HKDSE Physics Position and movement by topic - HKEAA

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DSE Physics - Section B : MC. PB-FMl-M/03 DSE Physics - Section B : M.C. PB-FMl-M/04 FMl : Position and Movement FM1 : Position and Movement 9. <HKCE1986Paper1I-1> 13. < HKCE1989 Paper II- 3 > .. lcm2cm 3cm A particle is thrown vertically upwards. When the particle is at the maximum height, its acceleration is A. zero. B. changing from upwards to downwards . C. pointing upwards. D. pointing downwards. The diagram above shows a ticket-tape produced by a trolley being pulled by a rubber band. Which of the following statements about the trolley is/are true ? (1) Its displacement in.creases unifon:nly with time. 14. <HKCE 1989Paper II- 4> (2) Its velocity increases uniformly with time. (3) Its acceleration increases unifonnly with time. A coin and a f.eather are allowed to fall in a long vertical glass tube from which the air has been evacuated. Which one of the A. (1) only following combinations best descnOes the motion of the coin andthe feather ? B. (2) only Coin Feather C. (1) & (2) only D. (2) & (3) only A. uniform speed same uniform speed B. uniform acceleration same uniform acceleration C. uniform acceleration smaller uniform acceleration 10. <HKCE1987Paperll-6> D. uniformacc eleration greater uni.form acceleration displacement/m 4 --------------------- 15. < HKCE 1990 Paper II - 1 > The graph shows how the square of velocity of an object undergoing uniform acceleration varies with displacement. The object is initially at 2 rest and travels along a straight line. The acceleration of the object is 4 A. OSms- 2 O .fL'--+----+----+----i--- (tima) / s B. l.0ms-2 2 2 2 0 2 3 4 C. 2.QmS-2 ""'-----+---+--••Im An object is accelerated from rest along a straight line. The above graph shows the variation of its displacement with the D. 4.0ms-2 0 2 3 square of time. What is the acceleration of the object? A. o.sms-2 B. l.Oms-2 16. < BK.CE 1991Paper II - 4 > C. 2.0ms""2 D. 4.0ms-2 11. < HKCE 1988 Paper II - 1 > ' g: KL M N 16 I 0 36 I 64 The above figure shows the strobe photog;raph of a ball rollin8 down a slope. The stroboscope is flashing at a frequency of 5 Hz. Find the acceleration ofth e ball A. 0.20ms-2 The above figure shows the stroboscopic photograph of a ball rolling down a slope. If the stroboscope m akes 2 flashes per B. 0.50 m s-1 second, in which region does the ball have an average speed of40cms- 1 ? C. 0.67mS-2 A. LM D. l.OOms- 1 B. MN C. NO D. OP 17. <HKCE1991PaperII-5> velocity/ m s-1 The figure shows the ve,ocity-time graph of an object. 12. < BKCE 1989 Paper II - 2 > Whlch of the following statements about the object An object is falling from rest with an acceleration of9.8 m s-2. Which of the following statements is/are correct? is/are true? (1) It falls with a constant spe ed of9.8 ms- 1. (1) Its acceleration in the first 10 sis 2 m s 2- • (2) It falls. 9.8 m every sec ond. (2) The total distance travelled is 250 m. (3) Ithasaspeedof19.6ms- 1 after2s. (3) Itretums to its starting point after 25 s. 10 A. (1) only A. (1) only B. (3) only B. (2) only C. (1) & (2) only c. (1) & (3) only �---1---4--�-.....--~time/ s D. (2) & (3) only D. (2) & (3) only 0 10 20 30 DSE Physics - Section B : M.C. PB-FMl-M/05 DSE Physics - Section B : M.C. PB-FMl-M/06 FMl : Position and Movement FMl : Position and Movement 18. < HKCE 1992 Paper II - I> 21. <HKCE 1993 Paper II -3 > Which of the following displacemenMime graphs descnOes the motion of a particle moving in a straight line with uniform An object is thrown vertically upwards from a point A. It travels to the highest point B and then falls back to A. Neglecting deceleration? air resistance, which of the following statements i s/are true? A. B. (I) The total displacement of the object is zero. (2) The acceleration of the object is constant throughout themotion. (3) The time for the upwardmotion is longer than the time for the downwardmotion. A. (!) only B. (3) only C. (!) & (2) only D. (2) & (3) only LL. C. D. 22. <HKCE1993Pap erll-2> The paper tape shown is obtained from. a trolleymoving with uni£onn acceleration. The frequency of the ticker-tape timer is 50 Hz. Flnd the acceleration ofthe trolley. A 0..2lm� B. o.1omr I:: 4 I.5 on I�•---- 3.6cm c. 0.73mr D. I.05m r :A. 19. < HJ(CE 1992 Paper II- 2 > B 23. < BKCE 1993 Paper IT- 5 > '"' A man takes 2 s to walk from point A to point B, and then takes 3 s to walk froll1 point B to point C, where ABC is an The above diagram shows the variation of the acceleration of an object which is initially at rest Which of the following equilateral triangle of side 3 m. Find the magnitude ofhis average VELOCITY :fromA to C. velocity-time graphs correctly dcscnOes themotion ofthe object? A. 0.60ms- 1 A v B. .lee, B. l.00mS-1 C. 1.20ms-1 D. 1.25ms-1 20. < BK.CE 1992Paper Il- 4 > ,. t Velocity/m s"" 1 120 C. D. Tim.els -60 24. <HKCE 1994 Paper Il - 5 > Aman takes 30 s to walk 80 mtowards the east. He then takes 10 s to run 60 m towards the south. Which ofthe following -120 statements is/are correct? (1) Themagnitude of the resultant displacement of the man is 140 m. The above figure shows a man near the edge ofa cliff projecting a stone vertically upwards. The stone reaches the sea after (2) The average speed of the IllllD is 4.3m s-1• 18 s. The graph shows the velocity~ time for the motion of the stone. Find the height of the cliff. (Take g = 10 ms-2.) (3) Themagnitude ofthe average velocity ofthe man is 2.5 m s- 1• A. 180m A. (!) only B. 540m B. (3) only C. 720m C. (I) & (2) only D. 900m D. (2)& (3)onJy DSE Physics - Section B : M.C. PB-FMl-M/07 DSE Physics - Section B : MC. PB-FMl-M/08 FMl : Position and Movement F.M:1 : Position and Movement 25. <HKCE 1995 Paper11-S> 29. < HKCE 1997 Paper Il - 2 > Velocity / m S-1 A student walks along a curve ABC, which is made up of two semi-circular parts AB and BC of :radius 3 m and 4 m 20 respectively. He takes 2 s to walk from.A to Band 5 s from B to C. Find the magcitude of the average velocity of the B C -«e---➔ 10 student from A to C A 3m 4m A. l.0ms-1 0 t-----t------'0,.,.-::---t--- Time/ s B. 2.0ms-1 2 C. 2.3 ms-1 -10 D. 3.lms-1 The above diagram. shows the variation of the velocity of an object with time:. What is the distance travelled by the object in the fust 3 seconds ? 30. < BKCE 1997 Paper II- 6 > A. Sm B. 15m A particle is released from rest and falls vertically under gravity. Ifthe distance travelled by the particle in the 1st second is x C- 25m and that travelled in the 2nd second isy, find the ratio x: y. D. 30m A. 1: 1 B. 1 :2 26. < BKCE 1995 Paper 11 - 4 > c. 1 :3 D. 1 :4 An object starts from rest and moves with unifutm acceleration along a straight line. Which of the graphs below conceming the motion of the object is/axe correct 'l (s = displacement, v =velocity, t = time) 31. < HK.CE 1998 Paper II- 2 > A carUI!dergoes uniform deceleration along a straight road. Its velocity decreases from 30 m s-1 to 20m S-1 after travelling a distance of 100m. How much further will the car travel before it co mes to a rest '1 A. 50m "' o�, "' '�" O B. 80m C. lSOm A. (1) only D. 200m B. (!) & (2) only C. (2) & (3) only D. (!), (2) & (3) 32. < HK.CE 1998 Paper II - 4 > 27. <BKCE 1996Paperil-2 > A man walks 40m towards the west. He then walks 40 m towards the south and lastly walks 70m towards the east Find the magnitude of the resultant displacement of the man. o)_/_,_,_,_\_,_, tis A. 30m 10 20 30 � B. 40m C. 50m E=_,_ D. 70m The velocity..fune graph of a car travelling along a straight horizontal road is shown above. Which ofthe following graphs shows the variation of the acceleration a of the car with the time t '1 :ZS. <BKCE 1996Paperll-4> A. a B. a An object mc:ives with uniform acceleration along a straight line. Which of the following graphs correctly descnCe(s) the motion ofthe object? ,,, ,�- "',�- 0 f---+---+-->---+--+tls f----1---+--+--+-->-tls 10 20 30 40 10 20 c. 0 D. a A. (1) oilly f---+---+---+--+-+-->- ti, f"---+--+--+--+-->-tl, "' 3P, :,' 40 i B. (2) only 10 20 10 20 C. (!) & (3) o,!y ' D. (2) & (3) o,!y '--' DSE Physics - Section B : M.C. PB-FMI-M/10 FMl : Position and Movement 36. < HKCE 2000 Paper ll- 3 > A raoing car accelerates fromrest to a speed ofl00 lanh- 1 in 3.2 s. Find the average acceleration ofthe car. A. 4.34ms-2 B. 8.68 ms-2 C . 15.63 m s-2 D. 31.25 ms--2 37. <BKCE2001Paperil-2> A girl walks along a straight road from. a point A to a point B wit:b. an average speed I m s- 1• She then retums from. B to A along the srune road with an average speed 2 ms- 1• Find the average speed of the girl for the whole joumcy. A. ""'· B. 0.67mS-1 C. 1.33 ms-1 D. l.SOms-1 38. <BKCE2001PaperII-1> Distance from P A car travels along a straight road. The variation of the distance of the car from a fixed point P on the road with time is shown above. Which ofthe following statements is correct? A The speed of the car is decreasing. B . The car is movin.gtovmds P. C. There is an unbalanced force acting on the car. D. The area under the graph denotes the total distance travelled by the car. 39. < IIKCE 2002 Paper II - I> slm The figure above shows the distance-time graphs of two toy cars P and Q moving along linear track. Which of the following statements is/are correct? (I) Car.Pwillreach the20m-markfust. 2 ( ) Car Pis overtaking car Qatt=S s. (3) The average sp® of car Pin the first 5 sis smaller than that of car Q. A. (!) & (2) only B. 0) & (3) only C. (2) & (3) only D. (!), (2) & (3) DSE Physics - Section B : M.C. PB-FMl-M/11 DSE Physics - Section B : M.C. PB-FMl-M/12 FMl : Position and Movement FMl : Position and Movement 40. <HKCE2002Paperll-3> 44.. < BKCE 2003 Paper n- 4 > A pie<:e of stolle is hung from a balloon, which is rising vertically upward. If the string connecting the stone and the balloon A plane starts from rest and accelerates at 2 m s-2• Iftbe mimmum take-off speed i s 60 m s- 1, find the minimum distance suddenly breaks, which of the following velocity-time graphs represents the subsequent motion of the stone ? travelled by theplane before it takes off. (Note : Velocity pointing upward is takentobe positive.) A. 450m B. 900m c. 1800m D. 3600m 45. < HKCE2004 Paper n- 3 > Questions 41 and 42 : The figure shows the velocity-time graph of a car travelling along a straightroad. " The figure shows the veloci ty-time graphs of two students P and Q running along a straight road. They start at the same point Which of the following statements is/are correct? (1) TheaveragespeedofPbetween t"'O and t=t1 islargerthanthatofQ. (2) At t =ti, P and Q reach the same point (3) At t""t2 ,QisaheadofP. A. (1) only 41. <BKCE2003Paperll-1> B. (3) only C. (1) & (2) only What physical quantity does the area of the shaded re gion represent? D. (2) & (3) only A. ""'"' B .. momentum Questions 46 and 47 : C. acceleration Patrick is drivingalon g a straight horizontal road. At time t = 0, he observesthat an accident bas happened. He then applies the D. displacement brakes to sto p his car with uniform deceleration. The graph shows the variation of the speed of the car with time. v/ms-1 42. < BKCE 2003 Paper II - 2 > Wlrich of the statements are correct? 14 (1) Th.ecarchanges itsdirectionoftravelat t = t1. (2) The caris farthest away from the starting po:int at t = t1. (3) Thecarretumstoitsstartingpointat t = tz, A. (1) & (2)onl y B. (1) & (3) only C. (2) & (3) onl y 46. < BKCE 2005Paper Il-1 > D. (1),(2)&(3) What is thereaction time of Patrick? A. - B. 0.8 s G 43. < HK.CE 2003 Pa per n- 3 > C. 4.2 s A car starts at point A andtravels along a circular path of radius 30 m. A D. 5.0s After 15 s, the car retums topoint A. Find the average speedof the car - within 1hispcriodoftizne. 