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Physics Module Form 4                  Chapter 2 – Forces & Motion                           GCKL 2011    2.1           L ...
Physics Module Form 4             Chapter 2 – Forces & Motion                               GCKL 2011     v u            ...
Physics Module Form 4                Chapter 2 – Forces & Motion                                GCKL 2011Comparisons betwe...
Physics Module Form 4                 Chapter 2 – Forces & Motion                             GCKL 2011Example 1          ...
Physics Module Form 4                Chapter 2 – Forces & Motion                           GCKL 2011                      ...
Relating displacement, velocity, acceleration and time using ticker tape.                          VELOCITY               ...
- Distance between the dots decrease uniformly                                                       - The velocity of the...
Physics Module Form 4                Chapter 2 – Forces & Motion                             GCKL 20112.2        M      O ...
Physics Module Form 4           Chapter 2 – Forces & Motion                GCKL 2011   GRAPH              s versus t      ...
Physics Module Form 4                  Chapter 2 – Forces & Motion                               GCKL 2011  Example 1     ...
Physics Module Form 4               Chapter 2 – Forces & Motion                               GCKL 2011 2.3          I    ...
Physics Module Form 4          Chapter 2 – Forces & Motion                             GCKL 2011                          ...
Physics Module Form 4                 Chapter 2 – Forces & Motion                              GCKL 2011       2.4        ...
Physics Module Form 4       Chapter 2 – Forces & Motion                           GCKL 2011                               ...
Physics Module Form 4                Chapter 2 – Forces & Motion                             GCKL 2011                    ...
Physics Module Form 4                    Chapter 2 – Forces & Motion                                    GCKL 2011Example  ...
Physics Module Form 4              Chapter 2 – Forces & Motion                           GCKL 2011  2.5            F      ...
Physics Module Form 4               Chapter 2 – Forces & Motion                               GCKL 2011  Experiment to Fin...
Physics Module Form 4              Chapter 2 – Forces & Motion                                GCKL 2011     Procedure :   ...
Physics Module Form 4                Chapter 2 – Forces & Motion                             GCKL 2011    1. What force is...
Physics Module Form 4         Chapter 2 – Forces & Motion                             GCKL 2011 2.6              IMPULSE  ...
Physics Module Form 4                 Chapter 2 – Forces & Motion                               GCKL 2011                 ...
Physics Module Form 4                Chapter 2 – Forces & Motion                               GCKL 2011   2.7            ...
Physics Module Form 4                 Chapter 2 – Forces & Motion                              GCKL 2011   2.8            ...
Physics Module Form 4               Chapter 2 – Forces & Motion                              GCKL 2011                    ...
Physics Module Form 4                 Chapter 2 – Forces & Motion                          GCKL 2011                      ...
Physics Module Form 4              Chapter 2 – Forces & Motion                               GCKL 2011 2.9         F O R C...
Physics Module Form 4               Chapter 2 – Forces & Motion                          GCKL 2011  Two forces acting at a...
Physics Module Form 4        Chapter 2 – Forces & Motion        GCKL 2011  Find the resultant force          17 N         ...
Physics Module Form 4             Chapter 2 – Forces & Motion                    GCKL 2011Lift       Stationary Lift      ...
Physics Module Form 4                  Chapter 2 – Forces & Motion                           GCKL 2011  2.10           W O...
Physics Module Form 4                Chapter 2 – Forces & Motion                             GCKL 2011      Concept       ...
Physics Module Form 4                Chapter 2 – Forces & Motion                             GCKL 2011      Principle of C...
Physics Module Form 4                   Chapter 2 – Forces & Motion                   GCKL 2011  Example 8  A trolley is r...
