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Chapter 1: Introduction to Physics
1.1 Understanding Physics
explain what physics is recognize the physics in everyday objects and natural phenomena
1. A phenomenon is an occurrence that can be perceived by our senses.
2. In physics, we study natural phenomena, such as the eruption of volcano, rain fall, formation of
rainbow and the properties of matter, such as length, temperature, volume
3. There are many fields of study in physics, including force, motion, heat, light, waves, electricity,
electromagnetism, electronics and nuclear physics.
1.2Understanding Base Quantities and Derived Quantities explain what base quantities and derived quantities are list base quantities and their units list some derived quantities and their units. express quantities using prefixes.
express quantities using scientific notation express derived quantities as well as their units in terms of base quantities and base units. solve problems involving conversion of units
1. A physical quantity is a physical characteristic that can be measured.2. Base quantities are physical quantities that cannot be defined in terms of other base quantities.
There are five base quantities: length, mass, time, current and temperature.
Physical Quantity Base S.I. Unit
Base Quantity Quantity Symbol S.I. Unit Unit symbol
Length l metre m
Mass m kilogram kgTime t second s
Electric Current I ampere A
Temperature T kelvin K
Table 1Notes for teachers:
Symbol is a short form of a quantity. Example: A boy by the name Ahmad is called as Mad; a girl by thename Mary Jane is called MJ; a pet by the name cute-cute is called cc.
Unit is similar to the penjodoh bilangan in the Bahasa Melayu.For person, we say seorang or duaorang; but for a pet like hamsters, we say seekor or dua ekor.
The unit ampere and kelvin are the names of scientists we use to remind us of their contributions to therespective fields. However, when we write the unit fully, we write all in small letters, example: 1.2 ampere,
5.0 kelvin; when we write shortly, we write the first alphabet of the name in capital letter, example: 1.2 A, 5.0
K
3. Derived quantities are physical quantities consisting of combinations of base quantities., by
multiplication, division, or both operations.
4. Derived quantities as well as their units are expressed in terms of base quantities and base S.I.units as follows:
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2Notes for teachers:
Example: Given that velocity =time
ntdisplaceme. Express the unit for speed in base units.
Solution:
SI unit for velocity =for timeunitSI
ntdisplacemeforunitSI
=s
m
= ms-1
(read as metre per second)
Given that l : length, m : mass, t: time, I: electric current, T: temperature.
Derived quantities
(symbol)
Expressed in base quantities Derived units
Area
(A)
A = l x l Unit A = m x m
= 2m
(read as square metre)
Volume
(V)
V = l x l x l Unit V = m x m x m
= 3m (read as cubic metre)
Density
( ) =
V
m Unit =
3m
kg
= 3mkg
(read as kilogram per cubic metre)
Speed
(v)v =
t
l Unit v =
s
m
=1sm
(read as metre per second)Work or Energy
(W or E)
W = Fs
F = force
s = displacement
Unit W = kg2sm x m
= kg22 sm
= N m
= J
(read as joule)
Power
(P)P =
t
E
t
W Unit P =
s
J
=1sJ
= W
(read as watt)Velocity
(v)v =
t
l Unit v =
s
m
=1sm
(read as metre per second)
Acceleration
(a)a =
t
u-v
u = initial velocity
v = final velocity
t = time taken
Unit a =s
ms 1
=2sm
(read as metre per second per second)
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Force
(F)
F = ma
m = mass
a = acceleration
Unit F = kg x 2
sm
= kg 2sm
= N
(read as newton)
Impulse
(Ft)
Ft = change of momentum
= mvmu
m = mass
u = initial velocity
v = final velocity
Unit Ft = kg x 1sm
= kg1
sm = N s
(read as newton second)
Momentum
(p)
p = mv
m = mass
v = velocity
Unit p = kg x 1ms
= kg 1sm
= N s
(read as newton second)
Pressure
(P)
P =
A
F
F = force
A = area
Unit P =2
m
N
= 2mN
= Pa
(read as pascal)
Specific heat
capacity
(c)
c =m
Q
Q = heat energy
m = mass
= change in temperature
Unit c =Ckg
Jo
= 101
CkgJ
=kgK
J
=11
KkgJ (read as joule per kilogram per kelvin)
Frequency
(f)f =
T
1
T = period of swing; unit:
second (s)
Unit f =s
1
=1s
= Hz
(read as hertz)
Electrical charges
(Q)Q = It
I = electric current
t = time
Unit Q = A s
= C
(read as coulomb)Resistance
(R)R =
I
V
V = voltage; unit: volt (V)
I = electric current
Unit R =A
V
=1
AV
=
(read as ohm)
Table 2
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5. Prefixes are used to express some physical quantities that are either very big or very small.
