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JPN Pahang Physics Module Form 4Teachers Guide Chapter 1 : Introduction To Physics
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CHAPTER 1 : INTRODUCTION TO PHYSICS
1.1 Understanding Physics
1
Mechanical Energy
PHYSICS
Study of the natural phenomena and the
ro erties of matter.
Solid
Liquid
Gas
Mechanical Energy
Heat Energy
Light Energy
Wave Energy
Electrical Energy
Nuclear Energy
Chemical Energy
Relationship
with
matter
Properties of
Ener
Relationship
with
energy
Properties of
Matter
formsstates
Matter Energy
Mechanics
Properties
of matter
Heat
Light
Wave
in the fields of
Electricity &
ElectromagnetismAtomic Physics
& Nuclear
Electronics
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1.2 PHYSICAL QUANTITIES
Base quantity
1 A physical quantity is ..
2 Examples of scientific instruments:
3 A base quantity is a physical quantity which cannot be defined in terms of other physical
quantities.
4 Study the following picture and list the physical quantities that can be measured.
5 List of 5 basic physical quantities and their units.
Base quantity Symbol S.I. Unit Symbol for S.I. Unit
Length
Mass
Time
CurrentTemperature
6. Two quantities that have also identified as basic quantity. There are:
i) ..unit .. ii) . unit
..
The list of physical quantities :
1. .
2. .
3. .
4. .
5. .
6. .
7. .
8. .
9. .
batterybattery
any quantity that can be measured by a scientific instrument.
Stopwatch, metre rule balance, thermometer, ammeter
etc.
Height,
mass,
size,
age,
temperature,
current
Power,
Thermal energy
Pressure
l meter m
m kilogram kg
t second s
I Ampere A
T Kelvin K
Light intensity candela Amount of substance mol
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Standard Form
1 Standard form = A x 10n , 1 < A < 10 and n = integer
2 Standard form is used to ...
3 Some physical quantities have extremely small magnitudes. Write the following
quantities in standard form :
a. Radius of the earth = 6 370 000 m =.
b. Mass of an electron = 0.000 000 000 000 000 000 000 000 000 000 911 kg =...
c. Size of a particle = 0.000 03 m =
b. Diameter of an atom = 0.000 000 072 m = ...
c. Wavelength of light = 0.000 000 55 m = ..
Prefixes
1. Prefixes are usually used to ...
2. It will be written
3. The list of prefixes :
Tera (T)
Giga (G)
Mega (M)
kilo (k)
mili (m)
micro ()
nano (n)
pico (p)
1012
109
106
103
100
10-3
10-6
10-9
10-12
Hekto (ha)Deka (da)
desi (d)centi (c)
102
101
10-1
10-2
Eg :
1 Tm = .
3.6 mA = .
How to change the unit ;
Eg :1. Mega to nano
2. Tera to micro
3. piko to Mega
simplify the expression of very large and small numbers
6.37 x 106m
9.11 x 10-31 kg
3.0 x 10-5 m
7.2 x 10-8 m
5.5 x 10-7m
represent a large physical quantity or extremely small quantity in S.Iunits.before the unit as a multiplying factor.
1 1012 m
3.6 10-3A
1.33 MA = 1.33 106A
= 1.33 10 6-(-9) nA
= 1.33 10 -15 nA
1.23 Tm to unitm unit
1.23 Tm = 1.23 x 10 12m
= 1.23 x 10 12 (-6)m
= 1.23 x 10 18m
5456 pA to MA unit
5456 pA = 5.456 x 10 3 + (-12) pA
= 5.456 x 10 -9pA
= 5.456 x 10 -9 (6) MA
= 5.456 x 10 -15 MA
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4. Some physical quantities have extremely large magnitudes. These extremely large and
small values can be written in standard form or using standard prefixes. Write the
quantities in standard prefixes:
a. Frequency of radio wave = 91 000 000 Hz = .
