ASRJC 2024 JC2 H2 Physics Prelim P2 MS

1 9749/02/ASRJC/2024PRELIM [Turn Over Anderson Serangoon Junior College 2024 H2 Physics Preliminary Examination Mark Scheme Paper 2 (80 marks) 1a The net momentum of a system remains constant provided no external resultant force acts on the system. B1 B1 1bi (total initial momentum in the direction at right−angles to the direction of the initial path of ball A = 0) (taking ↑ as positive) 0 = 4.0 × 6.0 sin θ – 12 × 3.5 sin 30° θ = 61 ° C1 A1 1bii Considering momentum along the direction of the initial path of ball A: By Conservation of linear momentum, total initial momentum = total final momentum 4.0 × v = 4.0 × 6.0 cos 61° + 12 × 3.5 cos 30° v = 12 m s−1 C1 A1 1biii total initial k.e. = ½ (4)(12)2 = 288 J total final k.e. = ½ (4)(6.0)2 + ½ (12)(3.5)2 = 145.5 J Since the total kinetic energy before and after collision is different, the collision is inelastic. M1 A1 2ai take any two sets of coordinates to determine a constant value (F/x) F/x constant hence obeys Hooke’s law M1 A1 2aii 1 2 4.5 1.5 250 Nm(1.8 0.6) 10k − − −== − EP = area under graph or ½Fx or ½kx2 = 0.5 × 4.5 × 1.8 × 10–2 or 0.5 × 250 × (1.8 × 10–2)2 = 0.041 J (0.0405 J) C1 A1 2b Loss in KE of cart = Gain in EPE of spring KE = ½mv2 = 0.0405 v = (2 × 0.0405 /1.7)½ = 0.22 m s–1 C1 A1

2 9749/02/ASRJC/2024PRELIM 3a Intensity is proportional to (amplitude)2 Since amplitude is A cos θ, the graph of amplitude is a cosine function. Hence I = kA2 cos2  (Malus’ Law) Shape: M1 Axes values: A1 M1 A1 3bi I = kA2 cos2 ϕ , and Io = kA2 Intensity after passing through polaroid R = Io cos2 30o = 0.75 Io Intensity after passing through polaroid Q = (0.75 Io) cos2 (90°− 30°) = 0.19 Io C1 A1 3bii Working below is not required: Intensity after passing through polaroid R, IR = Io cos2 ϕ Intensity after passing through polaroid Q = IR cos2 (90° - ϕ ) = Io cos2 ϕ sin2 ϕ = Io (0.5 sin 2ϕ)2 where sin 2θ = 2 sin θ cos θ = 0.25 Io sin2 2ϕ Shape: sine-square graph with 4 cycles, starting with zero labelling of every 45° correct maximum value – 0.25 Io M1 A1 B1 3c Longitudinal waves cannot be polarised because the oscillation of particles in the wave is parallel to its direction of energy transfer. A1 θ/° intensity 0.25Io ϕ/o 0 30 45 90 180 270 360 P Q R ϕ 90– ϕ 0.19Io

3 9749/02/ASRJC/2024PRELIM [Turn Over 4a progressive waves transfer/propagate energy and energy in a stationary waves is localised/stored, OR amplitude constant for progressive wave and varies (from max/antinode to min/zero/node) for stationary wave, OR for stationary waves, particles within a loop/segment are in phase and for progressive wave particles within a wavelength are out of phase Any 2 of the above. B2 4bi wave/microwave from source/S reflects at reflector R reflected and (further) incident waves overlap/meet/superpose waves have same frequency/wavelength/period/speed and (amplitude), (so stationary waves are formed) B1 B1 B1 4bii detector/D is moved between reflector/R and source/S maximum and minimum/zero observed on meter/readings/measurements/recordings B1 B1 4biii determine/measure the distance between adjacent nodes/minima or maxima/antinodes or across specific number of nodes/antinodes wavelength is twice distance between adjacent nodes/minima or maxima/antinodes (or other correct method of calculation of wavelength from measurement) B1 B1 4c Waves from the two sources travel the same distance to the detector so the path difference is zero, constructive interference happens so the reading is maximum. M1 A1 5ai Since the field strength is zero at a point between the spheres / there are two sections of the graph that are of opposite signs hence opposite direction of E-field therefore the charges are the same sign. M1 A1 5aii At x = 0.08 m, the electric field strength due to sphere A cancels out the electric field strength due to sphere B OR net E is zero. EA = EB ( ) ( ) AB 22 oo4 0.080 4 0.040 QQ   = 2 A B 0.080 4.00.040 Q Q == M1 M1 A1

