Direction E.
Now suppose that AIR RESISTANCE CANNOT BE NEGLECTED.
Q41: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?
A: Direction F.
A
B
A
B
H
G
C
D
Direction of Motion
H
G
C
X
F
D
E
Y
F
E
Motion - Revision Questions
Question type: Projectile Motion
A bushwalker is stranded while walking. Search and rescue officers drop an
emergency package from a helicopter to the bushwalker. They release the package
when the helicopter is a height (h) above the ground, and directly above the
bushwalker. The helicopter is moving with a velocity of 10 ms–1 at an angle of 30° to
the horizontal, as shown in Figure 1. The package lands on the ground 3.0 s after its
release. Ignore air resistance in your calculations.
Figure 1
Q42: What is the
value of h in Figure
1?
A: Can approach in a
number of ways.
One way is: Find how long it takes to rise and fall back to same vertical height (h)
Then subtract from total time to give a time for vertical fall under gravity with an initial
velocity of 10 ms-1 directed at 30o below the horizontal
Motion - Revision Questions
Question type: Projectile Motion
Figure 1
1. Time to get back to h
Since this part of the journey is symmetrical time
to reach max height = ½ time for this part of
journey.
h
2. Value of h
again analyse vertical motion
Analyse vertical motion
u = 10 Sin 30o
2
s
=
ut
+
½
at
v
=
?
o
-1
u = - 10 Sin 30 ms
+ve a = 10 ms-2
h = (5)(2) + ½ (10)(4)
v=0
v = u + at
h = 30 m
s=h
-2
+ve a = 10 ms
0
0 = -10 Sin 30 + 10t
t = 2.0 s
s=?
t = 5/10 = 0.5 sec.
t=?
So, total time to get back to h = 1.0 s
Motion - Revision Questions
Question type: Projectile Motion
Figure 1
Q43: Assuming that the helicopter continues to fly with its initial
velocity, where is it when the package lands?
Which one of the statements below is most correct?
A. It is directly above the package.
B. It is directly above a point that is 15 m beyond the package.
C. It is directly above a point that is 26 m beyond the package.
D. It is directly above a point that is 30 m from the bushwalker.
Both the package and the helicopter have the same (constant) horizontal velocity,
so the package will always be directly below the helicopter.
So, alternative A is correct
Motion - Revision Questions
Question type: Sketch Graphs
Q44: Which of the graphs below best represents the speed of the package
as a function of time?
A: The key to the question is SPEED, this is the sum of both vertical and horizontal.
Speed never falls to zero (there is always a horizontal component), so C and D are out
A shows the object’s acceleration falling as it nears the ground – not so !
So B must be the correct answer
Motion - Revision Questions
Question type: Projectile Motion
Fred is playing tennis on the deck
of a moving ship.
He serves the ball so that it leaves
the racket 3.0 m above the deck
and travels perpendicular to the
direction of motion of the ship.
The ball leaves the racket at an
angle of 8° to the horizontal.
At its maximum height it has a
speed of 30.0 ms-1. You may ignore
air resistance in the following
questions.
Q45: With what speed, relative to the
deck, did the ball leave Fred’s racket?
Give your answer to three significant
figures.
A: Assuming projectile behaviour:
VHORIZONTAL = 30 ms-1 throughout the flight
30 ms-1
x = initial speed
x
8o
30 ms-1
thus, Cos 8 = 30/x
x = 30/Cos 8
= 30.3 ms-1
Motion - Revision Questions
Question type: Projectile Motion
Q46: At its highest point, how far was the
ball above the deck?
30 ms-1
x
8o
VVERT
30 ms-1
A: Using the equations of motion for the balls flight from racquet to point
of max height, where VVERT = 0
+ve
u = VVERT = 30.3 Sin 8
v=0
a = -10 ms-2
s=?
t=?
v2 = u2 + 2as
0 = (30.3 Sin 8)2 + 2(-10)s
s = 0.89 m
So Height above the DECK = 3.0 + 0.89 = 3.89 m
Motion - Revision Questions
Question type: Relative Motion
The ship is travelling straight ahead at a
velocity of 10 ms-1
Q47: When the ball is at its highest point at
what speed is it moving relative to the
ocean?
