2025 EJC H2 Prelim P3 MS

©EJC 2025 9749/J2H2PRELIM/2025 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2025 9749 PHYSICS MARK SCHEME Qns Answer Marks 1(a) (incident) wave reflects at end/top of tube (incident) wave and reflected wave superpose waves have same frequency, wavelength and speed B1 B1 B1 1(b)(i) line has maximum value of amplitude at h = 0 and h = 0.60 m only AND line has minimum/zero value of amplitude at h = 0.30 m only modulus cosine graph (Sharp at h = 0.30 m) M1 A1 1(b)(ii) line has maximum value of amplitude at h = 0 and h = 0.60 m only AND line has minimum/zero value of amplitude at h = 0.30 m only AND modulus cosine-squared graph B1 1(c)(i) vertical/along length of tube/along axis of tube B1 1(c)(ii) phase difference = 0 A1 1(c)(iii) phase difference = 180o A1 1(d) v = fλ f = 340 / (2 × 0.60) C1

2 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks = 280 Hz A1 1(e)(i) f = 340 / 0.60 = 570 Hz (or 2 x 280 = 560 Hz) A1 1(e)(ii) 0.60 / 4 = 0.15 m A1

3 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 2(a)(i) 2 62 32 (1.1 10 )(120 10 ) 1.1( 10 )2 1.38898... 1.4 (2 s.f.) LLR A r    −− − == =  = =  M1 2(a)(ii) Let potential at Y be 5.0 V RPQ//1.8 Ω = 1.4 1.8 1.4 1.8  + = 0.7875 Ω PD across RPQ 0.7875 5.00.7875 2.5= + = 1.198 V C1 PD across XY = 5.0 – 1.198 2 = 4.4 V C1 A1 2(b)(i) Current in wire A is in magnetic field produced by current in wire B. Wire A experiences a magnetic force towards wire B according to Fleming’s Left Hand Rule. B1 From Newton’s third law, an equal and opposite force acts on wire B. (or apply FLH rule again on wire B) B1 2(b)(ii)1. IA = −3.0 cos (200πt) = −3.0 cos (200π(6.5 x 10-3)) = 1.76 A From graph, IB = 1.2 A C1 For both IA and IB ( ) 0 7 6 2 4 10 (1.2) 4 8 102 0 05 BB d .( . )     − − =  = =  I F per unit length on wire A = BBIAL / L = BBIA = 64 8 10 1 76( . )( . ) − C1 = 68 4 10. − N m-1 A1

4 ©EJC 2025 9749/J2H2PRELIM/2025 2(b)(ii)2. • positions and shape of graph during intervals of non-zero force • positions of intervals of zero force B1 B1 2(b)(ii)3. 2 2 30 (15)2 34 W rms .PR ==  = I A1 force per unit length / N m –1 t / ms

5 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 3 (a)(i) (W =) 2.60 ×105 × (3.80 – 2.30) × 10–3 = 390 J A1 3(a)(ii) no (total) change (in internal energy) gas returns to its original temperature/ state/ no change to pressure and volume B1 B1 3(b) A to B row all correct (1370, – 390, 980) B to C row all correct (0, 550, 550) C to A row: ΔU adds to the other two ΔU values to give zero C to A row: w = 0 and q adds to w to give ΔU value complete correct answer: B1 B1 B1 B1 3(c) when vapourising, greater change in separation of atoms or molecules/ bonds are completely broken (compared to partially broken), hence a greater change in potential energy and internal energy greater change in volume, hence a greater work done B1 B1

