The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

## Form 4 Chapter 1 to Chapter 5

Related Publications

Discover the best professional documents and content resources in AnyFlip Document Base.

# Physics Module

### Form 4 Chapter 1 to Chapter 5

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics Date:
Guru Cemerlang Physics MSAB
CHAPTER 1: INTRODUCTION TO PHYSICS

1.1 UNDERSTANDING PHYSICS

1. The aim of physics is to explain the natural phenomena and the properties of matter of the

universe by using the simplest explanations.

2. Fields of study in physics are mechanics (force and motion), heat, light, waves, electricity,

electromagnetism, electronic and nuclear physics.

1.2 PHYSICAL QUANTITIES
1. A physical characteristic that is measurable.
2. Categorised into base quantities and derived quantities.
3. Also can classification into scalar quantities and vector quantities.

1.2.1 Base Quantities

1. Base quantities are quantities that cannot be defined in terms of other physical quantities.

Base quantity Symbol S.I. Unit Symbol for S.I. Unit

Length l Metre m

Mass m Kilogram kg

Time t Second s

Electric Current I Ampere A

Temperature T Kelvin K

1.2.2 Derived Quantities

1. Derived quantities are quantities that combining base quantities through multiplication,
division or both.

Derived quantity Formula Derived unit Name of derived unit

area area = length  width m  m = m2 –
volume volume = length  width  height m  m  m = m3 –

density density  mass kg  kg m3 –
volume m3

velocity velocity  displaceme nt m  m s1 –
momentum time s –

Acceleration momentum = mass  velocity kg m s-1 –
acceleration  change in velocity
m s1  m s-1 s1
time s
 m s2

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 1

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Formula Derived unit Date:
Derived quantity
Name of derived unit

Force force = mass x acceleration kg m s-2 Newton (N)
pressure N  kg m s2 Pascal (Pa)
pressure force m2 m2
weight area Newton (N)
work = kg m-1 s-2 Joule (J)
power Weight = mass x gravitational kg m s -2 Watt (W)
acceleration
N m = kg m2 s-2
work = force x displacement
pow er  w ork J s-1 = kg m2 s-3
time

kinetic energy K.E  1  mass  velocity2 kg m2 s-2 Joule (J)
potential energy 2 kg m2 s-2 Joule (J)

P.E = mass x gravitational
acceleration x height

Note that the physical quantities such as width, thickness, height, distance, displacement,
perimeter, radius and diameter are equivalent to length.

1.2.3 Standard Form (Scientific Notation)

1. Standard form is used to simplify the expression of very large or very small numbers.

2. Standard form  A  10n 1 < A < 10 n = integer

i. Radius of the earth = 6 370 000 m = 6.73  106 m

ii. Size of a particle = 0.000 03 m = 3.0  10-5 m

 

iii. Diameter of an atom = 0.000 000 072 m = 7.2  10-8 m

1.2.4 Prefixes
1. A Prefix is a letter written in front of the word or the symbol for the unit without space.
2. It used to represent a large physical quantity or extremely small quantity in S.I units.
3. A unit may take only one prefix. E.g. Mm  kkm, 103 kg  kkg.

Prefix Symbol Power Factor Prefix Symbol Power Factor
giga- G 109 centi- c 10-2
mega- M 106 milli- m 10-3
kilo- k 103 micro-  10-6
deci- d 10-1 nano- n 10-9

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 2

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:

1.2.5 Method of Expressing Prefix and Unit

1. Replace and Invite “- ve” (R & In)

2. Prefix – Prefix (P – P)

3. Examples:

i. cm  m ii. cm2  m2
786 cm = 786  10-2 m = 7.86 m 330  102 cm2 = 330  102  10-2(2) = 3.3 m2

iii. m  Mm iv. cm2  km2
2 660 m = 2 660  10(-) 6 = 2.66  10-3 Mm 35 cm2 = 35  10-2(2)  10-3(2) = 3.5  10-9 km2

v. kg  g vi. m3  mm3
3.7 kg = 3.7  103 X 10(-) -6 = 3.7  109 g 289 m3 = 289  10(-) -3(3) = 2.89  1011 mm3

km h-1  cm s-1

36 km h-1 = 36 km  3.6  1013()(2) cm  3.6  106 cm  103 cm s1
h 3600 s 3.6  103 s

2nd edition © 2011| chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 3

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:

1.3 SCALAR AND VECTOR QUANTITIES

1. Scalar quantities are quantities that have only magnitude or size without any direction.

 Examples: mass, length, distance, speed, volume

2. Vector quantities are quantities which has magnitude or size and direction

 Examples: velocity, force, displacement, acceleration

Description of events Magnitude Direction

The temperature in the room is 25 0C 

The location of Ayer Hitam is 60 km to the north-west of Johor Bahru  

The power of the electric bulb is 80 W 

A car is travelling at 80 km h-1 from Johor Bahru to Kuala Lumpur 

“- ve” value of vector quantities are represent the opposite of
positive direction. E.g.: If velocity of car 10 m s-1 is to the right,
than –10 m s -1 means that car move to left.

1.4 MEASUREMENTS

1.4.1 Using Appropriate Instruments to Measure

1. Choosing an appropriate instrument to measure a physical quantity depends on the size of the

measurement and the accuracy needed.

2. The magnitude of the quantity should not exceed the maximum capacity of the instrument

3. The instrument must be sensitive enough to detect and give a meaningful measurement of the

quantity.

Measuring instrument Range of measurement Smallest scale division
(Sensitivity and Accuracy)

Measuring tape Up to a few meters 1 cm or 0.1 cm

Meter rule 1 m 0.1 cm

Vernier calliper 10 cm ~ 12 cm 0.01 cm

Micrometer screw gauge less than 25 mm 0.01 mm

Thermometer -10 0C to 110 0C 1 0C
0 0C to 360 0C 2 0C

Stopwatch (analogue) - 0.1 s or 0.2 s

Stopwatch (digital) - 0.01 s

Ammeter // *(Voltmeter) - 0.1 A or 0.2 A
0.1 mA or 0.2 mA

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 4

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB

Beaker

Inside jaws
(For measuring internal diameter)

Main scale Tai

Vernier scale For

(Can be slide) measuring

Outside jaws depth

(For measuring external diameter or length of an obj

 Vernier callipers is used to measure: 0 cm
i. Small object
ii. Depth of a hole 0
iii. External diameter of a cylinder or pipe No
iv. Internal diameter of a pipe or tube.

 Gives readings to an accuracy of 0.01 cm.

= (1) + (2)
= 3.2 cm + 0.04 cm
= 3.24 cm

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Date:

(2) The fourth mark on the vernier scale
in line with main scale  0.04 cm

(1) The ‘0’ mark on the vernier scale acts as
il pointer for main scale reading  3.2 cm

ject)

1 0 cm 1

5 10 05 10
ot zero error
Positive zero error = 0.07 cm

Method 1: – 0.1 + 0.07 = – 0.03 cm
Method 2: 0.6 – (0.09  7) = – 0.03 cm

Method 3: Reverse the reading on the
vernier scale. The ‘7’ mark on the venier
zero error is – 0.03 cm.

