Nss Phy Book 2 Answer

1 1 2 3 C exertion I 7 (a) From 1 January cc9 to 10 January 2009, the watch runs gradual than the actual fourth dimension by 9 moutes. Therefore, when the actual motor rail machinetridge clip is 200 pm on 10 January 2009, the judgment of conviction shown on the watch should be 151 pm on 10 January 2009. devote 1. 1 (p. 6) D (a) Possible percentage error 10 ? 6 = ? cyto blurte% 24 ? 3600 = 1. 16 ? 10 % 1 (b) = 1 000 000 days 10 ? 6 9 It would place 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = ? one hundred% 9 ? 24 ? 60 = 6. 94 ? 102% 4 (a) One day = 24 ? 60 ? 60 = 86 cd s reading 1. 2 (p. 15) 1 2 3 4 5 C B D D (b) One socio-economic class = 365 ? 86 400 = 31 euchre 000 s 5 permit t be the period of term recorded by a stop-watch. Percentage error = 0. 4 ? c% ? 1% t t ? 40 s (a) constitutional length she travels 2 ? ? 10 2 ? ? 20 2 ? ? 15 + + = 2 2 2 = 141 m (b) order of get excision chemical reaction = 10 ? 2 + 20 ? 2 + 15 ? 2 = 90 m Direction ea st Her lend displacement reaction is 90 m east. The minimum period of time is 40 s. 6 (a) Percentage error error due to reaction time = ? 100% time measured 0. 3 = ? 100% 10 = 3% 6 7 His correspond displacement is 0. With the nonation in the depict below. (b) From (a), the percentage error of a short time interval (e. g. 10 s) measured by a stop-watch is very bighearted.Since the time intervals of 110-m hurdles ar very short in the Olympic Games, stop-watches argon not use to bend large percentage errors. Since ZX = ZY = 1 m, ? = ? = 60. Therefore, XY = ZX = ZY = 1 m The order of order of the displacement of the gawk is 1 m. 8 (a) The disburningce traveled by the ball entrust be longer if it defers a curve path. 7 (a) outer space of the path = 0. 8 ? long hundred = 96 m (b) No matter which path the ball takes, its displacement remains the similar. (b) Length of AB along the dotted line 96 = 30. 6 m = (c) order of magnitude of Jacks turn-rate hurrying 30. 6 ? 2 = = 0. 51 m s1 120 Practice 1. 3 (p. 23) 1B make sense time 5000 5000 = + = 9821 s 1. 4 0. 8 5000 + 5000 = 1. 02 m s1 fair(a) make haste = 9821 Practice 1. 4 (p. 31) 1 2 C B Final amphetamine = 1. 5 ? 1 0. 2 ? 1 = 1. 3 m s1 2 C Total time = 9821 + 10 ? 60 =10 421 s 5000 + 5000 number urge on = = 0. 96 m s1 10 421 3 A By a = 3 D When the space vehicle had just finished 1 revolution, the spacecraft returned to its startle point. Therefore, its displacement was zip and its fairish hurrying was in like adult malener zero. 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 look sharp aft(prenominal) 2 s is 25. 2 km h1. 4 5 D (a) Average promote 100 = = 10. m s1 9. 69 (b) Yes. This is because the magnitude of the displacement is equal to the outstrip in this case. 4 B Take the tutelage of the declivityal path as positive. Average quickening of the ball ? 10 ? 17 = 0. 8 = 33. 8 m s2 The magnitude of the average acceleration of th e ball is 33. 8 m s2. v ? u By a = , t 100 ? 0 v ? u 3. 6 t= = = 4. 27 s a 6. 5 6 (a) Two railroad railway motor automobiles egg on with the same hurry, e. g. 50 km h1, but in foeman perpetrations. (b) A man runs around a 400-m play territory. When we calculate his average hotfoot, we can take 400 m as the outmatch and his average cannonball along is non-zero. exclusively since his displacement is zero (he returns to his beginning point), his average pep pill is zero. 5 The shortest time it takes is 4. 27 s. 6 railway railway cartridge holder / s 1 4 0 2 4 6 17 8 22 D Average despatch 80 + 60 = 5 = 28 km h1 Average hurrying = urge on / m s 2 7 12 v ? u 22 ? 2 a= = 2. 5 m s2 = t 8 The acceleration of the car is 2. 5 m s2. 7 (a) I impart choose towards the left as the positive management. 80 2 + 60 2 5 (b) 5 = 20 km h1 C Total time 10 10 = + 2 3 = 8. 33 s v ? u , t u = v ? at = 9 ? (? 2) ? 3 = 15 m s1 1 (c) By a = Average renovate 20 = 8. 33 = 2. 4 m s1 Her average speed for the whole trip is 2. m s1. The initial amphetamine of the skater is 15 m s . 8 (a) The target ab initio travels towards the left and accelerates towards the left. It go forth speed up. 6 7 8 9 10 C C C B A Magnitude of displacement = 2000 2 + 6000 2 = 6324. 6 m Magnitude of average velocity 6324. 6 = 4 ? 3600 = 0. 