47. < HKCE2005 Paper II-2> A. "'"' Find the distance travelled by the car from time t = 0 to 5.0 s. B. 2ms·1 A. 29.4m C. 6.3 m s·1 B. 40.6m C. 46.2m D. 12.6ms·1 D. 81.2m DSE Physics - Section B : M.C. PB-FMl-M/13 DSE Physics M.C. - Section B PB--FM1 -M /14 FMl : Position and Movement Force & Motion I Position and Movement 48. < BKCE 2006 Paper II - 1 > Velocity 52. The displacement-time graph of an object moving along a straight line is shown below. (07) s J ,/ -----;,(, -A ' Which of the following graphs best represents the relationship between the velocity and time ofthe object ? _ _______ ____ __ J A B. � D. o ""'= '------- L----,►Time V V V V T -�· -�· .pL, -�· Two cm A and B start from rest simultaneously and travel along the same sttaight road. The velocity-time graphs of the two cars are shown above. Which of the following statements about the motion of the two cars is/are always correct? (1) A andB have the same average velocity during the time interval Oto T. (2) A and B have the same average acceleration during the time interval Oto T. (3) A andB travel the same displacement during the time interval Oto T. A. (l)only B. (2) only C. (I) & Q) only D. (2) & (3) only 53. (07) 49. < HK.CE 2006 Paper II- 28 > A B A car travels along a straight road from A to B with a uniform acceleration. The speed of the car is v1 at the instant when m.1f of the journey time from A toB is elapsed and its speed is Vl at the mid-way ofA andB. Which of the following is correct? -ci----�-----..... ,lrn A. vi is always smaller than 111. The above graph shows the variation of the s111uare of velocity vl with the displacements of a particle moving along a straight B. vi is always greater than 112. line. What is the acceleration of the particle? C. vi and l'2 are always equal. A. 0.5 m s-2 B. lms-i D. Whether v1 is greater than or smaller than l-'2 depends on the initial velocity ofthe car at A. C. l.Sms- 2 D. 2ms-i 50, < HK.CE 2006 Paper II - 7 > The acceleration of objects due to gravity on the Moon is about 1/6 that on the Earth. Which of the following diagrams 54. A fish jumps up Vertically to a maximum height of0.5 m above the water surface. What is the speed when it just leaves the shows the con:ect velocity -time graphs for a free falling object dropping respectively on the Earth's surface and the Moon's (08) surface ? '"""''? A. 3.J3ms-1 A a c n B. 4.43 m s-1 - V I C. 6.26m s-1 /Moon D. 9.81 rn s- 1 -­ 0 // 55. (08) 0 0 0 0 t=?I 51. < HK.CE 2007 Paper II -1 > ground � Picture (a) Pieture(b) A bicycle finish.es a 100-metre jou:rney in 9.77 s. Assume that the bicycle starts from rest aoo moves with a llllifonn acceleration. What is the acceleration of the bicycle throughout the journey ? An experiment is conducted by releasing a stone from rest to the ground. At constant time interval T, the positions of the A. l.OSms-2 stone are recorded. Picture (a) shows its positions at different time. Which of the following changes will give a path of the B. 2.10 ms-2 stone as shown in Picture (b)? (Neglect air resistance.) A. A shorter time interval is used. C. 10.2ms-2 B. A longer time interval is used. D. 20.Sms-2 C. A lighter stone ls used. D. A heavier stone is used. , DSE Physics - Section B : M.C. PB-FMl-M/15 . DSE Physics - Section B : M.C. PB-FMl-M/16 FMl : Position and Movement FMl : Position and Movement 56. <HKCE 2008 Paper n- 6 > 60. < BKCE 2009 Paper Il -28 > v/ms-1 A diver jumps up vertically in the air from a high platform and falls into water. The v-t graph below shows the variation of 20,r---------,� 0� the velocity of the diver against time from the point he jumps (P) until he is at the lowest point (Q) in 1he water. platform not to scale p 3 11 .6 2 o"'------'10_____ 20'-+ ' 0 ,___,,,________1 _ --r -+ ti, 0.3 At t = 0, a car and a truck are at the same point on a horizontal straight road. Their velocity - time graph is shown in the figure above. Which ofthe following statements is correct? At t = 10 s, the caris 100 m behind the truck, ,.., At t = 10 s, the car catches up the truck. At t = 20 s, the car is 100 m behind the truck. At t = 20 s, the car catches up the truck. �=--------------- Q ,=:=:=:=:=:=:=:=:=:=:=: --- ----------- -1 3 57. <HKCE2009Paperll-5> John and Mary a:re driving two cars, P and Q, along a straight horizontal v/ms·1 Which of the following is correct ? road respectively. At time t = 0, they both see an obstacle and apply the brakes to stop the cars with uniform deceleration. The variation of total distance travelled from P to Q height of the platform above water surface velocity with time ofthe two cars is shown in the figure below. Which of A, s.9m Sm the following statements is/are correct? G,.. I0.6m !Om {l) The two cars have the same initial speeds. C.-. 11.5 m Sm (2) The reaction times ofJohn and Mary are the same. (3) The total stopping distances oftbe two cars are the same. D... 11.sm 10.6m (2) only (3) only 61. < BKCE 2010 Paper ll-1 > (!) & (2) only (!) & (3) only 58. < HKCE 2009 Paper Il- 1 > A car is travelling at a constant speed of50 kmh-1• How much time does it take to travel 500 m? 0.1 s 10, Mary walks-along a triangularpathXJ'Zwhere XY= YZ'= ZK. It takes her 10 s, 20 sand 10 s to travel tbroughXY, YZ' and ZX' 36' respeetively. Which of the following graphs best represents the variation of distance travelled with time? 360 s A, Distance/ m {) ... Distance / m 59. <HKCE2009Paperll-2> x&P, P, P, y The figure above shows three paths Pi, P2 and Pl fromXto Yon a horizontal plane. Three students take the same time to travel fromX to Y via the th.rec paths respectively. Which of the following physical quantities about their joumey is/are the C � Distance / m Distance /m same? (I) ru,p1'oonont (2) distance (3) average speed (1) only (2) only (I) & (3) only (2) & (3) only 0 10 20 30 40 DSE Physics M.C. - Section B PB - FMl - M / 17 Force & Motion I Position and Movement M62 An obje<:t of mass m, released from rest at height h above the ground, talces time t rorcach the ground. If another object of (11) mass 2 m is released from rest at the same height, how long does it take to reach the ground? (}feglect air resistance.) A. t B. C. I D. ✓2, M63. (11) "- The graph above shows the velocity�time graph of an object which is thrown vertically upwards under gravity. Jfthe object is thrown vertically upwards with a higher initial velocity, which ofthe following graphs (in dotted lines) best represents the expected result? (Neglect air resistance.) A. C v/ms-1 D. DSE Physics M.C. - Section B PB - FMl -M/ 19 DSE Physics M.C. - Section B PB - FMl- M / 2o Force & Motion I Position and Movement Force & Motion I Position and Movement M78. a ball is released from rest at a certain height above the ground. If air resistance is neglected, what is the ratio of M7 2 Peter walks along a straight road from point P to point Q with an average speed of2 m s-1• He then runs back from Q to P (6) the distance travelled by the stone in the second to that travel led in the third second? along the same road with an average speed of 4 m s- 1• Which of the following statements are correct? ( 1) The resultant displacement of Peter in the whole journey is zero. � 3 B. 5 (2) The average velocity of Peter in the whole journey is Om s- 1• C. 5 (3) The average speed of Peter in the whole journey is 3 m s- 1• D. 8 A. (l) & (2) only B. Cl) & (3) only C. (2) & (3) only D. (1), (2) & (3) PartC: M7 3. ln the figure shovm, one ball is released from rest at the top of a The following questions are designed to give supplemental exercise for this chapter. tower that is 100 m high. The other ball is released from rest at the 0 mid-point of the tower. Which of the following quantities is the tower M69. A car travels with a speed of l 8 m s-1 • The driver suddenly sees a girl standing at 36 m in front. If the reaction time of the same for both balls as they full in air? (Neglect air resistance.) driver is 0.5 s, what should be the minimum deceleration of the car in order to avoid collision with the girl? A. change of velocity just before reaching the ground 0 lOOm A. 3.0 m s- 2 8. 4.5 m s-2 B. acceleration during the fall 50m C. 6.0 m s- 2 C. finalvelocity justbeforereachingtheground D. 9.0ms- 2 D. time of travel in the journey M70. vim s-1 M74. A boy wants to measure the height of building. He releases a stone at the top of the building from "rt ._·---,.---... rest and starts to keep the time. If the stone takes 2 s to reach the mid- height of the building, f1-l -� ·� , . which of the following statements is/are correct? Take g to be 1oms-2. (1) The height of the building is 40m. (2) The stone takes 45 to reach the bottom of the building (3) The stone reaches the bottom of the building with a speed of 4oros-1 0 -10 �----H ____________ H _____ �: ,I. A. (1) only B. (3) only C. (I) & (2) only The figure shows the time variation of the velocity of a car travelling along a straight road, starting from rest at a c ertain point D. (2) & (3) only P. What is the maximum distance from the point P that the car would reach within the time shown in the figure? Velocity/ m s-1 A. !00m M7 5. The graph shows the velocity of a body travelling in a straight B. 150m C. 200 m line. What is the averagevelocity of the body during the first D. 250 m 3 s? A. 4m s- 1 M71. Displacement from S C. 9m s-1 Time/s D. 12ms-1 M7 6. A car takes 20 s to travel the first 80 m, and another l Os to travel a further 70 m. What is the average speed ? A. 2.5 m s-1 B. 4.0 m s~1 C. 5.0 m s-1 Two cars A andB move along the same straight road. The variations of their displacement from an oil Station Swith time are D. 5.5 m s-1 shown in the above figure. Which of the following statements is/are correct? (l) The cars travel with the same velocity. (2) At time to, the two cars meet each other. M77. Peter throws a ball downwards at an initial velocity of5 m s-1 from the top of a building.After3 s. the ball reaches the (3) The two cars have travelled the same dsi tance from t "" 0 to I"' t.,. ground. What is the height ofthe building? A. 29m A _ (2) only B. 44m B. (3) only C. (1) & (2) only C. 59 m D. (2) & (3) only D. 88m DSE Physics - Section B : M.C. PB-FMl-M/21 DSE Physics - Section B : MC. PB-FMl-M/22 FM:1 : Position and Movement FM:1 : Position and Movement 78. A feather is dropped downwards with an initial velocity of2 m s- 1 at a height of15 m above the surface of the M oon. It is 84. A car··1ni.vels with a constant speed of 50 km h-1 during a 1:fuie interval. Which of the following values CANNOT be the known that the acceleration due to gravity on the Moon's surface is 16% of that of the Earth. Calculate the speed of the possible average velocity of the car in this time interval? feather when it reaches the surface of the Moon. A. 0kmh-1 .:A 6.85m s-1 B 7.lSms-1 B. 25 kmh-1 e, 8.45ms-1 c. 50kmh-1 D 9.25ms-1 D. 75 kmh-1 79. Two balls of the same mass are dropped from the top of a tall building one after the other. Air resistance is negligible. The separation between the two balls 85. A bo y throws a small ball up wards with an initial velocity of 15m s-1 at the top ofa building. The hei ght of the building is 30 m. If air resistance is negligible, t A remains constant calculate the time taken for the ball to reach the ground. B decreases with time. A 222s t!. increases with time, B. 333 s D depends on the mass ofthe two balls. c. 4.448 D. 5.55 s 80. Two identical balls are held above the ground as shown. One ball is higher than the other ball by a separation As. Air resistance is neglig1ble, Suppose tbetwo balls are released at the same time. During the .fligb.t, $rir separation will 86. A particle moves with an initial velocity of 5 in s-1 on a straight line under a unifonn acceleration of 2 mS-2• What is the distance travelled by the particle in the fourth second? A remam cOllStant A 12m B decreases with time. B. 24m C. increases with time, C. 36m D increases and then decreases. D. 48m 81. A fish jumps up with a certain initial speed to leave the water surface, It reaches a maximum. height of 80 cm above the water A. ball is thrown vertically upwards with an :initial velocity of 16ms-1 • What is the total distance travelled by the ball when it returns to the o.'¾'ii.:t(),I p..k'"itr� · . 1 surl'a.ce and retum.s back to the water. Treat the fish as a particle and neglect the air resistance, what is the time interval that the fish is above the water surface? (Take gto be 10 ms-2 .) /h 13m A 02, lh. !Sm C· 26m is 0.4 s D ... 32m c 0.8 s D L6s 82. Ball Pis thrown vertically upwards from the ground with an initial velocity of25 m s-1• At the same time, ball Q is thrown Part D : HKDSE examination questions vertically downwards with an initial vel ocity of 15 m s-1 at the top of a building 80 m above the ground. Asswne air resistance is negligible and their motions are along the same vertical line, detennine the height that the two balls meet. 88. < HKDSE Sa mple Paper IA- 7 > Take the acceleration due to gravity to be 10 m s-2• A. 30m 14 B. 40m C. 50m D. 60m LL------"---* ti, 0 0.8 5.0 83. A particle accelerates from rest with a uniform acceleration a along a straight line. It travels a distance of x in the third second and travels a distance ofy in the fifth second. Find the ratio ofx toy. Patrick is driving along a straight horizontal road. At time t"' 0, he observes that an accident has happened. He then applies the brakes to stop his car with unifurm deceleration. The graph shows the variation of the speed of the car with time. Find A 3:5 the distance tni.velled by the car from time t "' 0 to 5.0 s. B. 5:9 A 29.4m C. 9: 16 B. 40.6m C. 46.2m D. 9:25 D. 81.2m DSE Physics - Section B : M.C. PB-FMl-M/23 . DSE Physics - Section B : M.C. PB-FMl-M/24 Fl\i.11 : Position and Movement FMl : Position and Movement � <BKDSE Sampl� Paper IA - 12 > 94. <BKDSE2014PaperJA-5:> , Two small ide ntical objects P and Qare released from the t op of a building 80 m above the ground . Q 1s released 1 s after P. A particle is moving along a straight line with uniform accele ration. It truces 4 s to travel adistance of 36 m and then 2 s to Negle cting air resistance, what is the maximum vertical separation between P and Q i n the air ? travel the next 36 m. Vlhat is its acceleration? A. Sm A, 2.5ms· B. 10m ·(3.... 