2024 TKGS P2 PRELIM MS

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 Comments Basic concept of stability. Increase base area and lower centre of gravity of the student. 3(a) Principle of conservation of energy state that energy cannot be created of destroyed but transferred from one store to another OR converted from one form to another [1]. The total energy of an isolated system is constant. [1] Do not accept “transferred from one form to another” and “converted from one store to another”. Marker’s Comments Students can give either one of the 2 definitions given below. But, the words in red cannot be mixed up between the definitions. Definition 1: The principle of conservation of energy states that energy cannot be created nor destroyed, but it can be converted from one form to another. The total energy in an isolated system is constant. Definition 2: [Energy stores and Energy transfers] The principle of conservation of energy states that energy cannot be created nor destroyed. It can be transferred from one energy store to another . process. The total energy of an isolated system is constant. (b) Total energy = GPE + KE = 0.30 x 10 x 45 + ½ x 0.30 x 82 [1] = 144.6 ~145 J [1] Marker’s Comments Students need to read the question carefully.

A ball of mass 0.30 kg is projected horizontally with a speed of 8.0 m/s from the top of a building AD which is 45 m above the level ground DC. The ball has an initial speed and height. So, it has BOTH gravitational potential store and kinetic store. Many students only focused on one type of store. (c) Energy is transferred through work done by gravity [1] from the gravitational potential store of the ball to the kinetic store of the ball [1]. Marker’s Comments There are 2 versions to the answer. Stick to one version of answering. Do not mix up the terminologies. 1) The ball has energy in both gravitational potential store and kinetic store at A. Energy is transferred through workdone by gravity from the gravitational potential store of the ball to the kinetic store of the ball. Work is also done against air resistance and the energy is transferred to the internal store of the surroundings. 2) The ball has both gravitational potential energy and kinetic energy at A. As it follows the parabolic path, gravitational potential energy is converted to kinetic energy. Work is also done to overcome air resistance. 4 Marker’s Comments There are 2 concepts being tested here. Thermal processes and Kinetic Model. The mention of colour must indicate that there is heat being ABSORBED by the matt black glass bulb. Since there is no contact, students should not mention ‘conduction’.

Once the tap is closed, we are also looking at the effect of temperature on a fixed mass of gas. (pg 143 and 144 of TB) (b) Matte black surface is better absorber of heat than a shiny one, thus heat is absorbed faster through radiation [1]. This causes the air on the left side to have more frequent collisions with greater force due to higher kinetic energy (more energy in the kinetic store) [1]. The larger air pressure on the left forces the liquid level lower [1]. Marker’s Comments A simple way to remember this: For a fixed mass of gas (enclosed container), Temperature increases → average kinetic energy of the particles increases → frequency of collisions with the inner walls and the liquid increases, hence increasing the average force exerted by the particles per unit area → Pressure increases. 5(a) λ = 9.5 cm Accept 9.6 cm Marker’s Comments Well attempted in general. The compression is a region where particles are close together while rarefaction is a region where particles are spread apart. A wavelength is from one compression to the next compression or one rarefaction to the next rarefaction. Some students did not know that the diagram was drawn to scale and so were not able to obtain the magnitude of the wavelength. A ruler should be used to measure the wavelength marked on the diagram. (b) v = fλ = 200 x 9.5 cm [1] = 1900 cm/s or 19 m/s [1] Accept 19.2 m/s for λ = 9.6 cm Allow ecf R C λ

Marker’s Comments A common mistake was to express the final answer with the wrong unit (m/s instead of cm/s, or vice versa). Some students incorrectly interpreted “oscillates 200 times per second” to mean that the period was 200 s. The correct interpretation is that the frequency of the oscillation is 200 Hz. 6(a) It should be increasing frequency OR decreasing wavelength Infra-red and microwaves are in the wrong order Visible light is missing from the spectrum Choose any two Marker’s Comments Very well attempted. (b) It causes the human cell to ionise resulting in cell damage/mutation [1]. Marker’s Comments Students who did not obtain full credit lacked the description on the ionising effect of gamma rays. (c) Speed decre