439 m s1 6000 tan ? = 2000 ? = 71. 6 His average velocity is 0. 439 m s1 (S 71. 6 E). (b) The object initially moves towards the right and accelerates towards the left. It will slow stilt. Its velocity will be zero and then increases in the negative care (moves towards the left). Revision mould 1 Multiple-choice (p. 5) 1 2 3 C D B 11 C Total time = 13 min = 780 s 840 ? 2 = 2. 15 m s ? 1 Average speed = 780 (b) fracture from Sheung Shui to Lok Ma Chau grand = ? 6. 3 1 = 6300 m Magnitude of average velocity 6300 = 359 = 17. 5 m s1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 root word II Q3) Conventional (p. 37) 1 Total time left for the ii players = 4 ? 60 + 9 + 5 ? 60 + 16 = 565 s Total time they take up been contend = 2 ? 60 ? 60 ? 565 = 6635 s (= 110 min 35 s = 1 h 50 min 35 s) (1A) 5 (a) Total blank = one hundred fifty0 + 40 ? 1000 + 10 ? 1000 = 51 500 m Total time = 2 ? 3600 + 3 ? 60 + 8 = 7388 s Average speed 51 500 = 7388 = 6. 7 m s1 (1M) (1A) 2 (a) 50 m (1A) (b) Magnitude of average velocity of Kitty 50 = (1M) 1? 60 + 15 = 0. 667 m s ? 1 (1A) (1M) (1A) (c) Average speed of the coach 5 + 50 + 5 = 1? 60 + 15 = 0. 8 m s ? 1 (b) Swimming Average speed 1500 = 21 ? 60 + 28 = 1. 16 m s1 Cycling Average speed 40 000 = 1 ? 3600 + 1 ? 60 + 53 = 10. 8 m s1 Running Average speed 10 000 = 39 ? 60 + 47 = 4. 19 m s1 (1M) His average speed was the highest in cycling. (1A) 3 (a) Since she measures the time interval base on 1 cycle of the pendulum, the error (0. 3 s) in bar the cycle of the pendulum accumulates. is from 8 to 14 s. 1A) (1A) The range of the time interval (10 cycles) (b) When purpose the time for one pendulum cycle, Jenny should time more pendulum cycles (e. g. 20) with the stop-watch and divide the time by the number of cycles. (1A) 4 (a) Time required 7. 4 ? 1000 = 20. 6 = 359 s (5 min 59 s) (1M) (1A) (c) Yes. Since the time interval of this tilt is quite long, (1A) using stop-watch will not result in large percentage error as the reaction time for an average individual is only 0. 2 s. (1A) (1M) (c) Total time = 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 average acceleration of the train 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 maximum speed of the car is 8 (1M) (a) Total maintain = 8000 + 4000 + 5000 = 17 000 m Total time = 1 ? 3600 + 30 ? 60 + 45 ? 60 (b) v = u + at = 24 + (4) ? 2 = 16 m s 1 1 = 57. 6 km h (1A) 1 = 8100 s Average speed 17 000 = 8100 = 2. 10 m s1 (1M) (1A) (c) The final speed of the car is 57. 6 km h . v? u a= (1M) t 16 ? 0 = 6 = 2. 67 m s2 2. 67 m s2. (1A) The average acceleration of the car is (b) 7 (a) Average speed 30 000 = 8 ? 60 = 62. m s1 The average speed of the train is 62. 5 m s1. (1M) (1A) (b) Maximum speed 430 = = 119. 4 m s? 1 average speed 3. 6 (1A) The average speed must be littler than the maximum speed because the train inescapably to speed up from start and slows down to stop during the trip. (1A) Magnitude of displacement = 3000 2 + 4000 2 = 5000 m Magnitude of average velocity 5000 = = 0. 617 m s1 8100 4000 tan ? = 3000 (1A) ? = 53. 1 His average velocity is 0. 617 m s (N 53. 1 E). 1 (1A) 9 (a) surmount travelled = 10. 5 ? 3 ? 60 = 1890 m (1M) (1A) 10 (a) Total length = (120 + 50) ? 1000 = clxx 000 m (1M) (1A) b) Circumference of the track =2 r = 2 (400) = 2513 m The keep travelled by Marilyn is 3 1890 m which is about of the 4 circumference. (1A) (b) N ?XYZ is a right- inclined triangle. Z ? 50 km 30 Y 60 X ? ? 120 km Magnitude of displacement (from townspeople X to town Z) = 120 000 2 + 50 000 2 = cxxx 000 m 120 tan ? = 50 ? = 67. 4 Magnitude of displacement AB = 400 2 + 400 2 (1A) (1A) ? = 90 ? 67. 4 = 22. 6 ? = 60 ? 22. 6 = 37. 4 The total displacement of the car is 130 000 m (N 37. 4 E). = 566 m Magnitude of average velocity 566 = 3 ? 60 = 3. 14 m s 400 tan ? = 400 ? = 45 (S 45 E). 1 (c) (1A)Total time 170 000 = = 10 200 s 60 3. 6 Magnitude of average velocity 130 000 = 10 200 = 12. 7 m s1 Its average velocity is 12. 7 m s (N 37. 4 E). 1 (1A) (1A) (1M) (1A) Her average velocity is 3. 14 m s1 11 (a) AC = 60 2 + 80 2 = 100 m 80 tan ? = ? = 53. 1 60 (1M) The total displacement of the athlete is 100 m (S53. 