2.0 forces and motion
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Transcript of "2.0 forces and motion"

  1. 1. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.1 L I N E A R M O T I O N Physical Quantity Definition, Quantity, Symbol and unit Distance is the total path length travelled from one location to another. Distance, s Quantity: scalar SI unit: meter (m) (a) The distance in a specified direction. (b) the distance between two locations measured along the shortest path Displacement, s connecting them in a specific direction. (c) The distance of its final position from its initial position in a specified direction. Quantity: vector SI unit: meter (m) Speed is the rate of change of distance Speed,v Dis tan ce Speed = time Quantity: scalar SI unit: m s -1 Velocity is the rate of change of displacement. Velocity, v Displacement Velocity = time Direction of velocity is the direction of displacement Quantity : Vector SI unit: m s -1 Average speed TotalDis tan t Example: A car moves at an average v= speed / velocity of 20 ms -1 TotalTime On average, the car moves a distance/ displacement of 20 m in 1 second for the whole journey. Average velocity Displacement v TotalTimeUniform speed Speed that remains the same in magnitude without considering its directionUniform velocity Velocity that remains the same in magnitude and directionAn 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 increases, the object is said to be accelerating. 2-1
  2. 2. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 v u Acceleration is defined as the rate of change of velocitya t Change in velocity Acceleration= Time taken Final velocity,v - Initial velocity,uUnit: ms-2 = Time taken,t The velocity of an object increases from an initial velocity, u, to a higher final velocity, vAcceleration is positiveDecelerationacceleration is negative. The rate of decrease in speed in a specified direction. The velocity of an object decreases from an initial velocity, u, to a lower final velocity, v.Zero acceleration An object moving at a constants velocity, that is, the magnitude and direction of its velocity remain unchanged – is not acceleratingConstant acceleration Velocity increases at a uniform rate. When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity increases by 5 ms -1 for every second that the car is in motion. 1. Constant = uniform 2. increasing velocity = acceleration 3. decreasing velocity = deceleration 4. zero velocity = object at stationary / at rest 5. negative velocity = object moves in opposite direction 6. zero acceleration = constant velocity 7. negative acceleration = deceleration 2-2
  3. 3. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011Comparisons between distance and displacement Comparisons between speed and velocity Distance DisplacementTotal path length The distance between Speed Velocitytravelled from two locations The rate of change of The rate of change ofone location to measured along the distance displacementanother shortest path Scalar quantity Vector quantity connecting them in specific direction It has magnitude but It has both magnitude no direction and direction -1 -1Scalar quantity Vector quantity SI unit : m s SI unit : m sIt has magnitude but no It has both magnitudedirection and directionSI unit meter SI unit : meterFill in the blanks: 1. A steady speed of 10 ms -1 = A distance of 10 m is travelled every second. 2. A steady velocity of -10 ms -1 = A displacement of 10 m is travelled every 1 second to the left. -1 3. A steady acceleration of 4 ms -2 = Speed goes up by 4 ms every 1 second. 4. A steady deceleration of 4 ms -2 = speed goes down by 4 ms-1 every 1 second 5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right. 2-3
  4. 4. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011Example 1 Example 2Every day Rahim walks from his house to the junction Every morning Amirul walks to Ahmad’s housewhich 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 They then walk towards their school which is 60 mis 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? Distance = (80 +60 ) m = 140 m Displacement = 100 m 80 tan θ = =1.333 θ = 53.1º 60 (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? 140m(a)What is Rahim’s displacement from his house Speed = =0.156 in ms-1 • when he reaches the junction. 1.5 km to the 15  60 s right • When he is at warung Pak Din. 0.5 km to the 100m Velocity = = 0.111 ms-1 left. 15  60 s(b)After breakfast, Rahim walks back to his house. w hen he reaches home, (i) what is the total distance travelled by Rahim? (1.5 + 1.5 + 0.5+0.5 ) km = 4.0 km (ii) what is Rahim’s total displacement from his house? 1.5 +( -1.5) +(- 0.5 )+0.5 km = 0 kmExample 3 Example 4Salim running in a race covers 60 m in 12 s. An aeroplane flies towards the north with(a) What is his speed in ms-1 a velocity 300 km hr -1 in one hour. 60m Then, the plane moves to the east with Speed = = 5 ms-1 12 s 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? Average speed = (300 km hr -1 + 4 00 km hr -1 ) / 2 = 350 km hr -1 distance covered = 40 s × 5 ms-1 = 200 m (b)What is the average velocity of the plane? Average velocity = 250 km hr -1 2-4
