# SCIENCECLINIC SMARTPREP GR12 DBE ENG 2023 V4.1 (1)

> Grade 12 question paper from the Physical Sciences, 68 pages. Read online or get the free PDF download, memo linked when available.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-question-paper/cf008a44-87a6-4e55-aef2-e8f1bb7dd779
- Updated: 2026-09-27
- Subject: Grade 12
- Grade: Multiple grades
- Document type: Question paper
- Exam period: Physical Sciences
- Curriculum: CAPS
- Pages: 68

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [SCIENCECLINIC SMARTPREP GR12 DBE ENG 2023 V4.1 (1)](https://www.hlayiso.com/library/grade-12-physical-sciences-question-paper/cf008a44-87a6-4e55-aef2-e8f1bb7dd779)

## About this document

Grade 12 · Physical Sciences. Question paper, 68 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

© Lesson videos for every section of these SmartPrep theory summaries available on www.-scienceclinic.co.za )

### Page 2

VIDEO LESSONS!

We have free video lessons for
every topic in this book! These
videos can only be accessed via
Science Clinic's website - follow
these steps to access this amazing
resource!

DOWNLOAD A BOOK

Visit our website to download a
SmartPrep book. We'll send you
an email with download links.

VIDEOS WEBPAGE

Go to our Videos webpage. Log
in with the email address that
you used to download a book.

SELECT YOUR TOPIC

From your videos page, select a
topic. You'll be redirected to a
playlist with video lessons.

ABOUT OUR SMARTPREP BOOKS

Our SmartPrep books provide crystal-clear summaries of the essential
theory for Grade 8-12 Core Mathematics and Grade 10-12 Physical
Science. Our books are available in English and Afrikaans.

COPYRIGHT NOTICE AND LICENSE

The content in this SmartPrep Book is the original work of Science

Clinic (Pty) Ltd. You may distribute this material, provided that you

adhere to the conditions in our copyright notice and license:

¢ If you copy and distribute this book electronically or in paper form,
you must keep this copyright notice intact.

¢ If you use questions or syllabus summaries from this book, you
must acknowledge the author.

¢ This Smart Prep Book is meant for the benefit of the community
and you may not use or distribute it for commercial purposes.

* You may not broadcast, publicly perform, adapt, transform, remix,
add on to this book and distribute it on your own terms.

° By exercising any of the rights to this Smart Prep Book, you accept
and agree to the terms and conditions of the license provided on
this link:

CONTENT ACKNOWLEDGEMENT

Thank you to all the amazing people involved in the production,

translation and moderation of this book: R_ Bartholomew, L
Couperthwaite, N Cullinan, W Cloete, C de Beer, S Dippenaar, T
Fairless, | Govender, C Hare, L Kroukamp, R Lodge, K Munnik, C
Orchison, M Peyper, S Pienaar, Y Patterson, R Ramsugit, S Roberts, C
Steyn, K Storm, S Sapsford, X Sithenjwa, A Theron, C Visser, B Ward

QD ww sciencectinie.coza © exciencectinic EB ew acodook.comisciencectiniesa

### Page 3

TABLE OF CONTENTS

                    CORE THEORY SUMMARIES

      Physics
            Physics Data Sheets                        1
            Physics Definitions                        4
            2D Vectors                                 5
            Newton’s Laws of Motion                    7
            Newton’s Law of Universal Gravitation     11
            Momentum and Impulse                      12
            Work, Energy and Power                    16
            Motion in 1D                              19
            Vertical Projectile Motion                20
            Electricity                               23
            Electrostatics                            25
            Electromagnetism and Electrodynamics      27
            Photoelectric Effect                      30
            Doppler Effect                            32

      Chemistry
            Chemistry Data Sheets                     33
            Chemistry Definitions                     37
            Organic Chemistry                         39
            Quantitative Aspects of Chemical Change   48
            Energy and Chemical Change                51
            Rates of Reactions                        52
            Chemical Equilibrium                      53
            Acids and Bases                           56
            Electrochemistry                          59
            Fertilisers                               63

www

### Page 4

Physics

(fw). scienceclinic.co.za © 2022@.:

### Page 5

DATA FOR PHYSICAL SCIENCES GRADE 12
                               PAPER 1 (PHYSICS)

                GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12
                           VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

          NAME/NAAM                SYMBOL/SIMBOOL    VALUE/WAARDE
Acceleration due to gravity
                                         g                9,8 m·s-2
Swaartekragversnelling
Universal gravitational constant
                                         G          6,67 x 10-11 N·m2·kg-2
Universele gravitasiekonstant
Speed of light in a vacuum
                                         c             3,0 x 108 m·s-1
Spoed van lig in 'n vakuum
Planck's constant
                                         h             6,63 x 10-34 J·s
Planck se konstante
Coulomb's constant
                                         k           9,0 x 109 N·m2·C-2
Coulomb se konstante
Charge on electron
                                         -e             -1,6 x 10-19 C
Lading op elektron
Electron mass
                                         me            9,11 x 10-31 kg
Elektronmassa
Mass of Earth
                                         M             5,98 x 1024 kg
Massa van Aarde
Radius of Earth
                                         RE             6,38 x 106 m
Radius van Aarde

### Page 6

TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING

vf        vi  a 't                                       Δx       v i Δt  21 a't 2 or/of Δy        v i Δt  21 a't 2
                                                                   § vi  v f ·                     § vi  v f ·
vf
     2        2
           v i  2a'x or/of v f
                                      2       2
                                             v i  2a'y   Δx       ¨          ¸ Δt or/of Δy         ¨          ¸ Δt
                                                                   © 2 ¹                            © 2 ¹

FORCE/KRAG

Fnet       ma                                             p mv
fs max = μsN                                              fk = μk N
Fnet Δt = Δp
                                                          w mg
Δp = mv f - mvi
           m1m 2                                  m1m 2             M                             M
F =G                     or/of            F =G            g =G                or/of           g =G 2
            d2                                     r2               d2                            r

WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING

W F'x cos T                                               U mgh                   or/of       EP     mgh

                                                          Wnet       'K           or/of       Wnet       'Ek
  1                                           1
K   mv2                 or/of         Ek        mv2
  2                                           2
                                                          'K       K f  Ki       or/of       'Ek      Ekf  Eki
                                                               W
Wnc        'K  'U or/of Wnc               'Ek  'Ep      P
                                                               't
Pav = Fvav           / Pgemid = Fvgemid

WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

                                                               1
v        fO                                               T
                                                               f
         v r vL                           v r vL                                          c
fL              fs               fL              fb       E hf           or /of    E h
         v r vs                           v r vb                                          O
E = Wo + Ek(max) or/of E = Wo + K max where/waar
                                                      1  2                1  2
E hf and/en W0                  hf0 and/en E k (max) = mvmax or/of K max = mvmax
                                                      2                   2

### Page 7

ELECTROSTATICS/ELEKTROSTATIKA

        kQ1Q 2                                                kQ
F                                                      E
          r2                                                  r2

        W                                                     F
V                                                      E
        q                                                     q
        Q                             Q
n=                   or/of       n=
        e                             qe

ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE

                                                       emf ( ε )= I(R + r)
        V
R
        I
                                                       emk ( ε )= I(R + r)

Rs       R1  R 2  ...
 1       1     1                                       q I 't
                  ...
Rp       R1 R 2

W = Vq                                                        W
                                                       P
                                                              Δt
W = VI ' t
                                                       P = VI
W = I2R ' t

             2
                                                       P = I 2R
         V Δt
W=                                                            V2
          R                                            P
                                                              R

ALTERNATING CURRENT/WISSELSTROOM

                                                       Pave Vrms Irms        /   Pgemiddeld Vwgk I wgk
            I max                              Imaks
I rms                        /         I wgk
                 2                               2           2
                                                       Pave Irms R           /   Pgemiddeld   I 2wgkR
             Vmax                              Vmaks
Vrms                         /        Vwgk                          2
                                                                  Vrms
                                                                                               2
                                                                                              Vwgk
                 2                                2    Pave              /       Pgemiddeld
                                                                   R                           R

### Page 8

Grade 12 Science Essentials                                                                                                                    SCIENCE CLINIC 2022 ©

