physics experiment 1 stpm sem 1

Physics Study Notes

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Physics STPM Past Year Questions With Answer 2007

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Pra U STPM 2021 Penggal 1 - Physics

Format: 190mm X 260mm Extent= 336 pgs (16.62 mm) 70gsm Status: Cover Master BI vervion_2nd imp PRE-U PELANGI BESTSELLER STPM Text PRE-U STPM Text Chemistry CC039242a PRE-U Term 1 specially designed for students who are sitting for the STPM Physics examination. The comprehensive PRE-U notes and practices are based on the latest syllabus and exam format set by Majlis Peperiksaan Malaysia. This book will provide you with the STPM Text necessary skills and strategies to excel in the subject. TERM 1 Our Pre-U & STPM Titles: STPM Text Physics › Success with MUET › MUET My Way FEATURES › Pengajian Am Penggal 1, 2, & 3 › Bahasa Melayu Penggal 1, 2, & 3 ■ Comprehensive Notes and Practices › Biology Term 1, 2, & 3 Poh Liong Yong › Physics Term 1, 2, & 3 ■ Useful Features like Concept Maps, 1 Learning Outcomes, Exam Tips, Info › Chemistry Term 1, 2, & 3 Chem and STPM Taggings › Mathematics (T) Term 1, 2, & 3 › Sejarah Penggal 1, 2, & 3 SCAN ME TERM ■ Summary › Geografi Penggal 1, 2, & 3 Physics › Ekonomi Penggal 1, 2, & 3 ■ STPM Practices › Pengajian Perniagaan Penggal 1, 2, & 3 For the 21st Century Learner ■ STPM Model Paper Term 1 ■ Complete Answers TERM 1 eBook! Available MORE THAN MORE THAN 1 COPIES COPIES W.M: RM37.95 / E.M: RM38.95 SOLD CC039242a SOLD ISBN: 978-967-2856-92-4 PELANGI 26/07/2021 12:39 PM CVR_PreU_STPM_2021 Physics (Penggal 1).indd 1-3 26/07/2021 12:39 PM CVR_PreU_STPM_2021 Physics (Penggal 1).indd 1-3 CONTENTS Chapter STPM Practice 6 152 1 PHYSICAL QUANTITIES AND UNITS 1 QQ6 155 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1.1 Base Quantities and SI Units 2 Chapter 158 STATICS 1.2 Dimensions of Physical Quantities   4 7 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1.3 Scalars and Vectors 13 7.1 Centre of Gravity 159 1.4 Uncertainties in Measurements 19 7.2 Equilibrium of Particles 162 STPM Practice 1 25 7.3 Equilibrium of Rigid Bodies 172 QQ1 28 STPM Practice 7 179 Chapter QQ7 182 2 KINEMATICS 32 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • DEFORMATION OF SOLIDS 184 2.1 Linear Motion 33 8 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2.2 Projectile 45 8.1 Stress and Strain 185 STPM Practice 2 53 8.2 Force–extension Graph and QQ2 56 Stress-strain Graph 187 Chapter 8.3 Strain Energy 198 3 DYNAMICS 59 STPM Practice 8 202 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 3.1 Newton’s Laws of Motion 60 QQ8 205 3.2 Friction 68 Chapter 208 KINETIC THEORY OF GASES 3.3 Conservation of Linear Momentum 70 9 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 3.4 Elastic and Inelastic Collisions 71 9.1 Ideal Gas Equation 209 3.5 Centre of Mass 79 9.2 Pressure of a Gas 216 STPM Practice 3 82 9.3 Molecular Kinetic Energy 219 QQ3 85 9.4 The r.m.s Speed of Gas Molecules 223 Chapter 9.5 Degree of Freedom and Law of 4 WORK, ENERGY AND POWER 88 Equipartition of Energy 226 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 4.1 Work 89 9.6 Internal Energy of an Ideal Gas 229 4.2 Energy 92 STPM Practice 9 232 4.3 Power 96 QQ9 234 STPM Practice 4 99 Chapter THERMODYNAMICS OF GASES 237 QQ4 101 10 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 10.1 Heat Capacities 238 5 CIRCULAR MOTION 104 10.2 Work Done by a Gas 240 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 5.1 Angular Displacement and Angular 10.3 First Law of Thermodynamics 244 Velocity 105 10.4 Isothermal and Adiabatic Changes 255 5.2 Centripetal Acceleration 107 STPM Practice 10 270 5.3 Centripetal Force 108 QQ10 274 STPM Practice 5 120 Chapter HEAT TRANSFER 278 QQ5 122 11 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 11.1 Conduction 279 6 GRAVITATION 125 11.2 Convection 304 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 11.3 Radiation 305 6.1 Newton’s Law of Universal Gravitation 126 11.4 Global Warming 309 6.2 Gravitational Field 127 STPM Practice 11 311 6.3 Gravitational Potential 133 QQ11 316 6.4 Satellite Motion in a Circular Orbit 138 STPM Model Paper (960/1) 320 6.5 Escape Velocity 148 Summary of Key Quantities and Units 329 vi PRELIM STPM PHY T1.indd 6 4/9/18 8:11 AM CHAPTER 2 KINEMATICS Concept Map Kinematics Linear Motion Projectile u Motion with Constant Acceleration u sin • v = u + at u cos • s = 1 (u + v)t 2 • s = ut + 1 at 2 2 • v = u + 2as Effect of Air Resistance 2 2 v Graphical Method s v 0 t a 0 t 0 t t 0 Bilingual Keywords Acceleration: Pecutan Resistance: Rintangan Displacement: Sesaran Speed: Laju Kinematics: Kinematik Velocity: Halaju Projectile: Luncuran 32 02 STPM PHY T1.indd 32 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 2.1 Linear Motion 2010/P1/Q3, 2016/P1/Q2,Q18 Learning Outcomes Students should be able to: • derive and use equations of motion with constant acceleration • sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration 2 1. Kinematics is the study of motion without considering the causes of the motion. Linear motion is motion along a straight line. 2. The displacement s of a body from a point O to another point P is the distance moved by the body along the straight line OP. Displacement s has both magnitude and direction, hence it is a vector. 3. The velocity v of a body is its rate of change of displacement. v ds O ds P Velocity, v = Figure 2.1 dt Velocity v is also a vector, its direction is along the direction of the change in displacement ds. (Figure 2.1). ds 4. The expression v = gives the instantaneous velocity. The instantaneous velocity of a car dt travelling along a straight road is shown by the speedometer. 5. If the time taken by a body to move through a displacement s is t, then Final displacement, s average velocity = Time taken, t Figure 2.2 The speedometer shows the instantaneous velocity 6. Acceleration a is the rate of change of velocity. dv Acceleration, a = dt Acceleration is also a vector. Its direction follows the direction of the change in velocity dv. When the velocity of a body changes either in magnitude, or in direction, or both, the body accelerates. Motion With Constant Acceleration 1. The acceleration of a body is uniform if the magnitude of the acceleration is constant and its direction remains unchanged. 33 02 STPM PHY T1.indd 33 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 2. For a body travelling along a straight line, if u = initial velocity, v = final Exam Tips velocity after a time t, and s = displacement in time t, There are five quantities, Change in velocity then uniform acceleration, a = u, v, t, s, a, in the four Time taken equations for motion v – u under constant = acceleration. Each of the t equations involves only v = u + at .............................. a four of the quantities. 2 In a question, usually Displacement, s = Average velocity × Time values for three of the quantities are provided, ( ) and you are asked to u + v s = 2 t .............................. b calculate the value of another of the quantities. ( u + u + at Use the equation 2 ) s = t that contains the four quantities mentioned in the question. 2 2 1 s = ut + at .............................. c Info Physics 2s From ①, v – u = at From ➁, v + u = t A380 Airbus take-off speed (v – u)(v + u) = at × 2s A wide-body aircraft such as t the Airbus A380 has a take- –1 off speed of 280 km h and v – u = 2as .............................. d ➃ requires a runway of length 2 2 3000 m. What then is its ac- celeration? Example 1 –1 A motorist travelling at 72 km h on a straight road approaches a traffic light which turns red when he is 55.0 m away from the stop line. The reaction time is 0.7 s. With the brakes applied fully, the vehicle decelerates at 5.0 m s . How far from the stop line will he stop and on which –2 side of the stop line? Solution: –1 72 km h = 72 000 m = 20 m s –1 3 600 s –1 The vehicle travels at constant speed of 20 m s during the reaction time of 0.7 s. Distance travelled during the reaction time = 20 × 0.70 = 14.0 m When the vehicle stops, final velocity v = 0 Acceleration = –5.0 m s (negative because of deceleration) –2 Using v = u + 2as 2 2 2 Distance travelled when the brakes are applied, s = v – u 2 2a = 0 – 20 2 2 (–5) = 40.0 m Total distance travelled = (14.0 + 40.0) m = 54.0 m Hence, the vehicle stops (55.0 – 54.0) m = 1.0 m before the stop line. 34 02 STPM PHY T1.indd 34 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Example 2 The table below is from the handbook of a car. On a dry road, the car driven by an alert driver will stop in distances as shown below. Speed/ Thinking Braking Overall stopping m s –1 distance/m distance/m distance/m 5.0 3.0 1.9 4.9 2 10.0 6.0 7.5 13.5 15.0 9.0 17.0 26.0 20.0 12.0 30.0 42.0 25.0 15.0 47.0 62.0 