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics 
Guru Cemerlang Physics MSAB
Ratchet knob
Anvil Spindle Lock

Sleeve Thimble (with thimble scale) 
(with 
main pressu
scale) 

= 7.0 mm + 0.35 mm
= 7.35 mm

No zero error Positive zero error = +
 Ammeter is use to measure electric current. ● Voltmeter is use to

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Date:

Micrometer screw gauge is used to measure:
i. Objects that are small in size (less than 20 mm)
ii. Diameter of a wire
iii. Diameter of small spheres such as ball bearings
Gives readings to an accuracy of 0.01 mm.
The ratchet knob is used to prevent the user from exerting too much
ure on the object.
i. The thimble is turned until the object is gripped very gently between
vil and the spindle.
ii. The ratchet knob is then turned until a ‘click’ sound is heard.

+ 0.04 mm Negative zero error = - 0.03 mm

o measure potential difference.
● Mirror is used to avoid parallax errors.

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:

1.4.2 Accuracy, Consistency and Sensitivity in Measurements

1. Accuracy: The ability of an instrument to measure nearest to the actual value.
2. Consistency: The ability of an instrument to measure consistently with little or no relative

3. Sensitivity: The ability of an instrument to detect a small change in the quantity measured.

1.4.3 Errors in Measurements

1. No measurement is exact. All measurements will have some degree of error.

2. The error is the difference between the actual value of a quantity and the value obtained in
measurement.

3. There are two main types of errors:

i. Systematic errors
 Systematic errors are errors in the measurement of a physical quantity due to the
instruments, the effects of surrounding conditions and physical constraints of the
observer.
 The main characteristic of systematic error is that its magnitude is almost constant.
The value of the measurement is always greater, or always less than the actual
value.
 Sources of systematic errors

 Zero or end errors
 Personal error of the observer  Reaction time.
 Errors due to instrument  A stopwatch which is faster than normal.
 Errors due to wrong assumption  Assumed that the value of the acceleration

due to gravity, g is 10 m s-2, but the actual value is 9.81 m s-2.
 Can be overcome by improving the procedure of taking the measurements (note the

zero error of instruments), using a different instrument or getting somebody else to
make the measurements.

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 7

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:
ii. Random errors

 The main source of random error is the observer.

 The characteristics of random errors are:

 It can be positive or negative. The obtained reading may be greater or less than
the actual value.

 Examples of random errors:

 Parallax errors  incorrect positioning of the eyes when taking the readings of
measurements.

 Changes in the temperature during an experiment.

 Mistake in counting.
 Different pressures are applied when closing the gap of the micrometer screw

gauge.
 The best way of minimising random errors is by repeating the measurements in order

1.5 SCIENCETIFIC INVESTIGATION 8
The following processes are involved in scientific investigations:
1. Observe situation and identify a question.
2. Identify all variables involved in the situation

 manipulated variable, responding variable and fixed variable.
3. Form a hypothesis.
4. Plan and carry out experiment.
5. Analyse data and make a conclusion.

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:

1.5.1 Plan and report an experiment

Steps Example

1 Aim To investigate the relationship between A and B when C is constant.

2 Inference B depends on A.

3 Hypothesis When A increases, B increases. Or When A increases, B decreases.

4 Variables Manipulated variable: A

Responding variable: B

Fixed variable: C

5 List of apparatus All apparatus in the list must be shown in the arrangement of the
and materials apparatus.

6 Arrangement of the The diagram shown in the figure must be labelled.
apparatus

7 Procedures The method of controlling the manipulated variable and the method of
measuring the responding variable.

8 Tabulate the data Manipulated Responding variable Derived from
variable B, (unit) the responding
variable, (unit)
A, (unit) B1 B2 Mean, B

The values are  All numerical value must be stated to the same
fixed at uniform number of decimal places in each column.

intervals  Unit of measurement should not repeat with
numerical value.

9 Analyse the data  Plotting a graph.
(Discussion)
 Interpretation of the graph.
10 Conclusion
 Discussion is depends on experiment.
11 Precautions
B is directly or inversely proportional to A. //

B is increases or decreases linearly to A.

Eyes must directly perpendicular to the reading to avoid parallax error.

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 9

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics
Guru Cemerlang Physics MSAB
Date:

Situation: A few children are playing on a different length of swing in a playground. It is found that

the time of oscillation for each swing is different.

1 Aim To investigate the relationship between length and period of a simple
2 Inference pendulum.
The period of the oscillation depends on the length of the pendulum.

3 Hypothesis When the length of the pendulum increases, the period of the
oscillation increases.

4 Variables Manipulated variable: The length of the pendulum.

Responding variable: Period

Fixed variable: The mass of the pendulum and the displacement.

5 List of apparatus Retort stand with clamp, two small pieces of plywood, protractor, 100

and materials cm thread, plasticine ball, metre rule and stopwatch.

6 Arrangement of
the apparatus

7 Procedures 1. Set up the apparatus as shown in the figure above.

8 Tabulate the 2. Measure the length of the pendulum, l = 60.0 cm by using a meter rule.
data
3. Give the pendulum bob a small displacement 300.

4. Time of 10 oscillations is measured by using a stopwatch and recorded.

5. Repeat the timing for another 10 oscillations. Calculate the average

time. Period = t10 / 10 oscillations
6. Repeat steps 2 to 5 using l = 50.0 cm, 40.0 cm, 30.0 cm and 20.0 cm

Length, l Time for 10 oscillation, t (s) Period, T (s)

(cm) t1 t2 Mean, t10 (T = t10 / 10)

60.0 15.8 15.7 15.8 1.58

50.0 15.0 15.0 15.0 1.50

40.0 13.1 13.1 13.1 1.31

30.0 11.9 11.9 11.9 1.19

20.0 9.9 9.9 9.9 0.99

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 10

Mr Ng Han Guan Form 4 Chapter 1: Introduction To Physics Date:
Guru Cemerlang Physics MSAB
T / s Graf of period, T vs
9 Analyse the data pendulum’s length, l

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0 10 20 30 40 50 60 l / cm

10 Conclusion The length of the pendulum is directly proportional to the period of
11 Precautions oscillation.

Hypothesis accepted.

Oscillation time is measured when the pendulum attained a steady state.

2nd edition © 2011 | chp 1 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan 11

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion Date:
Guru Cemerlang Physics MSAB
CHAPTER 2: Force and Motion

2.1 ANALYSING LINEAR MOTION

2.1.1 Distance, Displacement, Speed, Velocity, Acceleration and Deceleration

1. Distance is the total length of the path travelled by an object.

2. Displacement is the length between two points measured along a straight line joining them in a

specified direction.

3. Speed is the rate of change of distance.

4. Velocity is the rate of change of displacement.

5. Acceleration is the rate of increasing velocity.

6. Deceleration is the rate of decreasing velocity.

Quantity Symbol SI Unit
Distance / Displacement d/s meter (m)
v meter per second (m s-1)
Speed / Velocity a meter per second per second (m s-2)
Acceleration

Scalar Vector Quantity Average speed, v Total distance travelled =
Quantity Average velocity, v Time taken
Distance Displacement Acceleration, a
Velocity Displacement travelled =
Speed Time taken
Acceleration
Deceleration Final Velocity - Initial Velocity = −
Time of Change of Velocity

7. Deceleration is the moment when velocity is decreasing and acceleration is negative value in

calculation (a = negative magnitude).

8. Example: An aeroplane flies from A to B, which is located 300 km east of A. Upon reaching B, the

aeroplane then flies to C, which is located 400 km north. The total time of flight is 4 hours.