3.0ms-2 2 ·C.., 4.0ms- D. 45m C. 35m p, 4.5 ms·1 2 90. < HKDSE Practice Paper IA - 7 > 95. < HKDSE 2014 Paper IA- 9 :> A ston e falls from rest. Neglecting air resistance, th e ratio of the distance travel led by the stone in the 1st second to that A partic le is projected vertically downward with an initial speed of 2.0 m s-1 travelled in the 2nd second is o from the rooftop of a house. The particl e reache s the ground with a spe ed 2.0m ,-, fl p A. 1: 1 of 11 m s-1 as shown. Estimate the height ofthe h ouse. Neglect air resistance. i � B. 1 :1 A. 33m houso C. 1: 3 D. 1 :4 B. 6.0m t D llms-1 ! C. 6.5m <BKDSE Practice Paper IA- 6 > D. 12m A toycar travelled due east for 10m in 5 s, then :immediately 91. tumed north and travelled 5 m for l s. What was the average 96. <HKDSE2015PaperIA-4:> speed of the car? The figure shows the velocity-time (v-t) graph of two carsP and Q travelling A. 1.9ms-1 Smin ls along the same straight road. At t = 0, the cars are at the same position . B. 2.2 ms-1 Which ded uc tions about the cars between t = 0 and t= t2 are correct ? V C. 2.Sms-1 (1) P and Q are always travell ing in the same d:ircct:ion. D. 3.Sms-1 10min5s (2) At, t = ti. the separation betwee n P and. Q - is at a maximum. - (3) Att= t2,�ebindP. I/ A c1) & (2) only (1) & (3) only � A particle i s released fromres t atXas shown. It takes time t1 to fall :fromXto Yand 92. <HKDSE2013PaperIA-8> Q , \3 time tztofall fromYtoZ. IfXY: i'Z= 9: 16, :findt1: k, N eglec t air resistance. C. (2) & (3) on ly 0 :x A. 2 :3 +' y .a (1),(2)&(3) B. 3:4 '' <HkDSE 2015 Paper IA - 9 :> •z '-''��"""""" A particl e travels at2.0 ms·1 due e ast for 1.5 sand the n travels at 4.0m s·1 d ue north for LO s. What is th e magnitude of its c. 4:3 D. 3:2 .., ground 0,verage vel ocity for the wholejourney ? A: 2.0ms·1 93. <HKDSE 2013 PaperlA-11 > B ... 2.8ms·1 Two particles P and Q start from the same position and travel al ong th e '&.'" 3.0m s-1 same straight li ne. The above figure shows the velocity-time {v-t ) graph v'�5.0ms·1 for P and Q. Which of the followiog descriptions about their motion is/are oornct? 4 {1} At t- 1 s, P changesits direction of IllOtion. 98. <HkDSE2016Pape�IA-4> 3 (2) At t = 2 s, the separation between P and Qis 4 m. 2 Thespeedometer of a carshown indicates the car's / i , (3) Att=4 s,P and Qmeeteach other. · A.� 'ih.stantane ous sp eed. A. (1) only t I, i).., 1nstantaneous velocity. p B. (2) only 1 (... uwerage speed of the whole journey. c. (1) & (3) on ly -2 D. (2) & (3) only · l).,_ q..,rerage velocity ofthe whole journey. DSE Physics • Section B : MC. PB-FMl-M/25 DSE Physics - Section B : M.C. Solution PB-FMl -MS/01 FMl : Position and Movement FMl : Position and Movement HXEAA's Marlcir1g Scheme is pmpamd for the markets' reference. It should not be :rega?C}ed as a set of model answers. 99. <HKDSE2017PaperIA-5> Students and teachers who are not involved :in the marking process are advised to intcJ:pret the Marldng Scheme with care. Which ofthefollowing statenients about the motion of any two objects is correct? , The object that takes a shorter time to complete the same path. must have greater average speed. M.C. Answers The object that travels a greater distance in I s must have greater ave.rage velocity. The object with greater velocity must have greater acceleration. I. D II. B 21. C 31. B 41. D 101. A ff the two objects have the same acceleration, they must be moving in the same direction 2. A 12. B 22. D 32. B 42. A 3. D 13. D 23. B 33. C 43. D 100. <HKDSE 2019 Paper IA-4> 4. D 14. B 24. B 34. D 44. B 5. B 15. B 25. C 35. B 45. A 6. D 16. D 26. D 36. B 46. B 7. C 17. A 27. C 37. C 47. B 8. C 18. C 28. D 38. B 48. B 9. B 19. A 29. B 39. A 49. A I 10. C 20. B 30. C 40. C 50. C 51. B 61. C 71. D 81. C 91. C 52. D 62. C 72. A 82. A 92. D 53. A 63. A 73. B 83. B 93. B 54. A 64. B 74. A 84. D 94. B 55. A 65. A 75. B 85. C 95. B 101 <HKDSE 2020 Paper fA-4> 76. C 86. A 96. D 56. D 66. C 57. A 67. D 77. C 87. C 97. A 58. C 68. C 78. B 88. B 98. A 59. A 69. C 79. C 89. C 99. A 60. C 70. C 80. A 90. C 100. D l as shown. Given that A ca. taJccs g muwtes tG uave1 along a� OFQR on a borizottta _ � et• find the � ofthe avm,ge velocity of the car m tfusJQUffl OP .. PQ-2 Jan. A. 30kmh-1 ' M.C. Sohm.on B. 36kmX1 1 C. 41 kmh- !. D D. St kmb-i s = ut+ ½at 2 :, (4) = {0)+ ½g(1)2 g = Smir' s = ut+ ½gt2 = (0)+ ½(8)(1 +4)2 = 100m 2. A FromAtoB: s = ut+ fot 2 (100) = u {4) + f (9.81)(4) 2 u = 538 ms-1 FromOtoA: v = u+ot (5.38) = (0) + (9.81) t t = 0.548 "" 0.55 s DSEPhysics- Section B: MC. Solution PB-FMl-MS/02 DSE Physics - Section B : M.C. Solution PB-FM1-MS /03 FMl : Position and Movement FMl : Position and Movement 3. D C ✓ (I) Ullllonn velocity ⇒ acceleration a=0 s = ut+lat2 = (O)t+lat1 2 2 ✓ (2) Example : Ifa ball is projectedupwards, its velocity is zero at the highest point, but a= g ;:!: 0 slope of the graph = .!:.a ✓ (3) Example : If a car turns roundwith constant speed, as the direction is changing , the velocity varies. 2 4-0 =-a -- 1 :. a=2ms""2 4. D 4-0 2 4 OR s = areaofthegraph= (Z+4)x = 12m 2 At t 2= 4s2,di splacements=4m. r Bys=ut+ ½at2 (4) = (0) + ½a(4) 5. B In vacuum, there is no air resistance, the ping-pong ball wouldfall under the acceleratio n due to gravity. B v = u+at = O+gt V � I Time interval for each flash = .!. = 0.5 s v ~ t graph is a straight line through the origin AtMN,averagespeed: v = 36-16 = 40cms-1 0.5 6. D By v2 = u 2 + 2as 12. B 0= u2 + 2(-a)s :. u 2 =2as 2 It falls with the acceleration due to gravity, thus the speed is increasing. U CC S (1) :. sz = 29.4m (2) ½ A:fterl S, s = u r + at 1 = (0) + ,½ (9.8)(1)1= 4.9 m. Moreover, as it falls with acceleration, the distance travelled in everysecondshould be increaslllg. ✓ (3) l tion of 9.8 ms-2 means in each second, there is a change in velocity of 9.8 m s-1 . Acceera 7. C A:fter2s,speed= 9.8x2 = 19.6ms-1• 6-0 a= slope of the graph "" = 2m s-2 3-0 13. D At the maximum height, velocity is zero 8. C but the acceleration of the particle is still equal to the acceleration due to gravity which is pointing do'ND.wards. Bys = ut + ½at2 = (0) + ½(9.81)t2 "" 4.9t 2 After falling for 1 s : SI = 4.9 x (1)'- = 4.9 m 14. B Afterfallingfor2s: sz= 4.9x(2) = 19.6m 2 AE air has been evacuated, the tube is vacuum, thus there is no air resistance acting on the falliog object Coin : falls with uniform acceleration (as it falls 1.lllder gravity) After falling for 3s: 33= 4.9 x (3)2 = 44.1 m Feather : falls with same uniform acceleration (as it experiences the same a cceleration due to gravity if no all" resistance) Dista nce travelled in the first second = 4.9 m Distancetravelledinthethirdsecond= 44.1 - 19.6 = 24.Sm 15. B By v2=u2+2as 9, B .-. slopeofthegraph= 2a 4-0 = 2a (I) Displacement increases in: 1 cm , 3cm, 6cm, 10cm, 15 cm, 21 cm; not uniformly 2-0 ✓ (2) Length ofeach section of tape represents thevelocity OR :. Velocity increases in unit of: 1, 2. 3, 4, 5, 6; ie. increases llllifonnly At the point when s = 3m, v2= 6m1- s-2 (3) Since velocity increases unifonnly, the acceleration is constant and not increasing By v2=ii+2as :. (6)=(0)+2a( 3) :. a = lms-2 DSE Physics - Section B : M.C. Solution PB-FMl-MS/04 DSE Physics - Section B : M.C. Solution PB-FMl-MS /OS FMl : Position and Movement FMl : Position and Movement 16. D 22. D T:imcintervalbetween2fushes = .!. =0.2s Time interval between2 dots : 1 tick = ...!:.... = 0.02s 5 50 . 4 u = oo = 02ms-1 (occurattheinstantofthemidpointof4cm) u = O.OlS = O.lSms- 1 (occuratth einstant ofthemidpointofl.5cm) 0.2 Sx0.02 0. 2 36 v= 1 = 0.6ms- 1 (occurattheinstantofthemidpointofl2cm) v = 0.0 (occurattheinstantofthemidpointof3.6cm) = 036ms- 1 0.2 5x0.02 From the instant of u to the instant ofv, there are only 2 time intervals, that is, 2 x 0.2s, From the instant ofu to the instant ofv, there are 10ticks, that is, 10x 0.02s. a=�=�= lms- v-u 0.36-0.15 = 1.05ms- 2 2 t 2x0.2 a -,-= IOx0.02 17. A 23. B ✓ 20-10 Slope ofthe v- t graph = ac celeration (1) a=slope of the graph= -- = 2ms-1 10-0 Before t = t0 , accelerati on is positive and constant, v- t graph is astraight lin e that v increases from Oto v 1 X ( 2) s = area of the graph= (10+25)x20 == 350m After t = tO , acceleration is zero, thus v ~ t graph a horizontal lineand velocity continues from v and remains constant 2 X (3) At 25 s, the displacements is 350m. whi ch is not Om, thus it is not the starting point 24. B 18. C X (1) Displacement s =,,/so'- + 6� = 100m Slope ofs ~ t graph represents velocity. 80+60 X (2) speed= = 3.5ms ➔ 30+10 For a particle illOVingwi:th dece leration, its velocity is decreasing. ✓ v=�=2.5ms- 1 The s ~ t graph with decreasing slope represents uniform deceleration. (3) 30+10 19. A 25. C resultant displacement 3 v _ =_ _ =0_6 m 6-1 Distance travelled = total area of the graph between the line and thex-axis totalti me taken 2+ 3 As distattce is a scalar, the <fuection is notrelevtml 20. B Thus, the absolute value ofthe area r epresents the distance travelled. Displacement s = net area ofv-t graph :. d = ½ (20)(2) + ½ (3 -2)(10) =25 m = f (6)(60) - f (18 -6() 120) 26. D = -540m ✓ v = u+at = 0+at a; t .·. The displacement is 540m in downward direction (1) :. V .·. Height of the cliff is 540m ✓ (2) s = ut+ ½at 2= 0+ ½at 2 :. s a; t '­ ✓ @) v2=u2+2as = 0+2as :. v2 a; s 21. C ✓ Fallitlg back to A means returning to the original position :. s = O 27. C (1) The resultant displacements is pointing :froni ✓ (2) When moving inair, the accelerationis equal to the acceleration due t o gravity gvm.ichis cOllStant the starting position to the final position. X ( 3) Same acceleration in upward and downward motion if there is no air resistance ;. timeforupwardmotion = timefordownwardmotion s = .J(40)2 +(30)2 = 50m DSE Physics - Section B : M.C. Solution PB-FMl-MS/06 DSE Physics - Section B : M.C. Solution PB-FMl-MS/07 FMl : Position and Movement FMl : Position and Movement 28. D 34. D X (!) Slope ofs~tgraph = velocity. Asslop e of v - t graph= acce leration, Decreasingslope ⇒ decreasing velocity. When t=Ostot=lOs, a:(+)⇒ slopeo fv~tgraph:(+) .-. The s ~ t graph with decreasingslo pe represents unifonn d e celeration, not accele ration. Whe nt=l0stot=20s, a:(+)⇒ slopeofv~tgraph:O (aho r izontalline) ✓ (2) Slope ofv ~ t graph = acceleration. When t=20 s to t=30 s, a:(+) with larger value => slope o fv~ t graph:(+) with largerslo pe Slope ofv ~ t graph ispositive and is a straight line ⇒ a uniform accele ration. ✓ (3) Constant positive a cceleration ⇒ uniform a cceleration 35. B X (!) At T,,, the two cars do not havesame di sp lace ment, therefore , they do not meet each other. 29. B remltant displacement X (2) At T0 , the two cars have the same positive velocity, thus they must move in the same direction. Average velo city "" total time taken ✓ Car A travels with increasing velocity while car B travels with decreasing velo city (3) v = 3x2+4x2 _ Zms-t 2+5 36. B v = lOOxlOOO = 27_78 ms-t 30. C 3600 Bys = ½gt 2 = ½(10)t2 =5t2 (Takegto be 10 ms-2 forsimpli city.) :. a= 8.68ms-2 Byv = u+at :. {27.78) = (0) + a (3.2) Displacement in 1 s : s1 = 5 x (1'/ "" 5 m Displacement in2 s: S2 = 5 x (2)2 = 20 m 37. C Distance travelled in the 1st second = Sm Assume an arbitrary value (1£-:titti) for the distance betweenA andB, say 10 m. Distance travelled in the 2ndse cond = 20 - 5 = 15 m TimetakenfromAtoB = � = 10s Ratio = 5:15 = 1:3 I Time taken fromB to A = � = � 31. B By .; = u2+2as forthofirstjoumey Averagespeed = Total distancetravelled = 10+10 = 1.33 m 5_ 1 Total time taken 10+5 (20}2 = (30)2 + 2 a(100) :. a= -2.Sms-2 By v2 = i?