SPM PHYSICS FORM 4 forces and motion

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  • 1. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2. FORCE AND MOTION2.1 ANALYSING LINEAR MOTIONDistance and displacement1. Types of physical quantity: (i) Scalar quantity: …………………………………………………………………. (ii) Vector quantity: …………………………………………………………………2. The difference between distance and displacement: (i) Distance: ………………………………………………………………………… (ii) Displacement: ……………………………………………………………………3. Distance always longer than displacement.4. Example: The following diagram shows the location of Johor Bahru and Desaru. You can travel by car using existing road via Kota Tinggi, or travel by a small plane along straight path. Calculate how far it is from Johor Bahru to Desaru if you traveled by: a. The car b. The plane Kota Tinggi 41 km 53 km Solution: Johor Bahru 60 km DesaruHands-on Activity 2.2 pg 10 of the practical book.Idea of distance and displacement, speed and velocity.Speed and velocity1. Speed is ..…………………………………………………………………………………2. Velocity is: ..……………………………………………………………………………...3. Average of speed: ………………………………………………………………………4. Average of velocity: ……………………………………………………………………... 19
  • 2. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion5. Example: An aeroplane flies from A to B, which is located 300 km east of A. Upon reaching B, the aeroplane then flies to C, which is located 400 km north. The total time of flight is 4 hours. Calculate i. The speed of the aeroplane ii. The velocity of the aeroplane Solution:Acceleration and deceleration1. Study the phenomenon below; 0 m s-1 20 m s-1 40 m s-1 Observation: ………………………………………………………………………………2. Acceleration is, ………………………………………………………………………. Then, a = Or, a v – u t3. Example of acceleration; t=2s t=2s A B C 0 m s-1 20 m s-1 40 m s-1 20
  • 3. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionCalculate the acceleration of car; i) from A to B ii) From B to C4. Deceleration happens ...………………………………………………………………… ………………………………………………………………………………………………5. Example of deceleration; A lorry is moving at 30 m s-1, when suddenly the driver steps on the brakes and it stop 5 seconds later. Calculate the deceleration of lorry.Analysing of motion1. Linear motion can be studied in the laboratory using a ticker timer and a ticker tape. Refer text book photo picture 2.4 page 26. (i) Determination of time: . . . . . . . . (ii) Determination of displacement as the length of ticker tape over a period of time. . x . . . . . . . y (iii) Determine the type of motion; . . . . . . . . ……………………………………………………………………………………….. . . . . . . . . ...……….…………………………………………………………………………….. . . . . . . . . .…………………………………………………………………………………….. 21
  • 4. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion (iv) Determination of velocity . . . . . . . . displacement = ……………………… time = ……………………………….. Velocity, v = (v) Determine the acceleration Length/cm v 8 7 6 5 4 u 3 2 1 0The equation of motion ticks1. The important symbols : ……………………………………………………………….. ………………………………………………………………………………………………2. The list of important formula;3. Example 1 : A car traveling with a velocity of 10 m s-1 accelerates uniformly at a rate of 3 m s-2 for 20 s. Calculate the displacement of the car while it is accelerating. 22
  • 5. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExample 2 : A van that is traveling with velocity 16 m s-1 decelerates until it comes to rest.If the distance traveled is 8 m, calculate the deceleration of the van.Execise 2.1 Length / cm1. Figure 2.1 shows a tape chart consisting of 5-tick strip. Describe 16 the motion represented by AB and BC. In each case, determine the ; 12 (a) displacement 8 4 (b) average velocity Figure 2.1 0 A B C Time/s (c) acceleration2. A car moving with constant velocity of 40 ms-1 . The driver saw and obtacle in front and he immediately stepped on the brake pedal and managed to stop the car in 8 s. The distance of the obstacle from the car when the driver spotted it was 180 m. How far is the obstacles from the car has stopped. 23
  • 6. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.2 ANALYSING MOTION GRAPHS 0m 100m 200m 300m 400m 500m displacement 0s 10s 20s 30s 40s 50s timeThe data of the motion of the car can be presented………………………………….The displacement-time Graph a) displacement (m) Graph analysis: ……………………………………………………………… ……………………………………………………………… time (s) ……………...………………………………………………b) displacement (m) Graph analysis: ……..………………………………………………………… ………………………………………………………………… time (s) ……….………………………………………………………… c) displacement (m) Graph analysis: …….…………………………………………………………… ………………………………………………………………… time (s) ..………………………………………………………………… d) Displacement (m) Graph analysis: …………………………….……………………………………… ……………………………………………..……………………… time (s) ……………………………………………………………………… ………………………………………………………………… 24
