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 no