2023 SCGS 4 IP EOY Physics ANS

1 2023 4 IP EOY SUGGESTED ANSWER SCHEME PAPER 1 (40 marks) Qn Answer Explanation 1 A All three forces form the triangle of forces in the correct order ( tip to tail) 2 D Weight decreases with decreasing g ; mass of satellite remains constant 3 C Let the initial volume of air be V cm3 Mass of air = V x 0.0012 g. Density of air at the final volume = ( 0.0012 V g )/0.5 V cm3 = 0.0024 g/cm3 4 A Acceleration is the rate of change of velocity. Car accelerates when it starts from rest and reaches zero at constant speed. 5 D The area under v-t graph of car D is the smallest. 6 D Let the time taken to accelerate by t. Thus ½ 10ms-1( 11+ 11-t) = 100 m ⇒ t =2s. Acceleration = 10 /2 ms-2 = 5.0 ms-2 7 D When parachute opens fully, the upward force is greater than the weight and the resultant force is upwards causing a sharp deceleration. Since the resultant force is upwards, the acceleration is also upwards. 8 B At constant velocity, P = Fv = 2400 N x 90 ms-1 = 2.16 x 105 W or 2.2 x 105 W ( 2 s.f.) 9 A From Fnet= ma 2000 N – R = 750 x 2.0 ⇒ R = 500 N or 0.5 kN 10 A A has the broadest base and the lowest centre of gravity. 11 A Clockwise moment = 16.0 N x (50.0 cm sin 30 o) = 400 Ncm Anticlockwise moment = 5.0 N x 50.0 cm = 250 Ncm Resultant moment -= 150 Ncm or 1.5 Nm anticlockwise 12 C Total anticlockwise moment = F x 0.5 + F x 2.5 = 3.0 F D cannot produce a larger anticlockwise moment as the force on the left is not acting at the edge of the rod, thus that force produces a smaller anticlockwise moment. 13 D Total pressure= 100000 Pa + 15x1000x10 Pa = 250 000 Pa 14 B Pgas + 60 mmHg = 756 mmHg ⇒ Pgas = 696 mmHg 15 D Pressure is the same throughout the liquid. 16 C Oil : efficiency = 200/500x 100% = 40% ; Nuclear : efficiency = 40/200x 100% = 20% ; Oil : efficiency = 9/10x 100% = 90% 17 B Using h = gain in GPE / weight , Athlete B has the highest jump of 2.3 m 18 C Decreasing volume increases the frequency of collision with the inner walls. Molecules move at the same average speed because temperature remains unchanged. Pressure increases due to more particles hitting per unit area of the inner wall. 19 A Temperature increases ⇒ molecules gain thermal energy and move faster. They move further apart, the potential energy increases. 20 C Brownian motion is the result of the unequal bombardment on all sides of a smoke particle by air molecules moving at different velocities. 21 D Recall EM waves application

