[1] for eqn & pv = nrt definition of p, p = pressure, V = volume, T = thermodynamic temperature,

Transcription

1 1(a) for eqn & pv = nrt definition of p, p = pressure, V = volume, T = thermodynamic temperature, V, T & n. n = no. of moles, (R = molar gas constant) n = amount of gas, T = temperature Not accepted: n = amount of gas molecules Common mistakes/misconceptions: 1. Stating 1 st Law (U = Q + W) instead 1(b)(i) Mean kinetic energy = 3/ kt = 3/ (1.38 x 10-3 )( ) = 6.38 x 10-1 J MC Common mistakes/misconceptions: 1. Conversion of 35 to value in K before substitution. Use of wrong formulae 1(b)(ii) Mean kinetic energy = ½ m ( c ) = 6.38 x 10-1 J r.m.s speed x6.38x x10-1 c ms 3-1 MC Many students did not know that r.m.s speed is mathematically expressed as c and/or erroneously expressed it as < c >. 1(c)(i) Internal energy of a system is the sum of the random distribution of kinetic and potential energies associated with the molecules/particles of the system. MC Common mistakes/misconceptions: 1. Omission of the word random. Inclusion of the word mean 1(c)(ii) 1. This would cause the thermodynamic temperature to increase (or the random kinetic energy of the molecules will increase), hence the internal energy of fixed mass of ideal gas will increase. MC Common mistakes/misconceptions: 1. Higher temperature is due to heat supplied to system.. Discussion of potential energy for ideal gas. 3. Use of 1 st Law (U = Q + W) when there is no information on Q Page 1 of 10

2 1(c)(ii). At constant temperature, the speed of the gas molecules will remain unchanged (or the random kinetic energy of the gas molecules will also remain constant), hence the internal energy of the fixed mass of ideal gas is unchanged. MC Common mistakes/misconceptions: 1. Higher pressure results in (more frequent collisions of molecules, which is correct) higher velocity and hence, higher random kinetic energy.. Relating higher pressure to work done on system 3. Discussion of potential energy for ideal gas 4. Use of 1 st Law (U = Q + W) when there is no information on Q (a) The direction of the body changes continuously hence there must be a force acting on this body. To allow the speed of the body to remain constant, the force must always act perpendicularly to the motion which means it has to be directed towards the centre of the circle. MC Not accepted: direction of speed Accepted: 1 st mark: Change in velocity implies force acting on object. Many students did not explain the direction of the force. (b)(i) MC Nil. Angular velocity 3 v 0. x 10 3 r 1.5 x rad s -1 (b)(ii) Centre of the circle Lift force acting on aircraft Weight of aircraft [1 mark for the correct labelling and direction of each vector] Lift force labelled as L and F, or described as normal contact force on aircraft by air Common mistakes/misconceptions: 1. Inclusion of centripetal force. Lift force labelled as N and T, or described as upthrust, air resistance or engine force Page of 10

3 (b)(iii) 3 v 4 (0. x 10 ) 5 Resultant force = m ( ) 1.0 x10.7 x 10 N r 3 1.5x10 Accept resultant force =.5 x 10 5 N MC Common mistake: 1. Did not convert 0. km and/or 1.5 km to values in metres before substitution (b)(iv) Since the passenger will be turning in a horizontal circle along with the aircraft, a force must act on him to change his direction of motion. The force acting on the passenger is in the same direction as the lift force. (An upward component of this force which is perpendicular to the plane of the motion must be present to balance the weight of the passenger.) 1 st mark: The passenger is in contact with the seat. nd mark: Normal to the seat Common mistakes/misconceptions: 1. Missing out on the horizontal component of force on passenger by seat, which is required to provide the centripetal force for circular motion of passenger. The force on passenger by seat is the centripetal force. 3(a) Electric field strength at a point is defined as the force per unit positive charge acting on a small charge placed at that point. Electric potential at a point is defined as the work done per unit positive charge in bringing a small charge from infinity to that point. MC Common mistakes/misconceptions: 1. Omitted one or more word(s) for per unit positive 3(b) By symmetry, magnitude of resultant electric field = 0 Unit: N C -1 or V m -1 MC Not accepted: C F -1 m -1 Common mistakes/misconceptions 1. Adding up E due to each of the 3 point charges algebraically (not considering their respective directions) Page 3 of 10

