PHYSICS OCUS 58 So far we hae deal only wih syses in which he oal ass of he syse, sys, reained consan wih ie. Now, we will consider syses in which ass eners or leaes he syse while we are obsering i. The rae of change of ass of he syse, d, is posiie when ass eners he syse and is negaie when ass leaes he syse. igure 6.42(a) shows a syse of ass whose cenre of ass is oing wih elociy as seen fro a paricular reference frae. An exernal force ex acs on he syse. Y Y - u + c c c X O X O A a ie laer he configuraion has changed o ha shown in figure 6.42(b). A ass has been ejeced fro he syse and is cenre of ass is oing wih elociy u wih respec o he iniially chosen reference frae. The ass of he syse has reduced o - and he elociy of he cenre of ass of he syse is changed o. You ay iagine ha he syse of figure 6.42 represens a rocke. I ejecs ho gas fro is orifice a a fairly high speed, decreasing is own ass and increasing is own speed. In a rocke he loss of ass is coninuous during he burning process. The exernal force ex is no he of he rocke bu is he force of graiy on he rocke and he resising force of he aosphere. In such oion fuel is burned in he rocke and exhaus gas is expelled ou he back of he rocke. The force exered by he exhaus gas on he rocke (which is equal and opposie o he force exered by he rocke on he gas o expel i) propels he rocke forward and is called as force. To analyze he siuaion le us, for he ie being, define he syse o be one of he consan ass. This eans ha in figure 6.42(b), we shall include in our syse no only he - of he body bu also he ejeced ass. Therefore, he oal ass of he syse has reained as in he figure 6.42(a). Doing so we can apply he resuls we hae deried so far for syses of consan ass. We will see ha his approach leads us o he for of Newon s second law for syses in which he ass is no consan. ro equaion (6.19), we hae ex dp or he ie ineral, approxiaely, we can wrie, ex P Pf Pi in which Pf is he final oenu, as shown in figure 6.42(b) and Pi is he iniial oenu, as shown in figure 6.42(a). Therefore, we hae, ( )( ) u Pf ( )( ) u
PHYSICS OCUS 59 ex u ( )...(6.33) Now, if we le approach zero, he configuraion of figure 6.42(b) approaches ha of figure 6.42(a); ha is, approaches d, he acceleraion of he body in figure 6.42(a). The quaniy is he ass ejeced in ; his leads o a decrease in he ass of he original body. Since d, he change in ass of he body wih ie, is negaie in his case, he posiie quaniy is replaced by d as approaches zero. inally, goes o zero as approaches zero. aking hese changes in he las equaion, we ge d d d ex u d d ex ( u ) d ex d...(6.34a)...(6.44b) The quaniy u in equaion (6.34a) is jus, he aie elociy of he ejeced ass wih respec o he ain body, as wrien in equaion (6.34 b). d The las er in equaion (6.34b),, is he rae a which oenu is being ransferred ino (or ou of) he syse by he ass ha he syse has ejeced (or colleced). I can be inerpreed as he force exered on he syse by he ass ha leaes i (or joins i). or a rocke his er is called he and i is he rocke designer s ai o ake i as large as possible. Equaion (6.34) suggess ha his requires ha he rocke ejec as uch ass per uni ie as possible and ha he speed of he ejeced ass aie o he rocke be as high as possible. We can rewrie equaion (6.34) as d = ex +...(6.35) in which d is he reacion force exered on he syse by he ass ha leaes (or joins) i. Here you should noe ha he force acing on he syse d, is along if d is ne gaie, i.e., ass of he syse is being reduced. A achine gun is ouned on a car on a horizonal fricionless surface as shown in figure 6.43 (a). The ass of he syse (car + gun) a a paricular insan is. A ha sae insan he gun is firing bulles of ass whose elociy, in he reference frae shown, is u. The elociy of he car in his frae is and he elociy of he bulles wih respec o he car is u. The nuber of bulles fired per uni ie is n. Wha is he acceleraion of he car? u
PHYSICS OCUS 6 e us selec he car and gun as our syse. Since ass of he syse,, is ariable, we apply Newon s second law in he for gien in equaion (6.35), which is d ex Since no ne exernal force acs on he syse, we hae ex in aboe equaion. Hence, we hae d d Now, d is a, he acceleraion of he syse; is u, poining o he lef in figure 6.43(b), and d is n. Insering hese in aboe equaion, we ge u = d/ = n a d ( n) ( u )( n) In figure 6.43(c), he analogous siuaion for a rocke is shown. Here we will consider he syse fro he iewpoin of Newon s hird law and he oenu principle. u = d/ e us choose a fixed-ass syse (rocke + gas) and obsere if fro is C frae. The rocke forces a je of ho gases fro is exhaus; his is he acion force. The je of ho gases exers a force on he rocke, propelling i forward; his is he reacion force. These wo forces ake an acion-reacion pair and are inernal forces of he syse under consideraion. In he absence of exernal forces he oal oenu of he syse is consered (he cenre of ass, iniially a res, reains a res). The indiidual pars of he syse ay change heir oenu, howeer; he ho gases acquire oenu in he backward direcion and he rocke acquires an equal agniude of he oenu in he
