SJI 2022 Prelim P1 and P2 ANS

2022 Physics Prelim Exam Marker’s Report Paper 1 1 D 11 D 21 C 31 C 2 B 12 C 22 D 32 B 3 C 13 A 23 A 33 B 4 B 14 C 24 C 34 D 5 A 15 C 25 A 35 C 6 C 16 A 26 B 36 C 7 C 17 D 27 D 37 B 8 C 18 D 28 C 38 B 9 B 19 A 29 B 39 A 10 A 20 D 30 C 40 D Paper 2 1 Fig. 1.1 shows how the speed of SpaceX, a rocket varies with time as it enters the gravitational field of a new Planet Y with negligible atmosphere. SpaceX has a total mass of 1.8 x 106 kg. Once it enters the atmosphere of Planet Y, it undergoes free fall for 8.0 s before its engine fires a continuous thrust to bring it to a gentle upright landing on the surface of Planet Y. Fig 1.1 (a) Calculate the weight of Space X on Planet Y. a=(v-u)/t = (82-10)/8 = 9.0 m/s2 W=mxg = 1.8 x 106 kg x 9.0 N/kg =1.62 x 107 N weight = ……………………..[2]

2 (b) At 8.0 s, significant amount of liquid Oxygen fuel undergoes combustion in the rocket engine to produce a constant thrust, T to decrease space X’s speed of descent. (i) On Fig. 1.2, draw and label all the forces acting on SpaceX during its descent. You may ignore air resistance. [1] Fig. 1.2 (ii) Hence, calculate the magnitude of the thrust, T. Taking upward as the positive direction Fresultant = T – W 1.8 x 106 kg x 4.1 N/kg = T – (1.62 x 107N) R =23.58 x 106 N = 2.36 or 2.4 x 107 N ecf allowed from (a) for W T = ………………….. [2] (iii) In reality, from t = 8.0 s to t = 28.0 s, the deceleration of SpaceX is increasing. Assuming that the thrust produced by the engine is uniform, explain in terms of forces why this is so. As the shuttle lands, its mass/weight decreases due to the burning of fuel, hence the upright resultant force is increasing (as thrust is constant) and deceleration is increasing ………………………………………………………………………………[1] T W

3 2 A man standing in a hot air balloon dropped a ping pong ball as the balloon accelerated upwards at 2.0 m/s2. The speed of the balloon is 2.0 m/s at the moment the ping pong ball is released. Assuming that the ping pong ball has the same velocity as the balloon when it is released. Fig. 2.1 shows how the velocities of the h ot air balloon and the ping pong ball vary with time. Fig. 2.1 (a) (i) Explain why the acceleration of the ping pong ball at its maximum height is 10 m/s2. ● The ping pong ball is momentarily at rest, hence there is no air resistance acting on it. ● The only force acting on the ball is weight as such it experiences acceleration due to gravity only which is 10 m/s2. ………………………………………………………………………………[2] (ii) Describe how the acceleration of the ping pong ball changes from the moment it is released until it reaches its terminal velocity. ● The gradient of the graph decreases to 0 at 2.2 s. Showing that acceleration is decreasing throughout the motion. ● The acceleration is zero (a= 0 m/s2) at t=2.2 s. ………………………………………………………………………………[2] (b) Calculate the average velocity of the hot air balloon from t = 0 to t = 3.0 s. Average velocity =( ½ ( 2 + 8) x 3) /3= 5.0 m/s [1]

4 3 Combustion engines in cars work by mixing air and fuel in suitable quantities before the mixture is drawn into a combustion chamber. Fig. 3.1 shows the side view of a section of a combustion engine in a car. The section of the pipe connected to the fuel tank has a narrower diameter. The pressure of the air flowing into the horizontal section of the pipe decreases as the diameter of the pipe decreases at X. Fig. 3.1 Fig. 3.2 shows a mercury barometer used to measure the atmospheric pressure where the car is located. Fig. 3.2 The density of fuel is 850 kg/m3 and the density of mercury is 13600 kg/m3. Take gravitational field strength g to be 10 N/kg. air at atmospheric pressure

