2022 JSS Phy Prelims Ans

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 1 Section A Qn Answer Marks 1a force of sled and boy on slope that is equal in magnitude to N (and opposite in direction). B1 1b As velocity increases, the resistive force increases. His resultant force will decrease since the forward force is a constant. Therefore, his acceleration will decrease. B1 B1 1c Lie down on the sled to make himself more streamlined. Bend lower to reduce his surface area exposed to air resistance. Accept: Go up higher the slope / increase weight / push harder to give an initial acceleration Any 1 B1 1d - correct diagram - arrows showing correct direction of force R W N R = 350 N (345 – 355 N) C1 C1 A1 2a Pressure on the floor in Fig. 2.1 is smaller than in Fig. 2.2 because in Fig. 2.1, the boy is sitting on four legs of the chair compared to two in Fig. 2.2. For the same force, a larger contact area will produce smaller pressure, since p = F/A. Reject: surface area / area B1 B1 2b The line of action of the combined weight acting at the CG is lying outside the base of the chair (on the right side of the pivot). This causes an clockwise moment since there is a perpendicular distance between the line of action of the combined weight and the pivot, causing instability. B1 B1 2c p = 0.05 x 13 600 x 10 + 100 000 = 107 000 Pa (or 110 000 (2sf)) B1 for hpg A1 3a Ball loses gravitational potential energy as it falls and gains kinetic energy. On impact with the floor, some kinetic energy is converted to thermal and sound energy to the surroundings. B1 B1 3bi GPE = 1.5 x 10 x (2.2 – 0.4) = 27 J B1 for height A1 3bii power = 27 / 0.6 = 45 W A1

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 2 Qn Answer Marks 3c Since mgh = ½ mv2, the relationship between height and time is not linear. Or Since there is no air resistance, the acceleration is constant. Therefore, the velocity increases with time. A1 4a  The gas molecules are in continuous, random motion, colliding with the inner surface of the balloon.  An average force is exerted by the molecules on the inner surface of the balloon during the collisions. The force per unit area gives rise to the pressure exerted by the molecules on the inner surface of the balloon. B1 B1 4b  There are less air molecules in the chamber. There will be less bombardments / collisions of the air molecules on the outer surface of the balloon. External pressure decreases.  Pressure inside balloon is higher than pressure outside. B1 B1 5(a)  visible light / ultra-violet / X-rays / gamma rays B1 (b)  scanning of luggages at airports OR  scanning of the human body (for broken bones or tooth decay) OR  checking for tiny flaws in heavy metal equipment or aircraft OR  radiation therapy to kill cancer cells  destruction or mutation of living tissues and cells Any 1 B1 (c) Time taken for pulse to be emitted and return = 0.05 – 0.02 ms = 0.03 ms v =2d t 1500 =2d 0.03/1000 d = 0.0225 m C1 C1 A1 6(a)  curved field lines drawn symmetrical  correct arrow direction M1 M1

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 3 Qn Answer Marks (b) (i)  negative sphere repels electrons and positive sphere attracts electrons to move from bottom layer of sand to top layer B1 (b) (ii)  negatively charged sand grains and positively charged sticky paper on the side nearer to the sand grains / positively charged sphere cause electrostatic forces of attraction B1 (c) V = W Q 5000 =2.0 Q Q = 0.00040 C (or 0.000400 C) C1 A1 7(a) (i)  graph is a curve / the values of I are not directly proportional to I B1 (ii)  decreases B1 (iii) straight line extended from line between 0 and 0.5 V to a maximum of (-3,-5) M1 I / mA V / V -2-3 3-1 210 -30 -10 -20 30 20 10 0

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 4 Qn Answer Marks (b) From graph when I = 20 mA, V = 1.7 V R = V I =1.7 20/1000 R = 85  C1 A1 (c)  brightness decreases (to zero when slider is at A)  resistance of AJ decreases causing p.d. across AJ to decrease  p.d. across LED decreases B1 B1 8ai B1 8aii Period, T = 4 x 2 = 8 ms = 8 x 10-3 s Frequency = 1 / T = 1 / (8 x 10-3) = 125 Hz C1 A1 X

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 5 8aiii Amplitude doubled Frequency doubled B1 B1 8bi VP IP = VS IS 100 x 0.4 = 200 x IS IS = 40 / 200 = 0.20 A C1 A1 8bii  The changing magnetic flux generated by the primary coil is unable to link to the secondary coil (the further away from the primary coil, the weaker the magnetic field) to induce a significant current. Or  Magnetic field lines are unable to reach and cut the secondary coil B1

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 6 Section B Qn Answer Marks 9 (a)(i) ratio of speed of light in vacuum to speed of light in the medium B1 (ii) borosilicate glass B1 (iii) Angle of incidence at CD = 90 - 50 = 40 n = sin r sin i 1.52 = sin r sin 40 r = 77.7 C1 A1 (b) n =1 sin c = 1 sin 31.3° = 1.92 Flint glass C1 A1 (c) (i)  light rays drawn vertical and horizontal, 90 to each other  correct arrow direction B1 B1 (ii)  angle of incidence in glass of 45 is greater than critical angle of 31.3  light ray travelling from optically denser glass to optically less dense air B1 B1 periscope

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 7 Qn Answer Marks 10a Fuse – a thin wire that melts when excess current flows through it and breaks the circuit. It is connected to the live wire. Circuit breaker – consists of an electromagnet that breaks the circuit when excessive current flows through it. Earth wire – low potential wire connected to the metal body of an appliance that allows excess current to flow when the appliance becomes live. Any 1 (can also be switch, double insulation, three pin plug) 10bi 600/230 + 50*3/230 + 1200*2/230 + 20*15/230 = 15 A M1 for any 2 correct sets of calculation A1 10bii The range of operating currents for the appliances is too wide. For appliances that draw less current such as the standing fan or refrigerator, it can be dangerous if an electrical fault occurs and the appliance overheats before the fuse can melt. M1 M1 10ci AB ↑ CD ↓ A1 Force must be drawn on coil 10cii Fleming’s left-hand rule A1 10ciii When AB rotates clockwise past its vertical position due to inertia, the split ring commutator will switch contacts with the carbon brush nearer to the south pole of the magnet. The current will now flow from A to B and the force acting on AB will be downwards, causing AB to continue rotating clockwise. A1 A1

S4E Phy 6091 P2 GE2 2022 ANSWER SCHEME 8 Qn Answer Marks 11Ea Rate of change of displacement B1 11Eb  From t = 0 s to t = 20 s, car is moving with a constant acceleration in the negative direction of 3 m s-2 / velocity increases at a constant rate in the negative direction.  From t = 20 s to t = 50 s, it is moving with a constant deceleration in the negative direction of 2 m s -2 / constant negative acceleration in the negative direction / velocity decreases at a constant rate in the negative direction.  From t = 50 s to t = 90 s, it is moving in the opposite / positive direction with constant acceleration of 2 m s -2 / velocity increases at a constant rate in the opposite / positive direction. B1 B1 B1 11Ec acceleration of car, a = (v – u) / t = [0 – (-60)] / (50 – 20) = 60 / 30 = 2 m/s2 deceleration = -2 m/s2 C1 A1 11Ed Area under the velocity-time graph B1 11Ee Award 1 mark each for correct shape of graph for the 3 time intervals, no need labelling  From t = 0 s to t = 20 s, displacement = ½ x 20 x 60 = 600 m (negative direction)  From t = 20 s to t = 50 s, displaceme