30.0 18.0 68.0 86.0 35.0 21.0 92.0 113.00 The thinking distance is the distance travelled by the car during the driver’s reaction time. The braking distance is the distance travelled by the car before the car is stopped when the brakes are applied. (a) Explain why the thinking distance is directly proportional to the speed, whereas the braking distance is not. State the relationship between the braking distance and the speed. (b) What is the value of deceleration used in calculating the braking distance? –1 (c) Calculate the overall stopping distance for a car travelling at 40 m s . Solution: (a) During the reaction time, t of the driver, the car travels at a constant speed, u. Hence, thinking distance, s 1 = ut s 1 ∝ u The reaction time t of a driver is constant, The final speed after the brakes are applied = 0 2 If deceleration = a, using v = u + 2as 2 0 = u – 2as 2 2 Braking distance, s 2 = u 2 ∝ u 2 2a u 2 (b) Using braking distance, s 2 = 2a –1 When u = 10 m s , s 2 = 7.5 m u 2 Deceleration, a = 2s 10 2 = 2 × 7.5 = 6.67 m s –2 (c) When u = 40 m s –1 2 u Thinking distance, s 1 = 40 × Reaction time Braking distance, s 2 = 2a 40 2 3.0 = 40 × (5.0) = 2 × 6.67 = 24 m = 120 m Overall distance travelled = (24 + 120) m = 144 m 35 02 STPM PHY T1.indd 35 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Quick Check 1 1. A car starts from rest and accelerates at 2. A car decelerates uniformly from 30 m s to –1 2.0 m s . 15 m s in a distance of 75 m. Calculate the –2 –1 (a) Find its speed at (i) 1.0 s, (ii) 2.0 s and further distance travelled before the car comes (iii) 5.0 s. to a stop. 2 (b) What is the average speed in (i) the first second, (ii) 2.0 s and (iii) 5.0 s? 3. A car accelerates through three gear changes (c) Calculate the distance travelled in with the following speed: –1 (i) 1.0 s, (ii) 2.0 s and (iii) 5.0 s. 20 m s for 2.0 s 40 m s for 2.0 s –1 60 m s for 6.0 s –1 What is the overall average speed of the car? Motion Under Gravity 2009/P1/Q2, 2011/P2/Q1, 2017/P1/Q2 1. When a body is released from rest, it falls towards the earth with an acceleration. 2. In free space where there is no air resistance, all objects irrespective of their mass, fall with the same acceleration, known as the acceleration of free fall or acceleration due to gravity, g. 3. Close to the earth, the value of g can be as assumed constant. In this book, the value of g is assumed –2 to be 9.81 m s unless otherwise stated. 4. Since the value of g is constant, equations of motion under uniform acceleration such as: 1 2 v = u + at, s = ut + at and v = u + 2as 2 2 2 can be used for motion under gravity by substituting the acceleration a as +g, or as –g where applicable. 5. When the body is only falling downwards, it may be easier to use a = +g, since the direction of motion is in the direction of g, that is downwards. 6. When the body is projected vertically and upwards, if it is assumed that the velocity of the body is positive, it implies that the upwards direction is assumed positive. Since the direction of g is downwards, hence acceleration a = –g. 7. When the body moves in both directions, upwards then falls downwards or vice versa, it would be better to assume the upwards direction as positive. Hence, the acceleration a = –g. Upwards journey Downwards journey Displacement : + E E Displacement : + Velocity : + F Velocity : – Acceleration : a = –g D Acceleration : a = –g G C B H A Below the level AI I Displacement : – Velocity : – J Acceleration : a = –g Free Fall INFO Figure 2.3 36 02 STPM PHY T1.indd 36 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 8. Figure 2.3 shows a particle being projected upwards from the point A. It reaches the highest point E and then falls down to a point below A. Example 3 A bullet is fired vertically upwards from the ground with an initial velocity of u and takes a time 2 of t 1 to reach a point P at a height h. From then the bullet takes a further time of t 2 to move to the highest point and back to the ground. Find in terms of t 1 , t 2 and g (a) the initial velocity u, (b) height h, and (c) the greatest height H reached. Solution: (a) Assume the upwards direction as positive, then a = –g. H Total time taken for the bullet to move up and back to the ground, t = (t 1 + t 2 ) P When the bullet returns to the ground, H v t 1 displacement, s = 0 h 1 Using s = ut + at 2 2 (t + t ) 1 2 1 0 = u(t 1 + t 2 ) – g(t 1 + t 2 ) 2 2 1 u = g(t 1 + t 2 ) 2 (b) When s = h, t = t 1 (c) At the highest point, s = H, velocity = 0 2 Using v = u + 2as 2 1 Using s = ut + at 2 2 0 = u – 2gH 2 1 1 2 u = g(t 1 + t 2 ) 2 2 u 2 1 h = ut 1 – gt 1 H = u = g(t 1 + t 2 ) 2 1 1 2g = 2 2 g(t 1 + t 2 )t 1 – gt 1 1 1 ] 2 2 = g(t 1 + t 2 ) 1 2g[ 2 = 2 gt 1 t 2 1 = g (t 1 + t 2 ) 2 8 Quick Check 2 1. A ball is thrown vertically upwards from the 2. A spaceship descends at a constant velocity top of a building, 9.0 m above the ground with of 10 m s on the moon. When it is 120 m –1 a speed of 8.0 m s . The ball hits the ground from the moon’s surface, an object falls off the –1 after a time T. Calculate spaceship. If the acceleration due to gravity on (a) the time taken to reach the greatest height. the moon is 1.6 m s , what is the velocity of the –2 (b) the greatest height reached by the ball. object when it reaches the surface of the moon? (c) the time, T taken by the ball to reach the ground. 37 02 STPM PHY T1.indd 37 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 3. The acceleration of free fall is determined P by timing the fall of a ball bearing using photocells. The ball is released from the point Light X P, and passes the point X and Y at time t 1 and beam t 2 after being released. Find an expression for h Photo-cells the acceleration due to gravity in terms of h, t 1 Y and t 2 . Light 2 beam Q Graphical Methods The motion of a body can be analysed by studying the various graphs for the motion: (a) Displacement-time graph (b) Velocity-time graph Displacement-time Graph 2010/P1/Q2, 2014/P1/Q2 1. Informations that can be deduced from the displacement-time graph: (a) The instantaneous displacement. The displacement s at any time t can be obtained off the graph. (b) The velocity = Gradient of the graph. (c) The manner that the velocity changes with time from the shape of the graph. 2. The displacement-time (s-t) graphs of a few types of motion are as shown in Figure 2.4. (a) s (b) s t t 0 0 Constant velocity Increasing velocity Gradient = constant Gradient increasing (c) (d) s s 0 t 0 t T 2T Decreasing velocity Motion of a body projected vertically upwards and (deceleration) then falling back to the ground. When t = T, the body Gradient decreasing is at the highest point. When t = 2T, the body is back on the ground. Figure 2.4 Displacement-time graphs 38 02 STPM PHY T1.indd 38 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Exam Tips Note the difference between graph (a) and graph (b) when t = 0. v v 2 0 t 0 t (a) (b) Initial velocity u = 0 Initial velocity u ≠ 0 Gradient at t = 0 is zero. Gradient at t = 0 is not zero. Velocity-time Graph 1. Informations that can be deduced from the velocity-time graph: (a) The instantaneous velocity. t 2 (b) Between the time t = t 1 and t = t 2 , displacement = ∫ v dt = shaded area under the graph. t 1 dv (c) The acceleration, a = = Gradient of graph. dt (d) How the acceleration changes with time from the shape of the graph. 2. Figure 2.5 shows the velocity-time (v-t) graphs for various types of motion. (a) v (b) v (c) v u u 0 t t t 0 t t t 0 t t t 1 2 1 2 1 2 Uniform acceleration. Uniform deceleration. Increasing acceleration. Gradient = constant Constant negative gradient Gradient increasing (d) v (e) v 0 t 0 t t t t t 1 2 1 2 Decreasing acceleration. Constant velocity Gradient decreasing Figure 2.5 Velocity-time graphs 39 02 STPM PHY T1.indd 39 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Example 4 The graph shows the speeds of two cars A and B Speed / m s –1 which are travelling in the same direction over a period of 80 s. Car A travelling at a constant 30 B speed of 20 m s overtakes car B at time t = 0. –1 2 In order to catch up with car A, car B immediately 20 A accelerate uniformly for 30 s to reach a constant speed of 30 m s . –1 (a) How far does car A travel during the first 30 s? (b) Calculate the acceleration of car B in the first 30 s. 5 (c) What is the distance travelled by car B in this 0 Time / s time? 30 80 (d) What additional time will it take for car B to catch up with car A after car A passes car B? (e) How far would each car have travelled since t = 0? (f) What is the maximum distance between the cars before car B catches up with car A? Solution: (a) Distance travelled by car A during the first 30 s = 30 × 20 = 600 m (b) Acceleration of car B = Gradient of graph 30 – 5 = 30 = 0.833 m s –2 (c) Distance travelled by car B during