  5. 5. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 400 Tan θ = = 1.333 θ = 300 (c)What is the difference between average speed and average velocity of the plane? Average speed is a scalar quantity. Average velocity is a vector quantity Example 5 The speedometer reading for a car travelling due north shows 80 km hr -1. Another car travelling at 80 km hr -1 towards south. Is the speed of both cars same? Is the velocity of both cars same? The speed of both cars are the same but the velocity of both cars are different with opposite directionA ticker timer  Use: 12 V a.c. power supply  1 tick = time interval between two dots.  The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2 consecutive dots is 1/50 = 0.02 s.  1 tick = 0.02 s 2-5
  6. 6. 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 sTICKER TAPE AND CHARTS TYPE OF MOTION Constant velocity – slow moving Constant velocity – fast moving  Distance between the dots increases uniformly  the velocity is of the object is increasing uniformly  The object is moving at a uniform / constant acceleration. 2-6
  7. 7. - Distance between the dots decrease uniformly - The velocity of the object is decreasing Uniformly - The object is experiencing uniform / constant deccelerationExample 6The diagram above shows a ticker tape chart for amoving trolley. The frequency of the ticker-timerused is 50 Hz. Each section has 10 dots-spacing.(a) What is the time between two dots. Time = 1/50 s = 0.02 s(b) What is the time for one strips. 0.02 s × 10 = 0.2 s(c) What is the initial velocity 2 cm / 0.2 s = 10 ms-1(d) What is the final velocity. 12 cm / 0.2 s = 60 ms-1(e) What is the time interval to change from initialvelocity to final velocity? ( 11 - 1) × 0.2 s = 2 s(f) What is the acceleration of the object. vu 60  10 -2 a= = ms = 25 ms-2 t 2 THE EQUATIONS OF MOTIONv  u  at u = initial velocity 1 v = final velocitys  ut  at 2 t = time taken 2 s = displacementv  u  2as 2 2 a = constant acceleration 2-7
  8. 8. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 20112.2 M O T I O N G R A P H S DISPLACEMENT – TIME GRAPH Velocity is obtained from the gradient of the graph. A – B : gradient of the graph is positive and constant velocity is constant. B – C : gradient of the graph = 0 the velocity = 0, object is at rest. C – D : gradient of the graph negative and constant. The velocity is negative and object moves in the opposite direction. VELOCITY-TIME GRAPH Area below graph Distance / displacement Positive gradient Constant Acceleration (A – B) Negative gradient Constant Deceleration (C – D) Zero gradient Constant velocity / zero acceleration (B – C) GRAPH s versus t v versus t a versus tZerovelocityNegativeconstantvelocityPositiveConstantvelocity 2-8
  9. 9. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 GRAPH s versus t v versus t a versus tConstantaccelerationConstantdeceleration 2-9
  10. 10. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Example 1 Example 2Contoh 11 (a) Calculate the acceleration at: Based on the s – t graph above: (i) JK (ii) KL (iii) LM (a) Calculate the velocity at (i) 2 ms-2 (ii) -1 ms-2 (iii) 0 ms-1 (i) AB (ii) BC (iii) CD (i) 5 ms-1 (ii) 0 ms-1 (iii) - 10 ms-1 (b) Describe the motion of the object at: (b) Describe the motion of the object at: (i) JK (ii) KL (iii) LM (i) AB (ii) BC (iii) CD (i) constant velocity 5 ms-1 (i) constant acceleration of 2 ms-2 (ii) at rest / 0 ms-1 (ii) constant deceleration of 1ms-2 (iii) constant velocity of 10 ms-1in opposite (iii) (iii) zero acceleration or constant velocitydirection Calculate the total displacement. (c)Find: Displacement = area under the graph (i) total distance 50 m + 50 m = 100 m = 100 m + 150 m + 100 m + 25 m = 375 m (ii) total displacement 50 m + (- 50 m) = 0 (c) Calculate the average velocity. Average velocity = 375 m / 40 s (d) Calculate = 9.375 ms-1 100m (i) the average speed = 2.86 ms-1 35s (ii) the average velocity of the movingparticle. 0 2-10