PHYSICS OCUS 61 Sand drops fro a saionary hopper a a rae µ = d/ ono a coneyor bel oing wih elociy, as shown in figure 6.44(a). Wha force is required o keep he bel oing a a consan speed? (Assue ha sand drops on he bel fro he negligible heigh.) d This is a clear-cu exaple of a force associaed wih change of ass alone, he elociy being consan. e us ake our syse he bel of arying ass, so ha we can apply he equaion (6.35), which is d ex d d we us pu in he aboe equaion because he elociy of he bel is consan. Hence, we ge = d ex...(i) ex Hence o keep he bel oing wih a consan elociy an exernal force us be applied on i, which should be equal and opposie o he force, d, acing on i. To an obserer a res on he bel, he falling sand (and he hopper) would appear o hae a horizonal oion wih speed in a direcion opposie o ha shown for he bel. ore forally we can wrie, bu u, so ha u Using hese subsiuions in equaion (i), we ge ex d ( )( ) µ
PHYSICS OCUS 62 In his case, q d ( ) is posiie because he syse is gaining ass wih ie,. Hence force on he syse is acing along he direcion of, i.e., in he direcion opposie o ha of he oion of he bel. Therefore, o keep he bel oing wih consan elociy, exernal force on i us be applied in he direcion of is oion. ro he iew poin of newon s second law his siuaion can be explained as follows. As he sand has negligible speed before i falls on he oing bel, he bel has o apply a force on he falling sand along he direcion of is own oion o oe he sand wih i and hence he falling sand applies a reacion force on he bel in he direcion opposie o ha of he oion of he bel. Thrus force on he bel is basically his reacion force only. Therefore, o keep he bel oing wih consan elociy an exernal force us be applied on i along he direcion of is oion and he agniude of he exernal force should be equal o ha he force. or a one-diensional case, fro nex ie onwards we will use equaion (6.35) in a ore siple way. We can rewrie i as ex ± = a...(6.36) we will judge he direcion of he force fro he iewpoin of Newon s hird law and is agniude d can be wrien as.. If he force is along he direcion of acceleraion, a, hen use + sign, oherwise sign. A freigh car filed wih sand has a hole so ha sand leaks ou hrough he boo a a consan rae of agniude. has an uni of kg/sec. A consan force acs on he car in he direcion of is oion. A = if be he speed of he car and be is ass, hen, find is speed afer soe ie. Here, he firs hing you should noice is ha as sand is jus leaking fro he boo, i is separaing fro he car wih zero aie elociy. Hence, if we chose he car as our syse, he force on i is zero. A soe ie if be he elociy of he car and a be is acceleraion, as shown in figure 6.45, fro equaion (6.36), we hae a a ex
PHYSICS OCUS 63 ( ) d and ex d d ln( ) ln( ) ln. A rocke is fired erically upward near he surface of he earh where he free-fall acceleraion of graiy is g. Show ha if he rocke sars fro res, is final elociy is gien by uex ln( i f ) g, where u ex is exhaus speed of he gas wih respec o he rocke and is he ie for he fuel o burn. i is he ass of he rocke a he sar of he ineral and f is he ass a he end of he ineral. a e he rocke is fired a =. Siuaion a soe ie is shown in figure 6.46. As he rocke is ejecing gases in he erically downward direcion, force on he rocke acs in he erically upward direcion. Hence, using equaion (6.36), we ge, a ex g hru d d g uex d d g uex d d d is negaie, d d uex g f d d u g ex i f uex ln g T i