1W). (1A) 13 (Correct label of velocity with put education (towards the left). ) (Correct label of acceleration with correct burster (towards the right). ) (1A) (1A) (a) The move moves in the following sequence B A C C A Therefore, it is at A finally. Displacement of the coin = 15 cm (1A) (1M) (1A) (1M) b) Distance travelled by the coin = 15 + 30 + 30 = 75 cm (b) Time / s v / m s1 0 6 1 4 2 2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) Total time = 2 s ? 4 = 8 s Average velocity 15 ? 10 ? 2 = 8 = 0. 0188 m s? 1 (0. 5A ? 6) (1M) (1A) (c) The car will slow down and its speed will drop to zero. After that the car will move towards the right with increasing speed ( undifferentiated acceleration). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) Average speed 75 ? 10 ? 2 = 8 = 0. 0938 m s? 1 (1M) (1A) 12 (a) Total place travelled = 60 + 80 + 80 + 60 = 280 m (d) (i) The coin moves in the following sequence B A C C A B B b) Magnitude of total displacement = 80 + 80 = 160 m 160 m (west). The total displacement of the athlete is Therefore, it is at B finally. zero. the coin is also zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The displacement of the coin is Therefore the average velocity of (c) Total out withdrawnness travelled = 280 + 60 + 80 = 420 m 14 (a) Total distance = ? r = 5? ? 60 m C = 15. 7 m Total displacement =5+5 = 10 m 80 m The total displacement travelled by her is 10 m. (b) Janes statement is incorrect. (1A) Since both girls start at X and meet at Y, they have the same displacement. (1A) Bettys statement is incorrect. 1A) Since both girls return to their starting point, their displacements are zero. (1A) Physics in articles (p. 40) (a) From 19 January 2006 to 28 February 2007, (1A) It takes New Horizons spacecraft a total of 406 days to travel from the Earth to Jupiter. (1A) (b) (i) Average speed total distance travelled = total time of travel (1M) = 8 ? 108 406 ? 24 (1A) (1M) = 8. 21 ? 104 km h? 1 (ii) Average acceleration change in velocity = total time 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 Motion II 10 (a) The object moves with a everlasting elocity. Practice 2. 1 (p. 61) D B D D B 30 ? 10 = 10 m s1 v= 2 (b) The object moves with a similar acceleration from delay. (c) The object moves with a homogeneous retardant, starting with a certain initial velocity. Its veloc ity becomes zero finally. The velocity of the car at t = 2 s is 10 m s1. 6 7 C (d) The object first moves with a uni influence acceleration from rest, then at a constant velocity, and finally moves with a small uniform acceleration again. (a) Total displacement = 4 ? 5 + (? 5) ? (7 ? 5) = 10 m The total displacement from the staircase to her classroom is 10 m. (e)The object moves at a constant velocity and then suddenly moves at constant velocity of same magnitude in the face-to-face direction. (b) Classroom C 8 (f) The object moves with uniform retardation from an initial velocity to rest, and continue to move with the uniform acceleration of the same magnitude in opposite direction. 9 (a) The object accelerates. (b) The object first moves with a constant velocity. thusly it becomes stationary and finally moves with a higher constant velocity again. 11 (a) The object moves with zero acceleration (with constant velocity of 50 m s1). (b) The object moves with a uniform ccelerati on of 5 m s2. (c) 12 The object moves with uniform deceleration of 5 m s2. (c) The object decelerates to rest, and then accelerates in opposite direction to return to its starting point. (a) It moves remote from the demodulator. (d) The object moves with uniform velocity towards the origin (the zero displacement position), passes the origin, and continues to move away from the origin with the same uniform velocity. (b) (c) The greatest rate of change in speed 0 ? 3. 5 = 2 = 1. 75 m s2 (d) Total distance travelled = expanse under the interpret 3. 5 ? 2 2 ? 6 = + 2 2 = 9. 5 m Practice 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 minimum length of the runway is 3. 5 km. 2 B Cyclist X is base at constant speed. Time for cyclist X to sphere finish line displacement 150 = = = 30 s time 5 For cyclist Y u = 5 m s1, s = 250 m, (b) Total distance travelled = playing arena under 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) Average speed total distance travelled = time taken 27 = 3 t = 13. 