2 Qn Answer Explanation 22 C After the wave have move to the right by one wavelength, both buoys would have execute one oscillation and return to their original starting position. 23 A Ultrasound is a longitudinal wave and requires a medium to propagate 24 D An echo is the repetition of a sound due to the reflection of sound 25 D Electrons are transferred to X and it becomes negatively charged. It will then repel electrons to the right of sphere Y , leaving an equal amount positive charges induced to the left of sphere Y. 26 A Positive rod induced electrons to the right of P while earthing discharges the induced positive charges on the right of P. Removing the earthing and the rod will spread the electrons uniformly throughout P. Q remains neutral. 27 A Neutral object, whether a conductor or insulator, can be induced by a charged rod ( positive or negative) and will be attracted by it. 28 B Assigning 2Ω to P and 1 Ω to work out the difference between V1 and V2. 29 D I/V graph of resistor and lamp are standard graphs 30 C Light affects LDR not thermistor. When more light shines on LDR, resistance of LDR drops. Using potential divider equation, p.d. across resistor rises and the current flowing through the lamp increases. Lamp glows brighter. 31 A When galvanometer reading is zero, no current flows through it. There is no potential difference between across the galvanometer. P.D. across 4Ω is = 0.5 A x 4Ω = 2.0 V. Using potential divider eqn, 2𝑉 = 𝑅 𝑅+15 (12𝑉)5𝑅 = 15 Thus R = 3 Ω 32 B Total energy used in one month = 2 hrs x 2400 W x 3600 s = 17280000 J 33 A Magnetic shielding kept the magnetic field of the magnetic in the iron ring. The N -pole of the compass will be pointing to the geographical north. 34 A End X will be induced by the compass N-pole and becomes for S-pole. End X becomes an induced N-pole as it repels the N-pole of the compass. Since End X can be induced with either pole, it has to be a ferromagnetic material. 35 C The resultant magnetic field direction in between the two wires is to the left, using the right hand grip rule. 36 D Using Fleming’s left hand rule to determine the direction. 37 A The two ends of the iron rod facing each other will have opposite polarity : S -pole on the left and N-pole on the right. 38 D When Q is moved to the left, X will be a N-pole. End Y will be a S-pole. To have a S-pole at Y, the S-pole of magnet P must move towards Y. 39 D Recall question. Stronger magnet will have higher flux density (more lines per unit area) 40 A H(no car)= 4 div x 2.5 ms/div x 330 m/s= 3.3 m ; H (car) = 2.25 div x 2.5 ms/div x 330 m/s = 1.85625 m Height of car = 3.3 m – 1.85625 m = 1.44 m

3 PAPER 2 SECTION A ( 50 marks) Qn Part Answer Remark 1 (a)(i) Displacement is distance travelled in a certain direction / Displacement is the distance travelled by a moving object in a straight line in a specified direction. Distance is the total length covered by a moving object regardless of the direction of motion. (ii) Average speed = total distance/total time = 85 m / 40 s = 2.1 m/s (iii) Numerical : Average velocity = total displacement / time = 1.2 m/s OR : Since the total displacement is the straight line distance between A and D and is smaller than the total distance, the average velocity is smaller than the average speed. (b) ● Before reaching terminal velocity, the air resistance increases and the downward resultant force ( W -Fair) acting on the helicopter decreases. The helicopter is undergoing decreasing acceleration. ● When it reaches terminal velocity, W = Fair. The resultant force is equal to zero and the helicopter descend with a constant speed/zero acceleration. 2 (a) GPE = 0.12 kg x 10 N/kg x 0.45 m = 0.54 J (b) Let the speed at B be vB . From ½ (0.12 kg ) 𝑣𝐵 2 = 0.54 J  𝑣𝐵 = 3.0 m/s (c) Since the track friction and air resistance is ignored, by the conservation of energy principle, all the GPE at A must be equal to the GPE at D. To reach F, it must have extra kinetic energy to move to the highest point of the loop. Since GPE is only 0.54 J, it will not go beyond D. (d) The car will retrace its path and return to A. 3 (a) Pressure of trapped air = (0.57 x1.4 x 104 x 10 ) Pa + 1.0x 105 Pa = 1.8 x 105 Pa (b)(i) ▪ The trapped air particles move at higher average velocities and bombard the inner walls of the tube with a larger average force. ▪ In addition, the frequency of collisions with the walls also increases. The increase in pressure of the enclosed air pushes the liquid down the tube, thus increasing the value of x. (ii) ▪ Pgas is now larger. ▪ From Pgas = Patm + hg  h = (Pgas – Patm) / g. Since the difference in pressure now bigger, h will now be larger. 4 (a) ▪ The length will increase ▪ The thickness will decrease (b)(i) Since the force exerted is closer to the pivot, the perpendicular distance is smaller. From the principle of moments, a larger force needs to be exerted by the rubber band to provide the counterclockwise moment for the moment due to the mass