Download Kinematics - Vicphysics

Physics Unit 3 2009
Newton’s laws of motion
1 of 17
Kinematics
Study Design
Apply Newton’s three laws of motion in situations where two or more coplanar forces act along a
straight line and in two dimensions;
Introduction
……………………………….
You will be familiar with many of the concepts of motion from your Physics Unit 2 studies. You
must know the definitions of the key terms, such as distance, displacement, speed, velocity,
acceleration, force, momentum and energy. In Unit 3 you will be expected to use many of the same
techniques to solve problems as in Year 11. This repeated content is illustrated in the diagram
below.
……………………..
Momentum
 Conservation of
momentum
 Impulse
 Graphical Solutions
Kinematics
 x,u,v,a,t formulae
 graphs
Motion
Energy
 Types (kinetic,
gravitational
potential, elastic
potential, Work)
 Calculations using
Conservation of
energy
Forces
 Newtons Laws
 Vector in one and
two dimensions
When solving Year 12 problems, you will have to choose which concept of physics (Kinematics,
Momentum, Energy or Forces) will be most useful in solving the problems.
Some problems you will be only able to solve using one concept, others can be solved in more
than one way, but may be very simple using a particular technique. Hence you must be able to
solve problems using all of the concepts, and you must develop an instinct for choosing the most
efficient path.
You will also need to decide whether to use a graphical or numerical technique to solve problems.
In addition there are a few new topics, specifically motion in more than on dimension (projectile
motion), motion in different inertial frames, circular motion and Newton’s law of Universal
Gravitation.
……………………………….
Newton’s laws of motion
Physics Unit 3 2009
2 of 17
Definitions
From your prior studies of motion you should be familiar with the following kinematic definitions:
Distance Travelled
Displacement
Speed
Velocity
Acceleration
- How far an object has moved in total during its motion. (m).
– How far an object is at from a reference position. (m)
– How fast an object is moving. (m/s)
– How fast an object and what direction an object is moving in. (m/s)
– The rate at which the velocity of an object is changing i.e. how many (m/s)
the velocity of an object is changing by every second. Acceleration also has
a direction
These physical quantities can be divided into two categories, scalars and vectors.
Vectors:
Vectors are quantities that have a magnitude and a direction.
E.g. displacement, velocity, and acceleration.
Scalars:
Scalars are quantities that only have a magnitude,
E.g. speed and distance travelled.
Average Quantities
You will occasionally be asked to determine average quantities. For example, you may be asked to
determine the velocity, on average, at which a car was moving between two times. Average
velocity and acceleration are determined using the following formulae.
average velocity =
total displacement
time taken
average acceleration =
change in velocity
time taken
vav =
aav =
x 2 - x1
Δt
v -u
Δt
where
x2 is the final position,
v is the final velocity and
x1 is the initial position,
u is the initial velocity.
t is the time period,
Graphs
You need to be able to use a wide range of graphs. When given a graph in the exam, look for the
following on the graph before even reading the question:
 type of graph (F – d, F – v, Energy – distance, F – t etc.).
 the units on the axis.
 the limit reading on each axis.
 Look at the scale on both axes, be aware for anything non-standard
 Think about what is given by a direct reading from the graph, the gradient of the graph and the
area under the graph
In Year 11 Physics it is typical to restrict the types of graphs that you experience to those with
‘time’ usually on the horizontal axis. Expect to find ‘distance’ ‘speed’ and many others on the
horizontal axis in Year 12.
Physics Unit 3 2009
Newton’s laws of motion
3 of 17
Graphical Techniques
In kinematics you can be asked to interpret several graphs. Graphs can be used to determine
instantaneous quantities i.e. the value of a quantity at a specific time. For example, a velocity time
graph (v-t) can be used to determine how fast an object was moving at a specific time. It could also
be used to determine how far the object has moved up to that time (by finding the area under the
curve) or its acceleration (by determining the gradient at a specific point). The type of information
that can be determined from different graphs is summarised in the following table.
x-t
v-t
a–t
'x' at any 't'
't' at any 'x'
'v' at any 't'
't' at any 'v'
'a' at any 't'
't' at any 'a'
Instantaneous
velocity at any
point.