Prefix Symbol Value
Tera T 1210
Giga G 910
Mega M6
10 kilo k 310
desi d 110
centi c 210
mili m 310
mikro 610
nano n 910
piko p 1210
Table 3
6. Standard form or scientific notation:
A x 10n
where 1 A 10, n is an integer (integer positive or negative).
Physical Quantity Value Standard form or
Scientific notation
Mass of earth 6 020 000 000 000 000 000 000 000 kg kg241002.6
Diameter of an oil
molecule
0. 000 000 000 74 m m10104.7
Speed or light in the
vacuum
299 792 458 m s-
18100.3 sm
Radius of earth 6 370 000 m m61037.6
Mass of hydrogen
atom
0. 000 021 kg kg5101.2
Time of a day 86 400 s s41064.8
Temperature of the
centre of the earth
6 000 000 K K6100.6
Size of a flu virus 0.000 000 2 m m7
100.2
Table 4
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1.3Understanding Scalar and Vector Quantities define scalar and vector quantities give examples of scalar and vector quantities.
1. Scalar quantities are quantities that have magnitude but no direction.2. Vector quantities are quantities that have both magnitude and direction.
Scalar Quantities Vector Quantities
Distance Displacement
Speed Velocity
work Acceleration
Area Force
Length Momentum
Table 1.3.1
3.Distance(s) Displacement(s)
Total length of the pathtraveled
Distance between twopoints measured along a
specific direction
Scalar quantity Vector quantity
Speed Velocity
Rate of change ofdistance
Rate of change ofdisplacement
Speed = time
cedis tan
Velocity = time
ntdisplaceme
Scalar quantity Vector quantity
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4. Annie the ant is traveling down the road to buy an umbrella for these rainy days. She
walks from her nest, A to B, B to C in 10 minutes time as shown in the picture below:
(a) What is the distance she traveled?
(b) What is her displacement from A?
(c) What is her speed?
(d) What is her velocity?
Solution:
(a) Distance traveled = AB + BC
= 3 m + 4m
= 7 m
(b) Displacement of the object from A = 5 m towards the direction of AC
tan = 4
30.75
= 36.9 o
The displacement of the ant is 5 m in the direction of S 36.9 o E from A.
(c) Speed = 1012.06010
7
sm
(d) Velocity =1
0083.0
6010
5
sm towards the direction of AC.
A
BC
4 m
3 m
Annie the antU
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1.4Measuring Instruments Measure physical quantities using appropriate instruments Explain accuracy and consistency Explain sensitivity Explain types of experimental error Use appropriate techniques to reduce errors
Accuracy, Consistency and Sensitivity in measurement & Errors
Definitions:
1. Consistency in measurements refers to how little deviation there is among the measurements made
when a quantity is measured several times.
2. Accuracy of a measurement is how close the measurement made is to the actual value of the quantity.
3. Sensitivity of an instrument is its ability to detect a small change in the quantity to be measured in a
short period of time.
4. The diagram shows the result for four shooters A, B, C and D in a tournament. Every shooter shot five
times.
The table shows the conclusion:
Table 1
Figure 1
5. Error is uncertainty caused by measuring instrument or the observer or the physical factors of the
surroundings.
6. Two main types of errors : systematic error and random error.
Systematic Error Random Error
Caused by:i. Error in instrumentsii. Error in calibration
Caused by:i. Surroundings factors, such as
temperature and wind
ii. Carelessness of the observer Example
i. Zero error Example
i. Parallax errorii. Error in counting Cannot be reduced or overcome Can be reduced Way of correction
i. Take the error into account Ways of correctioni. Take several readings andcalculate the average value.
Table 2
Shooter Consistency Accuracy
A High Low
B Low High
C High High
D Low Low
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Parallax errors
Definition:
A parallax error is an error in reading an instrument because the observers eyes and pointer are not in
line / perpendicular to the plane of the scale.
Concept & Explanation:
1. Figure 2, 3 and 4 show the correct positioning of the observers eyes to avoid parallax errors.
2.
How to avoid parallax error?
(a) position of eyes must be in line/ perpendicular / 90owith the scale of the reading to betaken.
