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



 2.1          LINEAR MOTIONs

Physical Quantity           Definition, Quantity, Symbol and unit

                            Distance is the ……
Distance, s                 Quantity: …                          SI unit : ..

                            (a) The distance in .. ….
                            (b) the distance between ….
Displacement, s                    ….direction.
                            (c) The distance of its final …..
                                 specified ….
                            Quantity:                               SI unit:

                            Speed is the
Speed,v
                            Speed =

                            Quantity:                           SI unit:


                            Velocity is the
Velocity, v
                                   Velocity =

                            Direction of velocity is
                            Quantity :                           SI unit:




Average speed               v=                                       Example: A car moves at an average
                                                                     speed / velocity of 20 ms -1
                                                                     On average, the car moves a distance/
                                                                     displacement of

Average velocity                            Displacement
                                      v
                                             TotalTime




                                                       2-1
Physics Module Form 4              Chapter 2 – Forces & Motion                              GCKL 2011



Uniform speed               Speed that remains the same in

Uniform velocity            Velocity that remains

An object has a non-        (a) The direction of motion changes or the motion is not linear.
uniform velocity if
                            (b) The magnitude of its velocity changes.

Acceleration, a             When the velocity of an object

     v u                   Acceleration is defined as the
a
       t                                     Change in velocity
                            Acceleration=
                                                Time taken
                                             Final velocity,v - Initial velocity,u
Unit: ms-2                                 =
                                                        Time taken,t

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




Deceleration



acceleration is negative.   The rate of decrease in speed in a specified direction.




Zero acceleration           An object moving at a constants velocity, that is,



Constant acceleration       Velocity increases at a uniform rate.
                            When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity



                            1. Constant =
                            2. increasing velocity =
                            3. decreasing velocity =
                            4. zero velocity =
                            5. negative velocity = object moves at opposite direction
                            6. zero acceleration =
                            7. negative acceleration = deceleration



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




Comparisons between distance and displacement            Comparisons between speed and velocity

      Distance               Displacement                        Speed                     Velocity
Total path length          The distance between        The rate of change of       The rate of change of
travelled from             two locations               distance                    displacement
one location to            measured along the          Scalar quantity             Vector quantity
another                    shortest path
                           connecting them in
                           specific direction


Scalar quantity

It has magnitude but no
direction

SI unit                    SI unit :




Fill in the blanks:

     1. A steady speed of 10 ms -1 = A distance of __________________________________________
     2. A steady velocity of -10 ms -1 = A displacement of _________________________________
     3. A steady acceleration of 4 ms -2 = Speed _____________________________________________
     4. A steady deceleration of 4 ms -2 = ____________________________________________________________
     5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right.




                                                        2-3
Physics Module Form 4                 Chapter 2 – Forces & Motion                             GCKL 2011




Example 1                                                      Example 2
Every day Rahim walks from his house to the junction           Every morning Amirul walks to Ahmad’s house
which is 1.5km from his house.                                 which is situated 80 m to the east of Amirul’s house.
Then he turns back and stops at warung Pak Din which is        They then walk towards their school which is 60 m
0.5 km from his house.                                         to the south of Ahmad’s house.

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




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

                                                                    Speed =

(a) What is Rahim’s displacement from his house,
    • when he reaches the junction.
                                                                   Velocity =
    •   when he is at warung Pak Din.


(b) After, Rahim walks back to his house. breakfast
    When he reaches home,
    (i) What is the total distance travelled by
        Rahim?

    (ii) What is Rahim’s total displacement from
         his house?


Example 3                                                       Example 4
Salim running in a race covers 60 m in 12 s.
                                                                An aeroplane flies towards the north with a
(a) What is his speed in ms-1                                   velocity 300 km hr -1 in one hour. Then, the plane
                                                                moves to the east with the velocity 400 km hr -1 in
                                                                one hour.

(b) If he takes 40 s to complete the race, what is his
                                                                (a) What is the average speed of the plane?
     distance covered?


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




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




                                                                (c) What is the difference between average speed
                                                                    and average velocity of the plane?




   Example 5
 The speedometer reading for a car travelling due north
 shows 80 km hr -1. Another car travelling at 80 km hr -1
 towards south. Is the speed of both cars same? Is the
 velocity of both cars same?




A ticker timer


     Use:
     1 tick = time interval
     The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2
       consecutive dots is
     1 tick =




                                                       2-5
Physics Module Form 4                 Chapter 2 – Forces & Motion                              GCKL 2011



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

                          VELOCITY                                                       FORMULA
                                                                   Time, t = 10 dicks x 0.02 s
                                                                               = 0.2 s
                                                                   displacement, s = x cm

                                                                   velocity =



ACCELERATION
                                                                   Initial velocity, u =


                                                                   final velocity, v =

                                                                   acceleration, a =
Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or
           t = (50 – 10) ticks x 0.02 s = 0.8 s




TICKER TAPE AND CHARTS                               TYPE OF MOTION




                                                           Distance between the dots increases uniformly
                                                          




                                                    2-6
Physics Module Form 4                 Chapter 2 – Forces & Motion                          GCKL 2011



                                                       - Distance between the dots decrease uniformly




Example 6

The diagram above shows a ticker tape chart for a
moving trolley. The frequency of the ticker-timer
used is 50 Hz. Each section has 10 dots-spacing.

(a) What is the time between two dots?



(b) What is the time for one strips?



(c) What is the initial velocity?



(d) What is the final velocity?


(e) What is the time interval to change from initial
velocity to final velocity?



(f) What is the acceleration of the object?

          vu    2
     a=
           t

 THE EQUATIONS OF MOTION

v  u  at                                                   u=
         1                                                   v=
s  ut  at 2                                                t=
         2                                                   s=
v  u  2as
 2     2
                                                             a=


                                                       2-7
Physics Module Form 4           Chapter 2 – Forces & Motion                          GCKL 2011


 2.2        MOTION GRAPHS


   DISPLACEMENT – TIME GRAPH            Velocity is obtained from

                                        A – B : gradient of the graph is

                                        B – C : gradient of the graph =
                                               object is
                                        C – D : gradient of the graph
                                         The object



   VELOCITY-TIME GRAPH                  Area below graph

                                        Positive gradient

                                        Negative gradient

                                        Zero gradient




   GRAPH              s versus t               v versus t                  a versus t
Zero
velocity




Negative
constant
velocity




Positive
Constant
velocity




                                               2-8
Physics Module Form 4                 Chapter 2 – Forces & Motion                                   GCKL 2011




     GRAPH                   s versus t                v versus t                       a versus t
  Constant
  acceleration




  Constant
  deceleration




Example 1:                                              Example 2:

                                                            velocity/ m s-1


                                                              20


                                                              10
                                                                                                     time/
                                                               0       10     20   30     40
                                                                                                     s

    Based on the s-t graph above:                            (a) Calculate the acceleration at:
    (a) Calculate the velocity at                                (ii) JK     (ii) KL     (iii) LM
        (i) AB     (ii) BC     (iii) CD


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


    (c) Find                                                 (c) Calculate
        (i) total distance                                       (iii) The total displacement

        (ii) total displacement                                     (iv) The average velocity


    (d) Calculate
        (i) The average speed

        (ii) The average velocity of the
             moving particle




                                                      2-9
Physics Module Form 4               Chapter 2 – Forces & Motion                         GCKL 2011



 2.3          INERTIA

Inertia                       The inertia of an object is the tendency of the object

Newton’s first law            Every object


Relation between inertia      The larger the mass,
and mass

           SITUATIONS INVOLVING INERTIA
     SITUATION                                              EXPLANATION
                                                          EEEEEEEEJNVJLKN drops straight into
                             When the cardboard is pulled away quickly, the coin
                             the glass.                  DNFLJKVNDFLKJNB
                                                          VJKL;DFN BLK;XC
                                                                  NB[F
                                                         NDPnDSFJ[POJDE]O-
                                                         JBD]AOP[FKBOP[DF
                                                            LMB NOPGFMB
                                                              LKFGNKLB
                                                           FGNMNKL’ MCVL
      Paste a picture        Chilli sauce in the bottle can be BNM’CXLB out if the bottle is moved
                                                               easily poured
                                                         NFGNKEPLANATION
                             down fast with a sudden stop. The sauce inside the bottle moves
                             together with the bottle.

