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Mechanics
L2 NCEA
Achievement Standard 2.4
Text Book reference: Chapters 7-13
Scalars and Vectors
A scalar quantity is one that has a
size (or magnitude) only
Eg. Mass, energy, time
A vector quantity is one that has a
size and a direction
Eg. Force, velocity, momentum
Motion
Distance
(scalar)
Symbol: d
Displacement
(vector)
Symbol: d or s
How far in total the
object has moved
Unit: m
How far the object
ends up from it’s
starting position
Unit: m
Motion
Speed
(Scalar)
How fast an object
is travelling
Symbol: s
Unit: ms-1
Velocity
(Vector)
The speed and
direction that an
object is travelling
Unit: ms-1
Symbol: v
Motion
Velocity is calculated by
Dd
velocity(v) 
Dt
Where d is displacement, t is time and D
means “the change in”.
Velocity may refer to either average
velocity or instantaneous velocity.
Constant velocity means that neither the
speed nor the direction of the objects
motion is changing.
Motion
Acceleration
(Vector)
Symbol: a
The rate at which the
velocity of an object
is changing
Unit: ms-2
Acceleration can be calculated by
Dv
a
Dt
Acceleration is always in the direction of ……………
The concorde flies at an average velocity of
1440 km hr-1. How long in seconds does it take to
fly 100km?
1440km hr-1 is 400ms-1 and 100km is 100,000m
Using v=d/t  t=d/v  100,000/400 and t=250s
How far will the concorde fly in a minute?
24km
Anyone not sure how to get that?
A car travels 500m to the right turns around and
travels another 1000m to the left.The car travelled
with a uniform speed and the time taken was 150s.
Find:
total distance travelled
1500m
total displacement
-500m
average speed of the car 10ms-1
average velocity of the car -3.3ms-1
Motion
Acceleration is used to describe motion
where the object slows down as well as
when it speeds up.
Sometimes the word deceleration is used.
Acceleration is given a negative value
when the object is slowing down.
Objects are accelerating when their
direction changes, even though their
speed may remain constant.
A car accelerates from 10ms-1 to 20ms-1 in 4.0s
Calculate its acceleration
A=2.5ms-2
Does everyone know how that was solved?
The same car can brake from 20ms-1 to rest in 5.0s
Find the acceleration.
a= 4.0ms-2
What is wrong with that answer?
a=-4.0ms-2
Task
• Measure the speed and velocity of your birds.
• I need to see times – distances and calculations
• The birds seem to move between 3-12 cms-1
Eric the snail moves at 3.0mms-1 N sees a bird and takes
off West at 4mms-1. What is his change in velocity?
Vectors
A vector is drawn as a straight,
arrowed line.
The arrow points in the direction of
the vector
The length of the line represents the
size of the quantity
Vectors
Vectors can be multiplied or divided
by a scalar
This will change the length of the
vector
A negative scalar will reverse the
direction
Eg Force F=
So
2F=
1/2F=
-3F=
Vectors
Vectors can be added together.
This is done by drawing them “head to
tail”.
The result is a vector called a resultant.
The resultant has the same effect as the 2
vectors combined.
The order in which they are added does
not matter.
d1
d2
Eg d1+d2
d1+ d2
Sp p137
Vectors
Vectors can be subtracted.
This is done by adding a negative
vector
Order does matter.
Eg. v1- v2
v1
v1- v2
v2
-v2
DV
DV=Vf-Vi
try the ball thing
If the velocity changes this means the object is………
If the object is accelerating there must be a …………..
applied in the direction of the acceleration.
Working out DV when Non Linear
1. Draw a vector representing each motion.
2. Draw the –Vi vector.
3. Draw a vector diagram of
DV=Vf-Vi or DV=Vf +-Vi
4. Using trig and/or pythagaros
find the magnitude and
DIRECTION
ofDV
Example on the board
Graphs of Motion
Distance / diplacement versus time…
Displacement(m)
B
A
C
D
Time (s)
E
A=Constant velocity (slow)
B=Constant velocity (faster)
C=Stopped
D=Constant velocity (backward)
E=Constant velocity (backward past starting
point)
Graphs of Motion
Speed / velocity versus time
Velocity (ms-1)
B
A
C
D
Time (s)
E
…area under the graph?
