2.1 Position. 2.2 Rest and Motion.

Transcription

1 Moion In ne Dimenion. Poiion. ny objec i ied poin nd hree oberer from hree differen plce re looking for me objec, hen ll hree oberer will he differen oberion bo he poiion of poin nd no one will be wrong. ece hey re obering he objec from heir differen poiion. berer y : Poin i m wy in we direcion. berer y : Poin i m wy in oh direcion. berer C y : Poin i 5 m wy in e direcion. Therefore poiion of ny poin i compleely expreed by wo fcor: I dince from he oberer nd i direcion wih repec o oberer. Th i why poiion i chrceried by ecor known poiion ecor. Le poin P i in xy plne nd i coordine re (x, y). Then poiion ecor (r ) of poin will be xˆi yˆj nd if he poin P i in pce nd i coordine re (x, y, z) hen poiion ecor cn be expreed r xi ˆ yj ˆ zkˆ.. Re nd Moion. If body doe no chnge i poiion ime pe wih repec o frme of reference, i i id o be re. nd if body chnge i poiion ime pe wih repec o frme of reference, i i id o be in moion. Frme of Reference : I i yem o which e of coordine re ched nd wih reference o which oberer decribe ny een. penger nding on plform obere h ree on plform i re. when he me penger i ping wy in rin hrogh ion, obere h ree i in moion. In boh condiion oberer i righ. oberion re differen bece in fir iion oberer nd on plform, which i reference frme re nd in econd iion oberer moing in rin, which i reference frme in moion. So re nd moion re relie erm. I depend pon he frme of reference. W S C N E 5m m m Tree i re Tree i in moion Plform (Frme of reference) Moing rin (Frme of reference)

2 Moion In ne Dimenion. Type of Moion. ne dimenionl Two dimenionl Three dimenionl Moion of body in righ line i clled one dimenionl moion. When only one coordine of he poiion of body chnge wih ime hen i i id o be moing one dimenionlly. e.g.. Moion of cr on righ rod. Moion of freely flling body. Moion of body in plne i clled wo dimenionl moion. When wo coordine of he poiion of body chnge wih ime hen i i id o be moing wo dimenionlly. e.g. Moion of cr on circlr rn. Moion of billird bll. Moion of body in pce i clled hree dimenionl moion. When ll hree coordine of he poiion of body chnge wih ime hen i i id o be moing hree dimenionlly. e.g.. Moion of flying kie. Moion of flying inec.. Pricle or Poin M. The mlle pr of mer wih zero dimenion which cn be decribed by i m nd poiion i defined pricle. If he ize of body i negligible in comprion o i rnge of moion hen h body i known pricle. body (Grop of pricle) o be known pricle depend pon ype of moion. For exmple in plnery moion rond he n he differen plne cn be premed o be he pricle. In boe coniderion when we re body pricle, ll pr of he body ndergo me diplcemen nd he me elociy nd ccelerion..5 Dince nd Diplcemen. () Dince : I i he cl ph lengh coered by moing pricle in gien inerl of ime. (i) If pricle r from nd rech o C hrogh poin hown in he figre. Then dince relled by pricle C 7 m (ii) Dince i clr qniy. (iii) Dimenion : [M 0 L T 0 ] (i) Uni : mere (S.I.) () Diplcemen : Diplcemen i he chnge in poiion ecor i.e., ecor joining iniil o finl poiion. (i) Diplcemen i ecor qniy (ii) Dimenion : [M 0 L T 0 ] (iii) Uni : mere (S.I.) m C m (i) In he boe figre he diplcemen of he pricle C C () If S C m of he indiidl. ( ) ( C) ( )( C)co 90 o = 5m, S, S... S n re he diplcemen of body hen he ol (ne) diplcemen i he ecor S S S S... S () Comprion beween dince nd diplcemen : n (i) The mgnide of diplcemen i eql o minimm poible dince beween wo poiion.

3 So dince Diplcemen. (ii) For moing pricle dince cn neer be negie or zero while diplcemen cn be. (zero diplcemen men h body fer moion h cme bck o iniil poiion) i.e., Dince > 0 b Diplcemen > = or < 0 Moion In ne Dimenion (iii) For moion beween wo poin diplcemen i ingle led while dince depend on cl ph nd o cn he mny le. (i) For moing pricle dince cn neer decree wih ime while diplcemen cn. Decree in diplcemen wih ime men body i moing owrd he iniil poiion. () In generl mgnide of diplcemen i no eql o dince. Howeer, i cn be o if he moion i long righ line wiho chnge in direcion. (i) If r Y nd r re he poiion ecor of pricle iniilly nd finlly. Then diplcemen of he pricle r r r nd i he dince relled if he pricle h gone hrogh he ph P. r r r X Problem. Smple problem bed on dince nd diplcemen mn goe 0m owrd Norh, hen 0m owrd e hen diplcemen i [KCET (Med.) 999; JIPMER 999; FMC 00] ().5m (b) 5m (c) 5.5m (d) 0m Solion : () If we ke e x xi nd norh y xi, hen diplcemen 0 ˆi 0ˆj Problem. So, mgnide of diplcemen =.5 m. body moe oer one forh of circlr rc in circle of rdi r. The mgnide of dince relled nd diplcemen will be repeciely r (), r (b) r, Solion : () Le pricle r from, i poiion ecor rˆi fer one qrer poiion ecor r r ĵ. r r r (c) r, (d) r, r Y So diplcemen rˆ j rˆi Mgnide of diplcemen r. X Problem. Solion : (b) nd dince = one forh of circmference r r The diplcemen of he poin of he wheel iniilly in conc wih he grond, when he wheel role forwrd hlf reolion will be (rdi of he wheel i R) () R (b) R (c) R (d) R Horizonl dince coered by he wheel in hlf reolion = R Pnew R Piniil R