4 3(c)(i) V A = V B = V C V resultant QA = V A + V B + V C = 3 V A = 3 x 4 πε r = 3 x (π)( )(0.0) = x 10 6 Unit: J C -1 or V MC Common mistakes in computation 1. Omitted negative sign for Q in substitution. Did not multiply by 3 3. Wrong conversion of prefix for µc -6 o 3(c)(ii) W = Q( V) = (+ 5 x 10-6 ) x [( 6.7 x 10 6 ) 0] = 34 J MC Common mistakes in computation 1. Omitted negative sign for V in substitution. Did not convert 5.0 µc to value in C before substitution 3(c)(iii) The +5.0 C point charge loses electric potential energy as it is brought from infinity to the centre of the circle. OR The external force acting on the point charge is opposite to its displacement. MC Ecf from c(ii) 4(a) It is a motion where the acceleration of the object is proportional and opposite to its displacement from the equilibrium position. [] or zero MC Common mistake: Omitted from the equilibrium position 4(b) Amplitude = 0.15 m MC Nil. Page 4 of 10

5 4(c) T=π m k MC Nil. 75 T =π 4100 = 0.85 s 4(d) One-quarter of a period later, the diver would be halfway between the highest and lowest points, i.e at the equilibrium position where his speed is at the maximum v max = wx = (0.15)( ) 75 = 1.1 m s -1 Direction: upwards. direction MC Ecf from the calculation of T in (c) Some students overlooked the fact that diver started from the lowest position and applied x = x o cos t instead of x = - x o cos t. Some did not convert to radian mode in calculator before computing. 4(e)(i) Normal Reaction both correct forces Weight Normal contact force labelled as N or F Page 5 of 10

6 4(e)(ii) The resultant force at the highest point of motion is W R = ma When the diver just ceases to remain in contact with the platform, the force the board exerts on the diver, R = 0 R = 0 By NL, mg = m x0 g = x0 g = ( f ) x0 f lowest = 1 g x = π 0.15 = 1.8 Hz MC Common mistakes in computation 1. Use of wrong formula, v = f. Inconsistent signs for acceleration i.e. g is positive but x0 when both are downwards is negative 5(a) Frequency f, is the number of oscillations/waves per unit time. defn Period T, is the time taken to complete one oscillation, so it is related to frequency by f = 1 T Wavelength, is the distance travelled by a wave in the time of one period. Speed v is the distance travelled per unit time, defn defn which is expressed as v = T = f working Speed is rate of change of distance. Speed is d t or distance time. Wavelength, is the shortest distance between particles in the same phase. BUT no mark for working. Not accepted: in 1 sec replacing per unit time Many students used mainly equations throughout their answers and did not define clearly all 3 variables, as required in the question. Page 6 of 10

7 5(b) A sound note played from a trumpet occurs at its resonance frequency, with a resonant wavelength depending on the length of the air column in the trumpet. So this sound note a particular trumpet has a fixed resonant wavelength. (Justification that wavelength is constant) Since the velocity of sound is proportional to the square root of the temperature measured in kelvin, i.e. v T v = k T = f 1 v v = f 1 f = T1 T f1 = f 1 = 35 Hz MC Common mistakes in computation 1. Assuming k = 1 Ans 5(c)(i) Polarisation is a phenomenon in a transverse wave where the vibrations of the elements in the wave are restricted to a plane. MC Many omitted to mention: 1. transverse wave. vibrations of the elements/particles 5(c)(ii) In a longitudinal wave, the elements in the wave vibrate parallel to the direction of the wave. They cannot be restricted to vibrate in any plane and hence cannot be polarized. So only transverse waves can be polarised. OR In a transverse wave, the elements in the wave vibrate perpendicularly to the direction of the wave. They can be restricted to vibrate in any plane and hence can be polarized. MC Many students did not distinguish between the direction of wave and direction of oscillations of elements/particles in the wave. It is noted that answers for this part tend to be characterized by loose use of key terms. Page 7 of 10

