Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
88 AL Physics/Essay Marking Scheme/P.1 PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE 88’ AL Physics: Essay Marking Scheme 1. (a) (i) Force repulsion Molecules in surface layer more widely spaced and in state of tension separation attraction Inner molecules in liquid (optional) At the surface of the water molecules escape and in the vapour state are more widely separated (on average). Also at surface the molecular concentration decreases and molecular forces are strongly attractive producing a surface layer under tension. A needle is supported by this ‘elastic skin’, with repulsive forces from the water molecules underneath needle preventing sinking. ½ ½ ½ ½ ½ ½ 3 (ii) solid B A liquid R C At point B in liquid there is an attractive force (I) BC due to the liquid molecules (II) BA due to the neighbouring molecules in the pin. If forces BA > BC (as for water) the resultant BR will be as shown in diagram. The liquid surface has to be at 90° to this and so it curves up sides of pin. The resultant BR will be opposed by repulsive (liquid) molecular forces. (iii) Work has to be done in separating further the molecules, at a liquid surface, against the molecular attraction forces. This leads to an increase in molecular P.E. (or free surface energy of liquid). In nature this will be a minimum and so any created liquid film has a minimum area. ½ ½ ½ ½ ½ ½ 3 ½ ½ ½ ½ 2 88 AL Physics/Essay Marking Scheme/P.2 (b) (optional) (a) (b) Liquid (a) with an acute angle of contact spreads while liquid (b) with an obtuse angle of contact does not. ½ ½ (i) Good spreading, e.g. flux in soldering, wetting agents in paints, lubricating oils. (ii) Poor spreading, e.g. silicones for waterproofing fabrics. 1 1 3 (c) Capillary tube Travelling microscope h Pin Liquid 2. 1 Since contact angle ~ 0 2rT = r2hg T = ½hrg 1 (i) Travelling microscope focused on bottom of meniscus of water in tube and then moved to focus on pin which just touched surface of water in beaker (beaker removed) - hence measurement h. 1 (ii) Tube broken at meniscus level (in tube) and the diameter of bore measured in two directions at right angle - hence measurement r. 1 (iii) Cleanliness of apparatus - should be cleaned by immersing in caustic soda, dilute nitric acid and distilled water, in turn. 1 (a) R long distance moon L dollar O2 O1 a b image 1 5 88 AL Physics/Essay Marking Scheme/P.3 The size of the image formed on the retina of the eye only depends upon the angle subtended by the object at the eye. This angle is similar for the larger object (moon) at a long distance and the smaller object (dollar) at a nearer distance. ½ ½ 2 (b) L I’ h’ O’ I h F ’ O C D 1 Following from (a) larger objects at a fixed distance will appear larger since the angle subtended at eye is greater. Clearly, from ray diagram, the viewed image I formed at the distance of nearest clear vision D appears greater than the image I’ formed at infinity since > ’, and hence magnification is greater for the former arrangement. Magnification is limited by the focal length of the lens. ½ ½ ½ ½ 1 4 (c) D objective I2 Fo h O Fe I1 ” eyepiece h1 Fo L1 f1 h2 Clearly, from ray diagram, a magnified image I1 is produced for object O by lens L1. In this case the final image I2 of this ‘object’ I1 produced by lens L2 subtends an angle ”. Since h1 > h the angle ” > ’ and the magnification has been increased. (d) (i) The intensity of light collected by the camera aperture area, i.e. d2 and this determines the intensity of the image. (ii) The recorded brightness (and darkening of chemicals on film) depends upon the time integration of the light energy received, i.e. t. L2 f2 1 ½ ½ ½ ½ 1 1 3 88 AL Physics/Essay Marking Scheme/P.4 (iii) F1 d F2 f1 1 f2 For distant objects it can be seen, from ray diagram, that the size of the image is f and the area of image f 2 Brightness of image 1/f2 ½ ½ ½ (iv) The f-stop is defined as f-n, where the aperture diameter d = f/n, f being the focal length of camera lens. In a camera the effective aperture diameter can be varied by a mechanical ‘iris’ e.g. by decreasing the f-stop the time exposure can be reduced which could produce a shaper image of a moving object. (or any other reasoning) 3. 