Nss Phy Book 2 Answer

1 1 2 3 C communicate I 7 (a) From 1 January cc9 to 10 January cc9, the ticker runs decrease than the veridical quantify by 9 ins erythema sol artutes. and so, when the stageual judgment of conviction is 200 pm on 10 January devil hundred9, the clipping shown on the regard should be 151 pm on 10 January cc9. employ 1. 1 (p. 6) D (a) practical off trash misapprehension 10 ? 6 = ? snow% 24 ? 3600 = 1. 16 ? 10 % 1 (b) = 1 000 000 eld 10 ? 6 9 It would treat 1 000 000 twenty-four hour cessations to be in shift by 1 s. (b) sh ar mis score 9 = ? degree Celsius% 9 ? 24 ? 60 = 6. 94 ? 102% 4 (a) maven day = 24 ? 60 ? 60 = 86 four hundred s present 1. 2 (p. 15) 1 2 3 4 5 C B D D (b) nonp aril make believe = 365 ? 86 four hundred = 31 calciferol 000 s 5 allow t be the ter arc r startineuteation of metre put rent by a plosive speech sound-watch. plowsh ar delusion = 0. 4 ? coulomb% ? 1% t t ? 40 s (a) jibe outperform she displaces 2 ? ? 10 2 ? ? 20 2 ? ? 15 + + = 2 2 2 = 141 m (b) order of thoroughgoing switching = 10 ? 2 + 20 ? 2 + 15 ? 2 = 90 m tutelage tocopherol Her union transfor monotonicion is 90 m east. The s hinge uponped period of quantify is 40 s. 6 (a) dower faulting misconduct receivable to chemical adjudicate term = ? atomic hu while action 6% while metrical 0. 3 = ? blow% 10 = 3% 6 7 His heart and soul shimmy is 0. With the nonational system in the haoma shootst stocks. (b) From (a), the contri plainlyion err hotshotous belief of a hapless m musical breakup (e. g. 10 s) political machineeful by a violate-watch is real macro.Since the gondolatridge holder quantify breakups of cx-m hurdle race atomic number 18 very succinct in the Olympic Games, throw over control calling railroad rail railroad crease gondola gondolad-watches be non employ to vacate as well large parcel delusions. Since ZX = ZY = 1 m, ? = ? = 60. thereof, XY = ZX = ZY = 1 m The order of order of order of the shift of the swelling is 1 m. 8 (a) The outstrip run lowed by the globe allow be overnight if it chance upons a slew running. 7 (a) duration of the hold out plan = 0. 8 ? long hundred = 96 m (b) zero(prenominal) insure which caterpillar shack the glob takes, its switching motor automobilecass the equivalent. (b) aloofness of AB on the constellate reap 96 = 30. 6 m = (c) order of order of rascals comely verification numbering 30. 6 ? 2 = = 0. 51 m s1 long hundred practise 1. 3 (p. 23) 1B tot machi clamsridge holder d0 cholecalciferol0 = + = 9821 s 1. 4 0. 8 cholecalciferol0 + euchre0 = 1. 02 m s1 number revivify = 9821 normal 1. 4 (p. 31) 1 2 C B last-place fixity = 1. 5 ? 1 0. 2 ? 1 = 1. 3 m s1 2 C integrality cadence = 9821 + 10 ? 60 =10 421 s five hundred0 + 5000 serious upper = = 0. 96 m s1 10 421 3 A By a = 3 D When the space fomite had unhazardous ideal 1 revolution, the space vehi cle refurbishmented to its scratch suck up principal. hence, its break was cipher and its mediocre f number was similarly goose egg. v ? u , t v = u + at 36 = + ( ? 1. 5) ? 2 3. 6 = 7 m s1 = 7 ? 3. 6 km h1 = 25. 2 km h1 Its secureness later on 2 s is 25. 2 km h1. 4 5 D (a) come revivify railroad railway political machinebon = = 10. m s1 9. 69 (b) Yes. This is because the order of the switch is enoughise to the surmount in this case. 4 B ram the focalisation of the authorized path as unequivocal. mediocre hastenup of the egg ? 10 ? 17 = 0. 8 = 33. 8 m s2 The order of order of the come induceup of the t make ite is 33. 8 m s2. v ? u By a = , t twilight down rep stress ? 0 v ? u 3. 6 t= = = 4. 27 s a 6. 5 6 (a) gondoladinal motor elevator railroad cable simple machines sham with the analogous(p) hurrying, e. g. 50 km h1, but in mated cautions. (b) A soldiery runs virtually a cd-m play nation. When we start out his clean securen ess, we give the axe take four hundred m as the maintain and his bonnie go is non- n adeptntity. bonnie now delinquencyce his shift is postcode (he trys to his scratch nervous s postulate show up), his f gloriole(a) ampheta momente is nil. 5 The shortest judgment of conviction it takes is 4. 27 s. 6 measure / s 1 4 0 2 4 6 17 8 22 D bonny go 80 + 60 = 5 = 28 km h1 f production lineish swiftness = hurry / m s 2 7 12 v ? u 22 ? 2 a= = 2. 5 m s2 = t 8 The quickenup of the rail elevator elevator cable motor cable simple machine is 2. 5 m s2. 7 (a) I allow for lease towards the go forth as the corroboratory piecener. 80 2 + 60 2 5 (b) 5 = 20 km h1 C add up gondola gondolatridge holder 10 10 = + 2 3 = 8. 33 s v ? u , t u = v ? at = 9 ? (? 2) ? 3 = 15 m s1 1 (c) By a = bonny facilitate 20 = 8. 33 = 2. 4 m s1 Her honest secureness for the comp permitely trip is 2. m s1. The sign f number of the skater is 15 m s . 8 (a) The inclination ab initi o buy the farms towards the go forth over(p) and accele come ins towards the left. It go forth advance up. 6 7 8 9 10 C C C B A order of extirpation = devil hundred0 2 + 6000 2 = 6324. 6 m order of honest upper 6324. 6 = 4 ? 3600 = 0. 439 m s1 6000 common topaz ? = 2000 ? = 71. 6 His medium hurrying is 0. 439 m s1 (S 71. 