2023 AHS Prelim Exam P2 mark scheme with marker's comments

2023 S4 Physics Prelim Exam Paper 2 mark scheme & markers’ comments 1(a) w to x: displacement decreases at an increasing rate / decreases non-uniformly in the negative / downwards direction. x to y: displacement increases at an increasing rate / increases non-uniformly in the negative / downwards direction. All 4 points - 2 marks, 2-3 points - 1 mark, otherwise - 0 (b) 1 (c) 3

Marker’s comments a) Students need to know displacement is positive or negative based on position, but velocity is positive or negative based on direction. When the question asks for direction of motion of object, it is moving downwards or towards the negative direction. Many candidates did not answer the question - addressing the change in displacement and the rate but they wrote about velocity and acceleration. They need to know that we can’t tell the rate of change of velocity from the graph given. b) Height is a distance. In order to know the height, marking out displacement at zero will not help even though that corresponds to the top of his head. c) Many candidate did not label the area under the graph and some label wrongly. This area must be formed with the time axis. NOTE that if you are using the second answer, all 5 correct, 2 marks, 2(a) 80 J (minus 1 mark for wrong or no unit) 1 (bi) Weight 1 (bii) Any diagonal straight line with a less steep gradient. 1 (c) EK gained = EP lost or ½ m v2 = m g h or v = √(2gh) or v = 2 x 10 x 3.25 v = 8.06 m/s (3 s.f.) or 8.1 m/s (2 s.f.) minus 1 mark for no or wrong unit 1 1 Marker’s comments a) Common mistake is reading values off wrongly due to incorrect scale. Many were not aware that there are only 4 small divisions between 1.0 m. bi) Many students chose mass as their answer, which is incorrect. Some students are still unsure of the difference between mass and weight. ‘Force’ was accepted even though the gradient gives a specific quantity (from mgh/h = mg), which is weight.

ii) Very well done. c) Common mistake is to equate ½ v2 = 80, where 80 is the value of EP. Some used the gradient, which they thought was mass, in their calculations. Method mark was awarded. 3(a) It is the point through which its whole weight appears to act. 1 3(b) 2 (c) Upwards force / Tension / Spring force increases. Since the mass is in equilibrium (or Net force is zero) with an additional downwards force applied, the upwards force/tension must have increased. 1 1 (d) Upwards tension is more than downwards weight, Or As there is a net upwards force, from F = ma mass will accelerate upwards. 1 1 1 Marker’s comments a) Many cited ‘on the object’ and were penalised, as the c.g. may not be on the object. Many students clearly has not put in the effort to memorise their definitions, which is such a waste. b) Tension quite often incorrectly identified as normal reaction force. Marks were not awarded if the force drawn does not touch the top surface of the mass and does not pass along the vertical straight portion of spring. Weight must be drawn from the centre of gravity of the mass, vertically downwards. c) Many students were able to correctly conclude that the tension will increase. However, many cited Newton’s Third Law as the cause, which indicates poor understanding of the laws of motion. An action and reaction pair of forces act on different bodies. In this situation, we are focusing on the forces acting on the mass only. Instead, students are to conclude that for an object to be in equilibrium, the resultant force on it must be zero. So the total upward force on mass = total downward force on mass. d) This question requires the use of Newton’s 2nd Law to explain the immediate effect the change had on the mass. Saying that the mass ‘moves at high speed’, ‘springs up’, ‘bounces up quickly’, does not imply that the object is accelerating. Students need to choose their words and phrases carefully. Many students also went on to explain what will subsequently happen, which is not required. T drawn on spring or starting from spring and labelled [1] W drawn from c.g. (eye test) and labelled [1] Add one more mark for definitaion of c.g.

4(a)(i) Pressure is the force acting per unit area. 1 (ii) 𝐹𝐻 𝐴𝐻 = 𝐹𝐽 𝐴𝐽 or 350 25 = 𝐹𝐽 600 FJ = 8400 N (minus 1 mark for wrong or missing unit) 1 1 (b) liquids are incompressible. 1 (c) VH = VJ or 25 x 5 = 600 x d d = 0.208333… = 0.21 cm (2 s.f.) or 0.208 cm (3 s.f.) (minus 1 mark for wrong or missing unit) Work done at H = Work done at J 1 1 Marker’s comments a(i) generally well done, however a number of students mentioned on a /over a unit area which is not acceptable (ii) generally well done except for a number of students that used F x A instead of the correct formula for pressure (b) a number of students mentioned that liquids do not have a fixed shape. However, this property also applies to gases, and does not address the idea of effective operation of the hydraulic press. (c) generally well done. 5(a) (i)&(ii) normal drawn 90o to the mirror 1

angle of incidence drawn between incident ray and normal 2 rays each joining rays R and S respectively, converge at image, obeying laws of reflection Correct image position and labelled with letter I 1 2 marker’s comments (i) was generally well done with a small number of students drawing the normal perpendicular to the ground instead. (ii) large number of students were unable to do this question correctly as they failed to recognise that the image formed is behind the mirror. this led to them force fitting the reflected light rays to converge in front of the mirror, ignoring the laws of reflection. 1m for image behind mirror 1 m for both light rays obeying the laws of reflection (subtract 1 mark for failure to draw arrows or not using dotted lines for the construction lines) (b)(i) Note: the image is inverted 1 (ii) Larger / bigger in size / magnified Further from the lens / image distance is larger 1 1 Marker’s comments (i) this question was quite poorly done as students failed to visualise the orientation of the object and hence the image (as shown in the picture) (ii) a significant number of students did not address the idea of the difference in the image. 6a(i) The helicopter gains electrons from the air / electrons are transferred from the air to the helicopter. The helicopter has excess electrons / has more negative charges than positive charges. 1 1 (ii) The excess electrons will flow to the ground / out of the helicopter. The helicopter will be earthed / neutral / no longer live / the helicopter now has zero potential / is no longer at live potential / there is no potential difference between the helicopter and soldier B / ground. 1 1 b

(b) Q = It or 32 000 x 15 x 10-3 = 480 C 1 1 Marker’s comments (i) Some students wasted time explaining why electrons are transferred to the helicopter by friction which is not what the question is asking. A few students confused charging by friction with charging by induction. Students should note that the object to be charged (in this case the helicopter) is initially neutral. (ii) The term ‘neutrally charged’ is incorrect because being neutral means not charged. It’s like saying someone is deadly alive or lively dead. 7(a) Current is the rate of flow of (electric) charge. 1 (b) Switches and fuses should be connected to the live wire. We can deduce where the live wire is by looking at where