(b) When taking reading from an ammeter, we must make sure that the eyes are exactly infront of the pointer, so that the reflection of the pointer in the mirror is right behind the
pointer. In other words, the reflection of the pointer on the mirror could not be seen by the
observer, then it is free from parallax error.
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Measuring Instruments & Accuracy
Measuring Instruments:
Table 2
(A) Instruments measuring length
1. Metre Rule
Figure 5
2. Vernier Calipers
The same wire is measured by a vernier caliper. The reading is as follows:
Figure 6
3. Micrometer Screw Gauge
The diameter of the wire is measured by a micrometer screw gauge. The reading is as follows:
Figure 7
Physical Quantity Measuring Instrument
Length Metre-rule, vernier caliper, micrometer screw gauge
Current Ammeter
Mass Triple-beam-balanceTemperature Thermometer
Time Mechanical stopwatch, digital stopwatch
Voltage Voltmeter
Ruler A Ruler B
Sensitivity 0.1 cm 0.5 cm
Accuracy 0.1 cm 0.5 cm
Length of wire 4.8 cm 5.0 cm
Sensitivity 0.01 cmAccuracy 0.01 cm
Length of wire 4.78cm
Sensitivity 0.01 mm
Accuracy 0.01 mmDiameter of wire 6.5 +0.22
= 6.72 mm
4 5
0 5 10
20
250 5
wire
2 3 4 510Ruler A
2 3 4 501 Ruler B
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Vernier Calipers
Positive zero error Negative zero error
Figure 9
Positive zero error = + 0.08 cm
All measurements taken with this vernier calipers
must be corrected by subtracting 0.08 cm from the
readings.
Figure 10
Negative zero error = - ( 0.10.08 ) cm
= - 0.02 cm
All measurements taken with this vernier calipers
must be corrected by subtracting - 0.08 cm, which
is adding 0.08 cm to the readings
Eample
(i) Figure 11 (ii)
Zero error = + 0.04 cmVernier calipers reading = 0.4 + 0.01
= 0.41 cm
Corrected reading
= vernier calipers readingzero error
= 0.410.04
= 0.37 cm
Example
(i) Figure 12 (ii)
Zero error = -(0.10.07) cm= - 0.03 cm
Vernier calipers reading = 3.6 + 0.02
= 3.62 cm
Corrected reading
= vernier calipers readingzero error
= 3.62(-0.03)
= 3.62 + 0.03
= 3.65 cm
1) How to read from a vernier calipers?
Figure 8 shows the use of a vernier calipers
to measure the size of the inner diameter of
a beaker.
Inner diameter= main scale reading + vernier scale reading
= 3.2 + 0.04
= 3.24 cmFigure 8
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Exercise:
1 Write down the readings shown by the following
(a)
(b)
(c)
(d)
2. (a) The following diagram shows the scale of a vernier callipers when the jaws are closed.
Zero error = + 0.02 cm
(b) The following diagram shows the scale of the same vernier callipers when there are 40
pieces of cardboard between the jaws.
0 5 10
0 1
0 5 10
6 7
0 5 10
7 8
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 = 5.640.02 = 5.62 cm
0 5 10
4 5A B
QP
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Micrometer Screw Gauge
1) How to read from a micrometer screw gauge?
Figure 13Figure 13 shows the use of a micrometer screw gauge to measure the size of a spherical object.
Main scale reading = 5.5 mm
Thimble scale reading = 12 x 0.01
= 0.12 mm
Final reading = 5.5 + 0.12
= 5.62 mm
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2. Positive zero error and negative zero errorPositive zero error Negative zero error
Figure 14
Positive zero error = + 0.04 mm
All measurements taken with this micrometer
screw gauge must be corrected by subtracting
0.04 mm from the readings
Figure 15
Negative zero error = - 0.04 mm
All measurements taken with this micrometer
screw gauge must be corrected by subtracting -
0.04 mm, which is adding 0.04 mm from the
readings
Example
Figure 16
Zero error = + 0.01 mm
micrometer screw gauge reading
= 2.5 + 0.35
= 2.85 mm
Corrected reading= micrometer screw gauge readingzero error
= 2.850.01
= 2.84 mm
Example
Figure 17Zero error = - 0.03 mm
micrometer screw gauge reading
= 6.0 + 0.08
= 6.08 mm
Corrected reading= micrometer screw gauge readingzero error
= 6.08(-0.03)
= 6.08 + 0.03
= 6.11 mm
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Exercise:
1. Write down the readings shown by the following micrometer screw gauges.
(a) (b)
Answer: 6.5 + 0.28 = 6.78 mm Answer: 17.0 + 0.42 = 17.42 mm
(c) (d)
Answer:4.5 + 0.06 = 4.56 mm Answer: 9.0 + 0.32 = 9.32 mm
2. (a) Determine the readings of the following micrometer screw gauges.