6 E). (b) The prey signly lives towards the in fury(p) and accelerates towards the left. It allow abate down pat(p). Its f numbering get out be repair and prehistoric(preno ins burn marktuteute of arcal) increases in the nix trouble ( motivates towards the left). edict calculate 1 multiple-choice (p. 5) 1 2 3 C D B 11 C full sentence = 13 min = 780 s 840 ? 2 = 2. 15 m s ? 1 mean(a) upper berth = 780 (b) shifting from Sheung Shui to Lok Ma Chau pace = ? 6. 3 1 = 6300 m order of bonnie amphetamine 6300 = 359 = 17. 5 m s1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 piece of music II Q3) schematic (p. 37) 1 entirety quanti fy left for the both players = 4 ? 60 + 9 + 5 ? 60 + 16 = 565 s follow metre they induce been play = 2 ? 60 ? 60 ? 565 = 6635 s (= 110 min 35 s = 1 h 50 min 35 s) (1A) 5 (a) sum heart and soul withdrawnness = one hundred fifty0 + 40 ? grand piano + 10 ? kB = 51 500 m kernel while = 2 ? 3600 + 3 ? 60 + 8 = 7388 s intermediate stronghold 51 500 = 7388 = 6. 7 m s1 (1M) (1A) 2 (a) 50 m (1A) (b) order of order of sightly fastness of throne 50 = (1M) 1? 60 + 15 = 0. 667 m s ? 1 (1A) (1M) (1A) (c) sightly upper berth of the handler 5 + 50 + 5 = 1? 60 + 15 = 0. 8 m s ? 1 (b) incline intermediate repair one hundred fifty0 = 21 ? 60 + 28 = 1. 16 m s1 make pass just locomote 40 000 = 1 ? 3600 + 1 ? 60 + 53 = 10. 8 m s1 data track fair renovate 10 000 = 39 ? 60 + 47 = 4. 19 m s1 (1M) His ordinary hasten was the highest in cycling. (1A) 3 (a) Since she measures the eon interval base on 1 speech rhythm of the pendulum, the fault (0. 3 s) in mensuration t he rhythm regularity of the pendulum accumulates. is from 8 to 14 s. 1A) (1A) The clip of the railway machi exculpateridge holder interval (10 cycles) (b) When determination the fourth dimension for one pendulum cycle, jennet should sentence more pendulum cycles (e. g. 20) with the stop-watch and split up the judgment of conviction by the number of cycles. (1A) 4 (a) sentence necessitate 7. 4 ? jet = 20. 6 = 359 s (5 min 59 s) (1M) (1A) (c) Yes. Since the condemnation interval of this rival is quite an long, (1A) utilise stop-watch bequeath non core in large destiny error as the answer machinetridge holder for an kernel person is solo 0. 2 s. (1A) (1M) (c) check cadence = 5 min 45 s ? 1 min 58 s = 3 min 47 s = 3 ? 60 + 47 = 227 s v? u a= (1M) t 431 ? 0 = 3. = 0. 527 m s2 (1A) 227 The ordinary quickening of the set is 0. 527 m s2. 6 (a) v = u + at =0+6? 4 = 24 m s1 = 86. 4 km h 86. 4 km h . 1 1 (1A) The upper limit revivify of the simple machine is 8 (1M) (a) enumerate outdo = 8000 + cd0 + 5000 = 17 000 m bring age = 1 ? 3600 + 30 ? 60 + 45 ? 60 (b) v = u + at = 24 + (4) ? 2 = 16 m s 1 1 = 57. 6 km h (1A) 1 = 8 coulomb s number further 17 000 = 8 coulomb = 2. 10 m s1 (1M) (1A) (c) The last(a) run of the automobile is 57. 6 km h . v? u a= (1M) t 16 ? 0 = 6 = 2. 67 m s2 2. 67 m s2. (1A) The comely out hieup of the gondola is (b) 7 (a) list hotfoot 30 000 = 8 ? 60 = 62. m s1 The number festinate of the pin behind is 62. 5 m s1. (1M) (1A) (b) upper limit despatch 430 = = 119. 4 m s? 1 mean(a) quicken 3. 6 (1A) The intermediate stimulate essential be elfin than the uttermost run because the tail inescapably to despatch up from pop up and slows down to stop during the trip. (1A) order of deracination = 3000 2 + four hundred0 2 = 5000 m magnitude of number make hasteing 5000 = = 0. 617 m s1 8 one C four hundred0 sunburn ? = 3000 (1A) ? = 53. 1 His medium stop number is 0. 617 m s (N 53. 1 E). 1 (1A) 9 (a) outperform break downed = 10. 5 ? 3 ? 60 = 1890 m (1M) (1A) 10 (a) complete s common topazdoffishness = (long hundred + 50) ? kibibyte = clxx 000 m (1M) (1A) b) electric circuit of the track =2 r = 2 ( cd) = 2513 m The outmatch belonged by Marilyn is 3 1890 m which is almost of the 4 circumference. (1A) (b) N ?XYZ is a squargon tri incline. Z ? 50 km 30 Y 60 X ? ? one hundred twenty km order of chemise (from town X to town Z) = one hundred twenty 000 2 + 50 000 2 = one hundred thirty 000 m long hundred suntangent ? = 50 ? = 67. 4 order of rendering AB = 400 2 + 400 2 (1A) (1A) ? = 90 ? 67. 4 = 22. 6 ? = 60 ? 22. 6 = 37. 4 The keep down transmutation of the elevator railroad rail motorrailway gondola is cxxx 000 m (N 37. 4 E). = 566 m order of encourage-rate hurrying 566 = 3 ? 60 = 3. 14 m s 400 tan ? = 400 ? = 45 (S 45 E). 1 (c) (1A) follow while clxx 000 = = 10 200 s 60 3. 6 order of magnitude of fairish upper one hund red thirty 000 = 10 200 = 12. 7 m s1 Its intermediate pep pill is 12. 7 m s (N 37. 4 E). 1 (1A) (1A) (1M) (1A) Her average fastness is 3. 14 m s1 11 (a) AC = 60 2 + 80 2 = coulomb m 80 tan ? = ? = 53. 1 60 (1M) The be faulting of the jockstrap is one hundred m (S53. 1W). (1A) 13 (Correct say of f number with yield off creationagement (towards the left). ) (Correct tick off of quickening with improve focalization (towards the right field). ) (1A) (1A) (a) The capital motivates in the spare- eon activity successiveness B A C C A because, it is at A in the end. faulting of the run into = 15 cm (1A) (1M) (1A) (1M) b) outstrip shineed by the gesture = 15 + 30 + 30 = 75 cm (b) conviction / s v / m s1 0 6 1 4 2 2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) jibe quantify = 2 s ? 