2024 SCGS 4 IP PHYSICS P1+P2 ANS

2024 4 IP EOY SUGGESTED ANSWER SCHEME PAPER 1 (40 marks) Qn Answer Explanation 1 C Acceleration and Weight are vectors while the rest are scalars. 2 D T: Density = 10/ 8.0 = 1.25 g/cm3 U: Density = 10/ 12.0 = 0.83 g/cm3 V: Density = 20/ 16.0 = 1.25 g/cm3 W: Density = 20/ 8.0 = 2.5 g/cm3 3 B By completing the parallelogram, resultant will be the diagonal. 4 A As the ball drops vertically it undergoes constant acceleration due to free fall, hence the first part of the graph must be a straight line of constant gradient. When the ball hits the ground and rebounds in the opposite direction, its velocity drops to zero and as it rebounds in the opposite direction, the graph shows a drop to zero velocity and then a negative velocity. Subsequently, it goes upwards with a constant gradient. 5 B v / t = a v / 2.0 s = 10 m/s2 v = 20 m/s area under velocity – time graph gives you displacement hence h = ½ x 20 x 2.0 = 20 m 6 D a = (v – u )/t 2.0 m/s2 = (v - 5.0 m/s )/ 10s v = 25 m/s 7 B Fnet = ma 2N – 1N = 10a a = 0.1 m/s2 8 C Action and reaction pair act at the same time, on mutually different bodies, have the same magnitude but opposite directions. 9 C Since there is no air resistance, both divers will fall with the same acceleration due to gravity and hence will have the same velocity before they the water. Distance travelled is obviously the same as they both jump from the same height. As energy in kinetic store = ½ x m x v2; since their mass are different, their energy in kinetic store will be different. 10 A Rate of work done against air resistance = Force x Distance / time = 3.0 N x (1/2 x 5.0 m/s x 10s) /10s = 7.5 W

11 B P = mgh/t . Larger mass of ball lifted to a greater height will require greater power when time taken to pick the ball is the same for all four. 12 D Steam from the boiler turns the turbines. 13 C Applying principle of moments, 6.0 N x 10 cm + 4.0 N x 30 cm = 5.0 N x d  d = 36 cm The 5.0 N weight is acting at the 86 cm mark 14 B Bar 2 , Take moments about any one of two edges will result in either a clockwise or anticlockwise moment. 15 A Height change of P and Q = 20 m ; Height change of R and S = 40 m 16 D Since the mercury level in both arms are at the same level, pressure exerted by liquid X column = Pressure exerted by liquid Y column. From hXX = hYY, hX > hY Y > pX 17 B Height of the barometer is independent of the cross-sectional area. 18 A Molecules cannot expand (statement 1 incrorrect). They cannot move further apart as the volume if fixed ( statement 4 incorrect). 19 C A is radiation; B is conduction ; C is radiation and conduction 20 D Pressure of gas X > Patm to cause the piston to move. After opening its pressure will reduce to the same as Patm but not lesser for that will shift the piston inwards. Nor will the pressure be more as that will indicate that the piston will continue moving whilst it is already mentioned that it just moves slightly to the right. 21 C A is incorrect as the liquid is not boiling yet. B is incorrect evaporation only applies to molecules at the liquid’s surface but not ANY molecules. D is incorrect because it describes boiling. 22 D Liquid Y expands and contracts more per degree change. 23 C Z being a positively charged rod, will attract free mobile electrons to the right side of Y, inducing positive charges on the left side of X. When earthed, electrons will flow up to neutralise the positive charges on the left side of X. Hence X will be neutral and Y will be negatively charged. 24 C Electric forces are action and reaction forces. They will be of the same magnitude but opposite in direction. Hence F1 = F2 and x = y. 25 A When the switch is closed, current will take the upper loop path and by pass the lower path. Hence A1 = A2. 26 A Current that flows out of the cell will pass through R1 before splitting up to the branches of R2 and R3, R4 & R5. Since V = IR, R1 will have the biggest potential difference. 27 D When bulb 1 breaks, the overall resistance increases and hence current will decrease. Voltage across bulb 2 will be larger and the voltage across bulb 3 and 4 will be smaller. Hence, bulb 2 becomes brighter and bulb 3 and 4 becomes dimmer. 28 C P = VI 1500 W = 240 V x I I = 6.25 A Current flowing in the live wire must be the same as that in the neutral wire. Earth wire carries no current when the kettle is working normally. 29 A From R = l/A , the longest wire with the smallest diameter ( and hence the smallest cross- sectional area ) will have the greatest resistance. 30 B There is an attraction by BOTH the S and N poles of the two magnets X is a soft iron. A stronger magnet will induce and attract the soft iron more, thereby producing a smaller resultant downward force ( Weight – magnetic force = resultant downward force) 31 A Appling right hand grip rule, compass Q is pointing to the left. Compasses P and R is pointing due North as the magnetic field at the plane of the paper is perpendicular to the compass and does not affect it, thus both N-pole of the two compasses points due North 32 A When the magnet enters the coil, there will be a changing magnetic flux linkage with the conductor and an emf will be induced. When the magnet is halfway through the coil, the emf induced on both ends of the coil will cancel out each other; hence there will be no induced emf. When the magnet leaves the coil, the induced emf will be opposite to the initial emf due to Lenz’s law. 33 B Secondary voltage = 50 V x (100/10) = 5.0 V V = IR 5.0 V = I x 25 I = 0.20 A

34 B The next ¼ cycle will see Q at the equilibrium position and is continuing to move down while P is at the crest and is also on the verge of moving down. 35 D All EM waves travel at the same speed in vacuum. X rays have a higher frequency and hence a shorter wavelength compared to visible light 36 B Y amplitude is twice that of X. X executes 2 cycles within the same time as Y. Thus fx = 2 fy 37 C The wavelength of UV radiation (UVR) lies in the range of 100–400 nm; X-rays have a wavelength in the range of 0.01–10 nm ; Visible light’s wavelengths in the range of 400–700 nm; Infrared (IR) radiation’s wavelengths between 760 nm and 100,000 nm 38 B There is no -radiation as the count rate did not drop to below 25000 counts/min when aluminium is used. 39 D Ionising radiation can mutate cells. 40 B 38000 → 17000 ( 2 years) ; 26000 →12000 ( 2 years) PAPER 2 SECTION A ( 70 marks) Qn Part Answer Remark 1 (a) The ping pong ball is momentarily at rest at the maximum height, 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/s 2 since g = W/m. (b) The ball rises up with a decreasing deceleration from 0 to 0.2 s [1] The ball then falls with a decreasing acceleration from 0.2 s to 2.2s where it reaches terminal velocity. [1] (c) displacement =( ½ ( 2 + 8) x 3) m = 15 m [2] Average velocity = 15 m /3 s = 5.0 m/s [1] 2 (a) a=(v-u)/t = (82-10) ms-1/8.0 s = 9.0 m/s2 [1] W=m x g =1.8 x 106 kg x 9.0 N/kg [1] =1.6 x 107 N (b) (i) T drawn to be longer than W. [1] W should be drawn from the CG of object. T should be upwards. [1]

(b) (ii) 1 & 2 Fnet = T – W = ma T – (1.6 x 107N) = 1.8 x 106 kg x 4.1 N/kg T =23.58 x 106 N = 2.4 x 107 N [1] ecf allowed from (a) for W (b) (iii) 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 3 (a) Gain in GPE = mgh = 800 kg x 10 Nkg-1 x 8.0 m [1] = 64 000 J [1] (b) 80% = (ou