اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة هسابقة في هادة الفيسياء الوذة ثالث ساعات

وزارة التربية والتعلين العالي الوذيرية العاهة للتربية دائرة االهتحانات اهتحانات الشهادة الثانىية العاهة الفرع : علىم عاهة هسابقة في هادة الفيسياء الوذة ثالث ساعات االسن: الرقن: دورة العام 5 العادية الخويس حسي راى 5 Ths exa s fred f fur exercses n fur pages. The use f nn-prgraable calculatrs s recended. Frst exercse: (7 pnts) Mechancal scllatr The a f ths exercse s t study the free scllatns f a echancal scllatr. Fr ths a the scllatr s fred f a puck () f (C) () ass =.5 kg fxed t ne end f x' G a assless sprng f unjnted turns and f stffness k = N/; the O ther end f the sprng s attached t Fg. x a fxed supprt (C) fgure. () sldes n a hrzntal ral and ts center f nerta G can ve n a hrzntal axs x'x. t equlbru, G cncdes wth the rgn O f the axs x' x. t an nstant t the pstn f G s defned, n the axs (O, ), by ts abscssa x = OG ; ts velcty v = v where v = x' = dx. The hrzntal plane thrugh G s taken as reference level fr the gravtatnal ptental energy. Theretcal study In ths part we neglect all frces f frctn. ) a) Wrte dwn the expressn f the echancal energy f the syste [(), sprng, Earth] n ters f k,, x and v. b) Derve the dfferental equatn n x that descrbes the tn f G. ) The slutn f ths dfferental equatn s f the fr: x = X sn ( t + φ ) where X and φ are T cnstants and T s the prper perd f the scllatr. a) Deterne the expressn f T n ters f and k. Calculate ts value. b) t t =, G passes thrugh the pnt f abscssa x = c wth a velcty f algebrac value v =. /s, deterne the values f X and φ. x Experental study In ths part, the frctnal frce s gven by f v where μ s a pstve cnstant. n apprprate setup allw t recrd the curve gvng the varatns f x = f(t) (fg ) and thse gvng the varatns f the knetc energy KE(t) f G and f the elastc ptental energy PEe(t) f the sprng (fg 3). ) eferrng t fgure, gve the value f the pseud-perd T f the tn f G.Cpare ts value t that f the prper perd T. ) eferrng t fgures and 3, specfy whch curve r represents PEe(t). X (T) X (T) 3) a) Verfy that the rat a where a s cnstant t be deterned. X () X (T) T b) Knwng that a = e, calculate, n SI unt, the value f μ. ) On fgure 3 tw partcular nstants t and t are lcated.

a) eferrng t fgure 3 ndcate wth justfcatn at whch nstant, t r t, the agntude f the velcty f the puck s : ) axu; ) equal t zer. b) What can yu cnclude abut the frce f frctn at each f the abve nstants? c) Deduce arund whch nstant t r t, the echancal energy decreases by a greater aunt...75 x(c).5.5 t(s) KE, PE e (J) Fg..5 t t.5 Fg.3 t(s) Secnd exercse: (7 pnts) Characterstc f an electrc cpnent In rder t deterne the characterstc f an electrc cpnent (D), we cnnect up the crcut represented n fgure. Ths seres crcut s cpsed f: the cpnent (D), a resstr f resstance =, a cl (L = 5 H; r = ) and an (LFG) f adjustable frequency f antanng acrss ts ternals a snusdal alternatng vltage u = u M. (L, r = ) + N D Frst experent We cnnect an scllscpe s as t dsplay the varatn, as a functn f te, the vltage u M acrss the generatr n the channel (Y ) and the vltage u M acrss the resstr n the channel (Y ). Fr a certan value f f, we bserve the wavefrs f fgure. The adjustents f the scllscpe are: vertcal senstvty: V/dv n the channel (Y );.5 V/dv n the channel (Y ); hrzntal senstvty: s/ dv. M u M Fg. u M ) edraw fgure and shw n t the cnnectns f the scllscpe. ) Usng fgure, deterne: a) the value f f and deduce the value f the angular frequency ω f u M ; b) the axu value U f the vltage u M ; c) the axu value I f the current n the crcut; d) the phase dfference between u M and. Indcate whch ne leads the ther. 3) (D) s a capactr f capactance C. Justfy. ) Gven that: u M = U sn ωt. Wrte dwn the expressn f as a functn f te. Fg.

