This exam is formed of three exercises in three pages numbered from 1 to 3. The use of a non-programmable calculator is recommended.

Transcription

1 0 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات امتحانات الشهادة الثانوية العامة الفرع : علوم الحياة الدورة العادية للعام مسابقة في مادة الفيزياء المدة ساعتان االسم: الرقم: This exam is formed of three exercises in three pages numbered from to. The use of a non-programmable calculator is recommended. First exercise: (7 points) Horizontal elastic pendulum The aim of this exercise is to study the effect of the mass on the motion of a horizontal elastic pendulum. This pendulum is formed of: an elastic spring (R), of negligible mass and of stiffness k = 400 N/m,wound around a horizontal rod; a solid (B) considered as a particle of mass m = 00 g. The solid (B) is formed of two particles (B) and (B) stuck together and of respective masses m = 5 g and m = 75 g. The solid (B) can slide, without friction, on the rod (Fig. ). At equilibrium, (B) is at O, taken as an origin of abscissas of the axis xx. (B) is displaced by a distance Xm, from O, in the positive direction, and then released without initial velocity at the instant t0 = 0. The horizontal plane through (B) is taken as a gravitational potential energy reference. At the end of two complete oscillations, (B) is detached from (B) and the system [(R), (B)] continues its oscillations. Figure represents the variation of the abscissa x of the moving solid as a function of time in the two intervals [0, 0. s] and [0.s, 0.5 s]. Take π = 0. A Graphical study Referring to figure, give in each of the intervals [0, 0. s] and [0.s, 0. 5 s]: ) the value of the amplitude of the motion; ) the type of oscillations performed by the oscillator; ) the value of the proper period of the oscillations. B Theoretical study of the oscillations of (B) Consider the system [(R), (B), Earth]. ) Calculate, at t0 = 0, the value of the mechanical energy of the system. ) At an instant t, (B) has an abscissa x and a velocity v of algebraic value v = dx dt. Write, at an instant t, the expression of the mechanical energy of the system in terms of k, m, x and v. ) a) Derive the second order differential equation in x that describes the motion of (B). b) Deduce the expression of the proper period T of the oscillations. c) Calculate the value of T, and then compare it to the result obtained in part (A ). 4) The time equation of motion of (B) is of the form: x = Xm sin( t + ). T Determine the value of the constant.

2 C Theoretical study of the oscillations of (B) Consider the system [(R), (B), Earth]. ) Referring to figure, give the instant at which (B) is detached from (B). ) The mechanical energy of the system [(R), (B), Earth] is equal to that of the system [(R), (B), Earth]. Justify. ) When (B) passes through O, its speed is V and when (B) passes through O, its speed is V. Show that V = V. Second exercise: (7 points) Study of discharging a capacitor A capacitor of capacitance C is initially charged under a voltage E. At t0 = 0, we connect across the terminals of the capacitor a resistor of resistance R= k (Fig.). At an instant t, the armature A carries the charge q > 0 and the circuit carries a current i. A Theoretical study ) Write the relation between i and q. du C ) Show that the differential equation of the voltage uc= uab across the capacitor is u C 0. dt RC ) The solution of this differential equation is uc = D e. Determine the expressions of the constants D and in terms of E, R and C. 4) Show that, after a time t =, the voltage across the capacitor attains 7% of its maximum value E. B Determination of the capacitance C In order to determine the value of C, we use a convenient apparatus, which traces, during the discharging of the capacitor, the curves representing uc = g (t) (Fig.) and n(uc) = f(t) (Fig.) uc (V) n(uc ), (uc in V) t K M t (ms) 0-4 L Fig. - We proceed in the three following methods: Fig. ) First method Referring to the curve of figure : a) give the value of E; b) using the result of part (A 4) determine the value of and deduce the value of C. ) Second method The figure shows also the tangent KL to the curve at point M ( ms, V). a) Referring to this figure determine the slope of the tangent at point M. t(ms)

3 b) Determine the value of C. ) Third method: a) Determine the expression of n (uc) in terms of E, R, C and t. b) Show that the shape of the curve in figure is in agreement with the obtained expression of the function n (uc) = f(t). c) Referring to the curve of figure, determine again the values of E and C. Third exercise: (6 points) Iodine The aim of this exercise is to show evidence of some characteristics of iodine. Iodine ( 5 I ) is radioactive and is a - emitter. Its radioactive period (half-life) is 8 days. Given: Mass of an electron: me = u; MeV =.60 - J; u = 9.5 MeV/c. Element Iodine ( 5I ) 7 Cesium( 55Cs) Xenon( 54 Xe ) Mass of nucleus u u u A Disintegration of iodine ) Write down the equation of the disintegration of iodine and identify the daughter nucleus. ) The disintegration of iodine nucleus, is often, accompanied with the emission of rays. Due to what is this emission? ) Calculate the radioactive constant of iodine in day - and in s -. 4) Show that the energy liberated by the disintegration of one nucleus of iodine is Elib.= J. B Application in medicine During a medical examination of a thyroid gland of a patient, we inject this gland with a solution of iodine. The thyroid of this patient captures from this solution a number N = 0 of iodine nuclei. ) Calculate, in Bq, the activity A corresponding to these N nuclei knowing that A = N. ) Calculate, in J, the energy liberated by the disintegration of these N nuclei. ) Deduce, in J/kg, the value of the dose absorbed by the thyroid gland knowing that its mass is 5 g. C Contamination On the 6 th of April 986, an accident took place in the nuclear power plant of Chernobyl that provoked an explosion in one of the reactors. One of the many radioactive elements that were ejected to the atmosphere is the iodine. This element spread on the ground, absorbed by cows and contaminated their milk and then captured by the thyroid gland of consumers. Every morning, a person drank a certain quantity of milk containing N0 =.60 6 nuclei of iodine. We suppose that all these nuclei were captured by the thyroid of that person, and that the person drank the first quantity at the instant t0 = 0. ) Determine, in terms of N0 and (expressed in day - ), the number of iodine nuclei that remained in the thyroid, at the instant: a) t = day, (just after drinking the nd quantity of milk); b) t = days, (just after drinking the rd quantity of milk). ) Deduce, at the instant t = days just after drinking the 4 th quantity of milk that the number N of the iodine nuclei that remained in the thyroid is: N = N0 ( + e + e + e ) where is expressed in day -. ) Serious troubles in the thyroid gland will take place if the activity of the iodine exceeds Bq. Show that at the instant t, the person was in danger.