  11. 11. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.3 I N E R T I AInertia The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its motion.Newton’s first law Every object continues in its state of rest or of uniform motion unless it is acted upon by an external force.Relation between inertia The larger the mass, the larger the inertiaand 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 The inertia of the coin maintains its NB[Fat rest. state NDPnDSFJ[POJDE]O- The coin falls into the glass due to gravity. JBD]AOP[FKBOP[DF LMB NOPGFMB LKFGNKLB FGNMNKL’ MCVL 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, the sauce continues in its state of motion due to the effect of its inertia. Body moves forward when the car stops suddenly The passengers were in a state of motion when the car was moving. When the car stopped suddenly, the inertia in the passengers made them maintain their state of motion. Thus when the car stop, the passengers moved forward. A boy runs away from a cow in a zig- zag motion. The cow has a large inertia making it difficult to change direction. 2-10
  12. 12. 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. 1. Safety in a car: Ways to reduce the negative (a)Safety belt secure the driver to their seats. effects of inertia 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
  13. 13. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.4 M O M E N T U MDefinition Momentum = Mass x velocity = mv SI unit: kg ms-1Principle of Conservation of Momentum In the absence of an external force, the total momentum of a system remains unchanged. Elastic Collision Inelastic collision ƒ Both objects move independently at their ƒ The two objects combine and move together respective velocities after the collision. with a common velocity after the collision. ƒ Momentum is conserved. ƒ Momentum is conserved. ƒ Kinetic energy is conserved. ƒ Kinetic energy is not conserved. Total energy is conserved. ƒ Total energy is conserved.Total Momentum Before = total momentum after Total Momentum Before = Total Momentum After m1u1 + m2u2 = m1 v1 + m2 v2 m1 u1 + m2 u2 = ( m1 + m2 ) v Explosion Before explosion both object stick together and at rest. After collision, both object move at opposite direction. Total Momentum Total Momentum after before collision is collision : zero m1v1 + m2v2 2- 13
  14. 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 opposite direction EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum) When a rifle is fired, the bullet of mass m, moves with a high velocity, v. This creates a momentum in the forward direction. From the principle of conservation of momentum, an equal but opposite momentum is produced to recoil the riffle backward. Application in the jet engine: A high-speed hot gases are ejected from the back with high momentum. This produces an equal and opposite momentum to propel the jet plane forward. The launching of rocket Mixture of hydrogen and oxygen fuels burn explosively in the combustion chamber. Jets of hot gases are expelled at very high speed through the exhaust. These high speed hot gases produce a large amount of momentum downward. By conservation of momentum, an equal but opposite momentum is produced and acted on the rocket, propelling the rocket upwards. 2- 14
  15. 15. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 In a swamp area, a fan boat is used. The fan produces a high speed movement of air backward. This produces a large momentum backward. By conservation of momentum, an equal but opposite momentum is produced and acted on the boat. So the boat will move forward. A squid propels by expelling water at high velocity. Water enters through a large opening and exits through a small tube. The water is forced out at a high speed backward. Total Mom. before= Total Mom. after 0 =Mom water + Mom squid 0 = 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 ExampleCar A of mass 1000 kg moving at 20 ms -1collides with a car B of mass 1200 kg moving at Before collision After collision10 m s -1 in same direction. If the car B is MA = 4 kgshunted forwards at 15 m s -1 by the impact, MB = 2 kgwhat is the velocity, v, of the car A immediately UA = 10 ms -1 r i g h t UB = 8 ms -1 l e f t VB 4 ms-1 rightafter the crash?1000 kg x 20 ms -1 + 1200 kg x 10 ms -1 = Calculate the value of VA . 1000 kg x v + 1200 kg x 15 ms -1 [4 x 10 + 2 x (-8)]kgms -1 =[ 4 x v + 2 x 4 ] kgms -1 v= 14 ms -1 VA = 4 ms -1 right 2- 15
  16. 16. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011Example ExampleA truck of mass 1200 kg moving at 30 ms-1 collideswith a car of mass 1000 kg which is travelling in A man fires a pistol which has a mass of 1.5 kg.the opposite direction at 20 ms-1. After the If the mass of the bullet is 10 g and it reaches acollision, the two vehicles move together. What is velocity of 300 ms -1 after shooting, what is thethe velocity of both vehicles immediately after recoil velocity of the pistol?collision? 0 = 1.5 kg x v + 0.01 kg x 300 ms -11200 kg x 30 ms -1 + 1000 kg x (-20 ms -1) v = -2 ms -1 = ( 1200 kg + 1000kg) v v = 7.27 ms -1 to the right Or it recoiled with 2 ms -1 to the left 2- 16
