This exam is formed of 4 obligatory exercises in four pages numbered from 1 to 4 The use of non-programmable calculators is allowed

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1 وزارةالتربیةوالتعلیمالعالي المدیریةالعامةللتربیة داي رةالامتحانات امتحاناتشھادةالثانویةالعامة فرع العلومالعامة مسابقةفي ال فیزیاء المدة:ثلاثساعات دورةسنة الاسم : الرقم : 005 ا لعادیة This exam is formed of 4 obligaory exercises in four pages numbered from o 4 The use of non-programmable calculaors is allowed Firs Exercise: (7 ps) Sudy of a horizonal mechanical oscillaor A solid (S), of mass m = 40 g, may slide on a sraigh horizonal rack. The solid is conneced o wo idenical springs of un-joined urns, of negligible mass, fixed beween wo suppors A and B. Each of hese springs has a siffness (force consan) k = 0.60 N/m, and a free lengh 0. We denoe by O he posiion of he cener of mass G of (S) when he oscillaor [(S) + wo springs] is in equilibrium, each spring having hen he lengh 0 (figure ). x A (S) 0 0 G O i B x The solid is shifed from his equilibrium posiion along he direcion x x, by a disance of 4. cm, and hen released wihou iniial velociy a he insan o = 0. During is oscillaions, a any insan, he abscissa of G is x and he algebraic value of is velociy is V, O being he origin of abscissas. The horizonal plane hrough G is aken as a graviaional poenial energy reference. I Theoreical sudy In his par, we neglec fricion. The solid (S) performs,in his case, oscillaions of ampliude Xmo = 4. cm. ) a) Show ha he expression of he elasic poenial energy of he oscillaor is P.Ee = kx. b) Wrie he expression of he mechanical energy M.E of he sysem [oscillaor, Earh] as a funcion of m, V, x and k. )a) Derive he differenial equaion ha governs he moion of (S). b) Deduce he expression of he proper period To of he oscillaor in erms of m and k. c) Calculae he value of To. Take = 3.4 II Experimenal sudy In realiy, he value of he ampliude X m decreases during oscillaions, each of duraion T. Some values of X m are abulaed as below. Insan 0 T T 3T 4T 5T Ampliude X mo = 4.0 X m =.86 X m =.95 X m3 =.33 X m4 = 0.9 X m5 = 0.6 X m (cm)

2 ) Draw he shape of he curve represening he variaion of he abscissa x of G as a funcion of ime. Scale: on he axis of abscissas cm represens T and on he axis of ordinaes cm represens cm. ) The duraion of 5 oscillaions is measured and found o be 0.75 s. a) Calculae T. b) Compare T and T o. c) Wha is hen he ype of oscillaions? 3) The decrease in he mechanical energy of he sysem [oscillaor, Earh] is due o he exisence of a force of fricion of he form f - hvwhere V = Vi and h is a posiive consan. a) From he above able of values, verify ha: Xm X m.. A where A is a posiive consan. X X mo m b) Knowing ha A is given by he expression A = e m, calculae h. 4) In order o compensae for he loss in he mechanical energy of he sysem, an apparaus (D) allows, a regular ime inervals, o provide energy o he oscillaor. a) Deermine he average power furnished by (D) beween he insans = 0 and = 5 T. b) Wha is hen he ype of oscillaions? Second Exercise: (7 ½ ps) Flash of a camera In his exercise, we inend o show evidence of he funcioning of he flash of a camera. The simplified circui of he flash of a camera is formed of an apparaus aken as a source of DC volage of E = 300 V, a capacior of capaciance C = 00 F, a resisor E of resisance R = 0 k, a lamp ( L ), considered as a resisor of resisance r = and a double swich K ( figure ). I Charging he capacior The capacior is iniially neural. The double swich is urned o posiion a he insan o = 0. The capacior sars charging (figure ) ) a) Derive, a an insan,he differenial equaion ha governs he variaion of he volage u c = u MN,as a funcion of ime,during he charging of he capacior. b) The soluion of his equaion, a an insan, has he form: u c = A + B e where A, B and are consans. Deermine hese consans in erms of E, R and C. ) Calculae he energy W sored by he capacior a he end of charging. II- Discharging he capacior The capacior being compleely charged, he double swich is urned o posiion. The capacior sars o discharge hrough he lamp (L). The insan of closing he circui is aken as an origin of ime. A an insan, he volage across he capacior is uc = umn = E. Jusify he direcion of he curren in figure (3). dq. Knowing ha i = - d e ht and he circui carries hen a curren i ( figure 3 ). E R C q K Figure R K M N q Figure K N i L M L Figure 3

