TKGS_2024 Phy Prelim Paper 2_Answers

Solution to 2024 Sec 4 Physics Prelim Exam Paper 2 Qn Solution 1(a) 1m for labelling v, 8.4, t 1m for straight line Marker’s Comments Students need to read the question carefully. A ball is thrown vertically down at a speed of 8.4 m/s (initial velocity) from point A which is 5.0 m above the ground. Assuming that air resistance is negligible (object falls with acceleration due to gravity – constant acceleration), the ball hits the ground at a speed of v m/s (final velocity) after t seconds. Once you are clear of this: The concepts being tested on a v-t graph of: a = (v-u)/t Distance travelled = Area under the graph (this is a trapezium) If you have clarity of the concepts, attempting this question is straightforward.

Many students drew the graph for terminal velocity which showed lack of understanding. The question already mentioned that air resistance is negligible. 1(b)(i) a = (v-u)/t OR 10 = (v – 8.4)/t [1] t = (v – 8.4)/10 [1] Marker’s Comments If the question has been comprehended correctly, The gradient of the graph will be the acceleration due to gravity. (ii) Distance travelled = area under the graph = Area of trapezium = ½ x (sum of parallel sides) x (height) ½(8.4 + v)t = 5 [1] ½(8.4 + v)(v-8.4)/10 = 5 v = 13.1 m/s [1] Allow ecf Marker’s Comments If the question has been comprehended correctly, Area under the graph is the distance travelled which is from point A to the ground (5.0 m). Common mistake:

Many students used t = d/s to calculate the speed. This can only be done for objects/bodies moving at constant speed. This is the case where the velocity is increasing at a constant rate (constant acceleration). 2(a) Principle of moments states that when a body is in equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments [1] about the same pivot [1]. Marker’s Comments This is a recall question. However, many students could not get the full 2 marks. Physics Syllabus, 6091, y2024, Learning Outcome: 4(c) state the principle of moments for a body in equilibrium (b)(i) CW moments = ACW moments 700 x 1 + 500 x 3 = F2 x 4 [1] F2 = 550 N [1] Marker’s Comments Physics Syllabus, 6091, y2024, Learning Outcome: 4(d) apply the principle of moments to new situations or to solve related problems Students who did not get this question right, must take note that perpendicular distance is from the pivot to the line of action of force. (ii) F1 + F2 = 700 +500 F1 + 550 = 1200 [1] F1 = 650 N [1] Allow ecf Marker’s Comments

When a system is in equilibrium, you need to focus on 2 things: Equilibrium of moments (which was done in 2b(i)) And Equilibrium of forces (in this context, sum upward forces = sum downward forces) (c) Stand with feet apart OR stand with slightly bended knees [1] Marker’s