(SAJC) 2024 H2 Phy Prelim P2 soln for sharing

1 SAJC 2024 Prelims / 9749 [Turn Over JC2 Prelim (H2 Physics) Paper 2 Solutions 1(a)(i) a = dv/dt v = + a dt = area enclosed by a-t graph  v = area enclosed by F-t graph / m (as F = ma) = [ ½ (1980)(10) ] /1100 = 9.0 m s-1 v = vi - v = 18.0 – 9.0 = 9.0 m s-1 [1 – for p or v using area under graph] [1] (ii) The net force is the braking force [1] (b) A downward slope with speed starting from 18 m s-1 and end at 0 Gradient of curve starts from 0 at t=0 and increases to a max gradient at 10 s. Gradient of curve decreases gradually from max at 10 s to 0 at t=20 s [1] [1] [1] (c) Gradient of the curve decreases non-linearly from a max at t= 0 to 0 at t=20 s. Distance is 0 at t=0 and max at t=20 [1] [1] 2(a) sum / total momentum before (interaction) = sum / total momentum after or sum / total momentum (of a system of interacting objects) is constant [1] if no resultant force / for an isolated system [1] 9

2 SAJC 2024 Prelims / 9749 [Turn Over (b)(i) 3m x 4 = m x v sinϴ (v sinϴ = 12) 2m x 6 = m x v cos ϴ (v cos ϴ = 12) therefore, sin ϴ = cos ϴ or tan ϴ = 1 ϴ = 45° or a method using close triangle of the momentum vectors and trigo [1] [1] [1] (b)(ii) mv x cos 45° = 12m or mv x sin 45° = 12m or (mv)2 = (3m x 4)2 + (2m x 6)2 v = 17 m s–1 or use pyth theorem for momentum [1-ecf] [1-ecf] (c) Chemical energy = ½ mv2 (By conservation of energy) Chemical energy = 0.0050 x 700 or 5.0 x 0.700 = 3.5 3.5 = (0.5 x 3m x 42) + (0.5 x 2m x 62) + (0.5 x m x 172) 0.0050 x 700 or 5.0 x 0.700 = 3.5 = 204m m = 0.017 kg [1] [1-ecf] [1-ecf] 3(a) F = ρgV V = 4 / 3 × π × (2.1 × 10–3)3 = 3.88 × 10–8 m3 ρ = 4.8 × 10–4 / 9.81 × V = 1300 kg m–3 [1] [1] (b)(i) W downwards > U upwards > Fv upwards. total length up = total length down [1] (ii) FV = 7.2 × 10–4 – 4.8 × 10–4 = 2.4 × 10–4 (N) velocity = 2.4 × 10–4 / (17 × 2.1 × 10–3) = 6.7 × 10–3 m s-1 [1] [1]

3 SAJC 2024 Prelims / 9749 [Turn Over 4 (a) From infinity to Mars, |KE gain| = |GPE loss| Thus, | ½ mv2 – 0 | = | msurface - m | = | (− 𝐺𝑀𝑚 𝑅 ) – zero | speed of rock v = 2𝐺𝑀 𝐷 2 where M = 6.4 x1023 kg, D = 6.8 x 106 m = 5.0 x 103 m s-1 [1] [1] (b)(i) ½ m <c2> = 3/2 kT  temperature T = 𝑚<𝑐2> 3𝑘 where mass of 1 atom, m = 4𝑥10−3𝑘𝑔 𝑁𝐴 , <c2> = [Ans in (a) ]2 = 2.5 x 107 m2s-2 = 4.0 x 103 K ● [1 m] for mass of 1 atom, m = 4𝑥10−3𝑘𝑔 𝑁𝐴 ● [1 m, ecf v & m ] for correct caln of T [1] [1] (ii) 1. The speed in (a) is the escape speed fr surface of Mars. 2. Since surface temperature of Mars is (much) lower than the temperature in (b) (i), 3. rms speed of He-4 is (much) lower than the escape speed. Thus, helium-4 gas is found on surface of Mars. [1] [1] 5(a) Progressive: all particles have the same amplitude Stationary: maximum to minimum/zero amplitude Progressive: adjacent particles within one wavelength are not in phase Stationary: waves particles are in phase between adjacent nodes and/or wave particles between adjacent nodes are in antiphase to wave particles in the next adjacent nodes. [1] [1] (b)(i) wavelength = 1.2 m(zero displacement at 0. 0.60 m, 1.2 m, 1.8 m, 2.4 m) either peaks at 0.30 m and 1.5 m and troughs at 0.90 m and 2.1 m or vice versa (but not both) maximum amplitude 5.0 mm [1] [1] (ii) 180 or  rad [1] (iii) t = 0, particles has KE as particle is moving t = 5.0 ms, particle has no KE as particle is stationary so decrease in KE between t = 0 to 5.0 ms. [1] [1]