5 s or t = ? 18. 5 s (rejected) Y needs 13. 5 s to communicate finish line. Therefore, cyclist Y will win the race. 3B Since the biff start decelerates afterward shoot into the wall, we could just consider the displacement of the bullet in the wall. To prevent the bullet from penetrating the wall, the bullet must stop in the wall. = 9 m s1 14 (a) She moves towards the transaction sensing element. (b) The highest speed of the girl in the journey is 3. 5 m s1. By v2 = u2 + 2as, 0 = 500 + 2 ? (? 800 000) ? s 2 8 By v = u + at, 14 = u + 2 ? 5 u = 4 m s1 s = 0. 156 m = 15. 6 cm 15. 8 cm The minimum thickness 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 thief at t = 5 s, its total displacement is 30 m.The dog is sitting initially, so u = 0. 1 By s = ut + at2, 2 1 30 = 0 + a(5)2 2 The displacement of the girl is 45 m. 9 (a) v = u + at = 0 + 20 ? 0. 3 = 6 m s? 1 The horizontal speed of the ball travelling towards the goalkeeper is 6 m s? 1. a = 2. 4 m s2 Its acceleration is 2. 4 m s2. (b) By v2 = u2 + 2as, 02 ? 62 a= = 22. 5 m s? 2 2 ? 0. 8 The deceleration of the football 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 reaction time of the cyclist is 0. 5 s. s= v ? u = 2a 2 2 90 3. 6 36 3. 6 2 ? 1. 5 ? 2 2 = one hundred seventy-five m (b) Braking distance (2. ? 0. 5)? 15 = 11. 25 m = 2 Thinking distance = 15 ? 0. 5 = 7. 5 m Stopping distance = 11. 25 + 7. 5 = 18. 75 m child. 20 m The distance travelled by the motorcycle is 175 m and its acceleration is 1. 5 m s . 2 7 (a) Thinking distance = speed ? reaction time 108 = ? 0. 8 = 24 m 3. 6 Therefore, the bicycle would not give the (b) Since the car decelerates uniformly, braking distance 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 ball can r from each one the hole. 2 12 (a) (i) By v = u + at, 0 = u + (4)(4. 75) u = 19 m s1The initial velocity of the car is 19 m s1. (c) Stopping distance = idea distance + braking distance = 24 + 33 = 57 m (ii) By v2 = u2 + 2as, 0 = 19 + 2 ? (4) ? s s = 45. 1 m 2 3 C For option A, apply equation v2 = u2 2gs and take s = 0 (the ball returns to the second degree), v = u = 10 m s1 ( un slopedly descending(prenominal)ss) The displacement of the car in the lead it moolah in front of the traffic light is 45. 1 m. This is the same velocity as the initial velocity of option B. Therefore, in both slipway the ball has the same vertical speed when it reaches the ground. (b) By v = u + 2as, 17 = 0 + 2 ? 3 ? s s = 48. 2 m 2 2 2The displacement of the car in the midst of starting from rest and moving at 17 m s is 48. 2 m. 1 4 B Take the upward direction as positive. 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 speed is 10 m s . 1 (b) see to it the first section. By v = u + at, v? u t= a 10 ? 0 = 0. 1 = 100 s analyze the second section. 1 By s = ut + at2, 2 1 800 = 10t + ? 0. 5t2 2 t = 40 s or t = 80 s (rejected) The speed of the bullet is 150 m s1 when it is fired. 5 Speed of muffin Equation used t=1s t=2s t=3s t=4s v = u + at Distance travelled by the stone 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 Total time taken = 100 + 40 = one hundred forty s It takes 140 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 Practice 2. 3 (p. 83) 1 2 D D = 0 + 10(1. 41) = 14. 1 m s1 It takes 1. 41 s for a plumbers helper to drop from a 10-m platform. His speed is 14. 1 m s1 when he enters the irrigate. 7 Take the upward direction as positive. By v = u + 2as, 4 = 0 + (2)(10)s s = 0. 8 m 2 2 2 Besides, since Y spends a shorter time to reach its highest point, it should be fired after X. 1 0 (a) By s = ut + The highest position reached by the puppy is 0. m above the ground. 8 (a) Consider the boys downward journey. Take the downward direction as positive. 1 By s = ut + at2, 2 1 0. 5 = 0 + (10) t2 2 t = 0. 316 s 1 2 at , 2 1 120 = 8t + ? 10 ? t2 2 t = 4. 16 s or t = ? 5. 76 s (rejected) It takes 4. 16 s to reach the ground. (b) v = u + at = 8 + 10 ? 4. 16 = 49. 6 m s1 Its speed on collision the ground is 49. 6 m s1. 11 (a) Distance amidst the ceiling and her hands = 6 2 1. 2 = 2. 8 m Hang-time of the boy = 0. 316 ? 2 = 0. 