NTSS Prelim P2 Ans

NEW TOWN SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 Secondary 4 Express NAME CLASS INDEX NO. PHYSICS 6091 [Revised]/02 Paper 2 22 Aug 2024 1 hour 45 minutes Candidates answer on the Question Paper. 0800 – 0845 Additional Materials: NIL READ THESE INSTRUCTIONS FIRST Write your name, index number and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A (70 marks) Answer all questions. Section B (10 marks) Answer any one question. Candidates are reminded that all quantitative answers should include appropriate units. The use of an approved scientific calculator is expected, where appropriate. Candidates are advised to show all their working in a clear and orderly manner, as more marks are awarded for sound use of Physics than for correct answers. The number of marks is given in brackets [ ] at the end of each question or part question. For Marker’s Use Section A Section B Total This document consists of 24 printed pages and 0 blank page.

2 Section A (70 marks) Answer all the questions in the spaces provided. 1 A man stands at the edge of a vertical cliff and he projects a stone vertically upwards. The stone reached its greatest height, fell past the man and hit the bottom of the cliff. The motion of the stone is represented by the velocity-time graph shown in Fig. 1.1. Fig. 1.1 (a) Explain how Fig. 1.1 shows that air resistance has negligible effect (C) on the motion of the stone. Stone has a constant acceleration (E) indicating free-fall acceleration (R). OR Graph has a constant gradient indicating free-fall acceleration. -Accept: a= 10m/s2 which is acceleration due to gravity -Majority wrote ‘constant acceleration/ gradient’ (E) without the (R) [1] (b) Based on the data shown in Fig. 1.1, determine the greatest height achieved by the stone. Greatest height = area under v-t graph (first 2.0 s) = ½(2.0)(20) [1] = 20 m [1] Most did well greatest height = ………………………. [2] [Total: 3 marks]

3 2 Fig. 2.1 shows the vertical forces acting on a weather balloon just after lift-off. Fig. 2.1 The balloon experiences an upward force of 95 N. The total weight of the balloon and its contents is 61 N. (a) Determine the total mass of the balloon and its contents. The gravitational field strength g is 10 N/kg (F) m = W / g (S) = 61/ 10 (AU) = 6.1 kg [1] total mass = ………………………. [1] (b) Calculate the initial vertical acceleration of the balloon. Fnet = 95 – 61 = 34 N a = Fnet / m = 34/ 6.1 =5.57 m/s2 [1] acceleration = ………………………. [1] (c) As the balloon rises, it experiences air resistance. The upward force remains at 95 N. Explain, in terms of forces acting, why the balloon will reach a constant speed. As air resistance increases, the total downward forces equal the upward force. When resultant force acting on the balloon is zero [1], acceleration of the balloon will be zero based on Newton’s 2nd law F = ma [1]. Hence the balloon will reach a constant speed. -Majority cannot equate the balanced forces correctly/ stated the wrong Newton law -Common mistakes: upward force = air resistance, forgetting the weight Air resistance = weight, which both are acting downwards Newton 1st law was stated [2] [Total: 4 marks]

4 3 Fig. 3.1 shows a sphere of weight 6.5 N suspended by a wire from point X. The sphere is pulled to one side by a horizontal force F so that the wire makes an angle of 30 with the vertical. Fig. 3.1 (not to scale) Use a scale of 1.0 cm to represent 0.50 N, draw a scale diagram in the space below to determine the magnitude of (a) the tension (TW) in the wire, (b) the horizontal force F. Diagram = right-angled triangle and all forces must be correctly labelled and in correct direction. [1] Most did well TW = (7.5  0.1) N [1] F = (3.7  0.1) N [1] [3] [Total: 3 marks] TW 30 wire string F sphere X 30°

5 4 Fig 4.1 shows a sphygmanometer, a device used to measure blood pressure of a patient using mercury in a manometer. Fig. 4.1 Fig 4.2 shows an enlarged view of the manometer after the doctor has depressed the air bulb that is connected to the sphygmanometer. The right arm of the manometer is open and exposed to atmospheric pressure. P is the air pressure that is equal to the blood pressure of the patient. P0 = atmospheric pressure = 1.01 × 105 Pa. Fig 4.2 (a) The doctor observes that h = 115 mm. Calculate the blood pressure of the patient. Assume the density of mercury to be 13600 kg/m3. Blood pressure P = P0 + hpg = (1.01 x 105) + (115 x 10-3)(1.36 x 104)(10) [1 for equating P = P0 + hpg ] = 1.17 x 105 Pa / 117 000 Pa [1] -Many did not convert mm to m / give 3 sf pressure = ………………………. [2] P P0 h mercury manometer [mm Hg] air bulb 90 60 120 180 150

6 (b) The doctor proceeds to draw a liquid vaccine from a bottle. Fig 4.3 shows a syringe with the vaccine in it. The syringe has a piston A with cross-section area 0.80 cm2 and a nozzle B with a cross sectional area of 0.13 cm2. The liquid pressure of the vaccine is 1.31 × 105 Pa. Fig 4.3 Calculate the force FH required to just push in piston A. Assume atmospheric pressure = 1.01 × 105 Pa. FH + PatmA = PliquidA [1] correct eqn FH + (1.01 x 105)(0.80 x 10-4) = (1.31 x 105)(0.80 x 10-4) FH= 2.4 N [1] [award 1m if student omits PatmA in the eqn and gives FH = 10.5 N] [award 1m if student states the right eqn but converts the area wrongly] -Majority could not attempt this qn. Force FH = ………………… [2] (c) Explain, without calculation, how the magnitudes of the force at the piston A, FA differs from the force at the nozzle B, FB. Since PA = PB [1] due to Pascal’s Law FA/AA = FB/AB Since AA > AB, [1] FA > FN. -Majority cannot answer well. -Some can give right explanation but state FA is smaller. [2] [Total: 6 marks] FH piston A nozzle B liquid vaccine

7 5 Fig. 5.1 shows a roller coaster of weight 5000 N and initially at rest at a height of 15.0 m above the ground. It is then released from rest and it rolls along the track for 50.0 m before it comes to a momentary rest at point X which is 11.0 m above the ground. The track feels warm after the roller coaster rolled past it. Fig. 5.1 (a) Ignoring loss of energy to the surroundings through other means, determine the work done against friction as it rolls from its initial rest position to point X. Work done against friction = difference in GPE between initial position and at X = mg( h) = (5000)(15.0 – 11.0) [1] = 20 000 J [1] [award 1m if GPE at first and at X are correct] -Many state WD against friction = F x d = 5000 x 50 but F = 5000N is not a frictional force. -Some did not realise weight = mg in the formula and used m= 5000