SOLUTIONS TO CONCEPTS CHAPTER 3
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1 SOLUTIONS TO ONEPTS HPTER 3. a) Disance ravelled = = 0 m b) F = F = D = 50 0 = 30 M His displacemen is D D = F DF m In ED an = DE/E = 30/40 = 3/4 = an (3/4) His displacemen from his house o he field is 50 m, an (3/4) norh o eas.. O Saring poin origin. i) Disance ravelled = = 60 m ii) Displacemen is only O = 0 m in he negaive direcion. Displacemen Disance beween final and iniial posiion. 3. a) V ave of plane (Disance/Time) = 60/0.5 = 50 km/hr. b) V ave of bus = 30/8 = 40 km/hr. c) plane goes in sraigh pah velociy = V ave = 60/0.5 = 50 km/hr. d) Sraigh pah disance beween plane o Ranchi is equal o he displacemen of bus. Velociy = V ave = 60/8 = 3.5 km/hr. 4. a) Toal disance covered = 64 km in hours. Speed = 64/ = 3 km/h b) s he reurns o his house, he displacemen is zero. Velociy = (displacemen/ime) = 0 (zero). 5. Iniial velociy u = 0 ( sars from res) Final velociy v = 8 km/hr = 5 sec (i.e. ma velociy) Time inerval = sec. v u 5 cceleraion = a ave = =.5 m/s. 6. In he inerval 8 sec he velociy changes from 0 o 0 m/s. verage acceleraion = 0/8 =.5 m/s change in velociy ime Disance ravelled S = u + / a 0 + /(.5)8 = 80 m. 7. In s 0 sec S = u + / a 0 + (/ 5 0 ) = 50 f. 0 sec v = u + a = = 50 f/sec. From 0 o 0 sec ( = 0 0 = 0 sec) i moves wih uniform velociy 50 f/sec, 3. W N S E m 40 m 40 m 0 m D Iniial poin (saring poin) ( 0 m, 0) 4 8 Time in sec S (in f) 50 0 O Y 30 m E X (0 m, 0) Iniial velociy u = (sec)
2 Disance S = 50 0 = 500 f eween 0 sec o 30 sec acceleraion is consan i.e. 5 f/s. 0 sec velociy is 50 f/sec. = 30 0 = 0 s S 3 = u + / a = (/)( 5)(0) = 50 m Toal disance ravelled is 30 sec = S + S + S 3 = = 000 f. 8. a) Iniial velociy u = m/s. final velociy v = 8 m/s ime = 0 sec, 8 v u 8 acceleraion = = 0.6 m/s 6 4 a 0 b) v u = as v u Disance S = = = 50 m. a 0.6 c) Displacemen is same as disance ravelled. Displacemen = 50 m. 9. a) Displacemen in 0 o 0 sec is 000 m. ime = 0 sec. V ave = s/ = 00/0 = 0 m/s b) sec i is moving wih uniform velociy 50/.5 = 0 m/s. a sec. V ins = 0 m/s sec i is a res. (slope of he graph a = sec) V ins = zero. 8 sec i is moving wih uniform velociy 0 m/s V ins = 0 m/s sec velociy is negaive as i move owards iniial posiion. V ins = 0 m/s. 0. Disance in firs 40 sec is, O + D = = 00 m. verage velociy is 0 as he displacemen is zero. 5 m/s O 0 D 40 (sec). onsider he poin, a = sec = 0 ; s = 0 m and = sec s = 0 m So for ime inerval 0 o sec hange in displacemen is zero. So, average velociy = displacemen/ ime = 0 The ime is sec.. posiion insananeous velociy has direcion along. For average velociy beween and. V ave = displacemen / ime = ( / ) = ime y
3 We can see ha is along i.e. hey are in same direcion. The poin is (5m, 3m). 3. u = 4 m/s, a =. m/s, = 5 sec Disance = s = u a = 4(5) + / (.)5 = 35 m. 4. Iniial velociy u = 43. km/hr = m/s u = m/s, v = 0 a = 6 m/s (deceleraion) Disance S = v u = m ( 6) 3.3