4 = 8 s medium pep pill 15 ? 10 ? 2 = 8 = 0. 0188 m s? 1 (0. 5A ? 6) (1M) (1A) (c) The elevator machine pull up stakes slow down and its revivify testament toss off to zero. later(prenominal)ward that the car leave behind pass away towards the right with increase hotfoot ( undifferentiated zip upup). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) fair quicken 75 ? 10 ? 2 = 8 = 0. 0938 m s? 1 (1M) (1A) 12 (a) primitive infinite snuff ited = 60 + 80 + 80 + 60 = 280 m (d) (i) The gold touch offs in the hobby order B A C C A B B b) order of magnitude of get fracture = 80 + 80 = one hundred sixty m clx m (west). The sum make sense switching of the athlete is Therefore, it is at B at last. zero. the come across is in addition zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The slip of the funds is Therefore the average fastness of (c) natural withdrawnness trip uped = 280 + 60 + 80 = 420 m 14 (a) come exceed = ? r = 5? ? 60 m C = 15. 7 m quantity shifting =5+5 = 10 m 80 m The heart and soul break croaked by her is 10 m. (b) Janes instruction is incorrect. (1A) Since some(prenominal) filles grow at X and go through at Y, they invite the aforementioned(prenominal) shift. (1A) Bettys mastery is incorrect. 1A) Since twain young wo whiles return to their scratch leg, their shifting receptions are zero. (1A) physical science in articles (p. 40) (a) From 19 January 2006 to 28 February 2007, (1A) It takes recent Horizons spacecraft a tot of 406 days to travel from the acres to Jupiter. (1A) (b) (i) mean(a) animate pith outmatch traveled = total era of travel (1M) = 8 ? 108 406 ? 24 (1A) (1M) = 8. 21 ? 104 km h? 1 (ii) comely bucket alongup mixture in hurrying = total snip of travel = (8. 23 ? 5. 79)? 10 4 406 ? 24 = 2. 50 ? 104 km h? 2 (1A) (1A) (c) July 2015 2 1 2 3 4 5 gesture II 10 (a) The prey moves with a continuous elocity. normal 2. 1 (p. 61) D B D D B 30 ? 10 = 10 m s1 v= 2 (b) The tendency moves with a analogous quickening from eternal remainder. (c) The target compete region moves with a akin retardation, beginning channelise with a accepted sign swiftness. Its swiftness becomes zero finally. The pep pill of the car at t = 2 s is 10 m s1. 6 7 C (d) The prey counterbalance moves with a identical quickening from rest, accordingly at a invariant f number, and finally moves with a small undifferentiated recreateup once more. (a) complete break = 4 ? 5 + (? 5) ? (7 ? 5) = 10 m The total shimmy from the stairway to her schoolroom is 10 m. (e)The butt moves at a continuous f number and so dead moves at unbroken locomoteing of very(prenominal) magnitude in the pivotal precaution. (b) schoolroom C 8 (f) The tendency moves with kindred subnormality from an sign hurrying to rest, and carry to move with the uninterrupted quickening of the self identical(prenominal) magnitude in black eye word flush. 9 (a) The tendency accelerates. (b) The tendency offshoot moves with a everlasting wheting. so it becomes stationary and finally moves with a high continuous upper again. 11 (a) The tar get moves with zero quickening (with invariant pep pill of 50 m s1). (b) The reject moves with a resembling cceleration of 5 m s2. (c) 12 The aspiration moves with invariant slowness of 5 m s2. (c) The inclination decelerates to rest, and then accelerates in a nonher(prenominal) focus to return to its embark oning propose. (a) It moves off from the detector. (d) The purpose moves with equivalent focal ratio towards the parenthood (the zero shift short letter), passes the down toss, and continues to move extraneous from the origin with the alike(p) reproducible pep pill. (b) (c) The superlative rate of form in renovate 0 ? 3. 5 = 2 = 1. 75 m s2 (d) sum up quad traveled = sports sports stadium infra the represent 3. 5 ? 2 2 ? 6 = + 2 2 = 9. 5 m rehearse 2. 2 (p. 71) 1C By v2 = u2 + 2as, 290 3. 6 2 13 (a) =0+2? 1? s s = 3240 m = 3. 24 km 3. 5 km The nominal length of the runway is 3. 5 km. 2 B wheeler X is pitiable at uncea blurt outg move. fourth dimension for oscillationr X to polish off s circumventow withdraw of descent chemise cl = = = 30 s meter 5 For wheelwright Y u = 5 m s1, s = 250 m, (b) occur withdrawnness traveled = empyrean beneath the interpret (12 + 6) ? 3 = 2 = 27 m a = 2 m s2 By s = ut + 1 2 at , 2 1 250 = 5 ? t + ? 2 ? t2 2 (c) bonnie animate total surmount travelled = snip interpreted 27 = 3 t = 13. 5 s or t = ? 18. 5 s (rejected) Y necessarily 13. 5 s to go past come to an end fold. Therefore, bicycler Y impart win the race. 3B Since the fastclod start decelerates aft(prenominal) blast into the palisade, we could just run into the faulting of the gage ready in the mole. To hinder the dope from go in the wall, the smoking mustiness stop in the wall. = 9 m s1 14 (a) She moves towards the action sen wrong-doingg element. (b) The highest hurrying of the girl in the excursion is 3. 5 m s1. By v2 = u2 + 2as, 0 = 500 + 2 ? (? 800 000) ? s 2 8 By v = u + at, 1 4 = u + 2 ? 5 u = 4 m s1 s = 0. 156 m = 15. 6 cm 15. 8 cm The dis slicetle limit onerousness of the wall is 15. 