30 ms-1
10 ms-1
A: V BALL REL OCEAN = VSHIP + VBALL
VSHIP
=
+
VBALL
30
10
θ
VBALL REL OCEAN
VBALL REL OCEAN = √(30)2 + (10)2
= 31.6 ms-1
Q48: at what angle is the ball travelling
relative to the direction of the ship’s travel?
Tan θ = 30/10 = 3.0
θ = 71.60
6.10 Projection Angles
+ve
v
vVERT

vHORIZ
•
There are two basic types of
vHORIZ = v Cos 
vVERT = v Sin 
projectile motion:
This does not vary This does vary from
(a) Objects initially projected
during the flight
+v Sin  at t = 0
horizontally from some point
to zero at t = ½t
above the Earth’s surface. As
to - v Sin  at t = t
shown previously .
(b) Objects initially projected at
Flight can be broken up into
some angle above the
2 equal parts each taking exactly
horizontal as shown here.
½t to complete
v
Total flight time = t
At this point all velocity is horizontal
6.11 Projectiles – Time, Height & Range
v
h
θ
Range (R)
An object is fired from ground level with a
velocity v at an angle θ to the horizontal
It reaches a maximum vertical height, h
and finally lands at the same horizontal
level as it started, having covered a
horizontal distance of R
1. Time to reach Maximum Height (h)
Upward is +ve.
u = v sin θ use eqns of motion:
v=0
v = u + at
a = -g
0 = v sin θ – gt
s=?
t = v sin θ
t=?
g
2. Total Time of Flight
Total time = (time to reach h) x 2 = 2 v sin θ
g
3. Maximum Height
Upward is +ve
u = v sin θ use eqns of motion
v=0
v2 = u2 + 2as
a = -g
0 = v2 sin2θ – 2gh
s=h
h = v2 sin2θ
t = v sin θ
2g
g
4. Range of Projectile
Horizontally vH = dH / t
vH = v cos θ
R = 2v sinθ x v cos θ
dH = R
g
t = 2v sinθ
2 x 2 sin θcos θ
=
v
g
g
= v2 sin 2θ
g
5. Maximum Range
occurs when sin 2θ = 1
2θ = 900
or θ = 450
•
•
6.12
Real
Life
Projectiles
The mathematical analysis of projectile
motion ignores the effects of friction,
in particular air resistance.
The real world trajectory differs from
the theoretical trajectory in three main
ways.
(1) The actual range (horizontal
distance) covered in the real world
is less because of the reduction of
the horizontal component of
velocity due to air resistance.
(2) The actual height achieved will
be less, due to the effect of air
resistance on the upward vertical
velocity.
(3) The object will fall more steeply
than it rises, the path of the
projectile is no longer symmetrical
around its highest point.
Loss of height
Theoretical
Actual
Loss of Range
Path no longer
symmetrical
Chapter 7
Topics covered:
• Elastic Materials and Hooke’s Law.
• Elastic Potential Energy.
7.0 Elastic Materials &
Hooke’s Law
Elastic materials such
as springs display
elastic behaviour. This
means that when they
are deformed, stretched
or compressed, they
change their shape or
condition.
When the force causing
the deformation is
removed, they return to
their original shape or
condition.
Such materials are able to
store energy while
deformed and release that
energy when allowed to
return to their original
condition.
Mathematically
F = - k(∆x)
The negative sign indicates that
the restoring force F acts in the
opposite direction to the
extension or compression ∆x
The relation between
Force and the Extension
and/or Compression of
an elastic material is
called Hooke’s Law:
where
F = Restoring Force (N),
k = Spring Constant (Nm-1)
∆x = Extension and/or
Compression (m)
•
•
•
7.1 Potential Energy in
Elastic
Materials
Elastic materials store energy when
they are deformed and release that
energy when they return to their
original condition.
