EJC 2023 J2 H2 PRELIM P2 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2023 9749 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1(a)(i) Precision is determined by the range in the values. Range of Set A = 1.55 – 1.43 = 0.12 cm Range of Set B = 1.58 – 1.42 = 0.16 cm Range of Set C = 1.56 – 1.45 = 0.11 cm Therefore, Set C is more precise as the range of values are smaller than the other two sets. M1 for all correct values A1 1(a)(ii) Accuracy is determined by the closeness of the measured value to the true value. Average values for Set A = 1.52 +1.49 +1.43 +1.52 +1.55 =1.5025 = 1.50 Average values for Set B = 1.42 +1.45 +1.43 +1.51+1.58 =1.4785 = 1.48 Average values for Set C = 1.45 +1.56 +1.47 +1.53 +1.46 =1.4945 = 1.49 Therefore, Set A is more accurate as the average values are closer to the true value of the diameter. M1 for all correct values A1

2 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 1(b)(i) Alternative s-t graph For s-t graph: 1m – for correct sketch of graph for car B and correct displacement value Also, gradient of the line for first 4.0s must be greater than the gradient of the line for the last 9.0s For a-t graph: 1m – for correct sketch of graph for car B and correct acceleration value B1 B1

3 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 1(b)(ii) Minimum distance occurs when both cars have the same speed (v = 60 m s−1). ( ) ( ) ( ) m 1 Distance covered by Car A during the 6.0 s = × 25 + 60 × 6.0 = 255 m 2 1 Distance covered by Ca 4 r B during the 6.0 s : 80 × 4.0 + × 80 + 60 × 2.0 u = 460 m 2 Minim m distance = (500 + 255) - 60 = 295     B1 C1 for both correct answer A1

4 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 2(a) No resultant force (in any direction) No resultant moment (about any point) B1 B1 2(b) Let the length of the rod be L. Take moment about the wall hinge. Sum of clockwise moment = Sum of anticlockwise moment ( )180 sin60 sin502 ooL T L = T = 45.2 N Let the vertical force at hinge be Ry. Let the horizontal force at hinge be Rx. No resultant force in vertical direction: Ry + T cos (180o – 60o – 50o) = W Ry + (45.2) cos 70o = 80 Ry = 64.5 N No resultant force in horizotal direction: Rx = T sin (180o – 60o – 50o) Rx = (45.2) sin 70o Rx = 42.5 N Force by hinge on rod 22 xyR R R=+ 2242.5 64.5R=+ = 77.2 N Let θ be the angle of R from the wall tan x y R R = 42.5tan 64.5 = θ = 33.4o from wall M1 A1 C1 A1 A1

5 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 3a Let speed of X before hitting Y be vi Gain in KE = Loss in GPE 2 1 1 2 2 2 9.81 1.50sin40 4.35 m s ii ii i mv mgh v gh v − = = =    = B0 M0 A1 3bi Let speed of X after hitting Y be vf Loss in KE = Gain in GPE 2 1 1 2 2 2 9.81 0.80sin40 3.18 m s ff ff f mv mgh v gh v − = = =    = EITHER Since collision is elastic, Relative speed of approach = relative speed of separation vi = vy - vf 4.35 = vy - (- 3.18) vy = 1.17 m s-1 OR By conservation of linear momentum, ( )30 4.35 3.18 (1) X X X X Y Y YY m u m v m v mv =+ += By conservation of energy Loss in KE of X = Gain in KE of Y ( ) 2 2 2 2 2 2 1 1 1 2 2 2 30 4.35 3.18 (2) X X X X Y Y YY m u m v m v mv −= −= Taking (2) ÷ (1) 11.17 m sYv −= M1 M1 A1 (either eqn correctly written) M1 A1