                             When the bottle stops suddenly,




      Paste a picture        Body moves forward when the car stops suddenly The passengers were in a
                             state of motion when the car was moving.

                             When the car stopped suddenly,




      Paste a picture        A boy runs away from a cow in a zig zag motion. The cow has a large inertia




                                                         2-10
Physics Module Form 4           Chapter 2 – Forces & Motion                             GCKL 2011




                                  The head of hammer is secured tightly to its handle by
                                   knocking one end of the handle, held vertically, on a hard
                                   surface.

                                     This causes the hammer head to continue on its
                                     downward motion
                                     when the handle has been stopped, so that the top
                                      end of the handle is slotted deeper into the hammer
                                       head.




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

                                 • This is because the drop of water on the surface of the
                                 umbrella moves simultaneously as the umbrella is rotated.

                                 •   When the umbrella stops rotating, the inertia of
                                     the drop of water will continue to maintain its
                                     motion.
  Ways to reduce the negative    1. Safety in a car:
  effects of inertia                 (a)Safety belt secure the driver to their seats.
                                         When the car stops suddenly, the seat belt provides
                                         the external force that prevents the driver from
                                         being thrown forward.
                                     (b)Headrest to prevent injuries to the neck during rear-
                                         end collisions. The inertia of the head tends to
                                         keep in its state of rest when
                                         the body is moved suddenly.
                                     (c)An air bag is fitted inside the steering wheel.
                                         It provides a cushion to prevent the driver from
                                         hitting the steering wheel or dashboard during a
                                         collision.


                                 2. Furniture carried by a lorry normally are tied up together by
                                    string.

                                     When the lorry starts to move suddenly, the furniture are
                                     more difficult to fall off due to their inertia because
                                     their combined mass has increased.




                                            2-11
Physics Module Form 4         Chapter 2 – Forces & Motion                                 GCKL 2011



                                  •     Two empty buckets which are hung with rope from the
  Relationship between mass             ceiling.
  and inertia                     •     One bucket is filled with sand while the other bucket is
                                        empty.
                                  •     Then, both pails are pushed.
                                  •     It is found that




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

                                      • When both buckets are oscillating and an attempt is made
                                        to stop them, the bucket filled with sand offers

                              




                                               2-12
Physics Module Form 4                  Chapter 2 – Forces & Motion                              GCKL 2011



   2.4         MOMENTUM


 Definition                                              Momentum =
                                                         SI unit:

 Principle of                                            In the absence of an external force,
 Conservation of
 Momentum
                   Elastic Collision                                   Inelastic collision




    ƒ   Both objects move                                    ƒ The two objects

    ƒ   Momentum                                             ƒ   Momentum
    ƒ   Kinetic energy                                       ƒ   Kinetic energy.
        Total energy                                         ƒ   Total energy




 Total Momentum Before =                                 Total Momentum Before =

         m1u1 + m2u2 = m1 v1 + m2 v2                             m1 u1 + m2 u2 = ( m1 + m2 ) v

  Explosion
                    Paste a picture                      Before explosion both object




                                                         Total Momentum         Total Momentum after
                                                         before collision is    collision :
                                                          zero                       m1v1 + m2v2



                                                  2-13
Physics Module Form 4              Chapter 2 – Forces & Motion                            GCKL 2011



                                                     From the law of conservation of momentum:
                                                     Total Momentum = Total Momentum
                                                      Before collision               after collision
                                                                       0 = m1v1 + m2v2
                                                                    m1v1 = - m2v2

                                                     Negative sign means



    EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum)
              Paste a picture                  When a rifle is fired, the bullet of mass m,
                                                         moves with a high velocity, v. This creates a
                                                         momentum in the forward direction.
                                                            From the principle of conservation of
                                                         momentum,




                 Paste a picture                     Application in the jet engine:




                                                     The launching of rocket
                                                            Mixture of hydrogen and oxygen fuels



                                                        These high speed hot gases produce

                                                         By conservation of momentum,




                                              2-14
Physics Module Form 4                Chapter 2 – Forces & Motion                                 GCKL 2011



                   Paste a picture                          In a swamp area, a fan boat is used.
                                                             The fan produces a high speed movement of air
                                                            backward. This produces a large momentum
                                                            backward.
                                                            By conservation of momentum, an equal but opposite
                                                            momentum is produced and acted on the boat. So the
                                                            boat will move forward.


                   Paste a picture                          A squid propels by expelling water at high velocity.
                                                            Water enters through a large opening and exits
                                                            through a small tube. The water is forced out at a
                                                            high speed backward.
                                                                   Total Mom. before= Total Mom. after
                                                                            0 =Mom water + Mom squid
                                                                                 0 = mwvw + msvs
                                                                           -mwvw = msvs
                                                            The magnitude of the momentum of water and
                                                            squid are equal but opposite direction.
                                                            This causes the quid to jet forward.

 Example




                                                            Example

                                                            Before collision              After collision
 Car A of mass 1000 kg moving at 20 ms -1
                                                            MA = 4 kg
 collides with a car B of mass 1200 kg moving at            MB = 2 kg
 10 m s -1 in same direction. If the car B is               UA = 10 ms -1 r i g h t
 shunted forwards at 15 m s -1 by the impact,               UB = 8 ms -1 l e f t            VB 4 ms-1 right
 what is the velocity, v, of the car A immediately
 after the crash?                                           Calculate the value of VA .




                                                     2-15
Physics Module Form 4                 Chapter 2 – Forces & Motion                                  GCKL 2011



 Example




                                                             Example
 A truck of mass 1200 kg moving at
 30 ms-1 collides with a car of mass
 1000 kg which is travelling in the opposite                 A man fires a pistol which has a mass of 1.5 kg.
 direction at 20 ms-1. After the collision, the two          If the mass of the bullet is 10 g and it reaches a
 vehicles move together. What is the velocity of             velocity of 300 ms -1 after shooting, what is the
 both vehicles immediately after collision?                  recoil velocity of the pistol?




                                                      2-16
Physics Module Form 4              Chapter 2 – Forces & Motion                         GCKL 2011



  2.5            FORCE

  Balanced Force                             Example:
  When the forces acting on an object are
  balanced,



  Effect : the object
   [velocity     ]
   or
   moves
    [a= ]




  Unbalanced Force/ Resultant Force          When the forces acting on an object are not balanced,
                                             there must be
                                             The net force is known as

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


                                             Effect : Can cause a body to
                                                 -




  Newton’s Second Law of Motion              The acceleration produced by a force on an object is




                                             Force = Mass x Acceleration

                                             F = ma




                                              2-17
Physics Module Form 4             Chapter 2 – Forces & Motion                               GCKL 2011



  Experiment to Find The Relationship between Force, Mass & Acceleration

  Relationship                           a& F                                        a&
  between                                                                             m
  Situation




                        Both men are pushing the same mass         Both men exerted the same strength.
                        but man A puts greater effort. So he       But man B moves faster than man A.
                        moves faster.

