(SAJC) 2024 H2 Phy Prelim P3 soln for sharing

1 SAJC 2024 Prelims / 9749 [Turn Over JC2 Prelim (H2 Physics) Paper 3 Solutions 1 (a)(i) Period of Earth, T = 24 x 3600 s (= 86 400 s) Angular velocity = 2π/T = 7.27 x 10-5 rad s-1 (shown) [1] (a)(ii) a = r2 = ( 6.38 x 106 ) x [Ans in (b) (i)]2 = 0.0337 m s-2 [1] [1- ecf] (b)(i) ● There is circular motion at the equator but not at the poles, ● At the equator, part of the gravitational force is used to provide the centripetal force (required to keep any mass located there in circular motion, centripetal force is provided by the gravitational force of Earth) (Hence only the remaining part of the gravitational force is available to provide the acceleration of free fall. ) Or ● Gravi force of Earth = (mg + mac)equa = (mg + mac)pole. [1] ● Since ac at pole = 0, gpole > gequa , ie g is different for the 2 locations[1] Note: g = acceleration of free fall as defined by the question not gravitational field strength. Be careful in use of symbols. [1] [1] (b)(ii) ● Value of g, at the pole is 9.81 m s-2 ● Since centripetal acceleration at equator (in b(ii)) is 0.0337 m s-2 , compared to 9.81 m s-2, the difference in g at these 2 locations is small. [1] [1] (c)(i) The gravitational force of attraction between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation. [1] (c) (ii) (By Newton’s law), Gravitational force on a (point) mass m at Earth’s surface, F = 𝐺𝑀𝑚 𝑅2 Since field strength g is the gravitational force per unit mass at that point, g = F/m = 𝐺𝑀 𝑅2 [1] [1] (d)(i) ● { Since the gravitational force provides the centripetal force, } 𝐺𝑀𝑚 𝑟2 = mr2 {where r = dist of satellite fr Earth’s centre} [1]

2 SAJC 2024 Prelims / 9749 [Turn Over ( r3 = GM/2 , not assessed) -- (1) ● From (c), GM = gR2 -- (2) { R = Earth’s radius } (2) into (1): (r3 = gR2/2 , not assessed) ● Substituting, r = 4.23 x107 m {g = 9.81 m s-2, R= 6.38 x 106 m, = 7.27 x 10-5 rad s-1} [1] [1] (d)(ii) 1. Satellite lies on the equatorial plane of the earth 2. Satellite rotates/revolves from west to east or, following the Earth’s rotation [1] [1] 2(a) Volume decreases ; hence w is positive (or small volume change so w is 0) As the ice changes state, heat must have been supplied; hence q is positive. internal energy increases [1] [1] [1] (b) gas expands/increase in volume with work done by gas (against atmosphere) or w is negative the sudden increase leaves little time/no time for thermal energy to enter and leave the gas or q = 0 internal energy decreases [1] [1] [1] 3(a) molecule(s) rebound from wall of vessel / hits walls/ collide change in momentum gives rise to impulse / force average change in momentum/impulse gives rise to the average (constant) pressure/force. [1] [1] [1] (b) p =1/3 ρ <c2> 1.02 × 105 =1/3 × 0.900 × <c2> <c2> = 3.4 × 105 crms = 580 m s–1 [1] [1] (c) Straight line with positive gradient Line passing through origin (can never reduce the volume to zero thus the graph need not be shown passing through the origin) [1]

3 SAJC 2024 Prelims / 9749 [Turn Over 4(a) amplitude = 0.50 cm [1] b(i) T = 0.8 s ω = 2π / T = 7.85 rad s-1 v = ω √(x02 – x2) = 7.85 × √({0.5 × 10-2}2 – {0.2 × 10-2}2) = 3.6 cm s-1 [1] [1 – ecf for (a)] [1 – ecf for (a)] (ii) d = 15.8 cm [1] (c) ● same period { Accept: small increase in period) ● (For same period) : displacement of damped oscillator is always smaller than that on Fig.4.2 (ignore first T/4) ● peak decreases (exponentially) with time [1] [1] [1]