Vav between any
two points
Meaningless
Instantaneous 'a'
Average 'a'
Meaningless
x
v
Graph type 
Found from
Direct reading
Gradient
Area under graph
The gradient at a particular time is determined by drawing a tangent line to the curve at that point,
and then determining the gradient of the tangent line.
Constant Acceleration
Consider the following series of graphs. These illustrate the relationships mentioned in the table
above. Notice that the velocity – time graph is the gradient of the displacement – time graph, and
the acceleration – time graph is the gradient of the velocity – time graph.
Displacement
Velocity
Acceleration
v
g
u
time
time
t
time
Physics Unit 3 2009
Newton’s laws of motion
4 of 17
Examples
The figure below appeared in a newspaper featuring skydiving from an aircraft. In this particular
example the total mass of the skydiver and equipment is 100 kg. The skydiver jumps from a height
of 3000 m above the ground and reaches a constant terminal velocity of 190 km h-1 in a time of 15
s. She then falls at this constant speed of 190 km h-1 for a further 35 s before opening the
parachute.
Example 1
Question 1 (1998)
Convert 190 km h-1 into m s-1.
Example 2
Question 2 (1998)
On the set of axes, sketch a graph of the motion of the skydiver for the first 50 s of falling. (Air
resistance cannot be neglected.)
Example 3
Question 3 (1998)
Explain why the speed remains constant between 15 s and 50 s of the motion.
Newton’s laws of motion
Physics Unit 3 2009
5 of 17
An object moves along a straight path. Below is a graph of velocity verses time of the object’s
motion.
velocity
ms-1
2
1
2
0
1
2
3
4
5
6
7
8
-1
Example 4
Question 1 (1983)
What is the average speed of the object’s motion during the first 3.0 second?
Example 5
Question 2 (1983)
What is the object’s distance from the starting point at 9.0 second?
Example 6
Question 3 (1983)
What is the acceleration of the object at 8.5 second (magnitude and sign)?
9 time (s)
Newton’s laws of motion
Physics Unit 3 2009
6 of 17
Example 7
Which one or more of the following graphs (A, B, C, D, E and F) represents the motion under
constant non-zero acceleration?
(one or more answers)
velocity
acceleration
displacement
A
velocity
C
B
time
D
time
displacement
time
acceleration
E
F
Constant
time
time
time
Acceleration Formulae
The constant acceleration formulae only apply when the acceleration of the
object does not change during its entire motion. The most common
example is motion under gravity. The constant acceleration formulae are in
the box.
x is the displacement
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time period in question
Note that t is a time interval, not a specific time.
v  u  at
v 2  u 2  2ax
1 2
at
2
1
x  vt  at 2
2
(u  v)t
x
2
x  ut 
When using these formulae to solve problems it is best to write down
everything that you know from the question, and then write down the thing
that you wish to find and then find a formula that relates what you have to what you need. If you
cannot find such a formula directly, determine anything you can, and re-read the question to
ensure that you have not missed any vital information. Some other facts to consider are:
 t = 0 is the beginning of the time interval being considered, i.e. the instant at which 'u' occurs.
 a negative answer for 't' indicates a time previous to 't' = 0.
 x is not necessarily the same measure as the total distance travelled
 a body that is travelling in one direction and accelerating in the opposite direction is slowing
down.
 when given the distance travelled in a certain time interval, this distance is the instantaneous
velocity halfway through the time interval. E.g. If a body travels 14 m in the seventh second
('t' = 6 to 't' = 7 sec) then the actual velocity at 6.5 seconds is 14 m/s.
 for motion along the horizontal it is usual to take 'to the right positive' for vector sense
 for vertical motion (bodies projected vertically or dropped from rest) the direction of the initial
displacement is usually taken as positive
 for vertical motion, the acceleration (symbolised by 'g') is 10 m/s2 vertically downwards at all
times, even if the body is momentarily at the top of its vertical flight.
Newton’s laws of motion
Physics Unit 3 2009
7 of 17
The ‘standing 400 m’ time for a car is the time that it takes to travel 400 m on a level road,
accelerating from rest.
The standing 400 m time of a car was 16.0 s.