the first 30 s = Area under the graph from t = 0 to t = 30 s = 1 (5 + 30) × 30 2 = 525 m (d) After 30 s, car A is ahead of car B by (600 – 525) m = 75 m For car B to catch up with car A, Further distance travelled by car B – Further distance travelled by car A = 75 m 30t – 20t = 75 t = 7.5 s Hence, car B catches up with car A (30 + 7.5) s from when car A passes car B, i.e. 37.5 s. (e) Distance travelled by each car = 20 × 37.5 m = 750 m (f) The two cars are furthest apart when the two graphs intercept at P. At P, the difference (Area under graph A – Area under graph B) is maximum. Refering to the figure: x = 10 x + y = 30 Speed / m s –1 y 15 2 2 x = y y + y = 30 30 B 3 3 3 y = × 30 P 5 20 y A = 18 s x 15 Hence, maximum separation = Shaded area 1 = × 18 × 15 5 2 Time / s = 135 m 0 30 80 40 02 STPM PHY T1.indd 40 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Example 5 (a) A body accelerates uniformly from rest along a straight line. Sketch a graph to show the variation of displacement with time. How can the instantaneous velocity be deduced from the displacement-time graph? (b) A cricket player throws a ball vertically upwards and catches it 3.0 s later. Neglecting air resistance, calculate 2 (i) speed of the ball when it leaves the player’s hand, (ii) the maximum height reached by the ball. (c) Sketch a graph to show how the velocity of the ball varies with time. Mark each of the following instants on the graph: (i) The ball leaves the player’s hand (t 1 ). (ii) The ball at the maximum height (t 2 ). (iii) The ball returns to the player’s hand again (t 3 ). (d) You were told that the air resistance is negligible. In actual fact the ball experiences a retarding force during its motion. Without making any calculation, explain how the air resistance affects (i) the time taken to reach the highest point, (ii) the value of the maximum height reached. (e) Discuss, taking into account the air resistance, whether the time t u taken by the ball to reach the highest point is greater or smaller than t d , the time taken for the ball to drop. –2 (Assume g = 10 m s ). Solution: (a) Instantaneous velocity = Gradient of the graph at time, t s 0 t Note that the gradient of the graph = 0 when t = 0 because initial velocity = 0 (b) (i) Suppose u = velocity of ball when it leaves (ii) At the maximum height H, the player’s hand. velocity v = 0 Acceleration a = –g Using v = u + 2as 2 2 = –10 m s –2 0 = 15 – 2 × 10 × H 2 Time t = 3.0 s H = 11.25 m Displacement s = 0 1 Using s = ut + at 2 2 1 0 = 3.0 u – × 10 × (3.0) 2 2 u = 15 m s –1 41 02 STPM PHY T1.indd 41 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics (c) Velocity, v + 0 t t t Time, t 1 2 3 2 – (d) (i) Neglecting air resistance, the time taken to reach the highest point is given by v = u + at v = 0, a = –g 0 = u – gt u Time taken, t = g With air resistance, the retardation a 1 > g using v = u + at 0 = u – a 1 t 1 u u Time taken, t 1 = < (a 1 > g) a 1 g Hence, the time taken is shorter. (ii) Without air resistance, the greatest height reached = H At the greatest height, v = 0 Using v = u + 2as 2 2 0 = u – 2gH a = –g 2 Greatest height, H = u 2 2g With air resistance retardation a 1 > g. u 2 u 2 Hence, the greatest height, H 1 = < (a 1 > g) a 1 g Therefore, the greatest height reached is lower. (e) When the ball moves up, When the ball falls, initial velocity = u, initial velocity = 0, final velocity = 0, final velocity, v < u velocity of projection because of friction time = t u time = t d 1 1 Using s = (u + v)t using s = (u + v)t 2 2 1 1 H 1 = (u + 0)t u 2 H 1 = (0 + v)t d 2 2H 1 2H 1 t u = t d = > t u (v < u) u v Hence, the time taken to come down is greater than the time taken to move up. 42 02 STPM PHY T1.indd 42 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Quick Check 3 1. A body initially at rest moves with uniform 3. The velocity-time (v-t) graph of a vehicle acceleration. Which graph best represents the travelling along a straight line is shown below. variation of displacement s of the body with time t? A C 2 s s t 0 t 0 Which of the following is the displacement- time (s-t) graph of the vehicle? B D A s C s s s 0 t 0 t t 1 t 2 t 3 t 1 t 2 t 3 0 t 0 t B s D s 2. A brick is dislodged from a tall building and falls vertically under gravity. Which of the following curves represents the variation of 0 t t t t 0 t t t t its height h above the ground with time t if 1 2 3 1 2 3 air resistance is negligible? A C 4. A tennis ball is released, it falls vertically to h h the floor and bounces back. Taking velocity upwards as positive. Which of the following is the velocity-time (v-t) graph of the ball? A v C v + + 0 t 0 t 0 t 0 t – – B D h h B v D v + + 0 t 0 t – – 0 t 0 t 43 02 STPM PHY T1.indd 43 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 5. The figure below represents the motion of 8. A puck slides across an icy surface and travels a a ball rebounding from the floor after being distance x in time t under uniform retardation. released from a point above the floor. What is Which of the following when plotted to the quantity represented on the y-axis? represent the motion of the puck would give a straight-line graph? y A x against t C x against t 2 t 2 B x against t D x against t 2 t 9. A car starts from rest and accelerates at –1 –2 0 Time 2.0 m s when a lorry moving at 16 m s in the same direction passes it. What is the distance A Displacement C Acceleration travelled by the car when it overtakes the B Velocity D Kinetic energy lorry? A 16 m C 64 m 6. A falling stone strikes a soft ground at speed u B 32 m D 128 m and decelerates uniformly until it stops. Which one of the following graphs best represents the 10. The velocity-time graph of an object is as variation of the stone’s speed v with distance shown in the figure. s moved into the ground? A v C v Velocity u u s 0 s 0 0 Time B v D v t 1 t 2 Sketch u u (a) the displacement-time graph, (b) the acceleration-time graph. 11. A particle moves in a straight line with 0 s 0 s uniform retardation. At time t = 0, its speed is u and its displacement from the origin is zero. 7. The graph below shows the variation with time Sketch labelled graphs to show how t of the velocity v of a bouncing ball, released (a) the velocity v of the particle, from rest. Downwards velocities are taken as (b) the displacement s of the particle from the positive. At which time does the ball start to origin, leave the floor on the rebounce. vary with time t. Explain the relation between the graphs. v 12. A ball is released from rest from a height of 1.00 m. It falls freely to the floor and rebounces to a height of 0.80 m. Neglecting the time of 0 Time impact with the floor, sketch the velocity-time A B C D graph of the motion, taking upwards velocity as positive. 44 02 STPM PHY T1.indd 44 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 2.2 Projectile 2008/P2/Q1, 2013/P1/Q2, 2014/P1/Q16 Learning Outcomes Students should be able to: • solve problems on projectile motion without air resistance • explain the effects of air resistance on the motion of bodies in air 2 1. A projectile is a body that travels under the action of gravity after being projected at an angle to the horizontal. 2. The motion of a projectile consists of two components: (a) a vertical component which is motion with uniform acceleration, g the acceleration due to gravity, (b) a horizontal component which is motion with constant velocity. 3. The path of a projectile is a curve known as parabola. 10 m s –1 30 m s –1 20 m s –1 30 m s –1 30 m s –1 –1 30 m s –1 30 m s –1 –10 m s –1 30 m s –20 m s –1 30 m s –1 –1 40 m s –1 30 m s –1 50 m s –30 m s –1 –1 53° a = –g 30 m s 30 m s –1 –1 –1 –40 m s 50 m s Figure 2.6 Projectile 4. Figure 2.6 shows the motion of a tennis ball with an initial velocity of 50 m s at an angle of –1 projection 53° to the horizontal. An analysis of the motion shows that: (a) The horizontal component of velocity is constant at 30 m s , because there is no horizontal –1 force on the ball. (b) The horizontal displacements are the same for equal time intervals. (c) The vertical component of velocity continuously changes during the motion. (d) At the highest point of the trajectory, the vertical component of velocity is zero. (e) The acceleration a is constant and vertically downwards, a = –g. 5. To study the motion of a projectile, we consider the horizontal component and vertical components separately. u sin θ u a = –g H y θ O u cos θ x R Figure 2.7 45 02 STPM PHY T1.indd 45 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 6. Figure 2.7 shows a body projected with a velocity u at an angle θ to the horizontal. 