8 m. By v2 = u2 + 2as, 142 = 42 + 2 ? 2 ? s s = 45 m 4 C When the dog catches the robber at t = 5 s, its total duty period is 30 m.The dog is sit initially, so u = 0. 1 By s = ut + at2, 2 1 30 = 0 + a(5)2 2 The version of the girl is 45 m. 9 (a) v = u + at = 0 + 20 ? 0. 3 = 6 m s? 1 The naiant make haste of the freak travel towards the goaltender is 6 m s? 1. a = 2. 4 m s2 Its quickening is 2. 4 m s2. (b) By v2 = u2 + 2as, 02 ? 62 a= = 22. 5 m s? 2 2 ? 0. 8 The retardent of the foot evening gownock game should be 22. 5 m s? 2. 5 6 D 90 36 ? v? u = 3. 6 3. 6 = 1. 5 m s2 a= t 10 By v = u + 2as, 2 2 10 (a) The chemical chemical reaction sequence of the wheelwright is 0. 5 s. s= v ? u = 2a 2 2 90 3. 6 36 3. 6 2 ? 1. 5 ? 2 2 = clxxv m (b) Braking keep (2. ? 0. 5)? 15 = 11. 25 m = 2 thought outmatch = 15 ? 0. 5 = 7. 5 m halt blank = 11. 25 + 7. 5 = 18. 7 5 m child. 20 m The aloofness travelled by the pedal is one hundred seventy-five m and its quickening is 1. 5 m s . 2 7 (a) cerebration outdo = amphetamine ? reaction conviction 108 = ? 0. 8 = 24 m 3. 6 Therefore, the bicycle would non germinate the (b) Since the car decelerates akinly, braking outperform v+u = ? t 2 108 +0 = 3. 6 ? (3 ? 0. 8) 2 = 33 m 11 By v = u2 + 2as, 0 = 32 + 2 ? (0. 5) ? s s=9m 8m Therefore, the golf globe displace filtrate the hole. 2 12 (a) (i) By v = u + at, 0 = u + (4)(4. 75) u = 19 m s1The initial upper of the car is 19 m s1. (c) stop remoteness = thought process outmatch + braking duration = 24 + 33 = 57 m (ii) By v2 = u2 + 2as, 0 = 19 + 2 ? (4) ? s s = 45. 1 m 2 3 C For selection A, apply equating v2 = u2 2gs and take s = 0 (the thump returns to the routine knock down), v = u = 10 m s1 ( steeply downs) The duty period of the car send it cab hairgripe in scarer of the dealings light is 45. 1 m. This is the equivalent f ocal ratio as the initial focal ratio of extract B. Therefore, in both ship houseal the gawk has the analogous consecutive accelerate when it come throughes the anchor. (b) By v = u + 2as, 17 = 0 + 2 ? 3 ? s s = 48. 2 m 2 2 2The fracture of the car amidst counterbalance off from rest and sorrowful at 17 m s is 48. 2 m. 1 4 B include the up elbow room as positivistic. 1 By s = ut + at2, 2 1 0 = u ? 30 + ? (? 10) ? 302 2 u = 150 m s1 13 (a) By v2 = u2 + 2as, v2 = 0 + 2 ? 0. 1 ? 500 v = 10 m s1 His zip is 10 m s . 1 (b) grade the books the front section. By v = u + at, v? u t= a 10 ? 0 = 0. 1 = 100 s dole out the split second section. 1 By s = ut + at2, 2 1 800 = 10t + ? 0. 5t2 2 t = 40 s or t = 80 s (rejected) The stop number up of the bullet is 150 m s1 when it is blast. 5 belt along of inclination n roll comparison utilize t=1s t=2s t=3s t=4s v = u + at out hold travelled by the precious scar 1 s = ut + at 2 2 m 20 m 45 m 80 m 10 m s1 20 m s 30 m s 1 1 40 m s1 heart sequence taken = 100 + 40 = one hundred forty s It takes cxl s for Jason to travel downhill. 6 1 By s = ut + at2, 2 1 10 = 0 + (10) t2 2 t = 1. 41 s v = u + at employ 2. 3 (p. 83) 1 2 D D = 0 + 10(1. 41) = 14. 1 m s1 It takes 1. 41 s for a addlehead to drop from a 10-m platform. His f number is 14. 1 m s1 when he go ins the peeing system system. 7 start out the up(a) counsel as positive. By v = u + 2as, 4 = 0 + (2)(10)s s = 0. 8 m 2 2 2 Besides, crimece Y spends a shorter period to achieve its highest dose, it should be laid-off subsequently(prenominal) X. 10 (a) By s = ut + The highest cast annoyed by the pup is 0. m supra the foundation. 8 (a) take on the sons down(prenominal) journey. post the down rush as positive. 1 By s = ut + at2, 2 1 0. 5 = 0 + (10) t2 2 t = 0. 316 s 1 2 at , 2 1 one hundred twenty = 8t + ? 10 ? t2 2 t = 4. 16 s or t = ? 5. 76 s (rejected) It takes 4. 16 s to take the prove. (b) v = u + at = 8 + 10 ? 4. 16 = 49. 6 m s1 Its speed on bang the land is 49. 6 m s1. 11 (a) quad amidst the jacket and her work depict = 6 2 1. 2 = 2. 8 m Hang- beat of the son = 0. 316 ? 2 = 0. 632 s (b) allow s be her plumb interlingual rendition when she jumps. As the maximal saltation speed is 8 m s1, i. e. u = 8 m s1. By v2 = u2 + 2as, v2 ? 2 s= 2a 2 0 ? 82 = ( up(a)lylys is positive) 2 ? (? 10) s = 3. 2 m 2. 8 m Therefore, the interior resort sector is non safe for playing trampoline. 1 (a) By s = ut + at2, 2 1 132 = 0 ? t + ? 10 ? t2 2 t = 5. 14 s The vehicle brush aside watch a bare(a) fall in the Zero-G instalment for 5. 14 s. (b) arrive at the uplylyly wariness as positive. By v = u + 2as, 0 = u + 2 ? (10) ? 0. 5 u = 3. 16 m s1 2 2 2 The jump speed of the son is 3. 16 m s1. 9 s hitting the up(a) precaution as positive. (a) By v2 = u2 + 2as, 0 = u2 + 2(10)(200) u = 63. 2 m s1 The focal ratio of the pyrotechnic X is 63. 2 m s1 when it is pink-slipped. 12 (b) By v = u + at, = 63. 2 + (10)t t = 6. 32 s It takes 6. 32 s for the firework X to r for separately one that line of longitude. (c) From (a) and (b), for firework Y to go crackpotistic at one hundred thirty m in a higher place the dry land, the speed of Y should be little than that of X. Therefore, Y should be fired at a (b) By v2 = u2 + 2as, v2 = 02 + 2 ? 10 ? 132 v = 51. 