5) Shw that the expressn f the vltage acrss the capactr s:. u N = cs (t + ) (un n V ; C n F ; t n s) 5C 6) pplyng the law f addtn f vltages and by gvng t a partcular value, deterne the value f C. Secnd experent The effectve vltage acrss the generatr s kept cnstant and we vary the frequency f. We recrd fr each value f f the value f the effectve current I. Fr a partcular value f = f = Hz, we ntce that I adts a axu value. ) Nae the phenenn that takes place n the crcut fr the frequency f = f. ) Deterne agan the value f C. Thrd exercse: (6 pnts) Electrc crcut Study f a (L, C) crcut The (L,C) crcut f fgure, cpsed f a capactr f capactance C and f a cl f nductance L and f neglgble resstance and a swtch k. Intally the arature f C cares a charge Q >. The swtch s clsed at t =. t an nstant t the charge f arature s q and the current n the crcut s. ) a) Indcate the fr f the energy n the crcut at t =. b) Deduce that the value f the current at t = s. ) Usng the prncple f the cnservatn f the electragnetc energy, shw that the dfferental equatn n q has the fr f: q'' + q LC. 3) The slutn f ths dfferental equatn s f the fr q = Q cs( t + ) ; Q, and are cnstants and Q >. a) Deterne. b) Deterne the expressn f Q n ters f f Q and that f n ters f L and C. ) a) Deterne the expressn f as a functn f te. b) Trace the shape f the curve = f(t). Study f a (L, ) seres crcut We cnsder the adjacent crcut. It s fred f an deal DC pwer (L) supply f e..f. E, a resstr f resstance, a swtch k and a cl f nductance L and f zer nternal resstance. Fg. The swtch s clsed at t =. ) pply the law f addtn f vltages, establsh the dfferental equatn n the current. ) Deduce that the value f the current n the steady state s I = E. D P E q L C N F Fg. k D k t 3) The slutn f the dfferental equatn s f the fr: = + e. Deterne the expressn f the cnstants, and n ters f I, and L. ) When k s pened we bserve a spark n k. a) Nae the phenenn respnsble f ths spark. b) T elnate ths spark a neutral capactr s used. Hw shuld t be cnnected? 3

Furth exercse: (7 pnts) Nuclear reactns Gven: ass f a prtn: p =.73 u; ass f a neutrn: n =.87 u; ass f 35 9 9 U nucleus = 35.39 u; ass f 36Kr nucleus = 89.997 u; ass f 35 Z a nucleus =.96 u; lar ass f 9 U = 35 g/le; vgadr's nuber: N = 6. 3 l - ; u = 93.5 MeV/c =.66-7 kg; MeV =.6-3 J. Prvked nuclear reactn s a result f cllsn wth a theral neutrn, a uranu 35 nucleus underges the fllwng reactn: U 9 36 Kr + Z a + y n n + 35 9 ) a) Deterne y and z. b) Indcate the type f ths prvked nuclear reactn. ) Calculate, n MeV, the energy lberated by ths reactn. 3) In fact, 7% f ths energy appears as a knetc energy f all the prduced neutrns. a) Deterne the speed f each neutrn knwng that they have equal knetc energy. b) theral neutrn, that can prvke nuclear fssn, ust have a speed f few k/s; ndcate then the rle f the deratr n a nuclear reactr. ) In a nuclear reactr wth uranu 35, the average energy lberated by the fssn f ne nucleus s 7 MeV. a) Deterne, n jules, the average energy lberated by the fssn f ne kg f uranu 35 U. 9 b) The nuclear pwer f such reactr s MW. Calculate the te Δt needed s that the reactr cnsues ne kg f uranu 35 U. 9 Spntaneus nuclear reactn ) The nucleus kryptn 9 9 36Kr btaned s radactve. It dsntegrates nt zrcnu Zr, by a seres f dsntegratns. a) Deterne the nuber f dsntegratns. b) Specfy, wthut calculatn, whch ne f the tw nucldes 9 9 36Kr and Zr s re stable. ) Uranu 35 9 U s an etter. a) Wrte dwn the equatn f dsntegratn f uranu 35 9 U and dentfy the nucleus prduced. Gven: ctnu 89 c Thru 9 Th Prtactnu 9 Pa b) The reanng nuber f nucle f 35 9 U as a functn f te s gven by: N = N e -λt where N s the nuber f the nucle f 35 U at t 9 = and s the decay cnstant f 35 U. 9 ) Defne the actvty f a radactve saple. ) Wrte the expressn f n ters f, N and te t. c) Derve the expressn f ln() n ters f the ntal actvty, λ and t. ln(), n q d) The adjacent fgure represents the varatn f ln() f a saple f 35 9 U as a functn f te. ) Shw that the shape f the graph, n the adjacent fgure, agrees wth the expressn f ln(). ) Usng the adjacent fgure deterne, n s -, the value f the radactve cnstant λ. ) Deduce the value f the radactve perd T f 35 9 U..9 8.5 t ( 7 s)