7. Consider the vertical component of motion: (a) Initial velocity = u sin θ Acceleration = a = –g Suppose H = maximum height reached At the maximum height, vertical component of velocity = 0 2 2 2 Using v = u + 2as 2 0 = (u sin θ) – 2gH 2 2 Maximum height, H = u sin θ 2g (b) If t 1 = Time taken by the body to reach the maximum height, using v = u + at 0 = u sin θ – gt 1 t 1 = u sin θ g 1 (c) The instantaneous height y, at any time t is given by s = ut + at 2 2 1 y = (u sin θ) t – gt 2 2 (d) Let T = Total time of flight, the time taken by the body to travel up and fall back to the ground. When the body lands on the ground, the vertical displacement s = 0. 1 Using s = ut + at 2 2 1 0 = (u sin θ) T – gT 2 2 2u sin θ Time of flight, T = g = 2t 1 This means that the time taken by the object to go to its maximum height is the same as the time it takes to move from the maximum height to the ground. 8. Consider the horizontal component of motion: (a) Horizontal component of velocity = u cos θ = constant Instantaneous horizontal displacement at any time t, is x = (u cos θ) t (b) The range R of the projectile, R = (u cos θ) T g ) = (u cos θ) ( 2u sin θ 2 R = u sin 2θ g 46 02 STPM PHY T1.indd 46 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics u 2 (c) The maximum range is g and it occurs when sin 2θ = 1 2θ = 90° θ = 45° To obtain the maximum range for a particular speed of projection, the body must be projected at an angle of 45° to the horizontal. 9. Figure 2.8 shows three possible ranges for a body projected with the same speed but at different 2 angles of projection. Exam Tips At the highest point 60° 1. velocity of projectile is u cos θ horizontally and 45° not zero. 30° Maximum range 2. kinetic energy 1 = m (u cos θ) 2 The range is maximum when the angle of projection is 45° 2 Figure 2.8 Example 6 A motorcycle stunt-rider moving horizontally takes off from a point 5.0 m above the ground with a speed of 30 m s . How far away does the motorcycle land? –1 5.0 m x Solution: Consider the vertical component of motion: Initial velocity = 0 Acceleration a = +g (Consider downwards direction as positive) Displacement s = 5.0 m 1 Using s = ut + at 2 2 1 5.0 = 0 + (9.81)t 2 2  2 × 5.0 Time of flight, t = 9.81 = 1.01 s Consider horizontal component of motion: Horizontal displacement x = 30 × t = 30 × 1.01 = 30.3 m 47 02 STPM PHY T1.indd 47 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Example 7 An aircraft flies at a height h with a constant horizontal velocity u so as to fly over a cannon. When the aircraft is directly over the cannon, a shot is fired to hit the aircraft. Neglecting air resistance, find in terms of u, h and g, the acceleration due to gravity the minimum speed of the shell in order to hit the aircraft. 2 Solution: v If v = minimum speed θ = angle of projection u With this minimum speed, the shell hits the aircraft at the maximum h height reached by the shell, and since the shell is fired when the air- θ craft is above the cannon, horizontal component of shell velocity = speed of aircraft, u v cos θ = u .............................. ① v sin θ 2 2 Using maximum height, h = 2g 2 2 v sin θ = 2gh .............................. ② 2 2 ① + ②: v sin θ + v cos θ = 2gh + u 2 2 2 v = u + 2gh 2 Example 8 The diagram shows the path of a bullet fired 20.0 m s –1 horizontally with a velocity of 20.0 m s from a –1 height of 2.0 m. Calculate (a) the speed of the bullet v, 2.0 m (b) the angle θ when the bullet hits the ground. v θ Solution: The horizontal component of velocity v x = 20.0 m s (constant) –1 To find the vertical component of velocity when the bullet hit the ground, consider vertical component of motion: Initial vertical component of velocity = 0 Acceleration, a = g Vertical displacement, s = 2.0 m 1 Using s = ut + at 2 2 1 2.0 = 0 + × 9.81 × t 2 2 t = 0.639 s 48 02 STPM PHY T1.indd 48 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Vertical component of velocity when bullet hits the ground v y = u + at = 0 + 9.81 × 0.639 v x = 6.27 m s –1 2 2 Speed of bullet v = v x + v y θ = 20 + 6.27 2 2 v y v = 21.0 m s –1 2 v y tan θ = v x 6.27 = 20 θ = 17.4° Quick Check 4 1. A stone is thrown from O and follows a v/m s –1 h/m parabolic path. The highest point reached is P. A 20 15 Which of the following is zero at the highest B 20 25 point P? C 15 25 A Acceleration D 25 20 B Vertical component of velocity 4. An aeroplane flying in a straight line at C Kinetic energy constant height of 2 000 m with a speed of D Momentum 200 m s drops an object. The object takes –1 2. A projectile is fired with an initial velocity a time t to reach the ground and travels a –2 u at an angle θ to the horizontal. Neglecting horizontal distance s. Taking g as 10 m s and air resistance, the horizontal distance x it has ignoring air resistance, which of the travelled, and the height y reached after a following gives the values of t and s? time t are t s 1 A 200 s 10 km A x = ut sin θ, y = ut cos θ – gt 2 2 B 100 s 05 km 1 B x = ut sin θ – gt , y = ut cos θ C 020 s 04 km 2 2 D 010 s 02 km 1 C x = ut cos θ, y = ut sin θ + gt 2 2 1 5. A bullet is fired horizontally from the top D x = ut cos θ, y = ut sin θ – gt 2 2 of a cliff on the surface of the earth with a –1 speed of 40 m s . Assuming no air resistance, 3. A projectile is fired with a horizontal velocity what is the speed of the bullet 3 s later? v from a height h as shown in the figure. (g = 10 m s ) –2 It strikes the ground at an angle of θ to the A 30 m s C 50 m s –1 –1 horizontal. Which of the following values of v B 40 m s D 70 m s –1 –1 and h will give the greatest value of the angle θ? v 6. A projectile leaves the ground at an angle of 60° to the horizontal. Its initial kinetic energy is K. Neglecting air resistance, find in terms of h K its kinetic energy at the highest point of the motion. θ 49 02 STPM PHY T1.indd 49 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 7. A ball is thrown with a velocity of 8.0 m s at (b) Calculate the maximum range, R. –1 60° to the horizontal. (c) What is its maximum height above the (a) Draw on the same axes, a graph to ground when its range is maximum? represent the variation with time of (Neglect air resistance) (i) v H , the horizontal component of 9. An aeroplane is flying at a constant horizontal velocity, –1 (ii) v V , the vertical component of velocity. velocity of 50 m s at a height of 1 000 m. 2 Identity your graphs and show suitable What is its horizontal distance from a target values of velocity and time. on the ground, so that a parcel released from (b) Use your graph to find the maximum the plane will hit the target. height reached by the ball. 10. A coin is pushed off from the smooth horizontal surface of a table of height 2.0 m. 8. It falls and strikes the floor at a horizontal distance of 3.2 m from the edge of the table. v Calculate (a) the time taken by the coin to fall through θ the air, R (b) the speed at which the coin leaves the table, –1 A missile is fired with a speed of 500 m s from (c) the velocity of the coin when it strikes the the ground. floor. (a) What is the angle of projection θ for the missile to achieve the maximum range? Effects of Air Resistance 1. When a body moves through the air, the air resistance against the motion of the body is known as the viscous drag. 2. The viscous drag on a body depends on (a) shape of the body Objects which are streamlined experience less drag. (b) velocity of the body The viscous drag is proportional to the square of the velocity. 3. When a body is released from rest and falls through the air, its velocity initially increases. As its velocity increases, the viscous drag increases. The acceleration of the body decreases. Finally when the acceleration is zero, the velocity is constant. This maximum constant velocity is known as terminal velocity. 4. Figure 2.9 shows how the velocity of the body varies with time. Figure 2.10 shows the variation of acceleration with time. Velocity Acceleration g Terminal velocity 0 Time 0 Time Figure 2.9 Figure 2.10 50 02 STPM PHY T1.indd 50 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 2 Figure 2.11 Info Physics 5. To slow down his fall, a skydiver spreads himself wide Highest free-fall to increase the viscous drag (Figure 2.11). What happens October 14, 2012, Austria’s Felix Baumgartner to the skydiver when his parachute opens? jumped off a helium-filled ballon at an altitude of 39,045 m and free-fall for 4 minutes, He reached a speed of 1342.0 km h (1.24 times the speed –1 of sound) Example 9 A ball bearing is released from rest below the surface of a viscous liquid in a tall and wide container. Sketch a graph to show how the height h of the ball bearing from the base of the container varies with time. Explain the shape of the graph. Solution: z If z = Instantaneous displacement from the liquid surface. H = Height of liquid z = H – h h = height from base of container H h = H – z h dh = – dz dH = 0 dt dt dt = –v v = velocity of ball bearing The initial velocity of the ball bearing is zero. Hence, when t = 0, gradient of graph = 0 The velocity of the ball bearing increases until the terminal velocity is achieved. h H 0 l Since the gradient of h - t graph, dh = –v, the gradient becomes more negative until a constant dt negative value is obtained. When the terminal velocity is achieved, the h-t graph is straight. Hence, the h-t graph is as shown. 