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Physics Module Form 4 Chapter 1 – Introduction to physics GCKL 2011

1-1

U N D E R S T A N D I N G P H Y S I C S

P E N G E N A L A N K E P A D A F I Z I K

Mengenali konsep

fizik dalam

kehidupan harian

dan fenomena

semulajadi.

Recognise the

concepts of

physics in

everyday objects

and natural

phenomena

Tandakan () dalam pernyataan yang berkaitan dengan fizik. Tick () the statements that are related to physics.

Blood circulation in our body is controlled by heart. Peredaran darah dalam badan yang dikawal oleh jantung.

A large truck moves faster than a car because it has a more powerful engine. Lori yang besar bergerak dengan cepat berbanding kereta disebabkan enjin yang lebih berkuasa.

We need water in our digestion system. Kita memerlukan air dalam sistem pencernaan.

Orange juice is acidic because its taste is sour. Jus oren ialah asid disebabkan ia berbau masam.

An object on a high building has a large potential energy. Sesuatu objek di atas bangunan tinggi mempunyai tenaga keupayaan yang besar.

When we heat water, its temperature increased. Apabila kita memanaskan air, suhunya akan meningkat.

We are sweating when our body metabolism is high. Kita berpeluh apabila badan kita mempunyai metabolism yang tinggi.

Pure water has pH 7. Air yang tulen mempunyai pH 7.

We cannot see object in dark Kita tidak dapat melihat objek dalam keadaan gelap.

A ship is floating in water. Sebuah kapal sedang terapung di atas air.

Human body coordination is controlled by hormone system. Koordinasi badan manusia dikawal oleh sistem hormon.

Oxidation will act faster in acidic medium. Pengoksidaan bertindak dengan cepat dalam medium berasid.

U N D E R S T A N D I N G B A S E A N D D E R I V E D Q U A N T I T I E S

P E N G E N A L A N K UA N T I T I A S A S D A N K U A N T I T I T E R B I T A N

Recognise

physical quantity

and unit

1. Identify Physical quantities, Magnitude, Units and Measuring instrument from the

statements below. Write them into the table below (next page).

A Ismail weigh a wooden block that has mass of 500 gram using a lever beam balance.

B Ong Beng Hock measures the length of a building which is 100 meter long using a measuring tape.

C Siew Mei measures her body’s temperature using a digital thermometer and obtains

38C.

1.1

1.2


1-2

D Bathumalai determines the volume of water using a measuring cylinder and obtains 150 milliliter.

E Hanisah measures the diameter of a wire which is 1.26 millimeter using a micrometer screw gauge.

F Vinisha takes the time of 20 oscillations of a pendulum using a stopwatch and obtains 24.6 seconds.

Statement Physical quantity Magnitude Unit Measuring instrument

A Mass 500 Gram Lever beam balance

B Length 100 Meter Measuring tape

C Temperature 38 C Thermometer

D Volume 150 Milliliter Measuring cylinder

E Diameter 1.26 Millimetre Micrometer screw gauge

F time 24.6 Second stopwatch

Define base

quantities and

derived quantities

are

2. Identify base quantities and derived quantity from the equation below.

(a) Volume = length x length x length

Base quantity = (i) _________________ Derived quantity = (i) _________________

(b) Area = length x length

Base quantity = (i) _________________ Derived quantity = (i) _________________

(c)

Base quantity = (i) _________________ (ii) ____________________ Derived quantity = (i) _________________

(i) Base quantity is physical quantity that __________ be derived from any quantities.

(ii) Derived quantity is physical quantity that ___________________ from the base

quantities.

length

volume

length

Area

Mass

Density

length

cannot

is derived


1-3

List base quantities

and their S.I unit

3. Choose base quantities from the physical quantities given above and state their S.I

units.

No. Base Quantity S.I Unit

1. length meter

2. Mass kilogram

3. Time second

4. Electric current Ampere

5. temperature Kelvin

List some derived

quantities and their

S.I units

4. Write 5 derived quantities from physical quantities given in the box above (previous page) and state their S.I units. [*any five]

No. Derived Quantity S.I Unit

1. Pressure Pascal

2. Force / weight Newton

3. Work / energy Joule

4. Velocity m s-1

5. Area m2

6. Volume m3

\ Express quantities

using scientific

notation

5. Rewrite the values below in scientific notation (Standard notation)

No. Original value Scientific notation

1. 12 000 m 1.2 x 104 m

2. 3 000 000 000 s 3.0 x 109 s

3. 0.000 000 000 56 N 5.6 x 10-10

N

4. 0.000 78 J 7.8 x 10-4

J

5. 0.0034 A 3.4 x 10-3

A

PHYSICAL QUANTITY Pressure Time Current Length Area Temperature Weight Force Volume Work Energy Power Velocity Mass

S.I UNIT Second Newton Ampere Kelvin kilogram Pascal Joule m2 Watt m s-1 meter m3


1-4

Express quantities

using prefixes

6. Arrange the prefixes given below in ascending order. Then, state their multiple / sub-multiple.

No. Prefix Multiple /

Sub-multiple

No. Prefix

Multiple /

Sub-multiple

1. Tera 1012 7. pico 10

-12

2. Giga 109 8. nano 10

-9

3. Mega 106 9. micro 10

-6

4. kilo 103 10. milli 10

-3

5. hecto 102 11. centi 10

-2

6. deca 101 12. deci 10

-1

Solving problem

involving

conversion of units

1. Rewrite the values below using the suitable prefix.

(i) 4.1 x 1012 m = __________ (vii) 3.8 x 102 K = __________

(ii) 9.3 x 101 s = __________ (viii) 1.7 x 109 W = __________

(iii) 0.5 x 10-3 J = __________ (ix) 4.1 x 103 C = __________

(iv) 11.2 x 10-2 N = __________ (x) 9.5 x 10-6 A = __________

(v) 5.9 x 106 V = __________ (xi) 8.6 x 10-12 m = __________

(vi) 6.6 x 10-9 m = __________ (xii) 2.2 x 10-1 s = __________

2. Replaced the prefix in the values below with the correct multiple or sub-multiple.

(i) 4.1Tm = 4.1 x 1012 m (vii) 3.8 daK = 3.8 x 101 K

(ii) 9.3 ms = 9.3 x 10-3 s (viii) 1.7 GW = 1.7 x 109 W

(iii) 0.5 kJ = 0.5 x 103 J (ix) 4.1 hC = 4.1 x 102 C

(iv) 11.2 cN = 11.2 x 10-2

N (x) 9.5 A = 9.5 x 10-6 A

(v) 5.9 MV = 5.9 x 106 V (xi) 8.6 pm = 8.6 x 10-12 m

(vi) 6.6 dm = 6.6 x 10-1 m (xii) 2.2 ns = 2.2 x 10-9 s

PREFIXES Nano (n) kilo (k) pico (p) mega (M) centi (c) giga (G) deci (d) deca (da) tera (T)

hector (h) micro () milli (m)

MULTIPLE / SUB-MULTIPLE 103 109 10-2 101 10-12 106 10-6 102 10-1 10-9 10-3 1012

4.1 Tm

9.3 das

0.5 mJ

11.2 cN

5.9 MV

6.6 nm

3.8 hK

1.7 GW

4.1 kC

9.5 A

8.6 pm

2.2 ds


1-5

Check Yourself 1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Which of the following physical quantities is not a base quantity? A Weight C Temperature

B Time D Electric current

Which physical quantity has the correct S.I unit?

Physical quantity S.I unit

A Temperature Celcius B Time minute C Mass Newton D Length metre

Time, speed, density, and energy are .............. quantities. A base C vector B scalar D derived

Which of the following shows the correct

relationship between the base quantities for density?