4 Moion In ne Dimenion So he diplcemen of he poin which w iniilly in conc wih grond = ( R) (R) R..6 Speed nd Velociy. () Speed : Re of dince coered wih ime i clled peed. (i) I i clr qniy hing ymbol. (ii) Dimenion : [M 0 L T ] (iii) Uni : mere/econd (S.I.), cm/econd (C.G.S.) (i) Type of peed : () Uniform peed : When pricle coer eql dince in eql inerl of ime, (no mer how mll he inerl re) hen i i id o be moing wih niform peed. In gien illrion moorcycli rel eql dince (= 5m) in ech econd. So we cn y h pricle i moing wih niform peed of 5 m/. Dince Uniform Speed 5m 5m ec ec ec ec ec 5m/ 5m/ 5m 5m 5m 5m/ 5m/ 5m/ 5m m/ 5m/ (b) Non-niform (rible) peed : In non-niform peed pricle coer neql dince in eql inerl of ime. In he gien illrion moorcycli rel 5m in econd, 8m in nd econd, 0m in rd econd, m in h econd ec. Therefore i peed i differen for eery ime inerl of one econd. Thi men pricle i moing wih rible peed. Dince Vrible Speed 5m 8m ec ec ec ec ec 5m/ 8m/ 0m m 6m 0m/ m/ 6m/ 7m ec 7m/ (c) erge peed : The erge peed of pricle for gien Inerl of ime i defined he rio of dince relled o he ime ken. erge peed Dince relled ; ken erge peed : When pricle moe wih differen niform peed,,... ec in differen ime inerl,,,... ec repeciely, i erge peed oer he ol ime of jorney i gien Tol dince coered Tol ime elped d d d =...

5 Moion In ne Dimenion 5 Specil ce : When pricle moe wih peed po hlf ime of i ol moion nd in re ime i i moing wih peed hen Dince erged peed : When pricle decribe differen dince d, d, d,... wih differen ime inerl,,,... wih peed,,... repeciely hen he peed of pricle erged oer he ol dince cn be gien Tol dince coered Tol ime elped d d d d d d d d d When pricle moe he fir hlf of dince peed of nd econd hlf of he dince peed hen When pricle coer one-hird dince peed, nex one hird peed nd l one hird peed, hen (d) Innneo peed : I i he peed of pricle priclr inn. When we y peed, i lly men innneo peed. The innneo peed i erge peed for infinieimlly mll ime inerl (i.e., 0 Innneo peed lim 0 d d () Velociy : Re of chnge of poiion i.e. re of diplcemen wih ime i clled elociy. (i) I i clr qniy hing ymbol. (ii) Dimenion : [M 0 L T ] (iii) Uni : mere/econd (S.I.), cm/econd (C.G.S.) (i) Type ). Th () Uniform elociy : pricle i id o he niform elociy, if mgnide well direcion of i elociy remin me nd hi i poible only when he pricle moe in me righ line wiho reering i direcion. (b) Non-niform elociy : pricle i id o he non-niform elociy, if eiher of mgnide or direcion of elociy chnge (or boh chnge). (c) erge elociy : I i defined he rio of diplcemen o ime ken by he body Diplceme n r erge elociy ; ken (d) Innneo elociy : Innneo elociy i defined re of chnge of poiion ecor of pricle wih ime cerin inn of ime. r dr Innneo elociy lim 0 d

6 6 Moion In ne Dimenion () Comprion beween innneo peed nd innneo elociy () innneo elociy i lwy ngenil o he ph followed by he pricle. When one i hrown from poin hen poin of projecion he innneo elociy of one i, poin he innneo elociy of one i, imilrly poin nd C re nd repeciely. Direcion of hee elociie cn be fond o by drwing ngen on he rjecory gien poin. (b) pricle my he conn innneo peed b rible innneo elociy. Exmple : When pricle i performing niform circlr moion hen for eery inn of i circlr moion i peed remin conn b elociy chnge eery inn. (c) The mgnide of innneo elociy i eql o he innneo peed. (d) If pricle i moing wih conn elociy hen i erge elociy nd innneo elociy re lwy eql. (e) If diplcemen i gien fncion of ime, hen ime deriie of diplcemen will gie elociy. Le diplcemen x 0 dx d Innneo elociy ( 0 ) d d For he gien le of, we cn find o he innneo elociy. e.g. for 0,Innneo elociy nd Innneo peed (i) Comprion beween erge peed nd erge elociy () erge peed i clr while erge elociy i ecor boh hing me ni ( m/) nd dimenion [ LT ]. (b) erge peed or elociy depend on ime inerl oer which i i defined. (c) For gien ime inerl erge elociy i ingle led while erge peed cn he mny le depending on ph followed. (d) If fer moion body come bck o i iniil poiion hen 0 ( r 0 ) b 0 nd finie ( 0). (e) For moing body erge peed cn neer be negie or zero (nle ) while erge elociy cn be i.e. 0 while = or < 0. Smple problem bed on peed nd elociy Y C X Problem. If cr coer /5 h of he ol dince wih peed nd /5 h dince wih hen erge peed i [MP PMT 00] () (b) (c) (d) 5

7 Moion In ne Dimenion 7 Tol dince relled Solion : (d) erge peed = Tol ime ken = x ( / 5) x ( / 5) x 5 x (/5)x (/5)x Problem 5. cr ccelered from iniil poiion nd hen rerned iniil poin, hen [IEEE 00] () Velociy i zero b peed incree (b) Speed i zero b elociy incree (c) oh peed nd elociy incree (d) oh peed nd elociy decree Solion : () he ne diplcemen = 0 Hence elociy = 0 ; b peed incree. erge elociy Noe : erge peed. peed. elociy Problem 6. mn wlk on righ rod from hi home o mrke.5 km wy wih peed of 5 km/h. Finding he mrke cloed, he innly rn nd wlk bck home wih peed of 7.5 km/h. The erge peed of he mn oer he inerl of ime 0 o 0 min. i eql o () 5 km/h (b) km/h (c) km/h (d) km/h 8.5 Solion : (d) ken in going o mrke hr 0 min. 5 we re old o find erge peed for he inerl 0 min., o remining ime for coniderion of moion i 0 min. 0 So dince relled in remining 0 min = 7.5.5km. 60 Hence, erge peed = Tol dince Tol ime (.5.5) km 5 = km / hr. (0 / 60) hr. 8 Problem 7. The relion x 6 decribe he diplcemen of pricle in one direcion where x i in mere nd in ec. The diplcemen, when elociy i zero, i () mere (b) mere (c) 5 mere (d) Zero Solion : (d) x 6 x ( 6) x ( 6) x dx d = ( ) 6 d d If elociy = 0 hen, 6 0 ec Hence =, x = () () + = 0 mere. Problem 8. Solion : (b) The moion of pricle i decribed by he eqion x b where 5 cm nd b cm. I innneo elociy ime ec will be [MU (Med.) 000] () 6 cm/ec (b) 8 cm/ec (c) 6 cm/ec (d) cm/ec dx x b = d 0 b