51 02 STPM PHY T1.indd 51 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Quick Check 5 1. A small metal sphere is held just below the C Acceleration / m s –2 surface of a viscous liquid in a tall vessel. It is then released and its displacement s with 10 time t is plotted. Which of the following is the 2 s-t graph? A s C s 0 Time D Acceleration / m s –2 t t 10 0 0 B s D s 0 Time 4. A ball is thrown vertically upwards in a viscous t t medium. The time of flight for the upwards 0 0 motion is t u and for the downwards motion is t d . Which of the following is correct? 2. A steel sphere is held a short distance below the surface of a liquid in a deep vessel and A t d > t u because the ball moves faster on its then released. The terminal velocity of the downwards motion and the viscous drag sphere in the liquid is independent of is greater. A the height of the liquid in the vessel B t d > t u because the magnitude of the B the density of the liquid acceleration when the ball is moving C the diameter of the sphere down is smaller than the magnitude of the D the temperature of the liquid retardation during the upwards motion. C t d < t u because the viscous drag is the 3. Which one of the following graphs represents greatest at the moment of projection. the acceleration of a small rain drop falling D t d < t u because at a given speed the through the air. Acceleration due to gravity net accelerating force when the ball –2 g = 10 m s . is moving downwards is greater that the A Acceleration / m s –2 net retarding force when it is moving upwards. 10 5. Four spheres are released from the top floor of a skyscraper and fall through air to the ground. The spheres attain the terminal velocity before 0 Time reaching the ground. The sphere with the B Acceleration / m s –2 greatest terminal velocity is A steel sphere of radius 1 cm 10 B steel sphere of radius 2 cm C polystyrene sphere of radius 1 cm D polystyrene sphere of radius 2 cm 0 Time 52 02 STPM PHY T1.indd 52 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics Important Formulae 1. Equations for motion with constant accelaration 1 1 v = u + at s = (u + v)t s = ut + at 2 2 2 v = u + 2as 2 2 2. Projectile 2 2 u sin θ 2 Maximum height, H = 2g 2 Range, R = u sin 2θ g Range is maximum when = 45° At time = t, horizontal displacement, x = (u cos θ)t 1 vertical displacement, y = (u sin θ)t – gt 2 2 STPM PRACTICE 2 1. An object is projected vertically upwards from 3. An object starts from rest and its acceleration- the top of a building of height 12.0 m with a time graph is as shown below. –1 speed of 6.0 m s . What is the time taken by Acceleration / ms –2 the object to reach the ground? [Acceleration of free fall is 10.0 m s .] 4 –2 A 1.06 s 2 B 2.00 s 0 Time / s C 2.26 s 4 8 12 16 D 4.90 s –2 2. The displacement-time graph of a body is What is the maximum velocity of the object? shown below. A 16 m s –1 B 22 m s –1 Displacement C 24 m s –1 D 26 m s –1 Q 4. A vehicle decelerates uniformly from a speed P R u. It takes a time t, and travels a distance S d before stopping. Its speed, and distance 0 Time t travelled after a time is given by 2 Which statement about the motion of the body Speed Distance travelled is true? u d A At P, the body is accelerating A 2 less than 2 B At Q, the speed of the body is zero B u more than d C The velocities at P and R are the same 2 u d 2 D At S, the speed is constant C Less than less than 2 2 u d D More than more than 2 2 53 02 STPM PHY T1.indd 53 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 5. An object is projected vertically from the top 8. (a) An object is projected with a speed of v at of a building and it falls to the ground below. an angle θ to the horizontal. Derive the Neglecting air resistance, which is the graph Cartesian equation for the trajectory of of displacement-y against time t? the object. A y C y (b) The figure below shows a boat approaching a cliff of height 15.0 m at a constant speed of 12.0 m s . A bullet is fired -1 2 horizontally with a speed u when the boat 0 t 0 t is 800 m from the cliff. The bullet strikes the boat. B y D y Bullet v Cliff 15.0 m 0 t 0 t 800 m 6. A stone is thrown vertically upwards with (i) The bullet strikes the boat how many an intial speed of 24.0 m s from the top of a seconds after been fired? -1 tower 60.0 m above the ground. (ii) What is the distance of the boat (a) Determine the speed of the stone when it from the cliff when it is struck by the strikes the ground. bullet? (b) Calculate the time taken to reach the (iii) Find the initial speed v of the bullet. ground from the instant it is thrown. (iv) Determine the magnitude and direction of the velocity of the bullet 7. A stone is thrown from a cliff which is at when it strikes the boat. a height h above the water surface with a velocity v at an angle θ to the horizontal as 9. A ball is kicked horizontally from a cliff of shown in the figure below. After a time t, the height H with a velocity of 8.0 m s . The –1 stone is at P(x,y). ball hits the water at an angle of 60° to the y horizontal as shown in the diagram. v P(x, y) 8.0 m s –1 Clif θ Cliff f x Cliff H h 60º R (a) Neglecting air resistance, calculate (i) speed of the ball when it hits the (a) Write the expressions for the x- and water, y-coordinates of P. (ii) H, the height of the cliff. (b) The height h = 10.0 m, v = 12.0 m s and (b) If air resistance is not negligible, discuss –1 θ = 60 . Calculate the horizontal distance the change in the angle that the ball hits o R. the water. 54 02 STPM PHY T1.indd 54 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 10. (a) An object in linear motion has an initial 13. (a) Define acceleration. velocity, u. The object has an uniform Explain how it is possible for a body to acceleration of a. When the displacement undergo acceleration when its speed is s, the velocity is v. Define uniform remains constant. acceleration and deduce the equation (b) A ball falls 2.00 m from rest and a further 2 2 v = u + 2as distance of 0.10 m as shown in the figure. –1 (b) A car is travelling at 54 km h , and when 2 the car is 25 m from the stop line, the traffic light turns red. The reaction time of the driver is 0.60 s. 2.00 m (i) Find the minimum deceleration of the car for the car to stop behind the stop line. (ii) Sketch a graph to show how the displacement of the car from the stop 0.10 m line varies with time. (c) A ball is kicked from the ground and it follows the path shown in the figure Neglect air resistance, calculate below. The greatest height reached is (i) the speed of the ball after falling 2.00 m, 2.0 m and the range in 22.0 m. (ii) the time to fall a distance of 0.10 m. 2.0 m 14. (a) Define velocity. 22.0 m Explain how the displacement and the Determine acceleration of an object may be deduced (i) the velocity of projection of the ball, from a velocity-time graph. (ii) the time of flight. (b) A trolley is placed at the top of a slope as shown in the figure. A block is fixed 11. A car starts from rest and travels along a rigidly to the lower end of the slope. straight road. The acceleration-time graph of the car is as shown below. Acceleration / m s –2 4 2 0 Time / s At time t = 0, the trolley is released from 2 4 6 8 10 the top of the incline and its velocity v –2 varies with time t as shown below. v (a) What is the maximum speed of the car? A (b) What is the speed after 10 s? 12. A missile is to be launched from a point P, 500 m from the peak of a hill of height 2000 m into enemy territory. If the missile C is to just pass over the hill and strike a target O 1200 m from P, calculate (a) the velocity at which the missile should be launched, B (b) the time of flight of the missile. 55 02 STPM PHY T1.indd 55 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics (i) Describe qualitatively the motion of The total stopping distance is the sum of the the trolley during the periods OA, AB distance travelled by the vehicle during the and BC. reaction time of the driver and the distance (ii) Calculate the acceleration of the travelled when the brakes are applied. x trolley down the incline. Draw a graph of against v. Use the graph to (iii) Find the length of the incline. find v (iv) Calculate the distance the trolley (a) the total stopping distance if the velocity 2 moved up the incline on the of the vehicle is 35 m s . –1 rebounce. (b) the reaction time of the driver. 15. The table below shows the total stopping distance x for different velocities v of a vehicle. v/m s –1 10.0 15.0 20.0 25.0 30.0 x/m 20.0 37.5 60.0 87.5 120.0 ANSWERS 1 2. 