2024 SJI Y4OP Prelim P2 detailed ans

ST. JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2024 (YEAR 4) CANDIDATE NAME MARK SCHEME CLASS INDEX NUMBER PHYSICS Paper 2 Candidates answer on the Question Paper. No Additional Materials are required. 6091/02 9 September 2024 1 hour 45 minutes (08:05 – 09:50) READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Write your answers in the spaces provided. Section B Answer one question. Write your answers in the spaces provided. 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. Section A Section B 1 2 3 4 5 / 6 / 8 / 5 / 8 / 8 6 7 8 9 - / 5 / 10 / 10 / 10 - 10 11 / 10 / 10 This document consists of 41 printed pages and 1 blank page. For examiner’s use Section A 70 Section B 10 Total 80

2 Section A Answer all questions. 1 A car is speeding along a straight road that has a speed limit of 50 km/h. The car passes a policeman on a stationary motorcycle at t = 0. The policeman starts chasing the car, using his motorcycle, at t = 0.50 s. Fig. 1.1 shows the velocity-time graph of the car. Fig. 1.1 (a) Using information from Fig.1.1, show that the car has exceeded the speed limit. 50 km/h = 14 m/s Or 24 m/s = 86.4 km/h Since speed > 50 km/h therefore car is speeding 0 5 10 15 20 25 30 35 car time /s velocity m/s 1.0 2.0 3.0 4.0 5.0 6.0 7.0

3 (b) The velocity of the policeman’s motorcycle increases uniformly from rest at t = 0.50 s at a rate of 6.0 m/s2. (i) On Fig 1.1,  draw the velocity-time graph of the policeman’s motorcycle from t = 0 until t = 6.0 s  indicate the velocity of the motorcycle at t = 6.0 s. Show all relevant working. a=v−ut v=u+at =0+(6.0)(5.5) =33.3 m/s straight line from 0 m/s to 33 m/s from t = 0.50 s to t = 6.0 s

4 (ii) At t = 7.5 s, the car notices the motorcycle and starts decelerating uniformly from 24 m/s to rest. The motorcycle also decelerates uniformly, and at t = 10.5 s, both come to a complete stop. Calculate the distance between them at t = 10.5 s. Distance travelled by car between t = 0s and 10.5s = ½ (10.5 + 7.5)×24 = 216 m Velocity of motorcycle at t = 7.5: a = (v – u)/t 6.0 = (v – 0)/(7.5 – 0.5) = v/7.0 v = 6.0m/s2 ×7.0s = 42 m/s Distance travelled by motorcycle = area under v-t graph = ½ × 10.0s ×42m/s = 210 m Distance apart = 6.0 m (iii) Sketch the displacement-time graph of the motorcycle on the axes provided in Fig. 1.2. 247.5s 10.5s42m/s0.5s

5 2 Fig. 2.1 shows a block of weight 50 N being pulled up a smooth slope at a constant velocity, using a wire attached to a motor. Fig. 2.1 (a) By drawing a scaled diagram, determine the tension in the wire. T = 17.5 N or (16.6 N – 18.4) N (b) (i) Calculate the useful power output of the motor in pulling the block up the slope, given that the block is moving with a constant velocity of 0.75 m/s. P = F x v = 17.5N x 0.75m/s = 13 W (ii) State two reasons why the rate of increase of energy in the gravitational potential store is equal to the useful power output of the motor. No work done against friction as slope is smooth, and energy in the kinetic store is unchanged as velocity is constant. 50 N smooth slope motor 20o block wire

6 (c) State and explain the change in the motion of the block if the gradient of the slope is lowered until it is horizontal, while the tension in the wire remains unchanged. The block will speed up/accelerate. The normal reaction force would be equal to the weight of the block when the surface is horizontal, hence the tension is the resultant force which causes the block to accelerate.

7 3 Fig. 3.1 shows a hand-operated hydraulic jack. A downward force of 10 N is applied on the handle, causing piston X to move downwards. This in turn causes an upward force to act on piston Y. A load placed on piston Y can then be raised. Fig. 3.1 (a) Calculate the magnitude of the minimum force exerted on piston X given that the downward force on the handle is 10 N. 10N x 50cm = F x 5cm F = 100 N (b) In another instance, a force of 50 N exerted on piston X results in an upward force of 200 N on piston Y. Given that piston X has a diameter of of 1.5 cm, calculate the diameter d of piston Y. px = pY 50/(𝜋1.52) = 200/(𝜋d2) d = 3.0 cm (c) Explain, with reference to the design of the hydraulic jack, why the force exerted on piston Y is larger than that exerted on piston X. (Pascal law) Pressure due the force on piston X is transmitted uniformly through the oil to piston Y. The increased force is the product of the pressure and the larger cross-sectional area of piston Y. (d) In one occasion, air bubbles are observed in the oil tank under piston Y. Explain how this affects the efficiency of the hydraulic jack. Some work will be done to compress the air bubble, causing the work done on piston Y to decrease. This reduces the efficiency of the system. valve B valve A 10 N piston Y piston X pivot oil 50 cm 45 cm handle oil reservoir

8 4 Fig. 4.1 shows the top view of successive crests of plane water waves at a particular instant. The waves are generated at a rate of 4.0 waves per second travelling in the direction shown. Fig. 4.1 (top view) (a) Determine the (i) wavelength, and Wavelength = 14cm ÷ 5 = 2.8 cm (ii) wave speed of the water waves. v = 4.0 x 2.8 = 11.2 cm/s (b) Fig. 4.2 shows a side-view of a section of the wave of amplitude 2.0 cm, at t = 0. X is a point on the wave. Fig. 4.2 (side view) Sketch the displacement-time graph of point X from t = 0 to t = 0.500 s on the axes provided in Fig. 4.3. direction of wave motion 14 cm

9 (c) Upon entering region B of different depth, the direction of travel of the plane waves turns through 20o as shown in Fig. 4.4. Fig. 4.4 (top view) (i) Complete Fig. 4.2 by drawing lines to represent the crests of waves X, Y and Z in region B. Solution region A region B 20o new direction of travel X Y Z

10 (ii) Compare the depths of the water in regions A and B, supporting your answer with a reason. By v = f𝜆, for the same frequency and shorter wavelength, the wave speed is lower, therefore the depth has decreased in B (more friction with bottom of tank). The water in region B is shallower.