2024 SCGS PRELIM P1 P2 MS

2024 PRELIM PHYSICS SUGGESTED ANSWER SCHEME PAPER 1 (40 marks) Qn Answer Explanation 1 B The clay will displace the same volume when it is placed in the bigger cylinder. 2 C Mass is unchanged, thus A & D is incorrect. Weight at both points are different as the acceleration of free fall is different, thus B is incorrect. 3 C By Pythagoras theorem, only C and D is possible. To construct a right-angle triangle at the point where the two cords meet, F2 has to be the larger force (8 N). 4 B Since the routes have different distances, average speeds for both are different. Displacement is the same from P to R even though distances are not equal. 5 A Calculating the area above the time axis: On Earth, height = ½ (1.2)(12)m=7.2 m ; On Mars, height= ½ (3.2)(12) m = 19.2 m Thus height difference = 12 m 6 B Total displacement = (+3.0 x 10)m + (-6.0 x 5)m = 0.0 m Average velocity = 0.0 m /15s = 0.0 ms-1 7 B A, B and C all demonstrates forces acting on the SAME body. 8 A Gravitational potential store = mgh⇒ h = Gravitational potential store / mg. Thus h ∝ 1/ m 9 C A and B are incorrect as there is no net work done since the suitcase is stationary. On the slide, work is not done on the suitcase – the suitcase slides down due to its own weight. C – work is done on it to raise it up by 10 m. 10 C Rate of increase in the kinetic store = (500 – 200)x 103 N x 40 ms-1 = 12 x 106 W 11 C Taking moments about the pivot, 0.43 N x (6 -2.4) cm = W x 1.2 cm W = 1.3 N 12 C Centre of gravity has to be at the pivot for the object to be at neutral equilibrium. 13 B Pressure of the gas = pressure due to piston + Atmospheric pressure 1.2 x 105 Pa = (4.0 x 10)/A + 1.0 x 105 Pa A = 2.0 x 10-3 m2 14 A Adding mercury to the trough will not affect the height h which is a measure of the atmospheric pressure. 15 C Since temperature remains constant, the average speed of the gas molecules remain constant. However, when compressed the density of the gas increases and hence the frequency of collisions of the gas molecules with the walls of the container increases. 16 B When air is sucked out, the pressure inside the packet drink decreased as the number of air molecules per unit volume in the packet decreased due to the decrease in the frequency of collision of the air molecules on the packet wall. 17 A B is incorrect as heat transfer is unidirectional and there is no such concepts as coldness. C’s description is heat transfer is via convection and not radiation. At thermal equilibrium there is no NET flow of thermal energy but thermal energy exchange is continuous. 18 B When water changes state from liquid to solid, the particles have to come closer together by losing internal energy and hence their average potential energy decreases due to the closer average distance between the molecules. 19 C Thermal energy supplied to X = Thermal energy supplied to Y (Cx) 𝜃 = (Cy) 2𝜃 Cx = 2 Cy 20 A Thermal energy lost by water = Thermal energy gained by ice 0.500 x 4200 x (30 - 𝜃) = (80 x 334) + (0.080 x 4200 x 𝜃) 𝜃 = 14.9oC 21 B P has the same brightness as Q but dimmer than R since the potential across both is lower than that across R. When filament in Q breaks, the potential difference across P is the same as R. The potential across P rises and the potential across R lowers. So P becomes brighter while R dims. 22 B Current passing through the 24.0Ω and 6.0Ω resistors = 15.0 V/30.0Ω = 0.5 A. Thus the current passing through the other arm = 1.0 Ω. 𝑉24 = 0.5(24)𝑉 = 12 𝑉 ; 𝑉6 = 1(6)𝑉= 6.0 V. The voltmeter thus registers a difference of 6.0 V.