  • 7. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion e) displacement (m) Graph analysis: ………………………………………………………….. ………………………………………………………….. ………………………………………………………….. time (s) ………………………………………………………….. f) displacement (m) Graph analysis: A B ………………………………………………………….. ………………………………………………………….. …………………………………………………………… O C time (s) time (s)The velocity-time Graph a) v/ m s-1 Graph analysis: ………………………………………………………….. ………………………………………………………….. …………………………………………………………… t t /s b) v/ m s-1 Graph analysis: ………………………………………..……………….. ………………………………………………………… ………………………………………………………… t t /s ………………………………………………………… c) v (m s-1) Graph analysis: …………………………………..…………………. ……………………………………………………… ……………………………………………………… t1 t2 t (s) 25
  • 8. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion d) v (m s-1) Graph analysis: ...…………………………………..……………….. …………………………………………………….. ……………………………………………………… t (s) ……………………………………………………… .……………………………………………………... e) v (m s-1) Graph analysis: ………..…………………………..……………….. ……….…………………………………………….. ……………………………………………………… t (s) ………………………………………………………Examples ………………………………………………………1. s/m Calculate:- (i) Velocity over OP, QR and RS (ii) Displacement P Q Solution : O R 0 2 4 6 8 t/s S2. v/m s-1 Calculate:- (i) acceleration,a over OP, PQ and QR (ii) Displacement P Q 10 Solution : 5 O R 0 2 4 6 8 10 t/s 26
  • 9. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExcercise 2.21. (a) s/m (b) s/m (c) s/m 10 t/s 0 2 4 t/s t/s -5 -10 Figure 2.21 Describe and interpret the motion of a body which is represented by the displacement time graphs in Figure 2.212. Describe and interpret the motion of body which is represented by the velocity-time graphs shown in figure 2.22. In each case, find the distance covered by the body and its displacement (a) v/m s-1 (b) v/m s-1 10 t/s 0 2 4 t/s -5 -10 Figure 2.22 27
  • 10. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.3 UNDERSTANDING INERTIAIdea of inertia1. ………………………………………………………………………………………………2. ………………………………………………………………………………………………3. ……………………………………………………………………………………………… Hand-on activity 2.5 in page 18 of the practical book to gain an idea of inertia4. Meaning of inertia : …………..…………………………………………………………………………………. ………………………………………………………………………………………………Mass and inertia1. Refer to figure 2.14 of the text book, the child and an adult are given a push to swing. (i) which one of them will be more difficult to be moved ……………………... (ii) which one of them will be more difficult to stop? …………………………….2. The relationship between mass and inertia : ……………………………….……………………………………………………………..3. The larger mass …………………………………………………………………………. ………………………………………………………………………………………………Effects of inertia1. Positive effect : ………………………………………………………………………… (i) ……………………………………………………………………………………… (ii) ……………………………………………………………………………………… (iii) ………………………………………………………………………………………2. Negative effect : …………………………………………………………………………. (i) ……………………………………………………………………………………... …………………………………………………………………………………….. (ii) ……………………………………………………………………………………… ……………………………………………………………………………………… (iii) ……………………………………………………………………………………… ……………………………………………………………………………………… (iv) ……………………………………………………………………………………… 28
  • 11. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExecise 2.31. What is inertia? Does 2 kg rock have twice the inertia of 1 kg rock? ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………2. Figure 2,3 A wooden dowel is fitted in a hole through a wooden block as shown in figure 2.31. Explain what happen when we (a) strike the top of the dowel with a hammer, ……………………………………………………………………………………… ……………………………………………………………………………………… (b) hit the end of the dowel on the floor. ……………………………………………………………………………………… ……………………………………………………………………………………2.4 ANALYSING MOMENTUMIdea of momentum1. When an object ic moving, …...…………………………………………………………2. The amount of momentum ...……………………………………………………………3. Momentum is defined……………………………………………………………………. ……………………………………………………………………………………………… 29
  • 12. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionConservation of momentum mg mb vb vg = 0 (mb + mg) Starting position before vb&g she catches the ball Receiving a massive ball vb vg mb mg Starting position before she throws the ball Throwing a massive ballThe principle of conservation of momentum :………………………………………………………………………………………………………………………………………………………………………………………………………………1. Elastic collision .………………………………………………………………………….. u1 m1 u2 v2 m2 m1 m2 Before collision after collision 30