A M

L L L

B M L M

T

C T T

L M

D M L

L L T

Which of the following is not a base S.I unit? A Gram C Ampere B Kelvin D Meter

A radio station airs its programmes by transmitting waves at a frequency of 102.3 MHz. How much is this frequency, in Hz?

A 1.023 x 102 B 1.023 x 105 C 1.023 x 106 D 1.023 x 108

Which of the following values is equal to 470 pF? A 4.7 x 10-10 F

B 4.7 x 1011 F C 4.7 x 10-7 F D 4.7 x 102 F

Hamid cycles at a velocity of 3.1 km h-1. What is this velocity, in m s-1?

A 0.09 C 1.16 B 0.86 D 11.61

Which measurement is the longest?

A 2.68 x 103 m B 2.68 x 10-1 mm

C 2.68 x 103 cm D 2.68 x 10-4 m

Which of the following conversion of unit is correct? A 24 mm3 = 2.4 x 10-6 m3 B 300 mm3 = 3.0 x 10-7 m3

C 800 mm3 = 8.0 x 10-2 m3

D 1 000 mm3 = 1.0 x 10-4 m3

Which of the following frequencies is the same as 106.8 MHz? A 1.068 x 10-4 Hz B 1.068 x 10-1 Hz

C 1.068 x 102 Hz D 1.068 x 106 Hz E 1.068 x 108 Hz


1-6

UNDE RS T A NDI NG S CAL A R AN D VE CT O R QUA NT I T IE S

Define scalar and

vector quantities. 1. Read the statements below to make a generalisation on scalar quantity and vector

quantity. Then classify the physical quantities into scalar quantity and vector quantity

in the table below.

A Hasan walks with a velocity of 2 m s-1 due West.

B Husna runs with a speed of 5 m s-1.

C Sangeetha walks along a displacement of 40 m due North.

D Jason runs along a distance of 30 m.

E Chin Wen push the table downwards with a force of 30 N.

F Wen Dee has a mass of 40 kg.

Scalar Quantity Vector Quantity

Speed Velocity

Distance Displacement

Mass Force

(i) Scalar quantity is physical quantity that has _______________________.

(ii) Vector quantity is physical quantity that has _______________________ and

________________________.

Differentiate

between distance

and displacement.

2. Underline the correct physical quantity.

(i) (Distance / Displacement) is the total length travelled by an object.

(ii) (Distance / Displacement) is the shortest length measured between the initial

point and the final point.

1.3

magnitude

magnitude

direction


1-7

EXAMPLE:

Check Yourself 2

1.

2.

3.

4.

A B

C

4 m

3 m

A boy walks from A to C through B.

(i) Distance of the boy = 4 m + 3 m = 7 m

(ii) Displacement of the boy = 42 + 32 = 5 m

Diagram below shows the path travelled by a

car from P to S.

What is the displacement of the car?

A 5.0 km C 8.2 km B 6.8 km D 9.0 km

Which quantity is a vector quantity?

A Area C Distance B Length D Displacement

Which of the following is group of vector quantities? A Velocity, mass, displacement B Speed, time, acceleration C Force, velocity, displacement

D Area, temperature, momentum

Which of the following quantities is a vector quantity?

A Energy C Force B Power D Pressure

5 km

3 km

1 km

P Q

S R


1-8

U N D E R S T A N D I N G M E A S U R E M E N T S

Recognise

appropriate

instrument for

physical quantities

1. State the suitable measuring instrument for the physical quantities in the table below.

No. Physical Quantity Measuring Instrument

1. Temperature Thermometer

2. Length Metre rule

3. Time Stopwatch

4. Mass Lever balance

5. Electric current Ammeter

6. Voltage Voltmeter

7. Density Hydrometer

8. Atmospheric pressure Barometer

9. Pressure Bourdon gauge

10. Force Spring balance

11. Volume Measuring cylinder

12. Diameter of tube Vernier calliper

13. Diameter of wire Micrometer screw gauge

Measure physical

quantity using

vernier calliper.

2. Label the part of vernier calliper below.

1.4

MEASURING INSTRUMENT Metre rule Barometer Thermometer Lever balance Spring balance Hydrometer Measuring cylinder Bourdon gauge Ammeter Voltmeter Stopwatch Vernier calliper Micrometer screw gauge

PART OF VERNIER CALLIPER

Inner jaws Outer jaws Main scale Vernier scale

0 1 2 3 4 5 6

Inner jaws Main scale

Outer jaws

Vernier scale


1-9

3. Take the reading from a vernier calliper:

EXAMPLE:

4. Read the vernier calliper below.

(i) (ii)

(ii) (iv)

(i) Read the main scale.

Main scale reading = 0.70 cm

(ii) Read the vernier scale.

Vernier scale reading = 0.02 cm

(iii) Total up the readings.

Actual reading = 0.72 cm

3 4

0

Main scale = ....................

Vernier scale = ....................

Actual reading = ....................

2 3

0

Main scale = ....................

Vernier scale = ....................

Actual reading = ....................

8 9

0

Main scale = ....................

Vernier scale = ....................

Actual reading = ....................

0 1

Main scale = ....................

Vernier scale = ....................

Actual reading = ....................

0 1 2

0

(i) 0.70

cm

(ii) 0.02

cm

3.10 cm

0.03 cm

3.13 cm

2.10 cm

0.06 cm

2.16 cm

8.50 cm

0.06 cm

8.56 cm

0.20 cm

0.04 cm

0.24 cm


1-10

Measure physical

quantity using

micrometer screw

gauge.

5. Label the part of micrometer screw gauge below.

6. Take the reading from a micrometer screw gauge. EXAMPLE:

7. Read the micrometer screw gauge below: (i) (ii)

(iii) (iv)

PART OF MICROMETER SCREW GAUGE Anvil Spindle Sleeve Thimble Ratchet

(i) Read the sleeve scale (main scale).

Sleeve scale reading = 3.00 mm

(ii) Read the thimble scale.

Thimble scale reading = 0.44 mm

(iii) Total up the readings.

Actual reading = 3.44 mm

0 50

40

(i) 3.00 mm

(ii) 0.44 mm

0 30

20

Sleeve scale = ....................

Thimble scale = ....................

Actual reading = ....................

0 20

10

Sleeve scale = ....................

Thimble scale = ....................

Actual reading = ....................

0

20

10

0 60

50

Anvil Spindle Thimble Ratchet

Main scale

5.00 mm

0.24 mm

5.24 mm

3.00 mm

0.16 mm

3.16 mm


1-11

Explain sensitivity. 8. Table below shows readings from three instruments J, K, and L that are used in measuring the mass of a Petri dish.

Instrument J Instrument K Instrument L

20 g 19.4 g 19.42 g

A piece of dried leaf of mass 0.05 g is then put in the Petri dish.

(i) Which instrument is able to detect the small change of the mass? [...............]

(ii) Which instrument is the most sensitive? [................]

(iii) Which instrument has the highest sensitivity? [................]

(iv) Sensitivity of instrument is the capability of the instrument to ...............................

..................................................................................................................................

(v) Which instrument gives reading in the most decimal place? [.................]

(vi) The ...................... the decimal place, the ........................... sensitivity of the instrument.

Explain accuracy. 9. Table below shows readings from three instruments P, Q, and R that are used in measuring the length of a wire. The actual length of the wire is 10.0 cm.

Instrument P Instrument Q Instrument R

10.1 cm 10.4 cm 9.6 cm

(i) Which instrument gives the closest reading to the actual length of the wire?

[...............]

(ii) Which instrument gives the most accurate reading? [...............]

(iii) Which instrument has the highest accuracy? [...............]