632 s (b) Let s be her vertical displacement when she jumps. As the maximum jumping speed is 8 m s1, i. e. u = 8 m s1. By v2 = u2 + 2as, v2 ? 2 s= 2a 2 0 ? 82 = (upwards is positive) 2 ? (? 10) s = 3. 2 m 2. 8 m Therefore, the indoor playground is not off the hook(predicate) for playing trampoline. 1 (a) By s = ut + at2, 2 1 132 = 0 ? t + ? 10 ? t2 2 t = 5. 14 s The vehicle can experience a free fall in the Zero-G facility for 5. 14 s. (b) Take the upward direction as positive. By v = u + 2as, 0 = u + 2 ? (10) ? 0. 5 u = 3. 16 m s1 2 2 2 The jumping speed of the boy is 3. 16 m s1. 9 Take the upward direction as positive. (a) By v2 = u2 + 2as, 0 = u2 + 2(10)(200) u = 63. 2 m s1 The velocity of the firework X is 63. 2 m s1 when it is fired. 12 (b) By v = u + at, = 63. 2 + (10)t t = 6. 32 s It takes 6. 32 s for the firework X to reach that superlative. (c) From (a) and (b), for firework Y to explode at 130 m above the ground, the speed of Y should be smaller 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 before it comes to a stop is 51. 4 m s? 1. lower speed. (c) Take the upward direction as positive. By v = u + at, v = v gt 2v = gt If the stone is projected with a speed of 2v, let the new time of travel be t?. (2v) = (2v) gt? v t? = 4 ( ) g = 2t Its new time of travel is 2t. 6B Take the upward direction as positive. 1 s = ut + at2 2 1 = (10)(4) + (10)(4)2 2 = 40 m The distance mingled with the sandbag and the ground is 40 m when it leaves the balloon. Revision exercise 2 Multiple-choice (p. 87) 1 D By v2 = u2 + 2as, 0 = 102 + 2a(25 10 ? 0. 2) a = 2. 17 m s2 His minimum deceleration is 2. 17 m s2. 2 3 D B Consider the rock released from the 2nd floor. By v2 = u2 + 2as, v2 = 2as floor. Note that s2 = 3. 5s. (v2)2 = 2as2 = 3. 5(2as) = 3. 5v2 v2 = 1. 87v (as u = 0) Then consider the rock released from the 7th 7 8 D C Take the downward direction 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 rockets should be fired for at least 9. 75 s. Both C and D satisfy this requirement. plainly for D, after firing for 10. 2 s, v = u + at = 200 + (20)(10. 2) = 4 m s1 i. e. it flies away from the bootleg with 4 m s1 upwards. It cannot land 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 opposite directio n). 9 10 D D 11 12 13 (HKCEE 2006 Paper II Q1) (HKCEE 2007 Paper II Q2) (HKCEE 2007 Paper II Q33) (b) (i) Conventional (p. 89) 1 (a) The reaction time of the driver 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 straight line with slope = 0. 35 m s1 (A straight line with slope = 0. 35 m s1 (1A) (1M) = 4 m s2 The acceleration of the car is 4 m s2. (c) The lemniscus distance of the car is the area under graph. Stopping distance 12 ? (3. 6 ? 0. 6) =12 ? 0. 6 + 2 = 25. 2 m The stopping distance of the car is shorter than 27 m. The driver will not be supercharged with driving past a red light. (1A) (1A) (1M) (ii) 2 (a) The object moves away from the motion sensor with uniform velocity 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 object moves with negative acceleration. (1A) Then, from t = 1. 45 s to 1. 50 s, the object changes its moving direction a nd moves towards the motion sensor again with a uniform velocity of 0. 35 m s1. (1A) (Correct axes with labels) (1A) (Correct graph with the acceleration of ? 0. 35 ? 0. 35 about 1. 40 ? 1. 30 = 7 m s2 at t = 1. 30 s to 1. 40 s) (1A) 3 (a) (b) Total displacement of the car = area bound by the v? t graph and the time axis vertebra 1 1 = (5 ? 5) ? (20 ? 3) 2 2 = ? 17. 5 m (1M) (1A) (c) Yes, the car moves 12. 5 m in front from t = 0 to t = 5 s. Therefore, it hits the roadblock. 1A) 5 Take the upward direction as positive. (a) From point A to the highest point (Correct axes with labels) (Correct shape of minibus graph) (Correct shape of sports cars graph) (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 above the trampoline. (1A) The maximum height reached by him is (1M) (b) From the graph in ( a), the deuce vehicles have the same velocity at t ? 2. 3 s after passing the traffic light. (1A) (1M) (c)The area under graph is the displacement of the cars. Consider 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 will take the lead 3 s after passing the traffic light. (1A) (b) Height of point A above the trampoline (1A) = 3. 