2024 JPJC H1 Prelim P2 solutions

2024/JPJC/Prelim/8867/02 Answers to 2024 JC2 H1 Preliminary Examinations Paper 2 Suggested Solutions: No. Solution Remarks 1(a)   4 4 P kAT Pk AT I      2 3 3 4 2 4 2 4 S base units of kg m sW kg s K m K m K k [1] for correct base units of power [1] for correct answer 1(b) 3 3 133 84 10 53.133 m s0.56 P v b Pv b      3 3 1 1 1 1 3 3 1 1(0.05) (0.07)53.133 3 3 2.1253 m s 2 m s P v b Pv b v P b v P b v v v                [1] for correct value of v [1] for correct substitution [1] for correct answer to 1 s.f. 2(a) Using 21 2s ut at  for vertical direction 21 2 11 2 0 (1.5) ( 10)(1.5) (10)(1.5) 7.5 m s u u       OR Using v u at  for reaching maximum height, 1 2 11 2 0 ( 10)( 1.5) (10)( 1.5) 7.5 m s u u         [1] substitution [1] answer 2(b) [1] line starts at 0 m [1] ends at 15 m w.r.t. label at y-axis 1.5 t / s Sh / m 10 20

2 2024/JPJC/Prelim/8867/02 2(b) [1] horizontal line at 10 m s1 w.r.t. label at y-axis 2(b) [1] line starts at 7.5 m s1 and cuts x-axis at 0.75 s [1] ends at 7.5 m s1 w.r.t. label at y-axis 3(a) The rate of change of momentum of a body (system) is proportional to the resultant force that acts on it and the momentum change takes place in the direction of the force. [1] 3(b) Impulse is defined as the product of force and time of impact. [1] 3(c)(i) [1] for drawing and labelling   2 sin35M m g Or in words [1] for drawing and labelling T 3(c)(ii) Trolley A:       2 sin35 2M m g T M m a ------ (1) Trolley B:   sin35T Mg Ma ------(2) (1)+(2):       2 sin35 sin35 2 2M m g Mg M m a   2 sin35 2 2mg M m a [1] for correct equation [1] for correct equation [1] for showing 1.5 t / s 1.5 t / s vh / m s1 vv / m s1 10 10 10 35° T (M + 2m) g sin 35°

3 2024/JPJC/Prelim/8867/02 [Turn over   sin35mga M m      2 sin35 2 2 mg M m a 3(c)(iii)          225 30 9.81 sin35 0.338 m s100 95 30a    219.0 0.3382 7.3 s t t [1] for correct acceleration [1] for correct substitution [1] for correct answer 3(c)(iv) The flexible buffer stops the trolley over a prolonged period of time. Applying Newton’s second law, this means that the rate of change of momentum of the trolley is reduced . The resultant force acting on the trolley is therefore reduced. [1] [1] 4(a) [2] all correct Or [1] any 3 correct (ignore length of arrows) 4(b)(i) Taking moments about point A, sum of anti-clockwise moments = sum of clockwise moments 4.0 4.0 9.81 1.0 25.0 9.81 2.0 60.0 9.81 3.6 662 N B B T T            [1] sub [1] ans 4(b)(ii) Sum of forces acting on scaffold = 0 N Upward forces = downward forces 4.0 25.0 60.0 89 9.81 662 211N A B A T T g g g T         [1] correct substitution [1] ans 4(c) As the painter walks towards point A, the tension in rope A will increase and the tension in rope B will decrease. By considering the moment of the forces about point A, as x decreases, the sum of clockwise moment decreases. (Since the beam is in equilibrium ), tension in rope B must decrease also. Hence the tension in rope A must increase , as total downward force is still the same. [1] [1] 5(a) Since the direction of motion of the body is always changing, its velocity is not constant and thus a resultant force acts on the body. Since the speed of the body is constant, the resultant force acts perpendicular to its velocity and thus, direction of motion, which is towards the centre of the circle. [1] [1] A B TA TB Wscaffold Npail Npainter