+Zas fo rtheseco ndjoumey 38. B (0)2 = (2 0)2+2(-2.5)s :. s = 80m X A. Since the slope represents the speed, a straight line indi cates a constant spee d without change. ✓ B. Since distance from P is decreasing, it is moving towards P. 32. B • c i zero and thus no unbalanced (n et) force acting on the car. Since the speed is constant, acceleration s slope ofv ~ t graph = a X D. Area under a distance-time graph has no physical meaning. Fo rt=0stot=l0s,slopeofv~tgraphis(+) ⇒ ais(+) Fort=10stot=30s,slope ofv~tgraphis0 ⇒ a=0 39. A For t = 30 sto t = 40 s, slope ofv~t graph is(-) => ais (-) ✓ (!) Fr omthe graph, when s=20m, car Phasa smaller value oft. Thus, Preaches the mark with ashorter ti me t. 33. C ✓ From the graph, before t= 5 s, car Q has a larger value ofs, car Pis behind car Q. (2) s 120+100 v=-=--- Att= 5 s, car P andQmeetand car Pi s overtaking car Q. t 30+20 X (3) Average spe e d = distance / time :. v = 4.4ms-1 In the :first 5 s, car P and car Q travel thesame distance, so they have thesame average speed. DSE Physics · Section B : M.C. Solution PB-FMl-MS/08 DSE Physics - Section B : MC. Solution PB-FMl-MS/09 FMl : Position and Movement FMl : Position and Movement 40. C 47. B When the string is broken, the stone has the same initial velocity as the balloon, thus it moves upward at t = 0. Distance travelled = area llJ1der the graph from O to 5 s So the stone first moves upward (vis positive), at the highest point, it is momentarily at rest (v =Om s-1), = ½(5.0+0.8)x 14 and then falls down (vis negative). = 40.6m During the whole motion of falling, the stone experiences the same acceleration due to gravity g, thus the slope ofthe graph is constant and equal to -g. 48. B e l cement (!) Average velocity = r su tant displa 41. D totaltime Since the area under the v-t graph represents the displacement, Area ofa velocity-time graph represents the displacement of the car. as the areas under the graph for A and B are different, th ey have different displacement from O to T, 42. A thus their average velocity must not be the same. ✓ (!) At t = t1, v = 0, the car reaches the extreme point and is momentarily at rest, ✓ (2) Average acceleration = v- u it then reverses its direction of travel ✓ (2) After t = ti, the car reverses its direction and travels backwards. Thus the car is farthest away at ti . As both cars have the same initial velocity u and final velocity v in time T, they must have the same average acceleration. (3) The area from t= 0 to t= t1 represents the distance travelled in forward direction while the area from t = t1 to t=ti represents the distance travelled in backward direction. (3) Since the areas under the two graphs A and B are different, As the two areas are not equal, the car does not return to its starting point at t2. their displacement must not be the same. 43. D 49. A displacement Distance travelled: d =211: x (30) = 188.5 m s ---------------------- Average speed=!!.. = 188·5 = 12.6ms-1 t 15 ½s --------------- l'l 44. B v, By v2=u2+1as '-='"---.-'-----'---+time (60)2 = (0) + 2 (2) s 0 ½t s = 900m The displacement-time graph of a car lllldergoing acceleration is a quadratic curve shown as above. At half of the time ofthe journey, fue velocity is vt that is represented by the slope at that point. Athalf of the displacement ofthe journey, the velocity is vz that is represented by the slope at that point. 45. A Consider the slope, v1 must be smaller than vz. ✓ (!) Area under a v-t graph represents the displacement. OR At time t1, area ofP is greater, thus P travels a greater distance, therefore, P has a greater average speed. As the car is accelerating, the speed is increasing, thus l'l is greater than v1• (2) At time ti, P and Q have different areas, thus they have different displacements, therefore, they must be at different points. 50. C (3) At time t2, the area under the graph ofP is greater, thus P should have travelled a greater distance, Slope of the v-t graph is the acceletation due to gravity g. therefore, P should be ahead of Q. Since the acceleration due to gravity on the Moon is smaller, the slope should also be smaller, thus option C is the answer. 46. B Reaction time is the time taken before Patrick takes action to brake the car, 51. B thus it is the time interval ofthe horizontal line. Bys = ut+ ½at 2 (100) = (0) + ½a(9.77)2 a = 2.lOms-2 DSE Physics - Section B : M.C. Solution PB-FMl-MS/10 DSE Physics - Section B : MC. Solution PB-FMl-MS/11 FMl : Position and Movement FMl : Position and Movement 52. D 58. C Slope ofth e velocity�time graph = acceleration The speed of SO kmli1 should be changed.into the SI unit of soxl OOO ms- 1• Straight line indicates constant slope, and thus constant acceleration. 3600 The first part is a constant acceleration. By s = ut (500) = (SOx lOOO)t :. t = 36s 3600 The second part is zero acceleration. The third part is a constant acc el eration with values less than the first part, as the slope is smaller. 59. A ✓ (1) Disptaeettleirt is the distance between the starting point and the ending point. 53. A They have same displacement. By v2 = u2+2as (2) Distanc e travelled dep ends on the path, thus the three students have different distance travelled. When s=O,tr'=l and when s=l,v2=2 Average speed is the distance travelled per time taken. (3) (2) = (1)+2a(l) :. a= 0.5ms-2 Although they take the same time, they have different distanc e, thus they have different average speed. 54. 60. C (0) - ,i' + 2 (-9.81) (0.5) Total distance travelled fromP to Q = total area under the graph from Os to 2 s = ½(0.3)x(3) + ½(2-03)x(l3) = 11.Sm 55. A Displacement from.P to the water surface = n et area under the graph fromO s to 1.6 s Bys = ut+ ½at 1 = ½(0.3)x(3) - ½O,6-0.3}x(13) = -Sm ✓ A. As the positions of the stol:les are closer, i.e. sis smaller, thus a smaller time interval t should be used. Thus, height of the platform. above water suiface = 8 m B. If a longer time interval is used. the positionsof th e stone should be more separat ed. C. Ifa lighter stone is used, the positions should be unchanged since g is independent of the mass of stone. 61. C Slope ofthe distance-time graph represents the speed. D. If a heavier stone is used, the positions should be unchanged sinc e g is independent of the mass of stone. The speed throughXY and ZX are the same since th ey take the same time. The speed through fZis smaller as it takes longer time, thus th e slope through yz is lower. 56. D Area under av - t graph = displacement 62. C Att=lOs: Since th e acceleration due to gravity g is independent of mass, displacemem ofthe car "' ½(10)(10) = 50 m thus the heavier body of mass 2 m falls with the same g to reach the ground, therefore, the time taken is the same, t. displacement ofthetruck = (10)(10) = 100m :. the car is 50 mbelrind the truck. 63. A Att=20s: Since the slope of the v -t graph is equal to the acceleration due to gravity g, displaceme11t ofthe car "" ½ (20) (20) = 200m and g is not affecied by the initial velocity, thus the slope should be th e same, that is, the dotted line should be parallel to the original line. displac ement ofthetruck = (1 0)(20) = 200m :. the car catches up the truck. 64. B As slope ofs~tgraph = velocity, 57. A From the graph, the initial speed ofP is about two times ofthat of Q. Attbe 1st ¼ cycle, vf and vis(+) => slope ofs-t graph t and the slope is(+) (1) ✓ (2 ) The time interval of the horizontal line is the reaction time, which are the same. At the 2nd ¼ cycle, v-1- and vis(+) => slope ofs-t graph.I, and the slope is(+) (3) The total stopping distance is represented by the area under the graph, Attbe 3rd ¼ cycle, vf and vis(-) => slope ofs-t graph t and the slope is(-) thus the total stopping difflllce of P is greater than that of Q. At the 4th ¼ cycle, v-1- and vis(-) => slope ofs-t graph .J, and the slope is(-) DSE Physics - Section B : M.C. Solution PB -FMl-MS/12 DSE Physics - Section B : M.C. Solution PB-FMl-MS/13 FMl : Position and Movement FMl : Position and Movement 65. A 69. C By v2 = u2+Zas Distancetravelledbythecarin 0.5s = 18x05 = 9m Let the speed of the object when it passes Y be v and let the distance betweenXYbe d. Distance between Y.Z is also d. Distance ofthe girl from the car when thebrake i sapplied = 36 - 9 = 27m X➔Y: v2 = (10)2 + Za(d) By v2 = u 2 + 2as :. {O) = (18)2 + 2a(27) :. a= -6ms-2 Y➔ Z: (20)2 = v2 + Za(d) :. Minimum deceleration = 6 m s-2 Combining the above two equations: v2-(10)2 = (20)2 -v2 2v2 = (10)2 +(20)2 70. C ;. V = 15,8mS-1 MaximumdistancefromP = area:from0to30s 66. C = ½x{10+30)xl0 = 200m For simplicity, takegto be 10ms~ • 2 Asboth ballii arc released from rest, u = 0. 71. D DisplacementofAattimeta!terAisreleased: sA = ut + ½af = ½(10)t 2 = St2 (1) Since carA is moving towards S but car B is moving away fromS, they move in opposite directions. As their dkections are not the same, thcir velocities must not be the same. DisplacementofB attimeta:fterAis released: sa = 5 (t-Z'f (Ball B falls at 2 s later) ✓ (2) At to, since they are at the same position, they meet each.other. Maximum separation occum when Ajust reaches the ground, that is, Afrulsfor 180m ✓ (3) Since they have the same speed, they must have travelled the same distance inthe same time. .·. (180) = 5t 2 :, t = 6s WhenAjustreachestheground: sa = 5(6-2) 2 = 80m M3:ximumseparation = 180 - 80 = 100m n. A ✓ (1) As Peter returns to the original position,the resultant displacement is zero . 67. D ✓ (2) Average velocity is defined as the total displacement per total time taken. Let the height of the building be hand the time taken to reach the ground bet. As the to1al displacement is zero, the average velocity ttlUSt be zero. Bys = ½ g t2 ifthemotionstartsfromrestatthe top • (3) Assume the distance between P and Q is D. Motionfromtoptotbe midpoint: (½h)= ½z{t-0.4)2 (l) Time taken to travel fromP to Q = !!.. h Motionfromtop totheb ottom: (h) = ½g t 2 (2) Time taken to travel from Q to P = !!.. (1). 1 _ (t-0.4) 2 t2 = 2(t-0.4)2 0.4s (2)° 2---,,- Average speed in whole journey = D +D = 2.67ms-1 D/2+D!4 Both sides talce squareroot : t = 1.414 (t-0.4) 1.414t-t = 1.414x0.4 t = 1.37 s 73. B Allbodies fall down with the same acceleration gin air where gis the acceleration due to gravity, 68. C Takeg = lOms-2. if air resistance is neglected. Bys = ut+ ½at2 = ½Cl0) 2 =5t2 74. A Afterls,s = Sx(1) 2 = Sm ✓ (1) Bys = ½gt2 = ½(10)(2) 2 :. s = 20m After2s,s = 5x(2)2 = 20m :. midheight = 20 m Atter3s,s = 5x(3) 2 = 4Sm :. height ofthe building = 2 x20 = 40 m Distance travelled in the second sec = 20-5 = 15 m Distance travelled in the third sec = 45 -20 = 25 m (2) Bys = ½gt2 :. (40) = f (IO) t 2 :.- t = 2.83s -1 Ratio = 15:25=3:5 (3) By v = gt = (10) (2.83) = 28.3ms DSE Physics - Section B : M.C. Solution PB-FMl-MS/14 DSE Physics - Section B : M.C. Solution PB-FMl-MS/15 FMl : Position and Movement FMl : Position and Movement 75. B 82. A Displacement during the first 3 seconds = area under the graph = ½ x 3 x 12 = 18 m ForballP s1 = 25! + f(-10)t 2 ForballQ:si = I5t + ½(+10 )t 2 Averagevelocity = � = !! = 6ms- 1 I 3 By si+si = 80 :.25t+l5tz80 :. t = 2s 76. C Height above the grmmd = s1 = 25 x (2) + ½(-10 ) (2 )2 = 30m Average speed= !!.. = � = 5.0ms-1 t 20+10 83. B Bys = ½ at 2 (u = 0) 77. C Since the ball falls down with acceleration, a= +g. (jJ X = ½a(32 -22) Bys = ut+fat 2 <l) y = ½a (52 -42) :. s = (5)x(3)+ ½(9.81) x(3)2 = 59.145"' 59m x:y = (32 -22):(52 - 42) = 5:9 78. B 84. D There is no air on Moon, thus there is no air resistance on the feather. During this time interval, the car travels a total distance d where d = v t. By v 2 =u2 +2as If the car travels in strai ght line, then the resultant displacements is equal to d, and the average velocity is SO kmh- 1 y 2 = (2)2 +2 (9.81 X 16% ) X {15) If the car does not travel in straight line, the resultant displacements must be less than d, v = 7.15 ms-2 average velocity may be25 km h-1• If the car returns to the starting point :finally , thens = 0, and the average velocity is O km.h-1 . 79. C The resultant displacements can never be greater than d, thus the average velocity cannot be 75 kmh-1. bs = SI - Sz = ½gt2 - ½z(t-to)Z = g•t-/0 - ½g•t.Z As t increases, l::,s increases. 85. C Thus the separation /J.s increases with time. Bys = ut+ ½at 2 80. A :. (-30) = (15) t + ½( - 9.81) t 2 Since the two balls are released at the SaIIre time, their displacemcnts are always the same, :. t=4.44s thus, their separation remains unchanged throughout the flight. 