4 5. Iniial velociy u = 0 cceleraion a = m/s. Le final velociy be v (before applying breaks) = 30 sec v = u + a = 60 m/s a) S = u a = 900 m when breaks are applied u = 60 m/s v = 0, = 60 sec ( min) Declaraion a = (v u)/ = = (0 60)/60 = m/s. v u S = = 800 m a Toal S = S + S = = 700 m =.7 km. b) The maimum speed aained by rain v = 60 m/s c) Half he maimum speed = 60/= 30 m/s 3.4 v u 30 0 Disance S = = = 5 m from saring poin a When i acceleraes he disance ravelled is 900 m. Then again declaraes and aain 30 m/s. u = 60 m/s, v = 30 m/s, a = m/s v u Disance = a = ( ) = 350 m Posiion is = 50 =.5 km from saring poin. 6. u = 6 m/s (iniial), v = 0, s = 0.4 m. v u Deceleraion a = = 30 m/s. s v u 0 6 Time = = = 0.05 sec. a u = 350 m/s, s = 5 cm = 0.05 m, v = 0 v u 0 (350) Deceleraion = a = = s 0.05 Deceleraion is. 0 5 m/s. 8. u = 0, v = 8 km/hr = 5 m/s, = 5 sec a = v u 5 0 = m/s. 5 =. 0 5 m/s. s = u a =.5 m a) verage velociy V ave = (.5)/5 =.5 m/s. b) Disance ravelled is.5 m. 9. In reacion ime he body moves wih he speed 54 km/hr = 5 m/sec (consan speed) Disance ravelled in his ime is S = 5 0. = 3 m. When brakes are applied, u = 5 m/s, v = 0, a = 6 m/s (deceleraion)
5 S = v u a 0 5 = 8.75 m ( 6) Toal disance s = s + s = =.75 = m. 3.5
6 0. (deceleraion on hard braking = 6 m/s ) (deceleraion on hard braking = 7.5 m/s ) a = 0 5 ( 6) = 9 m Driver X Reacion ime 0.5 Speed = 54 km/h raking disance a= 9 m Toal sopping disance b = m Speed = 54 km/h raking disance e = 5 m Toal sopping disance f = 8 m 3.6 Driver Y Reacion ime 0.35 Speed = 7 km/h raking disance c = 33 m Toal sopping disance d = 39 m. Speed = 7 km/h raking disance g = 7 m Toal sopping disance h = 33 m. So, b = = 33 m Similarly oher can be calculaed. raking disance : Disance ravelled when brakes are applied. Toal sopping disance = raking disance + disance ravelled in reacion ime.. V P = 90 km/h = 5 m/s. V = 7 km/h = 0 m/s. Police =0 In 0 sec culpri reaches a poin from. Disance convered by culpri S = v = 0 0 = 00 m. culpri ime = 0 sec he police jeep is 00 m behind he culpri. Time = s/v = 00 / 5 = 40 s. (Relaive velociy is considered). In 40 s he police jeep will move from o a disance S, where S = v = 5 40 = 000 m =.0 km away. The jeep will cach up wih he bike, km far from he urning.. v = 60 km/hr = 6.6 m/s. v = 4 km/h =.6 m/s. Relaive velociy beween he cars = (6.6.6) = 5 m/s. Disance o be ravelled by firs car is 5 + = 0 m. Time = = s/v = 0/5 = sec o cross he nd V car. In sec he s 5 m car moved = 6.6 = 33. m H also covered is own lengh 5 m. Toal road disance used for he overake = = 38 m. 3. u = 50 m/s, g = 0 m/s when moving upward, v = 0 (a highes poin). a) S = v u a 0 50 = 5 m ( 0) maimum heigh reached = 5 m b) = (v u)/a = (0 50)/ 0 = 5 sec c) s = 5/ = 6.5 m, u = 50 m/s, a = 0 m/s, V 5 m efore crossing V = 0 sec V 0 m fer crossing P
