ASRJC Prelim H2P2 MS Final (sharing)

1 9749/02/ASRJC/2023Prelim [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Prelim P2 Exam Mark Scheme Paper 2 (80 marks) E – Easy, A – Average, D – Difficult ECF Error carried forward SF Significant figures error M0 No A marks awarded AE Arithmetic error BOD Benefit of doubt ^ More is needed in answer POT Power of ten error CON Contradictory response XP Wrong physics TE Transcription error IR Irrelevant (part) response 1a = = = –1 0.180 9.81 0.036 4 m9 N  Fk e A C1 A1 1b = Fk e   = + = + 2 0.0236 k F e k F e  = +  = = –1 2( 0.02) 49 . N 6 37 4  m 3k Accept −= max min 2 kkk A C1 A1 1ci F is weight of the column of liquid above the area A    == = = = () () fluid fluid fluid fluid mgFp AA Vg A hA g A hg A B1 B1 A0 1cii U + ke = mg U = mg − ke = (0.180)(9.81) − 49(0.030) = 0.2958 N      = = = = -3 –50.2958 (2.0 10 )( 0 9.81) 15 0 kg m p A h A g U V g A C1 C1 A1 2a The total momentum before collision is non-zero. By COM, (total) momentum is never zero, so not possible for both blocks to be at rest simultaneously. A M1 A1

2 9749/02/ASRJC/2023Prelim bi By COLM, taking rightwards as positive (3M x 0.40) – (M x 0.25) = (3M x 0.20) + Mv v =0.35 m s-1 E A1 bii To right / away from block A, as direction to the right is taken as positive in (b)(i) E B1 c relative speeds of approach is non-zero, and relative speeds of separation is zero Relative speed of approach is not equal the relative speed of separation, Hence inelastic collision. A M1 A1 3ai work done per unit mass bringing (small test) mass from infinity (to the point) E B1 3aii (near Earth’s surface change in) height ≪ radius or height much less than radius potential inversely proportional to radius and radius approximately constant (so potential approximately constant) A B1 B1 3b curve from r to 4r, with gradient of decreasing magnitude and starting at (r, ±) and line passing through (2r, ±0.5) and (4r, ±0.25) line showing potential is negative throughout A B1 B1 3ci Gain in KE = loss in GPE ½m v2 = 0−(– GMm / R) At distance R = 3r, –11 24 6 6.67 10 6v r .0 10 6 2G 2 .4 13 0 M 3     == = 6.46 × 103 m s-1 At distance R = 4r, 6 –11 242GM 2v 6.67 10 6.0 10 614r .4 04 =  =  = 5.59 × 103 m s-1 Change in speed = 6.46 × 103 −5.59 × 103 = 8.7 × 102 m s-1 D C1 C1 A1 × × ×

3 9749/02/ASRJC/2023Prelim [Turn Over 4a a = –  2x a = acceleration, x = displacement from equilibrium position and  = angular frequency E B1 4bi  = 2 / T = 2 / 4.0 = 1.57 = 1.6 rad s–1 A B1 A0 4bii E = ½m2x02 Or E = max Ek = ½mv2max= ½m2x02 = ½ × 36 × 1.62 × 0.0802 = 0.29 J A B1 C1 A1 4c dome-shaped curve, starting and ending at EK = 0 maximum EK shown as 0.29 J, position of peak shown at h = 10.0 cm line intercepts h-axis at h = 2.0 cm and at h = 18.0 cm D B1 B1 B1 5a The field strength at a point equals the negative of the potential gradient there. i.e. the electric potential gradient is the electric field strength the direction of the field is the same as the direction of decreasing potential. A (B2) B1 B1 5bi Straight line vertically upward E B1 5bii E = V/d = 75/(1.2 ×10-2) = 6250 V m-1 E A1 5biii gain in kinetic energy (= loss in potential energy) = charge × p.d. or qV = ½mv2 because separation not in expressions so v is independent of separation D B1 A0 5biv (at x = 0.40 cm), potential = (–) 75 × 0.40 / 1.2 (= (–) 25 V) ½mv2 = qV ½ × 4 × 1.66 × 10–27 × v2 = 2 × 1.60 × 10–19 × 25 Or a = Vq / dm and v2 = 2as D C1 C1 × × ×

