اهتحانات الشهادة الثانىية العاهة الفرع: علىم الحياة هسابقة في هادة الفيسياء الرقن:

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1 وزارة التربية والتعلين العالي الوذيرية العاهة للتربية دائرة االهتحانات الرسوية اهتحانات الشهادة الثانىية العاهة الفرع: علىم الحياة دورة العام 70 العادية الخويس 0 حسيراى 70 االسن: هسابقة في هادة الفيسياء الرقن: الوذة: ساعتاى This exam is formed of three exercises in four pages. The use of non-programmable calculator is recommended. Exercise : (6 points) Mechanical oscillator onsider a mechanical oscillator formed of a massless spring of stiffness k and a solid (S) of mass m = 0.4 kg. The aim of this exercise is to determine the stiffness k of the spring by (S) two different methods. For this aim, the spring is placed horizontally, G fixed from one of its extremities to a fixed support and (S) is attached to x O x the other extremity. (S) may slide without friction on a horizontal rail A B AB and its center of inertia G can move along a horizontal axis x'x. At equilibrium, G coincides with the origin O of the axis x'x (Doc. ). Doc. At the instant t 0 = 0, G is at rest at O, (S) is launched with an initial velocity in the positive direction along x'x. Thus, (S) performs mechanical oscillations. x (cm) At an instant t, the abscissa of G is x = OG and the 4 dx algebraic value of its velocity is v =. The horizontal plane passing through G is considered as the reference level for gravitational potential energy. Take = 0. First method t (s) A convenient apparatus is used to trace the curve of the abscissa x as a function of time (Doc. ). -) Referring to the graph of document, indicate: --) the type of the oscillations of (S). Justify. --) the value of the amplitude X m of the oscillations; 4 --) the value of the proper period T0 of the Doc. oscillations. -) Indicate the nature of the motion of G and choose, from table below, the differential equation in x which describes the motion of G. Equation Equation Equation x' m k x = 0 x'' m k x = 0 x'' m k x' = 0 -) Determine the value of the stiffness k of the spring. / 4

2 Second method -) The mechanical energy of the system [(S), spring, Earth] is conserved. Why? -) The expression of the kinetic energy of (S) can be written in the form: KE = A k x, where A is constant. What does A represent? Justify. -) A convenient apparatus is used to trace the curve of the kinetic energy of (S) as a function of x (Doc. ). Using the graph of document, determine: --) the value of A; --) the value of the amplitude X m of the oscillations; --) the value of the stiffness k KE ( J) Doc. x ( m ) Exercise : (7 points) harging and discharging of a capacitor The aim of this exercise is to determine the capacitance of a capacitor by two different methods. onsider the circuit represented in document. It is formed of an ideal generator that maintains across its terminals a constant voltage of value E, a capacitor of capacitance, two resistors of resistances R = 0 k, R = 0 k and a double switch K. harging the capacitor The capacitor is initially neutral. At the instant t 0 = 0, we put K in position (); the charging phenomenon of the capacitor starts. -) Theoretical study --) Show that the differential equation that describes the E () K q A () D Doc. R R B variation of the voltage u = u BD across the capacitor has the form: E = R --) The solution of this differential equation has the form: u = A( e ). Determine the expressions of the constants A and in terms of E, R and. --) Dece that u = E at the end of charging of the capacitor.. u / 4

3 -) Experimental study In order to determine the value of, we use a n( E u convenient apparatus, which traces, ring the ) ; [E & u in V] charging of the capacitor, the curve representing ln(e u ) = f(t) (Doc.). [ln is the natural logarithm] --) Determine, using the solution of the obtained differential equation, the expression of ln (E u ) in terms of E, R, and t. --) Show that the shape of the curve in document is in agreement with the obtained expression of ln (E u ) = f(t) ) Using the curve of document, determine the values of E and. Doc. Discharging the capacitor The capacitor being fully charged. At an instant taken as a new origin of time t 0 = 0, the switch K is placed at position (); thus the phenomenon of discharging of the capacitor starts (Doc. ). -) Theoretical study --) Show that the differential equation in the voltage u = u BD across the capacitor has the form: u = 0; where is a constant to be determined in terms of R, R and. q --) The solution of this differential equation has the form: u = E where is constant. Show that =. -) Experimental study The variation of the voltage u across the capacitor as a function of time is represented in document 4. u (V) --) Determine, using document 4, the value 0 of the time constant of the discharging circuit. --) Dece the value of e K D A () R R B Doc. i t (s) Doc. 4 t (s) / 4