Zero error = - 0.02 mm Zero error = + 0.02 mm
(b) Determine the readings of the following micrometer screw gauges.
25
300 5
40
5 10 1545
0 0
45
5
0
0
5
0
0 0 5
15
20
30
350 5
Zero error = + 0.03 mm Reading shown = 6.5 + 0.18
= 6.68 mm
Corrected reading = 6.68(+0.03)
= 6.65 mm
5
100
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(B)Instrument Measuring Current : Ammeter
Ammeter ranged 0.0 A5.0 A
Sensitivity = 0.1 A
Accuracy = 0.1 A
Figure 18
Doubled ranged ammeter
Upper scale ranged 0.0 A5.0A;
Sensitivity = 0.1 A ; accuracy = 0.1 A
Lower scale ranged 0.00A1.00A;
Sensitivity = 0.02A ; accuracy = 0.02A
Reading = 0.30 A
Figure 19
Miliammeter 0 mA50 mA
Sensitivity = 1 mA
Accuracy = 1 mAReading = 15 mA
Figure 20
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(C) Instrument Measuring Temperature
Thermometer
Figure 21 Accuracy = 1oC
(D) Instrument Measuring Time
Mechanical StopwatchAccuracy = 0.2 s; Reading = 8.2 s
Digital Stopwatch
Accuracy = 0.01s
Reading = 3 minutes 55.62 s
Figure 22: Mechanical stopwatch
Figure 23: Digital stopwatch
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1.5 Scientific Investigation
Identify variables in a given situation Identify a queation suitable for scientific investigation Form a hypothesis Design and carry out a simple experiment to test the hypothesis Record and present data in a suitable form Interpret data to draw a conclusion Write a report of the investigation
Clone of SPM Try Exam of the Perak State year 2003: Paper 3 / Section B/ Question 2
Notes: MV -manipulated variable; RV-responding variable; C- constant
Two twin brothers, Micheal and Jackson, of thesame size, are swinging happily on the swings at a
playground as shown in the figure above.
However, the ropes that is holding the swing where Micheal is sitting islonger thanJacksons. And,
Micheal notices that his swing is swinging slower than his brother, Jackson.Using this information;
(a) make a suitable inference, [1 mark]
(b) state one appropriate hypothesis that could be investigated, [1 mark]
(c) describe how you would design an experiment to test your hypothesis using abob, strings and other
apparatus.
In your description, state clearly the following:
(i) aim of the experiment(ii) variables in the experiment(iii) list of apparatus and materials(iv) arrangement of the apparatus(v) the procedure of the experiment, which includes the method of controlling the manipulated
variable and the method of measuring the responding variable.
(vi) the way you would tabulate the data(vii) the way you would analyze the data [10 marks]
Keywords to indicate
C is mass
Keywords to indicate
MV is lengthKeywords to indicate RV is time of
making a complete swing
Keywords to indicate
the must-use-
apparatus and hinting
on the Pendulum
experiment
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Answer:
(a) Length of ropes influences time of making a complete swing
(b) When the length of pendulum increases, the period of swing increases.
(c)
Marks
1st
mark
/1
Aim To investigate the relationship between length of pendulum and
period of swing.
2n
mark
/2
MV:length of pendulum, l RV:period of swing, T
3r
mark
/3
C : mass of bob
4th
mark
/4
List of apparatus& materials
metre-rule, , retort standand clamp,split cork,
5th
mark
/5
Arrangement of
apparatus
6th
mark
/6
Method to control
MV
Measure l = 10.0 cm by using ametre-rule.
(Notes: Active or passive sentences are acceptable. Must have a
value + measuring instrument)
7th
mark
/
7
Method to control
RV
Measure time for 20 swings,t20 by using astop-watch.
Calculate period of a swing, T as follows: 20
20t
T
8th
mark
/8
Repetition Repeat the experiment with l= 20.0 cm, 30.0 cm, 40.0 cm, 50.0 cm
using the same bob.
9th
mark/9
: Tabulate data
l (cm) T (s)
10.0
20.0
30.040.0
50.0
10th
mark/10
: Analyze data
Plot graph T(s) against l (cm)
T s
l cm