2 0. 8 = 2. 4 m (1M) (1A) 6 (a) initial velocity v = 90 km h1 90 = m s1 3. 6 = 25 m s1 Thinking distance =v? t = 25 ? 0. 2 =5m The thinking distance is 5 m. (1A) (1M) 4 (a) The car moves forward with uniform acceleration at ? 1 m s? 2 from t = 0 s to t = 5 s. (1A) (1A) Then the car changes its moving direction.From t = 5 s to t = 8 s, it moves backwards with a uniform acceleration 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 slope of the graph is the ma gnitude of the acceleration of the apple. speed / m s? 1 7. 75 (1A) (1A) Hence, the deceleration 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 time / s (Correct labelled axes) (2A) (1A) (Straight line with a slope of 10 m s? 2) = 37. 5 m Braking distance = 37. 5 m Stopping distance = 37. 5 + 5 = 42. m (1M) (d) The two graphs have no difference. (1A) (1A) 8 (a) Take the downward direction as positive. By v2 = u2 + 2gs, v = u + 2 gs 2 The driver could not stop before the traffic light. Therefore, his claim is incorrect. (1A) (1M) 7 (a) Take the downward direction 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 cushion 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 landing 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 time 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 time of contact is 0. 221 s. (c) (b) Slope 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 ball due to gravity is 8. 21 m s2. (1M) (1A) (c) (Correct labeled axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) t = 2 s Displacement of the cable tramway = 0. 7 ? 0. 15 = 0. 55 m t = 3. 4 s (1A) Displacement of the trolley = 1. 175 ? 0. 15 = 1. 025 m t = 4. 9 s 1A) Displacement of the trolley = 0. 6 ? 0. 15 = 0. 45 m (1A) (b) It moves away from the motion 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 regularity does not work Then it rests momentarily at t = 3. 4 s. After that, it moves towards the motion since ultrasound will be reflected by the transpare nt plastic 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 ball is held 0. 15 m from sensor before being released. The ball hits the ground which is 1. m from the sensor. (1A) (1A) Therefore, the ball drops a height of 0. 95 m. which are 0. 45 m, 0. 65 m and 0. 775 m from the sensor in its first 3 rebounds. (1A) The acceleration of the trolley is ? 0. 507 m s? 2. (ii) The ball rebounds to the positions 10 (a) The motion sensor is protruded outside the table to avoid the reflection of ultrasonic signal from table. (1A) At the 1st 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 f irst 3 rebounds. (3A) 3. 9 (1M) 0. 95 ? 0. 55 (1A) 1 1 t3 t1 t2 t4 t5 t/s (ii) Acceleration = slope of graph = = 9. 75 m s2 ?6. 32 (3 straight lines) (Correct slopes) (1A) (1A) 12 Take the downward direction 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 time and velocity)(1A) 13 (a) Speed v = 70 km h1 70 = m s1 3. 6 = 19. 4 m s1 d Reaction time = v 6 = 19. 4 = 0. 309 s The reaction time of the man was 0. 09 s. (1M) (b) At 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 deceleration of the car was (c) (1A) Speed v = 80 km h1 80 = m s1 3. 6 = 22. 2 m s1 = 66. 6 m s2 Thinking distance = vt = 22. 2 ? 0. 309 = 6. 86 m By v = u + 2as, braking distance s v2 ? u2 = 2a 2 0 ? 22. 2 2 = 2 ? ? 3. 92) 2 2 (1A) Take the upward direction as positive. 1 s = ut + at2 (1M) 2 1 = 7 ? 1. 75 + ? (10) ? 1. 752 2 = 3. 06 m (negative means the water is below the origin board) The spring board is 3. 06 m above the water. Alternative method (1A) = 62. 9 m Therefore, the stopping distance = 6. 86 + 62. 9 = 69. 8 m (1A) Consider the upward motion and downward motion separately. For the upward motion, she takes 0. 7 s to reach the highest point from the spring board. Take the upward direction 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. Take the downward direction as positive. By s = ut + 1 2 gt , 2 1 s2 = 0 + ? 10 ? 1. 052 = 5. 51 m 2 (1A) This stopping distance is greater than the initial distance in the midst of the car and the boy. (1A) Ther efore, the car would have knocked down the boy if the car had travelled at 80 km h? 1 or faster. (d) A drunk has a longer reaction time. (1A) This means that the thinking distance, and thus the stopping distance (sum of thinking distance and braking distance), increases. (1A) (1M) (1A) 14 (a) Take 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 height of the spring board above the water = s2 s1 = 5. 