Example 8
Question 1 (2000)
Calculate the acceleration of the car, assuming constant acceleration for the entire journey.
A man at the top of a building 20m high releases a stone from rest; 0.60 second later he throws a
marble vertically downwards with an initial velocity of 8.0 ms-1.
Example 9
Question 1 (1980)
How long does it take the stone to reach the ground?
Example 10 Question 2 (1980)
Which of the following best represents the velocity-time graphs for the stone (S) and the marble
(M)?
velocity
velocity
A
velocity
B
C
M
M
M
S
S
S
time
velocity
time
time
velocity
D
E
M
M
S
S
time
time
***Example 11 (difficult)
Question 3 (1980)
How long after the stone was dropped does the marble pass the stone?
Newton’s laws of motion
Physics Unit 3 2009
8 of 17
Graphs of velocity versus time are shown below for a car and a motorcycle travelling along the
same road. The car passes the stationary motorcycle at t = 0.
Velocity
km hr-1
Motor cycle
Car
100
80
60
40
20
0
10
20
30
40
Time (sec)
Example 12 Question 1 (1978)
What is the average acceleration (in km hr-1 s-1) of the motorcycle during the first 15 seconds?
Example 13 Question 2 (1978)
At the instant t = 10 sec, the motorcycle’s
A acceleration increases, velocity decreases
C acceleration and velocity both increase
B acceleration decreases, velocity increases
D acceleration and velocity both decrease
***Example 14
Question 3 (1978)
At what time does the motorcycle overtake the car?
Forces
The relationship between a force and the acceleration it causes was first understood by Isaac
Newton(1672 – 1727). Newton summarised all motion by three laws:
Types of forces
Forces can be divided into two major categories, field forces and contact forces
.
Forces that act at a distance are called
Forces created by travelling bodies are
FIELD FORCES, (gravitational,
called CONTACT FORCES.
electrical or magnetic)
Physics Unit 3 2009
Newton’s laws of motion
9 of 17
Newtons First Law
An important consequence of this law was the
realisation that an object can be in motion without a
force being constantly applied to it. When you throw a
ball, you exert a force to accelerate the ball, but once
it is moving, no force is necessary to keep it moving.
Prior to this realisation it was believed that a constant
force was necessary, and that this force was supplied by that the air pinching in behind the ball.
This model, first conceived by Aristotle, proved tenacious, and students still fall into the trap of
using it.
Newtons 1st law of motion
If an object has zero net force acting
on it, it will remain at rest, or continue
moving with an unchanged velocity.
Newton’s first law is commonly tested on the exam. This is achieved by the inclusion of statements
such as “An object is moving with a constant velocity” within questions. Whenever you see the
key words constant velocity in a question, you should highlight them. The realisation that the
object is travelling at a constant velocity, and hence that the net force on the object is zero, will be
essential for solving the problem.
Newton’s Second Law
Newtons 2nd law of motion
This law relates to the sum total of the
forces on the body ( ΣF ) the body's mass
(m) and the acceleration produced (a)
ΣF = ma.
F
a=
m
In words, Newton’s Second Law states that a force on
an object causes the object to accelerate (change its
velocity). The amount of acceleration that occurs
depends on the size of the force and the mass of the
object. Large forces cause large accelerations.
Objects with large mass accelerate less when they
experience the same force as a small mass. The
acceleration of the object is in the same direction as
the net force on the object.
Note ΣF must have the same direction
as 'a'.
Newton’s Third Law
Newtons 3rd law of motion
For every action force acting on one object,
there is an equal but opposite reaction force
acting on the other object.
This law is the most commonly misunderstood.
You need to appreciate that these action/reaction
forces act on DIFFERENT OBJECTS and so you
do not add them to find a resultant force. For
example, consider a book resting on a table top as
shown in the diagram below. There are two forces
acting on the book: Gravity is pulling the book downward and the tabletop is pushing the book
upwards. These forces are the same size, and are in opposite directions but THEY ARE NOT a
Newton’s thirds law pair, because they both act on the same object.
N
W
The best way of avoiding making a mistake using Newton’s third law
is to use the following statement.
