2025 EJC H2 Prelim P2 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2025 9749 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1(a) k = [(3)(1.29) (3.3 × 102)2] / (9.9 × 104) = 4.3 A1 1(b) Method 1: fractional uncertainty = 8 0.092 0.07 0.30100 1.29 + + = 0.3 4.2 1.3 1k =  = = (allow 1 s.f.) Method 1: ( ) ( ) ( ) ( ) ( ) 22 max 4 22 max 4 3 3.3 10 1.08 1.29 0.09 5.7129.9 10 0.93 3 3.3 10 0.92 1.29 0.09 3.1329.9 10 1.07 1 5.712 3.132 1.29 12 k k k   + ==    − ==   = − = = 41k = C1 C1 C1 (calc of both kmax and kmin) C1 A1 1(c) unit of fA = unit of speed unit of A = unit of (speed/f) unit of A = 1 1 m s s − − = m M1 A1

2 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 2(a)(i) 21 2s ut at=+ (Taking upwards as positive) ( ) ( ) 217.8 5.9sin60 9.81 2 1.9 s (or 1.89 s) o tt t − = + − = C1 A1 2(a)(ii) Vertically: (Taking upwards as positive) v2 = u2 + 2as vv 2 = (5.9sin60)2 +2(–9.81)(1.2–9.0) vv = -13.4 m s-1 or v = u + at = -5.9 sin60 + 9.81 (1.885) = -13.4 m s-1 (Correct substitution and answer) horizontally: vh = 5.9cos60 vh = 2.95 (Correct substitution and answer) 13.4tan 2.95 78o   = = C1 C1 A1 2(b)(i) speed decrease, (So) viscous force / drag (force) decrease (resultant force decreases as) upthrust and weight remain the same B1 B1 2(b)(ii) U Vg= 2(1000)(9.81)(7.5 10 )−=U = 736 N = 740 N M1 2(b)(iii) resultant force = 740 + 950 – (78)(9.81) = 925 acceleration = F / m = 925 / 78 = 12 m s–2 (vertically) upwards C1 A1 A1

3 ©EJC 2025 9749/J2PRELIM/2025 Qns Answer Marks 3(a) For a system in equilibrium, sum of clockwise moments about a (or any) point equals sum of anticlockwise moments about the same point B1 B1 3(b) Let the length of the rod be L. Take moment about the point of contact between the rod and the ground. sum of clockwise moments 1000 cos302   oL OR 1000 sin602   oL sum of anticlockwise moments ( )sin17oTL OR ( )cos73 oTL OR sum of clockwise moments ( )1000 cos30 cos47 sin302  + o o oL TL OR ( )1000 sin60 sin43 cos602  + o o oL TL sum of anticlockwise moments ( )sin47 cos30ooTL OR ( )cos43 sin60ooTL T = 1500 N (or 1480 N) C1 C1 (C1) (C1) A1

4 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 4(a) By Principle of Conservation of Linear Momentum (0.50)(2.0) + (0.20)(0) = (0.50)(v1) + (0.20)(v2) [1] By Relative speed of approach = relative speed of separation u1 − u2 = v2 − v1 2.0 − 0 = v2 − v1 [2] Solve equations [1] and [2], v2 = 2.857 = 2.9 m s−1 C1 C1 A1 4(b) By Principle of Conservation of Energy, Loss in kinetic energy = Gain in Gravitational Potential Energy ( ) ( ) ( ) 2 2 1 02 1 2.857 9.812 0.416 m Consider maximum angle cos 1 cos 0.416 1.5 1 cos 43.7 mv mgh h h h L L L    −= = = = − = − =− = C1 C1 A1 4(c) There is an external resultant force acting on the bob provided by the tension in the string and weight. B1

5 ©EJC 2025 9749/J2PRELIM/2025 Qns Answer Marks 5(a) Rate of change of angular displacement B1 5(b)(i) At minimum speed, normal contact force = 0 AND Weight provides centripetal force B1 2mvmg r= 2 9.81 0.12 v= C1 11.08 or 1.1 m sv −= A1 5(b)(ii) Loss in EPE = Gain in GPE + Gain in KE 2211 22kx mgh mv=+ 22 -1 11 (0.16) (0.076)(9.81)(0.24) (0.076)(1.1)22 17.6 18 N m k k =+ == OR 22 -1 11 (0.16) (0.076)(9.81)(0.24) (0.076)(1.08)22 17.4 17 N m k k =+ == B1 C1 A1 (C1) (A1) 5(c) Loss in KE = Work done against resistive force 0.23 = (average resistive force) ( )0.30 0.25 2 (0.12)++ average resistive force = 0.18 N B1 A1