Calculate:

i. The average speed of the aeroplane

Average speed, v = Total distance travelled = 300+400 = 175 km h-1
Time taken 4

ii. The average velocity of the aeroplane

Average velocity, v = Displacement travelled = 500 = 125 km h−1
Time taken 4

9. Example: A lorry is moving at 30 m s-1, when suddenly the driver steps on the brakes and it stop

5 seconds later. Calculate the deceleration of lorry

Acceleration, a = Final Velocity - Initial Velocity = − = 0−30 = −6 m s-2
Time of Change of Velocity 5

Deceleration = 6 m s-2

1

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.1.2 Ticker Timer and Ticker Tape

1. A ticker timer is a device used to study linear motion (constant acceleration).

2. A ticker timer will lodge dots on the ticker tape at the rate of 50 dots per second (50 Hz).

3. The time interval between two dots (one tick) is:

 t  1s  0.02 s
50 dots

4. The distances between two dots can be different but the time interval between two dots is

always the same (0.02 s).

5. We can make measurement on the tape to find out the displacement, velocity and acceleration.
6. The type of motion can be study through the ticker tape are:

i. Constant velocity

ii. Increases velocity (constant acceleration)

iii. Decreases velocity (constant deceleration)

iv. Acceleration then uniform velocity

v. Ticker tape chart

Constant Velocity Constant Acceleration Constant Deceleration

Acceleration then uniform velocity Uniform velocity then deceleration

2

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.1.2.1 The Working Steps to Study Linear Motion of Ticker Tape and Ticker Tape Chart

1. Ticker tape

2. Ticker tape chart

3

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
2.2 ANALYSING MOTION GRAPHS
2.2.1 The Displacement – Time Graph (s – t Graph) (Velocity)
0
1. The type of s – t Graph:

Graph Displacement

Constant

Increases Constant +ve

OA Increases Constant +ve

AB Decreases Constant –ve
OA Increases (reversed direction)
AB Constant
BC Decreases Constant +ve
0
Non-uniform
increases Constant –ve
(reversed direction)

Increases
(accelerating)

Non-uniform Decreases
increases (decelerating)

2. Example: Velocity v – t Graph
s – t Graph

OP v  s  30
t2

 15 m s-1

PQ v s  0
t3

 0 m s-1

QR v  s   45
t3

 15 m s-1

Average speed after 8 seconds, v  Total distance travelled  30  30  15  9.375m s-1
Time taken 8

Average velocity after 8 seconds, v  Displaceme nt travelled  15  1.875ms1
Time taken 8

4

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion

Guru Cemerlang Physics MSAB Date:

2.2.2 The velocity – Time Graph (v – t Graph)

1. The type of v – t Graph:

Graph Velocity Graph gradient Area under the
(Acceleration) graph

(Displacement)

Constant a=0 +ve
s  v  t1

+ve

Increases Constant +ve s  1 v  t1
2

+ve

OA Increases Constant +ve s1  1 v  t1
2
Constant –ve
(deceleration) +ve

AB Decreases Constant +ve s2  1 v  (t2  t1)
2
a=0
+ve
Constant –ve
OA Increases (deceleration) s1  1 v  t1
Constant –ve 2
(accelerating in
AB Constant reversed direction) +ve

Constant +ve s2  v  (t2  t1)
(deceleration in +ve
reversed direction)
BC Decreases s3  1 v  (t3  t2)
2

Increases –ve
(reversed
OA direction) s1  1 v  t1
2

Decreases –ve
(reversed
AB direction) s2  1 v  (t2  t1)
2

Non-uniform Increases -
increases (accelerating)

Non-uniform Decreases -
increases (accelerating)

5

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
2. Example:
v – t Graph Displacement Acceleration

s – t Graph (Area under the graph) (Graph gradient)

OP sOP  1  30  2  30 m aOP  30  0  15 m s-2
PQ 2 2
QR
sPQ  30  (5  2)  90 m aPQ  30  30  0m s-2
RS 5  2

OS sQR  1  30  (7  5)  30 m aQR  0  30  15 m s-2
OS 2 75

sRS  1  15  (8  7) aRS  15  0  15 m s-2
2 87

 7.5 m Velocity increases
(–ve: reversed direction)
(reversed direction)

Total displacement Average velocity
= 30 + 90 + 30 - 7.5 = 142.5 m
= 142.5  17.81m s-1
8

Total distance Average speed
= 30 + 90 + 30 +7.5 = 157.5 m
= 157.5  19.69 m s-1
8

a – t Graph

Average speed after 8 seconds, v  Total distance travelled  30  30  15  9.375m s-1
Time taken 8

Average velocity after 8 seconds, v  Displaceme nt travelled  15  1.875ms1
Time taken 8

2.2.3 The Equations of Motion

For an object in linear motion with uniform acceleration, problems involving:

i.) displacement, s ii.) velocity, v iii.) acceleration, a iv.) time of motion, t

Can be solved by using the equations of motion:

i.) v  u  at ii.) s  1 (u  v)t iii.) s  ut  1 at 2 iv.) v2  u2  2as
2 2
6

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
1. Worked example:

A car accelerates from rest at 2.5 m s-2 for 10 s. Find

i.) the velocity reached and

Given: u = 0 m s-1, a = 2.5 m s-2, t = 10 s, v =?

Applying: v  u  at  v  0  2.5(10)  25 m s-1

ii.) the distance travelled after 10 s.
Given: u = 0 m s-1, a = 2.5 m s-2, t = 10 s, s =?

Applying: s  ut  1 at 2  s  0(10)  1 (2.5)(10)2  125 m
22

iii.) the driver then steps on the brake pedal with a constant force. The car stops after
travelling a distance of 50 m. Calculate the deceleration of the car after the brakes are
applied.
Given: u = 25 m s-1, v = 0 m s-1, s = 50 m, a =?

02  252  2a(50)
Applying: v2  u2  2as  a   252   625  6.25 m s-2

2(50) 100

* The negative sign shows deceleration. Deceleration = 6.25 m s-2

2. A driver accelerates his car from 20 m s-1 with an acceleration of 2 m s-2. What is his velocity
after 8 seconds?
Given: u = 20 m s-1, a = 2 m s-2, t = 8 s, v =?

Applying: v  u  at  v  20  2(8)  36 m s-1

3. A cyclist starts from rest and reaches a velocity of 20 m s-1 in 8 s. Calculate his acceleration.
Given: u = 0 m s-1, v = 20 m s-1, t = 8 s, a =?
20  0  a(8)

Applying: v  u  at  a  20  2.5 m s-2
8

4. A car was moving with constant velocity of 30 m s-1. The driver saw an obstacle in front and
he immediately stepped on the brake pedal and managed to stop the car in 6 s. The distance
of the obstacle from the car when the driver spotted it was 100 m. How far the obstacle from
the car is after the car has stopped?
Given: u = 30 m s-1, v = 0 m s-1, t = 6 s, s =?
Applying: s  1 (u  v)t  s  1 (30  0)6  90 m
22
The obstacle from the car after the car has stopped is: 100 m – 90 m = 10 m

7

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.3 UNDERSTANDING INERTIA

2.3.1 Concept of Inertia

1. An object in a state of motion tends to stay in motion in a straight line with same speed.

2. Only an external force can cause a change to the state of motion of an object.

3. However, the object will resist any attempt to change its original state of motion.

4. The tendency of the object to remain at rest or, if moving, to continue its uniform motion in a

straight line is called inertia.

5. The concept of inertia was explained by Sir Isaac Newton in the First Law of Motion:
 Every object continues in its state of rest or of uniform motion unless it is acted upon by

an external force.

6. Relationship between inertia and mass

 The larger the mass, the large its inertia.