A=Constant acceleration (low)
B=Constant acceleration (high)
C=Constant velocity
D=Constant deceleration to stop
E=Constant acceleration in opposite direction
Kinematic Equations
To solve problems involving objects moving in
straight lines with constant acceleration.
Terms used:
d=distance/displacement (m)
vi=initial velocity (ms-1)
vf=final velocity (ms-1)
a=acceleration (ms-2)
t=time (s)
Kinematic Equations
 vi  v f
d  
 2
v f  vi  at

t

If you know 3 variables
you can work out the other
2
2
1
d  vi t  at
2
v  v  2ad
2
f
2
i
Tricks of the Trade
It is assumed you know gravity in
any problem which involves rising or
falling.
Look out for Vi=0 or Vf=0 in other
words from rest or stops.
Make sure you get the signs correct.
A rising object will have
–acceleration due to gravity acting
in the opposite direction to motion.
• A grasshopper’s legs extend by 2.0cm in
0.020s when jumping from rest. Assuming the
jump is vertical:
• What is the average acceleration of the
grasshopper while extending it’s legs?
• With what velocity does the grasshopper
leave the ground?
• What is the maximum height the grasshopper
can jump?
• A flea takes 1.0 millisecond to reach take off
speed of 1.2 ms-1 in a jump.
• What is it’s average acceleration?
• Assuming vertical take off how high does the
flea reach?
• A jet plane lands on one end of a runway 1.0km
long. It’s maximum stopping acceleration is
-4.0ms-2 and it takes 20s to come to rest.
Does the plane stop in time?
Vectors
Vectors can be resolved into
components.
This is done using SOHCAHTOA
and/or a2+b2=c2
Vertical
F
Component
40°
Horizontal component
A ship sails from Lyttelton and sets a straight
course of 130km in a direction N230E from the
New Brighton pier.
How far North of the pier is the ship?
130 cos 230 =
120km
How far east of the pier is the ship?
130 sin230 =
51km
A supermarket trolley is pushed with a force of
200N acting at an angle of 400 to the ground. Find
the effective horizontal force pushing the trolley
along.
200N
400
FH=Fcos
=200 cos 400
= 153N
How fast is the ball rising after being hit?
How fast is the ball moving horizontally?
Vv = V sin 47.10
= 52.0 sin 47.10
= 38.1ms-1
52.0ms-1
47.10
VH=Vcos 47.10
=52.0 cos 47.10
= 35.4ms-1
Kiwi Bobsled
When a green light shows the team
accelerates at 2.0ms-2 for 5.0s and then
they all jump in. Acceleration still the
same.
How fast is it going after 5.0s?
What is the distance after 5.0s?
What is the average speed @ 5.0s?
What distance is covered when v=40ms-1
R.McLeod the Cyclist
• If he rides at 6.0ms-1 for 6.0s then 12ms-1
for 12s what is his average speed.
•Clue: the answer is not 9.0ms-1
10ms-1
• Mr KK runs athletically up the stairs at 5.5ms-1.
A bunch of chemists are lazily traveling on the
esculator at 2.3ms-1. What is the relative speed
of Mr KK w.r.t. :
The chemists?
The ground?
A group of shoppers going
down a similar esculator?
• A train goes by at 95ms-1
• A Man is walking forward
at 1.2ms-1
• How fast will the man be
moving to an observer on
the ground? In the train?
• A train goes by at 95ms-1
• The Man is walking towards the
back now at 1.2ms-1
• How fast will the man be
moving to an observer on
the ground? In the train?
• A train goes by at 95ms-1
• A Man is walking forward
at 1.2ms-1
• How fast will a bird flying 10ms-1
in the same direction see the man
moving? How fast will the man see
the bird flying?
• A train goes by at 95ms-1
A bird is flying 10ms-1 in the same
direction as the train. How fast
will these people see the bird
flying?
• A train goes by at 95ms-1
A bird is flying 10ms-1 in the
opposite direction as the train.
How fast will these people see
the bird flying?
Relative Velocity
The velocity of one object in relation to
another object.
The velocity an object appears to move
at may change if the object measuring
is also moving.
The velocity of B relative to A can be
calculated by doing this vector
subtraction….
vBrelA  vB  v A
(Do Page 49 Questions 3B)
Projectile Motion
Projectile motion is a parabolic
shaped motion experienced by
moving objects that have only the
force due to gravity acting on them.