                                                                      Grade 12 Physics Definitions
                           Normal force (FN): the force or the component of a force which a surface exerts on an object with which it is in contact, and
                           which is perpendicular to the surface
                           Frictional force (Ff): the force that opposes the motion of an object and which acts parallel to the surface
                           Static frictional force (fs): the force that opposes the tendency of motion of a stationary object relative to a surface
                           Kinetic frictional force (fk): the force that opposes the motion of a moving object relative to a surface
                           Newton's first law of motion: a body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/
  Newton’s Laws            net force acts on it
    Revision               Newton's second law of motion: when a resultant/net force acts on an object, the object will accelerate in the direction of the
                           force at an acceleration directly proportional to the force and inversely proportional to the mass of the object
                           Newton's third law of motion: when object A exerts a force on object B, object B SIMULTANEOUSLY exerts an oppositely
                           directed force of equal magnitude on object A
                           Newton's Law of Universal Gravitation: each body in the universe attracts every other body with a force that is directly
                           proportional to the product of their masses and inversely proportional to the square of the distance between their centres
                           Weight: the gravitational force the Earth exerts on any object on or near its surface

                           Momentum: the product of an object's mass and its velocity
                           Linear momentum: a vector quantity with the same direction as the velocity of the object
                           Newton’s second law in terms of momentum: the net force acting on an object is equal to the rate of change of momentum
                           of the object in the direction of the resultant/net force
                           Impulse: the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object. It is a
 Momentum and
                           vector quantity
   Impulse
                           Isolated system (in Physics): a system on which the resultant/net external force is zero. An isolated system excludes external
                           forces that originate outside the colliding bodies, e.g. friction. Only internal forces, e.g. contact forces between the colliding
                           objects, are considered
                           Principle of conservation of linear momentum: the total linear momentum of an isolated system remains constant (is
                           conserved)

     Vertical              Projectile: an object which has been given an initial velocity and then moves under the influence of the gravitational force only
Projectile Motion          Free fall: motion during which the only force acting on an object is the gravitational force

                           Work done on an object by a constant force F: is FΔxcosθ , where F is the magnitude of the force, Δx the magnitude of the
                           displacement and θ the angle between the force and the displacement
                           Work-energy theorem: The work done on an object by a net force is equal to the change in the object's kinetic energy OR the
                           work done on an object by a net force is equal to the change in the object's kinetic energy. (Wnet = ΔK = Kf - Ki)
   Work, Energy
                           Conservative force: is a force for which the work done in moving an object between two points is independent of the path taken
    and Power
                           Non-conservative force: is a force for which the work done in moving an object between two points depends on the path taken
                           Principle of conservation of mechanical energy: the total mechanical energy (sum of gravitational potential energy and
                           kinetic energy) in an isolated system remains constant
                           Power: is the rate at which work is done or energy is expended

                           Doppler Effect: the change in frequency (or pitch) of the sound detected by a listener, because the sound source and the listener
  Doppler Effect
                           have different velocities relative to the medium of sound propagation

                           Coulomb's law: the magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is
                           directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r)
                           between them
   Electrostatics          Electric field: a region of space in which an electric charge experiences a force. The direction of the electric field at a point is the
                           direction that a positive test charge would move if placed at that point
                           Electric field at a point: The electric field at a point is the electrostatic force experienced per unit positive charge placed at that
                           point

                           Ohm's law: the potential difference across a conductor is directly proportional to the current in the conductor at constant
                           temperature
 Electric Circuits
                           Power: the rate at which work is done
                           Emf: the maximum energy provided by a battery per unit charge passing through it

                           rms value: the DC potential difference/current which dissipates the same amount of energy as AC
                           rms potential difference: the AC potential difference which dissipates/produces the same amount of energy as an equivalent DC
 Electrodynamics
                           potential difference.
                           rms current: the alternating current which dissipates/produces the same amount of energy as an equivalent direct current (DC).