6 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 3bii Gain in KE of Y = Loss in GPE of X ( )( ) ( ) 2 2 2 1 2 2 2 0.030 9.81 sin40 1.50 0.80) 1.17 0.192 kg 190 g Y y X X Y y Y Y Y m v m g h m g hm v m m m = = −= = = OR From (b)(i) Using the law of conservation of momentum in (1) and energy in (2) ( ) 22 2 2 230 4.35 3.18 (1)YYmv+= ( ) 2 2 230 4.35 3.18 (2)YYmv−= Taking (1)2 ÷ (2) 193 gYm = = 190 g (2 s.f.) B1 M1 A0 B1 M1 A0 3ci The collision is inelastic because the relative speed of approach is non-zero but the relative speed of separation is zero, which means that they are not equal. A1 3cii By conservation of linear momentum, ( ) ( )( ) -1 -1 ' ' 0.030 4.35' 0.030 0.190 ' 0.5932 m s ' 0.59 m s X i X Y Xi XY m v m m v mvv mm v v v =+ = + = + = = M1 A0

7 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 3ciii Loss in KE = Gain in GPE + work done against friction ( )( ) ( ) ( )( ) ( )( ) 2 2 2 1 ' sin502 1 ' sin502 1(0.030 0.190)(0.59 )2 (0.030 0.190) 9.81 sin50 2.2 0.0099 m X Y X Y X Y X Y m m v m m gd fd m m v m m g f d d d + = + + + = + + + = +   + = OR Resultant force along the slope = friction + component of weight 𝑚𝑎 = 2.2 + 𝑚𝑔 sin 50° 𝑎 = 2.2 0.220 + (9.81) sin 50° 𝑎 = 17.51 m s−2 where 𝑎 is directed downslope. Using 𝑣2 = 𝑢2 + 2𝑎𝑠 , where s is the distance travelled up the slope, d. 0 = (0.59)2 − 2(17.51)𝑑 𝑑 = 0.0099 m B1 M1 A1 (M1) (C1) (A1)

8 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 4a Electric force per unit positive charge on a small stationary test charge at that point. B1 4bi Considering the distance between the nearest two equipotential lines around point A (4V and 0V) d = 3.5 cm E = V/d = (4−0)/(3.5×10−2) F = eE = 1.6 × 10−19 × (4−0)/(3.5×10−2) = 1.8 × 10−17 N M1 A1 4bii From point B to C, V = −6 −4 = −10V Change in EP = eV = −1.6 × 10−19 (−10) = 1.6 × 10−18 J (Gain) By Principle of Conservation of Energy, Loss in Kinetic energy = Gain in Electric Potential Energy ½ mu2 − ½ mv2 = 9.6 × 10−19 ½ (9.11 × 10−31)(5.3 × 106)2 − ½ (9.11 × 10−31)v2 = 1.6 × 10−18 v = 5.0 × 106 m s−1 C1 C1 A1 c The change in potential from P to Q can be determined by estimating the area under the graph, 1 square : (0.5 × 103) V m−1 × 10−2 m = 5 V 18 squares : 18 × 5 = 90 V Change in potential energy = 1.6 × 10−19 × 90 = 1.44 × 10−7 J C1 C1 A1

9 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 5(a) 3 0.56 93 6.0 10 diode VR I −= = =  A1 5(b) 3 3 3 3 1.2 0.56 (6.0 10 ) 110 diode RE V V R R − =+ = +  = C1 A1 5(c) 12 3 1.2 50 200 4.8 10 A EI RR − = + = + = C1 A1 5(d) 3 3 26.0 10 + 4.8 10 1.1 10 AtotalI − − −=   =  2 2 (1.1 10 )(1.2) 1.3 10 W totalP I E P − − = =  = C1 A1 5(e) 50 1.0 250 x = 0.20 mx = C1 A1

10 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 6(a)(i) The magnetic force is downwards (by FLHR, B points into the paper and current is to the left) in the region between the plates, hence the electric force must be upwards. Since the particles are negatively charged, the electric field must be downwards (opposite to the electric force). Hence, P is positive. (Plate Q is grounded, hence is at 0 V.) B1 B1 6(a)(ii) To move undeflected, the net force must be zero. EBFF qE qvB Ev B = = = M1 A0 6(a)(iii) Ev qvB qEB   , or FB > FE. Since the magnetic force FB points downwards, the particles will be deflected downwards. (Note that the path is not parabolic.) B1 B1 6(b)(i) The magnetic force is always perpendicular to the velocity. Hence, it does no work on the particles. The kinetic energy, hence the speed, of the particle remain constant. B1 B1 6(b)(ii) The magnetic force provides the centripetal force. 2 2 With , v q vqvB m r m Br Ev B qE m Br =  = = = M1 M1 A0 undeflected path