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93 AL Physics/Essay Marking Scheme/P.1 PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE 93’ AL Physics: Essay Marking Scheme 1. (a) Steady flow - liquid elements which start at a given point always follow the same path and have the same velocity at each point on the path. 1 ½ Turbulent flow - liquid elements which start at a given point take random paths and their velocities vary in magnitude and direction 1 ½ 3 (b) Consider a cross-section of the pipe, the liquid layer touching the pipe wall is always stationary due to adhesive force between the liquid molecules and pipe wall. The velocity of liquid is greatest at the centre. Internal friction exists between liquid layers with different velocities because of intermolecular forces so velocity falls off gradually as the pipe wall is approached ½ ½ ½ ½ ½ ½ 3 (c) Liquid in contact with the bottom surface of the block moves at the block’s velocity v Liquid in contact with the floor is stationary. v Thus a velocity gradient is set up in the liquid of thickness t. t So, force required = liquid frictional force = Area Velocity gradient Av = t Assumptions : - uniform thickness of liquid - the liquid is Newtonian ANY - constant velocity gradient across the thickness of liquid THREE - only the liquid lying below the block moves (ignore edge effect) (d) Marker A falling ball-bearing long glass tube containing liquid Marker B ½ ½ ½ ½ ½ ½ ½ ½ ½ 4 93 AL Physics/Essay Marking Scheme/P.2 - The diameter of the tube is large compared with the diameters of ball-bearings so that streamline conditions are satisfied. - Marker A is far enough below the liquid surface for the ball-bearing to have its terminal velocity at A - Dip the ball-bearing in the liquid and thereby coated, before dropping so as to reduce the chance of air bubbles adhering to the falling ball-bearing - Avoid using ball-bearings of large radii as their terminal velocities are high and vortices may form - Release the ball-bearing at the centre of the tube to reduce the effect of the wall of the tube on the streamlines - Marker B is at a considerable distance from the bottom of the tube so as to reduce the effect of the bottom of the tube on the streamlines ANY FOUR 2. (a) progressive wave - waveform advances as time goes on - energy is transmitted along the direction of travel of the wave - particles within one wavelength have ANY different phases FOUR - all particles are vibrating - all vibrating particles have the same amplitude stationary waves - waveform does not advance - energy is confined within the region of the stationary wave - all particles between two adjacent nodes are in phase - some particles (at nodes) have no vibration - different particles have different amplitudes, in particular, amplitude is maximum at anti-nodes A stationary wave is formed when there is superposition of two waves of nearly equal amplitude and equal frequency travelling in opposite directions 1 ½ 1 ½ 1 ½ 1 ½ 1 ½ 1 ½ ½+½ ½+½ ½+½ ½+½ ½+½ ½ ½ ½ ½ (b) (i) At T1, the waves arrive in phase to produce a loud sound Phase difference then increases between the waves due to different frequencies. At T2, the waves are completely out of phase, little or no sound is heard. Later at T3, the waves are in phase again and a loud note is heard. ½ ½ ½ ½ y1 T2 T1 T3 time ½ y2 T1 T2 T3 time Variation of amplitude y3 resultant T2 1 T1 T3 T Beats (not to scale) time 6 6 93 AL Physics/Essay Marking Scheme/P.3 (ii) Suppose the beat period = T, then in time T number of cycles of f1 = f1T number of cycles of f2 = f2T Assume f1 greater than f2, then f1T - f2T = 1 1 f1 - f2 = T 1 beat frequency = = f1 - f2 T 3. ½ ½ ½ 1 ½ ½ 3½ (c) Radar installed near the road sends microwaves of frequency f1 to a travelling car, then the microwaves are reflected back to the radar. Due to Doppler effect, the observed frequency f2 of the reflected microwaves is slightly different from f1. Hence by comparing the transmitted and reflected microwaves, beats are formed. As the beat frequency (= f1 - f2) depends on the car speed, the car speed can be checked. ½ ½ ½ ½ ½ ½ 3 (a) (i) average drift velocity: ~10-4 m/s (ii) speed of electrical signal: ~ 108 m/s In a current-carrying conductor, electrons tend to accelerate along the opposite direction of the electric field inside. Due to collisions between electrons and atoms (ions) of the conductor, electrons move in zig-zag paths and drift with small displacement in unit time. ½ ½ ½ ½ ½ ½ electric field direction with electric field e- without electric field small displacement 1 Electric field travels at an extremely high speed in a circuit so electrons at every point of the circuit are influenced by the electric field nearly simultaneously as the switch is closed, electrical signal results at once. (b) v = average drift velocity e = electronic charge A = area of cross-section of the wire n = no. of conduction electrons per unit volume ½ mark for ANY TWO of them 1 ½ ½ 5 1 Suppose electrons drift a distance l along a wire in time t, the charge q flows through a cross-section of the wire is q = nlAe ½ Drift velocity v = l/t ½ current in the wire i = q/t = (nlAe)/(l/v) ½ = nAve 3 93 AL Physics/Essay Marking Scheme/P.4 (c) When a current flows through the meter, the coil experiences a couple, the coil then turns until it is stopped by the increasing tension in the springs. Thus the larger the current through the meter, the greater the forces on the coils, and a greater angular deflection results. To achieve a linear scale: Set up a radial magnetic field then flux density B of the field is nearly constant at the coil The coil stops rotating when coil = spring NBAI = k ( = angular deflection) hence I (N, B, A & k are constants) 2½ 2½ ½ ½ ½ ½ ½ 1 ½ ½ ½ ½ 8 93 AL Physics/Essay Marking Scheme/P.5 4. 