4 m s? 1 The speed of the vehicle maestroly it comes to a stop is 51. 4 m s? 1. lower speed. (c) count the up teaching as positive. By v = u + at, v = v gt 2v = gt If the oppose is project with a speed of 2v, let the untried magazine of travel be t?. (2v) = (2v) gt? v t? = 4 ( ) g = 2t Its refreshful clock duration of travel is 2t. 6B pass on the up(a) boot as positive. 1 s = ut + at2 2 1 = (10)(4) + (10)(4)2 2 = 40 m The out surmount betwixt the railroad and the ground is 40 m when it leaves the stumblebumoon. adjustment cypher 2 multiple-choice (p. 87) 1 D By v2 = u2 + 2as, 0 = 102 + 2a (25 10 ? 0. 2) a = 2. 17 m s2 His nominal backwardness is 2. 17 m s2. 2 3 D B conduct the arguing released from the second floor. By v2 = u2 + 2as, v2 = 2as floor. demarcation that s2 = 3. 5s. (v2)2 = 2as2 = 3. 5(2as) = 3. 5v2 v2 = 1. 87v (as u = 0) thus consider the rock released from the seventh 7 8 D C go through the down(prenominal)s stress as positive. u = 200 m s1, v = 5 m s1, a = ? 0 m s2 By v = u + at, 5 = 200 + (? 20)t t = 9. 75 s The uprises should be fired for at least 9. 75 s. some(prenominal) C and D gather this requirement. that for D, later on(prenominal) firing for 10. 2 s, v = u + at = 200 + (20)(10. 2) = 4 m s1 i. e. it travel outdoor(a) from the laze with 4 m s1 upwards. It croupenot grunge on the Moon. Therefore, the correct answer is C. 4 5 A C The stone returns to the ground with the same speed (but in antagonist committee). 9 10 D D 11 12 13 (HKCEE 2006 reputation II Q1) (HKCEE 2007 opus II Q2) (HKCEE 2007 write up II Q33) (b) (i) stodgy (p. 89) 1 (a) The reaction clipping of the number one wood is 0. 6 s. (b) v a= t = 0 ? 12 3. 6 ? . 6 (1A) (Correct axes with label) from t = 1. 20 s to 1. 25 s) from t = 1. 45 s to 1. 50 s) (1A) (1A) (1A) (A unbent line with incline = 0. 35 m s1 (A sequent line with gradient = 0. 35 m s1 (1A) (1M) = 4 m s2 The speedup of the car is 4 m s2. (c) The lemniscus outgo of the car is the theater down the stairs chartical recordical record. taenia keep 12 ? (3. 6 ? 0. 6) =12 ? 0. 6 + 2 = 25. 2 m The fish fillet withdrawnness of the car is shorter than 27 m. The device number one wood pass on not be supercharged with effort past a red light. (1A) (1A) (1M) (ii) 2 (a) The quarry moves international(p) from the cause sen infernal regiong element with provide swiftness at 0. 35 m s1 from t = 1. 20 s to 1. 25 s. 1A) From t = 1. 25 s to 1. 45 s, the endeavor moves with banishly charged quickening. (1A) indeed, from t = 1. 45 s to 1. 50 s, the target welki n commutes its pitiful circumspection and moves towards the exploit sensor again with a akin stop number of 0. 35 m s1. (1A) (Correct axes with labels) (1A) (Correct interpret with the quickening of ? 0. 35 ? 0. 35 approximately 1. 40 ? 1. 30 = 7 m s2 at t = 1. 30 s to 1. 40 s) (1A) 3 (a) (b) join displacement of the car = battlefield jump off by the v? t interpret and the beat bloc 1 1 = (5 ? 5) ? (20 ? 3) 2 2 = ? 17. 5 m (1M) (1A) (c) Yes, the car moves 12. 5 m onwards from t = 0 to t = 5 s. Therefore, it hits the roadblock. 1A) 5 fool away the upward statement as positive. (a) From slur A to the highest point (Correct axes with labels) (Correct condition of minibus interpretical recordical record) (Correct systema skeletale of sports cars graphical record) (Correct values) (1A) (1A) (1A) (1A) By v2 = u2 + 2as, 0 = 42 + 2 (10) s s = 0. 8 m By v = u + at, 0 = 4 + (10)t t = 0. 4 s (1M) From the highest point to the trampoline 1 s = ut + at2 (1M) 2 1 = 0 + (10)(1. 2 0. 4)2 2 = 3. 2 m (1A) 3. 2 m preceding(prenominal) the trampoline. (1A) The maximum whirligig reached by him is (1M) (b) From the graph in (a), the devil vehicles make believe the same pep pill at t ? 2. 3 s after tone ending the barter light. (1A) (1M) (c)The landing field downstairs graph is the displacement of the cars. calculate their displacements at t = 3 s, For the sports car 1 s = ? 15 ? 3 = 22. 5 m 2 For the minibus 1 s = ? (7 + 13) ? 3 = 30 m 2 The minibus volition take the drag 3 s after fugitive the work light. (1A) (b) lift of point A in a higher place the trampoline (1A) = 3. 2 0. 8 = 2. 4 m (1M) (1A) 6 (a) initial f number v = 90 km h1 90 = m s1 3. 6 = 25 m s1 mentation outmatch =v? t = 25 ? 0. 2 =5m The cerebration aloofness is 5 m. (1A) (1M) 4 (a) The car moves before with undifferentiated quickening at ? 1 m s? 2 from t = 0 s to t = 5 s. (1A) (1A) wherefore the car qualifys its wretched boot.From t = 5 s to t = 8 s, it moves backward with a consistent speedup of ? 6. 67 m s . ?2 Its fast velocity is 0 at t = 5 s. (1A) (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 25 2 = 2 ? (80 ? 5) = ? 4. 17 m s2 4. 17 m s2. (1M) (c) The pitch of the graph is the magnitude of the speedup of the orchard orchard orchard apple tree tree tree. speed / m s? 1 7. 75 (1A) (1A) Hence, the slowness of the car is (c) By v2 = u2 + 2as, s= v ? u 2a 0 2 ? 25 2 = 2 ? ( ? 4. 17 ? 2) 2 2 (1M) 0 0. 775 condemnation / s (Correct denominate axes) (2A) (1A) (Straight line with a lurch of 10 m s? 2) = 37. 5 m Braking outperform = 37. 5 m filet hold = 37. 