Eg. Bullets,shotputs,netballs, water
jets, rugby balls
Projectile Motion
When dealing with projectiles, the horizontal
and vertical components are treated
separately.
The horizontal motion is constant velocity (as
there are no forces acting in this direction).
The vertical motion is constant acceleration
of 10ms-2 due to the force of gravity.
Kinematic Equations for vertical motion
Do Page 89 Questions 6B
The canon ball travels 25ms-1 when fired
horizontally from the top of a 45m cliff.
t=0s
t=1.0s
t=2.0s
For each position find the horizontal,
vertical and resultant velocity.
t=3.0s
t=1.0s
t=0s
t=2.0s
t=3.0s
Type 1 Questions
What is a projectile path?
A projectile path is the movement of an object
under the action of gravity only.
Explain the motion of the golf ball in
the vertical direction.
Give a reason for your explanation
Type 1 Questions
The ball is moving with a constant acceleration
acting vertically downwards
(constantly decelerating upwards).
The golf ball’s acceleration is due to gravity.
The golf ball’s weight is the unbalanced force
acting on it.
Back to Golf-Still Type 1
20 m s-1
32 m s-1
Draw clearly on the diagram a velocity vector
to represent the size and direction of the
initial velocity (U) of the golf ball.
Golf Analysis
-1
20 m s
U
32 m s-1
What is the size and direction of the balls
speed ?
Quote the answer to the correct number of
significant figures.
Golf Analysis
u2 = 202 + 322
 u = 37.735925 = 38 m s1
tan  = 20/32   = 320.
What is the time taken for the ball to reach the top
of its flight?
vf = vi + at
 0 = 20 – 10t
 t = 2.0 s
Golf Analysis
Calculate the maximum vertical height (H)
reached by the golf ball at the top of its flight.
Acceleration due to gravity is 10 m s-2.
vf 2 = vi 2 + 2ad
 02 = 202 + 2 x(- 10)H
 H = 20 m
Golf Analysis
Explain the motion of the golf ball in
the horizontal direction,
Give a reason for your explanation.
Motion is constant velocity (speed and direction)
as there is no unbalanced force acting on the golf
ball in the horizontal direction.
What is the velocity of the ball at the top
of its flight?
32ms-1 horizontal
Golf Analysis
Calculate the horizontal distance (R)
travelled by the golf ball.
v = Dd / Dt
 32 = R/2t where Dt = 2t
 R = 2 x 32 x 2 = 128m
What is the velocity of the ball when it lands?
38ms-1 @320 to the ground
What is a force?
(type 1)
Forces
A force causes the motion or shape
of an object to change.
Force is a vector quantity so must
have both a size and a direction
Force is measured in Newtons N.
A resultant (or net) force is
produced when 2 or more forces act
on an object. These forces can be
added to find the resultant.
Forces
Newtons First Law Of Motion:
An object will remain in it’s current state
of motion until a force acts to change it.
Newton’s Second Law Of Motion:
The acceleration of an object is
proportional to the net force applied.
Law 2 can be written like this for short:
Fnet  ma
Forces
Newton’s Third Law Of Motion:
For every action there is an equal
and opposite reaction.
Not What it seems!!!
A 500kg hot air balloon rises at a rate of 0.75ms-2
in a cool Christchurch morning air.
What is the total force lifting the balloon?
5375N
5000N to overcome gravity
+
375N to cause the acceleration
Type 1
Draw a force diagram of an aeroplane accelerating
in level flight.
Draw a force diagram of Jim standing with both
feet on a skate board traveling between B and
D block.
What will be happening to a cyclist experiencing
no net force?
Forces
Friction:
Friction occurs when two surfaces
move past each other. One of these
surfaces could be air – eg air
resistance is a frictional force.
Friction is a force that always
opposes the direction of the motion.
Friction is sometimes called: drag,
water resistance, air resistance or
the retarding force.
Forces
Tension:
This is the force that occurs in
connecting strings and ropes
Tension pulls in both directions along
the string or rope.
Weight:
This is the force of gravity pulling
downwards on an object.
Weight can be calculated by:
Fw  mg
g is acceleration due to gravity and
has a value of 10ms-2 on Earth
(Do Page 55 Questions 4A)
Torque
Torque causes things to spin.