5 (a) Explain how Fig. 3.1 shows that the air pressure in the pipe decreases with the diameter of the pipe. Height of fuel in vertical pipe is higher than fuel in the tank. this shows that the pressure at X is lower than atm P in tank (b) (i) Calculate the value of atmospheric pressure in Pa using Fig. 3.2. Patm = 𝜌𝑔ℎ = 13600 × 10 × 0.76 = 103 𝑘𝑃𝑎 atmospheric pressure = ………………………… [1] (ii) Hence, determine the gas pressure at point X. Pgas = Patm – P3.0cm = 103 𝑘𝑃𝑎 − 0.29(10)(850) = 101 𝑘𝑃𝑎 gas pressure = …………………….. [2] (c) Suggest how the combustion engine makes use of the pressure difference in the different sections of the pipe to mix fuel and air. ● The section with lower pressure is connected to the fuel tank. ● The lower pressure in the pipe results in a net upward force to act on the fuel causing it to rise up and enter the horizontal section of the pipe. ..…………………………………………………………………………………….... [1] 4 A student fills up an ice-cube tray with 200 g of water at 31oC and placed it in the freezer unit of a refrigerator. It takes 20 minutes for the temperature of water in the tray to drop to its freezing point. Specific heat capacity of water is 4200 J/kg oC and the specific latent heat of fusion of water is 330 kJ/kg. The heat capacity of the ice-cube tray is 120 J/oC. (a) Calculate the average rate of thermal energy lost by the water and the ice -cube tray as it cools down from 31oC to its freezing point. Give your answer in Watt.

6 Pxt = mxcx∆θ +Cx∆θ P x (20 x 60)s = 0.2 kg x 4200 J/kg oC x 31 oC + 120 J/oC x 31 oC P= 29760 J/1200s = 24.8 W rate of thermal energy lost = ………………….. [2] (b) Assuming that the rate of thermal energy lost by the water (and ice cube tray) in the freezer unit is constant, calculate how much additional time is needed for the water in the ice-cube tray to be completely frozen. ecf allowed from (a) 24.8 W x t = 0.2 kg x 330 000 J/kg t = 2661 s = 44 minutes Also accepted: P x (20 x 60)s = 0.2 kg x 4200 J/kg oC x 31 oC P=21.7W 21.7 W x t = 0.2 kg x 330 000 J/kg t = 3041 s = 50.7 minutes time taken = ………………………………. [2] (c) In reality, the time taken for the water in the ice-cube tray to be completely frozen is longer than the calculated time in (b). Explain why is this so. The rate of thermal energy lost decreases as the difference in the temperature of the water in the tray and its surrounding decreases. …………………………………………………………………………..……………[1]

7 5 One method of making sandpaper is by passing a roll of paper through nylon friction pads. An aerosol sprays positively charged fine droplets of glue onto the paper, spreading it evenly on the surface of the paper. The resulting sticky paper is then pressed over a flat table covered with sand grains. Fig. 5.1 (a) (i) Explain how the paper becomes negatively charged after it passes through the nylon friction pads. Due to friction electrons are transferred from the pads to the paper, charging it negatively. …………………………………………………………………………………[1] (ii) Explain how this method allows the glue droplets to spread out and stick to the paper easily. The glue droplets are charged oppositely to the paper (positively ) and unlike charges attract hence the droplets are attracted paper easily The droplets also repel each other, as like charges repel hence it spread out and stick to the paper evenly. …………………………………………………………………………………[2] (b) Draw in Fig. 5.2 the electric field pattern between two identical glue droplets. [1] Same number of field lines and spacings, direction outwards. Shape of field. Fig. 5.2

8 6 A girl lost one of her earrings when she swam in a swimming pool at midnight. She brought a torchlight and shine it into the pool to search for her lost earring. Fig 6.1 shows how the narrow beam of light from her torchlight was incident at a point 1.5 m away from the edge of t