8 8 Moion In ne Dimenion Problem 9. = ec, = = 8 cm / ec ( b cm ) rin h peed of 60 km/h for he fir one hor nd 0 km/h for he nex hlf hor. I erge peed in km/h i [JIPMER 999] () 50 (b) 5. (c) 8 (d) 70 Solion : (b) Tol dince relled = km nd Tol ime ken = hr hr hr 80 erge peed 5. km/h Problem 0. peron complee hlf of i hi jorney wih peed nd re hlf wih peed. The erge peed of Solion : (b) he peron i [RPET 99; MP PMT 00] () (b) (c) In hi problem ol dince i diided ino wo eql pr. So d d = d d d d d / d / = (d) Problem. cr moing on righ rod coer one hird of he dince wih 0 km/hr nd he re wih 60 km/hr. The erge peed i () 0 km/hr (b) 80 km/hr (c) 6 km / hr (d) 6 km/hr Solion : (d) Le ol dince relled = x nd ol ime ken +.7 ccelerion. erge peed x ( / ) x ( / ) x = 6 km / hr x / x / 0 60 The ime re of chnge of elociy of n objec i clled ccelerion of he objec. () I i ecor qniy. I direcion i me h of chnge in elociy (No of he elociy) () There re hree poible wy by which chnge in elociy my occr When only direcion of elociy chnge ccelerion perpendiclr o elociy When only mgnide of elociy chnge ccelerion prllel or niprllel o elociy When boh mgnide nd direcion of elociy chnge ccelerion h wo componen one i perpendiclr o elociy nd noher prllel or ni-prllel o elociy e.g. Uniform circlr moion e.g. Moion nder griy e.g. Projecile moion () Dimenion : [M 0 L T ] () Uni : mere/econd (S.I.); cm/econd (C.G.S.) (5) Type of ccelerion : (i) Uniform ccelerion : body i id o he niform ccelerion if mgnide nd direcion of he ccelerion remin conn dring pricle moion.

9 Moion In ne Dimenion 9 Noe : If pricle i moing wih niform ccelerion, hi doe no necerily imply h pricle i moing in righ line. e.g. Projecile moion. (ii) Non-niform ccelerion : body i id o he non-niform ccelerion, if mgnide or direcion or boh, chnge dring moion. (iii) erge ccelerion : The direcion of erge ccelerion ecor i he direcion of he chnge in elociy ecor d (i) Innneo ccelerion = lim 0 d () For moing body here i no relion beween he direcion of innneo elociy nd direcion of ccelerion. Y g g g X e.g. () In niform circlr moion = 90º lwy (b) In projecile moion i rible for eery poin of rjecory. F (i) If force F c on pricle of m m, by Newon nd lw, ccelerion m d d x dx (ii) y definiion d d d i.e., if x i gien fncion of ime, econd ime deriie of diplcemen gie ccelerion (iii) If elociy i gien fncion of poiion, hen by chin rle d d d dx dx d d. dx (ix) If pricle i ccelered for ime by ccelerion nd for ime by ccelerion hen erge ccelerion i (x) If me force i pplied on wo bodie of differen me m nd m eprely hen i prodce ccelerion nd repeciely. Now hee bodie re ched ogeher nd form combined yem nd me force i pplied on h yem o h be he ccelerion of he combined yem, hen F F m m m F m dx d

10 0 Moion In ne Dimenion m m F F F F So, (xi) ccelerion cn be poiie, zero or negie. Poiie ccelerion men elociy increing wih ime, zero ccelerion men elociy i niform conn while negie ccelerion (rerdion) men elociy i decreing wih ime. (xii) For moion of body nder griy, ccelerion will be eql o g, where g i he ccelerion de o griy. I norml le i 9.8 m/ or 980 cm/ or fee/. Smple problem bed on ccelerion Problem. The diplcemen of pricle, moing in righ line, i gien by where i in mere nd in econd. The ccelerion of he pricle i [CPMT 00] () m/ (b) m/ (c) 6 m/ (d) 8 m/ Solion : (b) Gien d elociy () nd ccelerion () d d d () 0 m / Problem. The poiion x of pricle rie wih ime x b. The ccelerion of he pricle will be zero ime eql o [CSE PMT 997; HU 999; DPMT 000; KCET (Med.) 000] Solion : (c) () b Gien (b) b (c) b (d) Zero dx d x b elociy ( ) b nd ccelerion () = 6b. d d When ccelerion = 0 6b 0. 6b b Problem. The diplcemen of he pricle i gien by y b c d. The iniil elociy nd ccelerion re repeciely [CPMT 999, 00] () b, d (b) b, c (c) b, c (d) c, d Solion : (c) Gien y b c d = Ping 0, iniil = b So iniil elociy = b Now, ccelerion () Ping = 0, iniil = c d d dy d 0 c d 0 b c d Problem 5. The relion beween ime nd dince x i x x, where nd re conn. The rerdion i ( i he elociy) [NCERT 98] () (b) (c) d Solion : () differeniing ime wih repec o dince x dx So, ccelerion () = d d d dx. = dx d d dx. (x ) (d).. dx d x Problem 6. If diplcemen of pricle i direcly proporionl o he qre of ime. Then pricle i moing wih [RPET 999] () Uniform ccelerion (b) Vrible ccelerion (c) Uniform elociy (d) Vrible ccelerion b niform elociy