1 ½ 1 7 (a) (i) (ii) (iii) (iv) 2 (v) (i) (ii) (i) Both ends of the spring are momentarily moved sideways at the same time. (ii) One end of the spring is attached to a rigid support and the other end given a momentarily ‘flick’ as in (i). After reflection amplitude of return pulse is reversed and a further pulse is produced from the free end as in (i). 2 (b) When the pulses pass through each other : (i) they remain unaffected (in amplitude). 1 4 88 AL Physics/Essay Marking Scheme/P.5 (ii) Where they cross the total displacement is the vector sum of the individual displacements. 1 (c) Following from the principle of superposition, in a region where 2 wave-trains meet the amplitudes either reinforce or cancel, depending upon whether they are in-phase or anti-phase - giving rise to an interference pattern (or fringes). 2 1 (i) Sound Displacement f2 f1 0 Time (a) A B Resultant Displacement C Variation of amplitude 0 Time (b) Beat period = T Hence, Beat frequency = 1/T = f1 – f2 e.g. beats produced between two sources of slightly different frequencies - as in music if one instrument is slightly out of tune. OR any other example 1 1 (ii) Light eye N (i) A i M (ii) D O C E r B t r t 1 e.g. different colours observed as eye is moved at different angles, viewing a thin oil film on water. Different path distances for white light (i) and (ii), 2t cos r = n for re-inforcement. varies for different coloured light. OR any other example 1 5 88 AL Physics/Essay Marking Scheme/P.6 (d) (i) At observing location the interfering waves must have a constant (or slowly observable changing) phase difference. 1 (ii) Interference between any 2 sound sources is possible if condition (i) holds good. 1 (iii) Interference for light waves only possible from single source via 2 different path distances. 1 (iv) Light is not coherent (normally) since waves are emitted in random quanta or wave packets. To maintain phase relation path distance has to be < wave packet length. 1 1 5 [Bonus mark 1 - if laser mentioned, then path distances longer.] 4. (a) + ½ E A R B (i) Electric field, E, is the force per coulomb of charge acting on the free electrons which move in a net direction along the connecting wires to . Direction of electric field is to . 1 ½ (ii) Potential difference, VA - VB, is the energy converted to other forms, from the electrical P.E., per coulomb of charge passing from B to A. 1 (iii) E.m.f., , is the energy imparted by the battery per coulomb of charge passing internally between its plates. 1 (b) The moving free electrons, under the influence of the electric field suffer collisions with the atoms in the resistor. They loss part of their K.E. which is transformed into making the atoms vibrate, generating heat. (c) (i) According to (a) (ii) for a charge Q the converted energy in R is (VA - VB)Q. dQ Energy rate = (VA - VB) or (VA - VB)I dt V VB = I2R, since by Ohm's law R = A I If all the energy is converted to heat, rate of heat dissipation = I2R. 4 ½ ½ ½ ½ ½ ½ ½ 2 88 AL Physics/Essay Marking Scheme/P.7 (ii) I V ½ IR In an a.c. circuit containing reactance : V I= , where V is the voltage across the 2 R2 circuit [V, I are r.m.s. values] Rate of heat dissipation, still, = I2R ½ ½ (d) The power is transmitted as an a.c. high voltage (132 kV) along transmission lines over long distances from the power stations to sub-stations where it is transformed down to 6 kV, - to 220 V for domestic consumption. (i) Reason for high voltage Transmitted power, P = IV Power loss in transmission line = I2R = (P/V)2R - by increasing V a significant reduction in power loss (ii) Reason for use of a.c. It is relatively easy to transform the voltages up or down using transformers. (e) RL Vm 3 ½ ½ ½ ½ ½ ½ 1 4 I R V ½ There is a voltage line drop due to the resistance of mains cables RL and current I taken by appliance V = Vm - IRL. ½ (i) Length (and condition) of cable leads from transformer station will vary RL for any particular location. 