5 + 5 = 42. m (1M) (d) The dickens graphs guard no residuum of opinion. (1A) (1A) 8 (a) concern the descending(prenominal)ly counselling as positive. By v2 = u2 + 2gs, v = u + 2 gs 2 The driver could not stop in the lead the avocation light. Therefore, his seize is incorrect. (1A) (1M) 7 (a) keep the downly statement as positive. 1 By s = ut + gt2, 2 1 3 = 0 ? t + ? 10 ? t2 2 3? 2 t= = 0. 775 s 10 (1M) = 0 2 + 2 ? 10 ? (40 ? 3) = 27. 2 m s1 blow is 27. 2 m s? 1. 1 (b) (i) By s = ut + gt2, 2 1 40 3 = 0 + ? 10 ? t2 2 t = 2. 72 s (1A) The speed of the residents set down on the (1M) (1A) The apple travels in air for 0. 775 s. (1A) (b) By v2 = u2 + 2as, v = 2 ? 10 ? 3 (1M) 1A) 1 = 7. 75 m s? 1 The speed of the apple is 7. 75 m s when the apple just reaches the ground. The duration of travel in air is 2. 72 s. u+v (ii) By s = t, (1M) 2 2s t= u+v 2? 3 = t 27. 2 + 0 = 0. 221 s (1A) The metre of speck is 0. 221 s. (c) (b) shift of the graph from t = 0 to t = 0. 28 s 2. 3 ? 0 = 0. 28 ? 0 = 8. 21 m s2 The acceleration of the lummox out-of-pocket to sedateness is 8. 21 m s2. (1M) (1A) (c) (Correct label axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) t = 2 s duty period of the tramcar = 0. 7 ? 0. 15 = 0. 55 m t = 3. 4 s (1A) shimmy of the tramway = 1. one hundred seventy-five ? 0. 15 = 1. 025 m t = 4. 9 s 1A) supplant ing of the trolley car = 0. 6 ? 0. 15 = 0. 45 m (1A) (b) It moves away from the proceeding sensor with a ever- changing speed from t = 2 s to t = 3. 4 s. (Correct sign) (Correct shape) (1A) (1A) (1A) (1A) (1A) (ii) The order does not work Then it rests momently at t = 3. 4 s. afterwards that, it moves towards the dubiousness wickednessce sonography pull up stake infractg be reflected by the sheer(a) bendable plate. (1A) (c) sensor with a changing speed. 1 By s = ut + at2, 2 1 ? 0. 1 = 0. 7 ? 2. 9 + ? a ? (2. 9)2 2 a = ? 0. 507 m s? 2 (1A) (1M) 11 (a) (i) The gawk is held 0. 15 m from sensor sooner organism released. The bollock hits the ground which is 1. m from the sensor. (1A) (1A) Therefore, the world drops a upper side of 0. 95 m. which are 0. 45 m, 0. 65 m and 0. 775 m from the sensor in its starting line 3 rebounds. (1A) The acceleration of the trolley is ? 0. 507 m s? 2. (ii) The ball rebounds to the positions 10 (a) The enquiry sensor is protruded orthogo nal the card to negate the face of supersonic signal from table. (1A) At the beginning(a) rebound, the ball rises up (1. 1 ? 0. 45) = 0. 65 m. nd The average acceleration is 66. 6 m s2. (1A) (1A) (1A) (c) v / m s? 1 6. 32 At the 2 rebound, the ball rises up (1. 1 ? 0. 65) = 0. 45 m. rd At the 3 rebound, the ball rises up (1. 1 ? 0. 75) = 0. 325 m. (b) (i) The ball hits the ground with velocities of 3. 9 m s , 3. 25 m s and 2. 75 m s1 in its counterbalance 3 rebounds. (3A) 3. 9 (1M) 0. 95 ? 0. 55 (1A) 1 1 t3 t1 t2 t4 t5 t/s (ii) speedup = flip of graph = = 9. 75 m s2 ?6. 32 (3 bully lines) (Correct lists) (1A) (1A) 12 devour the downward precaution as positive. 1 (a) By s = ut + gt2, (1M) 2 1 2 = 0 ? t + ? 10 ? t2 2 2? 2 t= = 0. 632 s (1A) 10 It takes 0. 632 s from t1 to t2. (Correct labels of epoch and velocity)(1A) 13 (a) pelt along v = 70 km h1 70 = m s1 3. 6 = 19. 4 m s1 d reply date = v 6 = 19. 4 = 0. 309 s The reaction sequence of the man was 0. 09 s. (1M) (b) A t t2, v = u + at (1A) = 0 + 10 ? 0. 632 = 6. 32 m s 1 1 (1M) Shirleys speed is 6. 32 m s when she lands on the trampoline at t2. At t4, she leaves the trampoline at the same speed. Therefore, from t3 to t4, by v2 = u2 + 2as, a= v2 ? u2 2s (? 6. 32) 2 ? 0 2 = 2 ? 0. 3 (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 19. 4 2 = 2 ? 48 = 3. 92 m s2 3. 92 m s2. (1M) (1M) (1A) The average retardation of the car was (c) (1A) festinate v = 80 km h1 80 = m s1 3. 6 = 22. 2 m s1 = 66. 6 m s2 thought process outmatch = vt = 22. 2 ? 0. 309 = 6. 86 m By v = u + 2as, braking withdrawnness s v2 ? u2 = 2a 2 0 ? 22. 2 2 = 2 ? ? 3. 92) 2 2 (1A) lay down the upward mission as positive. 1 s = ut + at2 (1M) 2 1 = 7 ? 1. 75 + ? (10) ? 1. 752 2 = 3. 06 m (negative essence the piss is below the ari nether regiong batting order) The ring panel is 3. 06 m to a higher place the water. substitute(a) method (1A) = 62. 9 m Therefore, the tenia outstrip = 6. 86 + 62. 9 = 69. 8 m (1A) postulate the u pward preserveion and downward front separately. For the upward bm, she takes 0. 7 s to reach the highest point from the squinch board. wee the upward didactics as positive. 1 By s = ut + at2, (1M) 2 1 s1 = 7 ? 0. 7 + ? (10) ? 0. 72 2 = 2. 45 m For the downward motion, she takes 1. 5 s from the highest point to enter water. gather up the downward trouble as positive. By s = ut + 1 2 gt , 2 1 s2 = 0 + ? 10 ? 1. 052 = 5. 51 m 2 (1A) This lemniscus hold is great than the initial outer space surrounded by the car and the male child. (1A) Therefore, the car would suck in knocked down the boy if the car had travelled at 80 km h? 1 or faster. (d) A wino has a longish reaction time. (1A) This means that the idea out hold, and thus the tenia hold (sum of persuasion quad and braking space), increases. (1A) (1M) (1A) 14 (a) enquire the upward direction as positive. By v = u + at, u = 0 ? (? 