  17. 17. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.5 F O R C E Balanced Force Example: When the forces acting on an object are balanced, they cancel each other out. The net force is zero. Effect : the object is at rest [velocity = 0] or moves at constant velocity [ a = 0] Unbalanced Force/ Resultant Force When the forces acting on an object are not balanced, there must be a net force acting on it. The net force is known as the unbalanced force or the resultant force. Weight, W = Lift, U Thrust, F = drag, G Effect : Can cause a body to - change it state at rest (an object will accelerate - change it state of motion (a moving object will decelerate or change its direction) Newton’s Second Law of Motion The acceleration produced by a force on an object is directly proportional to the magnitude of the net force applied and is inversely proportional to the mass of the object. The direction of the acceleration is the same as that of the net force. Force = Mass x Acceleration F = ma 2 - 17
  18. 18. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Experiment to Find The Relationship between Force, Mass & Acceleration Relationship between a& F a& 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 The acceleration produced by an object depends on the net force object depends on the mass applied to it. Hypothesis The acceleration of the object The acceleration of the object increases when the force applied decreases when the mass of the increases object increases Variables: Manipulated : Force Mass Responding : Acceleration Acceleration Constant : Mass Force Apparatus and Ticker tape, elastic cords, ticker timer, trolleys, power supply, friction Material compensated runway and meter ruler. 2 - 18
  19. 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. The elastic cord is stretched manipulated until the end of the trolley. The until the end of the trolley. The trolley variables. trolley is pulled down the runway is pulled down the runway with the with the elastic cord being kept elastic cord being kept stretched by stretched by the same amount of the same amount of force 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 -Repeating Repeat the experiment by using two Repeat the experiment by using two, experiment. , three, four and five elastic cords three, four and five trolleys. 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. 20. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 1. What force is required to move a 2 kg 2. Ali applies a force of 50 N to move a 10 kg object with an acceleration of 3 m s-2, table at a constant velocity. What is the if frictional force acting on the table? (a) the object is on a smooth surface? (b) The object is on a surface where the Answer: 50 N average force of friction acting on the object is 2 N? (a) force = 6 N (b) net force = (6 – 2) N = 4N 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? D Find (a) the average deceleration, (b) the average braking force. (a) u = 20 ms -1 v = 0 s = 30 m a=? a = - 6.67 ms-2 (b) force = 1200 x 6.67 N = 8000 N 2 - 20
  21. 21. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.6 IMPULSE AND IMPULSIVE FORCE Impulse The change of momentum mv - mu m = mass Unit : kgms-1 or Ns u = initial velocity v = final velocity Impulsive Force The rate of change of momentum in a t = time collision or explosion Impulsive force = change of momentum mv  mu  time t Effect of time Impulsive force Longer period of time →Impulsive force decrease Unit = N is inversely proportional to Shorter period of time →Impulsive force increase time of contact Situations for Reducing Impulsive Force in Sports Situations Explanation Thick mattress with soft surfaces are used in events such as high jump so that the time interval of impact on landing is extended, thus reducing the impulsive force. This can prevent injuries to the participants. Goal keepers will wear gloves to increase the collision time. This will reduce the impulsive force. A high jumper will bend his legs upon landing. This is to increase the time of impact in order to reduce the impulsive force acting on his legs. This will reduce the chance of getting serious injury. A baseball player must catch the ball in the direction of the motion of the ball. Moving his hand backwards when catching the ball prolongs the time for the momentum to change so as to reduce the impulsive force. 2 - 21
