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87 AL Physics/Essay Marking Scheme/P.1 PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE 87’ AL Physics: Essay Marking Scheme 1. (a) (i) The rate of change of momentum produced in a body is proportional to the resultant force acting on it and occurs in the direction of the force. ½ ½ (ii) The body will move with an acceleration a in the direction of the resultant force F and a = F/m, where m is the body mass. ½ ½ (b) (i) The mass m of a body is constant and is a measure of its inertia to any enforced change of state, either stationary or moving. For a force F the body experiences an acceleration 1/mass. ½ (ii) The weight W of a body can be variable and is the force which would act if the body were allowed to fall freely under the influence of gravity. W = mg, where g is the local 'free-fall' acceleration due to gravity. ½ (iii) In a lift moving downwards with an acceleration (or for a lift moving up, suddenly stops), there will be less reaction between the weight (force) of the passenger and the floor of the lift, giving the impression of weightlessness. 2 ½ ½ ½ ½ 1 1 5 (c) P F v C A constant (centripetal) force F acts towards the center O of the circle. This provides an acceleration in this direction for the rotating body. Only the speed v of body is constant, its velocity (and momentum) are continuously changing - due to the accelerating force F. ½ ½ 1 2 87 AL Physics/Essay Marking Scheme/P.2 (d) l glass tube paper marker rubber bung screw nuts 1 The glass tube is held vertically, the bung is whirled around above his head by one student and the speed of bung is increased until the marker is just below tube. Another student times, say, 50 revolutions of the bung. By moving marker the length l of the string can be varied and the relation between l and the angular velocity () obtained. Since T (tension provided by screwnuts*) and mg are constant from T cos = mg it is clear that is constant. Also T sin = mr², 1/r. 1 ½ 1 ½ 1 1 l r T mr2 mg *However there is friction between string and glass rod which may vary throughout experiment. 2. (a) (i) In a gas the molecules move randomly with high speeds throughout all available space and are comparatively far apart (~ 10 molecular diameter). Thus normally the forces between molecules are very small - except that on collision with one another large repulsive forces will occur. Molecules will give rise to a pressure (force) on the sides of a container due to impact of the moving molecules. When there is an increase in temperature the average kinetic energy of the molecules will increase and (1) in free space gas will expand with molecules moving into cooler gas or (2) if contained there will be an increase in pressure on the sides of the container. 1 7 1 1 ½ ½ ½ ½ 4 87 AL Physics/Essay Marking Scheme/P.3 (ii) Resultant force Repulsion Force Separation of atoms r0 Attraction 1 Most solids are crystalline with molecules formed of atoms at ‘fixed’ separations in a lattice. The separations r0 are very close (< 1 molecular diameter) and if atom nearer a repulsive force predominates while farther away force is predominantly attractive. The atoms at any particular temperature vibrate about this mean separation r0. At higher temperatures there is asymmetrical vibration with the displacement greater on the extension side - hence solid expands. 1 1 1 4 (b) (i) Ionic Bonding Cl ions Na ½ e.g. NaCl consists of +ve Na ions and -ve Cl ions formed because an electron is easily transferred from the Na to the Cl atom. The ions are held together in the lattice by attractive electrostatic forces. ½ ½ 1½ (ii) Metallic Binding Electron sea Positive metal ion nucleus plus inner electrons (Optional) Electron Metal atoms have ‘outer’ electrons which are loosely held and can become detached, drifting around randomly. All atoms share the ‘free’ electrons. The electrostatic attraction ½ ½ 1 87 AL Physics/Essay Marking Scheme/P.4 constitutes the metallic bond. (iii) Covalent Binding C C C Carbon atom Outer