2.3.2 Situations Involving Inertia EXPLANATION
SITUATION The inertia of the coin maintains its state at
When the cardboard is pulled rest. The coin fall into the glass due to
away quickly, the coin drops gravity.
straight into the glass.

Sauce in the bottle can be The sauce inside the bottle moves together

easily poured out if the bottle with the bottle. When the bottle stops

is moved down fast with a suddenly, the sauce continues in its state

sudden stop. of motion due to the effect of its inertia.

The cow has a large inertia making it
difficult to change direction.

A boy runs away from a cow in a zig zag motion.

The head of hammer is secured tightly to its This causes the hammer head to continue

handle by knocking one end of the handle, held on its downward motion when the handle

vertically, on a hard surface. has been stopped, so that the top end of the

handle is slotted deeper into the hammer

8

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.3.3 Ways to Reduce the Negative Effects of Inertia

METHODS EXPLANATION

Safety belt secures the driver to their seats. When the car stops suddenly, the

Safety belt seat belt provides the external force that prevents the driver from being thrown

forward.

Headrest Headrest to prevent injuries to the neck during rear-ends collisions. The inertia
of the head tends to keep in its state of rest when the body is moved suddenly.

An air bag 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.

Furniture carried by a When the lorry starts to move suddenly, the furniture is more difficult to fall off

lorry normally is tied due to their inertia because their combined mass has increased.

up together by string.

2.4 Analysing Momentum

1. When an object is moving with a speed in a straight line, we say that it has linear momentum.

2. Linear momentum is defined as the product of mass and velocity.

 Momentum = Mass x Velocity p=mv

 SI unit is kg m s-1. It can also be written as N s (newton second).

3. Momentum is a vector quantity, so its direction must be taken into consideration (direction of

motion).

2.4.1 Principle of Conservation of Momentum
1. In the absence of an external force, the total momentum of a system remains unchanged.
2. When objects collide in the absence of external force,

 Total momentum before collision = Total momentum after collision
3. When objects collide and bounce perfectly, the collision is said to be an elastic collision.

Before collision After Collision

 m1 u1 + m2 u2 = m1 v1 + m2 v2

4. Inelastic collision is said to occur whenever colliding objects get attached to one another or

coupled together after collision.

 m1 u1 + m2 u2 = (m1 + m2) v 9

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

5. Momentum is also conserved in an explosion. The momentum before explode is zero.

After explode, the total momentum is still zero because the forward momentum is equal in

magnitude to the backward momentum.

 0 = m1 v1 + m2 v2  m1 v1 = - m2 v2

“-ve” sign means opposite direction

2.4.2 Applications of Conservation of Momentum

SITUATIONS EXPLANATION

 When a rifle is fired, the bullet of small mass,

m2, moves with a high velocity, v2. 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.

 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.

Aircraft with jet engines 10

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB Date:

 Fuel burns explosively in the combustion

chamber.
 Hot gases expelled at high speed have a

large momentum downwards.
 The gain of momentum of the rocket is equal

and opposite to the momentum of the ejected

fuel.

2.5 Understanding the Effects of a Force
1. Forces can make things move or slow things down. Forces also hold things still and keep things

balanced.
2. Balanced forces on an object:

 When the forces acting on an object are balanced, they cancel each other out. The net force
acting on the object is zero.

 Balanced forces do not changed the state of rest or state of motion (constant velocity) of
an object.

 When an object is at rest or moves with constant velocity, it means that there are no forces

acting on the body or that all the forces acting on the body are balanced (Fnet = 0 N).

3. Unbalanced forces on an object:

 When the forces acting on an object are not balanced, there must be net force acting on it.

 The net force is known as the unbalanced force or the resultant force.

 The unbalanced force or resultant force can cause a body:

- to change its state of rest - to change its state of motion

11

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.5.1 Relationship between Force, Mass and Acceleration (F =ma)

1. Newton’s Second Law of Motion:

 The acceleration produced by a net 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.

Acceleration α Net force

a∝F

Acceleratio ∝ 1  a α ∝  F∝mxa

a ∝ 1

F = kma the constant k = 1

 Thus, the relationship between force, F, mass, m, and acceleration, a is written as

F = ma

2. The SI unit of force is newton, N.

3. We define one newton (1 N) as the force needed to a 1 kg object an acceleration of 1 m s-2.

4. A force, F acting on a mass, m. the mass moves with a constant acceleration, a.

 If the force acting on the same mass is doubled, its acceleration is also doubled.
 If the same force, F is acting on a mass of 2m, the acceleration, a is reduced by half.

2.5.2 Worked Example Balanced or Unbalanced Acceleration and
1. forces / Net Force
Unbalanced forces, direction
Problems Fnet = 3 – 2 = 1 N

Unbalanced forces, a = = 0.33 m s-2
Fnet = 15 – 3 = 12 N

Balanced forces to the left 
Fnet = 0 N
a = = 6 m s-2

to the right 

a = 0 m s-2

2. An object of mass 2 kg is pulled on the floor by a force of 5 N and having a constant velocity.
(a) What is the frictional force between the object and the floor?

(b) Calculate the acceleration of the object if the object is pulled by a 17 N force?

12

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

3. A 1 000 kg car is travelling at 72 km h-1 when the brakes are applied. It comes to a stop in a

distance of 40 m. What is the average braking force of the car?

4. A trolley of mass 20 kg is pulled along the ground by a horizontal force of 30 N. the opposing
frictional force is 20 N. Calculate the acceleration of the trolley.

2.6 Analysing Impulse and Impulsive Force

1. The large force that acts over a very short period of time during collisions and explosions is

known as impulsive force.

2. The magnitude of the impulsive force is usually not constant during that very short time.

Therefore, it is impossible to measure directly the magnitude of the impulsive force during this

time interval.

3. The product of the impulsive force, F and the time of impact, t is defined as the impulse of the

force.

4. Therefore, impulse, Ft is equal to the change of momentum brought about by the impulsive

force. (unit for impulse is N s or kg m s-1)
 Ft = Change of momentum

5. Impulsive force, F can be written as

 F =Change of momentum = −

6. Therefore, impulsive force can be defined as the rate of change of momentum in a collision or

explosion.

7. The magnitude of the impulsive force that acts to change the momentum is inversely

proportional to the time of impact.

8. A shorter time of impact will produce a larger impulsive force while a longer time of impact

will result in a smaller impulsive force.

13

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.6.1 Situations Involving Impulsive Forces

SITUATIONS EXPLANATION SITUATIONS EXPLANATION

Thick mattresses with a Hammer and nail are

soft surface are used in made of metal. When a

events such as the high hammer hits at a nail, the

jump so that the time collision time is small

interval of impact on and hence the impulsive

landing is increased, thus force is great. So the nail

reducing the impulsive penetrates a wooden

force. material easily.

In games like baseball, An exponent of karate has
cricket and softball, the the capability to split a
catcher always pulls his thick wooden slab by
hand backward when bringing his hand down
catching the ball to hard onto the surface of
lengthen the collision the wood. The momentary
time and hence reduces contact produces a great
the impact of the impulsive force. impulsive force which
splits the wooden slab.

A high jumper will bend his legs upon landing. Pestle and mortar are made of
This is to increase the time of impact in order to stone. When a pestle is used
reduce the impulsive force acting on his legs. to pound chillies the hard
This will reduce the chance of getting serious surfaces of both the pestle
injury. 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.

2.7 Being Aware of the Need for Safety Features in Vehicles

1. Crumple zones

 In a crash, the bonnet and boot of the car is designed to crumple, making the collision last a

slightly longer time. The chassis contains parts that have grooves or beads cast into them. In

a collision, these beads act as weak points in the members, causing them to crumple in a

concertina shape. The force exerted on the car (and on its passengers) is then smaller.