1 (ii) Even at a particular location variation of current taken will vary mains voltage across appliance. 1 3 88 AL Physics/Essay Marking Scheme/P.8 5. (a) v m1 u q m1v OR s m2w m2 w p m1u ½ (i) Electric collision (I) Total kinetic energy conserved, 1 1 1 m1u2 m1v2 m2w2 2 2 2 (II) Total momentum (vector) is conserved, p q s or m1u m1v m2w ½ ½ (ii) Non-elastic collision (I) Total kinetic energy is not conserved some of energy appearing in another form due to work done against an internal force, increasing the internal energy such as e.g. heat. (II) Total momentum (vector) is conserved. ½ ½ ½ 3 (b) (i) -particle scattering Alpha particles +Ze +2e P Nucleus ½ -particle lose their K.E. on approaching the +ve charged nucleus, being repelled by an electrostatic force. At P, distance of closest approach K.E. lost = P.E. due to -particle location in electric field of nucleus. -particle is then ‘reflected’ away from nucleus and finally acquires the same K.E. as it had initially. Collision is elastic. (ii) ½ ½ ½ ½ 1 235 0 n/ 92U collisions Neutrons penetrate the 235 92 U nuclei and cause fission, 148 85 1 1 ( 235 92 U + 0 n 57 La + 35 Br + 3 0 n ) (equation optional) There is an apparent loss of mass ½ ½ ½ 2½ 88 AL Physics/Essay Marking Scheme/P.9 resulting a release of energy, according to relation E = mc2. Collision is inelastic. ½ ½ (iii) electron/Xe collisions On collision some of the K.E. of the electron will be ‘absorbed’ by atom, exciting the atoms to a higher internal (quanta) energy level and the electron will continue with a reduced K.E. Thus K.E. is not conserved and collision is non-elastic. 2½ ½ ½ ½ ½ 2 (c) Franck-Hertz experiment C Gas I G A Electrons I xenon-filled thyratron V1 0 - 25 V Q G V2 V2 ~ 1.5 V P At low voltages V1 electrons repelled by -V2 do not reach anode. As V1 increases K.E. of electrons increase and more electrons are collected by A, increasing current I. Collisions with xenon atoms are elastic. At VC some electrons have sufficient Kinetic energy to excite the atoms to their first higher internal energy level, electrons lose energy and current decreases (velocity reduced, less no. of electrons reaching A per sec.) On further increasing V1 electrons (which have lost energy on collisions) now have sufficient energy to reach A and current increases. Importance of experiment is the demonstration of internal discrete (quanta) energy for atoms. 6. S Galvo reading 1+½ R VC V1 ½ ½ ½ ½ ½ ½ ½ ½ ½ (a) B r O x N = 2r 1 v = r P A Consider the projection of point P2 on the diameter AB point N. Acceleration of P2 = = 2r along PO. ½ 6 88 AL Physics/Essay Marking Scheme/P.10 Acceleration of N is d2x (or x ) = 2r cos , directed dt 2 towards O. Displacement of N from O is x = r cos . Hence the relation, x = - 2x ½ ½ ½ (b) Simplest method is to attach spring to a rigid support at one end and at the other to the object resting on a smooth table then pull object, extending spring. Object will oscillate about unstretched length of spring with S.H.M. since extension of spring force applied. Restoring force = -kx, where k is force constant of spring [force/ extension]. [Note: for suspended object, S.H.M. is about ‘normal’ extended end of spring] (c) (i) The force between atoms, F(x) = so, F dU dx 3 1 1 1 3 ½ a b 3 2 x x ½ F 0 (general shape) b/2a + x0 = b/a xm x P - Q For equilibrium location F = 0 b so x0 at P a At small value of ‘x’ the 2nd (+ve) term predominates while for large values of ‘x’ the 1st (-ve) term predominates. At Q the force F is a minimum corresponding to dF 0. condition dx 3b 2a 3b i.e. 3 4 0 or x m 2a xm xm (ii) Consider a small displacement dx from equilibrium position P, a b F = 2 (x0 dx ) (x0 dx )3 = a dx b dx (1 )2 3 (1 )3 x0 x0 x02 x0 a dx b dx (1 2 ) 3 (1 3 ) 2 x0 x0 x0 x0 using Binomial expansion = ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ 4½ 88 AL Physics/Essay Marking Scheme/P.11 1 1 3b a a4 mx 3 [2a ]dx 3 dx 3 dx , 2 x0 x0 x0 b since x0 = b/a i.e. F = 2 3½ (d) Clearly this represents S.H.M. and a4 b3 (ii) the oscillation period = T (i) the force constant, k = 2 m/2 mb3 2 a 4 / b3 2a4 1 1 2