(iv) Accuracy of instrument is the capability of the instrument to .................................. ...................................................................................................................................

(iv) Read the main scale.

Main scale reading = 0.70 cm

(v) Read the vernier scale.

Vernier scale reading = 0.02 cm

(vi) Total up the readings.

Actual reading = 0.72 cm

Sleeve scale = ....................

Thimble scale = ....................

Actual reading = ....................

Sleeve scale = ....................

Thimble scale = ....................

Actual reading = ....................

5.50 mm

0.19 mm

5.69 mm

3.00 mm

0.56 mm

3.56 mm

L

L

L

L

detect small

changes.

more higher

P

P

P

give reading

close to the actual size.


1-12

Explain

consistency

(Precision)

10. Table below shows four readings from three instruments X, Y, and Z that are used in measuring the length of a wire. Each instrument repeats the measurement for four times.

Instrument X 10.0 cm 10.1 cm 10.1 cm 10.0 cm

Instrument Y 10.1 cm 10.4 cm 10.2 cm 9.8 cm

Instrument Z 9.8 cm 9.6 cm 9.9 cm 9.5 cm

(i) Which instrument gives readings with the smallest deviation (difference)?

[...............]

(ii) Which instrument gives the most consistence readings? [...............]

(iii) Which instrument has the highest consistency? [...............]

(iv) Consistency of instrument is the capability of the instrument to .............................

...................................................................................................................................

Explain type of

experimental error.

11. In an experiment, the readings of measurement taken may have slightly difference due to some mistakes. The difference in the readings is called as .........................................

12. These errors can be caused by the change of environment, human factors or the

deficiency of measuring instrument.

13. Error that is caused by environment and human usually is (constant / changeable)*.

14. Error that is caused by the instrument is always (constant / changeable)*.

15. Type of Error:

Random Error Difference Systematic Error

Human factor and

environment Cause Instrument

Random Magnitude /

value Constant

Parallax error Example Zero error

Take few readings and find

the average reading

Method to reduce the

error

Add or deduct the zero

error from the reading.

Use different instrument

while taking readings and

find the average

X

X

X

give reading

with small deviation/difference.

error


1-13

Check Yourself 3

1. 4.

.

2. 5.

3.

6.

A, B, C, and D shows the shooting marks on a target. Which marks can explain the concept of precision of a measurement? A C

B D

Diagram below shows the target board in a game.

Which result is consistent but not accurate? A C

B D

The diagram shows the scale of a micrometer screw gauge.

What is the reading of the micrometer?

A 7.02 mm C 7.03 mm B 7.52 mm D 7.58 mm

A, B, C, and D show parts of four different

balance scales. Which balance is the most sensitive? A C

B D

Target Target board

Table below shows the readings of the thickness of a board which are taken by four students.

Student Reading/cm

1 2 3 4

A 2.50 2.50 2.50 2.50

B 2.53 2.53 2.53 2.53

C 2.52 2.53 2.54 2.53

D 2.71 2.73 2.74 2.74

The diagrams show the scales on a pair of vernier callipers and a metre rule.

Which comparison is correct about the

sensitivity of the vernier callipers and the metre rule when measuring the thickness of a wire? Vernier callipers Metre rule

A Low sensitivity Low Sensitivity B Low sensitivity High sensitivity

C High sensitivity Low sensitivity D High sensitivity High sensitivity

Vernier calliper Metre rule


1-14

Four students, A, B, C, and D use a micrometer screw gauge, a metre rule, and a vernier calliper to measure the thickness of a board. Which student records the reading correctly?

Micrometer Metre Vernier screw rule/mm calliper/mm

gauge/mm

A 11.1 11 11.13 B 11.13 11.1 11.128 C 11.128 11.1 11.13 D 11.13 11 11.1

7.

12.

8.

12. 9.

10.

Each student made four measurements. If the actual thickness of the board is 2.53 cm, which of the students A, B, C, and D made the measurements that are accurate but not consistent?

The diagram shows the scale of a vernier calliper.

What is the reading of the vernier calliper?

A 2.16 cm C 1.86 cm B 2.06 cm D 1.76 cm

Atmospheric pressure can be measured by using

A hydrometer B Bourdon gauge and manometer

C Bourdon gauge and mercury barometer D manometer and mercury barometer

The diagram shows the scale of a micrometer screw gauge.

What is the reading of the micrometer?

A 4.95 mm C 4.50 mm B 4.55 mm D 4.45 mm

11. Diagram (a) shows the reading of a vernier calliper while its jaws are closed. Diagram (b) shows the reading of the vernier calliper when a metal sheet is placed between the jaws.

(a) (b) What is the thickness of the metal sheet? A 0.46 cm C 0.38 cm

B 0.42 cm D 0.32 cm

Which of the following statements is correct

about zero error?

A Can be reduced by determining average reading.

B The magnitude of error increases when the value of the reading increases.

C Exist either in positive or negative.

D The magnitude of error increases if the range of scale is large.

Diagram below shows two types of ammeters,

X and Y, that can be used to measure electric current.

(a) Which ammeter is more sensitive?

...................................................................

(b) State one reason for your answer above.

...................................................................

................................................................... ...................................................................

Ammeter Y

Ammeter Y has smaller division of

scale


1-15

13. 13.

13.

14. 16.

18.

Which of the following ways can reduce the

parallax error while taking reading of current from an ammeter?

A Use a higher sensitivity ammeter. B Repeat the measurement and calculate

the average reading. C Take the reading using a magnifying

glass. D Use ammeter that has plane mirror

below the pointer.

What is the function of the plane mirror under the pointer in an ammeter?

A To increase the consistency of the

measurement. B To increase the accuracy of the

measurement. C To avoid parallax error. D To prevent zero error.

Figure below shows the scale of an ammeter.

(a) Name the physical quantity measured by

the ammeter. ...................................................................

(b) What is the value of the smallest division

on the scale? ...................................................................

(c) State the function of the mirror located

under the scale. ................................................................... ...................................................................

Mirror

(a) The external diameters of the cylinder at four different places are shown in the table below.

External diameter/cm Relative

deviation/%

2.04 2.05 2.04 2.06 0.37

(i) Why is the external diameter

measured four times? .............................................................

.............................................................

(ii) What is the purpose of calculating the relative deviation? ............................................................. .............................................................

.............................................................

Figure below shows the meniscus of oil in a measuring cylinder. P, Q and R are three eye

positions while measuring the volume of the oil.

(a) Which position of the eye is correct while taking the reading of the volume of oil? ...................................................................

(b) Give one reason for the answer above. ...................................................................

...................................................................

Electric current

0.1 A

To avoid parallax error

To get average reading / To find

relative deviation

To determine the consistency of the

measurement

Q. (but the direction must be 90)

Position of eyes is at the level of the

meniscus of the oil


1-16

17.

19.

Figure below shows a vernier calliper used to measure external diameter of a hollow cylinder.

(b) Name the part labelled X.

...................................................................

(c) What is the function of X? ...................................................................

...................................................................

A student is assigned to measure the thickness of a metal sheet. The student is provided with a vernier calliper.

(a) The student uses the vernier calliper to

measure the thickness of the metal sheet. Figure (i) shows the scale of the vernier calliper while the jaws are closed. Figure (ii) shows the scale of the vernier calliper when the metal sheet is put between the

jaws.

(ii)

(i)

(i) What is the zero error of the vernier calliper? .............................................................

(ii) Calculate the thickness of the metal sheet.

Thickness = .................................

