Transcript Slide 1

P2.1.3 Terminal Velocity
P2 Physics
Mr D Powell
Connection
•
•
•
Connect your learning to the
content of the lesson
Share the process by which the
learning will actually take place
Explore the outcomes of the
learning, emphasising why this will
be beneficial for the learner
Demonstration
• Use formative feedback – Assessment for
Learning
• Vary the groupings within the classroom
for the purpose of learning – individual;
pair; group/team; friendship; teacher
selected; single sex; mixed sex
• Offer different ways for the students to
demonstrate their understanding
• Allow the students to “show off” their
learning
Activation
Consolidation
• Construct problem-solving
challenges for the students
• Use a multi-sensory approach – VAK
• Promote a language of learning to
enable the students to talk about
their progress or obstacles to it
• Learning as an active process, so the
students aren’t passive receptors
• Structure active reflection on the lesson
content and the process of learning
• Seek transfer between “subjects”
• Review the learning from this lesson and
preview the learning for the next
• Promote ways in which the students will
remember
• A “news broadcast” approach to learning
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Index
P2.1.4 Terminal Velocity
a) The faster an object moves through a fluid the greater the frictional
force that acts on it.
b) An object falling through a fluid will initially accelerate due to the
force of gravity. Eventually the resultant force will be zero and the
object will move at its terminal velocity (steady speed).
c) Draw and interpret velocity-time graphs for objects that reach
terminal velocity, including a consideration of the forces acting on the
object.
d) Calculate the weight of an object using the force exerted on it by a
gravitational force: W = mg (F = ma)
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Index
a) The faster an object moves through a fluid the greater
the frictional force that acts on it.
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Newtons Law Number 1
Copy this law in your own
words...
If the resultant force is zero…
The object will stay still OR, if already moving, move at a constant speed
in a straight line.
Force from
muscles
Air resistance
There is a zero resultant force so the athlete runs at a constant speed.
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Newtons Law Number 2
Copy this law in your own
words...
If there is a resultant force on an object it will accelerate (speed up, or
slow down).
mass
The resultant forward force causes the puck to accelerate.
Force = mass x acceleration
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Newtons Law Number 3
Copy this law in your own
words...
For every ACTION there is an equal and opposite REACTION (think of an
action as a sort of impact or hitting force)
The tennis racket gives the ball a forward force (action) … but
the ball also gives the racket a backward force (reaction).
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Investigating TV with Oil.
1.
Pour 100ml of oil into the cylinder and get a spare
cylinder to pour into.
2.
Get 2 stopwatches,3 dice and some paper to prevent
spillage.
3.
Get two people down to eye level to time the fall
between 90-70 & 50&30ml. Each reading should be a
paired reading.
4.
Record the times in your table.
5.
Repeat 5 times in total. Enter results in table
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Example Results
1ooMl Cylinder - Trying to Establish the Terminal Velocity of Oil
1st Timing 1st Distance Velocity
/s
/m
m/s
Velocity
2nd
2nd Distance Velocity
Difference
Timing /s
/m
m/s
m/s
1
0.53
0.034
0.06
0.53
0.034
0.06
0
2
0.56
0.034
0.06
0.69
0.034
0.05
0.01
3
0.55
0.034
0.06
0.57
0.034
0.06
0
4
0.53
0.034
0.06
0.52
0.034
0.07
-0.01
5
0.56
0.034
0.06
0.54
0.034
0.06
0
Dev
0.000
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b) An object falling through a fluid will initially accelerate due to
the force of gravity. Eventually the resultant force will be zero
and the object will move at its terminal velocity (steady speed).
Why is this the case?
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c) Terminal Velocity
Terminal Velocity
Velocity of a Skydiver as she falls from a Plane
Velocity in meters per second (m/s)
60
50
40
?
30
20
10
0
0
5
10
Time in seconds (s)
15
Time (s)