86. A SJ. C Bys = ut+ ½ at2 From the instant of leaving the water surface to that reachlng the maximUDl height : By v 2 = u 2 + 2as :. (0) = u 2 + 2(-10) (0.8) 87. C u=4ms-1 Maximum height reached by the ball : From the instant ofleaving water surface to that of returning back to water surface By v 2 "'u 2 +2as By s = ut+ fat 2 :. (0) = (16)2 + 2 (-9.81) s (Q) = (4) t + ½ (-10) t 1 :.s = 13.0m t = 0.8 s Total distance travelled = 13.0 x 2 = 26 m DSE Physics - Section B : M.C. Solution PB-FMl-MS/16 DSE Physics - Section B : M.C. Solution PB-FMl-MS/17 FMl : Position and Movement FMl : Position and Movement 88. B 94. B Distancetravelled = areaunderthegraphfrom0to5s = ½{5.0+0.8)xl4 = 40.6m Let the initial velocity of the particle be u and the acceleration be a. Bys = ut+ fat 2 89. C Consider the first 4 s : For simplicity, takegto be 10 m s""2• :. (36) = u(4) + ±a(4) 2 ForP, Sf= ut+ faP= ½(IO)t 2= 5t 2 :. 36=4u+8a ForQ, SQ= S(t-1)2 Consider the whole journey of6 s: hs = 5t 2 - S(t-1)2 = lOt-5 :. (36+36) = u(6) + ½a(6)2 Maximum separation occurs when P just reaches the gro!llld. :. 72 = 6 u + 18a (80)=5t2 Combine the two equations: a "" 3 m s-2 :. t==4s ds,.,.,.= 10x4 - 5 = 35m 95. B 90. C Since the particle falls down with acceleration, a = + g Bys = fgt 2 = ½(10)t2=5t2 (Take gto be 10 m s-2 for simplicity.) Byv2=u2+2as Displacement in 1 s: s1 = 5 x (1)2 = 5 m :. (11)2 = (2)2 + 2 (9.81) s :. s = 6.0m Displacementin2 S: S2 = 5 X (2 )2 "' 20 ID Distance travelledin the 1st second = 5 m Distance travelled in the 2nd second = 20-5 = 15 m 96. D Ratio = S:15=1:3 ✓ (1) As the signs of the velocity v of carP and car Q are always positive, both car P and car Q are always travelling in forward direction. 91. C ✓ (2) From t=0to ti, speed ofP is greater than that of Q. Thus, their separation is increasing. Average speed=!!. = lO+S = 2.Sm s-1 From t = tr to t2, speed ofPis smaller than that of Q. Thus, their separation is decreasing. t 5+1 Therefore, at /2, the separation between two cars is at a maximum. ✓ (3) The area under v-t graph represents the displacement of the moving body. 92. D From t= 0 to t2, the total area under the graph P is greater than that ofQ. FromXtoY: XY= ½gtf Therefore, the displacement of car Pis greater than that ofQ. Thus, car Q�ehind carPat time t2 • FromXtoZ: XY+l'Z= ½ g (ti+t£f �--'- (_ 97. A XY+ yz - (t1 +t2 ) 2 Displacement ofthe wholejourney: s = .J(2.0x1.5)2 +(4.0xl.0)2 = Sm Take root ofboth sides : Average velocity = ! = ____ill_= 2.0 m s- 1 I=_,,_ t (1.5+1.0) 5 ti +t2 98. A 93. B At,.._. 1 s, the velocity ofP still remafus negative, thus it is still moving in backward direction. Speedometer shows the speed of the car at an instant since speed is a scalar that has magnitude only. (1) ✓ (2) At t = 2 s, displacement ofP is -2 m and displacement of Q is + 2 m, thus their separation is 4 m. Speedometer can also show the magnitude of the velocity at an instant. (3) At t = 4 s, since the area under the two graphs are not the same, their displacements are not the same, Since velocity is a vector that has both magnitude and direction, thus, they cannot meet each other. a speedometer cannot show the velocity of the car since it does not indicate the direction of the car. DSE Physics - Section B : M.C. Solution PB-FMl-MS/18 DSE Physics - Section B : Question PB-FMl -Q / 01 FMl : Position and Movement FMl : Position and Movement 99. A The following list of formulae may be found useful : ✓ A By average speed = total distance travelled / time taken For uniformly accelerated motion v=u+at For the same path 'With same distance travelled, shorter time gives great er average speed. s '-' ut+½af B. Average velocity = resultant displacement/ time taken An object may have greater distance travelled but smaller resultant displacement ifit has direction changed. Equation of a straight line y = mx+c C. The objectwithgreatervelocity may not have greater acceleration, it may have even zero acceleration. D. For the same acceleration in forward direction, Use the following data wherever necessary : an object have travel forward with acceleration. Acceleration due to gravity g"' 9.81 ms-2 (close to the Earth) but another obj ectmay travel backward with deceleration. Part A : HKCE examination questions 1. <HKCE1987Paperl-1> The figure below shows an experimental set--up to find the acceleration due to gravity g. The ticker-tape timer produces 50 dots per second. A heavy ball attached to a paper tape is released from rest. direction ofmotion o f paper tape s f f f ! . . . .r 0.318m 0.132m (a) The paper tape obtained from the experiment is shown above. Find the average speed of the heavy ball in the interval AB and CD. Hence calculate th e value o fthe acceleration due to gravity g obtained in this experiment. (4 marks) IU 1.-&..v-l � :½11 ""'1 -e... l,j' " -f, ' _ e,ll' /1-w..,,1""- (b) State TWO pr=rutions thatshould be taken in this experiment. (2 marks) (c) How would the result ofg obtained from the experiment be affected if the metal ball was replaced by a ping•pong ball ? Explain briefly. (2 marks) DSE Physics • Section B : Question PB-FMl-Q/02 DSE Physics · Section B : Question PB-FMl-Q/03 FMl : Position and Movement FMl : Position and Movement 2. < HKCE 1992 Paper I-1 > 3. < HKCE 1993 Paper 1-1 > A car is travelling at a unifonn speed of 10 m s-1• The driver sees a wanting signal and applied the brakes to bring the car to A helicopter is initially at rest at a certain level above the ground. It accelerates uniformly and vertically upwards for 75 m rest with uniform deceleration. The figure below sho'W5 the speed-time graph of the car, starting from the instant the driver and reaches a speed of 15 m s-1• Assume the air resistance is negligible. first sees the signal. (a) Calculate the acceleration of the helicopter. speed/m s·1 20 (b) At this moment, an object is released from the helicopter. The object reaches the ground after 6 s. The figure below shows the velocity-ti.me graph of the object, starting from the instant the object is released. (g is taken to be IO ms-2.) 15 Velocity/ms-1 10 s Time/s time/ s 2 4 s (a) Write down the reaction time ofthe driver, Le. the time lapse between seeing the signal and starting to apply the brakes. (1 mark) (b) F.ind the area under the graph in the above figure. State its physical meaning. (4 m.ui<.s) (i) Write down the velocity ofthe object when it reaches the ground. (c) If there is an obstacle 20 m ahead when the driver first sees the signal, would the car hit the obstacle? Explain your answer. (2 marks) (ii) State the physical meaning of the area of the shaded region in the figure above. (2 m.ui<.s) (tit) Using the above figure, :find the height ofthe object above the ground when it is released. (3 m.ui<.s) (d) Assume that the reaction time of the driver and the deceleration of the car remain unchanged. (i) In the figure above, draw a speed-time graph for the car if it is initially travelling at 20 m s·1• (3 m.ui<.s) (ii} A student says 'lfthe initial speed of the car is doubled, the stopping distance of the car would also be doubled.' State whether his statement is true or false and explain briefly. (3 marks) (iv) Comment on the folfow:ing two statements : (4 m.ui<.s) Statement I : At time t = 1.5 s, the acceleration ofthe object is zero. Statement 2 : If the object is replaced by a heavier one, it would take the same time to reach the ground. (e) Suggest TWO factors that would affect the deceleration ofa car. (2 marks) DSE Physics - Section B : Question PB-FMl -Qi 04 DSE Physics - Section B : Question PB-FMl-Q/05 FMl : Position and Movement FMl : Position and Movement 4. < HK.CE 1996 Paper I- 2 > 5. < HKCE 1997 Paper I - 1 > Susan takes part in a 100 mrace at an athletic meet. She starts at time t= Os and accelerate at a uniform rate of 1.6 m �2 for 5 s. She then maintains a uniform speed afterwards and reaches the :fuwhing line at t = 15 s. (a) (i) Find the speed of Susan at t= 5 s. (uJ Find Susan's average speed for the whole journey. (2maw) time/ s (b) (i) On the graph paper, draw the graph of Susan's speed against time from t = 0 to 15 s. (4maw) A boat starts from rest at time t = 0 s and travels along straight line. The Figure above shows the velocity-time graph of the boatfrom t=0to300s. (a) Descn'be the motion ofthe boat from t= 0 to 300 s. c•mawl (b) Find the acceleration of the boat in the first 50 s. {2 marks) (c) Draw the acceleration-time graph of the boat from t = D to 300 s. a (ii) State the physical meaning of the area under the graph. (c) Mary also takes part in the same race. She first accelerates: at a un:ifonn rate of 1.5 m s-z for 6 s and then maintains a uniform speed afterwards. Explain whether Susan or Mazy will reach the finishing line first. c•maw) (d) Find the distance travelled by the boat in the first 50 s. (2maw) (e) A buoy is located 900 m ahead of the starting point of the boat Explain whether the boat will pass the buoy during its motion as shown in the Figure above. (3 marks ) DSE Physics - Section B : Question PB-FMl-Q/06 DSE Physics - Section B : Question PB-FMl-Q/07 -- FMl : Position and Movement FMl : Position and Movement ��,- 6. <HKCE 1999 Paper I- 7 > 7. <l:IKCE2D00Paperl-7> Susan uses the following method to exanilile John's reaction time : Traffic light f<----- 42m ---->< A lony is travelling at a uniform speed of 16 m s-1 along a straight road. At time t = 0, the driver observes that a traffic light, which is at a distance of 42 m from the lorry, is turning red. The driver applies the brake at t = 0.5 s. The lorry then decelerates uniformly and comes to a rest at t = 4.5 s. (a) Sketch the speed-time graph of the lorry from t = 0 to 4.5 s. (3 marks) Figure l She holds a graduated mler upright with the zero mark starting at the bottom. John lines up his fingm near the bottom of the ruler. (See Figure l.) Without any warning, Susan releases the ruler and John grips the ruler with his finger as fast as possible. It is found that John grips at the 20 cm mark of the ruler. (See Figure 2.) Take the acceleration due to gravity g to be lOm�- {a) Show that John's reaction time is 0.2 s. (b) If a heavier ruler is used, how would the result of the above test be affected 'l Explain your answer. (2m m) (b) Find the deceleration of the lorry from t= 0.5 to 4.5 s. (1 mark) (c) Susan marks the other side of the ruler as shown in Figure 3 so that the reaction time can be read directly. (c) Explain whether the lorry will stop in front of the traffic light. (3 """"') Susan's scale for 0.30 0.25 0.20 0.15 0.10 0.05 0 reaction time/ s Graduated scale 30 25 20 15 10 5 0_ on the ruler/ cm ....,., Explain whether Susan's scale for the reaction time is correct or not (3 """"') . DSE Physics - Section B : Question PB-FMl-Q/08 DSE Physics - Section B : Question PB-FMl -Q /09 FMl : Position and Movement FMl : Position and Movement 8. < HKCE 2002 Paper I- 8 > 8. (c) A boy was hit by the car when he was crossing a zebra-crossing. Figure 2 below shows a sketch of the accident dravm A car is travelling with a speed u on a road. The stopping distance ofth.e car includes two parts : by the police. Let dbe the distance between the car and the boy at the moment the driver first observed the boy. The driver applied the brakes and a skid mark 36.0 m long was left on the road. After hitting the boy, the car travelled a distance of 19 .7 m before coming to rest. You may neglect the change in speed ofthe car during the impact. © the thinking distance .f. (Le. the distance travelled after the driver has seen a danger and before the brakes are on). i-1,�,.1 @ the braking distance :s 1&1,§ll/!iil (ie. the distance travelled after the brakes have been put on). Figure 1 shows the variations between P. ands with u. Distance/m l+--19.7m- d Fli;urc:Z (i) Write down the braking distance of the car. (1 mark) (ii) Using Figure 1, estimate the value of u. Speed u Im s- 1 (in') Estimate the thinking distance and the value of d. (3 m,,,ks) Flgurcl (a) Find the slope of the straight line in Figure 1 and state its physical meaning. (3 m,,,ks) (iv) The speed limit of the road is 50 kmh- 1 (i.e. 13.9 m s-1). If the car is travelling at this speed. explam whether it would bit the boy. (3 marks) (b) Assume that the deceleration a of the car rcmalllS unchanged at different speeds. Write down an equation relating u, s 9. < HKCE 2005 Paper 1-1 > and a. Using Figure 1, find the value of a. (3 marks) x/m 0 10 20 DSE Physics - Section B : Question PB-FMl-Q/10 . DSE Physics - Section B : Question PB-FMl-Q/11 FMl : Position and Movement FMl : Position and Movement 9. A car moves along a straight road. The Figure above shows the variation of the displacementx of the car from a certain point 10. (a) Describe the motion ofthe dog between t = O and 5.5 s. (3nwks) onthe road with time t. {a) Describethemotion ofthecarftom t=O to 40s. (3nmks) (b) The dog reaches the ball at t= 5.5 s. How far did the ball roll? (b) Findtheaveragevelocityofthecarfrom t=O to 40s. (2nmks) 10. < HKCE 2011 Paper I- 2 > (c) John rolls the ball with an initial velocity of 6 m s·1 and it decelerates uniformly afterwards. Draw the velocity-time graph of the ball ni the above figure. (2 marks) t= On a horizontal grassland, John rolls a ball for bis dog to catch. At time 0, John stands side by sidewith the dog and rolls the ball forward ni a straight line. The dog immediately starts to run towards the ball. The figure below shows an instant at which the dog is running towards the ball. Part B : Supplemental exercise 11. (a) If you are given a stopwatch and an iron ball, descnOe how you and your partner can find the height of a building, assuming that the value ofg is already known. (2 marks) The ball stops after a while, and the dog reaches the ball some time later. The velocity-time graph of the dog is shown in the figure below. (b) Ifthe ball is released from the top of the building and the time for it to reach the ground si 2 s, estimate the height of the buil<ling. (2 nmks) (c) If there is air resistance, what would be the effect o n the timetaken to reach the groundwhen the ball is released from the top ofthe building? Explain briefly. (2 marks) DSE Physics - Section B : Question PB-FMl-Q/12 DSE Physics - Section B : Question PB-FMl-Q/13 FMl : Position and Movement FMl : Position and Movement 12. 