  22. 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 large impulsive force on the wooden slab. A massive hammer head moving at a fast speed is brought to rest upon hitting the nail within a short time interval. The large change in momentum within a short time interval produces a large impulsive force which drives the nail into the wood. A football must have enough air pressure in it so the contact time is short. The impulsive force acted on the ball will be bigger and the ball will move faster and further. Pestle and mortar are made of stone. When a pestle is used to pound chillies, the hard surfaces of both the pestle and mortar cause the pestle to be stopped in a very short time. A large impulsive force is resulted and thus causes these spices to be crushed easily. Example 1 Answer: (a) Impulse = 60 kg x ( 6 ms-1 - 0 ) A 60 kg resident jumps from the first floor of a burning house. = 360 Ns His velocity just before landing on the ground is 6 ms-1. 360 Ns (a) Calculate the impulse when his legs hit the ground. (b) Impulsive force = 0.5s (b) What is the impulsive force on the resident’s legs if he =7200 N bends upon landing and takes 0.5 s to stop? (c) He experienced a greater (c) What is the impulsive force on the resident’s legs if Impulsive force of 7200 N and he might injured his legs he does not bend and stops in 0.05 s? (d) Increase the reaction time so as to (d) What is the advantage of bending his legs upon landing? reduce impulsive force Example 2 Rooney kicks a ball with a force of 1500 N. The time of (a) Impulse = 1500N x 0.01 s contact of his boot with the ball is 0.01 s. What is the impulse = 15 Ns delivered to the ball? If the mass of the ball is 0.5 kg, what is the velocity of the ball? 15 Ns (b) velocity = = 30 ms-1 0.5kg 2 - 22
  23. 23. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.7 S A F E T Y V E H I C L E Safety features in vehicles Head rest Crash resistant door Windscreen pillars Crumple zones Anti-lock brake system (ABS) Traction control Front bumper Air bagsComponent FunctionHeadrest To reduce the inertia effect of the driver’s head.Air bag Absorbing impact by increasing the amount of time the driver’s head to come to the steering. So that the impulsive force can be reduceWindscreen To protect the driver (shattered proof)Crumple zone Can be compressed during accident. So it can increase the amount of time the car takes to come to a complete stop. So it can reduce the impulsive force.Front Absorb the shock from the accident. Made from steel, aluminium, plastic orbumper rubber.ABS Enables drivers to quickly stop the car without causing the brakes to lock.Side impact bar Prevents the collapse of the front and back of the car into the passenger compartment. Also gives good protection from a side impactSeat belt To reduce the effect of inertia by avoiding the driver from thrown forward. 2 - 23
  24. 24. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.8 G R A V I T Y Gravitational Objects fall because they are pulled towards the Earth by the force of gravity. Force This force is known as the pull of gravity or the earth’s gravitational force. The earth’s gravitational force tends to pull everything towards its centre. Free fall  An object is falling freely when it is falling under the force of gravity only.  A piece of paper does not fall freely because its fall is affected by air resistance.  An object falls freely only in vacuum. The absence of air means there is no air resistance to oppose the motion of the object.  In vacuum, both light and heavy objects fall freely.  They fall with the same acceleration i.e. The acceleration due to gravity, g. Acceleration due to  Objects dropped under the influence of the pull of gravity with gravity, g constant acceleration.  This acceleration is known as the gravitational acceleration, g. -2  The standard value of the gravitational acceleration, g is 9.81 m s . -2 The value of g is often taken to be 10 m s for simplicity.  The magnitude of the acceleration due to gravity depends on the strength of the gravitational field. Gravitational field The gravitational field is the region around the earth in which an object experiences a force towards the centre of the earth. This force is the gravitational attraction between the object and the earth. The gravitational field strength is defined as the gravitational force which acts on a mass of 1 kilogram. F -1 g= Its unit is N kg . m 2 - 24
  25. 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. 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) t = 2 s u = 0 g = 10 v=? A coconut takes 2.0 s to fall to the ground. What v = u + gt = 0 + 10 x 2 = 20 ms-1 is (b) s = ut + ½ at2 = 0 + ½ (10) 22 = 20 m (a) its speed when it strikes the ground (b) ) the height of the coconut tree 2 - 26