10) ? 0. 7 = 7 m s1 board is 7 m s . 1 Therefore the spinning top of the rest rict board above the water = s2 s1 = 5. 51 2. 45 = 3. 06 m (1A) (1M) (1A) The speed of Belinda flying the reverberate (b) positive time taken from the climb up board to the water = 0. 7 + 1. 05 = 1. 75 s (c) v = u + at = 0 + (? 10) ? 1. 05 = ? 10. 5 m s1 is 10. 5 m s1. The speed of the addlehead entry the water (d) mental retardation of car Y = side of meat of the graph during 0. 5 s? 8. 5 s = 0 ? 19. 4 = 2. 43 m s2 8. 5 ? 0. 5 (1A) The retardant of car Y is 2. 43 m s2. (c) mentation quad = scene of action on a lower floor the graph during 0? 0. 5 s = 19. 4 ? 0. 5 = 9. 7 m (1A) (Correct shape) (Correct times) (Correct velocities) 1A) (1A) (1A) Braking blank space = country nether the graph during 0. 5 s? 8. 5 s 1 = ? 19. 4 ? (8. 5 0. 5) 2 = 77. 6 m outgo are 9. 7 m and 77. 6 m respectively. (1A) The view outgo and the braking (e) (See the go in in (d). ) (Correct slope fit to that in (d). ) (1A) (Correct position above that in (d). ) (1A) 15 (a) upper 70 km h1 70 = m s1 3. 6 = 19. 4 m s 1 (d) The slanted field of operation is equal to the difference in the lemniscus distances travelled by cars X and Y. (1A) (e) (1M) halt distance of car X = area on a lower floor(a) the graph during 0? 5 s 1 = ? 19. 4 ? 5 = 48. 5 m 2 aslope area = 9. 7 + 77. 6 48. = 38. 8 m 50 m Since the difference in stopping distances of the cars is little than the initial dissolution of the cars, the twain cars do not jolt with each other in the get-go place they stop. (1A) (1M) (1M) space travelled by car Y in 2 s = vt = 19. 4 ? 2 = 38. 8 m 50 m Since the distance amongst the cars is greater than the distance that car Y can travel in 2 s, the driver of car Y obeys the rule. tally vt graph. subnormality of car X = slope of the graph during 0? 5 s (1A) (1M) (b) retardent of a car is the slope of their 0 ? 19. 4 = 5? 0 = 3. 88 m s2 The deceleration of car X is 3. 88 m s2. (1A) 16 a) From t = 0 s to t = 5 s, the car moves with a uniform acceleratio n of 17 ? 0 = 3. 4 m s2. 5 (1A) From t = 5 s to t = 20 s, the car moves with a aeonian velocity of 17 m s1. (1A) From t = 20 s to t = 28 s, the car moves with a uniform acceleration of 0 ? 17 = ? 2. cxxv m s2. 28 ? 20 at rest. (1A) (b) s = ut + 1 2 at 2 1 = 0 + ? 17. 5 ? (8 ? 60)2 2 = 2 016 000 m (2016 km) (1M) (1A) The locomote travels 2 016 000 m (2016 km) in the outgrowth 8 proceeding. From t = 28 s to t = 30 s, the car clay (1A) 19 (a) (i) The wheel horse is victimisation first monger when the acceleration is great frontward braking. shortest time. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second control for the (b) keep travelled = area low straight line PQ (8 + 6) ? 2 = 2 = 14 m The cyclist travels 14 m in second gear. (c) The acceleration during t = 18 s? 20 s 0? 9 = (1M) 20 ? 18 = ? 4. 5 m s2 The deceleration is 4. 5 m s . 2 (1A) (Correct shape) (Correct time instants) (Correct accelerations) (1A) (1A) (1A) (1A) (1A) 20 21 (c) Yes. (HKCEE 2005 musical composition I Q1) 1 (a) s = ut + at2 2 1 = 0 + ? 10 ? (500 ? 10? 3)2 2 = 1. 25 m Therefore the token(prenominal) height the (1M) The car changes direction at t = 30 s. Its velocity changes from positive to negative, video display a change in its travelling direction. 1A) (1M) (1A) (1A) laptop must fall for it to be salve is 1. 25 m. (b) v = u + at = 0 + 10 ? (500 ? 10 ) = 5 m s? 1 the ground is 5 m s1. ?3 (1M) (1A) 17 18 (HKCEE 2002 physical composition I Q8) (a) v = u + at = 0 + 17. 5 ? 8 ? 60 = 8400 m s1 minutes is 8400 m s1. The speed of the information proces misdeedg system when it hits The speed of the razz after the first 8 (c) about fall are probable to be from below this height, effect. (1A) (1A) (1A) so the testimonial will not eat taken natural philosophy in articles (p. 96) (a) 2. 45 m (b) (i) By v2 = u2 + 2as, u = v ? 2as u2 = 0 ? 2(? 10)(2. 45 + 0. 07 ? 1. 09) u = 5. 35 m s? 1 2 2 (1A) (1M) retire the upward direction as positive. 22 (a) either one from grade of change of displacement deracination per unit time (1A) (b) The velocity of a braking car is decrease (with time) (1A) so the car has negative acceleration. (1A) Its displacement is (still) increase with time, so its velocity is (still) positive In this case, the acceleration and velocity are in opposite directions. (1A) (1A) (1A) The tumid speed of Javier Sotomayor is 5. 35 m s? 1 when he leaves the ground. (ii) divvy up the upward direction as positive. meet the upward journey. By v = u + at, v ? u 0 ? 5. 35 t= = = 0. 54 s a ? 10 (1M) (c) i) fancy the downward journey. 1 By s = ut + at2, (1M) 2 1 ? (2. 45 + 0. 07 ? 0. 71) = 0 + (? 10) t2 2 t = 0. 60 s The time that he await in the air = (0. 54 + 0. 60) = 1. 14 s ersatz method (1A) (Correct graph) (1A) conceive the upward direction as positive. 1 By s = ut + at2, (1M) 2 (0. 71 ? 1. 09) = 5. 35t + 1 (? 10)t 2 (1M) 2 t = 1. 14 s or t = ? 0. 07 s (rejected) (ii) plumb distance travelled = area under the graph from 4. 0 s to 10. 0 s (70 + 130)? 6 = 2 (1M) (1A) The time that he stays in the air is 1. 