11 Solion : () Gien h x or x K (where K= conn) dx d Velociy () K nd ccelerion () K d d I i cler h elociy i ime dependen nd ccelerion doe no depend on ime. So we cn y h pricle i moing wih niform ccelerion b rible elociy. Moion In ne Dimenion Problem 7. pricle i moing ewrd wih elociy of 5 m/. In 0 ec he elociy chnge o 5 m/ norhwrd. The erge ccelerion in hi ime i [IIT-JEE 98] () Zero (c) m/ owrd norh-e Solion : (b) 5 o co owrd norh- erge ccelerion we ( cler from he figre). 5 0 m/ (b) (d) m/ m/ Problem 8. body r from he origin nd moe long he x-xi ch owrd norh-we owrd norh-we h elociy ny inn i gien by ( ), where i in econd nd elociy i in m/. Wh i he ccelerion of he pricle, when i i m from he origin? 5m / 90 o 5m / () 8 m/ (b) m/ (c) m / (d) 0 m / Solion : (b) Gien h x d, x C, 0, x 0 C 0 When pricle i m wy from he origin 0 ( ) ( ) 0 ec d d d d ec for m/ Problem 9. body of m 0 kg i moing wih conn elociy of 0 m/. When conn force c for ec on i, i moe wih elociy m/ec in he oppoie direcion. The ccelerion prodced in i i [MP PET 997] () m/ (b) m/ (c) 0. m/ (d) 0. m/ Solion : (b) Le pricle moe owrd e nd by he pplicion of conn force i moe owrd we 0 m/ nd m/. ccelerion Chnge in elociy ( ) (0) m/

12 Poiion Moion In ne Dimenion.8 Poiion Grph. Dring moion of he pricle i prmeer of kinemicl nlyi (,,, r) chnge wih ime. Thi cn be repreened on he grph. Poiion ime grph i ploed by king ime long x-xi nd poiion of he pricle on y-xi. Le i poiion-ime grph for ny moing pricle Velociy = Chnge in poiion ken y (i) y y y D P P T T = 0 o o = 0 i.e., line prllel o ime xi repreen h he pricle i re. = 90 o o = i.e., line perpendiclr o ime xi repreen h pricle i chnging i poiion b ime doe no chnge i men he pricle poee infinie elociy. Prciclly hi i no poible. y C x P T = conn o = conn, = 0 i.e., line wih conn lope repreen niform elociy of he pricle. P T i increing o i increing, i poiie. i.e., line bending owrd poiion xi repreen increing elociy of pricle. I men he pricle poee ccelerion. P i decreing o i decreing, i negie T i.e., line bending owrd ime xi repreen decreing elociy of he pricle. I men he pricle poee rerdion. P conn b > 90 o o will be conn b negie T i.e., line wih negie lope repreen h pricle rern owrd he poin of reference. (negie diplcemen). P C Srigh line egmen of differen lope repreen h elociy of he body chnge fer cerin inerl of ime. S T P T Thi grph how h one inn he pricle h wo poiion. Which i no poible. P T The grph how h pricle coming owrd origin iniilly nd fer h i i moing wy from origin.

13 Dince Moion In ne Dimenion From ringle C n C C D C y.(ii) y y compring (i) nd (ii) Velociy = n = n I i cler h lope of poiion-ime grph repreen he elociy of he pricle. Vrio poiion ime grph nd heir inerpreion Noe : If he grph i ploed beween dince nd ime hen i i lwy n increing cre nd i neer come bck owrd origin bece dince neer decree wih ime. Hence ch ype of dince ime grph i lid p o poin only, fer poin i i no lid hown in he figre. For wo pricle hing diplcemen ime grph wih lope nd n poee elociie nd repeciely hen n Smple problem bed on poiion-ime grph Problem 0. The poiion of pricle moing long he x-xi cerin ime i gien below : () 0 x (m) Which of he following decribe he moion correcly () Uniform, ccelered (b) Uniform, decelered (c) Non-niform, ccelered (d) There i no enogh d for generliion Solion : () Innneo elociy x, y ing he d from he ble 0 ( ) m /, 6m/ nd 0 m/ i.e. he peed i increing conn re o moion i niformly ccelered. Problem. Which of he following grph repreen niform moion () (b) (c) (d) Solion : () When dince ime grph i righ line wih conn lope hn moion i niform. Problem. The diplcemen-ime grph for wo pricle nd re righ line inclined ngle of 0 o nd 60 o wih he ime xi. The rio of elociie of : i [CPMT 990; MP PET 999; MP PET 00] Solion : (d) () : (b) : (c) : (d) : n from diplcemen grph. So n 0 n 60 o o / Problem. From he following diplcemen ime grph find o he elociy of moing body

14 Velociy Diplcemen (ec) Moion In ne Dimenion () (b) m/ (c) (d) m/ m/ o Solion : (c) In fir inn yo will pply n nd y, n 0 m/. i i wrong bece forml n i lid when ngle i mered wih ime xi. Here ngle i ken from diplcemen xi. So ngle from ime xi o Now n 60 o o o Problem. The digrm how he diplcemen-ime grph for pricle moing in righ line. The erge elociy for he inerl = 0, = 5 i x () 0 (b) 6 m (c) m (d) m Solion : (c) erge elociy = Tol diplceme n Tol ime = 5 = m/ Problem 5. Figre how he diplcemen ime grph of body. Wh i he rio of he peed in he fir econd nd h in he nex wo econd Y () : (b) : (c) : (d) : Solion: (d) Speed in fir econd = 0 nd Speed in nex wo econd = 5. So h rio :.9 Velociy Grph. The grph i ploed by king ime long x-xi nd elociy of he pricle on y-xi. Dince nd diplcemen : The re coered beween he elociy ime grph nd ime xi gie he diplcemen nd dince relled by he body for gien ime inerl. Then Tol dince = ddiion of modl of differen re. i.e. d Tol diplcemen = ddiion of differen re conidering heir ign. i.e. r d here nd re re of ringle nd repeciely nd i he re of rpezim. ccelerion : Le i elociy-ime grph for ny moing pricle y ccelerion = Chnge in elociy ken (i) o Diplcemen (meer) D 5 C X x

15 From ringle C, n y compring (i) nd (ii) ccelerion () = n C C D C.(ii) I i cler h lope of elociy-ime grph repreen he ccelerion of he pricle. Moion In ne Dimenion 5

