form 4 physics chapter 1 - teacher's copy

7/30/2019 Form 4 Physics Chapter 1 - Teacher's Copy

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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


MV is lengthKeywords to indicate RV is time of

making a complete swing


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

form 4 physics chapter 1 - teacher's copy

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