22 m s –1 2h 1. (a) Use v = u + at 3. g = 2 2 t 2 – t 1 (i) 2.0 m s –1 (ii) 4.0 m s –1 (iii) 10.0 m s –1 3 1 1. C 2. B 3. C 4. C 5. A (b) Use average speed = (u + v) 2 6. C 7. C 8. B 9. C (i) 1.0 m s –1 (ii) 2.0 m s –1 10. (a) Displacement (ii) 5.0 m s –1 (c) Use s = average speed × time (i) 1.0 m (ii) 4.0 m (iii) 25.0 m 2. 25 m Time 3. 48 m s –1 0 t 1 t 2 (b) Acceleration 2 1. (a) Use v = u + at, t = 0.815 s (b) Use v = u – 2gH Time 2 2 H = 3.26 m above the building. 0 t 1 t 2 = 12.26 m from the ground 1 2 (c) Use s = uT – gT , s = –9.0 m 2 T = 2.40 s 56 02 STPM PHY T1.indd 56 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics 11. (a) v (b) s 2. B : Velocity = gradient of graph u At Q, gradient = 0. 3. C : Increase in velocity = area between graph and time-axis 1 = (12)(4) m s = 24 m s –1 –1 0 t 0 t 2 4. B : Shaded area under graph > total area under graph ds Speed 2 v = = gradient of (s-t) graph dt Area = distance or s =v dt = area under (v-t) graph. u travelled in time t/2 12. v u/2 0 t/2 t Time 0 t 1 2 5. C : y = ut – gt 2 Gradient of y-t graph decrease until zero at the 4 highest point. 2 2 6. (a) v = u + 2as 1. B 2. D 3. C 4. C 5. C = 24.0 + 2(–9.81)(–60.0) 2 1 v = 41.9 m s –1 6. K 4 1 (b) s = (u + v)t 7. (a) v /m s –1 2 1 –60.0 m = (24.0 + (–41.9)t 6.9 2 t = 6.70 s 4.0 v H Alternative: 1 0 t /s s = ut + at 2 1.41 2 1 –60.0 m = (24.0)t + (–9.81)t 2 2 Solving t = 6.70 s –6.9 v V 7. (a) x = (v cos θ)t 1 (b) Maximum height = 2.43 m s = ut + at 2 2 8. (a) 45° 1 2 4 (b) 2.55 × 10 m = (v sin θ)t – gt 2 (c) 6.37 × 10 m (b) Vertical motion, 3 1 9. 714 m –10.0 = (12.0 sin 60 )t – (9.81)t 2 o 1 2 2 10. (a) 0.639 s (Use s = ut + gt for vertical motion) 2 2 9.81t – 20.8t – 20 = 0 –1 (b) 5.01 m s (Use 3.2 m = vt) 20.8 ± 20.8 – 4(9.81)(–20) 2 –1 (c) v x = 5.01 m s , v y = 0 + (9.81)(0.639) t = 2(9.81) = 6.27 m s –1 = 2.84 s –1 2 V = v x + v y = 8.03 m s Horizontal motion, 2 o R = (12.0 cos 60 )(2.84) m v y At an angle to the horizontal, tan ( ) = 51.4° = 17.0 m –1 v x 8. (a) 5 v sin θ v 1. D 2. A 3. C 4. B 5. B P(x, y) y STPM Practice 2 θ v cos θ x 1 1. C : s = ut + at 2 x 2 1 –12.0 m = (6.0)t – (10.0)t 2 2 t = 2.26 s 57 02 STPM PHY T1.indd 57 4/9/18 8:19 AM Physics Term 1 STPM Chapter 2 Kinematics x 10. (a) Refer to page 34 x = (v cos θ)t, t = v cos θ –1 –1 –1 1 (b) (i) 54 km h = 54 × 10 3 m s = 15 m s 60 × 60 y = (v sin θ)t – gt 2 Distance travelled under deceleration 2 1 x ( x = (v sin θ) v cos θ ) – g ( v cos θ ) 2 = (25 – 15 × 0.60) m = 16 m 2 2 2 v = u + 2as gx 2 = x tan θ – 2 2 2 2 2v cos θ a = 0 – 15 m s = –7.0 m s -2 -2 2(16) (b) (i) Vertical motion (ii) 1 s (m) s = ut + at 2 2 25 1 15.0 = 0 + (9.81)t 2 16 2 2(15.0) t = s = 1.75 s 9.81 0 t (ii) Distance of boat from the cliff u sin θ 2 2 = 800m – (12.0)(1.75) m (c) (i) H = 2.0 m = 2g = 779 m u sin2θ 2 R = 22.0 m = g (iii) v(1.75) = 779 m Solving, θ = 10.3 and u = 35 m s –1 o v = 445 m s –1 22.0 (ii) T = = 0.64 s (iv) V x = 445 m s –1 ucos θ 1 –1 V y = 0 + (9.81)(1.75) m s –1 11. (a) v max = (8)(4) m s = 16 m s –1 = 17.2 m s –1 2 1 (b) Speed = 16 – (2)(2) m s = 14 m s –1 –1 2 V = 445 + 17.2 m s –1 2 2 –1. –1 = 445.3 m s . 12. (a) 140 m s at 19° to the horizontal V y (b) 9.1 s -1 At an angle tan ) to the horizontal = 2.2° ( V x 13. (a) Acceleration is the rate of change of velocity. Eg. A body accelerates from rest. 9. (a) (i) v x = 8.0 m s –1 (b) (i) 6.26 m s –1 v = 8.0 ms –1 x 60º (ii) 0.0157 s 14. (a) Refer to page 33 v v y (b) (i) OA – constant acceleration AB – constant deceleration v x BC – constant acceleration Cos 60° = v (ii) 1.67 m s –2 (iii) 1.20 m 8.0 v = m s = 16.0 m s –1 (iv) 0.65 m –1 cos 60° (ii) v y = v x tan 60° = 13.9 m s –1 15. (a) 157.5 m Vertical motion: (b) 1.0 s 2 v y = 0 + 2(9.81)H 2 13.9 2 H = m = 9.85 m 2(9.81) (b) If there is air resistance, path of the ball is shown in the diagram. New path Cliff θ The angle θ  60° 58 02 STPM PHY T1.indd 58 4/9/18 8:19 AM STPM Model Paper (960/1) Paper 1 Kertas 1 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fifteen questions in Section A. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam Bahagian A. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. The following measurements are obtained in an experiment. L = (10.0 ± 0.1) cm and H = (2.0 ± 0.1) cm. Which expression has the greatest percentage uncertainty? Bacaan berikut diperolehi dalam suatu eksperimen. L = (10.0 ± 0.1) cm dan H = (2.0 ± 0.1) cm. Ungkapan yang manakah mempunyai peratus ketakpastian tertinggi? A (L – H) C LH L B (L + H) D H 2. A particle X moves from the point O to a point S via the point Q at constant speed. Another particle Y moves from point O to the point S via the point P with the same speed as shown in the diagram. Suatu zarah X bergerak dari titik O ke titik S melalui titik Q pada laju seragam. Suatu zarah yang lain Y bergerak dari titik O ke titik S melalui titik P pada laju yang sama seperti ditunjukkan dalam gambar rajah. Q S O P Which statement is correct about the magnitude and direction of the average acceleration of the particles X and Y? Pernyataan yang manakah yang betul tentang magnitud dan arah pecutan purata zarah X dan Y? Magnitude of acceleration Direction of acceleration Magnitud pecutan Arah pecutan A Same / Sama Same / Sama B Same / Sama Different / Berbeza C Different / Berbeza Same / Sama D Different / Berbeza Different / Berbeza 320 MODEL PAPER PHY T1.indd 320 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) 3. The displacement-time graph of a body is shown below. Graf sesaran-masa suatu jasad ditunjukkan di bawah. Displacement Sesaran Time, t 0 Masa t 1 t 2 t 3 When is the body (i) stationary, and (ii) accelerating? Bilakah jasad itu (i) pegun, dan (ii) memecut? Stationary Accelerating Pegun Memecut A t = 0 and t 0 t < t 3 1 B t = 0 and t t t < t 3 2 3 C t t < t 0 t < t 1 1 2 D t t < t t t < t 1 2 2 3 4. An object is dropped from height and as it falls it experiences air resistance. Which statement is correct about the acceleration of the object? Suatu objek dilepaskan dari suatu ketinggian dan semasa jatuh objek itu mengalami rintangan udara. Pernyataan yang manakah yang betul tentang pecutan objek itu? A Acceleration is constant but less than the acceleration of free fall. Pecutannya malar tetapi kurang daripada pecutan jatuh bebas. B Acceleration equals the acceleration of free fall and air resistance is constant. Pecutannya sama dengan pecutan jatuh bebas dan rintangan udara adalah malar. C Acceleration decreases because the weight of the object increases. Pecutannya semakin berkurang sebab berat objek semakin bertambah. D Acceleration decreases because air resistance increases. Pecutannya semakin berkurang sebab rintangan udara bertambah. 5. The force F on a body which is initially at rest on a smooth horizontal surface varies with time t as shown in the graph below. Suatu jasad yang pada awalnya pegun di atas permukaan ufuk yang licin dikenakan daya F yang berubah dengan masa t seperti di tunjukkan dalam graf di bawah. F 0 t Which graph correctly shows the variation of the velocity v of the body with time t? Graf yang manakah menunjukkan dengan betul ubahan halaju v jasad itu dengan masa t? A v B v C v D v 0 t 0 t 0 t 0 t 321 MODEL PAPER PHY T1.indd 321 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) 6. A load is dropped from rest from a height and falls onto a pile in the ground. The pile is driven a distance into the ground before it stops. Suatu pemberat dilepaskan dari suatu ketinggian dan jatuh di atas sebatang kayu yang tercecak pada tanah. Kayu itu bergerak suatu jarak tertentu ke dalam tanah sebelum berhenti. Load Pemberat Pile Kayu Ground Tanah Which statement is correct? Pernyataan yang manakah adalah betul? A Collision between the load and pile is elastic. Perlanggaran antara pemberat dan kayu adalah perlanggaran kenyal. B Momentum of load and pile after collision equals momentum of pile before collision. Jumlah momentum pemberat dan kayu selepas perlanggaran adalah sama dengan momentum pemberat sebelum perlanggaran C Kinetic energy of pile immediately after collision equals loss of potential energy of load. Tenaga kinetik kayu sebaik sahaja selepas perlanggaran adalah sama dengan kehilangan tenaga keupayaan pemberat. D Work done against resistance of the ground equals kinetic energy of load and pile immediately after collision. Kerja yang dilakukan menentang rintangan tanah sama dengan tenaga pemberat dan kayu baik sahaja selepas perlanggaran. 7. Two loads of 80 N and 50 N hang from a rope that passes over a pulley which is attached to a motor. The pulley has a diameter of 0.20 m and when it is being rotated at 5.0 rad s the loads –1 are stationary. Pemberat 80 N dan 50 N tergantung pada hujung tali yang melalui satu takal yang diputarkan oleh sebuah motor. -1 Diameter takal ialah 0.20 m dan apabila takal berputar pada 5.0 rad s , pemeberat-pemberat adalah pegun. 