16 Velociy Velociy Velociy Velociy Velociy Velociy Velociy Velociy 6 Moion In ne Dimenion Vrio elociy ime grph nd heir inerpreion = 0, = 0, = conn i.e., line prllel o ime xi repreen h he pricle i moing wih conn elociy. Velociy Velociy = 90 o, =, = increing i.e., line perpendiclr o ime xi repreen h he pricle i increing i elociy, b ime doe no chnge. I men he pricle poee infinie ccelerion. Prciclly i i no poible. =conn, o = conn nd i increing niformly wih ime i.e., line wih conn lope repreen niform ccelerion of he pricle. increing o ccelerion increing i.e., line bending owrd elociy xi repreen he increing ccelerion in he body. decreing o ccelerion decreing i.e. line bending owrd ime xi repreen he decreing ccelerion in he body Poiie conn ccelerion bece i conn nd < 90 o b iniil elociy of he pricle i negie. Poiie conn ccelerion bece i conn nd < 90 o b iniil elociy of pricle i poiie. Negie conn ccelerion bece i conn nd > 90 o b iniil elociy of he pricle i poiie. Negie conn ccelerion bece i conn nd > 90 o b iniil elociy of he pricle i zero. Negie conn ccelerion bece i conn nd > 90 o b iniil elociy of he pricle i negie. Smple problem bed on elociy-ime grph Problem 6. bll i hrown ericlly pwrd. Which of he following plo repreen he peed-ime grph of he bll dring i fligh if he ir reince i no ignored

17 (m/ ) Speed in km/hor Speed Speed Speed Speed Moion In ne Dimenion 7 () (b) (c) (d) Solion : (c) In fir hlf of moion he ccelerion i niform & elociy grdlly decree, o lope will be negie b for nex hlf ccelerion i poiie. So lope will be poiie. Th grph 'C' i correc. No ignoring ir reince men pwrd moion will he ccelerion ( + g) nd he downwrd moion will he ( g ). Problem 7. rin moe from one ion o noher in hor ime. I peed-ime grph dring hi moion i hown in he figre. The mximm ccelerion dring he jorney i [Kerl (Engg.) 00] D C L N M E in hor Solion : (b) () 0 km h (b) 60 km h (c) 00 km h (d) 0 km h Mximm ccelerion men mximm lope in peed ime grph. h lope i for line CD. So, lope of CD = mx km h Problem 8. The grph of diplcemen / ime i S I correponding elociy-ime grph will be [DCE 00] () (b) (c) (d) V V V V Solion : () We know h he elociy of body i gien by he lope of diplcemen ime grph. So i i cler h iniilly lope of he grph i poiie nd fer ome ime i become zero (correponding o he pek of he grph) nd hen i will be negie. Problem 9. In he following grph, dince relled by he body in mere i [EMCET 99] Y () () X

18 Velociy Velociy Velociy Velociy (m/) (m/ ) 8 Moion In ne Dimenion Solion : () (b) 50 (c) 00 (d) 00 Dince = The re nder grph S ( 0 0) 0 = 00 mere Problem 0. For he elociy-ime grph hown in figre below he dince coered by he body in l wo econd of i moion i wh frcion of he ol dince coered by i in ll he een econd [MP PMT/PET 998; RPET 00] () (b) (c) (d) Solion : (b) Dince coered in ol 7 econd = re of rpezim CD ( 6) 0 = 0 m Dince coered in l econd = re of ringle CDQ= 0 0 m So reqired frcion 0 0 Problem. The elociy ime grph of body moing in righ line i hown in he figre. The diplcemen nd dince relled by he body in 6 ec re repeciely Solion : () () (b) (c) (d) 8 m, 6 6 m, 8 6 m, 6 8 m, 8 m m m m re of recngle C = = 8 m re of recngle CDEF = ( ) = m re of recngle FGHI = = m Diplcemen = m of re wih heir ign = 8 + ( ) + = 8 m Dince = m of re wih o ign = = 6 m Problem. bll i hrown ericlly pwrd which of he following grph repreen elociy ime grph of he bll dring i fligh (ir reince i negleced) [CPMT 99; MU (Engg.) 000] D C P C Q 5 7 () F G E H I 6 (ec) D () (b) (c) (d)

19 Moion In ne Dimenion 9 Solion : (d) In he poiie region he elociy decree linerly (dring rie) nd in negie region elociy incree linerly (dring fll) nd he direcion i oppoie o ech oher dring rie nd fll, hence fll i hown in he negie region. c Problem. The ccelerion-ime grph of body i hown below The mo probble elociy-ime grph of he body i () (b) (c) (d) Solion : (c) From gien grph ccelerion i increing conn re d k (conn) k (by inegrion) d d k d kd d d k d i.e., i dependen on ime prboliclly nd prbol i ymmeric bo -xi. nd ddenly ccelerion become zero. i.e. elociy become conn. Hence (c) i mo probble grph. Problem 5. Which of he following elociy ime grph i no poible k () (b) (c) (d) Solion : (d) Pricle cn no poe wo elociie ingle inn o grph (d) i no poible. Problem 6. For cerin body, he elociy-ime grph i hown in he figre. The rio of pplied force for inerl nd C i () D (b) (c) 0 o 60 o C

20 Velociy (m/ec) Velociy Velociy Velociy Velociy Veloci 0 Moion In ne Dimenion Solion : (d) (d) Rio of pplied force = Rio of ccelerion = C n 0 n 0 = Problem 7. Velociy-ime grph of wo cr which r from re he me ime, re hown in he figre. Grph how, h () Iniil elociy of i greer hn he iniil elociy of (b) ccelerion in i increing leer re hn in (c) ccelerion in i greer hn in (d) ccelerion in i greer hn in Solion : (c) cerin inn lope of i greer hn ( ), o ccelerion in i greer hn Problem 8. Which one of he following grph repreen he elociy of eel bll which fll from heigh on o mrble floor? (Here repreen he elociy of he pricle nd he ime) () (b) (c) (d) Solion : () Iniilly when bll fll from heigh i elociy i zero nd goe on increing when i come down. J fer rebond from he erh i elociy decree in mgnide nd i direcion ge reered. Thi proce i repeed nill bll come o re. Thi inerpreion i well explined in grph (). Problem 9. The djoining cre repreen he elociy-ime grph of pricle, i ccelerion le long, nd C in mere/ec re repeciely Solion : () (), 0, 0.5 (b), 0, 0.5 (c),, 0.5 (d), 0.5, 0 ccelerion long m / (ec) 0 ccelerion long ccelerion long C m/.0 Eqion of Kinemic. Thee re he rio relion beween,,, nd for he moing pricle where he noion re ed : = Iniil elociy of he pricle ime = 0 ec = Finl elociy ime ec = ccelerion of he pricle = Dince relled in ime ec