4 2024/JPJC/Prelim/8867/02 5(b)(i) The net force towards centre of the circle provides the centripetal force required to move in a circular path. 2 2 cos cos , as . mvT mg r mvT mg r L L        [1] for statement and equation [1] for r = L 5(b)(ii) The point where the force in the rod changes from tension to compression is when T = 0 N.      2 2 22 cos 0 cos 2.0cos 0.51 9.81 0.80 120.6 120 mvT mg L mvmg L v gL                   [1] for T = 0 N [1] for correct substitution and answer 5(b)(iii)1 . [1] correct R and mg A B L  mg T cosmg  v rod R, force rod exerts on bob mg, weight

5 2024/JPJC/Prelim/8867/02 [Turn over 5(b)(iii)2 . For circular motion,         2 2 2 0.050 2.00.050 9.81 0.80 0.24 N mvmg R r mvR mg L        [1] for statement and equation [1] correct answer 5(b)(iv) [1] graph is cosine and cuts at about 120° [1] correct max. and min. T values 5(b)(v) The K.E. of the mass is constant but its G.P.E. is constantly changing. So as the mass is moving upwards, an external device is needed to supply energy to increase its G.P.E. As the mass moves downwards, its G.P.E. is decreasing and the external device must absorb the lost in G.P.E. [1] energy supply [1] energy absorb 6(a)      3 1 161 10average speed 4 60 48 60 20 9.3 m s       [1] for substitution [1] for correct answer 6(b) st 3 11 10 average speedFor 1 measurement, 5 m s200   3 11 10average speedFor 2nd measureme 3.3 m s300nt,   3 14 10average speedFor 3rd measuremen 5 m s800t,   3 13 10average speedFor 4th measureme 4.3 m s700nt,   3 11 10average speedFor 5th measuremen 5 m s200t,   3 14 10average speeFor 6th measurement, d 13.3 m s300   3 11 10average speedFor 7th measuremen 4 m s250t,   T  0

6 2024/JPJC/Prelim/8867/02 [1] for correct low average speed for 1 km to 2 km, 6 to 9 km, 14 to 15 km [1] for correct medium average speed for 0 to 1 km, 2 to 6 km, 9 to 10 km [1] for correct high average speed for 10 km to 14 km [1] all speed – distance graphs are horizontal 6(c)(i)   work done 24 3000 72 000 J Fd   [1] 6(c)(ii)    3 1 72 10power per unit mass 700 78 1.3 W kg    . [1] correct substitution [1] correct time of 700 s [1] correct answer 6(c)(iii) From the 6th km to the 9th km mark, the K.E. is constant. As the height increased, as such, energy is needed as there is an increase in G.P.E.      1 power per unit mass to increase G.P.E. 9.81 400 700 5.6 W kg mgh mt     1 1 1 total power per unit mass 1.3 W kg 5.6 W kg 6.9 W kg       [1] correct gain in height [1] correct 5.6 W kg−1 [1] correct answer 6(d)(i)    3 1 12000 10power per unit mass 70 24 60 60 1.98 W kg      [1] for correct answer

7 2024/JPJC/Prelim/8867/02 [Turn over 6(d)(ii) The cyclist needs higher power per unit mass as the cyclist has to provide energy as work against resistive force, as well as to move up hills while cycling, increasing the cyclist’s G.P.E. [1] any correct Physics relevant to difference 7(a)(i) At V = 12 V, I = 2.5 A 0.208V I At V = 6 V, I = 1.25 A 0.208V I At V = 9.6 V, I = 2.0 A 0.208V I Since ratio of I to V is constant, I is proportional to V. [1] show at least 2 values of ratio of V  or V  . [1] conclusion 7(a)(ii)1. Resistance of resistor X = 12 2.5 = 4.8  [1] answer 7(a)(ii)2. Current in wire AB = 9.0 4.0 5.0 = 1.0 A [1] answer 7(a)(ii)3. Current in resistor X = 9.0 4.8 2.7 = 1.2 A [1] answer 7(a)(ii)4. Resistance of wire AC = 0.70 4.01.0  = 2.8  Potential difference across AC = 2.8 9.04.0 5.0  = 2.8 V Potential differenc