6091/ 01/02 Nov 2024 Suggested Answers (Montfort Sec)

2024 GCE O Level Physics 6091 Propose Answer Paper 2 Section A Qn Answer Comments 1(a) 10 m/s2 B1 CAO 1(b)(i) Speed should be 30 m/s after 3.0 s with g = 10 m/s2 but with a speed of 15 m/s after 3.0 s indicate the presence of air resistance decreasing the rate of change of speed. B1 Must use data. 1(b) (ii) [1] Constant acceleration (straight line) from 0 to 15 for first 3.0 s [1] Ball hits floor and in contact for 0.2 s(3.2 s) [1] From -10 m/s to 0 m/s at 5.0 s [1] gradient should be the same as previous line. 1(c) The area of the velocity- time graph between t = 0 s and t = 3.0 s would give the height. B1

Qn Answer Comments 2(a) Inertia (depends on mass only) 2(b) i 5000 N [1] to the left acting on the road from the car [1] ii F = ma F = 5000 – 3500 = 1500 1500 = ma a = 1500 / 800 = 1.9 m/s2(2sf) C1 B1 iii 3(a) 2.0 cm2 = 2.0 x 10-4 m2 Pressure = Force / area = 60 / (2.0 x 10-4) Pressure = 3.0 x 105 Pa C1 B1 3(b) Oil is an incompressible liquid, hence pressure from piston P is transmitted to piston Q. Piston Q has a wider area than piston P, hence with the same pressure, force at piston Q will be smaller. B1 B1 3(c)i Strong internal forces between molecules of oil as well as closely packed molecules make it difficult for molecules to be compressed. B1 B1 Must describe structure rather than motion. 3(c)ii Some of the energy will be loss to compress the bubbles of air. B1

Qn Answer Comments 4(a)i Work done = force x distance Force = work done / distance Force = 1.6 x 105 / 210 = 760 N (2sf) C1 B1 4(a)ii Speed = distance / time = 210 / 150 = 1.4 m/s Ek = ½ mv2 = ½ x 160 x (1.4)(1.4) = 156.8 = 160 J (2sf) C1 B1 4(b) Non- renewable renewable B1 for non-renewable B1 for renewable nuclear fuel fossil fuels biofuel hydropower geothermal wind 5(a) v = l × f l = v / f = 3.0 x 108 / 2.5 x 109 = 0.12 m C1 B1 5(b)i Radio waves infrared ultraviolet B1 5(b)ii In remote controllers to control equipment/appliances Intruder alarms / thermal imaging B1 5(b)iii Artificial tanning in sunbeds / bank note authentications / disinfecting water/ surgical equipment B1 5(c)i High power radio waves can cause heating of the skin / body tissues, which can cause burns B1 5(c)ii High frequency electromagnetic radiation at low powers can have high ionising effect and can damage cells. Long term exposure may cause cancer. B1 B1 6(a)i [1] for correct rays [1] for correct position of image 6(a)ii The image of the can be obtained by first extending or extrapolating the 2 ray of reflections behind the mirror as virtual rays. The point of intersection between the 2 extended will lie the image. B1

Qn Answer Comments 6(b) Time between pulses = 1/8 = 0.125 s Max distance = ½ (speed x time) = ½ (320 x 0.125) = 20 m B1 C1 B1 7(a) total resistance in parallel = (!"#$!%+!&')-1 = 20 W Total resistance in circuit = 100 W + 20 W = 120 W C1 B1 7(b)i Current in circuit = 9.0 / 120 = 0.075 A p.d. across 100 W resistor = 0.075 x 100 = 7.5 V p.d. between A and B = 9.0 – 7.5 = 1.5 V C1 B1 7(b)ii P = IV = 0.075 x 7.5 = 0.5625 W = 0.56 W C1 B1 8(a) When current goes beyond 7.0 A, the greater the current, the shorter the time before the fuse melt. B1 8(b)i Standard fuse: Less than 1.0 s from table 0.99 s correspond to 12 A, hence half the current rating would be 6.0 A. B1 8(b)ii Fast blow fuse: 0.1 s would be 0.09 s from table correspond to 40 A; hence half the current rating would be 20 A. If its fuse current rating is 6.0 A, it would allow too much current to flow before the fuse melt which may cause damage/overheating (resulting in fires) to equipment. B1 B1 8(c)i P = I2R, E = Pt = I2Rt = (30)2(0.015)(0.16) = 2.16 J = 2.2 J (2sf) C1 B1 8(c)ii Energy dissipated = 9.6 – 2.16 = 7.44 J or 7.4 J (2sf) C1 8(c)iii Conduction – within the wires into the 2 metal terminals/connectors at both ends of the fuse Radiation – throughout the fuse including sealed glass cylinder and, in all direction. OR Convection – in the air inside the glass cylinder B1 B1 (c)iv 30A – No change in time as no energy is loss to the surrounding. All energy is used to melt fuse. Same as before. 8A – Time taken will be shorter as heat transfer through convection cannot take place in a vacuum hence no heat loss to the surroundings and more heat is used to melt fuse. B1 B1