5.0 rad s –1 50 N 80 N What is the output power of the motor? Berapakah kuasa output motor? A 15 W C 40 W B 30 W D 45 W 322 MODEL PAPER PHY T1.indd 322 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) 8. A uniform beam of weight W rests on a smooth peg P. The normal reaction of the peg P on the beam is N. The lower end of the beam is on a rough floor and the force on this end is F. Which diagram correctly shows the force W, N and F? Sebatang rod yang beratnya W terletak di atas kayu P yang licin. Tindak balas normal kayu P ke atas rod itu ialah N. Hujung bawah rod itu adalah di atas lantai yang kasar dan daya pada hujung bawah itu ialah F. Gambar rajah yang manakah menunujukkan dengan betul daya W, N dan F? A N C N F P F P Floor Floor W Lantai W Lantai B N D N F P P F Floor W Lantai Floor W Lantai 9 A stone of mass m at the end of a string of length l is projected from the point P with a velocity u as shown in the diagram. The stone then moves in a vertical circle. Seketul batu berjisim m diikat pada hujung seutas tali yang panjangnya l. Batu itu dilancarkan dari titik P dengan halaju u seperti ditunjukkan dalam gambar rajah di sebelah. Batu itu kemudian bergerak dalam bulatan tegak. P u T Q Which expression is correct for the tension T in the string when the stone is at the lowest point Q? Ungkapan yang manakah adalah betul bagi tegangan T dalam tali apabila batu di titik terendah Q? mu 2 mu 2 mu 2 mu 2 A + mgl B + 2mg C + 3mg D + 4mgl l l l l 10. The mass of the Earth is M and its radius is R. If G is the universal gravitational constant, which expression gives the gravitation field strength at a height h = 2R above the Earth? Jisim Bumi ialah M dan jejarinya R. Jika G ialah pemalar semesta kegravitian, yang manakah adalah ungkapan bagi kekuatan medan graviti pada tinggi h = 2R dari permukaan Bumi? A 2GMR B GM C GM D GM 2 9R 2 3R 2 2R 2 11. A copper wire is of length L and diameter d. A steel wire is of length 2L and diameter 2d. The Young’s modulus of steel is twice that of copper. Loads of the same mass hang from the lower ends of the wires. The extension of the copper wire is x, what is the extension of the steel wire? Panjang seutas dawai kuprum ialah L dan diameternya ialah d. Panjang seutas dawai keluli ialah 2L dan diameternya 2d. Modulus Young keluli adalah dua kali nilai modulus Young kuprum. Pemberat-pemberat yang sama digantung daripada hujung bawah kedua-dua dawai itu. Pemanjangan dawai kuprum ialah x, berapakah pemanjangan dawai keluli? A 1 x B x C 2x D 4x 4 323 MODEL PAPER PHY T1.indd 323 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) 3 –2 12. An ideal gas in a container of volume 2.0 × 10 m is at a temperature of 300 K and pressure 5 of 1.10 × 10 Pa. How many moles of the gas is in the container? –2 Suhu suatu gas unggul di dalam suatu bekas berisi padu 2.0 × 10 m ialah 300 K dan tekanan gas ialah 3 5 1.10 × 10 Pa. Berapakah bilangan mol gas di dalam bekas itu? A 0.44 moles B 0.88 moles C 1.13 moles D 2.27 moles 13. At room temperature, the molar heat capacity of a gas at constant volume, C is xR. If C is p,m V,m C the molar heat capacity at constant pressure, what is the value of the ratio p,m ? C v,m Pada suhu bilik, muatan haba molar pada isi padu malar C suatu gas ialah xR. Jika C ialah muatan haba molar p,m V,m C pada tekanan malar, berapakah nilai nisbah p,m ? A x – 1 B x + 1 C v,m C x D x x x x + 1 x – 1 14. Two different metal rods P and Q of the same cross sectional area of 5.0 cm are joined in series 2 and perfectly lagged. The lengths of the rods are 20 cm and 10 cm respectively. The thermal -1 conductivity of rod P is 200 W m K and the thermal conductivity of rod Q is 400 W m K . -1 -1 -1 o The temperatures at the ends of the jointed rod are 100 C and 30 C. o Luas keratan rentas dua batang logam yang berlainan P dan Q adalah sama, 5.0 cm . Dua batang logam itu 2 disambungkan secara siri dan ditebat dengan sempurna. Panjang P dan Q adalah masing-masing 20 cm dan 10 cm. -1 -1 -1 -1 Kekonduksian termal P dan Q adalah masing-masing 200 W m K dan 400 W m K . Suhu di hujung-hujung rod gabungan adalah 100 C dan 30 C. o o 100°C P Q 30°C 20 cm 10 cm What is the rate of heat flow along the joint rod? Berapakah kadar aliran haba sepanjang rod gabungan itu? A 4.7 W B 26.7 W C 28.0 W D 33.4 W 15. Material medium is not required if heat is transferred by Bahan antara tidak diperlukan jika haba dipindah melalui A conduction / kekonduksian B convection / perolakan C radiation / sinaran D convection and radiation / perolakan dan sinaran Section B (15 marks) Bahagian B [15 markah] Answer all the questions in this section. Jawab semua soalan dalam bahagian ini. 16. (a) An object is projected at an angle of 45 to the horizontal. Find the value of the ratio of o the maximum height reached H to the range R of the projectile. Suatu objek dilancarkan pada sudut 45 dari ufukan. Tentukan nilai nisbah tinggi maksimum H kepada julat o R luncuran itu. [4 marks / 4 markah] 324 MODEL PAPER PHY T1.indd 324 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) (b) A ball is kicked from a distance of 6.00 m from a wall of height 3.00 m. Sebiji bola ditendang pada jarak 6.00 m dari suatu dinding yang tingginya 3.00 m. (i) State the angle of projection so that the ball just clear the wall at the maximum point of its trajectory. [1 mark] Nyatakan sudut luncuran bola supaya bola itu melepasi dinding pada tinggi maksimum lintasan bola. [1 markah] (ii) Find the speed of projection of the ball. [3 marks] Cari laju bola diluncurkan. [3 markah] 17. (a) (i) Define ideal gas. [1 mark] Takrifkan gas unggul. [1 markah] (ii) State the quantities that determine the state of an ideal gas. [1 mark] Nyatakan kuantiti-kuantiti yang menentukan keadaan suatu gas unggul. [1 markah] (b) On the same axes, sketch graphs to show how the pressure p of an ideal gas varies with the volume V when the gas expands each time to twice its initial volume from the same initial state. Pada paksi-paksi yang sama, lakarkan graf untuk menunjukkan perubahan tekanan p suatu gas unggul dengan isi padunya V apabila gas itu mengembang kepada dua kali isi padu asal daripada keadaan asal yang sama. (i) in an isothermal expansion [2 marks] secara pengembangan isotermal [2 markah] (ii) in an isobaric expansion, and [1 mark] secara pengembangan isobarik, dan [1 markah] (iii) an adiabatic expansion. [2 marks] secara pengembangan adiabatik. [2 markah] Section C (30 marks) Bahagian C [30 markah] Answer any two questions in this section. Jawab mana-mana dua soalan dalam bahagian ini. -1 18. A car accelerates from rest to a speed of 20 m s in 5.0 s. The total mass of the car and driver is 1920 kg. Sebuah kereta memecut daripada pegun kepada laju 20 m s dalam 5.0 s. Jumlah jisim kereta dan pemandu ialah -1 1920 kg. (a) Name two forces that resist the motion of the car, and state how each force varies when the speed of the car increases. [3 marks] Nyatakan dua daya yang menentang pergerakan kereta, dan nyatakan bagaimana setiap satu daya itu berubah apabila laju kereta bertambah. [3 markah] (b) (i) Calculate the acceleration of the car. [2 marks] Hitungkan pecutan kereta. [2 markah] 325 MODEL PAPER PHY T1.indd 325 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) (ii) What is the resultant force on the car during the acceleration? [2 marks] Berapakan daya paduan pada kereta semasa kereta memecut? [2 markah] (iii) State and explain how the horizontal forward force on the car changes when it accelerates. [3 marks] Nyatakan dan jelaskan bagaimana daya mengufuk ke depan pada kereta berubah apabila kereta memecut. [3 markah] (iv) Draw a labelled diagram to show the directions of the horizontal forces on the car during its motion. [2 marks] Lukiskan gambar rajah berlabel untuk menunjukkan arah daya-daya mengufuk pada kereta semasa kereta bergerak. [2 markah] (c) When the car is travelling at a constant speed of 20 m s , its power output is 25 kW. Find -1 the total resistive force. [3 marks] -2 Apabila laju kereta adalah tetap 20 m s , kuasa output ialah 25 kW. Tentukan jumlah daya rintangan pada kereta itu. [3 markah] 19. (a) (i) Define Young’s modulus of a material. [1 mark] Takrifkan modulus Young suatu bahan. [1 markah] (ii) A wire is of length L and its cross sectional area is A. Derive an expression that relate the Young’s modulus Y of the material to the force constant k of the wire. [2 marks] Panjang seutas dawai ialah L dan luas keratan rentasnya A. Terbitkan ungkapan yang mengaitkan modulus Young Y bahan dawai itu dengan pemalar daya k dawai. [2 markah] (b) The natural length of a vertical wire is 3.00 m and its diameter is 1.20 mm. Loads are slowly added to the lower end. The extension-force graph