4 9749/02/ASRJC/2024PRELIM 5bi E = V/d = 100 / 0.050 = 2000 N C−1 Direction of electric field strength is upwards (high to low potential) B1 B1 5bii qEa m= = ((1.6 x 10−19) (2000) / (9.11 x 10−31) = 3.5 x 1014 m s−2 C1 A1 5biii VqU = = (– 1.6 x 10-19) [(– 100) – (– 150)] = – 8.0 x 10-18 J C1 A1 6a change in magnetic flux density ∆B = 5 – 50 = – 45 mT (from graph, accept positive value) average induced e.m.f. = ∆Φ / ∆t = ∆NBA / ∆t = NA ∆B / ∆t = 180  π (5.310-3/2)2 (– 4510-3) / 0.30 = – 5.95710-4 = – 6.010-4 V B1 C1 A1 6b As the coil moves away from the magnet, there is a rate of change of magnetic flux linkage through the coil due to the decreasing magnetic flux density resulting in an induced e.m.f. in the coil, according to Faraday’s law. With the resistor connected in a closed loop with the coil, a current is induced in the closed loop / coil that opposes the motion of the coil, according to Lenz’s law. As work is done (by external force)/ energy is required to move the coil away at constant speed, this gives rise to the heating effect of induced current (in resistor). B1 B1 B1 7ai As p.d. becomes negative, there are photoelectrons that reach collector C, contributing to the photocurrent, as they have kinetic energy greater than the work done against electric field between the metal plates. (As the p.d. becomes more negative, only the more energetic photoelectrons reach the collector C) Hence, the sloping section of the graph shows that photoelectrons are emitted from metal plate E with a range of kinetic energies. B1 B1 7aii From Fig. 7.2, the stopping potential, V = −2.2 V (Since stopping potential stops the most energetic photoelectrons emitted from metal plate E, by conservation of energy) loss in kinetic energy of most energetic photoelectrons = work done against e−field ½ mvmax2 − 0 = qV vmax = √ qV 1 2⁄ m = √ (−1.60×10−19)(−2.2) 1 2⁄ (9.11×10−31) = 8.79 × 105 m s−1 B1 M1 A1

5 9749/02/ASRJC/2024PRELIM [Turn Over 7aiii frequency is constant ➔ stopping potential is the same intensity is halved ➔ rate of photons incident on metal plate E is halved ➔ rate of photoelectrons emitted is halved ➔ photocurrent is halved at every respective value of p.d. B1 B1 7b Since E = hf or hc/λ, from work function energy, threshold wavelength = hc work function energy = (6.63 × 10−34)(3.00 × 108) 5.8 × 10−19 = 343 nm (For photoelectrons to be emitted, the wavelength of the radiation should be less than the threshold wavelength.) Since the wavelength of the radiation from the second lamp is longer than the threshold wavelength [appropriate comment comparing wavelengths/ energies/ frequencies], there will be no effect on photoelectric current (as this wavelength does not cause photoelectric effect of zinc). Note: Also accept analysis using threshold frequency or minimum energy for photoelectron emission. C1 A1 A1 8a Fast moving/ high speed/ high (kinetic) energy electrons moving towards the target metal (anode), undergo rapid/large deceleration/ acceleration as they interac