     Optical               Photoelectric effect: the process whereby electrons are ejected from a metal surface when light of suitable frequency is incident
 Phenomena and             on that surface
  Properties of            Threshold frequency (fo): the minimum frequency of light needed to emit electrons from a certain metal surface
    Materials              Work function (Wo): the minimum energy that an electron in the metal needs to be emitted from the metal surface

Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes.                    Page 4

### Page 9

Grade 12 Science Essentials
                                                                                                                       Vectors in 2D                                                                                                     SCIENCE CLINIC 2022 ©

RESOLVING INTO                                                                                   Fx                                    COMPONENTS ON A
COMPONENTS
                                                                                    x                                                  SLOPE
                                                                                                             θ                                                                 Fg // = Fg sin θ
Diagonal vectors can be broken into                                                                                                    When forces act on objects on a
components. When vectors are bro-                             Fx = F cos θ                                                             slope, it is useful to resolve vec-     Fg ⊥ = Fg cos θ            F g//
ken into the x- and y-components,                                                                F                      Fy             tors into components that are                                                                           Fg θ     F g⟂
we are determining the horizontal                             Fy = F sin θ                                                             parallel (//) or perpendicular (⟂)
(x-axis) and vertical (y-axis) effect                                                                                                  components. The most common                                                  Fg θ   F g⟂
of the vector. We can use sin or cos,                                                                                                  force resolved into components                                θ
provided that the angle is between                                                                                                     on a slope is weight (Fg).
the force F and the horizontal.                                                                                             y                                                                                                                           F g//
                                                                                                                     CONSTRUCTING FORCE TRIANGLE
  When forces are not co-linear, force triangles can be used to determine resultant forces or the equilibrant. When force triangles are formed, basic geometric rules can be used to determine vectors or resultants.
                               Tail-to-head                                                                                 Parallelogram                                                                    Manipulation
Used for consecutive vectors (vectors that occur in sequence).                           Used for vectors that act concurrently on the same object.                          The vector arrows can be manipulated to form a force triangle to determine the
                                                                                         The resultant is the diagonal of a parallelogram that originates                    resultant forces or an equilibrant. The vectors/arrows may only be moved if the
Eg. A boat travels 90 m east, and then moves 50 m north.
      y                                                                                  from the tail of the vectors (common origin).                                       magnitude and direction are both kept constant.
                                                                                                                 y                                                           When manipulating the vector arrows, the following has to remain the same:
                                                                                                                                                                             • Length of arrow (magnitude)
                                                                                                                                                                             • Angle of the arrow (direction)
                                                                                                                                                                             • The direction of the arrow head
                                                t
                                                                     Vector 2
                                          tan
                             su l
                           Re                                                                                                                                                Eg. An object is suspended from a ceiling by 2 cables. Below is a free body dia-

                                                                                                      Vector 2
                                                                                                                                 nt
                                                                                                                                l ta                                         gram as well as a force triangle that can be used to calculate the values of T1 and
                                                                                                                        Resu                                                 T2.
                                                                                                                                                                                      Free body diagram                               Force triangle
                                                                                x
                                          Vector 1
                                                                                                                                                          x
This principle can also be applied to more than 2 vectors                                                                       Vector 1
taken in order. The resultant is from the tail of the first vector
to the head of the last.                                                                 Eg. Two tugboats apply a force of 6 000N and 5 000N at bear-
                                     y                                                   ings of 60° and 120° respectively on a cargo ship.
                                                                                                                                                                                      T1
                                                      Vect
                                                                                                      y
                                                                                                                                                                                                         T2                                      T1
                                                          or 4
                               nt                                                                                                                                                                                                 Fg
                          Resulta
                                                                                                                       0N
                                                                    or 3                                         6   00
                                                                                                                        Resultant
                                                                   Vect
                                                                                                                                                                 x
                                                                           x                                50                                                                             Fg                                                      T2
                                                                                                              00
                                                                                                                 N

                               Vector 1
                                                          o   r2
                                                       ct
                                                     Ve

   Follow us on Instagram: @science.clinic                                  Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes.                              Page 5

### Page 10

Grade 12 Science Essentials
                                                        2D Vectors- Resultant and Equilibrant                                                                                                                          SCIENCE CLINIC 2022 ©