  27. 27. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.9 F O R C E S I N E Q U I L I B R I U MForces in Equilibrium When an object is in equilibrium, the resultant force acting on it is zero. The object will either be 1. at rest 2. move with constant velocity. rd Action is equal to reactionNewton’s 3 Law Examples( Label the forces acted on the objects) Paste more picture Paste more pictureResultant Force A single force that represents the combined effect of two of more forces in magnitude and direction. Addition of Forces Resultant force, F = F1 + F2 Resultant force, F = F1 + - F2 2 - 27
  28. 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 = …… A force F can be resolved into components Resolution of Forces 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 Fy = F sin θ Component of weight normal to the plane = mg cos θ 2 - 28
  29. 29. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Find the resultant force 17 N 5N (d) (e) 7N FR 2 - 29
  30. 30. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011Lift 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 40 -30 = 10 N 30-2 = 28 N 2. Find the moving mass, m 4 + 3 = 7 kg 3+ 4 = 4 kg 3. Find the acceleration, a 40 -30 = (3+4)a 30 -2 = (4+3 )a 10 =7a 28 = 7a a =10/ 7 ms-2 a = 4 ms-2 4. Find string tension, T T- 3 (10) = 3 a 30 – T = 3 (a) T = 30 + 3 (10/7) T =30- 12 =240 /7 N = 18 N 2 - 30
  31. 31. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 2.10 W O R K , E N E R G Y , P O W E R & E F FI C I E N CY Work Work done is the product of an applied force and the displacement of an object in the direction of the applied force W = Fs W = work, F = force s = displacement The SI unit of work is the joule, J 1 joule of work is done when a force of 1 N moves an object 1 m in the direction of the force The displacement , s of the The displacement, s of the object is in the direction of the force, F object is not in the direction of the force, FW = 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. 75 Nm What is the work done by the in pulling the crate? girl? 12 Nm 40 N cos 50º x 6 Nm 2 - 31
  32. 32. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011 Concept D Formula & Unit Power ef The rate at which work is done, W in P= or the amount of work done per t second. iti p = power, W = work / energy on t = time Energy  Energy is the capacity to do work.  An object that can do work has energy  Work is done because a force is applied and the objects move. This is accompanied by the transfer of energy from one object to another object.  Therefore, when work is done, energy is transferred from one object to another.  The work done is equal to the amount of energy transferred. Potential Energy Gravitational potential energy is m = mass the energy of an object due to h = height its higher position in the gravitational field. g = gravitational acceleration E = mgh Kinetic Energy Kinetic energy is the energy of an m = mass object due to its motion. v = velocity 2 E = ½ mv 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 No work is done when: The direction of motion of the object is perpendicular to that of movement of the satellite. The object is stationary. the applied force. A student carrying his bag while waiting at the bus stop A waiter is 32 2 - carrying a tray of food and walking
  33. 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 one form to another, total energy in a system is constant. Total energy before = total energy after Example 4 A worker is pulling a wooden block of weight W, with a force of P along a frictionless plank at height of h. The distance travelled by the block is x. Calculate the work done by the worker to pull the block. [Px = Wh] Example 5 A student of mass m is climbing up a flight of stairs which has the height of h. He takes t seconds. What is the power of the student? mgh [ t Example 6 Example 7 A stone is thrown upward with -1 initial velocity of 20 ms . What is the maximum height which can be reached by the stone? [ 10m ] A ball is released from point A of height 0.8 m so that it can roll along a curve frictionless track. What is the velocity of the ball when it reaches point B? [4 ms-1] 2 - 33
  34. 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? [ v2 = 30( ms-1)2] [ v = 5.48 ms-1] Example 9 A ball moves upwards along a frictionless track of height 1.5 m -1 with a velocity of 6 ms . What is its velocity at point B? [v2 = 30( ms-1)2 v = 5.48 ms-1] Example 10 A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope? [6 ms-1] 2 - 34

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