Inner jaws

To measure internal diameter of hollow

object

-0.04 cm

Zero error = - 0.04 cm

Reading = 3.62 cm

Actual reading = 3.62 – (-0.04) cm

= 3.66 cm

3.66 cm


1-17

U N D E R S T A N D I N G S C I E N T I F I C I N V E S T I G A T I O N

Identify variables

in a given situation 1. Identify and state the variables that can be investigated from the situations below.

EXAMPLE: The car moves faster when it is pushed harder.

Cause : pushed harder Manipulated variable : Force

Effect : moves faster Responding variable : Speed/Velocity/ Acceleration

No. Situation Manipulated

variable

Responding

variable

1. The temperature of smaller block rises faster when it is heated.

Mass Temperature

2. The pendulum system with longer string

takes longer time to stop. Length Time

3. The loaded lorry is harder to stop than the empty lorry.

Mass Time to stop

4. The trolley that falls from the higher place moves faster.

Height Speed

5. The spring becomes longer when it is

pulled harder. Force Length

1.5


1-18

Making inference 2. Write inference from the given variables.

EXAMPLE:

Manipulated variable : Length Responding variable: Time

Inference : The length affects the time taken.

No. Manipulated

variable

Responding

variable Inference

1. Force Acceleration The force affects the acceleration

2. Mass Temperature The mass affects the temperature

3. Force Extension The force affects the extension

4. Mass Time The mass affects the time

5. Force Pressure The force affects the pressure

6. Area Pressure The area affects the pressure

7. Temperature Volume The temperature affects the volume


1-19

Form hypothesis. 3. Write hypothesis from the given variables.

EXAMPLE:

Manipulated variable : Length Responding variable: Time

Hypothesis : The longer the length, the longer the time taken.

No. Manipulated

variable

Responding

variable Hypothesis

1. Force Acceleration The larger the force, the higher the

acceleration

2. Mass Temperature The larger the mass, the lower the

temperature

3. Force Extension The larger the force, the longer the

extension

4. Mass Time The larger the mass, the longer the time

5. Force Pressure The larger the force, the higher the

pressure

6. Area Pressure The larger the area, the lower the pressure

7. Temperature Volume The higher the temperature, the larger the

volume

Analyse the data. 4. Data obtained from an experiment can be analysed by plotting a line graph.

Manipulated variable is on the x-axis, and responding variable is on the y-axis.

The variables must be stated together with the correct unit.

EXAMPLE:

Manipulated variable : Mass

Responding variable : Time

Mass/kg

Time/min


1-20

5. Sketch a graph to analyse the following variables:

(i) Manipulated variable : Force (ii) Manipulated variable : Mass

Responding variable : Acceleration Responding variable : Temperature

(iii)

Manipulated variable : Force (iv) Manipulated variable : Mass

Responding variable : Extension Responding variable : Time

Force/N

Acceleration/m s-2

Mass/kg

Temperature/C

Force/N

Extension/cm

Mass/kg

Time/s


1-21

(v) Manipulated variable : Force (vi) Manipulated variable : Area

Responding variable : Pressure Responding variable : Pressure

Interpret data to

draw a conclusion.

6. The conclusion of an experiment is made based on the line graph obtained.

EXAMPLE:

Conclusion: Conclusion:

The time is directly proportional to the mass.

The pressure is inversely proportional to the area.

Conclusion:

The temperature is linearly

increasing with the time.

Mass/kg

Time/min

1

𝐴𝑟𝑒𝑎/

Pressure/Pa

m-2

Time/min

Temperature/C

Force/N

Pressure/Pa

Area/cm2

Pressure/Pa


1-22

Interpret data to

draw a conclusion.

7. Write a conclusion based on the line graphs below:

(i) (ii)


The square of period is directly

proportional to the length

The temperature is inversely

proportional to the mass

(iii) (iv)


The volume is linearly increasing

with the pressure

The extension is directly

proportional to the force

Length/cm

Period2/s2 Temperature/C

1

𝑀𝑎𝑠𝑠/ kg-2

Pressure/kPa Force/N

Volume/m3 Extension/cm


1-23

Check Yourself 4

1. 4.Which of the following graphs obeys the equation F = kx, where k is a constant? A C

B D

Diagram below shows an investigation about

the stretching of a spring. Babies of different masses are supported by identical springs.

Which of the following variables are correct?

Manipulated

variable

Responding

variable

Constant

variable

A Mass of the baby

Length of the spring

Diameter of the spring

B Length of the spring

Mass of the baby


C Diameter of

the spring

Length of

the spring

Mass of the

baby

D Mass of the baby


Length of the spring

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-24

.

2.

5.

3.

Table below shows the results of an experiment to investigate between load and extension when a spring is stretched.

Load, F/N 100 150 200 250 300

Extension, x/cm

1.0 1.5 2.0 2.5 3.0

The original length of the spring is l0 = 15.0 cm. What is the manipulated variable? A Load, F B Extension, x C Original length of the spring, l0

D Material used to make the spring

The graph shows the relationship between v and t.

The relationship between v and t is represented by the equation

A 𝑣 𝑝

𝑞𝑡 + 𝑝 C 𝑣 −

𝑝

𝑞𝑡 + 𝑝

B 𝑣 𝑝

𝑞𝑡 + 𝑞 D 𝑣 −

𝑝

𝑞𝑡 + 𝑞

The graph shows the relationship between physical quantities P and Q.

Which statements about the graph is correct?

A If Q = 1, then P = 2. B The gradient of the graph is 1. C P is directly proportional to Q. D The equation of the graph is P = 1 + 3Q


2-25

Physical Quantity Definition, Quantity, Symbol and unit

Distance, s

Distance is the total path length travelled from one location to another.

Quantity: scalar SI unit: meter (m)

Displacement, s

(a) The distance in a specified direction.

(b) the distance between two locations measured along the shortest path

connecting them in a specific direction.

(c) The distance of its final position from its initial position in a

specified direction.

Quantity: vector SI unit: meter (m)

Speed,v

Speed is the rate of change of distance

Speed = time

ceDis tan

Quantity: scalar SI unit: m s -1

Velocity, v

Velocity is the rate of change of displacement.

Velocity = time

ntDisplaceme

Direction of velocity is the direction of displacement

Quantity : Vector SI unit: m s -1

Average speed

v =TotalTime

tTotalDis tan

Example: A car moves at an average speed / velocity of 20 ms -1

On average, the car moves a distance/ displacement of 20 m in 1 second for the

whole journey.

Average velocity

Displacement

TotalTimev

Uniform speed Speed that remains the same in magnitude without considering its direction

Uniform velocity Velocity that remains the same in magnitude and direction

An object has a non-

uniform velocity if

(a) The direction of motion changes or the motion is not linear.

(b) The magnitude of its velocity changes.

Acceleration, a When the velocity of an object increases, the object is said to be accelerating.

Acceleration is defined as the rate of change of velocity


2-26

v ua

t

Unit: ms-2

Acceleration is positive

Change in velocityAcceleration=

Time taken

Final velocity,v - Initial velocity,u =

Time taken,t

The velocity of an object increases from an initial velocity, u, to a higher final velocity, v

Deceleration

acceleration is negative.

The rate of decrease in speed in a specified direction.

The velocity of an object decreases from an initial velocity, u, to a lower final velocity, v.

Zero acceleration An object moving at a constants velocity, that is, the magnitude and direction of its velocity remain unchanged – is not accelerating

Constant acceleration Velocity increases at a uniform rate.

When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity increases by 5 ms -1 for every second that the car is in motion.

