2020 YIJC H1 Physics Prelims P2 Answers

2020 JC2 H1 Physics Preliminary Examination Paper 2 Suggested Solutions (SECTION A) 1 (a) (i)  (Contact) force on plank by ground – correct direction and labelling  (Contact) force on plank by man – correct direction and labelling  Weight of plank (force on plank by Earth) – correct direction and labelling *Do not accept symbols. B1 B1 B1 Marker’s comment: Poorly attempted. Many students could only identify weight of the plank. Many wrongly labelled the contact force on plank by man as weight of man. (ii) The moment of a force about a point is defined as the product of the force and the perpendicular distance from the line of action of the force to that point. B1 Marker’s comment: Mostly well done. Some did not get any credit as their phrasing of “perpendicular distance” is vague and did not reflect the essence of definition of moment. (iii) Taking moment about A (anticlockwise moments) = (clockwise moments) FB × 5.0 = 880 × 0.5 + 200 × 2.5 FB = (440 + 500) / 5.0 = 188 N M1 A1 Marker’s comment: Mostly well done. Common mistakes include not factoring in the moment due to weight of man or weight of plank. (b) The clockwise moment will increase due to the increase in perpendicular distance (between CG & point A), B1

2 thus the anticlockwise moment due to FB will have to increase to maintain equilibrium as well and magnitude of FB will increase. B1 A0 Marker’s comment: Poorly attempted. While many could correctly state that the CW moment will increase, they did not explain the reason for that increase. Similarly, they did not mention the corresponding increase in ACW to maintain rotational equilibrium and just drew link between increase in CW moment and hence increase in FB. 2 (a) Area under F-x graph is work done in stretching bow. More potential energy is stored Arrows gain more kinetic energy, has less deviation from intended flight path / more accurate OR Arrows gain more kinetic energy, has further range OR Arrow hits target faster B1 B1 Mostly well done. Most could identify that in stretching the bow more potential energy is stored and in turn identify reasonable advantage of that. (b) (i) By counting squares under F-x graph, Elastic potential energy = area under F-x graph = 1.25 J × 65 = 81 J Acceptable to +/- 2 squares M1 A1 Marker’s comment: Mediocre attempt. While some students were able to correctly identify that the EPE is given by area under F-x graph, they were unable to give an accurate estimate as they did not take into consideration the scale of both axes. Some students also used the wrong method – using formula (EPE = ½ kx2) to calculate without realising that F is not a constant. (b) (ii) By Conservation of Energy, Loss in Ep = Gain in Ek for arrow   22181 2 3.5 10 v  v = 68.0 m s-1 M1 A1 Marker’s comment: Mostly well done. Ecf was giv