8 5(c)(iii) Let the image to be viewed by a left eye be polarised in a vertical plane, and image to be viewed by a right eye be polarised in a horizontal plane. The left spectacle of the viewer is also polarised in a vertical plane, and the right spectacle is polarised in a horizontal plane. (Through this spectacle, the right eye cannot view the image meant for the left eye because they are polarised perpendicularly, and vice versa. This would result in the viewing to be in 3D.) for images on screen to be polarized at right angles; for spectacles to be polarized at right angles; MC Some explanations lacked details such as images need to be polarized perpendicularly. same polarization planes btw corresponding images and spectacle 5(d)(i) The Principle of Superposition states that the resultant displacement at a point due to several waves of the same type is the vector sum of the displacements due to those waves acting individually. MC Some students replaced displacement at a point with amplitude, which is not accepted, or omitted of the same type. [] or zero 5(d)(ii) The superimposed images in (c)(iii) is plane polarised perpendicularly to each other. As such, their displacements cannot be added or subtracted. plane polarised perpendicularly need not be included if c(iii) is correct. 6 (a) Using x = d D x = λd d -9 ( )(.3) = = m MC Common mistakes in computation: 1. Wrong conversion of prefix for nm. Wrong formula used for eqn Page 8 of 10

9 6(b) The amplitude of wave from slit A, a is related to its intensity by I = k a A a A = I k The amplitude of wave from slit B, a B is given by 1 3 I = k a B I a B = 3 k The amplitude of the dark fringe near P is given by a = a A a B = I k I 3k The intensity of the dark fringe near P is given by I R = ka = k ( I k + I 3k I I k 3k ) = I ( ) = I for a working Ans MC Most students understood that subtraction is applicable for a dark fringe but many subtracted in terms of intensities, which is wrong. The correct method is to subtract 1 amplitude from the other and use this resultant amplitude to determine the resultant intensity. 7(a) Intensity required at 1 km away, P I eye A eye.5 10 = = W m Consider light guide, Plight I A P I. A light π W MC Common mistakes in computation 1. Wrong formula for 4r. Did not convert 0.50 cm to value in m before substitution value 7(b)(i) Shorter wavelengths means the anti-nodal lines will be closer to one another. Hence, aircrafts may lock on to the wrong line of maxima or difficult to identify the central line of maxima or difficult to differentiate the lines of maxima. lines closer wrong line of max Page 9 of 10

10 MC Common misconceptions: 1. Waves of longer wavelengths can be picked up further away.. Radiowaves travel slower than microwaves. Mistaken that frequency for both types of waves is the same, hence radiowaves have a higher and, based on v = f, higher v. 7(b)(ii) Since the two radio waves are in phase, path difference is always zero along centre-line. Hence constructive interference occurs. [] The entire centre anti-nodal line is at equal distance from the emitters. Phase difference is always zero. Common misconceptions: 1. Stationary waves are formed between emitters and aircraft. 7(c)(i) P is nearer to the aircraft. Hence intensity (or amplitude) of signal should be higher. MC Common mistakes: Misinterpreting Fig. 7.3 to conclude that signal B reaches aircraft first. Use of vague terms e.g. strong signals to indicate higher amplitude. 7(c)(ii) From Fig. 7.3, phase difference MC Not accepted: 90 rad 7(c)(iii) Since phase difference of signals path difference 4 (Distance from P to plane Distance from Q to plane λ Hence = m f c = 3.53 x 10 7 Hz, m m) value value MC Nil. Page 10 of 10