of the wire during the loading process up to the point Q is shown below. Panjang asal seutas dawai mencancang ialah 3.00 m dan diameternya 1.20 mm. Beban ditambahkan dengan perlahan di hujung bawah dawai. Graf pemanjangan-daya dawai itu semasa beban ditambah sehingga titik Q ditunjukkan di bawah. Extension (mm) Pemanjangan 3.0 Q P 2.0 1.0 O 0 8.0 16.0 24.0 Force Daya (N) (i) State the types of deformation experienced by the wire along OP, and along PQ. Describe in atomic terms the difference in deformations along OP and PQ. [4 marks] Nyatakan jenis canggaan yang dialami dawai itu sepanjang OP, dan PQ. Huraikan dalam sebutan atom- atom perbezaan canggaan sepanjang OP, dan PQ. [4 markah] (ii) Calculate the Young’s modulus of the material of the wire. [3 marks] Hitungkan modulus Young bahan dawai itu. [3 markah] (iii) Copy the graph and draw a line to show the variation of extension with force when the force is reduced from the point Q to zero. Hence estimate the permanent extension in the wire. [2 marks] Salin graf di atas dan lukiskan satu garis untuk menunjukkan ubahan pemanjangan dengan daya apabila daya pada dawai dikurangkan dari titik Q ke sifar. Seterusnys anggarkan pemanjangan kekal pada dawai. [2 markah] 326 MODEL PAPER PHY T1.indd 326 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) (iv) Estimate the energy dissipated from the wire in the loading and unloading process. [3 marks] Anggarkan tenaga yang terlesap daripada dawai semasa daya ditambahkan dan daya dikurangkan. [3 markah] 20. (a) (i) Write an equation to represent the first law of thermodynamics. Explain clearly the terms in the equation. [2 marks] Tuliskan persamaan yang mewakili hukum termodinamik pertama. Jelaskan sebutan-sebutan dalam persamaan. [2 markah] (ii) Use the first law of thermodynamics to explain that in an isothermal expansion, although heat is supplied to the gas, the temperature of the gas remains unchanged. What happens to the heat supplied? [2 marks] Gunakan hukum termodinamik pertama untuk menjelaskan bahawa dalam pengembangan isotermal, walaupun haba dibekalkan kepada gas, suhu gas kekal tidak berubah. [2 markah] (b) An ideal gas has a volume of V when its temperature is T and pressure is p . The gas 0 0 0 undergoes the following processes: Pada suhu T dan tekanan p , isi padu suatu gas unggul ialah V . Gas itu melakukan proses berikut: 0 0 0 I: The pressure of the gas is increased to three times its initial pressure at constant volume. I: Tekanan gas ditambahkan sehingga tiga kali tekanan asal pada isi padu malar. II: The temperature of the gas is then reduced back to T at constant pressure. 0 II: Suhu gas kemudian dikurangkan sehingga T pada tekanan malar. 0 III: The gas expands isothermally until its volume is V and its pressure is p . 0 0 III: Gas itu mengembang secara isotermal sehingga isi padunya V , dan tekanannya p . 0 0 Determine in terms of p , V and T , 0 0 0 Tentukan dalam sebutan p , V dan T , 0 0 0 (i) the temperature of the gas after process I, [2 marks] suhu gas selepas proses I, [2 markah] (ii) the volume of the gas after process II, and [2 marks] isi padu gas selepas proses II dan [2 markah] (iii) the work done by the gas in the complete cycle. [4 marks] kerja yang dilakukan oleh gas itu dalam satu kitar yang lengkap. [4 markah] (c) Draw a p-V graph to illustrate the processes I, II and III. [3 marks] Lukiskan graf p-V untuk menggambarkan proses-proses I, II dan III. [3 markah] ANSWERS PAPER 1 (b) (i) 45 o 2 2 Section A (ii) H = v sin 45° = 3.00 m 1. A 2. B 3. C 4. D 5. C 2g 6. D 7. A 8. D 9. D 10. B Or maximum R = v 2 = 2(6.0) m 11. A 12. B 13. B 14. C 15. C g v = 10.8 m s -1 Section B 17. (a) (i) Ideal gas: Gas that obeys the ideal v sin 45° 2 2 16. (a) H = gas equation pV = nRT for all 2g temperatures and pressures. 2 R = 2v sin45° cos45° (ii) State of a gas is determine by its g temperature, pressure and volume. H 1 1 = tan 45° = R 4 4 327 MODEL PAPER PHY T1.indd 327 4/9/18 8:32 AM Physics Term 1 STPM STPM Model Paper (960/2) (b) p (iii) Extension (mm) (ii) 3.0 Q (i) P (iii) 0 V 2.0 V 0 2V 0 Section C 1.0 18. (a) Friction between the road and the tyres. Constant. 0.3 O Force (N) Air resistance. Increases as speed increases. 0 8.0 16.0 24.0 20 – 0 –2 (b) (i) a = m s = 4.0 m s –2 Permanent extension = 0.3 mm 5.0 (iv) Energy dissipated (ii) F = ma = 3.84 × 10 N 3 (iii) Forward force > friction + air = area between loading and unloading resistance graphs –3 = 5 × 10 J As the speed increases, air resistance increases. 20. (a) (i) Q = ∆U + W To maintain a constant acceleration Q : heat supplied, the forward force must increase. ∆ U : change in internal energy (iv) W : work done by the gas Air resistance (ii) Isothermal expansion, temperature constant. Forward force ∆ U = 0 Q = 0 + W Heat supplied Q used to do work Friction against external pressure when the (c) Power = Fv gas expands. 25 × 10 3 (b) (i) I: V = constant, p ∝ T F = N = 1.25 × 10 N 3 20 p = 3p , hence T = 3T 0 1 0 1 Stress (ii) II: p = constant, V ∝ T 19. (a) (i) Young’s modulus = Strain When T = 3T , V = V F/A (ke)L kL 1 0 0 V (ii) Y = = = Hence when T = T , V = 0 e/L Ae A 2 0 2 3 (b) (i) OP: Elastic deformation. (iii) I : ∆V = 0, W = 0 1 When the wire is stretched, atoms are V II : W = (3p )( 0 – V ) = – 2p V displaced from the mean positions. 2 0 3 0 0 0 When force is removed, atoms return V 2 to the original positions. III: W = nRT ln ( V ) and (p V = nRT ) 3 0 0 0 PQ: Plastic deformation. 1 V When the wire is stretched beyond = p V ln ( 0 ) = p V (ln3) 0 0 V /3 0 0 the elastic limit, atomic planes slide 0 over each other. When the force W = W + W + W 3 1 2 is removed, atoms do not return = (ln3 – 2) p V = – 0.90p V 0 0 0 0 completely to their original positions. 1 YA p (ii) k = = gradient L 3p II 20 20 0 Y = ( )( ) N m –2 –3 2 2.5 × 10 –3 π(0.60 × 10 ) III = 2.1 × 10 N m –2 I 10 p 0 V 0 V /3 V 0 0 328 MODEL PAPER PHY T1.indd 328 4/9/18 8:32 AM Format: 190mm X 260mm Extent= 336 pgs (16.62 mm) 70gsm Status: Cover Master BI vervion_2nd imp PRE-U PELANGI BESTSELLER STPM Text PRE-U STPM Text Chemistry CC039242a PRE-U Term 1 specially designed for students who are sitting for the STPM Physics examination. The comprehensive PRE-U notes and practices are based on the latest syllabus and exam format set by Majlis Peperiksaan Malaysia. This book will provide you with the STPM Text necessary skills and strategies to excel in the subject. TERM 1 Our Pre-U & STPM Titles: STPM Text Physics › Success with MUET › MUET My Way FEATURES › Pengajian Am Penggal 1, 2, & 3 › Bahasa Melayu Penggal 1, 2, & 3 ■ Comprehensive Notes and Practices › Biology Term 1, 2, & 3 Poh Liong Yong › Physics Term 1, 2, & 3 ■ Useful Features like Concept Maps, 1 Learning Outcomes, Exam Tips, Info › Chemistry Term 1, 2, & 3 Chem and STPM Taggings › Mathematics (T) Term 1, 2, & 3 › Sejarah Penggal 1, 2, & 3 SCAN ME TERM ■ Summary › Geografi Penggal 1, 2, & 3 Physics › Ekonomi Penggal 1, 2, & 3 ■ STPM Practices › Pengajian Perniagaan Penggal 1, 2, & 3 For the 21st Century Learner ■ STPM Model Paper Term 1 ■ Complete Answers TERM 1 eBook! Available MORE THAN MORE THAN 1 COPIES COPIES W.M: RM37.95 / E.M: RM38.95 SOLD CC039242a SOLD ISBN: 978-967-2856-92-4 PELANGI 26/07/2021 12:39 PM CVR_PreU_STPM_2021 Physics (Penggal 1).indd 1-3 26/07/2021 12:39 PM CVR_PreU_STPM_2021 Physics (Penggal 1).indd 1-3

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EXPERIMENT 1 : PHYSICAL QUANTITIES AND MEASUREMENT

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Measurement is an essential activity that need to be presented in the physics experiment; when the measurement is close to the perfect, the experiment is said to be success. In this second physics laboratorial activity, the students are undergoing five activities toward the concept of measurement. Determining length, sensing mass, calculating volume, comparing density, and checking time are what the student have done in laboratory; by using the concept of percent error and reading graph, the students are expected to be able to understand the basic concept of measurement and its SI measurement unit. Keyword: Length, Mass, Volume, Density, Time, Error, Slope, Graph.

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Introduction • Some examples of mechanical quantities that need to be measured include position (or displacement), angular velocity (rotation rate of a shaft), force, torque (moment), and shaft power. • Instruments of various kinds have been invented to measure each of these quantities. • In this learning module, several of these instruments are discussed; the list here is by no means exhaustive. • The purpose here is to give you a feel for how mechanical instruments work, and to make you aware of the variety of ways to measure things.

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