4 9749/02/ASRJC/2023Prelim v2 = (2 × 75 × 2 × 1.60 × 10–19 × 0.40 × 10–2) / (1.2 × 10–2 × 4 × 1.66 × 10–27) v = 4.9 × 104 m s–1 (C1) A1 6a A progressive wave is a wave in which energy is carried from one point to another by means of vibrations or oscillations within the wave. A transverse wave is a wave in which the oscillations of the particles in the wave are at right angles to the direction of transfer of energy of the wave. E B1 B1 6b Speed, v is defined as distance travelled divided by the time taken. From the definition of wavelength, λ, in one cycle of the source, the wave energy moves a distance λ. The time taken for one cycle is the time period T. Since f = 1 / T, wave speed, v = ( total distance / time taken ) f== T λv E B1 B1 A0 6ci Angle  amplitude intensity 180 A I 90 0 0 60 0.50A 0.25I intensity  cos2 θ A B1 B1 B1 6cii intensity angle  zero 90 maximum 0, 180 2 I 32.8, 147 Intensity after passing through polaroid Q, IQ = I cos2 θ Intensity after passing through polaroid R IR = IQ cos2 θ = I cos4 θ D B1 B1 7a When atoms are excited, the electrons move from a lower energy level to a higher energy level. When the electrons de-excite from a higher level to a lower level, they emit photons with energies corresponding to the differences in energy levels of the atoms , giving rise to line emission spectrum. Since the frequencies of the photons are fixed, the photons emitted have discrete amounts of energies. Hence, energy levels are discrete. Examiner’s comments: Many students did not explain how the energy of the photon is linked to the energy levels in atoms. A few students described the observations for absorption line spectrum. A M1 A1

5 9749/02/ASRJC/2023Prelim [Turn Over 7b Since d sin θ = λ for 1st order maxima, and Hence, 9 9 -9 ( ) 410 10 (1) 0.008( ) 434 10 (2) From Equation 1 and 2, =0.1367 m Using, 0.075( ) (3) From Equation 1 and 3, =635 10 m Ld D Ld D L LdX D X − − = + = + =  Or Using idea of d sin θ = λ and small angle approximation,  Since  ,   , L   434 410 410 0.8 8.2 656 nm −− = = X X D M1 A1 (M1) (A1) 7c Energy of photon with wavelength 410 nm 34 8 9 19 (6.63 10 )(3.0 10 ) 410 10 4.85 10 J hc  − − − = =  = 19 19 19 4.85 10 +(-4.08 10 ) 0.77 10 J −− − =   = Energy level D M1 M1 X 410nm 434nm 0.8 cm 7.4 cm 1st order maxima of 410 nm 1st order maxima of 434 nm 1st order maxima of X Diffraction Grating L D

6 9749/02/ASRJC/2023Prelim A1 7d The lowest energy level shown in Fig 7.3 may not be the energy level at ground state. D A1 7e When the electrons are directed to pass through a thin carbon film, concentric rings are observed It shows a diffraction pattern that is unique to wave behaviour after passing through the carbon film. A M1 A1 7f 2 k k of electron = 2 2= pE m p mE By de Broglie’s equation, 34 31 19 10 2 6.63 10 2(9.11 10 )(4.08 10 ) 7.69 10 m − −− − = = =  = k h p h mE  A C1 A1 8a greater lattice vibrations more frequent collision of electrons with lattice ions /lower drift velocity of the electrons A B1 B1 8bi connect cells in series E B1 8bii connect cells in parallel E B1 8c Active cooling could fail/active cooling needs energy input, increasing costs or decreasing system output/ difficult to eliminate passive cooling A B1 8d site panel so that there is an air gap around it e.g. mounts panels a small d