                   RESULTANT: The single vector which has the same effect as the                                                                  EQUILIBRANT: The force that keeps a system in equilibrium.
                        original vectors acting simultaneously on an object.
                                                                                                                                          The equilibrant is equal in magnitude but opposite in direction to the resultant force.
                                         PYTHAGORAS (90° ONLY)                                                                                                        COMPONENT ADDITION
Pythagoras can only be applied to vector triangles that are right angle triangles.                                            The resultant of diagonal forces can be determined using Pythagoras by determining the x-resultant and
                                                                                                                              y-resultant first. This is especially useful for determining resultants when more than 2 forces act on an
               FOR FINDING SIDES:                                                 FOR FINDING ANGLES:
                                                                                                                              object and a force triangle can not be used.
                                                                        o                          a                o
                    R2 = x2 + y2                                sin θ =                  cos θ =          tan θ =               EXAMPLE:
                                                                        h                          h                a
                                                                                                                                Three forces act on an object as shown in the diagram below. Determine the resultant force on the ob-
                                                                                                                                ject.
EXAMPLE:
                                                                                                                                1.Determine the x- and y-components of each force.
A boat travels 90 m due east, and then moves 50 m due north. Determine the displacement                                                             11N force:
of the boat.         y                                                                                                                                                     Fy      =           F sin θ
                                                                                                                                Fx    =         F cos θ
                                                                                                                                      =        11 cos 70                           =           11 sin 70                         11 N
                                                                                                                                      =    3,76 N right (90o )                     =       10,34 N up (0o )                             70°
                                                                                                                                                                                                                            40°           35°
                                                                                                                                                         30N force:
                                                                                                                                Fx    =          F cos θ                   Fy      =              F sin θ           30 N           20 N
                                                                tant                                                                  =         30 cos 40                          =             30 sin 40
                                                     su l                                                                                  22,98 N left (270o )
                                                   Re                                              50 m
                                                                                                                                      =                                            =       19,28 N down (180o )
                                                                                                                                                         20N force:
                                                                                                                                Fx    =          F cos θ                   Fy      =              F sin θ
                                                                                                                                      =         20 cos 35                          =             20 sin 35
                                     θ                                                                    x                           =    16,38 N right (90o )                    =       11,47 N down (180o )
                                                            90 m
                                                  R2        =            x2 + y2                                                2. Determine the x- and y-resultants of components.
                                                   R        =                 2
                                                                        90 + 50      2                                          Take left (270o) as positive               Take down (180o) as positive
                                                                                                                                                                                                                                  2,84 N
                                                   R        =          102,96 m                                                 Fx    =    −3,76 + 22,98 − 16,38           Fy      =       −10,34 + 19,28 + 11,47                         θ
                                                                                                                                      =     2,84 N left (270o )                    =        20,41 N down (180o )
                                                                              o
                                                tan θ       =

                                                                                                                                                                                                                       20,41 N
                                                                              a

                                                                      tan−1( 50                                                                                                                                                           R
                                                                             90 )
                                                                                                                                3. Find resultant-Pythagoras.             4. Find angle- trigonometry
                                                  θ         =                                                                                                                                    o
                                                                                                                                R2    =          x2 + y2                   tan θ       =         a
                                                  θ         =           29,05∘                                                                                                                     20,41
                                                                                                                                 R    =       2,842 + 20,412                 θ         =     tan−1 2,84
                                                                                                                                                                                                     ∘
                               Remember that θ calculated is relative to the x-axis,                                             R    =          20,61 N                     θ         =       82,08
                                                        ∘                 ∘               ∘
                                      ∴ bearing = 90 − 29,05 = 60, 95
                                                                                                                                ∴Resultant = 20,61 N at a bearing of 187,92°
                                ∴ Displacement = 102,96 m at a bearing of 60,95∘

   Follow us on Instagram: @science.clinic                          Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes.                          Page 6

### Page 11

Grade 12 Science Essentials
                                                                       Newton’s Laws of Motion                                                                                                                             SCIENCE CLINIC 2022 ©

                                                                                                          FORCES
    Non-contact force: A force exerted between             Contact force: A force exerted between               A force is a push or a pull action exerted on an object by another object. This action can be exerted
                                                                                                                while objects are in contact (contact force) or over a distance (non-contact force).
  objects over a distance without physical contact.       objects that are in contact with each other.
                                                                                                                Because forces have magnitude and direction, they are vectors. Force is measured in newton (N). 1 N is
  Electrostatic force (FE)                             Applied force (FA)                                       the force required to accelerate a 1 kg object at 1 m·s-2 in the direction of the force. We can therefore
                                                                                                                say that 1 N = 1 kg·m·s-2.
  Gravitational force (w/Fg)                           Tension (T or FT)
                                                                                                                           NOTE:
  Magnetic force                                       Friction (Ff or fs/fk)
                                                                                                                           Free-Body Diagram: forces drawn away from a dot which represents the object
                                                       Normal force (N/FN)