  • 13. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2. Inelastic collision :………………………………………………………………………... u1 v m u2 = 0 1 m2 m1 + m 2 Before collision after collision3. explosion : …….....…………………………………………………………………... (m1 + m2), u = 0 v1 v2 m2 Before explosion after explosionExample 1 : Car A Car BCar A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at20 m s-1 in front of it. Car A and B move separately after collision. If Car A is still moving at25 m s-1 after collision, determine the velocity of Car B after collision.Solution :Example 2 :Car A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at20 m s-1 in front of it. Car A is pulled by Car B after collision. Determine the common velocityof Car A and B after collision.Solution : 31
  • 14. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExample 3 :A bullet of mass 2 g is shot from a gun of mass 1 kg with a velocity of 150 m s-1 . Calculate thevelocity of the recoil of the gun after firing.Solution :Exercise 2.41. An arrow of mass 150 g is shot into a wooden block of mass 450 g lying at rest on a smooth surface. At the moment of impact, the arrow is travelling horizontally at 15 ms-1. Calculate the common velocity after the impact.2. A riffle of mass 5.0 kg fires a bullet of mass 50 g with a velocity of 80 m s-1 .Calculate the recoil velocity. Explain why the recoil velocity of a riflle is much less than the velocity of the bullet.2.5 UNDERSTANDING THE EFFECT OF A FORCEIdea of force1. What will happen when force act to an object? ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… 32
  • 15. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionIdea of balanced forces1. An object is said to be in balance when it is: ……………………………………………………………………………………………… ………………………………………………………………………………………………2. Stationary object ……………………………… explanation : Stationary object ……………………………………………… ……………………………………………… ……….…………………………………….. …………………………………………3. An object moving with uniform velocity …………………………….. explanation : …..……………. …………… …………………………………………….. …………………………………………….. …………………………………………….. ……………………………… ………..……………………………………. …………………………………………….. ……………………………………………..Idea of unbalanced forces1. A body is said to be in unbalanced..……………………………………………………2. ……………………….. Explanation; ……………………………………………… ……………………………………………… ……………………………………………… ……… …….. ………………………………………………Relationship between forces, mass and acceleration (F = ma)Experiment 2.2 page 29.Aim : To investigate the relationship between acceleration and force applied on a constantmass.Experiment 2.3 page 31Aim: To investigate the relationship between mass and acceleration of an object underconstant force. 33
  • 16. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion1. Refer to the result of experiment 2.2 and 2.3, ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………2. 1 newton (F = 1 N) is defined as the force required to produce an acceleration of 1 m s-2 (a=1 m s-2) when its acting on an object of mass 1 kg ( m = 1 kg) So, …………………………………………………………………………………………3. Example 1 : Calculate F, when a = 3 m s-2 dan m = 1000 kg Example 2 : m = 25 kg F = 200 N Calculate the acceleration, a of an object.Exercise 2.51. A trolley of mass 30 kg is pulled along the ground by horizontal force of 50 N. The opposing frictional force is 20 N. Calculate the acceleration of the trolley.2. A 1000 kg car is travelling at 72 km h-1 when the brakes are applied. It comes to a stop in a distance of 40 m. What is the average braking force of the car? 34
  • 17. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.6 ANALYSING IMPULSE AND IMPULSIVE FORCEImpulse and impulsive force1. Impulse is ……………………………………………………………………………….2. Impulsive force is ……………………………………………………………………… ………………………………………………………………………………………………3. Formula of impulse and impulsive force: Refer, F = maExample 1; v u wall If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 1 s Impulse, Ft = and impulsive force, F =Example 2; v u Wall with a soft surface If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 2 s Impulse, Ft = and impulsive force, F =4. The relationship between time of collision and impulsive force. ……………………………………………………………………………………………… ……………………………………………………………………………………………… 35