FA on B = - FB on A
In the example of the book on the table the Force Table on Book is a Newton third law pair with the
Force Book on Table. Notice the first force is on the book and the second force is on the table. They do
Newton’s laws of motion
Physics Unit 3 2009
10 of 17
not act on the same object. Similarly the weight force, which is the gravitational attraction of the
earth on the book, is a Newton third law pair with the gravitational force of the book on the earth.
The gravitational effect of the book on the earth is not apparent because the earth is so massive
that no acceleration is noticeable.
Drawing Force Diagrams
You will often be asked to draw diagrams illustrating forces. There are several considerations when
drawing force diagrams:
 The arrows that represent the forces should point in the direction of applied force. The
length of the arrow represents the strength of the force, so some effort should be made to
draw the arrows to scale.
 An arrow representing a field force should begin at the centre of the object.
 An arrow representing a contact force should begin at the point on contact where the force
is applied.
 All forces should be labelled.
Some sample force diagrams of common situations are drawn below.
Mass on a string
Mass in free flight
T
m
mg
mg
Velocity v = 0, so T = mg
Velocity v = constant upwards, so T = mg
Velocity v = constant downwards, so T = mg
F = mg = ma
Accelerating Upwards, T - mg = ma.
Acceleration Downwards, mg - T = ma.
Mass pulled along a plane
Smooth (No Friction)
Rough (Friction)
N
a
N
a
m
m
T
Fr
mg
T = ma, N + mg = 0
T
m
m mg
T - F = ma, N + mg = 0
Bodies with parallel forces acting
a
a
a
Newton’s laws of motion
Physics Unit 3 2009
F1
m
m
F1
11 of 17
m
m
F2
F2
F1
F2 – F1 = ma
F1 + F2 = ma
m
F2
m
F1 + F2 = ma
Bodies with non-parallel forces acting
a
a
a
F1
m
F2
F1
m
m
F2
F1
m
F2
F1 + F2 = ma
m
m
F1 + F2 = ma
F1 + F2 = ma
The vectors need to be resolved in order to solve for the acceleration.
Inclined planes
Another example of forces acting at angles to each other is an object on an incline plane. There
are only three different types of examples of a body on an incline plane without a driving force.
A body accelerating
The component of the weight force acting down the plane is larger then the frictional forces. (This
is also true if there are no frictional forces). For these situations you would take down the plane to
be positive, the reason for this is that the acceleration is down the plane.
Forces perpendicular to the plane
Fnet = mgcos  - N = 0

N
Forces perpendicular to the plane
Fnet = mgsin  - F = ma
mg

Thus the acceleration is down the
plane. If there is not friction then
the acceleration is gsin 

F
A body travelling at constant speed
This can be the when an object is not changing its speed whilst travelling down an incline or when
the object is at rest on the incline plane.
Forces perpendicular to the plane
Fnet = mgcos  - N = 0

N
Forces parallel to the plane
Fnet = mgsin  - F = 0
mg


Thus the acceleration is zero
F
Newton’s laws of motion
Physics Unit 3 2009
12 of 17
A body decelerating
For these situations you would choose up the plane to be positive, this is because this is the
direction of acceleration.
Forces perpendicular to the
plane
Fnet = mgcos  - N = 0

N
Forces parallel to the plane
mg
Fnet = F - mgsin  = ma


F
Thus the acceleration is up
the plane.
A recent Transport Accident Commission television advertisement explains the significant
difference between car stopping distances when travelling at 30 kmh-1 and 60 kmh-1.
Example 15 2000 Question 10
The stopping distance, from when the brakes are applied, for a car travelling at 30 kmh-1 is 10 m.
Which one (A – D) is the best estimate of the stopping distance for the same car, under the same
braking, but travelling at 60 kmh-1?
A. 20 m
B. 30 m
C. 40 m
D. 90 m
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
A car of mass 1300 kg has a caravan of mass 900 kg attached to it. The car and caravan move
off from rest. They have an initial acceleration of 1.25 m s-2.
Example 16 2000 Question 11
What is the net force acting on the total system of car and caravan as it moves off from rest?
Example 17 2000 Question 12
What is the tension in the coupling between the car and the caravan as they start to accelerate?