                                            Normal force (FN)                                                                                                           Friction (Ff or fs/fk)
The perpendicular force exerted by a surface on an object in contact with it.                                     Frictional force due to a surface is the force that opposes the motion of an object in contact
                                                                                                                  with it, parallel to the surface.
                                                                                     FN                           Friction is the parallel component of the contact force on an object by the surface on which
The normal force is equal to the perpendicular component of
gravity if there are no other forces acting on the object.                                                        it rests. The friction between the contact surfaces is determined by the properties of both the contact
                                                                                                                  surfaces of the object and surface. The coefficient of friction (µs/µk) is a description of the roughness of
                                                                                                                  the surface. The rougher the surface, the greater the coefficient of friction.
                                               FN = Fg
                                                                                                                                 Static friction (fs)                                                       Kinetic friction (fk)
                                                                                     Fg
                                                                                                                  Static friction is the frictional force on a sta-                          Kinetic friction is the frictional force on a mov-
If alternative forces act on the object, the normal force will change depending on the direction and mag-         tionary object that opposes the tendency of                                ing object that opposes the motion of the
nitude of the applied force. All vertical forces must be balanced if there is no acceleration in the vertical     motion of the object. The magnitude of the                                 object. The magnitude of the kinetic friction is
plane.                                                                                                            static friction will increase from 0N as the parallel                      constant for the specific system at all velocities
                                   FA                                                       FA                    component of the applied force is increased, until                         greater than zero, and irrespective of the applied
                       FN                                                       FN                                maximum static friction is reached. fs(max) is the                         force.
                                    θ                                                        θ                    magnitude of friction when the object just starts
                                                                                                                  to move.

                                                                                                                                     fs(max) = μs FN                                                             fk = μk FN
                        Fg                                                      Fg
                                                                                                                          fs(max) = maximum static friction (N)                                     fk = kinetic friction (N)
                                                                                                                          μs = coefficient of friction (no unit)                                    μk = coefficient of friction (no unit)
              FN + FAy − Fg = 0                                       FN − Fg − FAy = 0
                                                                                                                          FN = normal force (N)                                                     FN = normal force (N)
              FN + FA sin θ = Fg                                      FN = Fg + FA sin θ
                                                                                                                         If the applied force is greater than the maximum static friction, the object will start to move.

Objects suspended from a rope/string/cable have no                                                                                                                                             fsmax
                                                                                FT
normal force, as there is no surface on which the object rests.

                                                                                                                                              Fric%on (N)
The tension is equal to the perpendicular component of                                                                                                                               )
                                                                                                                                                                                  (f s
gravity if there are no other forces acting on the object                                                                                                                     n
                                                                                                                                                                        c   $o
OR the full magnitude of Fg for vertically suspended
                                                                                                                                                                 c   fri                   Kine%c fric%on (fk)
objects that are stationary/moving at constant velocity.                                                                                                      a$
                                                                                                                                                            St
                FT + (−Fg ) = 0
                                                                                Fg
                                                                                                                                                                                   Applied force (N)

  Follow us on Instagram: @science.clinic             Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes.                                       Page 7

### Page 12

Grade 12 Science Essentials
                                                                                          Newton’s Laws of Motion                                                                                                             SCIENCE CLINIC 2022 ©