23 C Total kWh (unit) consumed = 0.1 kW x 21 x 5 hr = 10.5 kWh. Total cost = 10.5 kWh x $0.24 /kWh = $2.52 24 C Fuse prevents excessive current from passing through , thus preventing excessive joule heating due to I2R. 25 A Since all are metals and, when in contact, they all become a single conductor. The positive charges will be ‘distributed’ with the movement of the electrons throughout the spheres and the rod. 26 A B & C are incorrect as positive charges do not move. Without induction by the positively-charged rod and the migration of the electrons from the gold leaf and metal rod, excess positive charges cannot occurred at these two locations. Thus A precedes D. 27 A In order for the comb to attract the iron fillings, electrostatic induction must take place. Electrons in the iron filings are induced by the charged comb which is electrically charged by friction. 28 B P’s end facing Q is a S-pole. R’s end facing Q is also a S-pole. Both repel each other producing the field lines as shown. There are no field lines in between them so Q must not be a magnetic material 29 D Applying Fleming’s LHR, all are correct. For statement 3, an alternating current passing through the wire means that the direction is changing every half-cycle of the current. 30 D All three conditions are needed for a larger force for a quicker rotation. 31 B A slower moving magnet will produce a smaller induced emf. Since the N-pole is moved away from the same end of the coil, the induced emf will be opposite to that when it is moved in by Lenz’s Law. 32 D Vs/Vp = 240 V/ 15 V = 16. D has the correct turns ratio. 33 A Angle of incidence at the boundary = 90 o – 38 o = 52 o The critical angle must be less than 52 o for TIR to take place. n = 1/ sin 𝑐 1.24 = 1/ sin c 𝑐 = 54 o Hence 1.24 cannot be the refractive index of the medium. 34 B n = 1/ sin 𝑐 1.4 = 1/ sin c 𝑐 = 45.6 o Hence the minimum value of 𝜃 is 46 o 35 C C is on the right of the page so it is laterally inverted. The letter i is symmetrical so there is no lateral inversion. 36 A Object at B produces an enlarged image; Object at C produces image at infinity, image size is undefined ; Object at D produces a virtual enlarged image. 37 B PQ – shows the point 3 moving to a position of maximum displacement, hence PQ is the amplitude. RS – is the distance between the centre of compression to the centre of rarefaction which is half of a wavelength. 38 A T = 2.5 ms x4 = 0.01s f = 1/T = 100 Hz v = f x λ = 100 x (2 x 7.5) v = 1500 m/s distance= 1500 x 0.10 = 150 m 39 D Frequency of the sound does not change as the instruction source did not change. As sound enters into water, it speeds up and since v = f x λ, λ increases. 40 C Few cm of lead must be used to contain gamma rays as they are the most penetrating.

PAPER 2 SECTION A ( 50 marks) Qn Part Answer Remark 1 (a) There is a reaction time between seeing the obstruction and stepping on the brakes. This translates to some distance moved by the car before decelerating. (b) It travels at a constant speed ( for 1.0 s) followed by a decreasing speed ( for 3.5 s ) and then comes to complete rest (for the last 0.5s) (c) Speed = 20 m/0.9 s = 22 ms-1 ( 2 s.f.) (d)(i) There is an additional resistive force, air resistance. From Newton’s 2nd law, deceleration = F+Fair / m will be greater than F/m . (ii) Air resistance decreases as speed decreases. From F+Fair / m , a decreasing Fair will a non-constant deceleration. 2 (a)(i) Moments of the sign’s weight about P = 3.4 x 103 kg x 10 Nkg-1 x 0.5 m = 1.7 x 104 Nm No mark for non- conversion of mass to weight (ii) Taking moments about P and applying P.O.M. Wblock x 2.0 m = 1.7 x 105 Nm  Wblock = 8500 N (b) The sign will rotate clockwise about P/topple as there is no anticlockwise moment to prevent the rotation/ a net clockwise moment (c) The CG of the block is now nearer the pivot P, thereby reducing the anticlockwise moment. There will be a net clockwise moment and the sign will rotate about P. The suggestion is thus incorrect 3 (a) From C to D the cord starts to stretch, causing the tension in the cord to increase. Tension is greater than the weight, resulting in an upward resultant force causing him to decelerate. As the