3 a) deermine he expression of he curren i as a funcion of ime, b) calculae he maximum value of i, c) deermine he duraion a he end of which he curren reaches 70 % of is maximum value. d) calculae, a he insan, he volage u C across he capacior. 3. a) Assuming ha he energy released by he capacior by he end of he duraion is convered oally ino ligh in he lamp, deermine he average power received by he lamp during. b) The flash lamp emis ligh as long as he average power i receives is greaer or equal o W. Knowing ha he duraion of he flash is, jusify he emission of he flash beween he insans 0 and. Third exercise : (7 ps) Phooelecric Effec The experimens on phooelecric emission performed by Millikan around he year 95, inended o deermine he kineic energy K.E of he elecrons emied by meallic cylinders of poassium (K) and cesium (Cs) when hese cylinders are illuminaed by monochromaic radiaion of adjusable frequency ν. The objec of his exercise is o deermine, performing similar experimens, Planck s consan (h), as well as he hreshold frequency ν o of poassium and he exracion energy W o of poassium and ha of cesium. I - ) Wha aspec of ligh does he phenomenon of phooelecric effec show evidence of? ) A monochromaic radiaion is formed of phoons. Give wo characerisics of a phoon. 3) For a given pure meal, he inciden phoons of a monochromaic radiaion provoke phooelecric emission. Give he condiion for his emission o ake place. II- In a firs experimen using poassium, a convenien apparaus is used o measure he kineic energy K.E of he elecrons corresponding o frequency νof he inciden radiaion. The obained resuls are abulaed in he following able: ν(hz) K.E (ev) Given : ev = J. - Using Einsein s relaion abou phooelecric effec, show ha he kineic energy of an exraced elecron may be wrien in he form : K.E = a ν+ b. - a) Plo, on he graph paper, he curve represening he variaion of he kineic energy K.E versus ν, using he following scale: on he axis of abscissas: cm represens a frequency of 0 4 Hz on he axis of ordinaes: cm represens a kineic energy of 0.5 ev. b) Using he graph, deermine: i) he value, in SI, of h, he Planck s consan. ii) he hreshold frequency νo of poassium. 3- Deduce he value of he exracion energy W o of poassium. III- In a second experimen using cesium, we obain he following values: K.E = ev for ν= 70 4 Hz. ) Plo, wih jusificaion,on he preceding sysem of axes, he graph of he variaion of K.E as a funcion of ν. ) Deduce from his graph he exracion energy W o of cesium. 3

4 Fourh exercise : (6 ps) Fuel and a power plan The objec of his exercise is o compare he masses of differen fuels used in power plans producing he same elecric power. The power plan, of elecric power P = W, has an efficiency supposed o be 30 % whaever he naure of he fuel used be. A. Energy furnished by he fuel Calculae, in J, he energy furnished by he fuel during day. B. I. Thermal power plan The power plan uses fuel-oil. The combusion of kg of his fuel-oil liberaes J of energy. Calculae, in kg, he mass m of fuel-oil consumed during day. II. Power plan using nuclear fission In he power plan, we use uranium enriched wih 35 U. One of he fission reacions is : 35 9 U n 38 Sr + 54 Xe + 0 n ) In order ha fission reacion may ake place, he neuron used mus saisfy a condiion. Wha is i? ) The fission of uranium 35 nucleus liberaes energy of 89 MeV. a) In wha form does his energy appear? b) Calculae, in kg, he mass m of uranium 35 necessary for he power plan o funcion during day. III. Power plan using nuclear fusion? The hermonuclear fusion reacion has no ye been conrolled. If such conrolling becomes wihin reach, we may provoke reacions like hose aking place in he Sun. The balanced fusion reacion of hydrogen in he Sun may be wrien as : 4 4 H He + A Z X ) Idenify he paricle A Z X specifying he laws used. ) Wha condiion mus be saisfied for his fusion o ake place? 3) Deermine, in J, he energy liberaed in he formaion of a helium nucleus. 4) Calculae, in kg, he mass m 3 of hydrogen necessary for he power plan o funcion day. C. Sugges he mode ha is he mos convenien for he producion of elecric energy for a counry. Jusify your answer. Given : u = 93.5 MeV/c = kg ; MeV = J ; Masses of nuclei and paricles 4 H :.0078 u ; He : u ; A 35 Z X : u ; U : u. 4

5 Soluion Firs exercise I- -a) PEe= /kx + /kx = kx (/p) b) M.E = K.E + P.Ee = ½ mv + kx (/p) - a) M.E = ce => dm. E d k = 0 = mv x+ kxv => x+ x m = 0 (/p) b) The differenial equaion is of he form x+ x = 0 where o = of he moion. The proper period is T o = c) T o = 0,4 =,45 s. (/p) 0,6 II- ) The shape of he curve (/p) m =. (p) o k o k is he proper angular frequency m 0,75 ) a) T =,50 s (/p) 5 b) T=,50 s e T o =,45s => T > T o ( /4p c) Oscillaions are free damped (/4p),86 3) a) A = = 0,68 (/p) 4, h b) ln A = - T. ( /p) Thus : h = 0,05 Kg/s (/p) m 4- a) M.E = ½ mv + kx. A maximum abscissa, V = 0, hus M.E = k(xm) A = 0, M.E0 = k(xm0) =, J. A = 5T, M.E 5 = k(x m5 ) = 0, J. The decrease in he mechanical energy of he sysem is M.E =, , =, J. Thus : P av =. 3 M E,03530 = = 0, W. ( /4 p) 5T 5,5 b) The oscillaor performs driven oscillaions. (/4 p) 5