Velocity m/s
0
0
1
10
2
20
3
30
4
36
5
40
6
44
7
48
8
50
9
52
10
54
11
55
12
55
13
55
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Progress Check…
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Terminal Velocity
Time (s) Velocity m/s
0
0
1
10
2
20
3
30
4
36
5
40
6
44
7
48
8
50
9
52
10
54
11
55
12
55
13
55
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Summary Questions
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Multichoice...
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Matchup...
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d) What is the link?
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d) Investigation...
We can use the apparatus above to
accelerate a trolley with a constant
force. Use the newtonmeter to pull
the trolley along with a constant force.
You can double or treble the total
moving mass by using double-deck
and triple-deck trolleys.
A motion sensor and a computer
record the velocity of the trolley as it
accelerates.
Here is a typical set of results which
show that for the same force the
acceleration reduces as mass
increases.
Can you work out a simple formulae
which relates force, mass &
acceleration?
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d) Weight or Gravity?
7.36 × 1022 kilograms
¼ radius of Earth
 A lot of people slip into lazy English and
try and use the ideas of Weight and
Gravity interchangeably.
 However, they are not the same thing.
m
g 2
r
 Gravity is a special force which acts
upon mass and is measured in Newtons
per kilogram.
 Weight is measured in Newtons and is
a unit of forces
 Gravity is on Earth is derived from the
total mass of Earth. For this mass the
Force of Gravity is 10N/kg. On the moon
it is only 1.6N/kg as the Earth has 81.2
more mass and 4 times the radius.
5.9742 × 1024 kilograms
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d) Comparison
Body
Sun
Mercury
Venus
Earth
Moon
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Multiple of
Earth gravity
27.90
0.3770
0.9032
1 (by definition)
0.1655
0.3895
2.640
1.139
0.917
1.148
0.0621
m/s²
274.1
3.703
8.872
9.8226  10
1.625
3.728
25.93
11.19
9.01
11.28
0.610
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d) Gravity....
So the force of gravity pulls down on masses accord to gravitational field
strength. This varies with height but near to the Earth is a constant 10N/kg.So
1kg would weigh;
Weight (N) = mass (kg) x Gravitational Field (g) (N/kg)
W = mg
W = 1kg x 10N/kg
W = 10N
The weight feels a force of 10N
800g would weigh;
W = 0.800kg x 10N/kg
W = 8N
1.
2.
3.
4.
5.
(W=mg)
5N
3N
3.4kg
28kg
0.01kg
Using the above formula (on the earth) find the following;
1) W? if m = 0.5kg
2) W? if m = 300g
4) m? if W = 280N
5) m? if W = 0.1N
3) m? if W = 34N
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F  ma
or
F
a
m
F
m
a
d) Acceleration again....
We can actually consider the gravity acting on objects as a
form of acceleration. However, this time the units are
different and we can quote the acceleration as the Force per
kilogram or N/kg instead of m/s2. In fact both are the same;
10N/kg = 10m/s2 = 10ms-2,
In which case we actually find that one Newton of force can
be defined as the force required to give a mass of 1kg, an
acceleration, of 1 m/s2 or 1ms-2
Force = mass x acceleration
Work out the following;
1. m= 50kg, a = 10N/kg, F =
2. m= 100kg, a = 5N/kg, F =
3. F= 50N, a = 10N/kg, m =
4. F= 30N, m = 5kg, a =
5. F= 28N, a = 2.5ms-2, m =
1.
2.
3.
4.
5.
500N
500N
5kg
6N/kg
11.2kg
F
a
m
3N
 3 Nkg 1
1kg
3N
 1.5 Nkg 1
2kg
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d) Applications to F=ma
 Complete the table below showing the resultant force, mass and
acceleration of objects in different situations.
Resultant force
(in
(innewtons)
newtons)N
Mass
Mass
(in kilograms)
(in kilograms)
kg
Acceleration
(in m/s2 or N/kg)
a) Athlete
accelerating at
start of 100 m
race
560
70
8.0
b) Car accelerating
3000
1200
2.5
c) Lorry braking
16 000
20000
0.8
d) Plane taking off
40,000
8000
5.0
F  ma
or
F
a
m
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d) Exam Questions...
A vehicle of mass 1500 kg braked to a standstill from a velocity of 24 m/s in 12 s.
1) Show that the deceleration of the vehicle was 2.0 m/s2.
(0- 24)ms-1 /12s = -2.0 ms-2
2) Calculate the resultant force on the vehicle.
F = ma = 1500kg x -2.0 ms-2 = 1500kg x -2.0 Nkg-1 = -3000N
F  ma