Symbol: t (Greek letter Tau)
Units: Nm
The size of a torque depends on the
size of the force and the
perpendicular distance from the
pivot to where the force is applied.
t  Fd
Equilibrium
An object is at equilibrium if it is at
rest or moving uniformly (First Law)
Two conditions apply:
(Do page 63 Questions 4B)
Equilibrium
All the forces acting on the object
must add to zero
SF=0
(Do page 63 Questions 4B)
Equilibrium
All the torques acting on the object
must add to zero.
St=0
(Do page 63 Questions 4B)
Momentum
The amount of “oooomph” an object
has.
Momentum depends on the mass of
an object and it’s velocity.
Symbol: p
Unit: kgms-1
Momentum is a vector.
Momentum can be calculated using:
p  mv
Momentum
If a force acts on an object, it’s
momentum will change.
The change in momentum can be
calculated by subtracting vectors.
Change in momentum =final
momentum – initial momentum.
Dp  p f  pi
Impulse
When a resultant force acts on an object,
the amount it changes the object’s
momentum by depends on how long the
force acts for.
The force multiplied by the time it acts for
is called impulse.
Units: Ns
Impulse equals the change in momentum.
FDt  Dp
(Do Page 73 Questions 5A)
Conservation of Momentum
The conservation of momentum
principle states: Momentum is
conserved in collisions and
explosions as long as there is no net
external force acting.
This means the momentum before
equals the momentum after.
m1vi1  m2vi 2  m1v f 1  m2v f 2
Conservation of Momentum
The same principle applies in 2
dimensions.
The vector representing the sum of
the momentums before must be the
same vector as the one representing
the sums after.
Do Page 79 Questions 5B
Circular Motion
Period of Rotation T - time it takes to
make one rotation (revolution, cycle)
Measured in seconds s.
Frequency f – number of rotations
completed per second.
Measured in Hertz Hz or s-1
T and f are inverses of each other.
1
T
f
Circular Motion
Circumference – distance travelled
in one rotation (m)
Circumference  2r
The speed of an object moving in a
circle can be calculated by:
d 2r
Speed (v)  
t
T
Circular Motion
An object moving in a circle may be
travelling at constant speed, but
because its direction is always
changing, its velocity is changing….
If velocity is changing, the object is
accelerating….
If an object is accelerating, there
must be a net force acting on it….
Circular Motion
The force acting on an object in
circular motion is in towards the
centre of the circle, changing the
objects direction but not its speed.
This is called centripetal force.
This force causes a centripetal
acceleration towards the centre of
the circle.
Circular Motion
Centripetal force and
acceleration can be
calculated using the
following formulae:
v=speed(ms-1)
r=radius of motion
Do Page 98 Questions 7A
2
v
ac 
r
2
v
Fc  m
r
Solve these
The minute hand of the clock is 10cm,
second hand 9.0cm, hour hand 7.0cm.
How fast is the tip of each hand going?
A Swift, the world’s fastest bird, of mass 0.10kg
is seen to complete a circular turn of radius 10m
without changing speed of 72kmhr-1.
i) Find the centripetal acceleration of the bird
and compare it to gravity.
ii) Find the centripetal force on the bird.
Solve These
A record player has speeds of 33rpm, 45rpm and
78rpm.
Find the period of revolution for a record played
at each speed.
If a fly lands 30cm from the centre of rotation.
What is the tangential velocity if the record is
going 33rpm?
Type 1 Questions
What are the units for momentum, torque,
tangential speed, work, acceleration?
A cricket ball is thrown from the boundary.
Describe the path it takes.
Draw a force diagram for the ball travelling
through the air.
What is the direction of the net force on the ball?
Timy and Cameroon sit on a bench outside A2
at playtime. Draw a diagram showing their weight
the weight of the bench and the support forces
Through the bench legs
Solve This
A stone of mass 750g is tied to the end of a
string and spun. The string has a breaking strain
of 35N and is 1.0m long. It is spun in a plane
horizontal to the earth at a rate of 60 times
a minute.
i) What is the tangential velocity of the stone?
ii) What is the centripetal acceleration of the stone?
iii)Show whether the string will break.
iv)If the stone is now spun in a vertical plane at
the same speed show whether the string will
break now.
Motion due to Gravity
All objects accelerate towards the
ground at (-) 10ms-2 because of
gravity when dropped.
This acceleration is fairly constant
at the Earth’s surface, but varies at
great altitudes or on other planets.
Gravity is always an attractive
force unlike magnetism or electric
forces.