  • 18. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExercise 2.61. A force of 20 N is applied for 0.8 s when a football player throws a ball from the sideline. What is the impulse given to the ball?2. A stuntman in a movie jumps from a tall building an falls toward the ground. A large canvas bag filled with air used to break his fall. How is the impulsive force reduced?2.7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES Safety features in vehicles Reinforced passenger compartment Crash resistant door Head rest pillars Windscreen Crumple zones Anti-lock brake system (ABS) Traction control bumpers Air bags 36
  • 19. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionImportance of safety features in vehicles Safety features ImportancePadded dashboard Increases the time interval of collision so the impulsive force produced during an impact is thereby reducedRubber bumper Absorb impact in minor accidents, thus prevents damage to the car.Shatter-proof windscreen Prevents the windscreen from shatteringAir bag Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers. Prevents the passengers from being thrown out of the car. SlowsSafety seat belt down the forward movement of the passengers when the car stops abruptly. Prevents the collapse of the front and back of the car into theSide bar in doors passenger compartment. Also gives good protection from a side-on collision.Exercise 2.71. By using physics concepts, explain the midifications to the bus that help to improve that safety of passengers and will be more comfortable. 37
  • 20. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.8 UNDERSTANDING GRAVITYCarry out hands-on activity 2.8 on page 35 of the practical book.Acceleration due to gravity.1. An object will fall to the surface of the earth because………………………………...2. The force of gravity also known ………………………………………………………...3. When an object falls under the force of gravity only, ………………………………... ………………………………………………………………………………………………4. The acceleration of objects falling freely ………………………………………………5. The magnitude of the acceleration due to gravity depends ………………………... ………………………………………………………………………………………………Gravitational field1. The region around the earth is ………………………………………………………….2. The object in gravitational field …………………………………………………………3. The gravitational field strength is defined ……………………………………………..4. The gravitational field strength, g can be calculate as;5. At the surface of the earth, …………….………………………………………………………………………………..6. This means ……………………………………………………………………………………………..7. Example 1. Can you estimate the gravitational force act to your body? mass = 60 kg, g = 9.8 N kg-1, F = ? Example 2, A satellite of mass 600 kg in orbit experiences a gravitational force of 4800 N. Calculate the gravitational field strength. 38
  • 21. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion Example 3, A stone is released from rest and falls into a well. After 1.2 s, it hits the bottom of the well. (a) What is the velocity of the stone when it hits the bottom? (b) Calculate the depth of the well.Weight1. The weight of an object is defined ……………………………………………………..2. For an object of mass m, the weight can be calculate as : Example : The mass of a helicopter is 600 kg. What is the weight of the helicopter when it land on the peak of a mountain where the gravitational field is 9.78 N kg-1?Exercise 2.81. Sketch the following graphs for an object that falling freely. (a) Displacement-time graph, (b) Velocity-time graph (c) Acceleration-time graph 39
  • 22. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2. The following data was obtained from an experiment to measure the acceleration due to gravity. Mass of steel bob = 200 g, distance covered = 3.0 m, time of fall = 0.79 s. Calculate the acceleration due to gravity of steel bob. Give the explanation why your answer different with the constant of gravitational acceleration, g = 9.8 m s-2.2.9 IDEA OF EQUILIBRIUM FORCESAn object is in equilibrium when :1. ………………………………………………………………………………………………2. ……………………………………………………………………………………………… stationary object An object moving with uniform velocity 40
  • 23. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionAddition of Force1. Addition of force is defined as ...…………………………………………………….. ……………………………………………………………………………………………… ……………………………………………………………………………………………… Examples : the forces are acting in one direction F1 = 10 N F2 = 5 N Resultant force, F Example : the forces are acting in opposite directions F1 = 10 N F2 = 5 N Resultant force, F Example : the forces are acting in different directions F2 = 5 N 500 F1 = 10 N Parallelogram method: 1. Draw to scale. 2. Draw the line parallel with F1 to the edge of F2, and the line parallel with F2 to the edge of F1 3. Connect the diagonal of the parallelogram starting from the initial point. 4. Measure the length of the diagonal from the initial point as the value of the resultant force. 41