6 Second exercise I- - a) E = Ri + u C, avec i = dq/d = Cdu C /d ; we have : E = R Cdu C /d + u C. ( p) b- u C =A + B e B ; du C /d = - e B => E = -RC e + A + B e => Be RC RC (- ) + (A- E) = 0 => (- ) = 0 =>= RC e A E = 0 =>. A= E. On he oher hand a = 0, uc = 0 => A + B = 0 => B = - E (/p) - W= ½ C (uc) = ½ CE => W = 9 J (/p) II- ) u MN > 0 => he curren has he direcion from high o low poenial. (/ p) dq du ) a) i = - = - C C E E = C e = e (/p) d d r b) I max = r E = 300 A. (/p) E E c) A he end of he duraion, we have : i = 0,7 I max = 0,7 => e r r = 0,7 r E => e = 0,7 ou = 0,356 => = s (p) d) if = = s, he volage across he capacior is : u C = E e = 300e -0,35 =,4 V. ( p) 3- a) The energy sored in he capacior a he insan is hen : W = ½ C (uc) = 0-4 (,4) = 4,5 J. W = W W = 4,5 J (/p) W P m = = 6,4.0 4 W. b) The lamp receives a power equal o he power of is normal funcioning, i produces hen a flash. (/p) 6

7 Third exercise -) Corpuscular aspec. (/p) ) Zero mass; speed in vacuum = c ; zero charge ; energy = h ν.(/p) 3) If h WO or O or S) (/p) II-- Einsein s relaion gives : h ν= W O + K.E We can wrie : K.E = h ν- W O = a ν+ b où a = h e b = -W O (/p) - a) Represenaion (p).5.5 Series Series Series b) i- K.E = f(ν) is a sraigh line no passing hrough he origin having a slope h.. 9 K. E K. E (,85 0,5),6.0 h = = 6, J.s (/p) ii- If he elecron is exraced wihou velociy (K.E = 0), The meal is illuminaed wih a radiaion of frequency WO equal o hreshold frequency O =. The hreshold frequency corresponds o he inersecion of he obained h line wih he axis of abscissa. Graphically we find O = 5,5.0 4 Hz. (p) WO 3) On a : O = h => W O = h O = 6, ,5.0 4 = 3,5.0-9 J =, ev (/). III- ) In order o plo he curve for cesium, i is enough o locae he poin (7.0 4 Hz ; ev ) in he sysem of axes hen draw a sraigh line parallel o he previous line. (/p) ) In order o deermine he exracion energy of cesium, we produce he line unil i mees he axis of K.E; we find W O =,9 ev. (/p) 7

8 Fourh exercise A- The energy consumed by he plan during s is: = 0 0 W 30 The energy consumed by he plan during day is: E = =864.0 J. (/p) B - I The mass of fuel-oil needed for funcioning day is : m = = 9,.0 6 kg. (/p) II. ) The energy of he neuron is of he order of 0. ev (or slow neuron or hermal neuron) (/4p). ) a) The energy liberae appears in he form of kineic energy of he neurons and he nuclei. (/4p) b) In order o liberaed a nuclear energy of 89 MeV, we need a mass of uranium equal o 35,0439 u. In In order o liberaed a nuclear energy E, we need a mass of uranium 7 35,0439, m = = kg.(p) 3 89,6.0 III-) The laws of conservaion of Z and of A give : Z = 0 and A =. The emied paricle is posiron. (3/4p) ) The nuclei mus have a large kineic energy (in he order of 0. MeV or a emperaure of he medium abou 0 8 K). (/4p) 3) The energy liberaed is given by E 3 = m.c m = 4,0078 4,005-0,00055 = 0,065 u m = 0,06593,5 MeV/c = 4,70338 MeV/c. hus : E 3 = 4,7 MeV = 39,5.0-3 J. (p) 4) In order o liberae nuclear energy of 39,5.0-3 J, We need a mass of hydrogen equal o 4,0078 u. In order o liberae he energy E, we need a mass of hydrogen. 7 4,0078, m 3 = =,5 kg.(p) 3 39,5.0 C- m3 < m < m : for he same producion of energy, we find ha he consumpion of hydrogen is 7 imes less han ha of uranium and imes less han fuel-oil. Fusion does no resul in radioacive nuclei _ Hydrogen is much more abundan in naure hen uranium _ Fusion is more energeic han fission _ Fusion does no produce oxic gases (/p) 8