BBSS 4E O Prelim 2024 Physics 6091 P2 MS

Bukit Batok Secondary School GCE O LEVEL PRELIMINARY EXAMINATION 2024 SECONDARY 4 EXPRESS PHYSICS 6091/02 MARK SCHEME General marking instructions: 1. If students leave their answers in 1sf, penalise one mark for the entire question. 2. If students leave out the units, penalise one mark for every question. 3. Exercise ecf for calculation questions which require the answer from the previous section. Students will receive full marks if the substitution and calculation is correct based on the wrong value in the previous section(s). 4. Mark Scheme Code a. B1: Independent mark b. C1: Compensation mark; given automatically if the answer is correct, i.e. the working need not be seen if the answer is correct; also given if the answer is wrong but the point is seen in the working. c. M1: Method mark: if not given subsequent A marks fall (up to next B, M or C mark). d. A1: Answer mark. e. ecf: error carried forward; it usually is applied even where not specifically indicated, i.e. subsequent working including a previous error is credited, if otherwise correct. Section A 1 (a) From t= 0 s to t = 18 s, the rocket is moving upwards and accelerating upwards. [A1] From t = 18 s to t = 50 s, rocket is moving upwards and accelerating downwards. [A1] Both motion and acceleration must be mentioned to award one mark [2] (b) 50 s [A1] [1] (c) Maximum is the area under the graph = ½ x 50 x 64 [M1] = 1600 m [A1] Allow ecf from (b) [2] Name: …………………………………………. Index no. ………. Class……….

BBSS/Preliminary Exam 2024/Physics 6091/Paper 2 Applying past knowledge to new situations 2 2 (a) gravitational field strength, g, is the gravitational force per unit mass placed at that point. [B1] [1] (b) m = W/g or 700 / 10 [C1] = 70 kg [A1] [1] (c) Fnet = ma or 700 – 20 = (70)(a) [C1] a = 9.71 m/s2 accept 2sf [A1] Allow for ecf from (b) [2] (d) There is an initial acceleration because the weight of the skydiver is greater than the air resistance which causes a decreasing acceleration in the same direction as the resultant force according to N2L. [B1] When the weight of the skydiver and air resistance is equal in magnitude and opposite in direction, there is no resultant force acting on the skydiver which resulted in no acceleration. He continues to fall at constant velocity. [B1] [2] (e) Force on Earth by the skydiver [A1] The force is equal in magnitude (700N) but mutually opposite in direction (towards the skydiver) [A1] [2] 3 (a) Pcushion = Pcolumn + Patm or 24 cmHg + 76 cmHg [C1] = 100 cmHg [A1] [2] (b) Pressure due to 50 kg mass is due to the pressure difference in liquid column = Pcolumn Hence, pressure due to 50kg mass = Pcolumn = 24cmHg or 0.24 [M1] x13 600x10 = 32640Pa [M1] F/A = 32640 A = 500 / 32640 = 0.0153m2 [A1] Allow for ecf from (previous working) [3] 4 (a) At point A, the ball possessed energy in the gravitational potential store and kinetic store since the initial speed is not zero. [B1] Conditional accept the term gravitational potential energy and kinetic energy. Students need to explain the presence of both energy stores in order to be awarded one mark. Important note to remind students to use the term stores in their answers. When the ball slides down from A, energy in the gravitational potential store is converted to the kinetic store. [B1] Energy from the kinetic store is lost to the surroundings due to work done against friction which causes the speed of the ball to decrease. [B1] No marks given if students mentioned loss to surrounding as thermal energy without mentioning how does the lost in energy comes about. At the same time, the friction along the inclined plane caused the energy in the gravitational and kinetics stores to be converted to the internal store of the ball. This caused the ball to heat up as it rolled down the plane. [B1] [4] (b) (i) GPS = mgh = 0.50 x 10 x 13 = 65 J [C1] [1] (ii) Assume no energy is loss and total energy is conserved, GPEC + Wfriction = total energy at A or 65 + 10.7 = ½ (0.50) (v02) +(0.50 x 10 x 7.5) [C1] v0 = 12.4 m/s [A1] Allow for ecf from (i) [2]

BBSS/Preliminary Exam 2024/Physics 6091/Paper 2 Applying past knowledge to new situations 3 (c) There is no work done against air resistance as the ball moves up to position C. or No sound energy is transferred out of the system. [1] 5 (a) For an object in equilibrium [B1], the sum of clockwise moments about a pivot is equal to the sum of anti-clockwise moments about the same pivot [B1]. [2] (b) Applying principle of moments, 0.45 x W + 19 x 1.3 +1.85 x W = 2.6 x 22 [M1] W = 14.1 = 14 N (2sf) [A1] [2] (c) Determine the magnitude and the direction of the force acting on the rod at end A. F + 22 = 19 +28 F = 25 N upwards [C1] Allow for ecf from (ii) even if students can’t prove its 14 N [1] 6 (a) Heater Y is more efficient as thermal energy loss to the surroundings is reduced. The plastic casing reduces thermal energy lost by conduction since plastic is a poor conductor of thermal energy. [B1] OR The air trapped between the plastic casing and the water tank reduces thermal energy loss by conduction better than X since air is a poor conductor of heat. [B1] OR The white casing reduces thermal energy loss by radiation since white is a poor emitter of radiant heat. [B1] Any 2 of the following points. No mark given to just stating Heater Y without explanation. [2] (b) (i) The change of its internal energy/store per unit mass for each unit change in its temperature. [accept internal energy since definition in textbook indicated as internal energy] [1] (ii) Loss of thermal energy of tea = mcΔT or = 0.25 x 4200 x (50.0- 5.0) [ΔT = 45, obtain from the graph] [C1] = 47250 J = 47300 (3sf) or 47000 J (2sf) [A1] do not penalise for sf, except a reasonable range of answers from 46200 (ΔT = 44) to 48300 (ΔT = 46) [2] (iii) Energy lost by hot water = Total energy gained by ice cubes to melt and increase temperature from 0 ºC to 5 ºC 47250 = m (336000) [C1] + m (4200) (5) [C1] m = 47250 / 357000 = 0.132 kg [A1] accept a range based on (bii) Allow ecf from b(ii) [3] 7 (a) Electrons is transferred to the glue from the nozzle and becomes charged via friction/contact. [B1] [1] (b) Like charges repel and hence the negatively charged glue droplets moved away from each other. [B1] [1] (c) Since unlike charges attract, the section of paper with the negatively charged glue attract the sand particles that are positvely charged and neutral . [B1]

BBSS/Preliminary Exam 2024/Physics 6091/Paper 2 Applying past knowledge to new situations 4 The attractive force is greater than the weight of the sand and the sand move upwards. [B1] [2] 8 (a) From graph, Current = 1.4 + 0.35 = 1.75 A / 1.8 A [A1] ( 2 sf) do not penalise for sf [1] (b) R = V/I = 4/1.75 or 1.8 = 2.29 Ω/ 2.3 Ω ( 2 sf) or 2.22 Ω/ 2.2 Ω [C1] [1] (c) They could have different cross-sectional areas. Except other reasonable answers. [1] (d) Q = I x t = 1.75 or 1.8 x 60 = 105 C or 108 C [C1] [1] (e) W = V x Q = 4.0 x 105 or 108 = 420 J or 432 J [C1] [1] 9 (a) As the light intensity increases, the resistance of the LDR decreases [B1]. Using the potential divider formula, the voltage across the fixed resistor will increase as follows V = (6000/6000+ RLDR decreases) [B1] [2] (b) Total resistance of buzzer and resistor 1/R