  Inference             The acceleration produced by an
                        object depends on the net force
                        applied to it.



  Hypothesis            The acceleration of the object
                        increases when the force applied
                        increases



  Variables:
  Manipulated :         Force
  Responding :
                        Acceleration
  Constant :            Mass

  Apparatus and         Ticker tape and elastic cords, ticker timer, trolleys, power supply and friction
  Material              compensated runway and meter ruler.




                                                2-18
Physics Module Form 4              Chapter 2 – Forces & Motion                               GCKL 2011



     Procedure :        An elastic cord is hooked over the       An elastic cord is hooked over a
  - Controlling         trolley. The elastic cord is stretched   trolley.
    manipulated         until the end of the trolley. The
      variables.        trolley is pulled down the runway
                        with the elastic cord being kept
                        stretched by the same amount of
                        force

  -Controlling          Determine the acceleration by            Determine the acceleration by analyzing
    responding          analyzing the ticker tape.               the ticker tape.
    variables.          Acceleration
                                                                                      v u
                                               v u                Acceleration a 
                           Acceleration     a                                          t
                                                 t
  -
                        Repeat the experiment by using two       Repeat the experiment by
  Repeating
    experiment.         , three, four and five elastic cords




  Tabulation of          Force, F/No of     Acceleration, a/ ms-2 Mass, m/   Mass,    1/m,    Acceleration/
                          elastic cord                              no of    m/g       g-1        ms-2
  data                         1                                  trolleys
                               2                                     1
                               3                                     2
                               4                                     3
                               5                                     4
                                                                     5

  Analysing
  Result




                                                2-19
Physics Module Form 4               Chapter 2 – Forces & Motion                            GCKL 2011



 1. What force is required to move a 2 kg object
                                   2                     2. Ali applies a force of 50 N to move a 10 kg
    with an acceleration of 3 m s- , if                      table at a constant velocity. What is the
    (a) the object is on a smooth surface?                   frictional force acting on the table?
    (b) The object is on a surface where the
         average force of friction acting on the
         object is 2 N?




   3. A car of mass 1200 kg travelling at 20 ms -1      4. Which of the following systems will
      is brought to rest over a distance of 30 m.          produce maximum acceleration?
      Find
      (a) the average deceleration,
      (b) the average braking force.




                                                 2-20
Physics Module Form 4                 Chapter 2 – Forces & Motion                               GCKL 2011


 2.6              IMPULSE AND IMPULSIVE FORCE

 Impulse                            The change of                                  m=
                                    Unit :                                         u =
                                                                                   v =
 Impulsive Force                   The rate of change                              t=

                                   change of momentum mv  mu
                                                     
                                           time          t
                                     Unit =

 Effect of time                     Impulsive force     Longer period of time →Impulsive force
                                    is
                                                        Shorter period of time →


                  Situations for Reducing Impulsive Force in Sports

       Situations                                                 Explanation

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




                                    Goal keepers will wear gloves to




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

                                    so as to




                                    A baseball player must catch the ball in the direction of the motion of
                                    the ball. Moving his hand backwards when catching the ball prolongs
                                    the time for the momentum to change so as to reduce the impulsive
                                    force.




                                                    2-21
Physics Module Form 4                 Chapter 2 – Forces & Motion                             GCKL 2011



                        Situation of Increasing Impulsive Force
        Situations                                                   Explanation
                                   A karate expert can break a thick wooden slab with his bare hand
                                   that moves at a very fast speed. The short impact time results in




                                   A massive hammer head moving at a fast speed is brought to rest
                                   upon hitting the nail within a short time interval.




                                   A football must have enough air pressure in it so




                                   Pestle and mortar are made of stone. When a pestle is used to pound
                                   chillies the hard surfaces of both the pestle and mortar cause the pestle
                                   to be stopped in a very short time. A large impulsive force is resulted
                                   and thus causes these spices to be crushed easily.



 Example 1
 A 60 kg resident jumps from the first floor of a burning house.
 His velocity just before landing on the ground is 6 ms-1.
 (a) Calculate the impulse when his legs hit the ground.
 (b) What is the impulsive force on the resident’s legs if he
       bends upon landing and takes 0.5s to stop?
 (c) What is the impulsive force on the resident’s legs if
     he does not bend and stops in 0.05 s?
 (d) What is the advantage of bending his legs upon landing?


 Example 2
 Rooney kicks a ball with a force of 1500 N. The time of
 contact of his boot with the ball is 0.01 s. What is the impulse
 delivered to the ball? If the mass of the ball is 0.5 kg, what is
 the velocity of the ball?




                                                   2-22
Physics Module Form 4             Chapter 2 – Forces & Motion                               GCKL 2011



   2.7         SAFETY VEHICLE


  Safety features in vehicles




Component                                                  Function
Headrest
Air bag



Windscreen

Crumple zone



Front              Absorb the shock from the accident. Made from steel, aluminium, plastic or
bumper             rubber.
ABS                Enables drivers to quickly stop the car without causing the brakes to lock.

Side impact bar



Seat belt




                                                    2-23
Physics Module Form 4                  Chapter 2 – Forces & Motion                               GCKL 2011




   2.8            GRAVITY



  Gravitational
  Force                 Objects fall because they are


                        This force is known as the


                        The earth’s gravitational force




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

                                  In vacuum,
                                  They fall with




  Acceleration due to           Objects dropped
  gravity, g
                               




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

                        The gravitational field strength is defined as the gravitational force which acts
                        on a mass of 1 kilogram.

                                   F                        -1
                        g=                  Its unit is N kg .
                                   m




                                                     2-24
Physics Module Form 4               Chapter 2 – Forces & Motion                              GCKL 2011



                                                                      -1
                            Gravitational field strength, g = 10 N kg
                                                                    -2
                            Acceleration due to gravity, g = 10 m s

                                                                                                         -2
                            The approximate value of g can therefore be written either as       10 m s
                                         -1
                            or as 10 N kg .

  Weight                    The gravitational force acting on the object.
                            Weight = mass x gravitational acceleration
                            W = mg            SI unit : Newton, N and it is a vector quantity

  Comparison                                Mass                                       Weight
  between weight            The mass of an object is the            The weight of an object is the force of
  &                         amount of matter in the object          gravity acting on the object.
  mass
                            Constant everywhere                     Varies with the magnitude of gravitational
                                                                    field strength, g of the location

                            A scalar quantity                       A vector quantity
                            A base quantity                         A derived quantity
                            SI unit: kg                             SI unit : Newton, N
  The difference
  between a
  fall in air and
  a free fall in a vacuum
  of a coin and a
  feather.

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

                            At vacuum state: There is no air         At normal state: Both coin and feather
                            resistance.                              will fall because of gravitational force.
                            The coin and the feather will fall       Air resistance effected by the surface area of
                            freely.                                  a fallen object.
                            Only gravitational force                 The feather that has large area will have
                            acted on the objects. Both will fall     more air resistance.
                            at the same time.                        The coin will fall at first.




