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