13. The figure below shows a data-logging experimenW set-up to investigate the motion of a trolley. A trolley is held on an inclined runway and a motion sensor is mounted on the top of the ·runway to record the velocity of the trolley. 0 0 Motion sensor 0 To a computer Fixed block A car is travelling at a speed of 13 m s-1 along a straight road. At time t = 0, the car is 27 mfrom a traffic light and the light turns from green to yellow at 1his moment. The driver applies the brakes to stop the car. It is known that the red light will be The trolley is released from rest so that it runs downwards along the runway. At the end of the runway, there is a fixed block on 3 s after the yellow light is on. The figure below shows the speed-time graph of the car. used to stop the trolley. The velocity-time graph captured by the motion sensor is shown in the :figure below. Speed/ms- 1 13t----.-... 0.1 ,7 L___i____________....:,�--.. Time/s 0 U U -j j =l=i= 02 0. 0.6 o., 1.0 12 (a) What is the reaction time ofthe driver? (1 nmk) :j:Tllil�{s): (a) Wbatis the time taken for the trolley to move from the starting point to reach the end oftbe runway? (1 nmk) (b) Find the deceleration of1he car from t= 0,5 to t = 3.5 s. (2,n,w) (b) What is the distance travelled by the trolley before it reaches the end ofthe runway? (2,n,w) (c) Will the driver be charged for rumring a red light ? Explain your answer. (3nwks) (c) Find the acceleration ofthe trolley. (2 marks) DSE Physics - Section B : Question PB-FMl -Q / 14 DSE Physics - Section B : Question PB-FMl-Q/15 FMl : Position and Movement FMl : Position and Movement Part C : BKDSE examination questions 15. < HKDSE 2014 Paper 1B - 3 > Two cars A and B initially at the same p osition, start to travel al ong the same straight horizontal road. The graph below 14. < HKDSE 2012 Paper m - 4 > shows how their velocities vary with time. Train A initially travels at a speed of 60 m S-1 along a straight horizontal railway. Another identical train B travels ahead ofA velocity/ m s-1 in the same direction on the same railway. Due to mechanical failure,Bis only travelling at 20 m s-1• 60ms-1 25 20 At time t = 0, A andBare x m apart, the captam ofA receives a stopping signal and immediate ly A decelerates at 4 m s-2 15 wbileBcontinues to travel at 20m s-1• A eventually collides withB after 5 s. Neglect air resistance. (a) Find the speed ofA just before collision. (2 marks) 5 time/ s (a) Descnbe the motion ofcarA alOllg the whole journey from t= 0 tot= 80s. (b) The graph below shows how the speed of B varies with time within this 5 s. Sketch on the same graph the variation of the speed ofA within the same period. (1 mark) speed/ ms -1 (b) (i) Which car attained the greatest acce leration throughout the journey? Find this acceleration. (2nwks) 60 l-+-+-1-+-+---+-+--+---, (ii) Sketch the acceleration-time (a - t) graph of carBfrom t= 0tot= 80 s. (2mm) 40 l-++-+--1-+--++-+--I 20 l-+-+-1-+--+J>-t-+--+---, I 0 I 3 . 5 ti' p L--'---'---'---'---'---'---'--'---'---+ time/S 2 4 6 -2 -3 (c) Based on the above information, determine the separationx of the two trains at t = 0. (3mm) (c) (i) At t = 20s, what is the separation between cars A andB? (2mm) (ii) Deduce the time at which car B catches up with car A. (2mm) DSE Physics - Section B : Question Solution PB-FMl-QS/01 DSE Physics - Section B ; Question Solution PB-FMl-QS/02 FM:1 : Position and Movement FMl : Position and Movement HKEAA's Mad:ing Scheme is� for the rnarkeis' YCference. It should not be regarded as a set of model answers. Students and teachen who are not involved in the marking process are ad.vised to int.crpret theMm:kiDa SCMm.C with cate. 2. (d) (ii) False (1) If initial speed is doubled, the area under the graph is not doubled. [2) Question Solution <OR> 1. (a) u = � = l.32ms-1 [l] False (1) 0.02x5 Area under the graph = stopping distance ofthe car = ½ (0.5 +4.5) x 20 = 50 m [l] v = � = 3.18ms-1 [l] 0.02x5 50 misnot doubled of the origuial 15 m. [l] = � = 3.18-1.32 g [l] (e) Any'.l'WOofthefollowing: [2] * t 0.02 xlO B:ralcing force of the car * - 9.3ms- 2 (1) Number of passengers in the car OR Mass ofthe car (b) © The ball shou ld be heavy enough so that air resistance is negligible. (1) * Nature of the road surface OR Tyre condition @ Place polystyrene tileon the· ground under the ball so that the ball would not bit the floor directly. (1) * Gradientoftheroad OR Angleofinclinationoftheroad (c) Theresultofgwou ldbeSlllaller. [l] Since the air resistance will become significant. [l] 3. (a) By v1-=u'-+2as [l] (15)1=0+2a(75) 2. (a) Reaction time= 0.5 s [l] a=l.5ms-2 [l] (b) Area = ½ (0.5+2.5)x10 [l) (b) (i) v -45ms-1 = [l] = 15m [l] (ii) The area represents the displacement(OR distance travelled) ofthe object (1) Area under the gtaph. is the stopping distance of the car. [2] in its upward motion. (1) (c) The car would nothit tb.e obstacle, (l] <OR> since the total stopping distance travelled is 15 m which is less than 20 m. [l] The area represents the upward displacement ofthe object [2) (d) (;) (fu) Height = Area of graph below t-axis- area of shaded region [l) = fx45x4.5-{xl5xl.5 [l) = 90m [l] <OR> Displacement = ½ x 15 x 1.5 - ½ x 45 x 4.5 [l) = -90m [l) Height= 90m [l] (iv) Statement 1 : Incorrect ( OR false) [l] The acceleration ofthe object at t= 1.5 sis equal to the acceleration due to gravity. [l] <OR> 2 4 The acceleration of the object at t = 1.5 s is equal to - 10 ms-2 from the graph, which is not zero. (1) < For same reaction time = 0.5 s > [l) Statement 2 : Correct ( OR true ) [l] < For sameslope> [l) Acceleration due to gravity is independent of the mass ofthe object. [l] < For stopping time at t= 4.5 s > [l) DSE Physics - Section B : Question Solution PB-FMl-QS/03 DSE Physics - Section B : Question Solution PB-FMl-QS/04 FMl : Position and Movement FMl : Position and Movement 4. (a) (i) SpeedofSusanat5s = at 5. (a) From t= 0 to SOs, the boat accelerates unifonnly. [1] =1.6x5 [1] From t= 50 to 150 s, the boat travels with a uniform velocity. [!] = 8ms-1 [l] Fromt= 150to 250 s, the boat decelerates unifonnly. [1] O From t= 250 to 300 s, the boat travels bacbwms. [!] (ii) AveragespeedofSusanforthewholejoumey = IO [1] 15 (b) AC-1elerotion = slope ofthegraph [1] = 6.67mS-1 [1] = 2..=o.Ims-2 [1] 50 (b) 0) speed/ms-' (o) a/ms-2 12 0.J IO o+----i---<---t--+---+---;-----... , 1, 50 -0.05 4 <For a=O.Ifromt=Oto50s> [1] <For a = 0 from t=50stol50s> [!] <For a = - 0.05fromt= 150 to300 s> [1] time/s (d) Distance travelled in the first 50 s = Area under the -v-t graph [!] 0 = fx50x5 = 125m [1] < Two axes labelled vvith unit correctly > [1] < From O s to 5 s, straight line with positive slope> ·[1] (e) The boat is farthest away fromthe starting point at t = 250 s. [1] < From5 s, horizontalline > [1] Att = 250 s, distance travelled = f x (250 + 100) x 5 = 875 m [1] < Whole graph is correct> [1] As the :farthest dist.ance is smaller than 900 m, the boat will not pass the buoy. [1] (li) The area tmder the graph represents the distance travelled by Susan. [l] OR 6. (a) Speed/ms-1 The area under the graph :represents the displacement of Susan. [1] 16 ,--.,-__ (c) Distance travelled byMary in the first 6 s = fa t 2 = t X (1.5) X (6) 2 = 27m [1] Speed ofMary at 6th second = 1.5 x 6 = 9 m s~ 1 �-�----------"-c-Time/s Time to covertberemamingjoumey = � "" 8.11 s [1] 0 0.5 4.5 9 < Two axes labelled with correct unit > [1] TotaltimetakenbyMary = 6+8.11 = 14.l s [!] < From Oto 0.5 s, a horizontal line> [!] :. Mruy will win in the race. [!] <From 0.5 to 4.5 s, a straight line of deceleration to rest> [I] DSE Physics - Section B : Question Solution PB -FMl-QS/ 05 DSE Physics - Section B : Question Solution PB-FMl-QS/06 FMl : Position and Movement FMl : Position and Movement 6. (b) deceleration = !! = 4 ms--2 [I] 9. {a) From!= 0to 10s, the car remains at rest [!] 4 From!= 10 to 20 s, the car moves withacceleration. [I] (OR acceleration= -4 m:r2 ) [I] From t = 20 to 40s, the car travels with a uniform velocity. [I] (c) Stopping distance = area under the gtaph [I] !.. 3oo-SO = 16x0.5 + ½xl6x4 = 40m [I] (b) Avera ge velocity = = [I] I 40 As the stopping distance < 42 m, the lony will stop in front ofthe traffic light. [I] = 6.25ms-1 [!] 7. (a) s = ut+ -fat2 [I] 10. (a) The dogacceleratesfromrestbetweent=0 tot= 1 s. [I] (0.2) = ½ (10) t2 :. t=0.2s [!] It then maintains a constant velocity between t= 1 to t = 4.5s. [!] lt then deceleratesand stops at t= 5.5 s. [I] (b) The result would not be affected [I] because the acceleration due to gravity is independent of the mass . [I] (b) Distance = area under the graph [!] = ½<3.5+5.5)x(2) = 9m [!] (c) Susan's scaleisnotcorrect [I] Since s = ½at2 [!] (c) [2] thus s should be proportional to t2• [I] 8. (a) Slope of the line "" 14 [I] 20 = 0.7s [I] The slope represents thereaction time ofthedriver. [I] (b) By v1-=r?+2as :. 0 = (u)2 + 2(-a)s .·. r? = 2as [!] FromFigurel, whens= 20m, u "" l6m s-1 :. (16)2 = 2a(20) [!] ,·+Y-r- :. a = 6.4ms--Z [I] " '...;:!: time/s (c) (i) Bra.king distance = 36 m [!] (nl From Figure 1, when s = 36 m, u = 21.5 m s-1 [I] 11. {a) The partner should release the iron ball at the top of the building and (llll FromFigurel,when u = 21.Sms-1, thethinkingdistance .e= 15m [I] Ishould record the time, t, taken for theball to fall to the ground. [I] From.Figure 2, d = 15 + 36-19.7 [!] = 31.3m [I] Then the height of thebuilding canbe calculated by s = ½g t2 [!] (iv) Ifu is equal to 13.9 m :r1, then the thinking distance I!, is equal to 10 m (b) Height ofbuilding= ½g t 2 = ½ (9.81) (2}2 [I] and the braking distance s is equal to 15 m from Figure 1. [I] = 19.6m [I] Thus the stopping distance is equal to 25 m. [I] {c) The time taken wouldbe longer [!] As the stopping distance is smaller than d, the car will not hit theboy. [I] since theacceleration wouldbecome smaller due to air resistance. [I] DSE Physics · Section B : Question Solution PB-FMl-QS/07 DSE Physics • Section B : Question Solution PB-FM1 -QS /08 FMl : Position and Movement FMl : Position and Movement 12. (a) Reaction time = 0.5 s [l] 15. (a) From I"" 0 to 10 s, carA accelerates unifbnnly. [l] From t= 10 to 80 s, car A travels with the uniform velocity. [l] (b) Deceleration = slope ofthe graph 13 (b) (i) Car B attuned the greater acceleration. [l] [l] 3.5-0.5 ZO-O = 2ms-2 a =slope= [l] 20-10 = 4.33 ms-2 [l] (ii) alms-2 (c) Stoppingdistance = Totalareaunderthegraph 3 = (13) X (0.5) + ½ (13) x (3.5 - 0.5) (l] 2 ' = 26m Ul l Since the stopping distance is less than 27 m, the driver will not be charged. [l] 0 -1 2 3 0 ' 0 -2 -3 13. (a) Thnetaken = 1.2 s [1] <Acceleration from 10 s to20 sand deceleration from 60 s to 80 s COII'ect> [l] < All correct > [l] (b) Distance travelled area from Oto 12 s t (0.54) X (1.2) [l] (c) (i) DisplacementofcarAatZOs = ½(lO+ZO)x(IS) = 225m (l] 0.324m (l] Displacement of car Bat 20 s = ½ (10) x (20) = 100 m (c) Acceleration = slope ofthe line Separation = 225 - 100 = 125 m [l] 0.54 [l] (ii) Car B travels faster than car A by 5 m s-1• I.2 .". (125) = (5) X /:J.t M =2Ss [l] The time that car B catches up with car A is at 45 s. [l] 14. (a) v = u + at ""(60) + (-4)(5) OJ = 40ms-1 [l] (b) speed/ms- 1 - 60 f--. 