                                                 2-25
Physics Module Form 4                 Chapter 2 – Forces & Motion                          GCKL 2011




                           (a) The two spheres are falling       The two spheres are falling down with
                           with an acceleration.                 the same acceleration
                           The distance between two              The two spheres are at the same level
                           successive images of the sphere       at all times. Thus, a heavy object and
                           increases showing that the two        a light object fall with the same
                           spheres are falling with increasing   gravitational acceleration
                           velocity; falling with an             Gravitational acceleration is
                           acceleration.                         independent of mass


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



  Motion graph for free fall object
        Free fall object                 Object thrown upward                 Object thrown upward and fall




     Example 1
  A coconut takes 2.0 s to fall to the ground. What
  is
      (a) its speed when it strikes the ground
      (b) ) the height of the coconut tree




                                                 2-26
Physics Module Form 4            Chapter 2 – Forces & Motion                   GCKL 2011



   2.9          FORCES IN EQUILIBRIUM

  Forces in                When an object is in equilibrium,
  Equilibrium




            rd
  Newton’s 3 Law


                   Examples( Label the forces acted on the objects)


      Paste more picture




                                                                      Paste more picture




  Resultant                    A single force that
  Force



  Addition of Forces



                                                 Resultant force, F =         +


                                                 Resultant force, F =         +




                                              2-27
Physics Module Form 4               Chapter 2 – Forces & Motion                          GCKL 2011




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




  STEP 1 : Using ruler and protractor, draw      STEP 3
  the two forces F1 and F2 from a point.         Draw the diagonal of the parallelogram. The
                                                 diagonal represent the resultant force, F in
                                                 magnitude and direction.




  STEP 2
  Complete the parallelogram


                                                 scale: 1 cm = ……




  Resolution of Forces                           A force F can be resolved into components
                                                 which are perpendicular to each other:
                                                 (a) horizontal component , FX
                                                 (b) vertical component, FY


                                                 Inclined Plane




  Fx = F cos θ
                                                 Component of weight parallel to the plane = mg sin θ
                                                  Component of weight normal to the plane = mg cos θ
  Fy = F sin θ



                                                2-28
Physics Module Form 4        Chapter 2 – Forces & Motion   GCKL 2011




  Find the resultant force




  (d)                                             (e)




                                        2-29
Physics Module Form 4             Chapter 2 – Forces & Motion                 GCKL 2011



Lift

       Stationary Lift             Lift accelerate upward            Lift accelerate downward




 Resultant Force =                Resultant Force =             Resultant Force =

 The reading of weighing          The reading of weighing       The reading of weighing
 scale =                          scale =                       scale =




              Pulley




 1. Find the resultant force, F



 2. Find the moving mass, m



 3. Find the acceleration, a




 4. Find string tension, T




                                             2-30
Physics Module Form 4                  Chapter 2 – Forces & Motion                           GCKL 2011




  2.10           WORK, ENERGY, POWER & EFFICIENCY



 Work                                         Work done is



                                              W = Fs         W=      ,F=        s=




 The SI unit of work is the joule, J
                                              1 joule of work is done when




                                                                              The displacement , s of the
      The displacement, s of the object is in the direction of the force, F   object is not in the direction of
                                                                              the force, F
W = Fs

                                                 s           F




                                        W= F s

 Example 1                              Example 2                             Example 3
 A boy pushing his bicycle with a       A girl is lifting up a 3 kg           A man is pulling a crate of fish
 force of 25 N through a distance       flower pot steadily to a height       along theW = (Fwith θ)force of
                                                                                       floor cos a s
 of 3 m.                                of 0.4 m.                             40 N through a distance of 6 m.




 Calculate the work done by the                                               What is the work done
 boy.                                   What is the work done by the          in pulling the crate?
                                        girl?



                                                     2-31
Physics Module Form 4     Chapter 2 – Forces & Motion                             GCKL 2011




      Concept                                D                           Formula & Unit
  Power                                      ef
                             The rate at which work is                   W
                                             in                    P=
                             done,or
                                                                        t
                                             iti                   p = power, W = work / energy
                                             on                     t = time
  Energy                           Energy is the capacity to do work.
                                




  Potential Energy                                                m=
                             Gravitational potential energy is    h=
                             the energy of an object due to
                                                                  g=
                             its higher position in the
                             gravitational field.                 E=




  Kinetic Energy             Kinetic energy is the energy of an   m=
                             object due to its motion.            v=


                                                                  E=

                                                                  No force is applied on the object
                                                                  in the direction of displacement
                                                                  (the object moves because of its
                                                                  own inertia)
                                                                  A satellite orbiting in space.
                                                                  There is no friction in space. No
                                                                  force is acting in the direction of
                             A waiter is carrying a tray of       movement of the satellite.
  No work is done when:      food. The direction of motion of
                             the object is perpendicular to
                             that of the applied force.




                             A waiter is carrying a tray of
                             food and2-32
                                      walking
Physics Module Form 4                 Chapter 2 – Forces & Motion                             GCKL 2011




    Principle of Conservation of         Energy can be changed from one form to another, but it cannot
    Energy
                                         be created or destroyed.
                                         The energy can be transformed from




                                         Example 4
                                         A worker is pulling a wooden block of weight, W, with a force
                                         of P along a frictionless plank at height of h. The distance
                                         travelled by the block is x. Calculate the work done by the
                                         worker to pull the block.




  Example 5
  A student of mass m is climbing
  up a flight of stairs which has
  the height of h. He takes t
  seconds..
  What is the power of the student?




  Example 6                              Example 7
  A stone is thrown upward with
  initial velocity of
         -1
  20 ms . What is the maximum
  height which can be reached by
  the stone?




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




                                                  2-33
Physics Module Form 4                 Chapter 2 – Forces & Motion                     GCKL 2011




  Example 8

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




  Example 9

 A ball moves upwards along a
 frictionless track of height 1.5 m
                          -1
 with a velocity of 6 ms . What is
 its velocity at point B?




  Example 10




  A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the
  slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope?




                                                 2-34
Physics Module Form 4          Chapter 2 – Forces & Motion                   GCKL 2011



 2.11         ELASTICITY


Elasticity                         A property of matter that enables an object




                                   No external force is applied.
                                   Molecules are at their equilibrium separation.
                                   Intermolecular force is equal zero.




                                   Compressing a solid causes its molecules

                                   Repulsive intermolecular force




                                   Stretching a solid




                                           Its molecules are
Stretching a wire by an external
force:
                                   When the external force is removed:
                                      The attractive intermolecular forces




                                                   2-35
Physics Module Form 4         Chapter 2 – Forces & Motion                                GCKL 2011



Hooke’s Law                       The extension of a spring




                                  F= k x       where
                                              F=
                                              x=
                                              k=




Force extension graph             Based on the graph:
                                   Relationship between F & x :

                                  The gradient of the graph represent =

                                  Area under the graph


                                                                                  2
                                     = elastic potential energy = ½ F x = ½ k x



The elastic limit of a spring     The maximum force that



                                  If a force stretches a spring beyond its elastic limit, the spring




Force constant of the spring, k   The force required to produce one unit of extension of the
                                  spring.