4 SAJC 2024 Prelims / 9749 [Turn Over 6(a)(i) R = ρ L / A P = I2 R Therefore rate of heat loss, P ρ L / A PX / PY = (x/y) x (Lx/LY) x (AY / AX) = (1/1.58) (1/1.50) (1.50) = 0.633 [1] [1] (ii) Wire X because there is lower rate of energy loss as thermal energy in the wire. [1] (iii)1. The cable is made up of 5 thin wires in parallel. Let resistance of a single thin wire be R. 1 / 0.0458 = 5 (1 / R) R = 0.229 Ω [1] [1] 2. Any one: Several thin wires are more flexible than a single thick wire Thin wires have more surface area and thus dissipate heat more effectively/better dissipate the heat (Note: both wires have the same heat production but heat dissipation from the wire can be different) Cable remains workable even when one wire is broken [1] (b)(i) Note: The line must not be horizontal at high voltage. Hence do not overdo the decreasing slope. The graph near the origin should be linear. [1] (ii) With increasing magnitude of V, more heat is dissipated and temperature of filament rises, Lattice metal ions vibration amplitude increase, Collision frequency between metal ions and electrons increase, Resistance increases (and the current increases at a decreasing rate.) [1] [1] [1]

5 SAJC 2024 Prelims / 9749 [Turn Over (iii) No, the equation is only true where the I-V graph is a straight line passing through origin (ohmic conductor) but this I-V graph is not. OR No, as the resistance is represented by the inverse of the gradient of a line connecting that point to the origin. It is not the inverse of the tangent at that point because the resistance is the ratio of the potential difference to the current. [1] 7(a)(i) Nuclear fusion is the process where two light/small nuclei are combined to produce a heavier/large nucleus with the release of energy. [1] (ii) line with a peak at A≈56 line with steep initial positive gradient on the left of peak and gentler negative gradient at all points to the right of peak and line does not return to 0 binding energy [1] (iii)1. X shown at value of A to the right of the peak [1] (iii)2. Y shown at value of A close to 1 [1] (iv) energy from 1 nucleus = (1.77  1013) / (6.02  1023) = 2.94  10–11 J nucleon number of Z = 93 + 139 + 2 – 1 = 233 Energy Released = Total Binding Energy Final – Total Binding Energy Initial 2.94  10–11 J = [(1.25 + 1.81)  10–10] - binding energy of Z binding energy of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 = 2.77  10–10 J binding energy per nucleon = (2.77  10–10) / (233  1.60  10–13) = 7.43 MeV [1] [1 – equation to for energy released] [1 – J to MeV] A

6 SAJC 2024 Prelims / 9749 [Turn Over (b)(i) A = N and  = ln 2 / T Since the number of moles of fluorine-18 is 𝑛, the initial number of fluorine-18 nuclei: N = n  NA A0 = 𝜆𝑁0 2 photons produced from each decay, so R0 = 2 A0 =2 𝜆𝑁0 = 2    n  NA = (2 ln 2) nNA / T [1] [1] (ii) exponential decay curve from t = 0 to t = 2T, starting at (0, R0) and with a negative gradient of continuously decreasing magnitude line with negative gradient passing through (T, R0 / 2) and (2T, R0 / 4) [1] [1] 8(a)(i) 1. Solar energy can be affected by cloud cover, OR is only present in the day. 2. Wind speeds can vary for wind power 3. It is difficult to efficiently store excess energy generated for use at a later timing when the supply of energy falls. (any 2 of the above) [2] (ii) γ-radiation is high energy/high frequency/short wavelength electromagnetic radiation / photons [1] [1] (iii) γ-radiation can penetrate the skin/go through human/has