This exam is formed of three exercises in three pages. The Use of non-programmable calculators is allowed.

Transcription

1 008 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات امتحانات الشهادة الثانوية العامة الفرع : علوم الحياة مسابقة في مادة الفيزياء المدة ساعتان االسم: الرقم: الدورة اإلستثنائية للعام This exam is formed of three exercises in three pages. The Use of non-programmable calculators is allowed. First exercise (7 points) Mechanical oscillator A spring of un-jointed loops, of stiffness constant k = 0 N/m and of horizontal axis, is fixed from one extremity to a fixed obstacle; the other extremity is attached to a puck M of mass m = 00 g. The center of inertia G of M can slide, without friction, along a horizontal axis xx of origin O and unit vector i. The horizontal plane passing through G is taken as a gravitational potential energy reference. x M G O i V M x At the instant t0 = 0, the puck M, initially at rest at O, is hit with another puck M of mass m = m moving initially with a velocity V = V i (V > 0). After collision, the puck M rebounds on M with a velocity V and the puck M moves with a velocity V 0 = V0 i, and performs oscillations with a constant amplitude Xm = 0 cm. ) Give the sign of V0. ) Let x and v be respectively the algebraic values of the abscissa and the velocity of G at an instant t after the collision. a) Write, in terms of x, m, k and v, the expression of the mechanical energy of the system (M, spring, Earth) at the instant t. b) Derive the differential equation of second order in x that describes the motion of M. c) The solution of this differential equation is of the form x = Asin( 0t + ). Determine the values of the positive constants A, 0 and. d) Deduce that the magnitude of the velocity V0 of M just after the collision is m/s. 3) Knowing that the collision between M and M is supposed to be perfectly elastic, determine: a) the value V of the velocity of M before collision; b) the velocity V of M just after the collision.

2 Second exercise (7 points) Determination of the capacitance of a capacitor A i In order to determine the capacitance C of a capacitor, we connect it in series with a resistor of resistance R = 0 Ω across the terminals of a low frequency generator (G) delivering across its terminals an alternating sinusoidal voltage ug = Um cosωt. The circuit thus constructed carries an alternating sinusoidal current i (Fig). Take =. and 0.3 =. (G) M Figure R C D ) Redraw the circuit of figure () and show the connections of the oscilloscope in order to display the voltages ug = uam across the generator and ur = udm across the resistor. ) Which of the two voltages, ug or ur, represents the image of the current i? Justify your answer. 3) In figure, the waveform () represents the variation of the voltage ug with time. a) Specify, with justification, which of the voltages ug or ur, leads the other. b) Determine the phase difference between the voltages ug and ur. () ) Using the waveforms of figure, determine the angular () frequency ω, the maximum value Um of the voltage ug and the maximum value Im of the current i. Horizontal sensitivity: 5 ms/div. Vertical sensitivity on both channels: V/div. 5) a) Write down the expression of the current i as a function of time t. b) Deduce the expression of the voltage uc = uad across the terminals of the capacitor as a function of C and t. 6) By applying the law of addition of voltages and giving the time t a particular value, determine the value of C. Figure Third exercise (6 points) Interference of light Consider Youngs experiment set-up that is formed of two very thin parallel slits F and F, separated by a distance a = mm, and a screen of observation (E) placed parallel to the plane of the slits at a distance D = m from the mid point I of FF and a thin slit F, equidistant from F and F, situated on the straight line (Δ) whose intersection with (E) is the point O. The object of this exercise is to study the interference pattern observed on the screen (E) in different situations. (Δ) F F I F (E) M O A First situation The slit F is illuminated with a monochromatic light of wavelength = 0.6 m in air. ) Describe the interference pattern observed on (E). ) Consider a point M on the screen at a distance d from F and d from F. Specify the nature of the fringe thus formed at point M in each of the following cases:

3 a) d d = 0 ; b) d d =.8 m; c) d d = 0.96 m. 3) F is moved along (Δ). We observe that the interference fringes remain in their positions. Explain why. ) F is moved perpendicularly to (Δ) to the side of F. We observe that the central fringe is displaced. In which direction and why? B Second situation Now the slit F is illuminated with white light. ) We observe at point O a white fringe. Justify. ) Specify the color of the bright fringe that is the nearest to the central fringe. C Third situation Consider two lamps (L) and (L) emitting radiations of same wavelength, we illuminate F by (L) and F by (L), we observe that the system of interference fringes does not appear on the screen (E). Why? 3

4 008 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات مشروع معيار التصحيح امتحانات الشهادة الثانوية العامة الفرع : علوم الحياة مسابقة في مادة الفيزياء المدة ساعتان االسم: الرقم: الدورة اإلستثنائية للعام First exercise (7 points) Part of the Q Answer Mark V0 < a 0.50 Mechanical energy: ME = PE + KE = k x m V.b Without friction Conservation of mechanical energy ME = k x m V = constant..00 By deriving with respect to time: dem dt kxx mvv 0 ; k x + x 0. m.c x = A sin( 0t + ) ; x = A 0 cos( 0t + ) and x = -A 0 sin( 0t + ) By replacing in the differential equation:.50 A 0 sin( 0t + ) + k m Asin( 0t + ) = 0 0 = k m = 0 0. = 00, 0 = 0 rd/s. For t0 = 0, x = Asin( ) = 0, then = 0 or and v = A 0 cos( ) = V0 < 0 ; as A > 0, then cos < 0 = rad and A = +0 cm..d v = x = - 0A cos( 0t) ; at t0 = 0, v = V0 = - 0xm = - m/s. 3 Conservation of linear momentum: P i P m V mv mv.00 0 In algebraic values: V = V +V0. (I) Elastic collision Conservation of KE: m(v - V ) mv 0 (II) (II) (I) V + V = V0 f m V m V mv 3 Substituting in(i) we obtain: V V0 =.5 V0 = -.5 m/s. V = V0 - V = - ( -,5) = 0.5 m/s V = 0.,5 i 0.00

5 Second exercise (7 points) Part of the Q Answer 0.5 Mark ur = Ri = ct i ur is the image of i a ur leads ug, because in this circuit the current always leads the voltage across the generator.(ur attains the maximum before). 3.b T div. π 0.5div=> rad. π π T = div 5 ms/div = 0 ms ω 00π rad/s. T 0.0 Um = div V/div = V. (UR)m =.8 div V/div =.8 V = V = R Im 5 a) Im = = 0. A. 0 π π 0.5 i = Imcos ( t + ) = 0. cos ( t + ) b) 6 dq du i = = C C 0. uc = primitive of i = dt dt C 00 C π ug = uc + ur ; ur = cos ( t + ) π sin( t + ) cos ωt = 00 C For t = 0, we have : = C = 0-6 F = F. π π sin( t + ) + cos ( t + ) C + 5

6 Third exercise (6 points) Part of the Q Answer Mark A. - Fringes are parallel to the slits - Fringes are alternately bright and dark - Fringes are equidistant A..a d-d = 0 = k with k = 0 ; M is a bright central fringe d-d =.8 m = k with k = ;M is a bright fringe of order. d-d = 0.96 m = (k + ) / with k = ; M is a dark fringe of order A.3 ax FF remains equal to FF, the optical path difference does not vary D thus the interfringe i does not vary. A. FF > FF ; the optical path FF M increases. To locate the central bright fringe O, we must have FFO = FFO, the optical path FO must increase the central fringe is displaced to the side of F. B. We see at O a white fringe since all the bright fringes corresponding to 0.5 different colors superpose at O. B. D x = k ; for k =, x is the smallest value corresponding to the a smallest wavelength we observe a violet bright fringe. C No, since the two sources are not coherent. 0.5 A..b A..c 6