4 SAJC 2024 Prelims / 9749 [Turn Over 5 (a)(i) light intensity has maximum value at 0°, 180°, 360° and zero intensity at 90°, 270° ‘sinusoidally-shaped’ curve of constant amplitude [1] [1] (a)(ii) I = I0 cos2ϴ 4.2 = 7.6 cos2ϴ ϴ = 42° [1] (b) wave passes (through) an opening/gap/aperture or, wave passes (by / through / around) an obstacle/edge wave spreads (into geometrical shadow) [1] (c) nλ = d sinϴ d = (3 x 4.3 x 10–7) / sin 68° = 1.4 x 10–6 m 1.4 x 10–6 x sin 68° = 2 λ or, 3 x 4.3 x 10–7 = 2λ λ = 6.5 x 10–7 m [1] [1] [1]

5 SAJC 2024 Prelims / 9749 [Turn Over 6(a)(i) From the graph, R = 80 Ω, T = 40 Ω at 120oC E = I RT 3.0 = I (80 + 40) I = 25 x 10-3 A = 0.025 A [1] [1] (ii) (When the temperature increases from 0 oC to 75 oC, magnitude of RATE of Increase of resistance of R is smaller than the magnitude of RATE of Decrease of resistance of T.} {Compare their gradients} Thus, effective resistance decreases at a decreasing rate with respect to temperature. [1] (iii) From graph at 30 C, resistance of R = 55  and resistance of T = 110  By potential divider rule, p.d. across T = (RT / Rtotal ) x 3.0 V = (110/110+55) X 3.0 = 2 [1] [1] (b) (i) faulty lamp: lamp E nature of fault: lamp fused/ fuse melt/broken filament [1] [1] (ii) Resistance of one non-faulty lamp = 30.0 / 2 = 15.0 [1] (iii) switch ohm-meter reading S1 S2 S3 / Ω open open open closed open open 30.0 closed closed open 25.0 closed closed closed 15.0 - 1 mark for each mist ake

6 SAJC 2024 Prelims / 9749 [Turn Over 7(a) Since F = - dEp/dx & dEp/dx is the gradient of graph. Hence force is proportional to the gradient of the curve. [1] [1] (b)(i) Since all values of Ep are negative ( Fig 7.2), & EP = Ep is negative only if the 2 pt charges are of opposite sign. or F = -dEp/dx Since dEp/dx is positive, (fr Fig 7.2), F is negative. Since a negative F indicates an attractive force between the charges, the charges are of opposite sign. [1] [1] (ii) When Qp is doubled: Since EP = , Ep is doubled at every x. Thus gradient would be doubled/increased/become steeper. or, since F = QpQQ / (40x2), F is doubled for every x. & F = - gradient of graph, gradient is doubled at every x. [1] [1] (c) Ep = (𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑓 𝑃)(𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑓 𝑄) 4𝜋𝜀𝑜𝑟 Since (charge of P) x (charge of Q) = constant Ep 1/r Ep = - 3.6 eV at x = 4 x 10-10 m Ep = - 5.1 eV at x = r 3.6/5.1 = r / 4 x 10-10 r = 2.82 x 10-10 m [1] [1]

7 SAJC 2024 Prelims / 9749 [Turn Over 8 (a)(i) 1 mark for each of 4 pts: 1. Correct directions of uniform E & uniform B fields for a charge particle of a stated sign 2. Correct directions of electric force FE & magnetic force FB (they are in opposite directions) 3. Stated FE = FB (mag) (to be undeflected & hence exit slit S3 & be selected) 4. ie 𝐵 𝑞 𝑣 = 𝑞 𝐸 { essential }  𝑣 = 𝐸 𝐵 { essential } [4] (b)(i) ● As coil rotates, there is a change in flux (linkage) ● due to a change in the area of the coil perpendicular to B field By Faraday’s law, an emf is induced (when there is a change in flux linking the coil) [1] [1] (ii) { For a coil rotating in a B field, instantaneous E = NBA cost . Hence max emf = NBA } Any 2 of 4: number of turns, (magnetic) flux density, area of coil & angular velocity/speed of