v  u  v
a

t
t
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Index
d) Exam Questions...
1) A cyclist accelerated along a flat road from a standstill to a
velocity of 12 m/s in 60 seconds. The mass of the cyclist and the
bicycle was 80 kg. Show that the acceleration of the cyclist was
0.2 m/s2.
12ms-1 /60s = 0.2 ms-2
Calculate the resultant force on the cyclist and the bicycle. On
reaching a velocity of 12 m/s, the cyclist in 1) stopped pedalling
and slowed down to a velocity of 8 m/s in 10 s, when she
started pedalling again.
Calculate:
2) The deceleration of the cyclist when she slowed down.
(8ms-1 -12ms-1) / 10s = -0.4 ms-2
F  ma

v  u  v
a

t
t
3) The size and direction of the resultant force on the cyclist
when she slowed down.
F = ma = 80kg x -0.4 ms-2 = -32N
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Summary Questions 1
RF
V
M
RF
V
RF
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d) Summary Questions 2
kgm/s2 = kgms-2 = N
a = F/m = 20N / 8 kg = 2.5 N/kg
a = F/m = 0.2N / 500g = 0.2N / 0.5kg = 2 N/kg
a = F/m = 5kN / 20 kg = 5000N / 20kg = 250 N/kg
F = ma = 5kg x 5 m/s2 = 25 kgm/s2 or 25 kgms-2 or 25N
F = ma = 15kg x 3 m/s2 = 45 kgm/s2 or 45 kgms-2 or 45N
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W = mg or F = ma
Resultant
force
(in newtons)
N
a) Athlete
accelerating
at start of
100 m race
b) Car
accelerating
c) Lorry
braking
d) Plane taking
off
Mass
(in
kilograms)
kg
70
3000
W = mg or F = ma
8.0
a) Athlete
accelerating
at start of
100 m race
0.8
b) Car
accelerating
c) Lorry
braking
d) Plane taking
off
1200
16 000
8000
Resultant
force
(in newtons)
N
Acceleration
(in m/s2 or
N/kg)
5.0
3000
Mass
(in
kilograms)
kg
Acceleration
(in m/s2 or
N/kg)
70
8.0
1200
16 000
0.8
8000
5.0
P2.1.4 Terminal Velocity
P2.1.4 Terminal Velocity
a) The faster an object moves through a fluid the greater the
frictional force that acts on it.
a) The faster an object moves through a fluid the greater the
frictional force that acts on it.
b) An object falling through a fluid will initially accelerate due
to the force of gravity. Eventually the resultant force will be
zero and the object will move at its terminal velocity (steady
speed).
b) An object falling through a fluid will initially accelerate due
to the force of gravity. Eventually the resultant force will be
zero and the object will move at its terminal velocity (steady
speed).
c) Draw and interpret velocity-time graphs for objects that
reach terminal velocity, including a consideration of the forces
acting on the object.
c) Draw and interpret velocity-time graphs for objects that
reach terminal velocity, including a consideration of the forces
acting on the object.
d) Calculate the weight of an object using the force exerted on
it by a gravitational force: W = mg (F = ma)
d) Calculate the weight of an object using the force exerted on
it by a gravitational force: W = mg (F = ma)
P2.1.4 Terminal Velocity
P2.1.4 Terminal Velocity
a) The faster an object moves through a fluid the greater the
frictional force that acts on it.
a) The faster an object moves through a fluid the greater the
frictional force that acts on it.
b) An object falling through a fluid will initially accelerate due
to the force of gravity. Eventually the resultant force will be
zero and the object will move at its terminal velocity (steady
speed).
b) An object falling through a fluid will initially accelerate due
to the force of gravity. Eventually the resultant force will be
zero and the object will move at its terminal velocity (steady
speed).
c) Draw and interpret velocity-time graphs for objects that
reach terminal velocity, including a consideration of the forces
acting on the object.
c) Draw and interpret velocity-time graphs for objects that
reach terminal velocity, including a consideration of the forces
acting on the object.
d) Calculate the weight of an object using the force exerted on
it by a gravitational force: W = mg (F = ma)
d) Calculate the weight of an object using the force exerted on
it by a gravitational force: W = mg (F = ma)