b. Diameter of the earth = 12 800 000 m =
c. Distance between the moon and the earth = 383 000 000 m =
d. Mass of the earth = 6 000 000 000 000 000 000 000 000 kg =
Derived quantities
1 A derived quantity is .
2 Determine the derived unit for the following derived quantities.
Derived
quantityFormula Derived unit
Name of
derived unit
Area area = length x width m x m = m2
Volume volume = length x width x height m x m x m = m3
Densityvolume
massd =ensity
3
3mkg
m
kg =
Velocity time
ntdisplaceme
v =elocity
1
sms
m =
Accelerationtime
velocityinchangeonaccelerati =
2
11-1
sm
ssms
sm
=
=
momentum momentum = mass x velocity kg m s-1
Force force = mass x acceleration kg m s-2 Newton (N)
pressurearea
forcepressure=
2
2
m
kgms kg m-1 s-2 (Nm-2) @ Pa
Weight weight = mass x gravitational acceleration kg ms -2 Newton (N)
Work work = force x displacement N mJoule (J)
Powertime
workpower= J s -1 Watt (W)
9.1 10 1MHz
12.8 Mm = 1.28 10 1 Mm
383 Mm = 3.83 10 2 Mm
6.0 10 15 Tg
a physical quantity which combines several basic quantities through
multiplication, division or both
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Derived
quantityFormula Derived unit
Name of
derived unit
kinetic energy2velocitymass
2
1K.E = kg m2s-2 Joule (J)
potential
energyP.E = mass x gravitational acceleration x height kg m2s-2 Joule (J)
charge charge = current x time Ampere second
(As)Coulomb (C)
voltagecharge
workvoltage = J C-1 Volt (V)
resistancecurrent
voltageresistance= VA-1 Ohm ()
Note that the physical quantities such as width, thickness, height, distance, displacement,
perimeter, radius and diameter are equivalent to length.
1.3 SCALAR AND VECTOR QUANTITIES
1 Scalar quantities are
Examples :
2 Vector quantities are...
Examples :
3 Study the following description of events carefully and then decide which events require
magnitude, direction or both to specify them.
Description of events Magnitude Direction
1. The temperature in the room is 25 0C
2. The location of Ayer Hitam is 60 km to the north-
west of Johor Bahru
3. The power of the electric bulb is 80 W
4. A car is travelling at 80 km h-1 from Johor Bahru
to Kuala Lumpur
1.4 MEASUREMENTS
Quantity which has only magnitude or size
Mass, Length, Speed, volume
Quantity which has magnitude or size and direction.
Velocity, Force, Displacement, Acceleration
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Using Appropriate Instruments to Measure
1 There are various types of.
2 We must know how to choose the appropriate instrument to ..
3 Examples of instrument and its measuring ability.
Measuring instrument Range of measurement Smallest scale division
Measuring tape
Meter rule
Vernier caliper
Micrometer screw gauge
4 Sample of measuring instruments:
4.1 Ammeter : ..
4.2 Measuring cylinder : ....................
4.3 Ruler :
wrong right wrong
10 11 12 13 14 15 Reading = cm
4.4 Vernier calliper
A venier calliper is used to measure:
a. b. .
mirror
pointer pointer mirror
Pointers image is behind the pointer
incorrect reading correct
reading
1 2 3
0 4
1 2 30 4
Pointers image can be seen
Right position of eye (eye are in a line perpendicular to the plane of
the scale)
wrong position of eye
wrong position of eyewater
is used to determine the volume of liquid.
is used to determine the length
Up to a few meters 0.1 cm
1 m 0.1 cm (0.01 m)
10 cm 0.01 cm
less than 2 cm (20 mm) 0.001 cm (0.01 mm)
is used to measure electric current
measuring instrument with different measuring capabilities.
measure a particular quantity.
small object depth of a hole
external diameter of a cylinder or pipe internal diameter of a pipe or tube
0.1 cm
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b. .
c.
Example :
4.6 Some others measuring instruments :
..
. ..
Hands-on activity 1.1 on page 1 of the practical book to learn more about choosing
appropriate instruments.
Exercise: Vernier Callipers
1. Write down the readings shown by the following(a)
One complete turn of the thimble
(50 division) moves the spindle by0.50 mm.
Division of thimble
= ..
= ..