                                  k is a measurement of the stiffness of the spring
                                        The spring with a larger force constant is
                                        A spring with a smaller force constant is




                                                  2-36
Physics Module Form 4              Chapter 2 – Forces & Motion                              GCKL 2011



Factors that effect elasticity
           Factor                       Change in factor                 How does it affects the elasticity

           Length                       Shorter spring
                                        Longer spring
           Diameter of spring wire      Smaller diameter
                                        Larger diameter
           Diameter spring              Smaller diameter
                                        Larger diameter
           Type of material             Springs made of different materials
                                        Elasticity changes according to the type of material



                                     Arrangement of the spring
                       In series                                                 In parallel




The same load is applied to each spring.                 The load is shared equally among the springs.
Tension in each spring = W Extension of                                             W
                                                         Tension in each spring =
each spring = x                                                                     2
Total extension = 2x                                                                  x
                                                         Extension of each spring =
If n springs are used: The total                                                      2
extension = n x                                          If n springs are used:
                                                                                  x
                                                         The total extension =
                                                                                  n

Example 1
The original length of each
spring is 10 cm.
With a load of 10 g, the extension
of each spring is 2 cm.
What is the length of the spring
system for (a),
(b) and (c)?




                                                  2-57
Physics Module Form 4               Chapter 2 – Forces & Motion                             GCKL 2011



 Example 2                                                             Using the diagram shown and the
                                                                      information given about the weight of
 Diagram below represent a                                            the two objects,
 50 cent coin and a leaf falling
 in a vacuum container.                                                compare the mass of the coin and the
 The coin is heavier than the                                         leaf, the time taken to fall, the position
 leaf.                                                                of the coin and the leaf and
                                                                      finally deduce the physical quantity
                                                                      which causes the objects to fall.


                                                 coin                                    leaf

Mass

of the taken to fall in a vacuum
Time coin

Position of the coin and the leaf


 Coin and leaf of different mass reach the bottom of the container at the same time.
Coin and leaf fall down due to gravitational force. The magnitude of gravitational pull is constant.
It does not depend on the mass


Example 3       Diagram 10 shows a student trying to launch a water rocket.
                You are required to give suggestion on how to design a water rocket for National
                Competition. Based on your knowledge on forces, motion and properties of materials,
                explain your suggestion based on the following aspect:
                (i)     material used
                (ii)    shape of the rocket
                (iii)    suitable angle to launch the rocket
                (iv)    volume of water in the rocket
                (v)     added structure for the motion of the rocket




                                                     2-58
Physics Module Form 4                  Chapter 2 – Forces & Motion                              GCKL 2011



Water Rocket

                 Aspect                              Structure                              Explanation

Acceration of the Rocket                Make from light material



Structure of the Rocket                 Aerodynamic



Upthrust force                          Fill with erated water or gasy
                                        drink with water


Stability of the rocket during flight   Use plasticine to make the head
                                        of the rocket

                                        Add fins at the back portion of
                                        the rocket

Example 4




                    Diagram 4.1 shows a cradle with a baby in it is oscillating vertically. Diagram 4.2
                    shows another identical cradle with a heavier baby in it is oscillating vertically. It
                    is observed that the cradle with a heavier mass baby oscillates at a higher
                    frequency.

                    Design an experiment to test the hypothesis using spring, slotted weight and other
                    suitable apparatus.

    4 (a)        Inference : The extension of the spring depends

                                                                                                                  1

      (b)        Hypothesis : As the                                                                              1

    c (i)        Aim : To investigate the relationship between                                                    1

                 Variables :

                                                          2-59
Physics Module Form 4             Chapter 2 – Forces & Motion                              GCKL 2011



c(ii)    Manipulated variable :                                                                          1

         Responding variable :

         Constant variable :                                                                             1

c(iii)   Apparatus : Metre rule, retort stand with clamp ,steel spring, slotted weight and               1
         pin.

           Set-up the apparatus



                                                                                                         1



c(iv)




c (v)    Method of controlling the manipulated variables :                                               1

         1. Arrangement the apparatus as shown in the diagram.

         2.Mark the initial




         Method of measuring the responding variables :

         Record                                                                                          1

         Extension of spring :

            Measure the

         Repeat the                                                                                      1




                                                2-60
Physics Module Form 4               Chapter 2 – Forces & Motion               GCKL 2011



c ( vi)   Tabulate Results                                                                   1

          Initial length , l0 =   cm

            Mass of the Slotted weight , m / g       40        80    120   160   200

            Weight of slotted mass / N               0.4      0.8    1.2   1.6    2.0

            Length of the spring ,l / cm

            Extension of the spring x = l- l0




(vii)                                                                                        1




