EJC 2023 J1 H2 Promo P2 MS

©EJC 2023 9749/J1H2PROMO/2023 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2023 9749 PHYSICS MARK SCHEME Paper 2 Structured Questions s/n Answer Marks 1(a) Velocity is the rate of change of displacement Examiner comments Well done by those who memorized the definition. Students are reminded to use definitions from EJC notes as extra words or lack of key words used may unintentionally convey a different meaning. B1 1(b) v2 = u2 + 2as = 2(4)(30) = 240 v = 15.49 = 15.5 m s -1 (3 s.f.) Examiner comments The given acceleration of 4 m s─1 along the slope is due to the resultant force of the weight of the car, the normal force on the car by the slope, and the friction on the car by the slope. Being a “show” question, it was crucial that the substitution to the correct formula is shown clearly. M1 Correct substitution A0

2 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 1(c) Method 1 ux = vx = 15.5 cos 27° = 13.81 m s-1 Taking downward as positive: vy2= uy2+ 2aysy = (15.5 sin 27°)2 + 2(9.81)(40) vy = 28.9 m s-1 2 2 -1 -113 8 28 8 32 m s or 32 0 m s =+=v. . . 1 28 9 64 or 64.5 to the horizontal 13 81θ == °°- .tan . Method 2 Gain in KE = Loss in GPE 22 22 1 1 ()2 1 ( 15 5 ) (9 81)(40)2 32 m s m v u mgh mv . m . v − −= −= = 1 15 5 cos 27cos 64 to the horizontal 32 o.θ −= = ° Examiner comments After point B, the only force acting on the car is its vertically downward weight. The direction of the velocity of the car at point B has both horizontal and vertical components. C1 C1 A1 A1 1(d) Consider vertical direction taking downwards as positive: s = ut + ½ at 2 40 = (15.5 sin 27°)t + ½ (9.81)t 2 t = 2.22 s or -3.66 (rejected) Examiner comments Since the horizontal component of the velocity of the car does not change and horizontal displacement is not given, candidates should use only the vertical component of the velocity to calculate time. M1 A1 32 15.5 cos 27o θ

3 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 1(e) Examiner comments Candidates should write 1.82 instead of 4sin27° on the ay axis. The positions of 9.81 and 1.82 on the ay axis should be roughly proportional to their magnitude. B1 1.82 m s-2 from B to C B1 9.81 m s-2 from B to C 1.82 9.81

4 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 2(a) Hooke’s Law states that the change in length of a material is directly proportional to the force applied on it, provided that the limit of proportionality is not exceeded. Marker’s Comments Most students were penalised for missing out / replacing “limit of proportionality” with other terms. B1 2(b) W = 0.5 Fx = 0.5 (290 000)(0.15) = 21 800 J Marker’s Comments Generally well done. It is possible to obtain the value of k and find EPE using the formula: EPE = ½ k x2. C1 A1 2(c) ( ) 4 0 712 8 10 568000 N 35 568000 291000 N 2 12 5 −−= = = ×= = = ° vuF ma m . cos . t T . T Marker’s Comments Generally well done. Common mistakes: 1. Assume velocity = 70 ms-1 at time = 2 s (Look carefully!) 2. Did not resolving the tension as shown in Fig. 2.1 M1 (acceleration) M1 (2Tcos12.5°) A1

5 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 2(d) Identify all forces 1. Normal contact force 2. Weight 3. F engine / Force by air on plane 4. Friction (on wheels) 5. Force by wire on plane (Tension) Correct direction & relative lengths & proper label for all Normal contact force, Weight (same length, opposite direction) Forward forces: Fengine OR Force by air on plane Backward forces: Air resistance, friction (on wheel), force by wire Marker’s Comments No marks awarded if any of the forces are missing. (optional: air resistance) Rationale for direction of forces: 1. Arrester wire system gives rise to tension (backwards) 2. F engine is forward because engines pushed to full power. 3. Wheels are free rolling, hence the friction on the wheel is backwards. (Refer to H204 Forces Notes: Pg 9 Example 5) 4. Normal contact upwards and weight downwards. 5. Air resistance is optional because it is unclear in the question regarding the speed of the aircraft when the force diagram was drawn. Air resistance can be negligible if the aircraft is travelling at a low speed. B1 B1 Fengine Weight Friction Tension Air resistance Normal contact force

6 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 3(a) The linear momentum of a body is the product of its mass and its velocity. Marker’s Comments Well done across cohort. B1 3(b)(i) Relative speed of approach = Relative speed of separation Taking to the right as positive, ( ) ( ) 12 21 2 -1 2 45 28 18 5 5 m s −=− −− = −− = uu v v . .v . v. To the right. Marker’s Comments Generally well done. A handful of students did not show the equation u1 – u2 = v2 – v1 As this is a “show” question, working must be very clear as that is what markers are looking out for. A1 mark for “Right” will be awarded regardless of the earlier proof. B1 A0 A1 3(b)(ii) By conservation of linear momentum, ( )( ) ( ) ( )( ) ( ) 1 12 2 1 12 2 22 2 00 5 0 45 28 00 5 0 18 55 38 g += + +−= −+ = mu m u mv mv . .m. . .m . m Marker’s Comments Generally well done. Students must be aware that momentum is a vector. B1 A1 3(c) By conservation of linear momentum, ( ) ( )( ) ( )( ) ( )( ) 1 12 2 12 -1 0 050 4 5 0 038 2 8 0 050 0 038 1.35 OR 1.4 m s += + + −= + = mu mu m m v . . . . . .v v Marker’s Comments Generally well done. Students must be aware of that the implication of a perfectly inelastic collision is the objects having the same final velocity. B1 A1

7 ©EJC 2023 9749/J1H2PROMO/2023 s/n Answer Marks 4(a) Velocity is a vector quantity and has both magnitude and direction. Since the object in a circular path keeps changing direction despite a constant speed, it has a changing velocity over time. Since accel eration is the rate of change of velocity, the object has acceleration. Marker’s Comments Unexpectedly well done! Skill demonstrated (or to learn): From definition of “acceleration”, work backwards to argue that velocity has changed due to the changing direction of the object. Then phrase answer from the starting condition in question (object moving in circular path) to the ending condition (has acceleration). B1 B1 4(b)(i)1. ( ) ( )( ) ( ) -1 0 10 L 98 i 1 G 2 E 00 oss n 5 P2 5 13 30 0 1 m s = ° = °= m . . .n g R sin si . Marker’s Comments Being a “show” question, it was crucial that the mathematics to the drop in height from P to Q is shown clearly. M1 4(b)(i)2. ( ) ( ) Work done against air resistance Force distance 60 2 J 600 14 2 0 360 0 022 O 3 R 0.022 1560 0 π π =  ×   ××  = ××   = =  RF . R . . Marker’s Comments Being a “show” question, it was crucial that the mathematics to the distance along the arc from P to Q was shown clearly. M1 4(b)(i)3. By principle of conservation of energy, loss in gravitational potential energy = gain in kinetic energy + work done against resistive forces ( ) 2 -1 10 15 0 100 0 02202 1.6 m s = + = . .v. v Marker’s Comments Part 4(b)(i)1,2 were meant to scaffold your an