1. Constant = uniform

2. increasing velocity = acceleration

3. decreasing velocity = deceleration 4. zero velocity = object at stationary / at rest

5. negative velocity = object moves in opposite direction 6. zero acceleration = constant velocity 7. negative acceleration = deceleration


2-27

Speed Velocity

The rate of change of

distance

The rate of change of

displacement

Scalar quantity Vector quantity

It has magnitude but no

direction

It has both magnitude and

direction

SI unit : m s-1

SI unit : m s-1

Comparisons between distance and displacement Comparisons between speed and velocity

Fill in the blanks:

1. A steady speed of 10 ms -1 = A distance of 10 m is travelled every second.

2. A steady velocity of -10 ms -1 = A displacement of 10 m is travelled every 1 second to the left.

3. A steady acceleration of 4 ms -2 = Speed goes up by 4 ms-1

every 1 second.

4. A steady deceleration of 4 ms -2 = speed goes down by 4 ms-1 every 1 second

5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right.

Distance Displacement

Total path length travelled

from

one location to another

The distance between two

locations measured

along the shortest path

connecting them in

specific direction

Scalar quantity Vector quantity

It has magnitude but no

direction

It has both magnitude

and direction

SI unit meter SI unit : meter


2-28

Example 1 Every day Rahim walks from his house to the junction

which is 1.5km from his house. Then he turns back and stops at warung Pak Din which is

0.5 km from his house.

(a)What is Rahim’s displacement from his house • when he reaches the junction. 1.5 km to the right

• When he is at warung Pak Din. 0.5 km to the

left.

(b)After breakfast, Rahim walks back to his house.

w hen he reaches home, (i) what is the total distance travelled by

Rahim? (1.5 + 1.5 + 0.5+0.5 ) km = 4.0 km

(ii) what is Rahim’s total displacement from his house? 1.5 +( -1.5) +(- 0.5 )+0.5 km = 0 km

Example 2

Every morning Amirul walks to Ahmad’s house

which is situated 80 m to the east of Amirul’s house. They then walk towards their school which is 60 m to the south of Ahmad’s house.

(a)What is the distance travelled by Amirul and his displacement from his house?

Distance = (80 +60 ) m = 140 m

Displacement = 100 m

tan θ = 60

80 =1.333 θ = 53.1º

(b)If the total time taken by Amirul to travel from his house to Ahmad’s house and then to school is 15 minutes, what is his speed and velocity?

Speed =140

15 60

m

s =0.156 in ms-1

Velocity =100

15 60

m

s = 0.111 ms-1

Example 3 Salim running in a race covers 60 m in 12 s. (a) What is his speed in ms-1

Speed = s

m

12

60= 5 ms-1

(b) If he takes 40 s to complete the race, what is his

distance covered?

distance covered = 40 s × 5 ms-1 = 200 m

Example 4

An aeroplane flies towards the north with a

velocity 300 km hr -1 in one hour. Then,

the plane moves to the east with the

velocity 400 km hr -1 in one hour.

(a)What is the average speed of the plane?

Average speed = (300 km hr -1 +

4 00 km hr -1 ) / 2 = 350 km hr -1

(b)What is the average velocity of the plane?

Average velocity = 250 km hr -1

Tan θ = 300

400 = 1.333 θ =


2-29

(c)What is the difference between average speed

and average velocity of the plane? Average speed is a scalar quantity. Average velocity is a vector quantity

Example 5 The speedometer reading for a car travelling due north

shows 80 km hr -1. Another car travelling at 80 km hr -

1 towards south. Is the speed of both cars same? Is

the velocity of both cars same?

The speed of both cars are the same but the velocity of

both cars are different with opposite direction

A ticker timer

Use: 12 V a.c. power supply

1 tick = time interval between two dots.

The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2

consecutive dots is 1/50 = 0.02 s.

1 tick = 0.02 s


2-8

Relating displacement, velocity, acceleration and time using ticker tape.

VELOCITY FORMULA

Time, t = 10 dicks x 0.02 s = 0.2 s

displacement, s = x cm

velocity =

2

ACCELERATION

Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or

t = (50 – 10) ticks x 0.02 s = 0.8 s

Initial velocity, u =

2

final velocity, v = 2

2

acceleration, a =

TICKER TAPE AND CHARTS TYPE OF MOTION

Constant velocity

– slow moving

Constant velocity – fast moving

Distance between the dots increases uniformly

the velocity is of the object is increasing uniformly

The object is moving at a uniform / constant

acceleration.


2-9

- Distance between the dots decrease uniformly

- The velocity of the object is decreasing Uniformly

- The object is experiencing uniform / constant

decceleration

Example 6

The diagram above shows a ticker tape chart for

a moving trolley. The frequency of the

ticker-timer used is 50 Hz. Each section has

10 dots-spacing.

(a) What is the time between two dots.

Time = 1/50 s = 0.02 s

(b) What is the time for one strips.

0.02 s × 10 = 0.2 s

(c) What is the initial velocity

2 cm / 0.2 s = 10 ms-1

(d) What is the final velocity.

12 cm / 0.2 s = 60 ms-1

(e) What is the time interval to change from

initial velocity to final velocity?

( 11 - 1) × 0.2 s = 2 s

(f) What is the acceleration of the object.

a = t

uv =

2

1060 ms-2 = 25 ms-2

THE EQUATIONS OF MOTION

2

2 2

1

2

2

v u at

s ut at

v u as

u = initial velocity v = final velocity t = time taken s = displacement a = constant acceleration

M O T I O N G R A P H S

2.2


2-10

DISPLACEMENT – TIME GRAPH

Velocity is obtained from the gradient of the graph.

A – B : gradient of the graph is positive and constant velocity is constant.

B – C : gradient of the graph = 0

the velocity = 0, object is at rest.

C – D : gradient of the graph negative and constant.

The velocity is negative and object moves

in the opposite direction.

VELOCITY-TIME GRAPH

Area below graph Distance / displacement

Positive gradient Constant Acceleration

(A – B)

Negative gradient Constant Deceleration

(C – D)

Zero gradient Constant velocity /

zero acceleration

(B – C)

GRAPH s versus t v versus t a versus t

Zero velocity

Negative

constant

velocity

Positive Constant

velocity

GRAPH s versus t v versus t a versus t


2-11

Constant acceleration

Constant deceleration


2-10

Example 1

Contoh 11

Based on the s – t graph above:

(a) Calculate the velocity at

(i) AB (ii) BC (iii) CD (i) 5 ms-1 (ii) 0 ms-1 (iii) - 10 ms-1

(b) Describe the motion of the object at: (i) AB (ii) BC (iii) CD

(i) constant velocity 5 ms-1 (ii) at rest / 0 ms-1 (iii) constant velocity of 10 ms-1in opposite

direction

(c)Find: (i) total distance 50 m + 50 m = 100 m

(ii) total displacement 50 m + (- 50 m) = 0

(d) Calculate

(i) the average speed s

m

35

100= 2.86 ms-1

(ii) the average velocity of the moving particle.

0

Example 2

(a) Calculate the acceleration at: (i) JK (ii) KL (iii) LM (i) 2 ms-2 (ii) -1 ms-2 (iii) 0 ms-1 (b) Describe the motion of the object at:

(i) JK (ii) KL (iii) LM

(i) constant acceleration of 2 ms-2

(ii) constant deceleration of 1ms-2 (iii) (iii) zero acceleration or constant

velocity

Calculate the total displacement. Displacement = area under the graph

= 100 m + 150 m + 100 m + 25 m = 375 m

(c) Calculate the average velocity. Average velocity = 375 m / 40 s = 9.375 ms-1


2-11

I N E R T I A

Inertia The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its motion.

Newton’s first law Every object continues in its state of rest or of uniform motion unless it is acted upon by an external force.