2. Strong steel struts or framework of the car

 The strong steel strut prevents the collapse of the front and back of the car into the passenger

compartment. Also gives good protection from a side-on collision.

14

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

 To reduce the risk of injury, the interior of the car must be made to absorb the impact. The car

passenger’s neck being damaged in near-end collision.

4. Collapsible steering wheels
 Steering wheel of a car is made of material soft enough to lengthen the collision time and to
cushion the momentum impact of the driver’s head during an accident.

5. Shatterproof windscreen glass.
 Usage of shatterproof windscreen to prevent the passengers from being injured by glass

pieces during accident because the glass pieces will not scattered easily.

6. Automatic air bag.
 An air bag in a car is designed to inflate only when the vehicle experiences a 20 km h-1 or
greater impact. The car’s computer controls make a decision in few milliseconds to detonate

the gas cylinders that inflate the air bag. As the driver lunges forward into the air bag, allowing

the driver to slow in a longer time. Injury is thus minimized. The air bag also spreads the

impact force over a larger area of the body.

7. Seatbelt
 An inertial reel seatbelt is taut but allows enough free movement during normal traveling.

When an accident occurs, as the car exceeds a certain deceleration value, a locking

mechanism will engage and stop the driver from crashing forward. The driver is not brought to

a stop immediately and this helps reduce the force acting on a driver.

8. Tyre design
 The tyre of a car should be broad and with friction grooves so as to control better the stability

and change of momentum of the car.

9. Safety helmets
 Safety helmets are compulsory for motor cyclists. There is a growing amount of evidence to

suggest that many injuries to pedal cyclists would also be much less serious if they wore
safety helmets too. The purpose of a safety helmet is to protect the wearer’s head from large

forces in an accident. It works on exactly the same principle as the seat belt and crumple

zone. Inside the helmet there is a layer of expanded from or other similar padding. In a crash,
if the motor cyclist’s head hits another vehicle or the road, the webbing and the padding inside

the helmet allow the head to move a short distance before distance. The time allowed for the
cyclist’s moving head to slow down and stop is longer. So the average force on the head is

smaller.

10. Accident avoidance systems. New vehicle safety technologies now concentrate on preventing

accidents rather than reducing their effects. These include:
 Antilock brake systems (ABS) – Usage of the ABS which will not immediately stop the car

once the brakes are applied. The car will be momentarily brought to rest so that the impulsive

15

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
force is smaller.

 Variable-ratio response steering systems

 Reverse collision warning systems

11. Protecting the pedestrian

 Vehicles are designed to protect pedestrians also.
 Vehicles are designed with a low, energy-absorbing bumper to reduce knee and hip damage

to pedestrians.

 Bumper bars are placed below waist level to reduce the like hood of the pedestrian being run

over.

 Bonnet of a car can reduce casualty in car accidents. When the car hits a pedestrian, the

bonnet of the car will automatically spring upwards. This will prolong the collision time and

hence reduces the impulsive force acting on the pedestrian.

12. Driver warning functions: Two different types of driver warning functions will be provided.

 Lateral Drift Warning Function. This Functionality will help drivers avoid inadvertent drift-off

that may result in striking another vehicle, roll-over or impact with a fixed object.
 Curve Speed Warning Function. This feature would warn drivers to slow down for an

upcoming curve so as not to lose control and depart the road.

2.8 Understanding Gravity
1. According to Sir Isaac Newton, all objects are pulled by the force of

gravity which causes them to fall to the surface of earth.
2. The pull of gravity causes objects to fall with acceleration. This

means that objects that fall are moving with increasing velocity.
3. The acceleration of objects falling freely is known as the acceleration due to gravity.
4. The magnitude of the acceleration due to gravity depends on the strength of the

gravitational field.

2.8.1 Gravitational Field

1. Gravitational field is a region in which an on object experiences a force due to gravity.
2. The earth’s gravitational field extends out into space, in all directions. This field gets weaker the

further you go out from the centre of the Earth.

3. At a pint in the gravitational field, the gravitational field strength, g is defined as the

gravitational force acting on a 1 kg mass placed at that point (force per unit mass), =

4. The gravitational field strength at the surface of the earth is 9.8 N kg-1. This means that an object

of mass 1 kg will experience a gravitational force of 9.8 N.
5. The gravitational field strength, g of the Moon is only 1.6 N kg-1 because the mass of the Moon is

about one-sixth that of the Earth’s.

16

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.8.2 The Value of Acceleration Due to Gravity, g

1. From the definition of the gravitational field strength, g = …..…… (1)

 According to Newton’s second law of motion, F = ma...……………. (2)

 Substituting for F, a = g = 9.8 m s-2

2. So, the acceleration due to gravity, g, is equal 9.8 m s-2 for Earth.

3. The acceleration due to gravity does not depend on the mass of the falling object.

4. All objects falling freely fall with the same acceleration.

2.8.3 Free fall
1. An object falls freely only in vacuum.
2. The absence of air means that there is no air resistance to oppose

the motion of the object.
3. In vacuum, both light and heavy objects fall freely. They fall with the

same acceleration.
4. A piece of tissue paper does not fall freely because its fall is

affected by air resistance.
5. A heavy golf ball can be considered to be falling freely because the air resistance is small

compared to the pull of gravity and therefore is negligible.
6. If the piece of tissue and the heavy golf ball fall from the same height in vacuum, both will fall

with the same acceleration and reach ground at the same time.

2.8.4 Weight
1. The weight of an object is defined as the gravitational force acting on the object.

 Weight, W = mg
 g = 9.8 m s-2 (this value normally is rounded up in SPM exam  10 m s-2)

2.9 Analysing Forces in Equilibrium
1. When a body is in equilibrium, the resultant force on it is zero.
2. When the equilibrium is reached, then the object is in two states, that is

i. remains stationary (if the object is stationary) (at rest)
ii. moves at a constant velocity ( if the object is moving)
3. Newton’s third law of motion states , “ To every action there is an equal but opposite direction”

 F1 = - F2

17

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.9.1 Examples of Forces in Equilibrium

Two Forces in Equilibrium (object stay at rest or moves with constant velocity)

Weight = Normal reaction Weight = Tension
W=R W=T

Weight = Buoyant Force Weight = Normal reaction
W = FB // W = U W=R

Weight = Normal reaction // W = R Weight = Lifting force // W = Flift
Pulling force = Frictional force // Fp = Ff Thrust = Drag // Fthrust = Fdrag

Weight = Normal reaction // W = R Weight of load + Weight of helium gas = Buoyant Force
Thrust = Air resistance + Frictional force //
Fthrust = Fresistance + Ffriction

18

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

Three Forces in Equilibrium (object stay at rest or moves with constant velocity)

FP + FQ + FR = 0 N T1 + T2 + W = 0 N

T1 + T2 + W = 0 N T1 + T2 + W = 0 N

2.9.2 Addition of Forces – Resultant Force, F

1. A resultant force is a single force that represents the combined effect of two or more forces in

magnitude and direction.

2. The resultant force is found by adding the magnitudes of the forces if the forces acting in the

same direction.

3. If the forces acting in opposite directions, the resultant force is found by determining the

difference in the magnitude of the two forces and acts along the direction of the larger force.

Forces Resultant Force

=

=

=

≠ 19

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

4. The resultant of two forces acting at a point at an angle to each other is found by a scaled

drawing of the parallelogram of forces.