electron Covalent bonds C C (optional) ½ C Electron sharing occurs between two or more atoms. e.g. 4 ‘outer’ electrons of C can be shared with 4 other C atoms to make 4 bonds - through interlocking electron ‘clouds’. Bonds are very strong. ½ ½ 1½ (c) (Optional) The glass fibres are strongly bound by covalent binding forces and are difficult to break. They are imbedded in a weaker yielding plastic material (matrix). The matrix (1) binds and holds the fibres together so that any applied load is transmitted to them, (2) it protects the surface of the fibres from scratches/and possibility of cracks/breaks occurring) and (3) if cracks do appear it should prevent them spreading from fibre to fibre - crack deflected along interface. Such a material combines great strength with low weight. 3. 1 1 1 1 4 (a) (i) Light waves E v B E v 1 B These are a particular electromagnetic wave of high (1014 Hz) frequency. They consist of an electric field E and magnetic field B component which undergo time variations in planes at right angles to each other and the direction of propagation. The changing magnetic field produces the changing electric field (and vice-versa). Waves do not require any medium in which to propagate with the relatively high speed of 3 108 m/s. ½ ½ 1 1 ½ ½ 5 87 AL Physics/Essay Marking Scheme/P.5 (ii) Sound waves Displacement (right) C R C R Distance 0 (b) (left) C R C Normal air pressure when wave absent R Pressure Distance (a) 2 These are of much lower frequency (50 - 2 104 Hz) and are due to momentum transfer of energy from pockets of air vibrating with S.H.M. about their mean locations. There are associated pressure changes of rarefactions R and compressions C. Waves propagate outwards from vibrating source and need a medium for propagation. Speed is much lower ~ 330 m/s in air. ½ ½ ½ ½ ½ ½ 5 (b) (i) Refraction NIGHT DAY v warmer air T cool air OR cool air warmer air v2 > v1 ground v1 > v2 1 ground Refraction is a bending of the propagation path of the waves due to change of speed experienced when either moving into another medium - or in the example given, due to a change of temperature. Snell’s Law of refraction holds good sin i / sin r = 1n2 = v1/v2. ½ ½ 2 (ii) Diffraction P obstacle 1 87 AL Physics/Essay Marking Scheme/P.6 Diffraction is a bending of the waves around the edge of obstacles. there are centres of disturbance on the wave front in plane of obstacle, such as P, and these send out waves (spherical) as shown propagating behind obstacle. (c) For observable diffraction the wavelength of the waves has to be comparable with the dimension of the opening. Briefly for a ‘slit’ maxima occur due to re-inforcement of waves from secondary sources at an angle to normal of slit when sin = /a. ‘a’ being width of slit. For sound ~ 1 m while for light ~ 6 10-7 m. Thus very narrow slits needed. 4. ½ ½ 2 ½ ½ ½ ½ 2 (a) Electromagnetic Induction takes place when (i) there is a relative motion between a magnetic field and an object which cuts the magnetic field and (ii) there is a magnetic flux change e.g. through a coil in a circuit, due to the changing current. electric field, or e.m.f. E is induced, E = -d/dt d/dt is the rate of change of magnetic flux. Direction of E is so as to oppose the magnetic field change (drive current/produce magnetic field for a conductor). (b) (i) ½ ½ ½+½ ½ ½ 3 L/R L R VL VR E S On closing switch S a back e.m.f. VL is produced across L due to changing I and magnetic field through coil. At first VL E (slightly less since a current has to flow) and there is a large rate of change of current dI/dt, though I is small. dI/dt reduces with time and also VL - leading to an increase in I. Current I is a maximum when dI/dt and the back e.m.f. VL = 0. ½ 1 1 1 I E/R ½ t 4 87 AL Physics/Essay Marking Scheme/P.7 (ii) C/R C R VC VR E S On closing switch S a back e.m.f. VC is set up across C due to the flow of charge Q to C from the battery. At first Q 0 and dQ/dt is large (no back e.m.f.). The back e.m.f. VC increases with time as the charge builds up on C and dQ/dt