Relation between inertia and mass

The larger the mass, the larger the inertia

SITUATIONS INVOLVING INERTIA

SITUATION EXPLANATION

EEEEEEEEJNVJLKNDNFLJKVNDFLKJNBVJKL;DFN BLK;XC

NB[F NDPnDSFJ[POJDE]O-JBD]AOP[FKBOP[DF

LMB NOPGFMB LKFGNKLB

FGNMNKL’ MCVL BNM’CXLB

NFGNKEPLANATION

When the cardboard is pulled away quickly, the coin drops straight into the glass.

The inertia of the coin maintains its state at rest. The coin falls into the glass due to gravity.

Chilli sauce in the bottle can be easily poured out if the bottle is moved

down fast with a sudden stop. The sauce inside the bottle moves together with the bottle.

When the bottle stops suddenly, the sauce continues in its state of motion

due to the effect of its inertia.

Body moves forward when the car stops suddenly The passengers were in a state

of motion when the car was moving.

When the car stopped suddenly, the inertia in the passengers made them maintain their state of motion. Thus when the car stop, the passengers moved forward.

A boy runs away from a cow in a zig- zag motion. The cow has a large inertia making it difficult to change direction.

2.3


2-23

The head of hammer is secured tightly to its handle by

knocking one end of the handle, held vertically, on a hard surface.

This causes the hammer head to continue on its

downward motion. When the handle has been stopped, so that the top

end of the handle is slotted deeper into the hammer head.

• The drop of water on a wet umbrella will fall when the boy rotates the umbrella.

• This is because the drop of water on the surface of the

umbrella moves simultaneously as the umbrella is rotated. • When the umbrella stops rotating, the inertia of

the drop of water will continue to maintain its motion.

Ways to reduce the negative effects of inertia

1. Safety in a car: (a)Safety belt secure the driver to their seats.

When the car stops suddenly, the seat belt provides the external force that prevents the driver from being thrown forward.

(b)Headrest to prevent injuries to the neck during rear-

end collisions. The inertia of the head tends to keep in its state of rest when the body is moved suddenly.

(c)An air bag is fitted inside the steering wheel.

It provides a cushion to prevent the driver from hitting the steering wheel or dashboard during a collision.

2. Furniture carried by a lorry normally are tied up together by string. When the lorry starts to move suddenly, the furniture are more difficult to fall off due to their inertia because their combined mass has increased.


2-24

M O M E N T U M

Relationship between mass and

inertia

• Two empty buckets which are hung with rope from the ceiling.

• One bucket is filled with sand while the other bucket is empty.

• Then, both pails are pushed.

• It is found that the empty bucket is easier to push.

Push and compared to the bucket with sand. • The bucket filled with sand offers more resistance to

movement.

• When both buckets are oscillating and an attempt is made to stop them, the bucket filled with sand offers more resistance to the hand (more difficult to bring

to a standstill once it has started moving) • This shows that the heavier bucket offers a greater

resistance to change from its state of rest or from its state of motion. An object with a larger mass has a larger inertia.

2.4


2-25

Definition Momentum = Mass x velocity = mv

SI unit: kg ms-1

Principle of Conservation of Momentum In the absence of an external force, the total

momentum of a system remains unchanged.

Elastic Collision Inelastic collision

ƒ Both objects move independently at their respective velocities after the collision.

ƒ Momentum is conserved.

ƒ Kinetic energy is conserved.

Total energy is conserved.

ƒ The two objects combine and move together with a common velocity after the collision.

ƒ Momentum is conserved.

ƒ Kinetic energy is not conserved.

ƒ Total energy is conserved.

Total Momentum Before = total momentum after

m1u

1 + m2u

2 = m1 v

1 + m2 v

2

Total Momentum Before = Total Momentum After

m1 u

1 + m

2 u

2 = ( m1 + m

2 ) v

Explosion

Before explosion both object stick together and at rest.

After collision, both object move at opposite

direction.

Total Momentum

before collision is

zero

Total Momentum after

collision :

m1v

1 + m2v

2


2-26

From the law of conservation of momentum:

Total Momentum = Total Momentum

Before collision after collision

0 = m1v

1 + m2v

2

m1v

1 = - m

2v

2

Negative sign means opposite direction

EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum)

When a rifle is fired, the bullet of mass m,

moves with a high velocity, v. This creates a

momentum in the forward direction.

From the principle of conservation of

momentum, an equal but opposite

momentum is produced to recoil the riffle

backward.

Application in the jet engine:

A high-speed hot gases are ejected from the back

with high momentum.

This produces an equal and opposite momentum

to propel the jet plane forward.

The launching of rocket

Mixture of hydrogen and oxygen fuels burn

explosively in the combustion chamber.

Jets of hot gases are expelled at very high speed

through the exhaust.

These high speed hot gases produce a large

amount of momentum downward.

By conservation of momentum, an equal but

opposite momentum is produced and acted on

the rocket, propelling the rocket upwards.


2-27

In a swamp area, a fan boat is used.

The fan produces a high speed movement of air

backward. This produces a large momentum

backward.

By conservation of momentum, an equal but opposite

momentum is produced and acted on the boat. So the

boat will move forward.

A squid propels by expelling water at high velocity.

Water enters through a large opening and exits

through a small tube. The water is forced out at a

high speed backward.

Total Mom. before= Total Mom. after

0 =Mom water + Mom squid

0 = mwv

w + msvs

- mwv

w = msvs

The magnitude of the momentum of water and

squid are equal but opposite direction. This

causes the quid to jet forward.

Example

Car A of mass 1000 kg moving at 20 ms -1 collides

with a car B of mass 1200 kg moving at 10 m s -1

in same direction. If the car B is shunted

forwards at 15 m s -1 by the impact, what is the

velocity, v, of the car A immediately after the

crash?

1000 kg x 20 ms -1 + 1200 kg x 10 ms

-1 =

1000 kg x v + 1200 kg x 15 ms

-1

v= 14 ms -1

Example

Before collision After collision M

A = 4 kg

MB

= 2 kg

UA = 10 ms -1 r i g h t

UB = 8 ms -1 l e f t V

B 4 ms-1 right

Calculate the value of VA .

[4 x 10 + 2 x (-8)]kgms

-1 =[ 4 x v + 2 x 4 ] kgms

-1

VA = 4 ms -1 right


2-28

Example

A truck of mass 1200 kg moving at 30 ms-1 collides

with a car of mass 1000 kg which is travelling in

the opposite direction at 20 ms-1. After the collision, the two vehicles move together. What is the

velocity of both vehicles immediately after collision?

1200 kg x 30 ms -1 + 1000 kg x (-20 ms

-1)

= ( 1200 kg + 1000kg) v

v = 7.27 ms -1 to the right

Example

A man fires a pistol which has a mass of 1.5 kg. If the mass of the bullet is 10 g and it reaches a velocity of 300 ms -1 after shooting, what is the

recoil velocity of the pistol?

0 = 1.5 kg x v + 0.01 kg x 300 ms

-1

v = -2 ms -1

Or it recoiled with 2 ms

-1 to the left

F O R C E

2.5


2-29

Balanced Force

When the forces acting on an object are

balanced, they cancel each other out. The

net force is zero.

Effect : the object is at rest

[velocity = 0]

or

moves at constant velocity

[ a = 0]

Example:

Weight, W = Lift, U Thrust, F = drag, G

Unbalanced Force/ Resultant Force

When the forces acting on an object are not balanced, there

must be a net force acting on it.

The net force is known as the unbalanced force or the

resultant force.

Effect : Can cause a body to

- change it state at rest (an object will

accelerate

- change it state of motion (a moving object

will decelerate or change its direction)

Newton’s Second Law of Motion

The acceleration produced by a force on an object is

directly proportional to the magnitude of the net

force applied and is inversely proportional to the

mass of the object. The direction of the acceleration

is the same as that of the net force.