Forces Resultant Force

=

or

=

or

=

or

Use a scale of 1 cm to represent 10 N to determine
the magnitude and direction of the resultant force on
the object.

=

2.9.3 Resolution of Forces in Equilibrium
1. A force can resolve into two perpendicular components.
2. With applying the concept of the parallelogram of forces, the two perpendicular components

20

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

are represented by the adjacent sides of a rectangle.

 The parallelogram of forces

Forces Resultant Force

=

 The resolution of forces Is same as Resolution of Forces
A Force



3. When an object being acted on by three forces is in equilibrium, the resultant of the three
forces is zero. This means that the resultant of any two of the forces is equal and opposite of
the third force.
 The resolution of forces for three forces in equilibrium
Three Forces in Equilibrium – Resultant Force is zero

T1 + T2 + W = 0 N  Break up or resolve T1 and T2 When an object is in equilibrium,
into two components (vertical and  the resultant of the
horizontal component). horizontal components of the
 W not needs to break up forces acting on it is zero.
because it is already in vertical  the resultant of the vertical
component. components is also zero.

21

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

Calculate the magnitude
of T1 and T2.

 Break up or resolve T1 into two
components (vertical and
horizontal component).
 W and T2 not need to break up
because it is already in vertical
component and horizontal
component respectively.

2.10 Understanding Work, Energy, Power and Efficiency

2.10.1 Work Done
1. Work is done when an applied force moves an object through

a distance in the direction of the force.
2. Work is defined as the product of the applied force and the

displacement of an object in the direction of the applied force.
 Work done = Force X Displacement in the direction of the force
 W=FXs
 SI unit for work is joule, J. (1 J = 1 N m = 1 kg m2 s-2)
3. When an object does not move in the direction of the applied force, the component (horizontal or
vertical) of the force in the direction of the displacement is used to calculate the work done.

W = (F cos ) X s = Fs cos  22
4. No work is done when

 a force is applied but no displacement occurs,
 an object undergoes a displacement with no applied force acting on it,
 the direction of motion is perpendicular to the applied force.

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.10.2 Principle of Conservation of energy

1. An object that can do work has energy.

2. Energy exists in different forms: kinetic energy, gravitational potential energy, elastic

potential energy, sound energy, heat energy, light energy, electrical energy and chemical

energy.

3. When work is done, energy is transferred from one object to another. The work done is equal to

the amount of energy transferred.

 Work done, W = Kinetic energy, Ek
 Work done, W = Gravitational potential energy, Ep
 Work done, W = Elastic potential energy, Ep

4. The principle of conservation of energy states that energy can be transferred from one to

another, but it cannot be created or destroyed.

5. This means that the total amount of energy always remains the same.

2.10.3 Work Done and the Change in Kinetic Energy
1. Kinetic energy is energy of an object due to its motion.
2. Consider a force, F acting on an object at rest of mass, m. The object will moves with an

acceleration, a over a displacement, s.

u = 0, v2 = 2as

3. The work done is the energy transferred to the object as kinetic energy. Therefore,
 Work done, W = Fs = Kinetic energy, Ek
 Substituting F = ma and s = v2 / 2a into Ek = Fs s = 2
 Thus, Ek = ½ mv2 2

2.10.4 Work Done and Gravitational Energy 23
1. Gravitational potential energy is the energy of an object due to its

higher position in the gravitational field.
2. The work done is transferred to the object as gravitational potential

energy.
 Work done = Gravitational potential energy, Ep = W = Fs
 Substituting F = Weight = mg and s = h into Ep = Fs
 Thus, Ep = mgh
3. When an object falls, gravitational potential energy is transformed into kinetic energy.
4. When an object is thrown upwards, its kinetic energy changes to gravitational potential
energy.

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
2.10.5 Power

1. If a man lift a barbell weighing 400 N through a vertical distance of 0.5 m at constant velocity, the

man do (400 N)(0.5 m) = 200 J of work whether it takes 1 second, 1 hour, or 1 year to do it.

2. But often we need to know how quickly work is done. We describe this in terms of power.

3. Power is defined as the rate at which work is done, or the amount of work done per second.

 Power = work done
time taken

 P = , SI unit is watt, W (1 W = 1 J s-1)

4. In mechanics we can also express power in terms of force and velocity.

2.10.6 Efficiency

1. The efficiency of a device is defined as the percentage of the energy input that is transformed

into useful energy.

 Efficiency = Useful energy output × % = × %
Energy input

2. The engine of the car transforms the chemical energy of the petrol into kinetic energy of the

vehicle.

3. The kinetic energy of the car is the useful energy output of the car engine while the thermal

(heat) and sound energy are unwanted energy.

4. Efficiency also can define as the percentage of the power input that is transformed into useful

power.

 Efficiency = out × 100%
in

2.11 Appreciating the Importance of Maximising the Efficiency of Devices
1. All the output energy and the unwanted energy are eventually lost to the surroundings.
2. Since the energy resources in this world are limited, it is very important that a device makes the

best possible use of the input energy.
3. Maximising the efficiency of devices helps to conserve energy resources.

2.12 Understanding Elasticity

1. Elasticity is the property of a material that enables it to return to its original shape and size

when the force that was acting on it is removed.

2. The elasticity of solids is due to the strong intermolecular forces between the molecules of the

solid.

3. When the external stretching force is removed, the strong attractive intermolecular force brings

the molecules back to their original positions.

4. The repulsive intermolecular force of solid under compression brings the molecules back to

their original positions. This is explaining solid under compression able to returns to its original

shape and size. 24

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:
2.12.1 Hooke’s Law

1. Hooke’s Law states that the extension of a spring is directly proportional to the applied force

provided that the elastic limit is not exceeded.

2. The elastic limit of a spring is defined as the maximum force that can be applied to a spring

such that the spring will be able to be restored to its original length when the force is removed.

3. If the elastic limit is exceeded, the length of the spring is longer than the original length even

though the force no longer acts on it. The spring is said to have a permanent extension.

4. The mathematical expression for Hooke’s law is
 ∝ , F = Stretching Force on the spring, x = Extension,

 = Extension-Original length

 Therefore, = , k = Force constant of the spring

 Force constant, = with units N m-1, N cm-1, or N mm-1

5. The force constant is a measure of the stiffness of the spring.

6. A spring with a larger force constant is harder to extend and is

said to be stiffer.

2.12.2 Elastic Potential Energy

1. When a force extends a spring, work is done.

2. The work done on the spring is the energy transferred to the

spring and stored as elastic potential energy.

3. The work done to stretch a spring by a length x is equal to the

triangular area under the graph of force versus extension.

 Triangular area under the graph = 1 = elastic potential energy
2

 From Hooke’s law, F = kx

 Therefore, Ep = 1 ( ) = 1 2
2 2

25

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 2: Force and Motion
Guru Cemerlang Physics MSAB
Date:

2.12.3 Factors that Affect Elasticity

Factor Change in factor Effect on elasticity Stiffness of the spring

Length Shorter Less elastic Stiffer spring
Longer More elastic Softer spring

Diameter of spring Smaller Less elastic Stiffer spring

coil Larger More elastic Softer spring

Diameter of spring Smaller More elastic Softer spring

wire Larger Less elastic Stiffer spring

Type of materials The elasticity changes with the type of materials

2.12.4 Problems Solving of Arrangement of the Spring

A spring Springs In Series Springs In Parallel

Lo = original length of the spring

Stretching Force = W = mg Stretching Force of each spring Stretching Force of each spring
Extension of spring = x = W = mg
Total length of system = Lo + x Extension of each spring = x = =
Total extension of system = 2x 2 2
Total length of system = 2Lo + 2x
Extension of each spring =

Total extension of system =

Total length of system = Lo +

Example Solution

The figure shown identical
springs with force
constant 5 N cm-1. The
original length of each
spring is 10 cm and a load
of 500 g.
Calculate the length, L of
the spring system.