21 n = Dince relled by he body in n h ec () When pricle moe wih zero ccelerion (i) I i nidirecionl moion wih conn peed. (ii) Mgnide of diplcemen i lwy eql o he dince relled. (iii) =, = [ = 0] () When pricle moe wih conn ccelerion Moion In ne Dimenion (i) ccelerion i id o be conn when boh he mgnide nd direcion of ccelerion remin conn. (ii) There will be one dimenionl moion if iniil elociy nd ccelerion re prllel or ni-prllel o ech oher. (iii) Eqion of moion in clr from Eqion of moion in ecor from... ( ) n (n ) n (n ) () Imporn poin for niformly ccelered moion (i) If body r from re nd moe wih niform ccelerion hen dince coered by he body in ec i proporionl o (i.e. ). So we cn y h he rio of dince coered in ec, ec nd ec i : : or : : 9. (ii) If body r from re nd moe wih niform ccelerion hen dince coered by he body in nh ec i proporionl o ( n ) (i.e. ( n ) n So we cn y h he rio of dince coered in I ec, II ec nd III ec i : : 5. (iii) body moing wih elociy i opped by pplicion of brke fer coering dince. If he me body moe wih elociy n nd me brking force i pplied on i hen i will come o re fer coering dince of n. 0, [ince i conn] So we cn y h if become n ime hen become n ime h of preio le. (i) pricle moing wih niform ccelerion from o long righ line h elociie nd nd repeciely. If C i he mid-poin beween nd hen elociy of he pricle C i eql o

22 Moion In ne Dimenion Smple problem bed on niform ccelerion Problem 0. body moe wih niform ccelerion nd zero iniil elociy. noher body, r from he me poin moe in he me direcion wih conn elociy. The wo bodie mee fer ime. The le of i [MP PET 00] () (b) (c) (d) Solion : () Le hey mee fer ime ''. Dince coered by body = nd. ; Dince coered by body = Problem. den i nding dince of 50mere from he b. oon he b r i moion wih n ccelerion of m, he den r rnning owrd he b wih niform elociy. ming he moion o be long righ rod, he minimm le of, o h he den i ble o cch he b i Solion : (c) () 5 m (b) 8 m (c) 0 m (d) m Le den will cch he b fer ec. So i will coer dince. Similrly dince relled by he b will be ( m / ) for he gien condiion d To find he minimm le of, 0, o we ge = 0 ec d hen = 0 m/. [KCET 00] Problem. cr, moing wih peed of 50 km/hr, cn be opped by brke fer le 6m. If he me cr i moing peed of 00 km/hr, he minimm opping dince i Solion : (d) () 6m (b) m (c) 8m (d) m ( = conn) m. Problem. The elociy of blle i redced from 00m/ o 00m/ while relling hrogh wooden block of hickne 0cm. The rerdion, ming i o be niform, will be [IIMS 00] Solion : (d) () 0 0 m/ (b) 00 m /, 00 m /, 0. m (00) (00) 0. 0 m/ (c) 5 0 m /.5 0 m/ (d) 5 0 m/ Problem. body r from re wih n ccelerion. fer econd, noher body r from re wih n ccelerion. If hey rel eql dince in he 5h econd, fer he r of, hen he rio : i eql o [IIMS 00] () 5 : 9 (b) 5 : 7 (c) 9 : 5 (d) 9 : 7 Solion : () y ing n S n, Dince relled by body in 5 h econd = 0 ( 5 ) Dince relled by body in rd econd i = 0 ( )

23 ccording o problem : 0 ( 5 ) = 0 ( ) 9 5 Moion In ne Dimenion Problem 5. The erge elociy of body moing wih niform ccelerion relling dince of.06 m i 0. m. If he chnge in elociy of he body i 0.8m dring hi ime, i niform ccelerion i[emcet (Med.) 000] () 0.0 m (b) 0.0 m (c) 0.0 m (d) 0.0 m Solion : (b) Dince erge elociy nd ccelerion ec Chnge in elociy m /. Problem 6. pricle rel 0m in fir 5 ec nd 0m in nex ec. ming conn ccelerion wh i he dince relled in nex ec () 8. m (b) 9. m (c) 0. m (d) None of boe Solion : () Le iniil ( 0) elociy of pricle = for fir 5 ec of moion 5 0 mere, o by ing 0 5 (5) 5. (i) for fir 8 ec of moion 8 0 mere 0 8 (8) (ii) y oling (i) nd (ii) 7 6 m / m / Now dince relled by pricle in ol 0 ec by biing he le of nd we will ge 8. m 0 So he dince in l ec = m Problem 7. body rel for 5 ec ring from re wih conn ccelerion. If i rel dince S, S nd S in he fir fie econd, econd fie econd nd nex fie econd repeciely he relion beween S, S nd S i 5 9 [MU (Engg.) 000] () S S S (b) 5S S S (c) S S S (d) S S S 5 5 Solion : (c) Since he body r from re. Therefore 0. S S (5) 5 00 (0) 00 S S S = 75 5 (5) 5 S S S S S S = Th Clerly S S S 5 Problem 8. If body hing iniil elociy zero i moing wih niform ccelerion 8 m / ec, he dince relled by i in fifh econd will be () 6 mere (b) 0 mere (c) 00 mere (d) Zero Solion : () S n (n ) 0 (8) [ 5 ] 6mere 5

24 Moion In ne Dimenion Problem 9. The engine of cr prodce ccelerion m/ec in he cr, if hi cr pll noher cr of me m, wh will be he ccelerion prodced [RPET 996] () 8 m/ (b) m/ (c) m/ (d) m / Solion : (b) F = m if F = conn. Since he force i me nd he effecie m of yem become doble m m m, m/ m m Problem 50. body r from re. Wh i he rio of he dince relled by he body dring he h nd rd econd. [CSE PMT 99] () 7/5 (b) 5/7 (c) 7/ (d) /7 S Solion : () S n ( n ), S. Moion of ody Under Griy (Free Fll). 7 5 The force of rcion of erh on bodie, i clled force of griy. ccelerion prodced in he body by he force of griy, i clled ccelerion de o griy. I i repreened by he ymbol g. In he bence of ir reince, i i fond h ll bodie (irrepecie of he ize, weigh or compoiion) fll wih he me ccelerion ner he rfce of he erh. Thi moion of body flling owrd he erh from mll lide (h << R) i clled free fll. n idel one-dimenionl moion nder griy in which ir reince nd he mll chnge in ccelerion wih heigh re negleced. () If body dropped from ome heigh (iniil elociy zero) (i) Eqion of moion : Tking iniil poiion origin nd direcion of moion (i.e., downwrd direcion) poiie, here we he = 0 = +g = g [ body r from re] [ ccelerion i in he direcion of moion] (i) h g (ii) gh (iii) h = 0 gh h g h g g g h n (n )...(i) (ii) Grph of dince elociy nd ccelerion wih repec o ime : n = g g (iii) h = (/)g, i.e., h, dince coered in ime,,, ec., will be in he rio of : :, i.e., qre of ineger.

