وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة مسابقة في مادة الفيزياء المدة ثالث ساعات االسن: الرقن: الدورة العادية للعام This exam is formed of four exercises i four pages umbered from to 4 The use of o-programmable calculator is recommeded First Exercise (7.5 poits) Mechaical oscillator Two solids (S ) ad (S ), of respective masses m l = g ad m = 5 g, ca slide o a horizotal table. The solid (S ) is fixed to oe ed of a sprig with u-joited turs ad of egligible mass ad of stiffess k = 5 N/m, the other ed A of the sprig beig fixed to a support as show i the figure below. (S ) is lauched towards (S ) ad attais, just before impact, the velocity V = V i where V =.48 m/s. Due to collisio, (S ) sticks to (S ) thus formig a sigle solid (S), just after collisio, at the istat t =, whose ceter of iertia G moves with the velocity V = V i. The horizotal plae through G is take as a gravitatioal potetial eergy referece. (S ) (S ) V x' O A- Theoretical Study i A Neglect all forces of frictio. ) Show that V =.4 m/s ) After collisio, (S), still coected to the sprig, cotiues its motio. At a istat t, we defie the positio of G by its abscissa x o the axis (O, i ), v = dx beig the algebraic measure dt of the velocity of G. The origi O of abscissas is the positio of G at the istat t =. a) Calculate the mechaical eergy of the system [(S), sprig, Earth] at the istat t =. b) Give, at the istat t, the expressio of the mechaical eergy of the system [(S), sprig, Earth] i terms of m, m, k, x ad v. c) Deduce that the abscissa of G is 6. cm whe v is equal to zero for the first time. 3) a) Derive the secod order differetial equatio of the motio of G. b) The solutio of this differetial equatio is of the form: x = X m si ( t + ). i) Determie the values of the costats X m, ad. ii) Calculate the value of the proper period T of oscillatios of G ad deduce the time t eeded by G to pass from O to the positio where v becomes zero for the first time. B- Experimetal study I fact, (S), agai shot with the velocity V at the istat t =, performs oscillatios of pseudoperiod very close to T. The velocity of G becomes zero for the first time at the istat t but the abscissa of G is just 6cm. ) Determie the eergy lost durig t. ) A apparatus (D), coveietly coected to the oscillator, provides eergy i order to compesate for the loss. Calculate the average power provided by (D). 3) The oscillator is at rest. The apparatus (D) ad the support are removed. The ed A of the sprig is coected to a vibrator, which vibrates alog the sprig, with a adjustable frequecy f. a) I steady state, (S) performs oscillatios of frequecy f. Why? b) For a certai value f of f, the amplitude of oscillatios of (S) attais a maximum value. i) Give the ame of the pheomeo that thus took place. ii) Calculate the value of f. x
Secod Exercise (7.5 poits) Duratio of chargig ad dischargig of a capacitor Cosider the circuit whose diagram is show i figure, where G is a geerator deliverig a square sigal (E, ) of period T (Fig. ), D a resistor of resistace R = k ad (C) a capacitor of capacitace C =. F. A oscilloscope displays the voltage u g = u AM across G ad the voltage u C = u BM across (C). A- Theoretical study Chargig of (C) Durig the chargig of (C), the voltage u g has the value E ad at a istat t, the circuit carries a curret i. duc ) Give the expressio of i i terms C ad. dt ) Derive, for t T, the differetial equatio i u C. i G K Fig. B A D C M Y A Y B 3) The solutio of this differetial equatio has the form: t u C = A(- e ), where A ad are costats. a) Determie, i terms of E, R ad C, the expressios of A ad. b) Draw the shape of the graph represetig the variatio of u C as a fuctio of time ad show, o this graph, the poits correspodig to A ad. Dischargig of (C) 4) Durig dischargig of ( C) the voltage u g =. We cosider the istat T as a ew origi of time. E u g T/ T Fig. t t Verify that u C = E e. 5) a) What must the miimum duratio of chargig be so that u C reaches practically the value E? b) What is the the miimum value of T? B- Experimetal study ) O the scree of the oscilloscope, we observe the waveforms of figure 3. a) Which curve correspods to the chargig of the capacitor? Justify the aswer. b) Calculate the value of E ad that of the period T of the square sigal. ) a) We icrease the frequecy of the voltage delivered by G. The waveforms obtaied are as i figure 4. Determie the ew period of the square sigal. Justify the shape of the waveform of u C displayed. b) We keep icreasig the frequecy of the voltage delivered by G. The waveform becomes almost triagular. Why? (3) Fig. 3 () () (4) S V = 5 V/div; S h = ms/div. Fig.4 S V = 5 V/div; S h = ms/div.