26

2nd edition © 2011 | chp 2 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure Date:
Guru Cemerlang Physics MSAB
CHAPTER 3: FORCE AND PRESSURE

3.1 UNDERSTANDING PRESSURE

3.1.1 What is Pressure?
1. The pressure acting on a surface is defined as a perpendicular force per unit area

F
2. Pressure, P = , [F = Force acting perpendicular on a surface in unit Newton (N), A = Surface

A
are in unit square metres (m2)]
3. The SI unit of pressure is Pascal (Pa).
4. The pressure of a given force increases as the surface area decreases.
5. Example 1: A wooden block is placed at different position on the surface of a piece of plasticine.
At what position is the pressure higher?

Wooden block

A B Plasticine

6. Example 2: Which shoe will exert a greater pressure on the when it is worn by the same women?

7. Example 3: The diagram below shows a wooden block of dimensions 8 cm × 10 cm × 12 cm. Its
weight is 12 N. On which side should the wooden block be placed to produce a maximum
pressure exerted on the table. What is value of this pressure?

1

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:
3.1.2 Application of Pressure

1. Tools like knives, chisels, axes and saws have sharp cutting edges. The surface area of contact

is small. When a force is applied on the tool, the small area of contact will produce a high

pressure to cut the material.

2. The flat base of each metal pole of a tent has a large surface area to deduce the pressure

exerted on the ground. The poles will not sink into the ground because of the flat bases.

3. .......………….…………….…………….…………….…………….…………….…………….…………….

…………….…………….…………….…………….…………….…………….…………….…………….…

………….…………….…………….…………….…………….…………….…………….…………….……

……….……………………….…………….…………….…………….………….…………….………….…

4. .......………….…………….…………….…………….…………….…………….…………….…………….
…………….…………….…………….…………….…………….…………….…………….…………….…

………….…………….…………….…………….…………….…………….…………….…………….……

……….……………………….…………….…………….…………….………….…………….………….…

Exercise
1. A table of mass 50kg has 4 legs is placed on a floor. Each legs has a cross sectional area of

25cm2. Find the pressure exerting on the floor (g=10ms-2)

2. The diagram below shows a concrete block of dimension 1.5m x 2.0m x 3.0m. Its weight is 60N.
Calculate (a) maximum pressure, (b) minimum pressure:

1.5m

3.0m 2.0m

a) maximum pressure b) minimum pressure

2

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:

3.2 UNDERSTANDING PRESSURE IN LIQUIDS

3.2.1 Density and Pressure in Liquids

1 Density (  ) is defined as mass per unit volume. The SI unit for density is kg m-3.

Density (ρ) = m
V

2 Pressure in liquids acts in all directions.
3 The formula of pressure in liquids can be derived from the following steps:

A cylinder of water i) Mass of a cylinder of water, m = ρ V = ρ A h
A ii) Weight of the cylinder of water, W = m g = ρ A h g

Volume h iii) The pressure of water at the base of the cylinder of
V = Ah
water is,

Water pressure, P = F = W = m g =  A h g =  h g
AA A A

4 Example 1: Given that the density of mercury is 13600 kg m-3. Calculate the pressure of mercury
at a point 25 cm from the mercury surface (g = 10 m s-2)
Solution:

5 Example 2: The figure shows a glass tube filled with 50 cm Liquid M 50cm
height of liquid M and 30 cm height of liquid N. The densities of 30cm
liquid M and N are 1000 kg m-3 and 2500 kg m-3 respectively. By Liquid N x
giving g = 10 m s-2, what is the pressure of y
a) liquid M at point x
b) liquid M and N at point y
Solution:

3

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:

3.2.2 Application of Pressure in Liquids

1. Water tank positioned high above the ground. A pipe connects the water to the house. The height

of the tank results in the high pressure of water at the lower end of the pipe. As the result, the

water rushes out at high speed when the tap is turned on.

2. The water at the bottom of the dam is at a higher pressure than

at the top. Hence, the wall of the dam has to be thicker at the

bottom to sustain this greater water pressure.

3. .......………….…………….…………….…………….…………….…………….…………….…………….

…………….…………….…………….…………….…………….…………….…………….…………….…

………….…………….…………….…………….…………….…………….…………….…………….……

……….……………………….…………….…………….…………….………….…………….………….…

.......………….…………….…………….…………….…………….…………….…………….…………….

…………….…………….…………….…………….…………….…………….…………….…………….…

………….…………….…………….…………….…………….…………….…………….…………….……

……….……………………….…………….…………….…………….………….…………….………….…

Exercise 3.2 Atmospheric pressure
1. A balloon is situated at 10 m below sea level, what is the total pressure at sea level:

experience by the balloon? [The density of seawater is 1100 kg m-3] Patm = 1.0 x105 Pa
Total Pressure, P = Atmospheric pressure + Liquid pressure

2. Water with density of 1 g cm-3 and oil are filled into a U-tube.
What is the density of the oil? *(Pressure at A = Pressure at B)

4

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:

3.3 UNDERSTANDING GAS PRESSURE AND ATMOSPHERIC PRESSURE

3.3.1 Gas Pressure
1 Air pressure is the force exerted on a surface by air molecules per unit area of the surface.
2 The gas pressure in a container is caused by the collisions of gas molecules with the walls of

the container.

3.3.2 Atmospheric Pressure
1 The atmosphere is a thick layer of air that surrounds the Earth.
2 The atmospheric pressure is caused by the downward force exerted by the air, which is the

weight of the atmosphere on the Earth’s surface.
3 Atmospheric pressure acts on every object on the surface of the Earth.
4 The average pressure produced by the atmosphere at sea level is 100 000 Pa (1 atm = 105 Pa).
5 Atmospheric pressure decreases with altitude. At higher altitudes, the air gets thinner and

thinner, the density and temperature of the air are lower. As a result, the frequency of collisions of
the molecules is lower.

3.3.3 Application of Atmospheric Pressure
1 When drinking with a straw, one has to suck the straw. This causes the pressure I the straw to

decrease. The external atmospheric pressure, which is greater, will then act on the surface of the
water in the glass, causing it to rise through the straw.
2 The figure shows two workers use suction cups to lift the
windscreen. When a suction cup is pressed on the windscreen,
air is being forced out of it resulting in a partial vacuum in the
cup. The surrounding atmospheric pressure forces the suction
cups against the windscreen and causes them to be stuck
tightly on the windscreen.
3 .......………….…………….…………….…………….…………….…………….…………….…………….
…………….…………….…………….…………….…………….…………….…………….…………….…
………….…………….…………….…………….…………….…………….…………….…………….……
………….…………….…………….…………….…………….…………….…………….…………….……
……….……………………….…………….…………….…………….………….…………….………….…
.......………….…………….…………….…………….…………….…………….…………….…………….
…………….…………….…………….…………….…………….…………….…………….…………….…
………….…………….…………….…………….…………….…………….…………….…………….……
………….…………….…………….…………….…………….…………….…………….…………….……
……….……………………….…………….…………….…………….………….…………….………….…

5

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:

3.3.4 Instrument for Measuring Gas Pressure and Atmospheric Pressure

3.3.4.1 Mercury Barometer

1 The barometer is made by filling a long glass tube with mercury. The tube is then turned upside

down in a bowl of mercury. The mercury level in the tube will drop until a level 76 cm above the

mercury level in the bowl.