25 (i) The dince coered in he nh ec, h n g (n ) So dince coered in I, II, III ec, ec., will be in he rio of : : 5, i.e., odd ineger only. () If body i projeced ericlly downwrd wih ome iniil elociy Eqion of moion : g h g Moion In ne Dimenion 5 gh h n g (n ) () If body i projeced ericlly pwrd (i) Eqion of moion : Tking iniil poiion origin nd direcion of moion (i.e., ericlly p) poiie = g [ ccelerion i downwrd while moion pwrd] So, if he body i projeced wih elociy nd fer ime i reche p o heigh h hen g ; (ii) For mximm heigh = 0 So from boe eqion nd = g, h g gh g h g ; gh ; h n (n ) h = 0 gh h g h g g (iii) Grph of dince, elociy nd ccelerion wih repec o ime (for mximm heigh) : ( /g) + (/g ) (/g) + (/g ) g I i cler h boh qniie do no depend pon he m of he body or we cn y h in bence of ir reince, ll bodie fll on he rfce of he erh wih he me re.

26 6 Moion In ne Dimenion () In ce of moion nder griy for gien body, m, ccelerion, nd mechnicl energy remin conn while peed, elociy, momenm, kineic energy nd poenil energy chnge. (5) The moion i independen of he m of he body, in ny eqion of moion, m i no inoled. Th i why hey nd ligh body when releed from he me heigh, rech he grond imlneoly nd wih me elociy i.e., ( h / g) nd gh. (6) In ce of moion nder griy ime ken o go p i eql o he ime ken o fll down hrogh he me dince. of decen ( ) = ime of cen ( ) = /g Tol ime of fligh T = + g (7) In ce of moion nder griy, he peed wih which body i projeced p i eql o he peed wih which i come bck o he poin of projecion. well he mgnide of elociy ny poin on he ph i me wheher he body i moing in pwrd or downwrd direcion. (8) bll i dropped from bilding of heigh h nd i reche fer econd on erh. From he me bilding if wo bll re hrown (one pwrd nd oher downwrd) wih he me elociy nd hey rech he erh rfce fer nd econd repeciely hen =0 (9) body i hrown ericlly pwrd. If ir reince i o be ken ino ccon, hen he ime of cen i le hn he ime of decen. > Le i he iniil elociy of body hen ime of cen nd h g ( g ) where g i ccelerion de o griy nd i rerdion by ir reince nd for pwrd moion boh will work ericlly downwrd. For downwrd moion nd g will work in oppoie direcion bece lwy work in direcion oppoie o moion nd g lwy work ericlly downwrd. So h ( g ) ( g ) ( g ) Compring nd we cn y h > ince (g + ) > (g ) ( g )( g ) (0) pricle i dropped ericlly from re from heigh. The ime ken by i o fll hrogh cceie dince of m ech will hen be in he rio of he difference in he qre roo of he ineger i.e. m m m m = 0,( ),( )...( ),...

27 Smple problem bed on moion nder griy Moion In ne Dimenion 7 Problem 5. If body i hrown p wih he elociy of 5 m/ hen mximm heigh ined by he body i (g = 0 m/ ) [MP PMT 00] ().5 m (b) 6. m (c).5 m (d) 7.6 m Solion : () (5) H mx. 5m g 0 Problem 5. body fll from re in he griionl field of he erh. The dince relled in he fifh econd of i moion i ( g 0m / ) [MP PET 00] () 5m (b) 5m (c) 90m (d) 5m g 0 Solion : (b) hn n h5 h 5 5 m. Problem 5. If bll i hrown ericlly pwrd wih peed, he dince coered dring he l econd of i cen i [CSE 00] () g (b) g (c) ( g) (d) Solion : () If bll i hrown wih elociy, hen ime of fligh g h ec elociy fer ec : g = g. g g ec So, dince in l '' ec : 0 ( g) ( g) h. g h g. Problem 5. mn hrow bll wih he me peed ericlly pwrd one fer he oher n inerl of econd. Wh hold be he peed of he hrow o h more hn wo bll re in he ky ny Solion : (d) ime (Gien g 9.8m / ) () le 0.8 m/ (b) ny peed le hn 9.6 m/ (c) nly wih peed 9.6 m/ (d) More hn 9.6 m/ Inerl of bll hrow = ec. If we wn h minimm hree (more hn wo) bll remin in ir hen ime of fligh of fir bll m be U greer hn ec. i.e. T ec or ec 9.6 m / g I i cler h for 9. 6 Fir bll will j rike he grond (in ky), econd bll will be highe poin (in ky), nd hird bll will be poin of projecion or on grond (no in ky). Problem 55. mn drop bll downide from he roof of ower of heigh 00 meer. he me ime noher bll i hrown pide wih elociy 50 meer/ec. from he rfce of he ower, hen hey will mee which heigh from he rfce of he ower [CPMT 00] () 00 meer (b) 0 meer (c) 80 meer (d) 0 meer Solion : (c) Le boh bll mee poin P fer ime. The dince relled by bll ( h ) g...(i) The dince relled by bll ( h) g...(ii) y dding (i) nd (ii) h h = 00 (Gien h h h 00. ) 00 / 50 8ec nd h 0 m, h 80 m P 00 m h h