Qn Answer Comments 9(a)i Alpha particles can be stopped by a piece of paper. The table shows a drop emission by 100 with a piece of paper. B1 9(a) ii Beta particles can be stopped, (500 to 0 count) by (5mm) a thin sheet of Aluminum as can be seen from table when it is used as an absorber. B1 9(a)iii Gamma radiation cannot be stopped by either paper or aluminum as it is transmitted through EM waves. From the table, count rate per min. remains at 0, hence no evidence of gamma radiation is observed. B1 9(b)i Natural occurring radiation from space or from environment (such as the soil in the ground or from equipment) B1 9(b)ii It can be measured by placing a detector (radiation) in the open, unobstructed and away from any known radioactive source B1 9(c)i 53 protons, 53 electrons and 131 – 53 = 78 neutrons B1 B1 9(c)ii Half-life is the time taken for half the nuclei of the nuclei of a sample to decay. B1 9(c)iii Half-life – 8.0 days 360 -> 180 -> 90 -> 45 Takes 3 half-lives for count rate to decrease to 45. Time taken = 8.0 x 3 = 24 days B1 B1

Qn Answer Comments 10(a) Iron is a of a soft magnetic material which can be magnetised and demagnetised easily (direction of magnetic field too). B1 10(b) 𝑵𝒔𝑵𝒑=𝑽𝒔𝑽𝒑 , 𝑵𝒔= 𝑽𝒔𝑽𝒑 × 𝑵𝒑 Minimum number if turns at 12 V: N,= 126.0 × 14=𝟐𝟖 𝐭𝐮𝐫𝐧𝐬 Maximum number of turns at 24 V: N,= 246.0 × 14=𝟓𝟔 𝐭𝐮𝐫𝐧𝐬 B1 B1 10(c) An alternating current produces changing magnetic field around the primary coil and hence a consistent change in magnetic flux linkage with the secondary coil inducing an emf in the secondary coil. B1 10(d)i B1 – curve B1 - positive 10(d)ii When diode is connected, current can only flow in the circuit when the voltage is positive, since current only flows only in 1 direction it is d.c. B1 10(e) Step-up transformer is used after the power station and generator stage to increase the output voltage and so as to decrease the output current before relaying into the transmission line to reduce power loss (Power loss = I2R). At the end of the transmission lines, step-down transformers are used to decrease the voltage to a safe level (240 V)for home, office and industries use. B1 B1 B1 B1

Qn Answer Comments 11(a) B1 – correct pattern B1 – right format – arrows, not touching each other, arrows correct direction 11(b)i When current is turned on, both rods will be magnetised by the magnetic fields from the coil. Both rods will be magnetised with the same polarity at both ends. Since liked poles repel, the two rods will move away from each other. B1 B1 11(b)ii When the current is removed, the iron rod which is a made of soft magnetic material will lose its magnetism but steel being a hard magnetic material will retain its magnetic field for some time attracting the non-magnetised iron rod. B1 B1 11(c)i Current in neutral – 5 A Current in earth – 0 A B1 11(c)ii When live wires touches the metal case, the circuit is complete with the low resistance earth wire. This increases the current in the circuit, strengthening the magnetic field of the solenoid on the Live wire and thus the electromagnet. This exerts a strong enough attractive force on the iron rod, attracting the rod resulting in a clockwise rotation, pushing up the contacts causing them to be separated. Circuit now is open current stops flowing. B1 B1 B1 B1