40 A 20 time/ s 0 2 4 6 < Straight line from 60 m s-1 to 40 m s- 1 during the 5s > [l] (c) DistancettavelledbyAduringthe5s = areaunderthegraph = ½(40+60)(5) = 250m (l] Distance travelled by B duringthe 5 s = 20 x 5 = 100 m [l] Separationx = 250-100 = 150m (l] Hon g Kong Diploma of Secondary Education Examination Physics- ColDpulsor y part (,>:,j ,l;<J! !l-) Section A-Heat and Gases (�;ftl ji.Jlt.) I DSE Physics • Section B : M.C. FM2 : Newton's Laws PB-FM2-M/Ol II 1. Temperature, Heat and Internal energy CU , #.>lo' M �&) 2. Transfer Processes (#,,tt.;@:i&a) The following list of formulae may be found useful : 3 . Change of State (,U.�fi!Ht'i!) 4 . Genm! a,, tAw UhU.lt:t#) For ucifonnly accelerated motion v = u+at 5. Kinetic Theory (¾.:f-;{-t;ttr) s = ut+½aP Section B -Force and Motion (tJ;fU�flJ) 1. Position.sndMovement(-bt•'fl'�H.IJ) � = u2+2as 2. Newton's Laws (4'-� ,t#) 3. MomentofForce (:h�) 4. Work.E nergyandPower(f'F;,'J, te.i:;fo#J$.) Use the following data wherever necessary : 5 . Momcntum(il : rl:) 6 . Proje ctile Motion (�d::ii.fh) Acceleration due to gravity g = 9.81 ms -2 (close to the Earth) 7 . CircularMotion(!lllilil.t:i) 8. Gravitation(�l:h) Section C - Wave Motion ()ltt;J) 1. Wavel'ropagation (it.lm4ilil!.) Part A : BKCE examination questions 2. WavePhenomeJ:1a(iA..:tilJ.!L'-) 3. Reflection and Refraction of Light (;\:,fl!J li.Mli.vi'M) 1. < HK.CE 1980 'Paper Il-10 > 4. Lenses CiUt) A spring balance suspended from the r:eiling of a lift re gisters the weight ofa 20 kg body as 1S0 N. The lift is probably 5. Wave Nann-e of Light (,t,lr;Ji'!tfh4f•l±) 6. Sound(-'!,¼) A. as=ullng with uniform velocity. Section D - Electricity and Magnetism ( 1f£ 'fl'4i) B. ascending with uni:foonacceleration. 1. Electrostatics C*'t'1j!,) C. descending with unifonn velocity. 2. Electric Circuits C1lt$) D. descending with uniform acceleration. 3. Domestic Electricity Clf:-%- m it) 4. MagneticField (:$Ji) 5. Electromagn etic Induction ( 1.1:•�.1&) I 2. < HKCE 1980 Paper II 7 > R 6. Altematin g Current (�lm.'It) Section E - Radioactivity and Nuclear Ene rgy (:tkAH .t .l::ftt-iS �) 1. Radiation and Radioactivity (l&Ahf•:!!l:.AtJJt$.) 2. Atomi.cModel()ff..f-�t!) 3 . Nuclear-Energy (alt/16) A horizontal force Fis applied to a block of mass Mon a rough horizontal surfu.ce. !he acceleration �f the block is a. Ifthe force is changed to 2Fand the :frictional force remains unchanged, then the acceleration of the block will be Ph ysics - Elective p art (l!!'i� !l-) A. greater than 2a. Elective 1 - Astronom y and Space Science (.1dt�:f'l':AA:.� fl-�) B. equal to 2a. 1. The universe seenindifterentscales (;f � $<'..f.'l�.§t'FV,i�'i!iiiitt.) C. between a and 2a. 2. Astronomy through history (�5C.�f.l{f-Al.t.) 3. Orbital motions under gravity (:if:/ r/r,J.fJl.i!.:i!.i'J) D. lesstb.!llla. 4. Statsandtheuniverse(•li.&.'f"<:f1i) Elective 2 - Atomic World CJ§.. 'T'"l!!:'.ff.) 3. < HK.CE 1980 Paper II 8 > R 1. Rufhetford'satomicmodel(.t;Ji,fUi.'T'. iJ!t!.) · 2. Photoelectric effect (,t11:iH:..!i) 3 . Bohr'satomic model of ydrogen (1/tJiil ll1J l.J.f 'T'llt!t!.) h 4. Particles or waves (;flr."T' �ib.) S. Probing into.nano scale (B..flti.#J;f<�.ff.) Elective 3 -Energy and Use of Energy (fl'g.-1:fi:i�i.l�-fl.lH) 1. Electricity at home ( U ffl 1:) 2. Encrgyefficie.ticyinbuilding (;it�tr!lflt.ilhl:.Jf-) In the figure shown,. Xand Yare blocks of mass 1 kg and 2 kg respeciively. Sis a spring balance of negligible mass and P is 3.. Energy efficiency in transporta tion (;l.�1\IIIJ�ii.iH:.$) a smooth pulley fixed at the top of two sttlooth inclined planes. What is the reading of S when Xis held stationary by an 4 . Non-renewable energy sources (;f;!J"�i'..flt.ii.) external force? 5. Renewable energy sources (;!J" -¥H.J6J1t.) A. 4.9N Elective4-MedicalPhysics (V-,!¥=4b.lJ£�) B. 9.8N 1. Making s-=nse of the eye (ntltr\l.G. ir) t C. 14.7N 2. Making senseof heear(.lf-lllJ�'B) 3. Medical imaging using non�ionizin g radiation rn it•$1M -1Hf,J$1t�) D. 19.6N 4 . Medical .imaging using ionizing radiation C 1uir..tut-1l�J$1'-4") 4. DSE Physics - Section B : M.C. FM2 : Newton's Laws <BKCE 1980 Pap er II-3 > PB-FM2- /02 I M I 7. <HKCE1981Paperll-6> DSE Physics - Section B : M.C. FM2 : Newton's Laws PB-FM2 -M /03 I � Two blocksA andB ofrnasses mi and·mz respectively m connected by a light spring on a horizontal frictionless table. The spring is stretched by moving the blooks apart. What is the ratio of the acceleration ofA to that ofB at the moment when they are released ? Five blocks of equal tnaSS A, B, C, D and E are connected by four identical strin� 81, Si, S3 and _S4 as shmm in the� A . m1:nn above. They are made to slide on a smooth horizontal sur-fAce by a steadily in creasmg force F applied to block A. Which of B. m2 :m1 the strings is most likel y to break first ? C. mi1·:mz2 A. 81 B. s, D. mi:m?- C. 83 D. S. s. <HKCE1980Paperll-4> 8. <.BKCE1981 Paper II-3 > A bod y of mass Mrests in equilibrium on a p lane inclined at an angle Bto the horizontal 'What is the frictional.force acting on the body? A. "'° B. Mg An object is resting on a rough plane inclined at an angle 0 to the horizontal, As &gradually increases, the frictional force C. Mgsln 0 acting on the object before sliding occurs is directly proportional to A l D. Mg cos 0 B. 0 C . sinB 9. < IIKCE 1981 Paper II- 4 > D. cos 0 A trolley of mass 1 kg placed on a smooth horizontal table is connected by two light strings to blocks A and B of masses 0.75 kg and 0.25 kg res pectively, as shown in the :figw-e. 6. < IIKCE 1981 Paper Il - 9 > Xand Yare frictionless pulleys. When the system is rdeased, what will be its acceleration ? A. Oms·2 B. 1.0 ms--2 C. 2 S . ms-2 D. 5 0 . ms...z An ob ject is pro jected cup a smooth inclined plane with an initial velocity u. Which of the followin g graphs best represents the variation ofthe displacements of the object along the plane with time t? 10. < BKCE 1982 Paper n - S> A ' B. ' A constant force directed to the left is acting on a body which is initially travelling to the right Which oftbe graphs below best represents the velocity-time grnph of the body? A. V B. V 0 0 0 0 c. ' D. s c. V D. V 0 0 0 0 11. DSE Physics - Section B : M.C. FM2 : Newton's Laws < HKCE 1982 Paper II- 7> PB-FM2-M/041 I 17. <HKCE1985Paperll-1O> DSE Physics - Section B : M.C. FM2 : Newton's Laws PB-FM2-M/0511 ft===1 A trolley on a smooth horizontal surface is pulled by two forces P and Q in the direction as shown in the figure. The magnitude of P and Q are 2 N and 1 N p respectively. If the mass of the trolley is 1 kg, the acceleration of the trolley is , A. 0.15m s-2 towards the left Q� B. 2.24 m s-2 towards the left ,L, ,C, ,k, � block C. 0.73 m� towards the right A block is given an initial velocity up a smooth inclined plane. Which graph below shows the acceleration of the D. 0.15 m s-2 towards the right against time ? {The upward direction of motion along the plane is taken as positive.) A. B. C. D. 12. <HKCE 1983 Paper II - 8 > The net force acting on a particle is zero. Which of the statements below concerning the motion of the particle may be 1rue ? A. The particle is swinging to and fro. B. The particle is decelerating in a straight line. C. The particle is moving in a circle with constant speed. D. The particle is moving with constant velocity. 13. < BKCE 1983 Paper II - 1 > A person in a lift:, which is ascending at a velocity of 10 m s-1, releases a ball. What is the velocity of the ball with respect to the earth at the moment when the ball is released ? 18. < HKCE 1986 Paper II- S > A. 10ms-1 upwaxds B. 10 m s-1 downwards C. 20 m s-1 upwards D. 20 m s-1 downwards s 14. < HKCE 1983 Paper n- 9 > The graph shows the variation of velocity v with time t when a metal ball v/mS-1 is released from rest and allowed to fall vertically under gravity through oil. 'Which of the following statements concerning the motion of the ball 10 ------------------- is/are correct? (1) The velocity of the ball decreases with time. (2) The acceleration ofthe ball decreases with time. (3) The ball stops falling after4 s. the A. (1) only The system of pulleys and blocks is at rest. What is the tension in string 8? (Neglect all :friction and the masses of strings and pulleys.) B. (2) only 0•--,---�2--,----4---a►t /s C. (!) & (3) only A. 9.8N D. (2) & (3) only B. 19.6N C. 29.4N 15. < HKCE 1984 Paper 11-1> D. 39.2N If the engine of a rocket travelling in space is turned o� the rocket will A. stop moving. 19. < HKCE 1986 Paper II- 3 > B. continue to move with unifonn velocity. C. continue to move with decreasing velocity. Which ofthe following motions ofa given mass requires the greatest force P ? (Assume that the surface is smooth.) A B. D. continue to move with uniform acceleration. upwnrd o.cceler.ition - o m s-2 upward acceleration - l m s· 2 16. < HKCE 1985 Paper n- 7 > T, [J T, 0 L L Two blocks A and B of masses 1 kg and 1.5 kg respectively are resting on a smooth horizontal surface and are linked by a c. D. upward accelemtion .. 1 m s-2 upward acccler.ition "' o m s-z string. They are pulled by a force of 1 ON as shown in the diagram. What are the tensions Ti and n ? Tension T1 TensionT2 A. !ON !ON B. !ON 6N C. !ON 4N D. 6N 4N DSE Physics - Section B : M.C. FM2 : Newton's Laws PB-FM2-M/06 I DSE Physics - Section B : M.C. FM2 : Newton's Laws PB-FM2 -Ml 07 II 20. < HKCE 1987 Paper lI - 3 > i 24. < HKCE 1991 Paper n � 9 > Which of the following statements is/are true? A man in a lift feels heavier when the lift is moving A force Fis applied to a block of mass 1 kg as shown below. The greatest (1) upwards with acceleration. value of F for the block to remam at rest is 11 N. What would be the (2) upwards with retardation. motion ofthe block if Fis not applied ? (3) downwards with retardation. (Take the acceleration due to gravity to be 10 m s-2.) A. {l) only A. remaining at rest B. (2) only B. sliding down with constant velocity C. (!) & (2) only C. sliding down with an acceleration of I m s- 2 D. (1) & (3) only D. sliding down with an acceleration of 5 m s-2 21. < HKCE 1987 Paper II -8 > 25. <HKCE1991Paperll-10> ,b_., A man ofweight W stands inside a lift which is moving upwards with a constant speed. ff the force exerted by the floor on the man is R, which ofthe below statements is/are correct ? (1) R is greater than Win magnitude. (2) R and Ware in opposite directions. (3) R and W form an action and reaction pair according to Newton's third law. T A. (1) only An object is subject to a resultant force (F) which varies with time (t) as shown in the diagram above. Which of the following B. (2) only .IL=.. ·.�.1�L. ·.�, graphs correctly shows the variation ofits speed (v) with time (t)? C. (1) & (3) only D. (2) & (3) only 26. <HKCE 1991 Paper II -2 > 12N A B T T T T miu:m IJl!I IJII UUII I I,m,M11mm:m1 'II M, ,,mi•mim; I nm I , I Two blocks of equal mass are placed on a smooth horizorital surface as shown above. A constant force of 12 N is applied to block A so that the two blocks move towards the right together. The force acting on A by B is 22. <HKCE 1989 Paper Il-1 > A. 6Ntotheleft. B. 6Ntotheright. C. 12Ntothe left. D. 12 N to the right Three blocks of equal mass are placed on a smooth horizontal surface as shown above. A constant force Fis applied to block A so that the three blocks move towards the right with the same acceleration. The resultant force acting on blockB is 27. < HKCE 1991 Paper n � 7 > A. 0 Force B. fF C. ½F D. fF 23. < BKCE 1990 Paper Il-4 > ,�---70------Time Yhas a linear relationship with time as shown. Y may represent The diagram above shows the variation of the net force acting on an object which is initially at rest Which of the following (1) the speed ofa body starting from rest under a constant force. velocity-time graphs correctly describes the motion of the object 7 A R C � (2) the distance travelled by a body at constant speed. (3) the acceleration of a body fulling from rest. A (I) only B. (3) only C. (1) & (2) only D. (2) & (3) only DSE Physics - Section B : M.C. PB-FM2-M/081 . DSE Physics - Section B : M.C. PB-FM2-M/09 FM2 : Newton's Laws FM2 : Newton's Laws I 28. < HKCE 1992 Paper n- 8 > 31. < HKCE 1994 Paper II- 6 > A man of mass 50 kg is standing in a lift. If the lift is falling freely, which of the following statements is/are true ? Smooth table (1) The weightoftheman isON. (2) The force acting on the floor of the lift by the man is 491 N. (3) The force acting on theman by the floor of the lift is ON. A. (1) only B. (3) only Two identical blocks X and Y are connected by a light string passing over a smooth pulley as shown above. The two blocks C. (l) & ( 2) only are released from rest. After a while, the string breaks. Which of the following correctly descn'bes the motion of the blocks D. (2) & (3) only immediately after the string breaks? (Take the acceleration due to gravity to be 10 m s-2.) X y accelerates at 5 m s- 2 32. < HK.CE 1994 Paper IT- 2 > �------------� - A stops moving B. moves with constant velocity accelerates at 5 m s- 2 C. moves with constant velocity accelerates at 10 m s-2 D. decelerates at 5 m s- 2 accelerates at 1O m s-2 A trolley is given a push and runs down a friction-compensated run.way. The motion of the trolley is recorded on the paper tape as shown above. Which ofthe following changes can enable the trolley to produce a paper tape as shown below: 29. <HKCE1992Paperil-9> B A R--�------.