4 0 وزارة التربية والتعليم العالي امتحانات الشهادة الثانوية العامة الدورة العادية للعام المديرية العامة للتربية الفرع : علوم الحياة دائرة االمتحانات مشروع معيار التصحيح First exercise: Horizontal elastic pendulum 7 points Part of the Q. Answer Mark A. X m of B is cm and X m of B is cm also A. The oscillations in each part are free un-damped oscillations. A. B. The periods are: T B = 0. s and T B = 0.05 s. M.E is conserved since the amplitude of the oscillations are constant M.E = K.E + P.E e + P.E g = m v + kx + 0, for x = X m,v = 0 thus M.E = P.E max = kx m. M.E = 00 (0.0) = 0.08 J. B. M E= mv + kx M.E = m v + kx. Being constant, its derivative is zero B..a thus: 0 = mvv + kxx v = x 0 and v = x we get x + ( k m )x = 0 B..b B..c B.4 The differential equation is of the form: x" + = k m T = = T = = m k m k = π 0π = 0. s ; x = 0 which is in agreement with the result obtained in part A. The time equation is x = X m sin ( t +φ), T at t 0 = 0, x = X m 4

5 X m = X m sin (φ) sin (φ) = φ = rd; C. B detached at t = 0. s C. C. M.E is the same in the two intervals since the M.E = [P.E el] max = k X, same K and same X m m At O, The M.E = kinetic energy becuase x = 0, the elastic potential energy is zero (P.E el = 0). For B : 0.08 = mv and for B : 0.08 = m V, m = 5 g and m = 00 g = 4m 4m v = m V V = 4 v and V = v. Part of the Q dq A. i = dt Second exercise: Study of discharging a capacitor Answer 7 points Mark dq du A. uc = Ri = R = RC C du ; u C + RC C = 0 dt dt dt du C = D A. dt e t D For t = 0, uc = E=D. t e + RC t D e = 0 A.4 uc = E t e, for t = uc = E e- = 0.7 E RC. B..a E = 8 V B..b uc = 0.7 E =0.7 8 =.96 V V. Graphically, we find for uc = V, t = =ms = 0 - s. =RC=0 C C = 0-6 F. 5

6 B..a slope = C B..b du.5 dt 0.00 = 04.6 V/s The differential equation is: u C = 04.6 = RC 0 C du C u C dt du = 0 C u C RC dt RC C= F. 0 C B..a ln (uc ) = ln(e t e ) ln (uc ) = lne t. RC B..b B..c ln (uc) = f(t) is a function of time: the shape of the curve is a straight line decreasing and not passing through the origin. For t =0, we have: ln (uc) =.08 = lne E = 8 V. And ln (uc ) = 0, for t = ms lne = ln8 thus E = 8 V 0 Now ln8 =.08 = C = F. 0 C Third exercise: Iodine 6 points Part of the Q. A. A Z 0 Answer Mark I X e 6

7 A. A. = ln A.4 conservation of mass number : A = conservation of charge number : 5 = Z - Z = 54 Daughter nucleus is Xenon 54 Xe The daughter nucleus Xe is in excited state and when it drops to the 54 ground state (lower state) emits the ray (photon) T = = days - and = = 0-6 s m = m before - m after = = u m = =0.978 MeV = J B. A= N=0-6 0 =0 5 Bq. B. The liberated energy is : E= = J E.560 B. The absorbed dose is: D= m J/kg After the duration t = day, according to the law of radioactive decay the C..a C..b C. C. t number of remaining nuclei is N =N 0 e = N 0 e ( in day - ) An additional number N 0 is taken when drinks the second quantity next morning. Thus the number of non-decay nuclei is then: N =N 0+ N 0 e = N 0 (+ e ) On the nd day, an additional N 0 from the third quantity and the number remaining from the previous milk is N e : N = N e + N 0= N 0(+ e ) e + N 0= N 0(+ e + e ) On the rd day, an additional N 0 from the fourth quantity and the number remaining from the previous milk is N e. N = N e + N 0= N 0(+ e + e ) e + N 0= N 0(+ e + e + e ) À t= days, the number N of nuclei is : N = N 0 (+ e e e The corresponding activity becomes : ) = nuclei. A = N = =9.7 GBq > 75G Bq At the instant t, the person is thus in danger. 7