reduces. Charge Q becomes constant (maximum) when VC = E and I = dQ/dt = 0 ½ 1 1 Theory VC = Q/C and I = dQ/dt = (E - VC)/R Q 1 dQ , where final charge Q0 = EC dt 0 ( Q0 Q ) CR ½ Integrating we obtain Q = Q0 (1 e t / RC ) 2 Q EC t (c) Charge max C L Charge max (i) (ii) Current max Charge max (iii) (iv) Current max Q (v) (not expected; for reference only) charge I time EM energy: electrical (E) only ME magnetic (M) only E only current ME EM M only The p.d. across the capacitor C. VC = Q/C and so voltage observed in C.R.O. will be of same wave form as that of Q. E only ½ 87 AL Physics/Essay Marking Scheme/P.8 (i) C, initially, has max. charge Q - ENERGY IN ELECTRICAL FIELD and I = 0. Current I starts to flow due to p.d. across C and back e.m.f. set up across L (greatest at first, and also dI/dt). 1 (ii) Q = 0, all charge transferred from one plate of C to the other. I is a maximum (no back e.m.f. VC = Q/C) - ENERGY IN MAGNETIC FIELD. 1 (iii) Q increases so that polarity of C (plates) is reversed I = 0 when back e.m.f. VL = VC ENERGY IN ELECTRICAL FIELD, Cycle continues. 1 (Deduct 1 mark if energy drawn not correct.) N.B. Due to dissipation of power in the resistance of the circuit oscillations will decay in amplitude with time. ½ 4 (Answer need not be in above form - an explanation of change of charge on C only required.) 5. (a) Disintegrations follow the statistical law of chance as to which particular nucleus at a particular time will disintegrate. At any time the rate of disintegrations is proportional to the remaining number of nuclei which have not yet undergone disintegration (N). dN N , Hence constant dt N dN t N dt ln N N 0 N0 N 0 i.e. 1 1 N N0 et 1 3 (b) Completed reactions are : 234 234 238 234 91 Pa 92 U 92 U 90Th 1 ’s mass is 4 (a.m.u.) and charge is +2 e ’s mass is negligible and charge is -e All reaction steps must show agreement in regard to conservation of mass and charge. 1 1 (c) +V (EHT) R source * GM tube The Geiger tube has to have a very thin mica window (or a spark counter could be used - better). C amplifier/scales or ratemeter 1 ½ 3 87 AL Physics/Essay Marking Scheme/P.9 (i) -particles. With source and Geiger tube almost touching the introduction of a sheet of paper will stop all -particle with a resulting slight drop in count rate. (ii) -particles. The insertion of increasing thickness of Al sheets will reduce the count rate, stopping for a thickness greater or ~ 5 mm - when all ’s will have been absorbed. (iii) -rays. These are very penetrating and a Pb sheet of thickness ~ 2.5 cm will be needed to reduce the count rate to ~ zero. (d) (i) Factors affecting dangerousness of a radioactive source are : (I) the activity strength (disintegrations per sec.) (II) -ray emitter has more penetrating power. (III) the greater the half-life, the longer it will be dangerous. (ii) Precautions are : (I) screening off of source by lead shielding (absorber). (II) monitor total radiation dose using photobadge or clip-on electrometer - check if reasonable. 6. dI = LI0 cos t dt or VL = LI0 sin (t + /2), i.e. VL has phase advance of /2 on current I Reactance = (VL)rms/(I)rms = L (a) (i) I = I0 sin t, VL = L (ii) I = I0 sint = I 0 dV dQ = C C dt dt ½ 1 ½ ½ ½ ½ 5 1 1 1 3 1 1 2 ½ ½ ½ ½ 2 ½ sin t dt = C dVC 1 1 cos t = sin(t ) C C 2 i.e. VC has a phase retardation of /2 on current I 1 Reactance = (VC)rms/(I)rms = C VC = ½ ½ ½ 2 87 AL Physics/Essay Marking Scheme/P.10 (b) VL - VC = (L - 1/C)I VL = LI OR VR = IR V I I VR = IR 1 VC = I/C Hence impedance, Z V / I R2 (L (c) At resonance 0L = R 2 (L I/I0 = [ 0 ]L 1 2 ) C and (e) (i) At resonance 2, = 2 ½+½ 1 ½ ½ 1 1 [1 ] =R 0C 0 [ 0 ]L simplifying, 1 1 and I0 = V/R 0C R since Power = I2R 1 hence, L =R C (d) 1 2 ) C 1 [1 ] =R 0C 0 [L 1 ]=R 02C = 3 ½ since << 0 1 1 L 02C 1 since R 2L ½ (VC)rms = Irms /(0C) (VC)rms = Vrms /(0RC) = QVrms ½ ½ ½ (ii) This is possible since an equal anti-phase voltage exists ‘across’ the inductance - these voltage contributions cancel - giving a much smaller voltage in complete circuit, Vrms = IrmsR. 1 ½ ½+½ 3 4