  • 24. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion F2 F1 Triangle method 1. Draw to scale. 2. Displace one of the forces to the edge of another force. 3. Complete the triangle and measure the resultant force from the initial point. Example 1: During Sport Day two teams in tug of war competition pull with forces of 6000 N and 5300 N respectively. What is the value of the resultant force? Are the two team in equilibrium? Example 2: A boat in a river is pulled horizontally by two workmen. Workmen A pulls with a force of 200 N while workmen while workmen B pulls with a force of 300 N. The ropes used make an angle 250 with each other. Draw a parallelogram and label the resultant force using scale of 1 cm : 50 N. Determine the magnitude of resultant force. 42
  • 25. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionResolution of a force1. Resolution of a force is …………………………………………………………………  Refer to trigonometric formula: Example : The figure below shows Ali mopping the floor with a force 50 N at an angle of 600 to the floor. F = 50 N Example of resolution and combination of forces F=? 200 N 400 43
  • 26. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionProblem solving1. When a system is in equilibrium, ……………………………………………………….2. If all forces acting at one point are resolved into horizontal and vertical components, ……………………………………………………………………………3. Example 1; Show on a figure; a) the direction of tension force, T of string b) the resultant force act to lamp 0 0 70 70 c) calculate the magnitude of tension force, T a) T b) T’ T mlamp = 1.5 kg Wlamp = 14.7 NExercise 2.91. Two force with magnitude 18 N and 6 N act along a straight line. With the aid of diagrams, determine the maximun possible value and the minimum possible value of the resultant force.2. A football is kicked simultaneously by two players with force 220 N and 200 N respectively, as shown in Figure 2.9. Calculate the magnitude of the resultant force. 220 N 900 200 N 44
  • 27. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.10 UNDERSTANDING WORK, ENERGY AND EFFICIENCYWork1. Work is done, …………………………………………………………………………….. ………………………………………………………………………………………………2. WORK is the product.……………………………………………………………………. ………………………………………………………………………………………………3. The formulae of work;4. Example 1; Force, F sExample 2; 80 N 600 s= 5m 45
  • 28. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExample 3;Example 4; F = 600 N S = 0.8 mEnergy1. Energy is .................................................................................................................2. Energy cannot be ....................................................................................................3. Exist in various forms such as …………………...…………………………………… ………………………………………………………………………………………………4. Example of the energy transformation; ……………………………………………………………………………………………… ………………………………………………………………………………………………5. ……………………………………………………………………………………………… Example : ……………………………………………………………………………………………… 46
  • 29. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionWork done and the change in kinetic energy Force, F s1. Kinetic energy is …………………………………………………………………………2. Refer to the figure above, Through, v2 = u2 +2as u=0 and, as = ½ v23. Example 1; A small car of mass 100 kg is moving along a flat road. The resultant force on the car is 200 N. a) What is its kinetic energy of the car after moving through 10 m? b) What is its velocity after moving through 10 m?Work done and gravitational potential energy h = 1.5 m1. Gravitational potential energy is………………………………………………………... ………………………………………………………………………………………………2. Refer to the figure above;3. Example; If m = 10 kg 47
  • 30. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionPrinciple of conservation of energyCarry out hands-on activity 2.10 on page 38 of the practical book.To show the principle of conservation of energy.1. Energy cannot be ……………………………………………………………………… ……………………………………………………………………………………………2. Example : a thrown ball upwards will achieve a maximum height before changing its direction and falls3. Example in calculation : A coconut falls from a tree from a height of 20 m. What is the velocity of coconut just before hitting the earth?Power1. Power is …………………………………………………………………………………2. A weightlifter lifts 180 kg of weights from the floor to a height of 2 m above his head in a time of 0.8 s. What is the power generated by the weightlifter during this time? g = 9.8 ms-2) 48