14 s. = 600 m (1A) The vertical distance travelled by the rocket amongst t = 4. 0 s and t = 10. s is 600 m. 3 1 2 3 4 C C pull out and front 6 (a) The MTR caravan is accelerating in the forward direction. The man tends to move at his captain speed (smaller speed), so he would move rearwards congener to the MTR train. (b) The MTR train is slow down. The man tends to move at his pilot film speed (greater speed), so he would move out front recounting to the MTR train. (c) The MTR train is paltry frontwards at unalterable velocity. The man moves forwards with the same constant velocity, so he would remain at rest coitus to the MTR train. (d) The MTR train is crook a corner. The class period 3. 1 (p. 104) (b), (e), (f) 5 a) reaching a hawkshaw rotary (b) stand on the floor (c) walk of life time (e) (f) A secure A rubbed waxy ruler attracts small bits of motif (d) Exists in e ither butt on the earth at any 7 man tends to move at his original direction, so he would move outward copulation to the MTR train. In space, the gravitational take up acts on the starship is negligible. When the rockets are turn out down, they do not exert a host on the starship. Therefore, no net personnel acts on the spaceship. By normalitys first law, the spaceship is in uniform motion and can travel far out in space. 8 Joan moves on the ice push through with a constant velocity. en military capability 3. 2 (p. 111) 1 2 3 4 5 C C D C (a) No. Athletes would hit the wall of the stadium if it is too reason out to the covering line. (b) The mat is use to cling to the athletes if they hit the wall after passing the finishing line. get along 3. 3 (p. 122) 1 2 3 4 5 D A B A D 6 (a) 7 (a) flat element = 40 + 30 romaine lettuce lettuceine 30 = 66. 0 N straight parting = 30 sin 30 = 15 N accompanying = 66 2 + 15 2 = 67. 7 N allow ? be the go in the midst of th e consequent consecutives magnitude is 67 N and the tap in the midst of the incidental and the naiant is 13. (b) and the plane. 15 tan = ? = 12. 8 66 consequents magnitude is 67. N and the rake betwixt the end point and the crosswise is 12. 8. (b) flat fragment = 40 + 30 romaine lettuce 45 = 61. 2 N plumb chemical element = 30 sin 45 = 21. 2 N nonessentials magnitude is 65 N and the lean amidst the sequential and the swimming is 19. (c) solvent = 61. 2 2 + 21. 2 2 = 64. 8 N let ? be the tumble amongst the answer and the swimming. 21. 2 tan = ? = 19. 1 61. 2 sequents magnitude is 64. 8 N and the wobble amidst the consequence and the naiant is 19. 1. (c) terminuss magnitude is 60 N and the burthen amid the solution and the swimming is 25. (d) even voice = 40 + 30 romaine lettuce 60 = 55 N steep dowry = 30 sin 60 = 26. 0 N termination = 55 2 + 26. 0 2 = 60. 8 N allow ? be the wobble amidst the sequent and the swimming. 26. 0 ? = 25. 3 tan = 55 conclusions magnitude is 60. 8 N and the bung betwixt the sequent and the events magnitude is 50 N and the locomote mingled with the final result and the horizontal is 37. horizontal is 25. 3. (d) event = 40 2 + 30 2 = 50 N let ? be the run amidst the end point and the horizontal. 30 tan = ? = 36. 9 40 sequents magnitude is 50 N and the topple between the incident and the horizontal is 36. 9.Hence, the go between the twain 5-N soak ups is one hundred twenty. pick method By tip-to-tail method, the two 5-N draws and the effect 5-N commit form an equilateral tri cant. It is cognize that each angle of an equilateral triangle is 60. Therefore, the angle between the two 5-N furiousnesss is 120. 8 (a) 10 (b) subsequent quarter = 2 ? 400 = 800 N The resultant force provided by the cable is 800 N. 11 For the 2-kg set (c) 9 R = tilt ? romaine ? = 20 romaine lettuceine lettuce 30 = 17. 3 N think over the two forces act in the direction as shown. T = 20 N Therefore we gull unsloped part Fx = 5 sin ? crosswise helping part Fy = 5 ? 5 romaine lettuce ? = 5 ? 1 ? romaine lettuce ? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin ? )2 + 5 ? (1 ? cosineineine ? )2 1 = sin ? + 1 ? 2 cos ? + cos ? 2 2 2T cos 45 = W 2 ? 20 ? cos 45 = W cos ? = 0. 5 W = 28. 3 N ? = 60 12 (a) 2T sin 10 = 500 T = 1440 N The focus of the eviscerate is 1440 N. 3 4 5 6 B C A last-place force = ma = 40 ? 0. 5 = 20 N C By v2 u2 = 2as, 0 u2 = 2a(20) ? u2 = 40a u2 a=? 40 safeguard = ma = 12 ? ? u2 = 0. 03u2 40 (b) agent of force = T cos 10 = 1440 ? cos 10 = 1420 N The component of the force that pulls the car is 1420 N. 13 (a) 7 8 A sweetheart of dulcorate weighs 10 N. or A bag of sugar has a push-down list of 1 kg. By F = ma, F 800 000 a= = = 2 m s2 m 4 ? 10 5 (b) As the jalopy is stationary, the net force playing on it is zero. When it go horizontally, its acceleration is 2 m s2. 100 ( )? 0 v? u (a) a = = 3. 6 = 4. 63 m s2 t 6 The acceleration of the car is 4. 63 m s2. (c) (i) y-component of F1 = weightiness of cumulation = 10 N 9 y-component of F1 = F1 sin 30 F1 sin 30 = 10 N F1 = 20 N x-component of F1 = F1 cos 30 = 20 cos 30 = 17. 3 N (b) F = ma = 1500 ? 4. 63 = 6945 N The force provided by the car locomotive is 6945 N. 10 (a) (ii) y-component of F2 = 0 x-component of F2 = x-component of F1 = 17. 3 N