The amount of energy stored can be
found from the Elastic Potential
Energy Formula:
ES = ½kx2
where
ES = Elastic P. E. (J )
k = Spring Constant (Nm-1)
x = extension and or
compression (m)
For materials which display
“irregular” behaviour, the Potential
Energy stored can only be found from
the area under the Force vs Extension
(or Compression) graph.
“REGULAR” ELASTIC BEHAVIOUR
Slope = Spring
Force
constant (k)
F
Extension
x
Area = ½Fx = ½kx2
= Energy stored
up to extension x
“IRREGULAR” ELASTIC BEHAVIOUR
Area = Energy stored up to x
Area is found by “counting squares”
Force
F
Extension
x
Chapter 8
Topics covered:
•Universal Gravitation
•Gravitational Attraction
•Circular Orbits
•Maths of Circular Orbits
•Kepler’s Laws
•Satellites in Space
•Energy Transfers
8.0 Law of Universal Gravitation
•
•
•
•
•
•
Gravity is the most well known of
all the Natural Forces.
We live with the effects of gravity
every day and would lead
completely different lives if
gravity was not present. It is
gravity alone which gives us our
sense of “up and down”.
Gravity is a FIELD force, it acts at
a distance.
It is ALWAYS an ATTRACTIVE
force.
Gravity arises from the interaction
of 2 masses.
The size of the gravitational force
is directly proportional to the
product of the two masses and
inversely proportional to the
square of the distance between
them.
Mathematically:
Fg = GM1M2
R2
where
Fg = Gravitational Force (N)
G = Universal Gravitational Constant
(6.67 x 10-11 Nm2kg-2)
M1, M2 = masses (kg)
R = Separation of the masses (m)
EACH of the bodies experiences the
SAME FORCE even though they may
have vastly different masses.
Motion - Revision Questions
Question type: Universal Gravitation
Newton was the first person to quantify the gravitational force
between two masses M and m, with their centres of mass
separated by a distance R as
F= GMm
R2
where G is the universal gravitational constant, and has a
value of 6.67 × 10-11 N m2 kg2.
For a mass m on the surface of Earth (mass M) this
becomes F = gm, where g = GM
R2
Q49: Which one of the expressions (A to D) does not describe the term g?
A. g is the gravitational field at the surface of Earth.
B. g is the force that a mass m feels at the surface of Earth.
C. g is the force experienced by a mass of 1 kg at the surface of Earth.
D. g is the acceleration of a free body at the surface of Earth.
A: B
Motion - Revision Questions
Question type: Universal Gravitation
The radius of the orbit of Earth in its circular motion around the
Sun is 1.5 × 1011 m
Q50: Indicate on the diagram,
with an arrow, the direction of the
acceleration of Earth.
Q51: Calculate the mass of the Sun.
Take the value of the gravitational
constant G = 6.67 × 10–11 N m2 kg–2.
(Figure 3).
Question assumes you know the period of rotation of the earth around the sun
ie. 365 days = 365 x 24 x 60 x 60 = 31536000 s (3.15 x 107 s)
F C = Fg
me4π2 R/T2 = Gmems/R2
ms = 4π2R3
GT2
ms =
4(3.14)2(1.5 x 1011)3
6.67 x 10-11(3.15 x 107)2
ms = 2 x 1030 kg
Motion - Revision Questions
Question type: Universal Gravitation
Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at
a constant speed at a radius R = 4.22 x 107 m from the centre of Earth.
Q52: What is the speed of Nato III in its orbit ?
A: The centripetal force (FC) required by Nato III to
complete its circular orbit is supplied by the force of
gravitational attraction (Fg) between Earth and Nato III.