                                                                                 Total   12




                                                  2-61

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2.0 forces and motion

  • 1. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.1 LINEAR MOTIONs Physical Quantity Definition, Quantity, Symbol and unit Distance is the …… Distance, s Quantity: … SI unit : .. (a) The distance in .. …. (b) the distance between …. Displacement, s ….direction. (c) The distance of its final ….. specified …. Quantity: SI unit: Speed is the Speed,v Speed = Quantity: SI unit: Velocity is the Velocity, v Velocity = Direction of velocity is Quantity : SI unit: Average speed v= Example: A car moves at an average speed / velocity of 20 ms -1 On average, the car moves a distance/ displacement of Average velocity Displacement v TotalTime 2-1
  • 2. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Uniform speed Speed that remains the same in Uniform velocity Velocity that remains An object has a non- (a) The direction of motion changes or the motion is not linear. uniform velocity if (b) The magnitude of its velocity changes. Acceleration, a When the velocity of an object v u Acceleration is defined as the a t Change in velocity Acceleration= Time taken Final velocity,v - Initial velocity,u Unit: ms-2 = Time taken,t The velocity of an object increases from an initial velocity, u, to a higher final velocity, v Acceleration is positive Deceleration acceleration is negative. The rate of decrease in speed in a specified direction. Zero acceleration An object moving at a constants velocity, that is, Constant acceleration Velocity increases at a uniform rate. When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity 1. Constant = 2. increasing velocity = 3. decreasing velocity = 4. zero velocity = 5. negative velocity = object moves at opposite direction 6. zero acceleration = 7. negative acceleration = deceleration 2-2
  • 3. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Comparisons between distance and displacement Comparisons between speed and velocity Distance Displacement Speed Velocity Total path length The distance between The rate of change of The rate of change of travelled from two locations distance displacement one location to measured along the Scalar quantity Vector quantity another shortest path connecting them in specific direction Scalar quantity It has magnitude but no direction SI unit SI unit : Fill in the blanks: 1. A steady speed of 10 ms -1 = A distance of __________________________________________ 2. A steady velocity of -10 ms -1 = A displacement of _________________________________ 3. A steady acceleration of 4 ms -2 = Speed _____________________________________________ 4. A steady deceleration of 4 ms -2 = ____________________________________________________________ 5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right. 2-3
  • 4. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Example 1 Example 2 Every day Rahim walks from his house to the junction Every morning Amirul walks to Ahmad’s house which is 1.5km from his house. which is situated 80 m to the east of Amirul’s house. Then he turns back and stops at warung Pak Din which is They then walk towards their school which is 60 m 0.5 km from his house. to the south of Ahmad’s house. (a) What is the distance travelled by Amirul and his displacement from his house? (b) If the total time taken by Amirul to travel from his house to Ahmad’s house and then to school is 15 minutes, what is his speed and velocity? Speed = (a) What is Rahim’s displacement from his house, • when he reaches the junction. Velocity = • when he is at warung Pak Din. (b) After, Rahim walks back to his house. breakfast When he reaches home, (i) What is the total distance travelled by Rahim? (ii) What is Rahim’s total displacement from his house? Example 3 Example 4 Salim running in a race covers 60 m in 12 s. An aeroplane flies towards the north with a (a) What is his speed in ms-1 velocity 300 km hr -1 in one hour. Then, the plane moves to the east with the velocity 400 km hr -1 in one hour. (b) If he takes 40 s to complete the race, what is his (a) What is the average speed of the plane? distance covered? (b) What is the average velocity of the plane? 2-4
  • 5. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 (c) What is the difference between average speed and average velocity of the plane? Example 5 The speedometer reading for a car travelling due north shows 80 km hr -1. Another car travelling at 80 km hr -1 towards south. Is the speed of both cars same? Is the velocity of both cars same? A ticker timer  Use:  1 tick = time interval  The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2 consecutive dots is  1 tick = 2-5
  • 6. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Relating displacement, velocity, acceleration and time using ticker tape. VELOCITY FORMULA Time, t = 10 dicks x 0.02 s = 0.2 s displacement, s = x cm velocity = ACCELERATION Initial velocity, u = final velocity, v = acceleration, a = Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or t = (50 – 10) ticks x 0.02 s = 0.8 s TICKER TAPE AND CHARTS TYPE OF MOTION  Distance between the dots increases uniformly  2-6
  • 7. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 - Distance between the dots decrease uniformly Example 6 The diagram above shows a ticker tape chart for a moving trolley. The frequency of the ticker-timer used is 50 Hz. Each section has 10 dots-spacing. (a) What is the time between two dots? (b) What is the time for one strips? (c) What is the initial velocity? (d) What is the final velocity? (e) What is the time interval to change from initial velocity to final velocity? (f) What is the acceleration of the object? vu 2 a= t THE EQUATIONS OF MOTION v  u  at u= 1 v= s  ut  at 2 t= 2 s= v  u  2as 2 2 a= 2-7
  • 8. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.2 MOTION GRAPHS DISPLACEMENT – TIME GRAPH Velocity is obtained from A – B : gradient of the graph is B – C : gradient of the graph = object is C – D : gradient of the graph The object VELOCITY-TIME GRAPH Area below graph Positive gradient Negative gradient Zero gradient GRAPH s versus t v versus t a versus t Zero velocity Negative constant velocity Positive Constant velocity 2-8
  • 9. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 GRAPH s versus t v versus t a versus t Constant acceleration Constant deceleration Example 1: Example 2: velocity/ m s-1 20 10 time/ 0 10 20 30 40 s Based on the s-t graph above: (a) Calculate the acceleration at: (a) Calculate the velocity at (ii) JK (ii) KL (iii) LM (i) AB (ii) BC (iii) CD (b) Describe the motion of the object at: (b) Describe the motion of the object at: (i) AB (ii) BC (iii) CD (ii) JK (ii) KL (iii) LM (c) Find (c) Calculate (i) total distance (iii) The total displacement (ii) total displacement (iv) The average velocity (d) Calculate (i) The average speed (ii) The average velocity of the moving particle 2-9
  • 10. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.3 INERTIA Inertia The inertia of an object is the tendency of the object Newton’s first law Every object Relation between inertia The larger the mass, and mass SITUATIONS INVOLVING INERTIA SITUATION EXPLANATION EEEEEEEEJNVJLKN drops straight into When the cardboard is pulled away quickly, the coin the glass. DNFLJKVNDFLKJNB VJKL;DFN BLK;XC NB[F NDPnDSFJ[POJDE]O- JBD]AOP[FKBOP[DF LMB NOPGFMB LKFGNKLB FGNMNKL’ MCVL Paste a picture Chilli sauce in the bottle can be BNM’CXLB out if the bottle is moved easily poured NFGNKEPLANATION down fast with a sudden stop. The sauce inside the bottle moves together with the bottle. When the bottle stops suddenly, Paste a picture Body moves forward when the car stops suddenly The passengers were in a state of motion when the car was moving. When the car stopped suddenly, Paste a picture A boy runs away from a cow in a zig zag motion. The cow has a large inertia 2-10
  • 11. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011  The head of hammer is secured tightly to its handle by knocking one end of the handle, held vertically, on a hard surface. This causes the hammer head to continue on its downward motion when the handle has been stopped, so that the top end of the handle is slotted deeper into the hammer head. • The drop of water on a wet umbrella will fall when the boy rotates the umbrella. • This is because the drop of water on the surface of the umbrella moves simultaneously as the umbrella is rotated. • When the umbrella stops rotating, the inertia of the drop of water will continue to maintain its motion. Ways to reduce the negative 1. Safety in a car: effects of inertia (a)Safety belt secure the driver to their seats. When the car stops suddenly, the seat belt provides the external force that prevents the driver from being thrown forward. (b)Headrest to prevent injuries to the neck during rear- end collisions. The inertia of the head tends to keep in its state of rest when the body is moved suddenly. (c)An air bag is fitted inside the steering wheel. It provides a cushion to prevent the driver from hitting the steering wheel or dashboard during a collision. 2. Furniture carried by a lorry normally are tied up together by string. When the lorry starts to move suddenly, the furniture are more difficult to fall off due to their inertia because their combined mass has increased. 2-11
  • 12. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 • Two empty buckets which are hung with rope from the Relationship between mass ceiling. and inertia • One bucket is filled with sand while the other bucket is empty. • Then, both pails are pushed. • It is found that Push and compared to the bucket with sand. • The bucket filled with sand offers more resistance to movement. • When both buckets are oscillating and an attempt is made to stop them, the bucket filled with sand offers  2-12