6 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 6(a)(i) (vertically) downwards B1 6(a)(ii) magnetic force (on sphere) is always perpendicular to its velocity no work done by force, hence no change in kinetic energy and speed. M1 A1 6(b) mg = Eq E = (1.6 × 10–10 × 9.81) / (0.27 × 10–9) = 5.8 N C–1 C1 A1 6(c) Magnetic force provides centripetal force Bqv = mv2 / r B = (1.6 × 10–10 × 0.78) / (0.27 × 10–9 × 3.4) = 0.14 T B1 C1 A1 Qns Answer Marks 7(a) Transition emits (one) photon with energy equal to the difference in energy between the two levels B1 photon energy = h x (frequency of radiation) B1 Thus, single frequency of radiation for each transition 7(b)(i) line to the left of the pair in Fig. 8.2, labelled A AND larger gap between line A and the nearest of the pair in Fig. 7.2, than between the lines in the pair B1 7(b)(ii) line to the left of both the pair in Fig. 7.2 and line A, labelled B AND larger gap between line B and line A, than between line A and the nearest one of the pair in Fig. 7.2 B1 7(c) E = Ephoton = hf (E3 − E1) = (E2 − E1) + (E3 − E2) = hfA + hfB E3 = E1 + h(fA + fB) C1 A1

7 ©EJC 2025 9749/J2PRELIM/2025 Qns Answer Marks 8(a) hc / λ = Φ + EMAX AND hc = gradient C1 gradient = e.g. [(0.40 – 0.20) × 1.60 × 10–19] / [(2.25 – 2.09) × 106] (working needed) (= 2.0 × 10–25) M1 h = (2.0 × 10–25) / (3.00 × 108) = 6.7 × 10–34 J s (both working and answer needed) A1 8(b) straight line with same gradient as the original AND straight line with x-axis intercept greater than 1.93 × 106 m–1 B1 8(c) (for EMAX = 0,) 1 / λ0 = 1.93 × 106 (m–1) f0 = (3.00 × 108)(1.93 × 106) = 5.8 × 1014 Hz C1 Φ = hf0 = ( 6.63 × 10–34) (5.8 × 1014) = 3.9 × 10–19 J = 2.4 eV M1 so potassium A1 8(d) more photons (per unit time) so (rate of emission) increases A1

8 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 9(a)(i) 11 10distance 0.28 AU (1.5 10 ) 4.2 10 m=   =  ( ) 26 22 10 4 4 2 3.8 10 4 4 4.2 10 1.714 10 1.7 10 W m PI r I   − ==  =  =  B1 B1 9(a)(ii) ()receivedP IA efficiency= 4(1.7 10 )(6.5)(0.28)receivedP = 43.1 10 WreceivedP = C1 A1 9(b)(i) 4 8 8 1.7 10 100 0.15 3.00 10 99.8 100 8.5 10 Pa r IPR c −  = + =     = M1 9(b)(ii) 21 2 kE mv p mv = = Showing BOTH the above to be awarded 1 mark 2 1 2 =  k pEm m Therefore 2 2 k pE m= M1 M1 9(b)(iii) 87(8.5 10 )(6.5) 5.5 10 NF PA −−= =  =  Using 2 2 pE m= , 2NNF p mEtt    =  =       7 27 6 195.5 10 2(4 1.66 10 )(5.0 10 1.6 10 )N t − − −  =      12 15.4 10 sN t −= C1 C1 A1 9(c) 5 4 count rate 3.6 10activity = sensitivity 1.2 10 30 GBq =  = C1 A1

9 ©EJC 2025 9749/J2PRELIM/2025 Qns Answer Marks 9(d)(i) 1 1 2 ln2 ln2 0.035 min20t  −=== 1 1 2 ln2 ln2 0.069 min10t  −=== A1 A1 9(d)(ii) 21 0.01 0.99 0.01 (0.035) 0.99(0.069) 6.87 10 min eff    −− =+ =+ = M1 9(d)(iii) ( ) 11 15 27 2.2 10 1.0 10 13 1.66 10 oN − − = =   ( ) 215 (6.9 10 )(15) 14 1.0 10 3.6 10 t oN N e e  − − − = = = C1 C1 A1 9(d)(iv) 13 13 0 7 6 1N C e neutrino→ + + (accept anti-neutrino and symbols , ) A1 9(e) Any of the following: relate distance from the Sun to intensity via the inverse-square law solar activity (e.g., solar flares) panel orientation, dust accumulation temperature effects on solar cell efficiency B1