�2N s (1) Giving the trolley a harder initial push. (2) Increasing the angle of inclination ofthe run.way. (3) Increasing the freqUency of the ticker-tape timer. In the diagram above, blocks A and B are connected by a light inextensible string and rest on a smooth horizontal table. The A. (1) only masses ofA andB are 2 kg and 3 kg respectively. Block A is pulled by a force of2 N. Find the tension in the string S. B. (3) only A. 0.4N B. 0.8N C. (1) & (2) only C. LON D. (2) & (3) only D. 1.2N m 33. < HK.CE 1994 Paper n- 4 > 30. < HKCE 1992 Paper Il-10 > Which of the following pairs of forces F1 and F2 is/are action and reaction pair(s) according to the Newton's thlrd law of motion? (1) F, Weight ofthe block Reaction from the floor Al (2) " 4N F, Weight ofthe ball Two blocks X and Y of weights 2 N and 8 N respectively are suspended by two light strings as shown in the diagram. A downward force of4 N is applied to X Find the tension Ti and T.1 in the two strings. Tension ofthe string T, T, A. 4N lON B,JJ B. 4N 14N C. 6N 12N F, D. 6N 14N (3 ) """"' Negative Positive charge F, Electric force acting on the negative charge :--0 34. < HKCE 1995 Paper II- 7 > A. (2) only 0-;, F, Electric force acting on the positive charge A car moves with a speed 30 km h-1• The driver applies the brake and the car is stopped in a distance of 12 m. If the car is moving at 60 km h-1, what is the stopping distance? Assume that the same constant braking force is applied in both cases. A. 12m B. (3) only B. 24m C. (1) & (2) only C. 48m D. (1) & (3) only D. 72m 35. DSE Physics - Section B : M.C. FM2 : Newton's Laws < HKCE 1995 Paper II- 6 > A trolley is placed on a horizontal ground . A force F inclined at an PB-FM2-M/10I I 39. <HKCE1996Paperil-8> DSE Physics - Section B : MC. FM2 : Newton's Laws PB-FM2-M/1111 angle 0 to the horizontal acts on the trolley. What is the horizontal component ofFthat pulls the trolley towards the right? A. F0 B. Fsin 0 Jo,,m:1 iii::,11 ,mru:11• '11, 1;. mimm11w,1 ·i 1 ; 1111J1!i'1 1 iu.immim111m, c. Peas 8 A block remains at rest on an inclini:dplane as shown above. Which of the following statements Ware true? D. FlsinB (1) The frictional force acting by the plane on the block.is zero. (2) The normal reaction acting by the plane on the block is zero. (3) The resultant force acting on the block.is zero. 36. < HKCE 1995 Paper II - 2 > A. (2) only B. (3) only C. (1) & (2) only D. (1) & (3) only Two blocks are connected together by a light string S placed on a smooth horizontal swface . They move with unifonn acceleration of2 m s·2 under theaction offorce F. What will theaccelerations of the blocks become ifS suddenly breaks? 40. <HKCE1996Paperil-6> 2 kg block 4 kg block coin A. 6m:rl Oms-2 B. 6ms·2 2m s·2 C. D. 2ms""2 Oms·2 Oms· 3ms·2 2 II vacuum 37. < HKCE 1995 Paper II - 11 > A coin and a featherare released from rest in a cylinder which is vacuum as shown. Which of the following is/are correct deductions from this experiment? W : the weight of the block (1) The masses of the coin and the feather are identical in vacuum. F : the gravitational force acting on the earth by the block (2) The coin and 'lh.e feather fall with the same acceleration in vacuum. (3) The forces acting on the coin and the feather in vacuum are identical. R : the force acting on the block by the ground A. (1) only B. (2)only C. (1) & (3) only D. (2) & (3) only The above diagram shows a block resting on the ground. Which of the following pairs of forces is/are action and reaction pair(s) according to Newton's third law of motion? 41. <HKCE 1996 Paper Il-10 > (1) Rand W (2) Wan.dF The below diagram shows a man lifting a ball vertically upwards with uniform acceleration. (3) FandR A. (1) only Let F1 be the force acting on the ball by the man, B. (2) only C. (3) only F2 be the force acting on the man by the ball, D. (1) & (2) only F; be the gravitational force acting on the ball. 38. < HKCE 1995 Paper Il- 5 > Two objects of difference masses are released from rest at thesame height. Assmne air resistance is negligible. Which of the following statements Ware correct? F, (1) A greater gravitational force is acting on the object v;ith greater mass. (2) They take the same time to reach the ground. Which of the following correctly describes the relation between the magnitudes of the forces? (3) They have equal velocities when they reach the ground. A. F1 =Fz>F3 A. (1) only B. Fi=Fo>F2 B. (1) & (3) only C. (2) & (3) only C. F1>F2=A D. (1), (2) & (3) D. F1>F2>F3 42. <BKCE 1996PaperII-5 > DSE Physics - Section B : M.C. FM2 : Newton's Laws PB-FM2-M/ 12 1 \I 47. DSE Physics - Section B : M.C. FM2 : Newton's Laws <HKCE 1998 Paper Il- 8 > PB- FM2- M/1 3 I Which of the following statements concerning the motion ofan object is/are correct? A girl in a lift uses a spring balance to measure the weigh t ofan object. The reading ofth� spring balance� 10 N when the (1 ) A constant unbalanced force is needed to keep an object moving with unifonn velocity. lift is at rest. When the lift is moving, the reading of the spring balance becomes 8 N. Which of the followmg descn"bes the (2) An increasing unbalanced furce is needed to keep an object moving vvith uniform acceleration. motion ofthe lift? (3) An object may remain at rest ifthere is no unbalanced force acting on it. A. moving downwards with a uniform velocity A. (2) only B. moving upwards with an acceleration B. (3) only C . (1) & (2) only C. moving downwards with an acceleration D. (!) & (3) only D. moving downwards with a deceleration 43. <HKCE 19.96Paperll-3 > 48. < HKCE 1998 Paper II· 7 > Which ofthe following statements about mass and weight is incorrect? A. Mass is measured in kilograms and weight in newtons. B. Mass is a measure ofthe inertia ofan object and w eight is a measure of the gravitational pull on it. ",, C. The weight ofan obj ect at a particular place is proportional to its mass. D. Both the mass and weight ofan object vary slightly at different places on the earth. Scale : 1 cm represents 1N pl -44. < HKCE 1997 Paper II- 9 > .......1-!2 Which ofthe following pan's of forces is/are action and reaction pair(s) according toNewton's third law ofmotion? (I) The weight ofa man standing on a chair. and The force acting on the man by the chair. t quilib rium, what should be the (2) The gravitational. force acting on the earth and The gravitational force acting on the moon Two forces F1 andF2 act on a particle Pas shown. Ifa third force FJ acts on Pto keep it in e by the moon. by the earth. magnitude of F3? (3 ) The force exerted by a swimmer on the and The force exerted by the water to push the A. 1.4N water to push the water backward. swimmer forward. B. 4.0N C. 4.2N A (1) only D. 4.5N B. (2) only c.. (1)&{3)only D. (2) & (3) only 49. <HKCE 1998 Paper n- 9 > Which of the following phenomena can be explained byNewton's first law ofmotion? 45. < HK.CE 1997 Paper n - 4 > (1) A passe nger in a car tends to move forward when the car suddenly stops. (2) A coin and a feather fall with the same acceleration in a vacuum. (3) The maximum mass an astronaut can lift on the moon is greater than on earth. A. (l)only B. (2) only A block is placed on a smooth inclined plane. A force P parallel to the inclined plane is applied to the block so that the block C. (1) & (3) only moves up the plane. Which of the following diagrams correctly shows all the forces acting on the block? D. (2) & (3) only A. B. C . Nomwl D. Normal l'll2ction p p SO. < HKCE 1999 Paper Il- 5 > # � Frioti, Friotl A B Weight Weight Weight 46. < HKCE 1998 Paper II- 6 > A broken-down car of mass 1000 kg is pulled by a tow-truck and moves at a constant velocity 8 m s-1 along a horizontal road. It is known that the frictional force acting on the caris 500 N. Find the tension in the cable connecting A light rope is fixed at two poles with the ends A andB at the same level A T-shirt ofweight2 N is hung at the midpoint C the truck and the car. of the rope. The rope depresses such that LACB = 150 . Find the tension in the rope. ° A. ON A 1.0N B. SOON B. 2.0N C. 8000N C. 3.9N D. 8500N D. 7.7N DSE Physics - Section B : MC. FM2 : Newton's Laws PB-FM2-M/14I I DSE Physics - Section B : MC. --- FM2 : Newton's Laws PB-FM2-M/1511 ==-===== 51. < HKCE 1999 Paper JI- 4 > : 55. <HKCE2000Paperll-7> A block is sliding down a :friction compensated runway as shown. Which of the following statements is/are correct ? (1) The speed of the block is increasing. (2) The normal reaction acting bytbe runway on the block is increasing. � ,L, (3) The net force acting on the block is zero. A block is placed on a rough inclined plane and then projected upwards along the plane. After reaching the highest point, A. (I) only the block slides down along the plane. Which of the following graphs shows the time variation of the velocity v of the B. (3) only block? C. (I) & (2) only A V B. V C. V D. V D. (2) & (3) only o�t o�, o�, 52. < HKCE 1999 Paper Il- 2 > A 2 kg steel sphere and a 1 kg wooden sphere are initially held at the same level above the ground and then released from rest simultaneously. Assume air resistance is negligible. Which of the following statements about the two spheres at any instant before they reach the ground is/are correct? . i (I) The speeds of the two spheres are equal. (2) The accelerations of the two spheres are equal. 56. < HKCE2000 Paper Il- 6> F, (3) The gravitational forces acting on the two spheres are equal. Three forces of magnitudes Ft, Fz and 10 N act on an object as shown. F,7 A. (1) only Ifthe object is in equihlrrium, find the force F2. B. (3) only C. (1) & (2) only A 5.0N D. (2) & (3) mtly B. 8.7N C. 11.SN 53. < HKCE2000 Paper II-2 > D. 173 N !ON It is said that Galileo Galilei (1564 - 1642), an Italian scientist, dropped a small iron ball and a large cannon ball from the top ofthe Leaning Tower of Pisa. He found that the two balls reached the ground at almost the same time. Which of the following is/are correctdeduction(s) from this experiment? 57. <HKCE2000Paperil-9> (1) The two balls fell with the same acceleration. (2) A body will maintain uniform motion if there is no external force acting on it. (3) The gravitational forces acting on the two balls were identical. uiU!lli A. (1) only A block is placed on a rough horizontal ground and a horizontal force acts on the block. If the magnitude of the force, F, is B. (3) only increased gradually, which of the following graphs shows the relation between F and the acceleration a of the block? C. (1) & (2) only A a B. 0 C. 0 D. a D. (2) & (3) only 54. < HKCE 2000 Paper II -8 > ,CF 1LF O�F ,L_F 58. < HKCE2000 Paper II-5> I An astronaut lands on the moon and finds that his weight is about one-sixth of that on the earth. Which of the following A uniform steel ball lies at rest on a horizontal ground and just touches a vertical wall as shown in the diagram. Which of deductions is/are correct? the following diagrams shows all the forces acting on the ball? (I) Ifhe throws an object upwards on the moon, it will reach a higher level than throwing the object with the same (Note ; W = gravitational force acting on the ball, R = normal reaction from the ground, speed on earth. F "" friction acting by the ground on the ball, N = normal reaction from the wall) (2) Ifhe releases an object on the moon, it will take a shorter time to reach the ground than releasing the object from the same height on earth. (3) The maximum weight he can lift on the moon is greater than on earth. A. (1) only B. (3) only C. {I) & (2) only D. (2) & (3) only