  • 13. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.4 MOMENTUM Definition Momentum = SI unit: Principle of In the absence of an external force, Conservation of Momentum Elastic Collision Inelastic collision ƒ Both objects move ƒ The two objects ƒ Momentum ƒ Momentum ƒ Kinetic energy ƒ Kinetic energy. Total energy ƒ Total energy Total Momentum Before = Total Momentum Before = m1u1 + m2u2 = m1 v1 + m2 v2 m1 u1 + m2 u2 = ( m1 + m2 ) v Explosion Paste a picture Before explosion both object Total Momentum Total Momentum after before collision is collision : zero m1v1 + m2v2 2-13
  • 14. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 From the law of conservation of momentum: Total Momentum = Total Momentum Before collision after collision 0 = m1v1 + m2v2 m1v1 = - m2v2 Negative sign means EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum) Paste a picture When a rifle is fired, the bullet of mass m, moves with a high velocity, v. This creates a momentum in the forward direction. From the principle of conservation of momentum, Paste a picture Application in the jet engine: The launching of rocket Mixture of hydrogen and oxygen fuels These high speed hot gases produce By conservation of momentum, 2-14
  • 15. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Paste a picture In a swamp area, a fan boat is used. The fan produces a high speed movement of air backward. This produces a large momentum backward. By conservation of momentum, an equal but opposite momentum is produced and acted on the boat. So the boat will move forward. Paste a picture A squid propels by expelling water at high velocity. Water enters through a large opening and exits through a small tube. The water is forced out at a high speed backward. Total Mom. before= Total Mom. after 0 =Mom water + Mom squid 0 = mwvw + msvs -mwvw = msvs The magnitude of the momentum of water and squid are equal but opposite direction. This causes the quid to jet forward. Example Example Before collision After collision Car A of mass 1000 kg moving at 20 ms -1 MA = 4 kg collides with a car B of mass 1200 kg moving at MB = 2 kg 10 m s -1 in same direction. If the car B is UA = 10 ms -1 r i g h t shunted forwards at 15 m s -1 by the impact, UB = 8 ms -1 l e f t VB 4 ms-1 right what is the velocity, v, of the car A immediately after the crash? Calculate the value of VA . 2-15
  • 16. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Example Example A truck of mass 1200 kg moving at 30 ms-1 collides with a car of mass 1000 kg which is travelling in the opposite A man fires a pistol which has a mass of 1.5 kg. direction at 20 ms-1. After the collision, the two If the mass of the bullet is 10 g and it reaches a vehicles move together. What is the velocity of velocity of 300 ms -1 after shooting, what is the both vehicles immediately after collision? recoil velocity of the pistol? 2-16
  • 17. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.5 FORCE Balanced Force Example: When the forces acting on an object are balanced, Effect : the object [velocity ] or moves [a= ] Unbalanced Force/ Resultant Force When the forces acting on an object are not balanced, there must be The net force is known as Weight, W = Lift, U Thrust, F = drag, G Effect : Can cause a body to - Newton’s Second Law of Motion The acceleration produced by a force on an object is Force = Mass x Acceleration F = ma 2-17
  • 18. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Experiment to Find The Relationship between Force, Mass & Acceleration Relationship a& F a& between m Situation Both men are pushing the same mass Both men exerted the same strength. but man A puts greater effort. So he But man B moves faster than man A. moves faster. Inference The acceleration produced by an object depends on the net force applied to it. Hypothesis The acceleration of the object increases when the force applied increases Variables: Manipulated : Force Responding : Acceleration Constant : Mass Apparatus and Ticker tape and elastic cords, ticker timer, trolleys, power supply and friction Material compensated runway and meter ruler. 2-18
  • 19. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Procedure : An elastic cord is hooked over the An elastic cord is hooked over a - Controlling trolley. The elastic cord is stretched trolley. manipulated until the end of the trolley. The variables. trolley is pulled down the runway with the elastic cord being kept stretched by the same amount of force -Controlling Determine the acceleration by Determine the acceleration by analyzing responding analyzing the ticker tape. the ticker tape. variables. Acceleration v u v u Acceleration a  Acceleration a t t - Repeat the experiment by using two Repeat the experiment by Repeating experiment. , three, four and five elastic cords Tabulation of Force, F/No of Acceleration, a/ ms-2 Mass, m/ Mass, 1/m, Acceleration/ elastic cord no of m/g g-1 ms-2 data 1 trolleys 2 1 3 2 4 3 5 4 5 Analysing Result 2-19
  • 20. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 1. What force is required to move a 2 kg object 2 2. Ali applies a force of 50 N to move a 10 kg with an acceleration of 3 m s- , if table at a constant velocity. What is the (a) the object is on a smooth surface? frictional force acting on the table? (b) The object is on a surface where the average force of friction acting on the object is 2 N? 3. A car of mass 1200 kg travelling at 20 ms -1 4. Which of the following systems will is brought to rest over a distance of 30 m. produce maximum acceleration? Find (a) the average deceleration, (b) the average braking force. 2-20
  • 21. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.6 IMPULSE AND IMPULSIVE FORCE Impulse The change of m= Unit : u = v = Impulsive Force The rate of change t= change of momentum mv  mu  time t Unit = Effect of time Impulsive force Longer period of time →Impulsive force is Shorter period of time → Situations for Reducing Impulsive Force in Sports Situations Explanation Thick mattress with soft surfaces are used in events such as high jump so that Goal keepers will wear gloves to A high jumper will bend his legs upon landing. This is to so as to A baseball player must catch the ball in the direction of the motion of the ball. Moving his hand backwards when catching the ball prolongs the time for the momentum to change so as to reduce the impulsive force. 2-21
  • 22. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Situation of Increasing Impulsive Force Situations Explanation A karate expert can break a thick wooden slab with his bare hand that moves at a very fast speed. The short impact time results in A massive hammer head moving at a fast speed is brought to rest upon hitting the nail within a short time interval. A football must have enough air pressure in it so Pestle and mortar are made of stone. When a pestle is used to pound chillies the hard surfaces of both the pestle and mortar cause the pestle to be stopped in a very short time. A large impulsive force is resulted and thus causes these spices to be crushed easily. Example 1 A 60 kg resident jumps from the first floor of a burning house. His velocity just before landing on the ground is 6 ms-1. (a) Calculate the impulse when his legs hit the ground. (b) What is the impulsive force on the resident’s legs if he bends upon landing and takes 0.5s to stop? (c) What is the impulsive force on the resident’s legs if he does not bend and stops in 0.05 s? (d) What is the advantage of bending his legs upon landing? Example 2 Rooney kicks a ball with a force of 1500 N. The time of contact of his boot with the ball is 0.01 s. What is the impulse delivered to the ball? If the mass of the ball is 0.5 kg, what is the velocity of the ball? 2-22
  • 23. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.7 SAFETY VEHICLE Safety features in vehicles Component Function Headrest Air bag Windscreen Crumple zone Front Absorb the shock from the accident. Made from steel, aluminium, plastic or bumper rubber. ABS Enables drivers to quickly stop the car without causing the brakes to lock. Side impact bar Seat belt 2-23
  • 24. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.8 GRAVITY Gravitational Force Objects fall because they are This force is known as the The earth’s gravitational force Free fall  An object is falling freely when it is falling under the force of gravity only.   An object falls freely only  In vacuum,  They fall with Acceleration due to  Objects dropped gravity, g  Gravitational field The gravitational field is the region around the earth in which an object experiences a force towards the centre of the earth. This force is the gravitational attraction between the object and the earth. The gravitational field strength is defined as the gravitational force which acts on a mass of 1 kilogram. F -1 g= Its unit is N kg . m 2-24
  • 25. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 -1 Gravitational field strength, g = 10 N kg -2 Acceleration due to gravity, g = 10 m s -2 The approximate value of g can therefore be written either as 10 m s -1 or as 10 N kg . Weight The gravitational force acting on the object. Weight = mass x gravitational acceleration W = mg SI unit : Newton, N and it is a vector quantity Comparison Mass Weight between weight The mass of an object is the The weight of an object is the force of & amount of matter in the object gravity acting on the object. mass Constant everywhere Varies with the magnitude of gravitational field strength, g of the location A scalar quantity A vector quantity A base quantity A derived quantity SI unit: kg SI unit : Newton, N The difference between a fall in air and a free fall in a vacuum of a coin and a feather. Both the coin and the feather are released simultaneously from the same height. At vacuum state: There is no air At normal state: Both coin and feather resistance. will fall because of gravitational force. The coin and the feather will fall Air resistance effected by the surface area of freely. a fallen object. Only gravitational force The feather that has large area will have acted on the objects. Both will fall more air resistance. at the same time. The coin will fall at first. 2-25