7 v u = as v = (u as) 50 ( 0)(6.5) = 35 m/s. 4. Iniially he ball is going upward u = 7 m/s, s = 60 m, a = g = 0 m/s s = u a 60 = 7 + / = 0 = ( 60) = aking posiive sign = = 4. sec ( ve) 0 Therefore, he ball will ake 4. sec o reach he ground. 5. u = 8 m/s, v = 0, a = g = 9.8 m/s a) S = v u a 0 8 (9.8) = 40 m v u 0 8 b) ime = =.85 a 9.8 =.85 = v = u + a = 8 (9.8) (.85) = 9.87 m/s. The velociy is 9.87 m/s. c) No i will no change. s afer one second velociy becomes zero for any iniial velociy and deceleraion is g = 9.8 m/s remains same. Fro iniial velociy more han 8 m/s ma heigh increases. 6. For every ball, u = 0, a = g = 9.8 m/s 4 h ball move for sec, 5 h ball sec and 3 rd ball 3 sec when 6 h ball is being dropped. For 3 rd ball = 3 sec S 3 = u a = 0 + / (9.8)3 = 4.9 m below he op. For 4 h ball, = sec S = 0 + / g = / (9.8) = 9.6 m below he op (u = 0) For 5 h ball, = sec S 3 = u + / a = 0 + / (9.8) = 4.98 m below he op. 7. poin (i.e. over.8 m from ground) he kid should be cached. For kid iniial velociy u = 0 cceleraion = 9.8 m/s Disance S =.8.8 = 0 m S = u a 0 = 0 + / (9.8) =.04 =.4. In his ime he man has o reach a he boom of he building. Velociy s/ = 7/.4 = 4.9 m/s. 8. Le he rue of fall be iniial velociy u = 0.8 0m 7m.8m 6 h 5 h 4 h 3 rd
8 cceleraion a = 9.8 m/s Disance S = / m S = u a. = 0 + / (9.8). = =.46 =.57 sec 4.9 For cade velociy = 6 km/hr =.66 m/sec Disance = v = =.6 m. The cade,.6 m away from ree will receive he berry on his uniform. 9. For las 6 m disance ravelled s = 6 m, u =? = 0. sec, a = g = 9.8 m/s S = u a 6 = u(0.) u = 5.8/0. = 9 m/s. For disance, u = 0, v = 9 m/s, a = g = 9.8 m/s v u 9 0 S = = 4.05 m a 9.8 Toal disance = = = 48 m. 30. onsider he moion of ball form o. jus above he sand (jus o penerae) u = 0, a = 9.8 m/s, s = 5 m S = u a 5 = 0 + / (9.8) = 5/4.9 =.0 =.0. velociy a, v = u + a = (u = 0) =9.89 m/s. From moion of ball in sand u = 9.89 m/s, v = 0, a =?, s = 0 cm = 0. m. v u 0 (9.89) a = = 490 m/s s 0. The reardaion in sand is 490 m/s. 3. For elevaor and coin u = 0 s he elevaor descends downward wih acceleraion a (say) The coin has o move more disance han.8 m o srike he floor. Time aken = sec. S c = u a = 0 + / g() = / g S e = u a = u + / a() = / a Toal disance covered by coin is given by =.8 + / a = / g.8 +a/ = 9.8/ = 4.9 a = 6. m/s = = 0.34 f/s. 3. I is a case of projecile fired horizonally from a heigh. 3.8 a m 6m.66 m/s.6m =0. sec 6 m 0cm 6f=.8m /a
9 h = 00 m, g = 9.8 m/s a) Time aken o reach he ground = (h / g) 00 = = 4.5 sec. 9.8 b) Horizonal range = u = = 90 m. c) Horizonal velociy remains consan hrough ou he moion., V = 0 m/s V y = u + a = = 44. m/s. Resulan velociy V r = ( 44.) 0 = 48.4 m/s. Tan = V y 44. =.05 V 0 = an (.05) = 60. The ball srikes he ground wih a velociy 48.4 m/s a an angle 66 wih horizonal. 33. u = 40 m/s, a = g= 9.8 m/s, = 60 ngle of projecion. a) Maimum heigh h = u sin g 40 (sin 60) = 60 m 0 00m b) Horizonal range X = (u sin ) / g = (40 sin (60 )) / 0 = 80 3 m. 0m/s V y V V r 3.9
10 34. g = 9.8 m/s, 3. f/s ; 40 yd = 0 f horizonal range = 0 f, u = 64 f/s, = 45 We know ha horizonal range X = u cos = 0 =.65 sec. ucos 64 cos 45 y = u sin () / g = 64 (3.)(.65) (.65) 3.0 = 7.08 f which is less han he heigh of goal pos. In ime.65, he ball ravels horizonal disance 0 f (40 yd) and verical heigh 7.08 f which is less han 0 f. The ball will reach he goal pos. 35. The goli move like a projecile. Here h = 0.96 m Horizonal disance X = m cceleraion g = 9.8 m/s. u Time o reach he ground i.e. = h 0.96 = 0. sec g 9.8 Horizonal velociy wih which i is projeced be u. = u u = 0. = 0 m/s. 36. Horizonal range X = = 6.7 f covered by e bike. g = 9.8 m/s = 3. f/s. g sec y = an u To find, minimum speed for jus crossing, he dich y = 0 ( is on he ais) g sec an = g sec g g u u an sincos sin u = (3.)