  • 31. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionEfficiency1. Defined……..…………………………………………………………………………….2. Formulae of efficiency :3. Analogy of efficiency; Device/ mechine Energy transformation4. Example; An electric motor in a toy crane can lift a 0.12 kg weight through a height of 0.4 m in 5 s. During this time, the batteries supply 0.8 J of energy to the motor. Calculate (a) The useful of output of the motor. (b) The efficiency of the motorCarry out hands-on activity 2.11 on page 39 of the practical book to measure the power. 49
  • 32. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExercise 2.101. What is the work done by a man when he pushes a box with a force of 90 N through a distance of 10 m? State the amount of energy transferred from the man to the force.2. A sales assistant at a shop transfers 50 tins of milk powder from the floor to the top shelf. Each tin has a mass of 3.0 kg and the height of thee top shelf is 1.5 m. (a) Calculate the total work done by the sales assistant. (b) What is his power if he completes this work in 250 s?2.11 APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY OF DEVICES1. During the process of transformation the input energy to the useful output energy,……………………………………………………………………………………..2. .……………………………………………………………………………………………..3. ……………………………………………………………………………………………… Example of wasting the energy; ………..………………… Input enegy output from the petrol energy …………………… ……………. ……………… ……………………. ..………………….. …………….. ………………….. ……………………. ..………………….. ……………. …………………. ……………………. 50
  • 33. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion4. The world we are living in face acute shortage of energy.5. It is very important that a device makes …………………………………………………………………Ways of increasing the efficiency of devices1. Heat engines ……………………..……………………………………………………… ………………………………………………………………………………………………2. Electrical devices. ...……………………………………………………………………... ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………Operation of electrical devices1. The electrical devices increase the efficiency………………………………….……2. Proper management ….....………………………………………………………………3. …………..……………………………………………………………………………… ……………………………………………………………………………………………… 51
  • 34. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2.12 UNDERSTANDING ELASTICITYCarry out Hands-on activity 2.12 page 40 of the practical book.1. Elasticity is ……………………………………………………………………………... ………………………………………………………………………………………………2. Forces between atoms ………………………………………………………………….. ………………………………………………………………………………………………3. Forces between atoms in equilibrium condition Force of attraction Force of repulsion Force of repulsion Explanation : ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………4. Forces between atoms in compression compressive force compressive force Force of repulsion Force of repulsion Explanation ; ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………5. Forces between atoms in tension force of attraction stretching force stretching force Explanation ; ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… 52
  • 35. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionCarry out Experiment 2.4 on page 41 of the practical bookTo investigate the relationship between force and extension of a springHooke’s Law1. Hooke’s Law states ……………………………………………………………………… ………………………………………………………………………………………………2. Elastic limit of a spring is defined………………………………………………………. ………………………………………………………………………………………………3. The spring is said to have a permanent extension,...………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………4. The elastic limit is not exceeded,…………………………………………….………… ……………………………………………………………………………………………… ………………………………………………………………………………………………5. Graf F against x F/ N Q P E 0 R x (cm)6. Spring Constant, k F/N 0.8 0 8 x/cm 53
  • 36. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExample 1; A spring has an original length of 15 cm. With a load of mass 200 g attached, the length of the spring is extend to 20 cm. a. Calculate the spring constant. b. What is the length of the spring when the load is in increased by 150 g? [assume that g = 10 N kg-1]Example 2; The graph shows the relationship between the Graph F against x of stretching force, F and the spring extension, x. F (N) spring P and spring Q (a) Calculate the spring constant of P and Q. (b) Using the graph, determine the 8 P stretching force acts to spring P and 7 spring Q, when their extension are 0.5 cm 6 5 Q 4 3 2 1 0 0.1 0.2 0.3 0.4 0.5 x (cm)Elastic potential energy1. Elastic potential energy ……………………………………………………………….. spring with the original length F compression x spring compressed x F x = compression x x F spring extended x = extension F, extension Other situation where the spring extended and compressed 54