6 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 4(a)(i) product of number of turns of coil, (magnetic) flux density and area area perpendicular to the (magnetic) field OR Magnetic flux linkage through a loop is product of magnetic flux through the loop and number of turns of wire in the loop. Magnetic flux is the product of an area and component of magnetic flux density perpendicular to that area. M1 A1 B1 B1 4(a)(ii) flux = B × πr2 = 0.17 × π 0.362 = 6.9 × 10–2 Wb time for one revolution = 1 / 25 s e.m.f. = rate of cutting flux or t   = 0.069 × 25 = 1.7 V C1 C1 A1 4(a)(iii) current (in disc) is perpendicular to magnetic field causes force to act on disc OR current (in disc) is perpendicular to magnetic field OR current causes force to act on disc force opposes rotation of disc Fleming’s left-hand rule indicates current is from rim to axle B1 B1 B1

7 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 4(b)(i) ring cuts (magnetic) flux and causes induced e.m.f. in ring (induced) e.m.f. causes (eddy/induced) currents (in ring) currents (in ring) cause magnetic field (around ring) two fields interact to cause resistive/opposing force OR current (in ring) is in a magnetic field which causes resistive force OR currents (in ring) dissipate thermal energy (thermal) energy comes from energy of oscillations B1 B1 M1 A1 (M1) (A1) (M1) (A1) 4(b)(ii) current cannot pass all the way around the ring (induced) currents smaller (OR no currents) smaller resistive force (so more oscillations) (OR no resistive force) or smaller rate of dissipation of energy (so more oscillations) B1 B1 B1

8 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 5(a)(i) energy required to separate the nucleons (in the nucleus) to infinity M1 M1 5(a)(ii) curve starting close to the origin and forming a single peak peak shown to left of centre, with steep line on LHS of peak and shallow line on RHS of peak B1 5(b)(i) Fusion B1 5(b)(ii) both particles have low A values or both particles are at left-hand end of graph He-3 has higher binding energy (per nucleon) than the total binding energy of both H-2 B1 B1 5(c) ∆m = [(2 × 2.014102) – (3.016029 + 1.008665)] u = 0.00351 u (Correct substitution and correct answer) E = ∆mc2 = 0.00351 × 1.66 × 10–27 × (3.00 × 108)2 = 5.24 × 10–13 J (Correct substitution and correct answer) 1.0 mol of deuterium forms 0.500 mol of helium-3 total energy = 0.500 × 6.02 × 1023 × 5.24 × 10–13 (Correct substitution) = 1.58 × 1011 J C1 C1 C1 C1 A1

9 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 6(a) upthrust is greater than weight (resultant force is) upwards B1 B1 6(b) At equilibrium, Upthrust = Weight  A =hg Mg ---(1) Taking downwards positive, at displacement x, Net force = mass x acceleration −=Mg Upthrust Ma −=Mg Vg Ma where V is the volume of water displaced ()− + =Mg A h x g Ma − − =Mg Ahg Axg Ma ----(2) Sub (1) into (2):  − − =Ahg Ahg Axg Ma   −= =−  Axg Ma Agax M OR Additional upthrust = net force Axg Ma = Since a and x are opposite in directions, Agax M =− Note: For “show that” question, the first method is preferred. M1 M1 M1 (M1) (M1) M1 6(c) A, ρ, g and M are constant, hence acceleration proportional to displacement negative sign indicates acceleration and displacement in opposite directions B1 B1 6(d)(i) ω = 2π / T ω = 2π / 1.3 = 4.8 rad s–1 C1 A1 6(d)(ii) ω2 = Aρg / M 4.82 = (4.5 × 10–4 × ρ × 9.81) / 0.17 ρ = 900 kg m–3 C1 C1 A1

10 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 6(d)(iii)1. Total energy = maximum Kinetic energy Max velocity max maxvx = = 2 2 2 2 max 1 1 1 ()2 2 2 oomv m x m x == algebra leading to final expression B1 M1 M1 6(d)(iii)2.   21(0.17) (4.8)(0.20)2 0.078 J= A1 6(d)(iv) 0.926 decrease in energy = 0.078 – (0.078 × 0.926) = 30.7 mJ C1 A1 6(d)(v) wave with same period (or slightly greater) AND peak height decreasing successively B1 B1