Thus Fg = FC
Thus GMem/R2 = mv2/R
So
v = √GMe/R
= √(6.67 x 10-11)(5.98 x 1024)
-------------------------------------(4.22 x 107)
= 3074 ms-1
= 3.1 x 103 ms-1
R
Earth
Nato III
Motion - Revision Questions
Question type: Circular Motion
Q53: Which ONE of the following statements (A - D) about Nato III is correct ?
A: The net force acting on Nato III is zero and therefore it does not accelerate.
B: The speed is constant and therefore the net force acting on Nato III is zero.
C: The is a net force acting on Nato III and therefore it is accelerating.
D: There is a net force acting on Nato III, but it has zero acceleration.
A: Alt C is correct
R
Earth
Nato III
8.1 Gravitational Attraction
What Gravitational Force do the Earth
and the Moon experience because of
their proximity to one another ?
Earth
FG
R = 384,000 km
= 3.84 x 108 m
FG
Moon
Note the distance (R) is measured
from the Centre of Mass of each body
Mass of Moon
MM= 7.34 x 1022 kg
FG = GMEMM
Mass of Earth
ME = 5.98 x 1024 kg
R2
= (6.67 x 10-11)(5.98 x 1024)(7.34 x 1022)
(3.84 x 108)2
= 1.99 x 1020 N
The Earth experiences an attractive force of 1.99 x 1020 N directed toward the Moon
while AT THE SAME TIME the Moon experiences the same force directed toward Earth.
Motion - Revision Questions
Question type: Gravitational Force
Q54: What is the magnitude of the force exerted by Earth on a
water molecule of mass 3.0 × 10-26 kg at the surface of Earth?
A: F = GMm
R2
= (6.67 x 10-11)( 5.98 x 1024)(3.0 x 10-26)
(6.37 x 106)2
= 2.95 x 10-25 N
Motion - Revision Questions
Question type: Gravitational Field Strength
Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a
constant speed at a radius R = 4.22 x 107 m from the centre of Earth.
R
Q55: Calculate the magnitude of the Earth’s
gravitational field at the orbit radius, R = 4.22 x 107 m,
of Nato III. Give your answer to 3 sig figs. You MUST
show your working. G = 6.67 x 10-11 Nm2kg-2 Me = 5.98
x 1024 kg.
Earth
Nato III
A:
g = GMe/R2
= (6.67 x 10-11)(5.98 x 1024)
-----------------------------------------(4.22 x 107)2
= 0.224 Nkg-1
8.2 Circular Orbits under Gravity
On the large scale of planets,
moons, stars and other bodies in
the universe, their motions are
determined by the gravitational
attractions between them.
If, for example, a moon travels in
a circular orbit around its host
planet, it must be subject to a
Centripetal Force.
This Centripetal Force must be
supplied by some kind of
interaction between the planet and
its moon.
This interaction is the
GRAVITATIONAL ATTRACTION
between the Planet and its Moon.
Thus the Centripetal Force (FC) needed
for circular motion is supplied by the
Gravitational attraction (Fg) between
the planet and its moon.
Planet
Moon
FG
FG
FC
8.3 Mathematics of Circular
Orbits
FC
FG
FG
Moon
Planet
The Centripetal Force (FC) the
moon is subject to is given by:
FC = MMv2/R
Remember the velocity of an
object travelling in a circle is:
v = 2R/T
 v2 = 4 2R2/T2
Substituting for v2 we get:
FC = MM 42R/T2
This Centripetal Force is supplied by the
Gravitational Force between the the Planet
and the Moon:
FG = GMMMp/R2
 GMMMP/R2 = MM42R/T2
Rearranging we get:
R3/T2 = GMP/42
The terms G, MP, and  are all constants
so their ratio is constant.
 The ratio R3/T2 is also a constant
8.4 Kepler’s Laws
Johannes Kepler (1570 - 1630)
discovered the laws governing
planetary motion which
describe the movement of the
planets in our solar system.