51 2. 45 = 3. 06 m (1A) (1M) (1A) The speed of Belinda leaving the spring (b) Total time taken from the spring 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 birdbrain entering the water (d) Deceleration of car Y = slope of the graph during 0. 5 s? 8. 5 s = 0 ? 19. 4 = 2. 43 m s2 8. 5 ? 0. 5 (1A) The deceleration of car Y is 2. 43 m s2. (c) Thinking distance = area under 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 distance = area under the graph during 0. 5 s? 8. 5 s 1 = ? 19. 4 ? (8. 5 0. 5) 2 = 77. 6 m distance are 9. 7 m and 77. 6 m respectively. (1A) The thinking distance and the braking (e) (See the figure in (d). ) (Correct slope analogue to that in (d). ) (1A) (Correct position above that in (d). ) (1A) 15 (a) Speed 70 km h1 70 = m s1 3. 6 = 19. 4 m s 1 (d) The coloured area is equal to the difference in the stopping distances travelled by cars X and Y. (1A) (e) (1M) Stopping distance of car X = area under the graph during 0? 5 s 1 = ? 19. 4 ? 5 = 48. 5 m 2 Coloured area = 9. 7 + 77. 6 48. = 38. 8 m 50 m Since the difference in stopping distances of the cars is smaller than the initial separation of the cars, the two cars do not collide with each other before they stop. (1A) (1M) (1M) Distance travelled by car Y in 2 s = vt = 19. 4 ? 2 = 38. 8 m 50 m Since the distance betwixt the cars is greater than the distance that car Y can travel in 2 s, the driver of car Y obeys the rule. corresponding vt graph. Deceleration of car X = slope of the graph during 0? 5 s (1A) (1M) (b) Deceleration 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 acceleration of 17 ? 0 = 3. 4 m s2. 5 (1A) From t = 5 s to t = 20 s, the car moves with a constant 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 Shuttle travels 2 016 000 m (2016 km) in the first 8 minutes. From t = 28 s to t = 30 s, the car remains (1A) 19 (a) (i) The cyclist is using first vend when the acceleration is greatest before braking. shortest time. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second gear for the (b) Distance travelled = area under s traight 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 Paper I Q1) 1 (a) s = ut + at2 2 1 = 0 + ? 10 ? (500 ? 10? 3)2 2 = 1. 25 m Therefore the minimum height the (1M) The car changes direction at t = 30 s. Its velocity changes from positive to negative, showing a change in its travelling direction. 1A) (1M) (1A) (1A) laptop must fall for it to be saved 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 Paper I Q8) (a) v = u + at = 0 + 17. 5 ? 8 ? 60 = 8400 m s1 minutes is 8400 m s1. The speed of the computer when it hits The speed of the Shuttle after the first 8 (c) Most falls are likely to be from below this height, effect. (1A) (1A) (1A) so the protection will not have taken Physics 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)Take the upward direction as positive. 22 (a) Any one from Rate of change of displacement Displacement 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) increasing with time, so its velocity is (still) positive In this case, the acceleration and velocity are in opposite directions. (1A) (1A) (1A) The vertical speed of Javier Sotomayor is 5. 35 m s? 1 when he leaves the ground. (ii) Take the upward direction as positive. Consider the upward journey. By v = u + at, v ? u 0 ? 5. 35 t= = = 0. 54 s a ? 10 (1M) (c) i) Consider 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 cincture in the air = (0. 54 + 0. 60) = 1. 14 s Alternative method (1A) (Correct gra ph) (1A) Take 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) tumid 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 between t = 4. 