(6.7) / (because sin 30 = /) u = 3.79 f/s = 3 f/s. 37. an = 7/8 = an (7/8) The moion of projecile (i.e. he packed) is from. Taken reference ais a. = 37 as u is below -ais. u = 5 f/s, g = 3. f/s, y = 7 f gsec y = an u 7 = (0.7536) g(.568) (5) = 0 = f (can be calculaed) 0.96m 7f 5f y 0 f.7f m 5 5 u 8f 5f 0 f
11 Horizonal range covered by he packe is f. So, he packe will fall = 9 f shor of his friend. 3.
12 38. Here u = 5 m/s, = 60, g = 9.8 m/s Horizonal range X = u sin (5) sin( 60) = 9.88 m g In firs case he wall is 5 m away from projecion poin, so i is in he horizonal range of projecile. So he ball will hi he wall. In second case ( m away) wall is no wihin he horizonal range. So he ball would no hi he wall. usin 39. Toal of fligh T = g change in displacemen H/ H H/ verage velociy = ime From he figure, i can be said is horizonal. So here is no effec of verical componen of he velociy during his displacemen. So because he body moves a a consan speed of u cos in horizonal direcion. The average velociy during his displacemen will be u cos in he horizonal direcion. 40. During he moion of bomb is horizonal velociy u remains consan and is same as ha of aeroplane a every poin of is pah. Suppose he bomb eplode i.e. reach he ground in ime. Disance ravelled in horizonal direcion by bomb = u = he disance ravelled by aeroplane. So bomb eplode verically below he aeroplane. Suppose he aeroplane move making angle wih horizonal. For boh bomb and aeroplane, horizonal disance is u cos. is ime for bomb o reach he ground. So in his case also, he bomb will eplode verically below aeroplane. 4. Le he velociy of car be u when he ball is hrown. Iniial velociy of car is = Horizonal velociy of ball. Disance ravelled by ball S b = u (in horizonal direcion) nd by car S c = u + / a 9.8 m/s where ime of fligh of ball in air. ar has ravelled era disance S c S b = / a. all can be considered as a projecile having = 90. u sin 9.8 = = sec. g 9.8 S c S b = / a = m The ball will drop m behind he boy. 4. minimum velociy i will move jus ouching poin E reaching he ground. is origin of reference coordinae. If u is he minimum speed. X = 40, Y = 0, = 0 sec Y = an g u cm/s ) = an u (because g = 0 m/s = 000 u m/s 0 cm 30 cm 0 cm E 0 cm 0 cm
13 u = 00 cm/s = m/s. The minimum horizonal velociy is m/s. 43. a) s seen from he ruck he ball moves verically upward comes back. Time aken = ime aken by ruck o cover 58.8 m. s 58.8 ime = = 4 sec. (V = 4.7 m/s of ruck) v 4.7 u =?, v = 0, g = 9.8 m/s (going upward), = 4/ = sec. v = u + a 0 = u 9.8 u = 9.6 m/s. (verical upward velociy). b) From road i seems o be projecile moion. Toal ime of fligh = 4 sec In his ime horizonal range covered 58.8 m = X = u cos u cos = 4.7 () Taking verical componen of velociy ino consideraion. y = 0 (9.6) ( 9.8) = 9.6 m [from (a)] y = u sin / g 9.6 = u sin () / (9.8) u sin = 9.6 u sin = 9.6 (ii) u sin ucos 9. 