Force = Mass x Acceleration

F = ma

Experiment to Find The Relationship between Force, Mass & Acceleration


2-30

Relationship between a & F a & m

Situation

Both men are pushing the same mass but man A puts greater effort. So he moves faster.

Both men exerted the same strength. But

man B moves faster than man A.

Inference The acceleration produced by an

object depends on the net force

applied to it.

The acceleration produced by an object

depends on the mass

Hypothesis The acceleration of the object

increases when the force applied

increases

The acceleration of the object decreases

when the mass of the object

increases

Variables: Manipulated

:

Responding :

Constant :

Force Acceleration

Mass

Mass Acceleration

Force

Apparatus and

Material

Ticker tape, elastic cords, ticker timer, trolleys, power supply, friction compensated

runway and meter ruler.


2-31

Procedure :

- Controlling

manipulated

variables. -Controlling

responding variables.

-Repeating

experiment.

An elastic cord is hooked over the

trolley. The elastic cord is stretched

until the end of the trolley. The trolley

is pulled down the runway with the

elastic cord being kept stretched by

the same amount of force

An elastic cord is hooked over a trolley.

The elastic cord is stretched until the

end of the trolley. The trolley is pulled

down the runway with the elastic cord

being kept stretched by the same

amount of force

Determine the acceleration by

analyzing the ticker tape.

Acceleration

Acceleration v u

at

Determine the acceleration by analyzing the

ticker tape.

Acceleration v u

at

Repeat the experiment by using two ,

three, four and five elastic cords

Repeat the experiment by using two,

three, four and five trolleys.

Tabulation of data Force, F/No of

elastic cord

Acceleration, a/ ms-2

1

2

3

4

5

Mass, m/

no of trolleys

Mass

, m

/

g

1/m,

g-

1

Acceleration/

ms-2

1

2

3

4

5 Analysing Result


2-32

1. What force is required to move a 2 kg

object with an acceleration of 3 m s-2,

if

(a) the object is on a smooth surface?

(b) The object is on a surface where the

average force of friction acting on the

object is 2 N?

(a) force = 6 N

(b) net force = (6 – 2) N

= 4 N

2. Ali applies a force of 50 N to move a 10 kg

table at a constant velocity. What is the

frictional force acting on the table?

Answer: 50 N

3. A car of mass 1200 kg travelling at 20 ms -1

is brought to rest over a distance of 30 m.

Find

(a) the average deceleration, (b) the average braking force.

(a) u = 20 ms -1 v = 0 s = 30 m a = ?

a = - 6.67 ms-2

(b) force = 1200 x 6.67 N

= 8000 N

4. Which of the following systems will

produce maximum acceleration? D


2-33

I M P U L S E A N D I M P U L S I V E F O R C E

Impulse The change of momentum mv - mu

Unit : kgms-1 or Ns

m = mass

u = initial velocity

v = final velocity

t = time Impulsive Force The rate of change of momentum in a

collision or explosion

Impulsive force =

change of momentum

time

mv mu

t

Unit = N

Effect of time Impulsive force is

inversely

proportional to

time of contact

Longer period of time →Impulsive force decrease

Shorter period of time →Impulsive force increase

Situations for Reducing Impulsive Force in Sports

Situations Explanation

Thick mattress with soft surfaces are used in events such as high jump so

that the time interval of impact on landing is extended, thus reducing

the impulsive force. This can prevent injuries to the participants.

Goal keepers will wear gloves to increase the collision time. This will

reduce the impulsive force.

2.6


2-34

A high jumper will bend his legs upon landing. This is to increase the time

of impact in order to reduce the impulsive force acting on his legs.

This will reduce the chance of getting serious injury.

A baseball player must catch the ball in the direction of the motion of the

ball. Moving his hand backwards when catching the ball prolongs the

time for the momentum to change so as to reduce the impulsive force.

Situation of Increasing Impulsive Force

Situations Explanation

A karate expert can break a thick wooden slab with his bare hand that

moves at a very fast speed. The short impact time results in a large

impulsive force on the wooden slab.

A massive hammer head moving at a fast speed is brought to rest upon

hitting the nail within a short time interval.

The large change in momentum within a short time interval produces a

large impulsive force which drives the nail into the wood.

A football must have enough air pressure in it so the contact time is

short. The impulsive force acted on the ball will be bigger and the

ball will move faster and further.

Pestle and mortar are made of stone. When a pestle is used to pound

chillies, the hard surfaces of both the pestle and mortar cause the pestle

to be stopped in a very short time. A large impulsive force is resulted

and thus causes these spices to be crushed easily.


2-35

Example 1

A 60 kg resident jumps from the first floor of a burning house.

His velocity just before landing on the ground is 6 ms-1.

(a) Calculate the impulse when his legs hit the ground.

(b) What is the impulsive force on the resident’s legs if he

bends upon landing and takes 0.5 s to stop?

(c) What is the impulsive force on the resident’s legs if

he does not bend and stops in 0.05 s?

(d) What is the advantage of bending his legs upon landing?

Answer: (a) Impulse = 60 kg x ( 6 ms-1 - 0 ) = 360 Ns

(b) Impulsive force = s

Ns

5.0

360

=7200 N

(c) He experienced a greater

Impulsive force of 7200 N and he might injured his legs

(d) Increase the reaction time so as to reduce impulsive force

Example 2

Rooney kicks a ball with a force of 1500 N. The time of contact of his boot with the ball is 0.01 s. What is the impulse

delivered to the ball? If the mass of the ball is 0.5 kg, what is the velocity of the ball?

(a) Impulse = 1500N x 0.01 s = 15 Ns

(b) velocity = kg

Ns

5.0

15 = 30 ms-1

S A F E T Y V E H I C L E

Component Function

Crash resistant door

pillars

Anti-lock brake system

(ABS)

Traction control Front bumper

Windscreen

Air bags

Head rest

Crumple zones

2.7

Safety features in vehicles


2-36

Headrest To reduce the inertia effect of the driver’s head.

Air bag Absorbing impact by increasing the amount of time the driver’s head to come to the

steering. So that the impulsive force can be reduce

Windscreen To protect the driver (shattered proof)

Crumple zone Can be compressed during accident. So it can increase the amount of time the car takes

to come to a complete stop. So it can reduce the impulsive force.

Front bumper Absorb the shock from the accident. Made from steel, aluminium, plastic or rubber.

ABS Enables drivers to quickly stop the car without causing the brakes to lock.

Side impact bar Prevents the collapse of the front and back of the car into the passenger compartment.

Also gives good protection from a side impact

Seat belt To reduce the effect of inertia by avoiding the driver from thrown forward.


2-27

G R A V I T Y

Gravitational Force

Objects fall because they are pulled towards the Earth by the force of gravity.

This force is known as the pull of gravity or the earth’s gravitational force.

The earth’s gravitational force tends to pull everything towards its centre.

Free fall An object is falling freely when it is falling under the force of gravity only.

A piece of paper does not fall freely because its fall is affected by air

resistance.

An object falls freely only in vacuum. The absence of air means there is no air resistance to oppose the motion of the object.

In vacuum, both light and heavy objects fall freely.

They fall with the same acceleration i.e. The acceleration due to gravity, g.

Acceleration due to gravity, g

Objects dropped under the influence of the pull of gravity with constant acceleration.

This acceleration is known as the gravitational acceleration,

g.

The standard value of the gravitational acceleration, g is 9.81 m s-2

.

The value of g is often taken to be 10 m s-2

for simplicity.

The magnitude of the acceleration due to gravity depends on the strength of the gravitational field.

Gravitational field The gravitational field is the region around the earth in which an object experiences a force towards the centre of the earth. This force is the gravitational attraction between the object and the earth.