1 2 mvm : maximum kinetic energy of photoelectrons 2 h : energy of incident photon : work function of metal - the work required to remove an electron from the metal surface (a) ½ ½ ½ ½ 2 ½ +½ (b) (i) Current i - Vs 0 p.d. V Photoelectrons emerge with different speeds (or K.E.). When V is positive, current is constant because all photoelectrons can reach electrode D Current falls when V is negative because the less energetic photoelectrons cannot overcome the potential barrier. When V reaches the stopping potential (V = -Vs) even the most energetic photoelectrons are repelled back so no current flows. ½ ½ ½ ½ ½ 3½ (ii) Current i increased intensity (I) ½+½ original curve ½+½ increased frequency (II) - Vs 0 p.d. V (I) increased light intensity constant frequency same maximum K.E. Vs remains unchanged intensity increased no. of photons increases no. of photoelectrons ejected increases current increases (II) increased light frequency frequency increased energy of each photon increases maximum K.E. of photoelectrons increases magnitude of Vs increases For the same intensity, if energy of each photon increases no. of photons decreases no. of photoelectrons ejected decreases current decreases ½ ½ ½ 3½ ½ ½ ½ ½ 2 93 AL Physics/Essay Marking Scheme/P.6 (c) Light is focused on the ‘soundtrack’ of a moving film. The ‘soundtrack’ varies the intensity of light falling on a photocell. The photocell creates a varying current, thus produces a voltage which is amplified for driving a loudspeaker. ½ ½ ½ ½ 2 (d) screen diffraction pattern carbon film + 4000 V 1 electron diffraction A beam of electrons strikes a thin film of graphite just beyond the anode. A diffraction pattern, consisting of concentric rings, is observed on the fluorescent screen showing the wave-like behaviour of electrons. 5. ½ 1 ½ 3 (a) (i) a.c. supply L, R C a.c. supply mA OR L R C mA LRC in series a.c. supply & milliammeter Set up the above circuit, set the signal generator output to a value, say 3 V, with a measurable current, and increase the frequency stepwise from a low value, say 10 Hz, check whether the output is constant at the previous setting, 3 V, then record the corresponding current readings on the a.c. milliammeter, when frequency increases, the current reading rises and then drops. ½ ½ ½ ½ ½ ½ 93 AL Physics/Essay Marking Scheme/P.7 (ii) (iii) smaller resistance (I) I EACH curve 1/2 mark larger resistance (II) f f0 f0 = resonant frequency current I and voltage across resistor VR are in phase, I = VR/R ½ mark for V0 lags behind VR0 ½ mark VL0 for VC0 > VL0 VR0 ½ ½ ½ For frequency < f0, V lags behind VR, so VR0 < V0 V V current I0 = R0 0 = Imax R R ½ V0 VC0 VL0 ½ mark for V0 , VR0 in phase ½ mark VR0 for VC0 = VL0 V0 For frequency = f0 V and VR in phase, so VR0 = V0 ½ V current I0 = Imax = 0 R ½ For frequency > f0, V leads VR, so VR0 < V0 V V current I0 = R0 0 = Imax R R ½ VC0 VL0 ½ mark for V0 leads VR0 ½ mark V0 for VC0 < VL0 VC0 VR0 11 93 AL Physics/Essay Marking Scheme/P.8 (b) (i) Radio signals from different transmitting stations induce e.m.f.s of various frequencies in the aerial, which cause currents flowing in the aerial coil. Then currents of the same frequencies are induced in coil L by mutual induction. If C is adjusted so that the resonant frequency of the LCR circuit equals the frequency of the wanted station, a large current and p.d. at that frequency only will develop across C This selected and amplified p.d. is then applied to the next stage of the receiver. (ii) Use an inductor of high L/R ratio so increases the resonant current and hence the voltage across C. 6. ½ ½ ½ ½ ½ ½ ½ 1 ½ 5 (a) (i) (ii) Umax P.E. K.E. P.E. Curve - shape correct - max. at xc - A and xc + A ½ ½ K.E. Curve - shape correct - zero at xc - A and xc + A ½ ½ xc marked ½ x xc - A xc xc + A In S.H.M., the total mechanical energy E is conserved At maximum displacement, K.E. = 0, P.E. = Umax Total mechanical energy of the system E = K.E. + P.E. = Umax (constant) so K.E. = Umax - P.E. for xc - A x xc + A (b) (i) mass of each spring element, dM = = 1 dl l2 M v2 2 2 L L Mv 2 l 2 dl 2L3 L Mv 2 Total kinetic energy of the spring = l 2 dl 0 2L3 Mv 2 1 3 = L 2L3 3 = = l L 1 ½ ½ ½ ½ Mv 2 6 (ii) Extension of spring, y = x - xc Total energy = K.E. of spring + K.E. of block + P.E. of spring Mv 2 mv 2 ky 2 6 2 2 dE M dv dv dy v mv ky dt 3 dt dt dt 4 1 kinetic energy of this spring element = ½ (dM) (V1)2 E= ½ dl M L speed of spring element at a distance l from the fixed end, V1 = v ½ ½ 4 ½ 1 ½ 93 AL Physics/Essay Marking Scheme/P.9 dx dy dv d 2 x d 2 y ; dt dt dt dt 2 dt 2 dE and =0 dt M d2y ( m) 2 ky 0 3 dt v or d2y dt 2 For S.H.M. and k ( M m) 3 d2y dt 2 k ½ ½ 1 y 2 y M ( m) 3 2 T = M ( m) = 2 3 k (c) Damped oscillation takes place, energy of the system decreases, amplitude of oscillation gradually decreases to zero, the frequency is smaller and the period is longer ½ ½ ½ ½ 6 ½ ½ ½ ½ 2