PHYSICS OCUS 64 f uex ln g T i In his case as he exernal force, ha is due o graiy, is acing in he opposie direcion of he acceleraion i is ake o be negaie and as he force is acing along he acceleraion i is aken o be posiie. A chain hangs on a hread and ouches he surface of a able by is lower end Aer he hread has been burned hrough, find he forces exered by he able on he chain. e be he oal lengh of he chain and be is oal ass. The par of he chain oing erically downwards is in free falling sae, hence, haing fallen by a disance x, he air falls wih speed 2 gx, as shown in figure 6.47(a). If we consider he par of he chain fallen on he surface of he able as our syse, hen, a he oen shown in he figure, ass of he syse is increasing and he syse is a res. In he nex ie ineral if lengh dx adds up in our syse hen increase in ass of he syse, x -x g = gx d ass of lengh dx of he chain (ass per uni lengh) lengh of eleen dx Rae of change of ass of he syse, d dx h x.g 2gx I can be explained in arious ways ha he adding ass exers force on he syse in he erically downward direcion, as shown in figure 6.47(b) Weigh of he syse and he noral reacion force acing on he syse fro he surface of he able are also shown in figure 6.47(b). Applying equaion (6.36) o his syse, we ge a ex...(i) ( g N) a and downward direccion is chosen as posiie direcion
PHYSICS OCUS 65 x g N d ass of he syse, ass of lengh x of chain N gx d gx d. 2gx 2 gx, using (i) As he syse is a res, aie elociy of adding ass is sae as is elociy. Therefore, gx N 2gx 2gx gx 2gx 2gx 3gx A he oen under consideraion he force exered by he able on he chain is hree ies he weigh of he chain resing on he able. Thrus force acing on he resing par of he chain is wice as grea as he weigh of his par. A soe ie if x be he lengh of he syse, hen, ass of he syse, ( x) ass per uni lenh lenh of he syse x If ass per uni lengh is defined as linear ass densiy and is denoed by, hen, ( x) x Rae of change of ass, d d ( x) dx for unifor disribuion of ass is consan. This proble can also be soled by reaing he falling par as syse wih decreasing ass. e us sole his proble using ipulse concep. Consider he siuaion a he oen shown in figure 6.47(a). In he nex ie ineral if dx lengh of he falling par of he chain his he surface of he able and coes o res, hen, change in oenu of ha eleen, dp p p f ( d ) i d is he ass of he
PHYSICS OCUS 66 Therefore, he aerage force aced on his eleen fro he par of he chain resing on he able is change in oenu lengh ie ineral dp d d 2 Using d Therefore, force aced on he par of he chain resing on he able by he adding ass or we can also say force acing on ha par, 2 2 gx 2 [Using 2gx ] As he ie ineral is infiniesially sall, his expression can be regarded as expression for insananeous alue of force.
PHYSICS OCUS 67 A girl can exer an aerage force of 2 N on her achine gun. Her gun fires 2-g bulles a 1 /s. How any bulles can she fire per inue? A fla car of ass sars oing o righ due o a consan horizonal force (ig.) Sand spills on he flacar fro a saionary hopper. The elociy of loading is consan and equal o µ kg/s. ind he ie dependence of he elociy and he acceleraion of he flacar in he process of lading. The fricions negligible sall. A rocke ejecs a seady je whose elociy is equal o u aie o he rocke. The gas discharge rae equals µ kg/s. Deonsrae ha he rocke oion equaion in his case akes he for. w = µu. Where is he ass of he rocke a a gien oen, w is is acceleraion, and is he exernal force. A rocke oes in he absence of exernal forces by ejecing a seady je wih elociy u consan aie o he rocke. ind he elociy of he rocke a he oen when is ass is equal o, if a he iniial oen i possessed he ass and is elociy was equal o zero. ake use of he forula gien in he foregoing proble. ind he law according o which he ass of he rocke aries wih ie, when he rocke oes wih a consan acceleraion w, he exernal forces are absen, he gas escapes wih a consan elociy u aie o he rocke, and is ass a he iniial oen equals. A cylindrical solid of ass 1 2 kg and cross -sessional area 1 4 2 is oing parallel o is axis (he x- axis) wih a unifor speed of 1 3 /s in he posiie direcion. A =, is fron face passes he plane x =. The region o he righ of his plane is filled wih saionary dienion of he cylinder reains pracically unchanged and ha he dus paricle of unifor densiy 1 3 kg/ 3. When a dus paricles collides wih he face of he cylinder, i sicks o is surface. Assuing ha he diensions of he cylinder reains pracically unchanged and ha he dus sicks only o he fron face of he x-co ordinae of he fron of he cylinder find he x-coordinae of he fron of he cylinder a =15 s. A chain of lengh weighing l per uni lengh begins o fall wih consan acceleraion hrough a hole in he ceiling. (a) When he lowes end of he chain has fallen hrough a disance x deerine is elociy (x). (b) ind he consan acceleraion of he falling chain. (c) ind he energy loss when he las link of he chain has lef he ceiling. The end of a chain of lengh and ass per uni lengh ha is piled on a plafor is lifed erically wih a consan elociy by a ariable force P. find P as a funcion of he heigh x of he end aboe he plafor. Also find he energy los during he lifing of he chain. P x x