Third Exercise (7.5 poits) Eergy levels of the hydroge atom The eergy levels of the hydroge atom are give by the relatio: 3.6 E = ; where E is expressed i ev ad is a o-zero whole umber. Give: Plack's costat: h = 6.6 34 J.s; ev =.6 9 J ; m = -9 m 8 visible spectrum i vacuum: 4 m λ 75 m; speed of light i vacuum c = 3 m/s. The Lyma series represets the set of radiatios emitted by the hydroge atom whe it udergoes a dowward trasitio from the level to the groud state =. ) a) The eergy of the hydroge atom is said to be quatized. What is meat by the term quatized eergy. b) Write dow the expressio of the eergy of a photo associated with a radiatio of wavelegth λ i vacuum. ) a) Show that the wavelegths λ i vacuum of the radiatios of the Lyma series are expressed i m by the relatio: λ = 9.3 ( ). b) i) Determie, i m, the maximum wavelegth λ of the radiatio of the Lyma series. ii) Determie, i m, the miimum wavelegth λ of the radiatio of the Lyma series. iii) Do the radiatios of the Lyma series belog to the visible, ultraviolet or ifrared spectrum? Justify your aswer. 3) A hydroge lamp illumiates a metallic surface of zic whose threshold wavelegth is = 7 m. a) Defie the threshold wavelegth of a metal. b) Electros are emitted by the metallic surface of zic. Why? c) The maximum kietic eergy KE of a electro emitted by a radiatio of the Lyma series is icluded betwee the values a ad b: a KE b. Determie, i ev, the values of a ad b. d) The maximum kietic eergy of these emitted electros is quatized. Why? 3
Fourth Exercise (7.5 poits) A uclear reactor The breeder reactor Read carefully the followig selectio:..the uclear reactors with fast eutros use uraium 38 or plutoium 39 (or both at the same time) as fuels.the priciple of a breeder reactor is to produce, startig from uraium 38 a amout of fissioable material that is equal or exceeds what the reactor cosumes sice the fial result would be the cosumptio of uraium 38 oly which is more abudat tha the uraium 35.. Give : Speed of light i vacuum: c = 3 8 m/s. Mass of eutro ( ):.87 u. u = 93.5 MeV/c =.66-7 kg ; MeV =.6-3 J. Elemet Tellurium Techetium Molybdeum Plutoium Neptuium 35 Nuclide Te Tc Mo 39 Pu 39 Np 5 43 4 94 93 Mass (u) 34.967.99.93 39.53 39.533 ) Draw from the selectio a idicator showig that producig a equal amout of eergy i a uclear power plat, uraium 38 has a advatage over uraium 35. ) I a breeder reactor, uraium 38 reacts with a fast eutro accordig to the followig reactio: 38 U + A X. (). 9 Z The ucleus A ZX obtaied is radioactive; it is trasformed ito fissioable plutoium accordig to the followig equatios A X e A Z + X +. () X A Z Z a) Idetify A X ad A Z X. e + 39 Pu + (3). 94 Z b) Write dow the global (over all) balaced equatio of the uclear reactio betwee a uraium 38 ucleus ad a eutro leadig to plutoium 39. [This reactio is deoted as reactio (4)]. c) Specify for each of the precedig reactios whether it is spotaeous or provoked. 3) The plutoium 39 94Pu may react with a eutro accordig to the followig reactio : 39 94 Pu + 4 Mo + 35 5 Te + 3. (5) a) Calculate, i MeV/c, the mass defect Δm i this reactio. b) Deduce, i MeV, the amout of eergy E liberated by the fissio of oe plutoium ucleus. c) Calculate, i joules, the eergy liberated by the fissio of oe kilogram of plutoium. 4) We suppose that each fissio reactio produces 3 eutros. Usig the precedig reactios, show that the role of oe of the three eutros agrees with the statemet of the selectio:. produce, startig from uraium 38 a amout of fissioable material that is equal or exceeds what the reactor cosumes 4
وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات هشروع هعيار التصحيح اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة مسابقة في مادة الفيزياء المدة ثالث ساعات االسن: الرقن: الدورة العادية للعام First exercise (7.5 poits) Part of the Q Aswer A. The liear mometum of the system is coserved durig the collisio m V + = (m + m ) V m V = V ; V = 5.48 =.4 m/s m m 6 Mark A..a ME = ½ Mv + ½ kx ; PE g =, t = x =, V = V ME(t = ) = ½ M V = ½.6(.4) =.48 J ; A..b ME = ½ (m + m ) v + ½ kx A..c A.3a) A.3.b.i No frictio ME = costat ½ (m + m ) v + ½ kx = ½ (m + m ) For v =, X m = M k V. V =.6 5 (.4) X m =.6 m = 6. cm. ME =½ (m + m ) v + ½ kx = costat at ay istat. dme (m + m ) vx" + kxv ad v durig oscillatios dt k x" + x =. (m m ) x = X m si( t + ) ; v = x' = X m cos( t + ) ; x" = - X m ). Replacig: - X m si( t + ) + k M X m si( t + ) = si( t + = k M k 5 The proper agular frequecy is = = 6.45 rd/s. M.6 At the istat t =, V =.4 m/s.4 = X m 6.45 cos() ad x = Asi = si = = or But cos() > = rd For =, X m =.4 =.6 m = 6. cm. 6.45 We obtai x (cm) = 6.si(6.45 t) A.3.b.ii The proper period T = t = T /4 =.43 s. 6,45 =,974 s
B. The loss of eergy is E = ME= ½ k( X - m X m) = 3.5-3 J. B. B.3.a B.3.b.i B.3.b.ii E Average power of the forces of frictio: P av = =.5 - W t The frequecy is f which is that of the vibrator, because the oscillator udergoes forced oscillatios It is the pheomeo of resoace of amplitude. T T ad f /T f =.3 Hz Secod exercise (7.5 poits) Part of the Q A. A. Aswer dq i = B du = C C. dt dt du For t T/, u g = E = u R + u C = Ri + u C RC C + u C = E. dt Mark A.3.a du C t t t = A e RC A e + A(- e ). = E dt t A e (RC -) + A E =, at ay time t A = E ad = RC. A.3.b See figure A.63A A.4 For T/ t T, = u R + u C = Ri + u C RC du C + u dt C =. u C = E t e du C dt = -E t e Replacig each quatity i the differetial equatio by its value, we obtai : t t -RC E e + E e = ; which is true, sice = RC. A.5.a The miimum duratio of chargig mode or dischargig mode so that u C reaches its steady state must be 5. A.5.b The miimum value of T must be B..a The curve (3) correspods to the chargig mode of the capacitor sice u C icreases with time. B..b E = 5 V/div div = V. The period T of the square sigal = ms/div div = ms. B..a The period T of the square sigal is ow: ms/div5 div = 5 ms. The duratio of the chargig ad of the dischargig is ow less tha 5. The capacitor has o more time to be completely charged ad discharged. B..b T<<<, the curve becomes liear (straight lie) durig chargig ad
dischargig. Third exercise (7.5 poits) Part of Aswer the Q.a The eergy of a atom ca take oly a certai umber of discrete values.b hc E = hν =. Mark.a hc 3.6 3.6 E E = hν = = - - (- ) = 3.6 ( - ) = 3.6 ( ) hc λ = ( ) 3.6. 34 8 6.6 3 λ= ( ) =.93 7 ( 9 ) m 3.6.6 λ = 9.3 ( ) m.b.i The eergy of a photo beig iversely proportioal to its wavelegth, miimum of eergy correspods to maximum λ = λ trasitio from = 4 λ = 9.3 ( ) = m 4.b.ii maximum of eergy correspods to the largest λ = 9,3 m..b.iii 9.3 λ(m) The Lyma s series spectrum belogs to ultra-violet domai 3.a Is the maximum wavelegth for the photoelectric effect to take place. 3.b The icidet wavelegth is such 9.3 λ(m), it is the smaller tha λ = 7 m, thus we have emissio of electros 3.c hc hc The relatio of Eistei gives: = + KE 3.d KE = hc - (KE) max = b b = hc( hc = hc( ); S S mi S 9 34 8 ) = 6.6 3 ( - 9 9.3 b =.75 J =.453 ev ( E C ) mi = a 34 8 a = hc( ) = 6.6 3 ( max S 9 a =.449 J =.8 ev. hc hc I the relatio KE = - S S 9 ; the values of are discrete the values KE form the a discotiuous successio the kietic eergy of the electros is the quatized - 7 7 9 9 ); ) ; 3
Fourth exercise (7.5 poits) Part of the Q Aswer startig from uraium 38 a amout of fissioable material that is equal or exceeds what the reactor cosumes sice the fial result would be the cosumptio of uraium 38 oly which is more abudat tha the uraium 35.. Mark.a 38 U + A X. () A = 39 ; Z = 9 doc A 39 9 Z ZX est U 9 39 9 U e.b () +() +(3) A + Z X +. () A A = 39 ; Z = 93, doc X est 39 N Z U P e (4) 38 39 9 94 U.c () : Provoked reactio ; () ad (3) : spotaeous; (4) : provoked. 3.a Δm =.86 u = 94.3 MeV/c 3.b E = m c = 94.3 MeV 3.c Mass of a plutoium 39ucleus is: 39 u = 39.665-7 kg = 396.86-7 kg Number of uclei cotaied i kg of plutoium 39 is : 4.5 uclei 7 396.86 93 P The eergy liberated:.5 4 94.3 = 4.9 6 MeV = 7.83 3 J. 4 A eutro iteracts with uraium 38 i order to form aother ucleus of plutoium. This shows that the plutoium uclei are i excess i the populatio of uclei. 4