In the previous slide we found the
ratio R3/T2 was a constant,
(its value is roughly 3.0 x 1018 ).
The 3rd law:
R3/T2 = GMP/42 can be
rewritten as:
R=
3
GMPT2
42
Notice that the radius of orbit of any
satellite, while it does depend on the
mass of the planet around which it
circulates, DOES NOT depend its own
mass.
This is true of any satellites whether
man made of naturally occurring.
In 1615 Kepler discovered that the
ratio of the of the cube of the
average sun - planet distance (R3)
to the square of its period (T2)
was a constant for ALL planets in
our solar system, this became
known as Kepler’s Third Law.
His other 2 laws establish
planets’ speeds and the
elliptical nature of planet orbits.
Motion - Revision Questions
Question type: Kepler’s 3rd Law
A satellite in a circular orbit of radius 3.8 × 108 m
around Earth has a period of 2.36 × 106 s.
Q56: Calculate the mass of Earth. You must show
your working.
A:
R3 = GMe
T2 4π2
Me =
(4π2) (3.8 x 108)3
(6.67 x 10-11) (2.36 x 106)2
= 5.83 x 1024 kg
8.5 Satellites in Space.
A satellite moving through space will
often use the gravitational field of a
planet, like Jupiter or star like our Sun to
help propel it through space.
A
At point A, the satellite comes under the
influence of the gravitational field of the planet.
The field does Work ON the satellite, accelerating
it toward the planet.
B
By the time it has reached B, the satellite has
increased its Kinetic Energy, and hence its
Speed, sufficiently to pass around, rather than
crash into, the planet.
Jupiter
The satellite flies past the planet leaving with
greater speed, having “stolen” some of the
energy stored in the planet’s field.
Path of Satellite
around Planet
This is the normal way of
sending satellites to the outer
planets and even outside our
solar system.
8.6 Energy Transfers in
Gravitational Fields
The Gravitational Field vs
Distance graph for Jupiter
showing positions A and B for the
satellite is shown.
Since Work Done = Energy
Transferred, the area under the graph
represents the work done by the field
on, and the change in energy
possessed by, “1 kg of satellite mass”
in moving from A to B.
In exam questions, the area
normally needs to be found by
the “counting the squares”
method.
For the satellite with a
mass >1 kg, total energy
possessed = area x mass
g(Nkg-1)
Area gives:
1. Work by the field on 1 kg of satellite moving from A to B.
2. The increase in Kinetic Energy possessed by 1 kg
of satellite in moving from A to B.
3. The loss in Potential Energy of 1 kg of
satellite moving from A to B
RJ
B
A
Distance R (m)
Motion - Revision Questions
Question type: Weight & Weightlessness
The Russian space station MIR (Russian meaning - peace) is in a circular
orbit around the Earth at a height where the Gravitational Field Strength is
8.7 Nkg-1
Q57: Calculate the magnitude of the gravitational force exerted by Earth
on the astronaut of mass 68 kg on MIR
A:
W = mg
= (68)(8.7)
= 592 N
When the astronaut wishes to rest he has to lie down and strap himself into bed.
Q58: What is the magnitude of the force that the bed exerts on the astronaut before
he begins to fasten the strap ?
A: 0 N
Motion - Revision Questions
Question type: Weightlessness
Newspaper articles about astronauts in orbit sometimes speak about zero
gravity when describing weightlessness.
Q59: Explain why the astronaut in the orbiting MIR is not really weightless.
A: Weight is the action of a gravitational field acting on a mass.
For true weightlessness to exist the gravitational field strength needs to be
zero.
There is no place in the universe where the gravitational field strength is zero,
so nothing is truly “weightless”.
The astronaut actually does have weight, as calculated earlier.
The astronaut’s “apparent weight” is zero because he is not subject to a
reaction force inside the spacecraft since both he and the spacecraft are in a
state of constant free fall.
Ollie Leitl 2009