  • 37. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionRelationship between work and elastic potential energy F/N Graph F against x F x x / cmExample ; 5 kg 15 cm 8 cmFactors that effect elasticityHands-on activity 2.13 on page 42 the practical book to investigate the factors that affectelasticity. Type of material different same same same Diameter of spring wire same different same same Diameter of spring same same different same Length of spring same Same same differentSummarise the four factors that affect elasticity Factor Change in factor Effect on elasticity Shorter spring Less elastic Length Longer spring More elastic Smaller diameter Less elastic Diameter of spring Larger diameter More elastic Smaller diameter More elastic Diameter of spring wire Larger diameter Less elastic Type of material the elasticity changes with the type of materials 55
  • 38. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionExercise 2.121. A 6 N force on a spring produces an extension of 2 cm. What is the extension when the force is increased to 18 N? State any assumption you made in calculating your answer.2. If a 20 N force extends a spring from 5 cm to 9 cm, (a) what is the force constant of the spring? (b) Calculate the elastic potential energy stored in the spring.Reinforcement Chapter 2Part A : Objective Questions1. When a coconut is falling to the 3. Calculate the weight of a stone with ground, which of the following mass 60 g on the surface of the quantities is constant? moon. (The gravitational acceleration of the A. Velocity moon is 1/6 that of the Earth.) B. Momentum C. Acceleration A. 0.1 N D. Kinetic energy B. 0.2 N C. 0.4 N2. In an inelastic collision, which of the D. 0.6 N following quantities remains E. 0.8 N constant before and after the collision? 4. The momentum of a particle is dependent on A. Total acceleration B. Total velocity A. mass and acceleration C. Total momentum B. weight and force D. Total kinetic energy C. mass and velocity 56
  • 39. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion5. Which of the following diagrams 8. m = 0.3 kg shows a body moving at constant velocity? 5m A. 2N 2N What is the momentum of the stone B. 12 N 7N just before it hits the ground? C. 12 N 14 N A. 0.15 kg m s-1 B. 0.3 kg m s-1 D. 20 N 17 N C. 1.5 kg m s-1 D. 3.0 kg m s-1 E. 15.0 kg m s-16. The graph below shows the motion of a trolley with mass 1.5 kg. Solution : -1 Velocity / ms 4 0 2 4 6 Time / s 9. A big ship will keep moving for some Calculate the momentum of the distance when its engine is turned off. trolley from t = 2s to t = 4s. This situation happens because the ship has A. 1.5 kg m s-1 B. 3.0 kg m s-1 A. great inertia C. 4.0 kg m s-1 B. great acceleration D. 6.0 kg m s-1 C. great momentum E. 7.5 kg m s-1 D. great kinetic energy 10. An iron ball is dropped at a height of7. This figure shows an aircraft flying 10 m from the surface of the moon. in the air. Lift Calculate the time needed for the iron ball to land. (Gravitational acceleration of the Thrust Air friction moon is 1/6 that of the Earth and g = 9.8 N kg-2) Weight A 0.6 s The aircraft above accelerates if B 1.4 s C 1.7 s A. Lift  Weight D 3.5 s B. Thrust  Lift E 12.0 s C. Lift  Air friction D. Thrust  Air friction 57
  • 40. JPN Pahang Physics Module Form 4 Chapter 2 : Force and MotionPart B : Structure Questions1. (i) Car A (ii) Car B Diagram 1.1 Diagram 1.1(i) and (ii) show two methods used by the mechanic to move a breakdown car. A constant force, F = 500 N is used to push and pull the car in method A and B. (a) (i) Which method is easier to move the car? ……………………………………………………………………………… (ii) State a reason for your answer in (a)(i). ……………………………………………………………………………… ……………………………………………………………………………… (b) The frictional force acting between the car and track surface in both methods is 200 N. Calculate, the (i) horizontal resultant force in method A. (ii) horizontal resultant force in method B. (iii) acceleration of the car in method B. (c) Suggest a method to move Car B so that the acceleration produced is equal to that of method A. ……………………………………………………………………………..……….. ……………………………………………………………………………………… 58