                   Newton’s First Law of Motion                                                                 Newton’s Second Law of Motion                                                Newton’s Third Law of Motion
An object continues in a state of rest or uniform (moving with                                     When a net force is applied to an object of mass, it acceler-               When object A exerts a force on object B, object B simulta-
constant) velocity unless it is acted upon by an unbalanced (net                                   ates in the direction of the net force. The acceleration is                 neously exerts an oppositely directed force of equal mag-
or resultant) force.                                                                               directly proportional to the net force and inversely propor-                nitude on object A.
Newton’s First Law is due to inertia- the resistance of an object to                               tional to the mass.
change its state of rest or constant/uniform motion.                                                                                                                           NB!
                                                                                                                                                                               Newton’s Third Law describes action-reaction force pairs. These
                                                                             −2                    Newton’s Second Law is dependent on the resultant force-
             Fnet = 0 N                               a = 0 m⋅s                                    The vector sum of all forces acting on the same object.
                                                                                                                                                                               are forces on different objects and can not be added or sub-
                                                                                                                                                                               tracted.
A 3kg object moves up an incline surface at an angle of 15º with a
constant velocity. The coe"cient of friction is 0,35. Determine                                                     Fnet = m a                    a ≠ 0 m ⋅ s −2                                      FA on B = − FB on A
the magnitude of the applied force.
                                                                      FA
                                              FN                                                    A 20 N force is applied to a 5 kg object. The object accelerates           Force pairs properties:
                                                                                                    up a frictionless incline surface at an angle of 15º. Determine
                                                                                                                                                                               • Equal in magnitude
                                                                                                    the acceleration of the object.
                                                                                                                                                                               • Opposite in direction
                            Fg//                                                                                                                      FA                       • Acts on different objects (and therefore DO NOT CANCEL each
                                                                                                                             FN                                                  other out)
                           fk
                                15°                 Fg   T                                                                                                                                                                 NOTE:
                                                                                                                                                                                                                           The force pairs shown
                       Take upwards as positive:                                                                Fg//                                                                                                       here are gravitational
                    Fnet⊥      =          0                                                                                                                                    Fman on earth                               forces.
                 FN + (−Fg ⊥ ) =          0
                       FN             =                  Fg ⊥
                                                                                                                   15°                 Fg                                                             Fearth on man        Gravity and Normal force
                                                                                                                                                                                                                           are NOT force pairs.
                                                                                                                                            T
                       FN             =               m g cos θ
                       FN             =            (3)(9,8)cos 15∘
                                                                                                                       Take upwards as positive:
                       FN             =               28,40 N
                                                                                                                   Fnet//      =            ma
                      ∴ FN            =       28,40 N ⊥ up from slope
                                                                                                                FA + (−Fg // ) =            ma                                                              Fman on wall
                                                                                                            20 − (5)(9,8)sin 15∘   =               5a
                 Fnet//                   =                      0                                              20 − 12,68         =               5a
          FA + (−Fg // ) + (−fk )         =                      0                                                                                 7,32
                                                                                                                       a           =
                    FA                    =                  Fg // + fk                                                                              5

                     FA                   =           m g sin θ + μk FN                                              ∴a            =   1,46 m ⋅ s−2 // up the slope
                     FA                   =    (3)(9,8)sin 15∘ + (0,35)(28,40)
                    ∴ FA                  =                17,55 N
                                                                                                   Effect of Newton’s Second Law on overloading:
                                                                                                   According to Newton’s Second Law, the acceleration of an object                                          Fwall on man
Importance of wearing safety belts:                                                                is directly proportional to the net force and inversely proportional
According to Newton’s First Law, an object will remain in motion at a                              to the mass of the object. If a vehicle is overloaded, the stopping
constant velocity unless a non-zero resultant force acts upon it. When a                           distance will increase which can lead to serious accidents. When            Newton’s Third Law during an accident
car is in an accident and comes to a sudden stop, the person inside the                            brakes are applied, the force (friction) remains the same, but the          According to Newton’s Third Law, the force that two objects exert
car will continue with a constant forward velocity. Without a safety                               increase in mass causes a decrease in negative acceleration, in-            on each other is equal in magnitude but opposite in direction. If
belt, the person will make contact with the windscreen of the car, caus-                           creasing the time (and distance) it takes for the vehicle to stop.          two cars are in an accident, they will both exert the same amount
ing severe head trauma. The safety belt acts as an applied force (new                                                                                                          of force on each other irrespective of their masses.
Fnet), preventing the forward motion of the person.

   Follow us on Instagram: @science.clinic                                Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes.                     Page 8

Text truncated here. Open the document page to read the rest.

## Collection

- [Past papers & memos](https://www.hlayiso.com/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