0 s and t = 10. s is 600 m. 3 1 2 3 4 C C Force and Motion 6 (a) The MTR train is accelerating in the forward direction. The man tends to move at his accepted speed (smaller speed), so he would move backwards sexual relation to the MTR train. (b) The MTR train is slowing down. The man tends to move at his original speed (greater speed), so he would move forwards relative to the MTR train. (c) The MTR train is moving forwards at constant velocity. The man moves forwards with the same constant velocity, so he would remain at rest relative to the MTR train. (d) The MTR train is turning a corner. The Practi ce 3. 1 (p. 104) (b), (e), (f) 5 a) Stretching a rubber band (b) Standing on the floor (c) Walking time (e) (f) A compass A rubbed plastic formula attracts small bits of paper (d) Exists in every object on the land at any 7 man tends to move at his original direction, so he would move outwards relative to the MTR train. In space, the gravitative ride acts on the spaceship is negligible. When the rockets are shut down, they do not exert a drive on the spaceship. Therefore, no net impression acts on the spaceship. By Newtons first law, the spaceship is in uniform motion and can travel far out in space. 8 Joan moves on the ice surface with a constant velocity.Practice 3. 2 (p. 111) 1 2 3 4 5 C C D C (a) No. Athletes would hit the wall of the field if it is too close to the finishing line. (b) The mat is used to protect the athletes if they hit the wall after passing the finishing line. Practice 3. 3 (p. 122) 1 2 3 4 5 D A B A D 6 (a) 7 (a) plane luck = 40 + 30 romaine lettuc e 30 = 66. 0 N Vertical share = 30 sin 30 = 15 N issue = 66 2 + 15 2 = 67. 7 N Let ? be the angle between the endpoint concomitants magnitude is 67 N and the angle between the sequent and the horizontal is 13. (b) and the horizontal. 15 tan = ? = 12. 8 66 sequents magnitude is 67. N and the angle between the conclusion and the horizontal is 12. 8. (b) Horizontal component = 40 + 30 romaine lettuce 45 = 61. 2 N Vertical component = 30 sin 45 = 21. 2 N Resultants magnitude is 65 N and the angle between the resultant and the horizontal is 19. (c) Resultant = 61. 2 2 + 21. 2 2 = 64. 8 N Let ? be the angle between the resultant and the horizontal. 21. 2 tan = ? = 19. 1 61. 2 Resultants magnitude is 64. 8 N and the angle between the resultant and the horizontal is 19. 1. (c) Resultants magnitude is 60 N and the angle between the resultant and the horizontal is 25. (d)Horizontal component = 40 + 30 romaine lettuce 60 = 55 N Vertical component = 30 sin 60 = 26. 0 N Resultant = 55 2 + 26. 0 2 = 60. 8 N Let ? be the angle between the resultant and the horizontal. 26. 0 ? = 25. 3 tan = 55 Resultants magnitude is 60. 8 N and the angle between the resultant and the Resultants magnitude is 50 N and the angle between the resultant and the horizontal is 37. horizontal is 25. 3. (d) Resultant = 40 2 + 30 2 = 50 N Let ? be the angle between the resultant and the horizontal. 30 tan = ? = 36. 9 40 Resultants magnitude is 50 N and the angle between the resultant and the horizontal is 36. 9.Hence, the angle between the two 5-N forces is 120. Alternative method By tip-to-tail method, the two 5-N forces and the resultant 5-N force form an equilateral triangle. It is known that each angle of an equilateral triangle is 60. Therefore, the angle between the two 5-N forces is 120. 8 (a) 10 (b) Resultant force = 2 ? 400 = 800 N The resultant force provided by the cable is 800 N. 11 For the 2-kg mass (c) 9 R = weight ? cosine ? = 20 cos 30 = 17. 3 N Suppose the two forces act in the direction as shown. T = 20 N Therefore we have Vertical component Fx = 5 sin ? Horizontal component Fy = 5 ? 5 cos ? = 5 ? 1 ? cos ? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin ? )2 + 5 ? (1 ? cos ? )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 tension of the string is 1440 N. 3 4 5 6 B C A winnings force = ma = 40 ? 0. 5 = 20 N C By v2 u2 = 2as, 0 u2 = 2a(20) ? u2 = 40a u2 a=? 40 Resistance = ma = 12 ? ? u2 = 0. 03u2 40 (b) Component 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 bag of sugar weighs 10 N. or A bag of sugar has a mass of 1 kg. By F = ma, F 800 000 a= = = 2 m s2 m 4 ? 10 5 (b) As the mass is stationary, the net force acting on it is zero. When it flies 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 = weight of mass = 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 engine is 6945 N. 10 (a) (ii) y-component of F2 = 0 x-component of F2 = x-component of F1 = 17. 3 N