AMKSS Prelim P2 Ans

SoluƟons for 6091(Revised) Paper 2 2024 Preliminary examinaƟon Pure Physics Paper 2 QuesƟon SoluƟon Mark allocaƟon 1(a) Scale 1cm: 2N (Do not accept ‘=’) Resultant force Accept between 31.5 N to 36.8 N DirecƟon: The resultant force makes an angle of 59° with the horizontal (Accept between 58° to 60°) Diagram: Directions indicated.  Single arrow for 30 N and 18 N force  double arrow for resultant force  Angle indicated.  Forces indicated(not length) Do not accept the following: DirecƟons (north east) Odd scales used for scale Eg 1: 3N; 1 cm : 6N There were students who reversed the scale order Eg 2N : 1cm and this is also unacceptable. [1] [1] [1] [1] [1] 1(b) Resultant force has increased Weight has decreased as the gravitaƟonal field strength on moon is weaker than the earth Many students missed out on menƟoning that the gravitaƟonal field strength is weaker on the moon compared to the earth. They had merely menƟoned that there is a difference in gravitaƟonal field strength and this is not to be accepted. [1] [1]

1(c) Increase surface area of balloon or Increase density of air/temperature of air (any one factor) Do not Accept: Increase the mass of the balloon. Students need to understand that most of the makeup of the balloon is air, which is light. The actual mass of the material of the balloon is very light and not feasible to increase this. Any slight increase in mass will only cause negligible change in the resisƟve force . Hence this factor will not be accepted. [1] 2(a)(i) Total clockwise moments 0.150 x 18 + 30 x 0.460 =2.7 + 13.8 =16.5 Nm Many students had only calculated one clockwise moment and hence lost this mark. There are also students who sƟll exhibited weak understanding of the perpendicular distance from the acƟng force to the pivot and subsƟtuted the wrong value for the perpendicular distance in their calculaƟons. Note: No half mark for the calculaƟon of just one clockwise moment. Students will also lose credit if they fail to state the formula before the subsƟtuƟon of values. [1] 2(a)(ii) Applying the principal of moments about the pivot, Sum of anƟclockwise moments = Sum of clockwise moments F x 0.04 = 16.5 F =412.5 N A sizable number of students did not write the statements in blue and hence lost credit for the failure to menƟon. [1] 2(b)(i) There is no change to the anƟclockwise moment as the total clockwise moments has not changed Not well done. There are students who tried applying the law of conservaƟon of energy to this quesƟon on moments and failed to explain accurately. [1] [1]

2(b)(ii) The perpendicular distance between F and pivot has decreased hence F has increased. This part was poorly aƩempted. Not many students realised that the perpendicular distance has reduced. Many sƟll saw that there was no change to the perpendicular distance. [1] [1] 3(a) (i) Increasing acceleraƟon (do not accept acceleraƟon only) (ii) constant deceleraƟon (do not accept deceleraƟon only) There are students who made contradictory statements and were not awarded credit Eg: The velocity is increasing at increasing rate, hence it experiences constant acceleraƟon Eg the velocity is decreasing at a constant rate, hence it experiences zero deceleraƟon. MisconcepƟon: acceleraƟon in negaƟve direcƟon [1] [1] 3(b) The area under the graph from t = 0 s to t= 2 s is smaller than t= 2 s to t = 7s. 3(c)(i) Note: Axis label must be wriƩen with units, else credit will not be given. There are also students who drew a curve instead of a straight line. [1] 3(c)(ii) Area = 0.02 = 0.5 x t x 0.26 t= 0.154 s This part is badly aƩempted. Students started solving by considering the gradient instead of the area. [1] 3(c)(iii) a = v-u/t = (0.26-0) /0.154 = 1.69 m/s2 Do note that mark will be deducted it formula and/or units are not given in the final answer [1]

4(a) Work done = mgh = 250 x 10 x 15 = 37500 J There are many students who calculated the total energy at A and did not read the quesƟon carefully. [1] 4(b) E(A) = E(B) Loss in GPE = gain in KE + WD (FricƟon) 37500 = 0.5 x 250 x (132 – 22) + WD (fricƟon) WD(fricƟon) = 37500 – 20625 = 16875 J This part is badly aƩempted as students forgot to take into consideraƟon that there is also kineƟc store at point A and hence was not able to get any credit for this part. [1] [1] 4(c) Any one The final speed at B will be lower (1). The total change in the energy in the gravitaƟonal store of the roller coaster is the same. Deepening the dip will lead to the fricƟonal forces acƟng over a longer distance. Thus, more energy is transferred to the internal store of the surroundings due to fricƟon and the final speed is lower (1). Or The speed of the roller coaster at the boƩom of the dip will be greater (1). There is a greater loss of energy in the gravitaƟonal potenƟal store of the roller coaster from the start of the ride to the boƩom of the dip. [1] Do not accept: The speed will change. This is a vague statement as there is no menƟon whether there will be an increase or decrease in the value of the speed. [2] 4(d) Energy cannot be created nor destroyed. It can only be converted from one form to another. The total energy in the system remains constant Many students missed out on the statement in blue and lost half a mark There are sƟll students who used ‘converted from one type to another’ instead of ’form ‘ [1] 5(a) For longitudinal waves, the direcƟon of wave travel is parallel to the direcƟon of the vibraƟon of the parƟcles. Sound wave There are many students who were confused between transverse and longitudinal wave and cited wrong examples eg electromagneƟc waves instead. Do not accept: longitudinal waves has compressions and rarefacƟons. [1] [1]

5(b)(i) Displacement = 3 x 0.1 m = 0.3 m There are only a handful of students who managed to get this right [1] 5(b)(ii) Speed = distance/ Ɵme =0.3/(0.882 x 10-3) =340 m/s Many students did not take into account the prefix for the Ɵme(milli) and hence obtained wrong answers for this part. There are also some students who were confused between milli and cenƟ. 6(a) More stable than compared to the whole cube Many students did not realise that the new structure of the cube shows a lighter top with a heavier boƩom and hence stability has been increased. [1] 6(b) PosiƟon of CG is lowered. [1] 7(a) Pressure(Q) =Pressure(P) P(atm) + (0.2) (1000)(10) =P(atm) + x(800)(10) X = 0.250 m This part is well done. Almost all the students managed to gain credit for this part. [1] [1] 7(b) No difference. X is independent of the gravitaƟonal field strength Some points for students to take note: GravitaƟonal field strength is not the same as gravitaƟonal force and hence should not be used interchangeably. Pressure is not the same on moon compared to on earth and hence this reasoning is also not acceptable. [1] 8(a)(i) 8(a)(ii) Angle x 1.5 = sin 50° /sin x X = 30.7° Angle y=30 +30.7 =60.7° [1] [1]

8(a)(iii) 8(b) N = 1/ sin c C = sin-1 (1/1.5) C = 41.8° The failure to gain full credit for this part was surprising as students should be well versed with the usage of this formula Total internal reflecƟon [1] [1] 9(a) I =0.48 – 0.16 =0.32 A [1] 9(b) E = 0.48 x 12 + 0.16 x 36 E= 11.5 V Majority of the students failed to consider the calculaƟons in blue and hence was unable to obtain the EMF of the system. There are also students who thought EMF was the total current. [1] [1] 9(c) EMF: Amount of work done by the electrical source to move a unit charge round the enƟre circuit P .D: Amount of work done to move a unit charge across an electrical component. Note: EMF is not charge. There are many students who associated EMF with charg