  • 41. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2. ceiling Tin P Q water M N  R hand (i) Diagram 2.1 (ii) a) Diagram 2.1(i) shows tin P that is empty and tin Q that is filled with water. A student find difficult to pushed tin Q. Write the inference about the observation. ……………………………………………………………………………………… b) Diagram 2.1(ii) shows a tin being released from the different positions M and N. The hand of a student at position R needs greater force to stop the motion of the tin falling from position M. Explain this observation. ……………………………………………………………………………………… ……………………………………………………………………………………… c) Based on the observation (i) and (ii), state two factors that affect the magnitude of the momentum of the object. ……………………………………………………………………………………… d) If water flows out from a hole at the bottom of the tin Q, how would the inertia of Tin Q depends on time ? ……………………………………………………………………………………3. 2 ms-1 P iron ball ( 2 kg ) S T 3.0 m smooth surface 1.0 m 2.0 m Q R Diagram 3 Rough surface The figure shows a iron ball that is rolled through PQRST. The rough surface of QR has frictional force of 4 N. a) Calculate (i) the kinetic energy of the iron ball at P. 59
  • 42. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion (ii) the potential energy of the iron ball at P. (iii) the total of energy of the iron ball at P. b) c) (i) Calculate the total of energy of the iron ball when it reaches at Q ? (ii) Calculate the work done against friction along QR. d) Calculate the total kinetic energy of the iron ball at S. e) Calculate the speed of the ball at position T.Part C : Essay Questions1. (i) (ii) Diagram 1.1 Diagram 1.1(i) shows the condition of a car moving at high velocity when it suddenly crashes into a wall. Diagram 1.1(ii) shows a tennis ball hit with racquet by a player. a) (i) What is the meaning of momentum? (ii) Based on the observations of Diagram (i) and (ii), compare the characteristics of car when it crashes into the wall and the tennis ball when it is hit with a racquet. Hence, relate these characteristics to clarify a physics concept, and name this concept. 60
  • 43. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion b) Explain why a tennis player uses a taut racquet when playing. c) In launching a rocket, a few technical problems have to be overcome before the rocket can move upright to the sky. By using appropriate physics concepts, describe the design of a rocket and the launch techniques that can launch the rocket upright.Answer a) (i) momentum is product of mass and velocity (ii) - The shape of car changed but the shape of wall remained. - The shape of ball remained but the shape of the racquet string was changed. (The racquet string is elastic but the wall is harder) - The time taken of collision between the ball and racquet string more than the time taken when the car hit the wall. - The impulsive force will decrease when the time of collision increased. - The concept is the impulsive force. b) - To decrease the time of collision between the ball and the racquet string. - Impulsive force will be increased. - The force act to the ball will be increased. - The velocity of ball will be increased. c) - Make a gradually narrower at the front shape (tapering) : To decrease air friction - Made by the high strength and high rigidity of materials : To decrease the probability to become dented (kemik). - Made by the low density of material. : To reduce the mass/weight - The structure is fractional engine : The mass will be decreased and the velocity will increase. - Made by the high of heat capacity of materials : It will be high heat resistance. 61
  • 44. JPN Pahang Physics Module Form 4 Chapter 2 : Force and Motion2. Properties Engine thrust Resistance force Brand Reaction time / s Mass / kg force / N /N A 0.3 1.5 10.0 4.0 B 0.5 1.8 12.5 2.4 C 0.2 0.9 6.5 2.2 D 0.6 2.5 16.0 6.5 In a radio-controlled car racing competition, 4 mini-cars branded A, B, C and D took part. The information of the 4 cars is given in the table above. Details of the above information are given as below; Reaction time - Duration between the moment the radio-controlled is switched on and the moment the car starts moving. Resistance - Average value of opposing forces includes the friction between wheels and track, and air resistance. (a) What is the meaning of acceleration? (b) Draw a graph of velocity against time that shows a car moving initially with constant acceleration, then moving with constant velocity and followed by constant deceleration until it stops. (c) Explain the suitability of the properties in the above table in constructing a radio- controlled car racing purpose. Hence, determine which brand of car will win the 50-metre race. (c) If Car B in the above table is moved up the plane at the angle of 30o to the horizon, (i) Show that the car is able to move up the plane. (ii) Determine the acceleration of the car. 62