A accuracy of micrometer
screw gauge = ..
Sleeve scale :
Thimble scale : .
Total reading : ..
Sleeve scale :
Thimble scale : .
Total reading : ...
0 5 10
7 8
2.00 mm
0.22 mm
2.22 mm
Analogue stopwatch digital stopwatch thermometer Ammeter
Measuring tape measuring cylinder beaker
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(b)
(c)
(d)
2. (a) The following diagram shows the scale of a vernier calliper when the jaws are closed.
Zero error = + 0.02 cm(b). The following diagram shows the scale of the same vernier calliper when there are
40 pieces of cardboard between the jaws.
0 5 10
0 1
0 5 1
6 7
0 5 10
4 5A B
QP
0 5 10
5 6
0 5 10
0 1
Answer: 7.89 cm..
Answer: 4.27 cm..
Answer: 6.28 cm..
Answer: 0.02 cm..
Reading shown = 5.64.cm
Corrected reading = 5.64 - (+ 0.02)= 5.62..cm
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3. Diagram 3 (a) shows the reading on a pair of vernier callipers when its jaws are closed
with nothing in between them. Diagram 3(b) shows the reading when it is used to
measure the thickness of a piece of wood.
What is the actual thickness of the wood?
Zero error = .. cm
Reading shown = ..cm
Actual thickness of the wood = .cm
Exercise: Micrometer Screw Gauge
1. (a) Determine the readings of the following micrometer screw gauges.
Zero error = -0.02.. mm Zero error = +0.03.. mm
(b) Determine the readings of the following micrometer screw gauges.
2. Write down the readings shown by the following micrometer screw gauges.
(a) (b)
0 0
45
5
0
0
5
0
0 5
15
20
Zero error = +0.03mm Reading shown = 6.67..mm
Corrected reading = 6.67-(+0.03)=6.64 mm
400 5
3
0 5 103
0 5 10
0 cm 1 2
(a)
0 5 10
4 5 6
(b)
-0.05
4.51
4.51- (-0.05) =4.56
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Answer: 6.88 mm Answer: ..12.32 mm
(c) (d)
Answer:4.71 mm Answer: 9.17 mm
Accuracy and consistency in measurements.1. Accuracy :
2. Consistency :
3. Sensitivity :
..
..
Hands-on activity 1.2 on page 2 of the practical book to determine the sensitivity of
some measuring instruments.
Errors in measurements
1. All measurements are values
2. In other word, it is a matter of
The ability of an instrument to measure nearest to the actual value
The ability of an instrument to measure consistently with little or no relative
deviation among readings.
The ability of an instrument to detect a small change in the quantity measured.
inaccurate but consistent consistent and accurate
Accuratebut not consistent inaccurate and not consistent
35
20
250
15
200 5
of approximation only.
how close the measurement is to the actual value.
error exist in all measurements.
Systematic errors
a weakness of the instrument
the difference between reaction time of the brain and the action.
zero error is when the pointer is not at zero when not in use.
Range of the measuring instrument absolute error .
Reaction time of the brain.
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3. This is because
4. Two main types of errors:
4.1
Occurs due to :
a)
b)
c)
Examples :
a)
b)
c)
Absolute error :
.
.
Parallax error :
Example :
Zero error : ...
Posit
Positive zero error Negative zero error
Zero error of screw meter gauge
Positive zero error
Zero error =
0 1 cm
0 1 2 3 4 5 6 7 8 9 10
Zero error =
0 1 2 3 4 5 6 7 8 9 10
0 1 cm
Refer to the smallest reading that can be measured by an instrument.
If, the smallest reading = 0.1 cmThen, Absolute error = 0.1 / 2 = 0.05 cm
It occurs because the position of the eye is not perpendicular to the scale of the
instrument.
wrong
position of the eye (no error)
wrong
where the pointer is not at zero when not in use
+0.03 cm - 0.04 cm
Correct reading = observed reading zero error
Zero error of Vernier calliper
Negative zero error
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4.2 ..
Occurs due to
a)
b)
c) Example :
a) ..
b) ..
.....................................................................................................................