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: Frst exercse (7 pnts) Part f the Q..a..b..a..b ME = v + kx nswer N frctnal frces (N nn-cnservatve frces), ME s cnserved dme = x'' + k x =. π x = X sn ( T t + φ ) π x' = X T cs ( π T t + φ ) ; x'' = X π sn ( T T t + φ ) Sub. n the dfferental equatn : π X sn ( T T t + φ ) + k X π sn ( T t + φ ) = T k = T = π k = 3..5 =.993 s s Fr t = : x = c = X sn φ v = c/s = X π cs φ The rat gves : tan φ =.68 φ =.56rd r.58 rad Fr φ =.56rd we get : X = 3.77 c < rejected Fr φ =.58rd we get : X = 3.77 c > accepted Therefre X = 3.77 c and φ =.58 rad. Mark. The graph gves T = s slghtly greater than T =.993 s.. On the graph f fgure, at t =, x s axu therefre PEe curve represents the varatns f PEe..3.a X (T).5 X (T) 7.5 = =.65 ; = =.6 a =.6 X () X (T).5 T.3.b T a = e =.6 = n.6 μ =.55 kg/s...a. t nstant t, the KE s axu therefre v has a axu value. +..a. t nstant t, the KE s equal t zer v has a zer value. +..b t nstant t, f has large apltude (v axu). t nstant t, f s equal t zer (v = )...c rund (t ) the frce f frctn s axu the echancal energy decreases by greater value. + +

Secnd exercse (7 pnts) Part f the Q nswer Mark Y (L, r = ) N + D. M Y..a..b T = 8 s f = 5 Hz. ω = πf = 5π rad/s. U = 3 = 6 V. U ()..c U () =.5 = V I = = -..d rad ; leads u M 8 leads u.3 M (D) s a capctr..5 sn(5t ) ( n and t n s) du = C N u N = C =.sn ( t ) C u N =. cs(5t ) 5C.6. U sn(t) = LI cs(t+ ) cs(5 t ) + sn (t+ ) 5C t = = LI. + 5C C =.6-6 F. Current resnance +. f = LC C =.6-6 F + +.5

Thrd exercse (6 pnts) Parte de la Q. Crrgé Nte..a Electrc energy +..b E ag = = L = t nstant t : E ttal = E elect + E ag = q /C + L = cnstant dq dq Wth = - = - q' and ' = - = - q''. de qq ' L ' C qq ' q L( q ')( q '') q '( Lq '') r q' q '' q C C LC q = Q cs( t + ) q = - Q sn( t + ).3.a.3.b..a q = - Q ( ) dq cs( t + ) ; = - = Q sn( t + ); fr t =, = = sn = ; = r rad; fr t= q = Q > = Q cs, wth Q > cs > =. Q = Q cs Q = Q. eplace q = Q cs t n the dfferental equatn: - Q ( ) cs( t ) + Q LC cs( t ) = dq = - = Q sn( t + ) = Q sn( t) = = LC LC..b t d. u D = u + u D E = + L,. d d E = + L, wth uc = L ; at steady state, = cnstant = I d E = E = I and I =. = +.e -λt t t = : + = = = - d t t t t e E e Le e (L ) E.3 E by dentfcatn : L = and = E = L E..a Self - nductn +..b crss the swtch k + 3

Furth exercse (7 pnts) Part f the Q nswer Mark..a Cnservatn f charge nuber: 9 + = 36 + z + thus z = 56 Cnservatn f ass nuber: 35 + = 9 + + y thus y =..b Fssn nuclear reactn. Δ = [ U + n ] [ Kr + a + n ] =35.39 [89.997 +.96 + 3.87] =.87 u E = Δc = [.87 93.5 Mev/c ] c = 69.53 MeV.3.a 7 69.53 K.E f each neutrn = =.96 MeV =.96.6-3 K.E =.739-3 J K.E = V then V = 3 KE =.739 7.87.66 V =.379 7 /s = 379 k/s..3.b deratr wll help n reducng ther speed s as t prvke re such reactns..a ass N = N lar ass = 35 6. 3 =.567 nucle. E = 7.6-3.567 = 6.97 3 J..b 3 6.97 E = P Δt Δt = = 6.97 5 s = 8 days 8 9..a 9 36 a =..b nn-stable nucleus decays nt a re stable ne thus 9 Zr s re stable 35..a 9 Z = 3 and Z = 9 X s thru..b. The actvty s the nuber f decays per unt te..b. = λn = N e -λt..c ln () = λt + ln( )...d. ln () = λt + ln( ) s a straght lne f negatve slpe cpatble wth the graph...d. λ = slpe f curve = 3. -7 s -,..d. λ = n() T T =.77 5 s = 7 8 years.