  • 26. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 (a) The two spheres are falling The two spheres are falling down with with an acceleration. the same acceleration The distance between two The two spheres are at the same level successive images of the sphere at all times. Thus, a heavy object and increases showing that the two a light object fall with the same spheres are falling with increasing gravitational acceleration velocity; falling with an Gravitational acceleration is acceleration. independent of mass Two steel spheres are falling under gravity. The two spheres are dropped at the same time from the same height. Motion graph for free fall object Free fall object Object thrown upward Object thrown upward and fall Example 1 A coconut takes 2.0 s to fall to the ground. What is (a) its speed when it strikes the ground (b) ) the height of the coconut tree 2-26
  • 27. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.9 FORCES IN EQUILIBRIUM Forces in When an object is in equilibrium, Equilibrium rd Newton’s 3 Law Examples( Label the forces acted on the objects) Paste more picture Paste more picture Resultant A single force that Force Addition of Forces Resultant force, F = + Resultant force, F = + 2-27
  • 28. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Two forces acting at a point at an angle [Parallelogram method] STEP 1 : Using ruler and protractor, draw STEP 3 the two forces F1 and F2 from a point. Draw the diagonal of the parallelogram. The diagonal represent the resultant force, F in magnitude and direction. STEP 2 Complete the parallelogram scale: 1 cm = …… Resolution of Forces A force F can be resolved into components which are perpendicular to each other: (a) horizontal component , FX (b) vertical component, FY Inclined Plane Fx = F cos θ Component of weight parallel to the plane = mg sin θ Component of weight normal to the plane = mg cos θ Fy = F sin θ 2-28
  • 29. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Find the resultant force (d) (e) 2-29
  • 30. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Lift Stationary Lift Lift accelerate upward Lift accelerate downward Resultant Force = Resultant Force = Resultant Force = The reading of weighing The reading of weighing The reading of weighing scale = scale = scale = Pulley 1. Find the resultant force, F 2. Find the moving mass, m 3. Find the acceleration, a 4. Find string tension, T 2-30
  • 31. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.10 WORK, ENERGY, POWER & EFFICIENCY Work Work done is W = Fs W= ,F= s= The SI unit of work is the joule, J 1 joule of work is done when The displacement , s of the The displacement, s of the object is in the direction of the force, F object is not in the direction of the force, F W = Fs s F W= F s Example 1 Example 2 Example 3 A boy pushing his bicycle with a A girl is lifting up a 3 kg A man is pulling a crate of fish force of 25 N through a distance flower pot steadily to a height along theW = (Fwith θ)force of floor cos a s of 3 m. of 0.4 m. 40 N through a distance of 6 m. Calculate the work done by the What is the work done boy. What is the work done by the in pulling the crate? girl? 2-31
  • 32. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Concept D Formula & Unit Power ef The rate at which work is W in P= done,or t iti p = power, W = work / energy on t = time Energy  Energy is the capacity to do work.  Potential Energy m= Gravitational potential energy is h= the energy of an object due to g= its higher position in the gravitational field. E= Kinetic Energy Kinetic energy is the energy of an m= object due to its motion. v= E= No force is applied on the object in the direction of displacement (the object moves because of its own inertia) A satellite orbiting in space. There is no friction in space. No force is acting in the direction of A waiter is carrying a tray of movement of the satellite. No work is done when: food. The direction of motion of the object is perpendicular to that of the applied force. A waiter is carrying a tray of food and2-32 walking
  • 33. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Principle of Conservation of Energy can be changed from one form to another, but it cannot Energy be created or destroyed. The energy can be transformed from Example 4 A worker is pulling a wooden block of weight, W, with a force of P along a frictionless plank at height of h. The distance travelled by the block is x. Calculate the work done by the worker to pull the block. Example 5 A student of mass m is climbing up a flight of stairs which has the height of h. He takes t seconds.. What is the power of the student? Example 6 Example 7 A stone is thrown upward with initial velocity of -1 20 ms . What is the maximum height which can be reached by the stone? A ball is released from point A of height 0.8 m so that it can roll along a curve frictionless track. What is the velocity of the ball when it reaches point B? 2-33
  • 34. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Example 8 A trolley is released from rest at point X along a frictionless track. What is the velocity of the trolley at point Y? Example 9 A ball moves upwards along a frictionless track of height 1.5 m -1 with a velocity of 6 ms . What is its velocity at point B? Example 10 A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope? 2-34
  • 35. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.11 ELASTICITY Elasticity A property of matter that enables an object No external force is applied. Molecules are at their equilibrium separation. Intermolecular force is equal zero. Compressing a solid causes its molecules Repulsive intermolecular force Stretching a solid Its molecules are Stretching a wire by an external force: When the external force is removed:  The attractive intermolecular forces 2-35
  • 36. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Hooke’s Law The extension of a spring F= k x where F= x= k= Force extension graph Based on the graph: Relationship between F & x : The gradient of the graph represent = Area under the graph 2 = elastic potential energy = ½ F x = ½ k x The elastic limit of a spring The maximum force that If a force stretches a spring beyond its elastic limit, the spring Force constant of the spring, k The force required to produce one unit of extension of the spring. k is a measurement of the stiffness of the spring  The spring with a larger force constant is  A spring with a smaller force constant is 2-36
  • 37. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Factors that effect elasticity Factor Change in factor How does it affects the elasticity Length Shorter spring Longer spring Diameter of spring wire Smaller diameter Larger diameter Diameter spring Smaller diameter Larger diameter Type of material Springs made of different materials Elasticity changes according to the type of material Arrangement of the spring In series In parallel The same load is applied to each spring. The load is shared equally among the springs. Tension in each spring = W Extension of W Tension in each spring = each spring = x 2 Total extension = 2x x Extension of each spring = If n springs are used: The total 2 extension = n x If n springs are used: x The total extension = n Example 1 The original length of each spring is 10 cm. With a load of 10 g, the extension of each spring is 2 cm. What is the length of the spring system for (a), (b) and (c)? 2-57
  • 38. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Example 2 Using the diagram shown and the information given about the weight of Diagram below represent a the two objects, 50 cent coin and a leaf falling in a vacuum container. compare the mass of the coin and the The coin is heavier than the leaf, the time taken to fall, the position leaf. of the coin and the leaf and finally deduce the physical quantity which causes the objects to fall. coin leaf Mass of the taken to fall in a vacuum Time coin Position of the coin and the leaf Coin and leaf of different mass reach the bottom of the container at the same time. Coin and leaf fall down due to gravitational force. The magnitude of gravitational pull is constant. It does not depend on the mass Example 3 Diagram 10 shows a student trying to launch a water rocket. You are required to give suggestion on how to design a water rocket for National Competition. Based on your knowledge on forces, motion and properties of materials, explain your suggestion based on the following aspect: (i) material used (ii) shape of the rocket (iii) suitable angle to launch the rocket (iv) volume of water in the rocket (v) added structure for the motion of the rocket 2-58
  • 39. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Water Rocket Aspect Structure Explanation Acceration of the Rocket Make from light material Structure of the Rocket Aerodynamic Upthrust force Fill with erated water or gasy drink with water Stability of the rocket during flight Use plasticine to make the head of the rocket Add fins at the back portion of the rocket Example 4 Diagram 4.1 shows a cradle with a baby in it is oscillating vertically. Diagram 4.2 shows another identical cradle with a heavier baby in it is oscillating vertically. It is observed that the cradle with a heavier mass baby oscillates at a higher frequency. Design an experiment to test the hypothesis using spring, slotted weight and other suitable apparatus. 4 (a) Inference : The extension of the spring depends 1 (b) Hypothesis : As the 1 c (i) Aim : To investigate the relationship between 1 Variables : 2-59
  • 40. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 c(ii) Manipulated variable : 1 Responding variable : Constant variable : 1 c(iii) Apparatus : Metre rule, retort stand with clamp ,steel spring, slotted weight and 1 pin. Set-up the apparatus 1 c(iv) c (v) Method of controlling the manipulated variables : 1 1. Arrangement the apparatus as shown in the diagram. 2.Mark the initial Method of measuring the responding variables : Record 1 Extension of spring : Measure the Repeat the 1 2-60
  • 41. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 c ( vi) Tabulate Results 1 Initial length , l0 = cm Mass of the Slotted weight , m / g 40 80 120 160 200 Weight of slotted mass / N 0.4 0.8 1.2 1.6 2.0 Length of the spring ,l / cm Extension of the spring x = l- l0 (vii) 1 Total 12 2-61