62 BPGH 2021 4E Prelim PHY P2 ANSWERS

1 Sec 4 Express Physics Prelim Examination 2021 – Marking Scheme Paper 2 Section A (50 marks) Qn Solutions Mark Allocation 1(a) Velocity is a vector quantity and hence it has a magnitude (= 3.0 m/s) and direction (which is due North). A1 1(b) scale : 1cm represent 0.5 m/s magnitude = 2.2 m/s (  0.2cm ) direction = 42 due west from the North ( 2) Alternative method: using cosine rule and sine rule respectively 𝑣 = √22 + 32 − 2(2)(3)𝑐𝑜𝑠45° = 2.1 m/s (sin  / 2.0 = sin(45) / 2.1   = 42  the following will result in loss of marks 1. wrong orientation of the vector diagram 2. labelling length instead of velocity 3. did not indicate angles on vector diagram 4. drew solid lines unnecessarily 5. resultant velocity was not indicated with a double arrow 6. inappropriate scale used, resulting in a small diagram B1 (vector diagram constructed & labelled correctly) A1 A1 2(a)(i) The moment of a force is the product of the force and the perpendicular distance from the pivot to the line of action of the force. B1 Any symbols used must be defined. 2(a)(ii) The filled buckets that cause moments about the axle are: buckets 2, 3 and 4. A1 45 v

2 2(a)(iii) The largest moment is provided by bucket 3 = (weight of water) x (perpendicular distance from the axle to the line of action of the weight) = (40)(10) x (1.6) = 640 Nm M1 A1 3(a) Insert the thermometer into a steam compartment with the bulb of the thermometer above boiling water. The stem of the thermometer with the marked upper fixed point should protrude from the top of the compartment. Use a manometer to check that the pressure inside the compartment is the same as the atmospheric pressure outside. When the level of the mercury in the t hermometer stem remains steady, it indicates the upper fixed point which is the steam point. B2 (deduct 1 mark for any omission of the three important points) 3(b) physical property that changes substance or object involved example volume a liquid resistance electromotive force (spelled out) a metal a thermocouple B1 (any one of the property and the corresponding substance) 3(c) mass of water = m Assumption: Heat lost by water = Heat gain by thermometer and no heat is lost from the water to the surrounding. m x 4.2 x (90-82) = 2.5 x (82-20) m = 4.6 g M1 A1 (a) B2 (correct construction of the 2 incident rays from the line deeper down below the image, incident rays bent towards the normal, arrow actual line on bottom of the floor

3 (Note: Due to refraction, the image line appears shorter than the actual line.) indicated on ray. Deduct 1 mark for any omission or mistake.) 4(b) Since light travels from water to air, Refractive index of water n is given by 1 𝑛 = 𝑠𝑖𝑛𝑠𝑖𝑛 𝑖 𝑠𝑖𝑛𝑠𝑖𝑛 𝑟 (i = angle of incidence in water, r = angle of refraction in air) 1 𝑛 = 𝑠𝑖𝑛𝑠𝑖𝑛 𝑖 𝑠𝑖𝑛 𝑟 sin i = 𝑠𝑖𝑛 (90°−40°) 1.33 i = 35 C1 A1 4(c) Light will bend towards the normal when it refracts at the water–oil boundary into the optically denser oil. It will then bend further away from the normal as it refracts at the oil-air boundary into air compared to the refracted ray emerging from a water-air boundary. This will cause the floor to look even shallower than before. Hence the image of the line will appear above level A. B1 B1 5(a)(i) Imaginary line on a wave that joins all adjacent points that are in phase. B1 5 (a)(ii) B1 wavefronts drawn correctly joining all crests (or troughs) across the waves wave fronts

4 5(b)(i) Frequency of the wave f = 5 cycles / 10 s = 0.5 Hz Wavelength of the wave  = 8.0 cm = 0.080 m Speed of the wave v = f  = (0.5)(0.080) = 0.04 m/s M1 A1 5(b)(ii) B2 (drawn 2 wavelengths, labelling the amplitude and wavelength. deduct 1 mark for any mistake or omission) 6(a) (i) The lighter part of the image on the paper reflects more light onto the drum and form a conducting area on the drum which cause the positive charges to be discharged. B1 B1 6(a)(ii) The darker areas of the image reflect less light. The corresponding regions on the drum remain insulating, and the positive charges remain on the surface of the drum. B1 B1 6(a)(iii) As opposite charges attracts, the toner powder is negatively charged and the paper is positively charged. B1 6(b)(i) As the paint leaves the nozzle of the spray gun, the paint droplets become charged by friction. B1 6(b)(ii) As like charges repel, the charged paint droplets will repel one another and spread out. B1 6b(iii) As the charged droplet spread out, it will form a uniform coat of paint on the car body. Or As the paint droplets are charged, it will reach the car body which is earthed (instead of falling on the floor) and in this way there will be less wastage of paint. B1 16.0 8.0 -2 .0 2.0 Distance / cm Displacement / cm

5 7(a) (Resistance is the ratio of the p.d. across the LED to the current flow through it.) From 0 V to 2.7 V, the current of the LED is zero, hence its resistance is infinitely hig h. From 2.7 V to 3.6 V , the current flow through the LED increases drastically to a very high value, which indicates that its resistance falls from an infinitely high value to zero. B1 B1 7(b)(i) The I-V graph for the filament lamp is not a straight line passing through the origin. B1 7(b)(ii) The resistance of the filament lamp changes (/does not remain constant) due to the rise in temperature of the filament a s current flows through it. B1 7(c) For the lamp, at V = 3.0V, I = 37 mA  resistance = 3.0 𝑉 37 ×10−3 𝐴 = 81  M1 A1 7(d) In a series circuit, the same current (37 mA) flows through the LED and the resistor. From the I-V graph for the LED, I = 37 mA, p.d. across LED, VD = 3.4 V Using V =IR, p.d. across R, VR = (37 x 10-3)(200) = 7.4 V Hence e.m.f., E = (p.d. cross lamp) + VD + VR = 3.0 V + 3.4 V + 7.4 V = 13.8 V C1 C1 A1 8(a)(i), (ii) B1 (C) B1 (F) 8(a)(iii) Using Fleming’s Left -Hand rule , the fore finger which indicates magnetic field, points to the left. The middle finger which indicates conventional current (flow of positive charges), points perpendicularly out of the plane of the paper and the thumb which indicates the force, points down. B2 (deduct 1 mark for any missing point) 8(b)(i) Clockwise rotation of the coil means that a pair of forces acts on the side AB and CD of the coil. Consider the side AB of the coil, the magnetic force acts upwards and the current flow from A towards B . Using B1 (explain using FLH rule on side C F

6 Fleming’s Left-Hand rule, with the thumb pointing in the direction of the force (up) and the mi ddle finger pointing in the direction of the current flow (A to B); the fore finger which indicates the direction of the magnetic field is found to point towards the magnetic pole P . Hence magnetic pole P is the South pole and magnetic pole Q is the North pole. AB or CD of coil) A1 8(b)(ii) As the coil rotates past the vertical position by half a turn , the current through side AB of the coil reverses direction (flow from B to A) as its split -ring commutator now makes contact with brush Y. By Fleming’s Left-Hand rule, the magnetic force on side AB acts down and co