28 8 Moion In ne Dimenion Problem 56. ery lrge nmber of bll re hrown ericlly pwrd in qick cceion in ch wy h he nex bll i hrown when he preio one i he mximm heigh. If he mximm heigh i 5m, he nmber of bll hrown per mine i (ke g 0m ) [KCET (Med.) 00] () 0 (b) 80 (c) 60 (d) 0 Solion : (c) Mximm heigh of bll = 5m, So elociy of projecion gh ime inerl beween wo bll (ime of cen) = ec min. g 60 So no. of bll hrown per min = m / Problem 57. pricle i hrown ericlly pwrd. If i elociy hlf of he mximm heigh i 0 m/, hen mximm heigh ined by i i (Tke g 0 m/ ) Solion : (b) () 8 m (b) 0 m (c) m (d) 6 m Le pricle hrown wih elociy nd i mximm heigh i H hen When pricle i heigh H /, hen i peed i 0 m /, 0 From eqion gh Mximm heigh H g H g H g g 00 g 00 = 0m 0 Problem 58. one i ho righ pwrd wih peed of 0 m/ec from ower 00 m high. The peed wih which i rike he grond i pproximely [MU (Engg.) 999] Solion : (b) () 60 m/ec (b) 65 m/ec (c) 70 m/ec (d) 75 m/ec Speed of one in ericlly pwrd direcion i 0 m/. So for ericl downwrd moion we will conider 0 m / gh ( 0) m / Problem 59. body freely flling from he re h elociy fer i fll hrogh heigh h. The dince i h o fll down for i elociy o become doble, i [HU 999] Solion : (b) () h (b) h (c) 6 h (d) 8 h Le poin iniil elociy of body i eql o zero For ph : 0 gh (i) For ph C : () 0 gx gx (ii) Soling (i) nd (ii) x h Problem 60. body liding on mooh inclined plne reqire econd o rech he boom ring from re he op. How mch ime doe i ke o coer one-forh dince ring from re he op () (b) (c) (d) 6 Solion : (b) S conn / Problem 6. one dropped from bilding of heigh h nd i reche fer econd on erh. From he me bilding if wo one re hrown (one pwrd nd oher downwrd) wih he me elociy nd hey rech he erh rfce fer nd econd repeciely, hen [CPMT 997; UPSET 00; KCET (Engg./Med.) 0 () (b) (c) (d) C = 0 h x

29 Moion In ne Dimenion 9 Solion : (c) For fir ce of dropping h g. For econd ce of downwrd hrowing h g g g( )...(i) For hird ce of pwrd hrowing h g( )...(ii) on oling hee wo eqion :. Problem 6. y which elociy bll be projeced ericlly downwrd o h he dince coered by i in 5h econd i wice he dince i coer in i 6h econd ( g 0 m / ) g () 58.8 m / (b) 9 m / (c) 65 m / (d) 9.6 m / 0 0 Solion : (c) y forml h n g(n ) [ 5 ] { [ 6 ]} 5 ( 55) 65m /. Problem 6. Wer drop fll reglr inerl from p which i 5 m boe he grond. The hird drop i leing he p he inn he fir drop oche he grond. How fr boe he grond i he econd drop h inn [CSE PMT 995] ().50 m (b).75 m (c).00 m (d).5 m Solion : (b) Le he inerl be hen from qeion For fir drop g () 5...(i) For econd drop 5 5 y oling (i) nd (ii) x nd hence reqired heigh h 5.75m. g x g...(ii) Problem 6. blloon i heigh of 8 m nd i cending pwrd wih elociy of m /. body of kg weigh i dropped from i. If g 0 m /, he body will rech he rfce of he erh in [MP PMT 99] ().5 (b).05 (c) 5. (d) 6.75 Solion : (c) he blloon i going p we will ke iniil elociy of flling body m /, h 8m, y pplying h g ; 8 (0) ec. 0 g 0m / Problem 65. pricle i dropped nder griy from re from heigh h( g 9.8 m / ) nd i rel dince 9h/5 in he l econd, he heigh h i [MNR 987] () 00 m (b).5 m (c) 5 m (d) 67.5 m Solion : (b) Dince relled in n ec = Dince relled in h n gn = h...(i) g 9h ec (n )...(ii) 5 Soling (i) nd (ii) we ge. h. 5 m.

30 0 Moion In ne Dimenion Problem 66. one hrown pwrd wih peed from he op of he ower reche he grond wih elociy. The heigh of he ower i () / g (b) / g (c) 6 / g (d) / g Solion : (b) For ericl downwrd moion we will conider iniil elociy =. 9 y pplying gh, () ( ) gh h. g Problem 67. one dropped from he op of he ower oche he grond in ec. The heigh of he ower i bo [MP PET 986; FMC 99; CPMT 997; HU 998; DPMT 999; RPET 999] () 80 m (b) 0 m (c) 0 m (d) 60 m Solion : () h g 0 80m. Problem 68. body i releed from gre heigh nd fll freely owrd he erh. noher body i releed from he me heigh excly one econd ler. The eprion beween he wo bodie, wo econd fer he relee of he econd body i ().9 m (b) 9.8 m (c) 9.6 m (d).5 m Solion : (d) The eprion beween wo bodie, wo econd fer he relee of econd body i gien by : = g( ) = 9.8 ( ). 5 m.. Moion Wih Vrible ccelerion. (i) If ccelerion i fncion of ime f() hen (ii) If ccelerion i fncion of dince f(x) hen (iii) If ccelerion i fncion of elociy = f () hen 0 d f( ) f( ) d nd f( ) dd x x 0 nd f( x) dx x x 0 d f( ) Smple problem bed on rible ccelerion Problem 69. n elecron ring from re h elociy h incree linerly wih he ime h i k, where Solion : () k m / ec. The dince relled in he fir econd will be [NCERT 98] () 9 m (b) 6 m (c) 7 m (d) 6 m x d d 9 m. 0 0 Problem 70. The ccelerion of pricle i increing linerly wih ime b. The pricle r from he origin wih n iniil elociy 0. The dince relled by he pricle in ime will be [CSE PMT 995] Solion : (c) () 0 b (b) 0 b (c) 0 b (d) 0 b 6 d = d = ( b ) d b

31 S b d 0 0 b 0 b d 0 b 6 Problem 7. The moion of pricle i decribed by he eqion fir econd Moion In ne Dimenion. The dince relled by he pricle in he () (b) (c) 6 (d) 8 [DCE 000] Solion : (d) d d d =