6 = an = = an (.333) = 53 gain u cos = u = = 4.4 m/s. ucos 53 The speed of ball is 4.4 m/s a an angle 53 wih horizonal as seen from he road. 44. = 53, so cos 53 = 3/5 Sec 35 m/s = 5/9 and an = 4/3 53 Suppose he ball lands on nh bench So, y = (n ) () [ball saring poin m above ground] g sec gain y = an u 0(0 y) (5 / 9) y = (0 + y)(4/3) (0 y) y From he equaion, y can be calculaed. y = 5 [ = 0 + n = 0 + y] n = 5 n = 6. The ball will drop in sih bench. 45. When he apple jus ouches he end of he boa. = 5 m, u = 0 m/s, g = 0 m/s, =? y
14 u sin = g 0 sin 5 = 5 = 0 sin 0 sin = / sin 30 or sin 50 = 5 or 75 Similarly for end, = 6 m Then = sin (g/u ) = sin (0.6) = 8 or 7. So, for a successful sho, may very from 5 o 8 or 7 o a) Here he boa moves wih he resulan velociy R. u he verical componen 0 m/s akes him o he opposie shore. Tan = /0 = /5 Velociy = 0 m/s disance = 400 m Time = 400/0 = 40 sec. b) The boa will reach a poin. In, an = 400 = 400/5 = 80 m. 47. a) The verical componen 3 sin akes him o opposie side. Disance = 0.5 km, velociy = 3 sin km/h Time = Disance 0.5 hr Velociy 3 sin 5 = 0/sin min. b) Here verical componen of velociy i.e. 3 km/hr akes him o opposie side. Disance 0.5 Time = 0. 6 hr Velociy hr = = 9.6 = 0 minue. 48. Velociy of man V m = 3 km/hr D horizonal disance for resulan velociy R. X-componen of resulan R = cos = 0.5 / 3sin which is same for horizonal componen of velociy. H = D = (5 + 3 cos ) (0.5 / 3 sin ) = For H o be min (dh/d) = 0 d 5 3cos 0 d 6 sin 8 (sin + cos ) 30 cos = 0 5 3cos 6 sin 30 cos = 8 cos = 8 / 30 = 3/ km/h 400m 3sin 500m 0 m/s 5 m m 0m/s 5km/h 3km/h 5km/h / m / m m/s W 3km/h N S 5km/h 5km/h 3sin 3km/h R E D
15 Sin = H = cos = 4/5 5 3cos 5 3( 3 / 5) = km. 6 sin 6 (4 / 5) In resulan direcion R he plane reach he poin. Velociy of wind V w = 0 m/s Velociy of aeroplane V a = 50 m/s In D according o sine formula 0 sin 50 sin30 sin 0 50 sin = sin (/5) a) The direcion is sin (/5) eas of he line. b) sin (/5) = = R = 50 0 (50)0 cos 3348 = 67 m/s. s Time = = 994 sec = 49 = 50 min. v Velociy of sound v, Velociy of air u, Disance beween and be. In he firs case, resulan velociy of sound = v + u (v + u) = v + u = / () In he second case, resulan velociy of sound = v u (v u) = v u = / () From () and () v = v = From (i) u = v Velociy of air V = nd velociy of wind u = = 5. Velociy of sound v, velociy of air u Velociy of sound be in direcion so i can reach wih resulan velociy D. ngle beween v and u is > /. Resulan D (v u ) Here ime aken by ligh o reach is negleced. So ime lag beween seeing and hearing = ime o here he drum sound. W N 30 0m/s 30 D 50m/s S vy v R R 50 v D V w u 0 0m / s v
16 = =. Displacemen velociy (v u)(v u) v u ( / )( / ) [from quesion no. 50] 5. The paricles mee a he cenroid O of he riangle. any insan he paricles will form an equilaeral wih he same cenroid. onsider he moion of paricle. any insan is velociy makes angle 30. This componen is he rae of decrease of he disance O. Iniially O = 3 a a Therefore, he ime aken for O o become zero. = a / 3 v cos30 a 3v 3 a 3v a 3. O * * * * 3.6
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