Do Page 86 Questions 6A
Energy
W  Fd
The three kinds of
mechanical energy are:
kinetic, gravitational and
elastic.
Work W is the process of
transforming energy from
one kind to another.
Energy E is measured in ……………..
d is the …………. moved in the direction
of the force.
1 of 6 on energy
Energy
If an object is lifted against gravity,
work is done transforming chemical
energy (muscles) into gravitational
potential energy.
The force needed is the weight force of
the object, the distance moved is the
change in height:
W  Fd
so

DE p  (mg )Dh
2 of 6 on energy
Energy
Any moving object has kinetic
energy.
Doubling the speed increases the
energy by four. (Squared
relationship)
When moving objects stop, this
energy is transformed into other
forms, eg sound, heat
Ek  mv
1
2
power
2
How many more man-3
Type 1 Questions
When Timy and Cameroon sat on a bench outside A2
at playtime you could work out the forces acting.
Discuss the Physics principles you would have
to assume to work out the forces?
Have to mention equilibrium
That means the forces add up to 0, nothing, zip,
nada
The torques clockwise equal the anticlockwise
torques
Type 1 Questions
Brody running with the union ball at 2.5ms-1
North changes direction and goes 3.2ms-1West to
avoid a tackler.
Draw labeled vector diagrams showing his initial
and final velocity.
Draw a vector diagram showing his change
in velocity.
He then crashes into another player and rebounds.
Discuss the Physics principle that will allow you to
calculate his rebound speed.
Power
Power P is the rate at which work is
done.
Measured in Watts W (or Js-1)
W DE
P 
t
t
conservation
4 of 6 get over it
Conservation of Energy
The conservation of energy principle
states: Energy cannot be created or
destroyed, only transformed from
one kind to another.
Efficiency/lost
One more after this
Energy Efficiency
Often some of the forms it is transformed
into are not useful. The energy is “lost” to us
The efficiency of an object is a measure of
the ratio of input energy to useful output
energy
Elastic collisions-stop talking and listen then
make a note
useful output energy
Efficiency% 
100
total input energy
End of the energy story for today
Solve This
80% of the electricity going into a light bulb gets
turned into heat.
How much energy does a 100W bulb use in
10 minutes and how much of this is turned into
light?
DE
DE
P
100 
t
10  60
DE  60,000
 60kJ
12kJ light
Solve This
A 60kg woman runs up a set of stairs in 15s. She
rises 10m in her climb. Calculate her power.
The important message with this problem is
to realise that the majority of energy is used
against gravity as opposed to the horizontal motion
DE
mgh
P
P 
t
t
60 x10 x10
P
15
P  400W
Solve This
At what speed must a 50 gram squash
ball travel if it is to have energy of 0.50J?
4.5ms-1
If the energy is doubled to 1.0J what
must the speed be?
6.3ms-1
Solve This
A sports car with mass ¾ of a tonne accelerates
to 108 kmhr-1 in 8.0s. Ignoring friction what power
is exerted by the motor?
1
2
W DE
P 
t
t
2
mv
P
t
1
2
x 750 x 30
2

8
 42187
42kW
Solve This
A 200W motor is used to lift a 15kg bucket of
cement 40m. How long will it take?
W DE
P 
t
t
mgh
P
t
15 x10 x 40
200 
t
t  30 s
Springs
Energy can be stored in a spring as
elastic potential energy.
Hookes Law: F=kx
F=force
k=spring constant (Nm-1) – a
measure of how stiff or soft a spring
is.
x=extension (m) – the amount a
spring is stretched or compressed
when the force is applied.
Springs
Hooke’s Law as a graph:
Force (N)
Gradient = k
Area under graph = energy
stored in spring
Extension (m)
Springs
Elastic potential energy can be found
by calculating the area under a
Hooke’s Law graph.
Area  b  h
1
2

 E  Fx
1
2
(  F  kx)
and
 E  kx
so
Do Page 107 Questions 8A
1
2
2
Solve This
A person sits in a car with a suspension of
spring constant 104 Nm-1. If the suspension is
compressed 1.0 cm how much energy is stored in
the springs?
Ep=1/2kx2
=0.5 x 104 x (0.01)2
=0.5 J
You know everything required to get
an excellence in Thursdays test now
But will you…………………….
Revise the problems in your text and
The 2.4 exam on the NCEA website
for 2004.