1.5 SCIENCETIFIC INVESTIGATION
Steps Explanation
1Making
observation
2Drawing
inferences
3
Identifying
and controlling
variables
4
Formulating a
hypothesis
rizontal reference Horizontal reference 3 divisions above
horizontal referencedivisions below
rizontal reference
ero error = + 0.02 mm Zero error = - 0.03 mm
Random error
carelessness in making the measurement.
parallex error , incorrect positioning of the eye when taking the readings.
sudden change of ambient factors such as temperature or air circulation.
Readings are close to the actual value but they are not consistent.
Can be minimized by consistently repeating the measurement at different places in
an identical manner.
Gather all available information about the object or phenomenon tobe studied.
Using the five senses, sight, hearing, touch, taste and smell.
A conclusion from an observation or phenomena using information thatalready exist.
Variables are factors or physical quantities which change in the courseof a scientific investigation.
There are three variables :i. Manipulated variables physical quantity which change according
to the aim of the experiment.ii. Responding variables physicals quantity which is the result of
the changed by manipulated variable.
iii. Fixed variables physicals quantities which are kept constantduringthe experiment.
Statement of relationship between the manipulated variable and theresponding variable those we would expect.
Hypothesis can either be true or false but in correct direction.
i. Conduct an experiment includes the compilation and
interpretation of data.ii. Making a conclusion regarding the validity of the hypothesis.
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5Conducting
experiments
Plan and report an experiment
Situation : A few children are playing on a different length of swing in a
playground. It is found that the time of oscillation for each swing is different.
Steps Example : refer to the situation above
1 Inference
2 Hypothesis
3 Aim
4 Variables
5 List of
apparatus and
materials
6 Arrangement of
the apparatus
The period of the oscillation depends on the length of the
pendulum.
When the length of the pendulum increases, the period of the
oscillation increases.
Investigate the relationship between length and period of asimple pendulum.
Manipulated variable : the length of the pendulum.Responding variable : Period
Fixed variable : the mass of the pendulum and the displacement.
l
Retort standprotractor
l
bob
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7 Procedures
8 Tabulate the
data
9
1
0
11
Analyse the
data
Discussion
Conclusion
T / s
1.4
1.2
1.0
0.8
0.6
0.4
0.2
Graf of period, T vspendulums
length, l
1. Set up the apparatus as shown in the figure above.2. Measure the length of the pendulum,l = 60.0 cm by using a meter
rule.3. Give the pendulum bob a small displacement 300.Time of
10 oscillations is measured by using a stop watch.4. Repeat the timing for another 10 oscillations. Calculate the average
time.Period = t10oscillations10
5. Repeat steps 2, 3 and 4 using l = 50.0 cm, 40.0 cm, 30.0 cm and20.0 cm
1.581.58
1.501.50
1.311.31
1.191.19
0.990.99
15.815.8
15.015.0
13.113.1
11.911.9
9.99.9
15.715.7
15.015.0
13.113.1
11.911.9
9.99.9
15.815.8
15.015.0
13.113.1
11.911.9
9.99.9
60.060.0
50.050.0
40.040.0
30.030.0
20.020.0
Period/ sPeriod/ s
(T = t(T = t1010/10)/10)AverageAverage
2211
Length,Length,ll //cmcm
Time for 10 oscillations / s
10 20 30 40 50 60 l / cm
Precautions :1. Oscillation time is measured when the pendulum attained a steady
state.2. Time for 10 oscillations is repeated twice to increase accuracy.3. Discussion (refer to given questions)
The period increases when the length of the pendulum increases.Hypothesis accepted.
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Reinforcement Chapter 1
Part A :Objective Question
1. Which of the following is a base SI
quantity?A Weight B Energy
C Velocity D Mass
2. Which of the following is a derived
quantity?
A Length B Mass
C Temperature D Voltage
3. Which of the following is not a basic
unit?
A Newton B kilogramC ampere D second
4. Which of the following quantities
cannot be derived?
A Electric current B Power
C Momentum D Force
5. Which of the following quantities is
not derived from the basic physical
quantity of length?