Newton’s laws of motion
Physics Unit 3 2009
13 of 17
After some time the car reaches a speed of 100 kmh-1, and the driver adjusts the engine power to
maintain this constant speed. At this speed, the total retarding force on the car is 1300 N, and on
the caravan 1100 N.
Example 18 2000 Question 13
What driving force is being exerted by the car at this speed?
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Anna is jumping on a trampoline. The figure below shows Anna at successive stages of her
downward motion.
a
b
c
d
Figure c shows Anna at a time when she is travelling downwards and slowing down.
Example 19 1999 Question 6
What is the direction of Anna’s acceleration at the time shown in Figure 4c? Explain your
answer.
Example 20 1999 Question 7
On Figure c draw arrows that show the two individual forces acting on Anna at this instant.
Label each arrow with the name of the force and indicate the relative magnitudes of the forces
by the lengths of the arrows you draw.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Newton’s laws of motion
Physics Unit 3 2009
14 of 17
Solutions
Example 1
53 m/s
To convert from km/hr to m/s you need to divide by 3.6. (This should be on your cheat sheet)
 190  3.6 = 52.8 m/s
Example 2
vel (km/hr)
200
100
0
10
20
30
40
50
60
to get full marks your graph had to show:
that the terminal velocity was reached after 15 sec
the velocity increased from 0 to 190 km/hr in the first 15 secs
that there was a smooth transition from acceleration to a terminal velocity where the
acceleration was zero.
Examiner’s comment Example 2
The 3 available marks were allocated as follows:
• 1 mark for the terminal velocity section after 15 s.
• 1 mark for a velocity increase from zero to 190 km h-1 between 0–15 s.
• 1 mark for a graph that shows a smooth transition from increasing velocity to terminal velocity.
The average mark for this question was 2.1/3, with the main error being a failure to recognise the
smooth transition, resulting in a graph as shown below. Such an answer scored only 2 marks.
Air resistance
Example 3
Constant speed (velocity) implies a net force of zero.
At terminal velocity, the air resistance (upwards) is equal
in magnitude but opposite in direction to the weight force (downwards).
It is always a good idea to include a diagram in this type of answer.
The weight force must come from the centre of mass.
The length of the vectors must be the same.
Weight = mg
Newton’s laws of motion
Physics Unit 3 2009
15 of 17
Examiner’s comment Example 3
The required explanation needed to cover the points:
• Constant speed (velocity) implies a net force of zero.
• At terminal velocity, the air resistance force (upwards) is equal in magnitude, but opposite in
direction, to the weight force (downwards).
Students who addressed this by using force diagrams were generally successful. However, a
number of written explanations simply gave the meaning of terminal velocity, rather than
addressing the physics of why the velocity remains constant. 32% of students were able to score
the full 3 marks, which is relatively low, considering the fairly straightforward nature of this
question.
Example 4
The average speed is always given by
In this case it is
area under the graph
=
time taken
distance travelled
.
time taken
1  0.5  2  2  1.5  1  1 2
2
2
3
=
4.5
= 1.5 ms-1.
3
Example 5
The distance from the starting point is called the displacement. Therefore this answer requires you
to consider vectors, so the direction of travel is important. In the first 3 seconds the object travels
4.5 m in one direction. From 3.5 (sec) to 9.0 (sec) it travels 21  1.0  1  1 3.5  21  1 1 = 4.5 m in the
opposite direction.
 after 9 (sec) the net displacement is zero.
Example 6
The acceleration is given by the gradient of the velocity – time graph. In this case the gradient is
positive, and its value is
1
= 1.
1
 +1 ms-2
Example 7
Constant non-zero acceleration means that the velocity – time graph is an oblique straight line.
 A and D
Example 8
The distance travelled was 400m, and it took 16.0 seconds. The initial speed was 0 m/s. You
need to find the acceleration.
This is given by substituting into the equation
x = ut + ½at2.
400 = 0  16 + ½  a  162
 400 = 128  a
a=
400
128
 a = 3.125 ms-2
This is probably best written as 3.13 ms-2 (correct to 3 sig figs)
Example 9
The time it takes for the stone to reach the ground is given by x = ut + ½at2
In this case u = 0, a = 10 and x = 20.