The gravitational field strength is defined as the gravitational force which acts on

a mass of 1 kilogram.

g = m

F Its unit is N kg

-1.

2.8


2-28

Gravitational field strength, g = 10 N kg-1

Acceleration due to gravity, g = 10 m s-2

The approximate value of g can therefore be written either as 10 m s-2

or as

10 N kg-1

.

Weight The gravitational force acting on the object.

Weight = mass x gravitational acceleration

W = mg SI unit : Newton, N and it is a vector quantity

Comparison

between weight &

mass

Mass Weight

The mass of an object is the amount

of matter in the object

The weight of an object is the force of gravity

acting on the object.

Constant everywhere Varies with the magnitude of gravitational

field strength, g of the location

A scalar quantity A vector quantity

A base quantity A derived quantity

SI unit: kg SI unit : Newton, N

The difference

between a

fall in air and

a free fall in a vacuum of a coin and a feather.

Both the coin and the feather are released simultaneously from the same height.

At vacuum state: There is no air

resistance.

The coin and the feather will fall

freely.

Only gravitational force

acted on the objects. Both will fall at

the same time.

At normal state: Both coin and feather will

fall because of gravitational force.

Air resistance effected by the surface area of a

fallen object.

The feather that has large area will have more

air resistance.

The coin will fall at first.


2-29

(a) The two spheres are falling

with an acceleration.

The distance between two successive images of the sphere increases showing that the two spheres are falling

with increasing velocity; falling with an acceleration.

The two spheres are falling down

with the same acceleration

The two spheres are at the same level at all times. Thus, a heavy object and a light object fall with the same gravitational acceleration

Gravitational acceleration is independent of mass

Two steel spheres are falling under gravity. The two spheres are dropped at the same time from the same height.

Motion graph for free fall object

Free fall object Object thrown upward Object thrown upward and fall

Example 1

A coconut takes 2.0 s to fall to the ground. What is

(a) its speed when it strikes the ground (b) ) the height of the coconut tree

(a) t = 2 s u = 0 g = 10 v = ?

v = u + gt = 0 + 10 x 2 = 20 ms-1

(b) s = ut + ½ at2 = 0 + ½ (10) 22 = 20 m


2-30

F O R C E S I N E Q U I L I B R I U M

Forces in Equilibrium When an object is in equilibrium, the resultant force acting on it is zero.

The object will either be

1. at rest

2. move with constant velocity.

Newton’s 3rd

Law

Action is equal to reaction

Examples( Label the forces acted on the objects)

Paste more picture

Paste more picture

Resultant Force A single force that represents the combined effect of two of more forces in

magnitude and direction.

Addition of Forces

Resultant force, F = F1 + F2

2.9


2-31

Resultant force, F = F1 + - F2


2-32

Two forces acting at a point at an angle [Parallelogram method]

STEP 1 : Using ruler and protractor, draw the

two forces F1 and F2 from a point.

STEP 3

Draw the diagonal of the parallelogram. The

diagonal represent the resultant force, F in

magnitude and direction.

scale: 1 cm = ……

STEP 2

Complete the parallelogram

Resolution of Forces

A force F can be resolved into components which are perpendicular to each other:

(a) horizontal component , FX

(b) vertical component, FY

Fx = F cos θ

Fy = F sin θ

Inclined Plane

Component of weight parallel to the plane = mg sin θ

Component of weight normal to the plane = mg cos θ


2-33


2-34

Find the resultant force

(d) (e)

17 N

5 N

FR

7N


2-35

Lift

Stationary Lift

Lift accelerate upward

Lift accelerate downward

Resultant Force = Resultant Force = Resultant Force =

The reading of weighing

scale =


scale =


scale =

Pulley

1. Find the resultant force, F 40 -30 = 10 N

30-2 = 28 N

2. Find the moving mass, m 4 + 3 = 7 kg 3+ 4 = 4 kg

3. Find the acceleration, a 40 -30 = (3+4)a

10 = 7 a

a =10/ 7 ms-2

30 -2 = (4+3 )a

28 = 7a

a = 4 ms-2

4. Find string tension, T T- 3 (10) = 3 a

T = 30 + 3 (10/7)

=240 /7 N

30 – T = 3 (a)

T =30- 12

= 18 N


2-36

W O R K , E N E R G Y , P O W E R & E F F I C I E N C Y

Work

Work done is the product of an applied force and the

displacement of an object in the direction of the applied

force

W = Fs W = work, F = force s = displacement

The SI unit of work is the joule, J

1 joule of work is done when a force of 1 N moves an object

1 m in the direction of the force

The displacement, s of the object is in the direction of the force, F

The displacement , s of the object is

not in the direction of the force,

F W = Fs

s F

W = F s

W = (F cos θ) s

Example 1

A boy pushing his bicycle with a force of 25 N through a distance of 3 m.

Calculate the work done by the

boy. 75 Nm

Example 2

A girl is lifting up a 3 kg flower pot steadily to a height of 0.4 m.

What is the work done by the girl? 12 Nm

Example 3

A man is pulling a crate of fish along the floor with a force of

40 N through a distance of 6 m.

What is the work done in pulling the crate?

40 N cos 50º x 6 Nm

Concept D

e

f

i

n

i

Formula & Unit

2.1

0


2-37

Power The rate at which work is

done, or the amount of

work done per second.

P = W

t

p = power, W = work / energy t = time

Energy Energy is the capacity to do work.

An object that can do work has energy

Work is done because a force is applied and the objects

move. This is accompanied by the transfer of energy

from one object to another object.

Therefore, when work is done, energy is transferred

from one object to another.

The work done is equal to the amount of energy

transferred.

Potential Energy

Gravitational potential energy is the energy of an object due to its higher position in the gravitational field.

m = mass

h = height

g = gravitational acceleration

E = mgh

Kinetic Energy

Kinetic energy is the energy

of an object due to its

motion.

m = mass

v = velocity

E = ½ mv2

No work is done when:

The object is stationary.

A student carrying his bag while waiting at the

bus stop

The direction of motion of the object is perpendicular to that of the applied force.

A waiter is carrying a tray

of food and walking

No force is applied on the object in the direction of displacement (the object moves because of its

own inertia)

A satellite orbiting in space.

There is no friction in space. No force

is acting in the direction of

movement of the satellite.


2-38

Principle of Conservation of Energy

Energy can be changed from one form to another, but it

cannot be created or destroyed.

The energy can be transformed from one form to another, total

energy in a system is constant.

Total energy before = total energy after

Example 4

A worker is pulling a wooden block of weight W, with a

force of P along a frictionless plank at height of h. The

distance travelled by the block is x. Calculate the work

done by the worker to pull the block.

[Px = Wh]

Example 5

A student of mass m is climbing up a flight of stairs which has the height of h. He takes t seconds.

What is the power of the student?

[ t

mgh

Example 6

A stone is thrown upward with initial

velocity of 20 ms-1

. What is the maximum height which can be

reached by the stone? [ 10m ]

Example 7

A ball is released from point A of height 0.8 m so that it can roll along a curve frictionless track. What is the velocity of the ball when it reaches point B? [4 ms-1]


2-39

Example 8

A trolley is released from rest at point X along a frictionless track. What is the velocity of

the trolley

at point Y?

[ v2 = 30( ms-1)2] [ v = 5.48 ms-1]

Example 9

A ball moves upwards along a frictionless track of height 1.5 m

with a velocity of 6 ms-1

. What is its velocity at point B?

[v2 = 30( ms-1)2

v = 5.48 ms-1]

Example 10

A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the slope, he does

work to overcome friction of 140 J. What is his velocity at the end of the slope? [6 ms-1]

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