A Electric charge B Density
C Velocity D Volume
6. Initial velocity u, final velocity v,
time tand another physical quantity kis related by the equation v - u = kt.The unit forkisA m s-1 B m-1 s
C m s-2 D m2 s-2
7. Which of the following has thesmallest magnitude?
A megametre B centimetre
C kilometre D mikrometre
8. 4 328 000 000 mm in standard form is
A 4.328 x 10-9 m B 4.328 x 10-6 m
C 4.328 x 106 m D 4.328 x 109 m
9. Which of the following measurements
is the longest?
A 1.2 x 10-5
cm B 120 x 10-4
dmC 0.12 mm D 1.2 x 10-11 km
10. The diameter of a particle is 250 m.
What is its diameter in cm? A 2.5 x 10 -2 B 2.5 x 10-4
C 2.5 x 10-6 D 2.5 x 10-8
11. Which of the following prefixes is
arranged in ascending order?
A mili, senti, mikro, desi
B mikro, mili, senti, desi
C mili, mikro, desi, senti
D desi, mikro, mili, senti
12. Velocity, density, force and energy are
A basic quantities
B scalar quantities
C derived quantities
D vector quantities
13. Which of the following shows the
correct conversion of units?
A 24 mm3 =2.4 x 10-6 m3
B 300 mm3=3.0 x 10-7m3
C 800 mm
3
=8.0 x 10
-2
m
3
D 1 000 mm3=1.0 x 10-4 m3
14. Which of the following measurements
is the shortest ?
A 3.45 x 103 m
B 3.45 x 104 cm
C 3.45 x 107 mm
D 3.45 x 1012m
15. The Hitz FM channel broadcasts radio
waves at a frequency of 92.8 MHz inthe north region. What is the frequency
of the radio wave in Hz?
A 9.28 x 104 B 9.28 x 105
C 9.28 x 107 D 9.28 x 1010
16. An object moves along a straight line
for time, t. The length of the line,s is
given by the equation2
2
1gts = . The
SI unit of g is
A m2 s2 B m s-2
C s-1 D s-2 m
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Part B : Structure Question
1. A car moves with an average speed of 75 km h-1 from town P to town Q in 2 hours as
shown in Figure 1. By using this information, you may calculate the distance between the
two towns.
P Q
Figure 1
(a) (i) Based on the statements given, state two basic quantities and their respective
SI units.
(ii) State a derived quantity and its SI unit.
(b) Convert the value 1 . m to standard form.
5 x 10-3
(c) Complete Table 1 by writing the value of each given prefix.
Table 1
(d) Power is defined as the rate of change of work done. Derive the unit for power in
terms of its basic units.
(e) Calculate the volume of a wooden block with dimension of 7 cm, 5 cm breadth and 12
cm height in m3 and convert its value in standard form.
Distance : m and time : s
Speed m s-1
= 0.2 x 103 m= 2.0 x 102 m
10-9
10-6
106
109
Power =time
work=
time
ntdisplacemeForceUnit =
s
mkgms 2= kg m2 s-3
Volume = (7 x 10-2) (5 x 10-2) (12 x 10-2)= 420 x 10-6
= 4.20 x 10-4 m3
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2. Figure 2 shows an ammeter of 03 A range.
Figure 2
(a) (i) Name component X. ...
(ii) What is the function of X? .
(b) Table 2 shows three current readings obtained by three students.
Table 2
(i) Did all the students use the ammeter in Figure2? ...
(ii) Explain your answer in (b)(i).
3. Figure 3 shows the meniscus of water in a measuring cylinder K, L, and M are three eye
positions while measuring the volume of the water.
(a) (i) Which of the eye positions is
correct while
taking the reading of the volume
of water?
.
Figure 3
(b) The water in the measuring cylinder is
replaced with 30 cm3 of mercury.
(i) In Figure 4, draw the meniscus of the
mercury in the measuring cylinder. Figure 4(ii) Explain why the shape of the meniscus of mercury is as drawn in (b)(i).
No
3rdreadings obtained by student 2 and 3 are out of the meter range.
L
The cohesive force is larger than the adhesive force
Mirror
To avoid parallax error
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