 20 = ½  10  t2
 20 = 5  t2
 4 = t2
 t = 2.
 The stone take 2 seconds to reach the
ground.
Physics Unit 3 2009
Newton’s laws of motion
16 of 17
Example 10
Since both the stone and the marble will both have the same acceleration, the gradient of the
graphs must be the same.
 Either A, C, D, E
The marble starts after the stone, so it must be either C or E.
The marble starts at v = 8.0 ms-1,  it must be graph C
***Example 11
For the marble to pass the stone, it must catch up with the stone. You need to equate two
equations of motion and solve for time (t).
If you let the marble start at time t = 0, then the time for the stone needs to be t – 0.6. This is
because it will have travelled for less time than the marble.
x = ut + ½at2 becomes x = 0  t + ½  10  t2
 x = 5t2
Marble
x = ut + ½at2 becomes x = 8  (t – 0.6) + ½  10  (t – 0.6)2
 x = 8(t – 0.6) + 5(t – 0.6)2
For the marble to pass the stone both equations must be equal.
 5t2 = 8(t – 0.6) + 5(t – 0.6)2
 5t2 = 8t – 4.8 + 5(t2 – 1.2t + 0.36)
 5t2 = 8t – 4.8 + 5t2 – 6t + 1.8
 0 = 2t – 3
2t = 3
 t = 1.5 secs.
Stone
Example 12
You must make sure that you read this question very carefully, because it is actually very easy.
The units for the answer are very unusual.
The average acceleration is given by
change in velocity
100
=
= 6.7 kmhr-1s-1
time taken
15
Example 13
B
At t = 10s, the gradient (acceleration) decreases but the velocity continues to increase.
***Example 14
The distance travelled by the two vehicles needs to be the same. The distance travelled is the
area under the graphs.
For the motorcycle the distance travelled in the first 15 seconds = 21  10  80  21 (80  100)  5 =
850m.
During the first 15 secs, the car travels 80  15 = 1200m. So at time t = 15, the car is 350m in front
of the motorcycle.
After t = 15, the motorcycle catches up at a rate of 100m every 5 seconds.
 A further 17.5 seconds to make up the 350m distance.
The motorcycle will catch the car at 15 + 17.5 = 32.5 seconds.
Example 15
C
This question was testing you understanding of how speed and stopping distance is related. Let’s
assume the mass of the car, and the force stopping it, are constant.
That is, if the car has a kinetic energy of 21 mv 2 and it requires a force by a distance to stop it then:
1
2
mv 2 = F  d if the speed of the car is doubled then the stopping distance in 4 times
larger. This is known as a square relationship.
For this question the speed of the car is 30kmh-1 and the stopping distance is 10m.
The speed of the car is then doubled to 60kmh-1 then because the speed has doubled the
stopping distance is now 4 times the 10m, which is 40m so C is the correct answer.
Physics Unit 3 2009
Example 16
Just Fnet = ma
Using the mass of the system.
Newton’s laws of motion
17 of 17
Fnet = (1300 + 900)  1.25 = 2.75  103N
Example 17
For this question think of the caravan as an object that is accelerating that is pulling it. So F = ma
= 900  1.25 = 1.13  103N
Example 18
If the car is travelling at a constant velocity then there is no net force acting on the car. Therefore
the magnitude of the driving force from the motor equals the retarding force of the car and the
caravan.
Fd = 1300 + 1100 = 2400N
Example 19
Up
At this time Anna is travelling downwards and slowing down. This means that her acceleration is
up, because it is opposing the motion (she is slowing down).
Example 20
Reaction force from the trampoline.
Weight = mg
The reaction force > weight, because she is slowing down, ie. a net upward force.
Examiner’s comment Example 20
Students were expected to draw two force arrows on Figure c. One arrow was the weight force,
acting through Anna’s centre of mass and the other arrow being the normal contact force, acting
at her feet. The arrow for the normal contact force should have been longer than the weight force
arrow.
The average mark for this question was 1.5/3, with the most common errors being in either
choosing the incorrect point of application for each force or in not indicating the relative
magnitudes as specifically asked in the question.