4 Exercise (7 points) The radioactive isotope phosphorus The radioactive isotope phosphorus ( 5 P ) is used in the diagnosing of cancer. Phosphorus, is injected into the human body, it decays and gives radiations. These radiations are detected by an appropriate device to create the image of the inside of the human body. The aim of this exercise is to determine the dose of radiation absorbed by a tissue of a patient ring 6 days. Phosphorus ( 5 P ) is a emitter; it disintegrates to give an isotope A ZS of sulfur. Given: mass of 5 P : u; mass of A Z S :.96 9 u; 4 mass of electron : u ; The radioactive period of 5 P : 4. days; u = 9.5 MeV/ ; MeV =.6 0 J. Energy liberated by the decay of phosphorus The disintegration of phosphorus nucleus is given by the following reaction: 5P A Z S 0 e -) Determine A and Z. -) Prove that the energy liberated by the above disintegration is E lib =.706 MeV. -) The sulfur nucleus is proced in the ground state. The emitted antineutrino carries energy of.0 MeV. --) The above disintegration of phosphorus is not accompanied with the emission of gamma rays. Why? --) alculate the kinetic energy carried by the emitted electron knowing that phosphorus and sulfur are considered at rest. Absorbed dose A patient is injected by a pharmaceutical proct containing phosphorus. The initial activity of phosphorus in the pharmaceutical proct at t 0 = 0, is A 0 =.60 6 Bq. -) alculate, in s -, the radioactive constant of phosphorus. -) Dece the number N o of nuclei of phosphorus present in the pharmaceutical proct at t 0 = 0. -) --) Determine the remaining number N of nuclei of phosphorus at t = 6 days. --) Dece the disintegrated number N d of nuclei of phosphorus ring the 6 days. --) The number of the emitted electrons is N = 6. 0 electrons ring the 6 days. Why? e -4) The emitted radiation is absorbed by a tissue of mass M = g. The antineutrino does not interact with matter, and suppose that the energy of the emitted electrons is completely absorbed by the tissue. -4-) alculate the energy E abs absorbed by the tissue ring the 6 days. -4-) The absorbed dose by the tissue is D = 0 0 E abs ring the 6 days. Dece the value of D in J/kg. M 4 / 4

5 وزارة التربية والتعلين العالي الوذيرية العاهة للتربية دائرة االهتحانات اهتحانات الشهادة الثانىية العاهة الفرع: علىم الحياة هسابقة في هادة الفيسياء الوذة: ساعتاى دورة العام 70 العادية الخويس 0 حسيراى 70 أسس التصحيح Exercise : (6 points) Mechanical oscillator Part Solution Mark - -- Free un-damped oscillation Since the amplitude is constant X m = 4 cm. -- T 0 = 0.8 s Nature simple harmonic motion or RSM Equation. The differential equation has the form : ; ; 5 N/m The mechanical energy of (S) is conserved e to the absence of friction Or : X m = constant Or: The work done by the non conservative forces is zero ME= KE PE e ; KE= ME - ½k (x) ; A is mechanical energy For x = 0 ; KE = ME = A = 0.0 J - -- KE = 0 ; x = X m, from the graph X m = 6 cm, then X m = 4 cm Slope = Or : choose a point on the graph for x = X m ; KE = 0J, therefore k = 5 N/m ; -.5 = - ½ k, then k = 5 N/m 0.75 /

6 Exercise : (7 points) harging and discharging of a capacitor Partie Solution Note -- u AD = u AB u BD, then E = R i u with i = we get : E = R u. -- A = e, replacing in the differential equation we get: A E = R e A( e ), then A = E and = R -- At the end of charging, t, then Or for t = 5 ; uc = 0.99 E = E e 0, thus u = E t u = E( e ) ; uc = E -E e ; E u c = E e ; ln (E u ) = ln (E e ) t ln (E u ) = n E R ln (E u ) has the form of y = at b its slope a < 0 ; Is in agreement with the shape of the curve which is a straight line of negative slope not passing through the origin The slope of this straight line is = = 50, then = 50 R 0.0 R = 0-6 F = μf and ln E =, thus E = 0 V Or : For t = 0, then ln (E u ) = ; = ln E, thus E = 0 V For ln (E u ) = 0, so t = 0.06 s, thertefore = 0-6 F. -- u = (R R ) i with i =, we get : u (R R ) u = 0, then = (R R ). = e Replacing u by u = E e E E ( e in the differential equation we get: ) = 0, therefore = For uc = 7.4 V, then t = 0.06 s ; = 0, thus = s Or: From the graph at t = 0.06 s, u = 7.4 = 0.7 0, so = 0.06 s..- = (R R ), then = 0-6 F = μf / 0.5

7 Exercise : (7 points) The radioactive isotope phosphore Part Solution Mark ν. By applying Soddy s laws: = A 0 0, Then A = ; 5 = Z 0, then Z = 6. m = = ( ) =.864 u m = MeV/c.706 MeV/c. = m. c =.7. c, then =.706 MeV Gamma rays are not emitted in the above decay since the daughter nucleus (sulfur) is proced in the ground state. =, so.706 =.0, therefore = MeV λ = =, therefore λ = = λ, =, therefore =.44 nuclei n = = = 0.495, N = =, therefore N = nuclei - -- = N =.44 -, therefore = 6. nuclei -- One electron is emitted in one decay of phosphorous-, so = Therefore, = = = J So = J -4- D = =, therefore D = 0.6 Gy = 0.6 J/kg. /

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

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