2 The level does not drop further because the mercury column

is held up by the atmospheric pressure.

3 The vertical height 76 cm mercury is known as Standard

Atmospheric Pressure or simply 1 atmospheric (at sea level).
4 By using the formula pressure in liquids, P = ℎ and given

that the density of mercury is 13 600 kg m-3, the 1 atm can be

determined in unit Pascal (Pa) as follows:

1 atm = 76 cm Hg (mercury)

= 0.76 m x 13 600 kg m-3 x 9.81

= 101 396.16 Pa

= 1.014 x 105 Pa

= 105 Pa

5 The vertical height of the column of mercury in a

mercury barometer will not vary when the glass

tube is tilted as shown in Figure (b) and (c).

3.3.4.2 Fortin Barometer
1 A Fortin Barometer is a type of mercury barometer which has a higher
accuracy.
2 A vernier scale helps to provide a more accurate reading.
3 The atmospheric pressure is measured in unit mm Hg.

3.3.4.3 Aneroid Barometer
1 An aneroid barometer can be used to measure atmospheric pressure.
2 The Aneroid barometer can be used as an altimeter (to determine
altitude) by mountaineers or by pilots to determine
an airplane’s altitude.
3 This is because the reading of the atmospheric
pressure corresponds to its height above sea level.

6

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:
3.3.4.4 Bourdon Gauge

1 The unit of measurement used in the Bourdon gauge is Pascal.

2 Bourdon gauges are more robust than manometers and more suitable

for measuring higher pressure. But they have to be calibrated before

they can be used.

3.3.4.5 Manometer
1 A manometer consists of a U-tube containing liquid (mercury, water, oil…).

2 The manometer is used to measure the difference in pressure between the two sides of the U-

tube as shown in figure (a) with containing mercury.

3 The mercury levels are the same with both sides of its U-tube

open to the atmosphere.

4 When the left side of the U-tube is connected to a vessel

containing a gas at a higher pressure than the atmospheric pressure, the mercury in arm B (left

side) will be pushed downwards and the mercury level at the right side of the U-tube will be rises.

5 Under equilibrium conditions, the pressure at B is equal to the pressure at C because both B and

C are at the same level (same liquids).

6 The gas pressure can be determined as follows:

Pressure at B = Pressure at C

Pressure at B = Gas pressure

Pressure at C = Patm + Pressure due to the h column of mercury

= 105 Pa + = 76 cm Hg + h cm Hg

 Gas Pressure = Pressure at B = Pressure at C = 76 cm Hg + h cm Hg

3.3.5 Various Units of Atmospheric Pressure

1 A height of 760 mm mercury (76 cm Hg) is known as Standard

Atmospheric Pressure (1 atm).

2 1 atmosphere is equal to 105 Pa (Pascal) with the density of

mercury is 13 600 kg m-3.

3 If water is used instead of mercury in the barometer. The vertical height of the water column in the

tube now is 10 m (1 atm = 10 m water = 76 cm Hg) because water less dense than water.

4 The SI unit for pressure in Pascal is too small a unit to be used in everyday life. Many common

situations involve pressure in the magnitude of 105 Pa use a bar as a unit of pressure. (1 atm =

105 Pa = 1 bar) 7

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure

Guru Cemerlang Physics MSAB Date:

3.4 APPLYING PASCAL’S PRINCIPLE

3.4.1 Transmission of Pressure in a Liquid
1 Liquids are practically incompressible.
2 When a force, F, pushes in the piston a pressure is applied to the water.

Pressure = F / A
where A is the cross-sectional area of the piston.
3 Water will spurt out through the openings (in all directions) with equal
velocity when the piston is pushed into the flask.

4 The applied pressure is transmitted uniformly throughout the
water (liquids) in accordance with Pascal’s principle.

3.4.2 Pascal’s principle
1 Pascal’s principle states that pressure applied to an

enclosed liquid is transmitted equally to every part

of the liquid.

2 Figure shows a basic hydraulic system. It operates on
Pascal’s principle.

3 Pressure transmitted equally from a smaller piston to a

larger one. Therefore,

P1 = P2

 F1  F2
A1 A2

4 The output force is given by

 F2  A2  F1
A1

Therefore, the hydraulic systems act as a force multiplier with the multiplying factor A2 .
A1

5 Example, if A2 = 6A1, then F2 = 6F1, where the multiplying factor A 2 = 6
A1

6 Example, In a hydraulic brake, a force of 60 N is applied to a piston with area of 3 cm2.

i) What is the pressure transmitted throughout the liquid?

Solution: P = F  60  20 N cm-2 = 2 105 Pa
A3

ii) If the piston at the wheel cylinder has an area of 8 cm2, what is the force exerted on it?

Solution: F = P X A = 20 N cm-2 X 8 cm2 = 160 N

8

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure

Guru Cemerlang Physics MSAB Date:

3.4.3 Applications of Pascal’s principle

3.4.3.1 Hydraulic Lift (Hydraulic Pumps)
1 The machine is equipped with a small cylinder connected to

a large cylinder. Both cylinders are filled with oil.
2 When input piston is pressed, the input piston exerts a

pressure on the surface of the oil.
3 The input pressure, Pin is equal to the output pressure, Pout,

we have Fin  Fout
Ain Aout

4 This pressure is transmitted by the oil and causing the large piston
(output piston) to produce a force large enough to lift a car.

Hydraulic Jacks for car lifting
3.4.3.2 Hydraulic Brake
1 The brakes of most cars are worked by hydraulic pressure.
2 When the driver’s foot presses the brake pedal, the piston in the master cylinder exerts pressure

on the brake fluid.
3 This pressure is transmitted to the wheel cylinder.
4 The pressure causes the pistons of the wheel cylinder to press the brake shoes.
5 The piston of the wheel cylinder is larger than that of the master cylinder so that larger force can

produce to press the brake shoes against the wheels to slow down the car.
6 When the brake pedal is released, a return spring restores the brake discs to their original

positions.

9

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure

Guru Cemerlang Physics MSAB Date:

3.5 APPLYING ARCHIMEDES’ PRINCIPLE

3.5.1 What is Buoyant Force?
1 The Buoyant Force is the upward force (up thrust) resulting from an

object being wholly or partially immersed in a fluid.
2 The upward force (up thrust) is equal to the weight of fluid

displaces (buoyant force).
3 Weight of fluid displaced depends on volume of fluid displaced and density of fluid.
4 When an object is immersed in a liquid (fluid), it displaces the liquid (fluid). The bigger the volume

of the object immersed, the bigger the volume of the liquid displaced.

10

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:
3.5.2 Archimedes’ Principle

1 Archimedes’ principle states that when an object is wholly or partially immersed in a fluid, it

experiences a buoyant force equal to the weight of the fluid displaced.

2 For a free floating object, the buoyant force is equal to the weight of the object. (W = FB)

3 If the weight of the object greater than buoyant force, the object will sink to the bottom. (W > FB)

4 If the weight of the object less than buoyant force, the object will rise. (W < FB)

3.5.2.1 A free Floating Object (W = FB)

3.5.2.2 A Sink Object (W > FB)

11

2nd edition © 2011 | chp 3 | Maktab Sultan Abu Bakar | Sekolah Kluster Kecemerlangan