2022 A Level H2 Physics P3 Soln

2022 A Levels H2 Physics 9749/02 Paper 3 suggested solutions 1a Inertia is the tendency for bodies to resist changes in its motion. 1b      11 22 1 22 6 6.67 10 7.35 10 1.62 N kg 1.74 10 GMg r        1ci Since a force of 10 kN is exerted on the rocket, by Newton’s 3rd law, a force of 10 kN is also exerted on the gases in the opposite direction.   on gas 1 10000 70.0 143 m s dmF v dt v v     1cii After 15.0 s, mass of rocket   4000 15.0 70.0 2950 kg        2 10000 2950 1.62 2950 1.77 m s netF ma T mg ma a a        1ciii Actual acceleration is greater. At a greater height, the gravitational field strength is lesser since gravitational field strength is inversely proportional to the square of distance from centre of planet. Thus the weight of the rocket will be lesser, resulting in a larger net force on the rocket. Hence its acceleration is greater. 2a 2 3 6 1.021.73 10 2 1.41 10 m R A R A                  l l 2bi time after being switched on / s ΔU q w 0-59 positive negative positive 60-100 zero Negative positive 2bii At constant temperature, ΔU = 0 work done by the power supply = Power x time Hence for 40 s, w = Pt = IVt = 12 x 230 x 40 =110400 = 1.10 x 105 J By 1st law of thermodynamics, q = -1.10 x 105 J 3ai -1 2 12002 ( ) 60 126 rad s f      3aii 2 2 21.20(0.230)(126) 1.2(9.81) 4370 N F W mr F mr W         

3aiii Washing machines vibrate vigorously when spinning. The large masses serve as a mass damper by reducing the amplitude of vibration. The masses also reduce the natural frequency so that it will be below the frequency of spinning, to prevent resonance from happening. The masses are also placed at the bottom to lower the centre of gravity of the washing machine, increasing the stability to prevent toppling when it vibrates. By Newton’s third law, the force on the towel by the machine (a)(ii) is equal in magnitude to the force on the machine by the towel. Hence, large masses placed at the bottom can also prevent such a large force from cracking the base of the machine. 3b At the holes, there is no normal contact force exerted on the water droplet, so the only force it experiences is its weight which acts vertically downwards. There is no normal contact force present at the holes to provide for centripetal force, hence the water droplet is not able to remain in circular motion. The velocity of the water droplets at the holes will be tangential to the drum. 4a The 2 spheres have the same sign. Based on Fig 4.2, the potential gradient of the graph changes from negative to zero before turning positive. This implies that electric field also changes direction since E=-dV/dx. Electric field between 2 charges will only be zero if the charges are of the same sign. Since potential between the 2 spheres is always positive and potential is a scalar, the 2 spheres are positively charged. 4b Using Fig 4.2, Find negative potential gradient , -dV/dx (=E) at x=0.15m, x=0.80m and 2 other points between x=0.15 m and 0.68m. Graph cuts through x axis at x=0.68 m since potential gradient at that point is zero. Join up the points with 1/x shape in mind. 4ci 81 1 10 599 V4 015 .o V    4cii Since there’s another charged sphere B nearby, the electric potential at the surface of sphere A will be the scalar addition of potential due to sphere A as well as sphere B. Thus, the potential on surface of sphere A shown in Fig 4.2 will be bigger. 5a When the switch is closed, the voltmeter will record a non -zero reading for a short time before becoming zero. When the switch is opened shortly after, the voltmeter will recording a non -zero reading in the opposite polarity which goes back to zero after a short time.

5b When the switch is open, there is no magnetic flux in the setup. When the switch is closed, there is a current flowing in the primary coil that produces a magnetic flux. The core confines the flux resulting in an increase in magnetic flux linkage in the secondary coil pointing downwards. By Faraday’s Law an e.m.f. proportional to the rate of change of flux linkage is produced. When current in primary coil is constant, there is no change in flux produced. Hence flux linkage in secondary coil remains constant and hence no e.m.f. is produced. When the switch is open there is a decrease in magnetic flux linkage in the secondary coil pointing downwards. By Faraday’s Law an e.m.f. will be produced. Since the magnetic flux linkage is now decreasing instead of increasing, the e.m.f. produce will be in opposite polarity. 6a From graph, when temperature is 20°C, R = 60 Ω Using the potential divider rule, 0.43 1.50 , where is the resistance of the fixed resistor60 24 (shown) fixed fixed fixed fixed R RR R      6b From graph, when temperature is 32°C, R = 40 Ω 1.50So current across thermistor = 0.0375 A40 1.50current across fixed resistor = 0.062524 current in cell = 0.0375 + 0.0625 = 0.10 A A   7a It means that the light waves have a constant phase difference and same frequency. 7b For the grating, 60.001 3.33 10 m300d          6 9 o o sin 3.33 10 sin 2 640 10 22.58 tan 22.58 2.1 0.873 m 87.3 cm d n y y               8ai The electric field strength at a point is the electric force per unit positive charge acting on a small stationary charge placed at that point. 8aii The electron gun must be in a vacuum so that the electrons emitted from the cathode will not collide wi th gas molecules in the gun. The ensures that there would be sufficient electrons accelerated towards anode and emerges as electron beam. 8aiii 16 19 gain in KE loss in EPE 2.48 10 1.6 10 1550 V eV V V        

8aiv Electrons may not be emitted horizontally from cathode (as shown in Fig.8.2). Electrons emitted may come from various positions below the surface of the cathode. Hence, they may collide with other atoms or electrons, losing some energy before emerging from the cathode with a range of speeds. As such, there would still be a range of speeds of electrons reaching the anode despite the same increase in KE. 8bi 0 7 3 2 4 10 120 3.5 0.30 1.8 10 T NIB r           8bii From the top view, current is clockwise. By right hand grip rule, the magnetic flux will be pointing into the paper. Hence, on Fig 8.4, the magnetic flux will be pointing downwards (within the col). Outside the coil, the field lines should continue and hence loops are observed. Note: 1) At least 5 lines with correct arrows 2) Lines must not cross. 3) Spacing of lines should be carefully considered while keeping symmetry. 8ci The initial direction of motion of electrons is perpendicular to the uniform magnetic field out of the page. By Fleming’s left hand rule, electrons will experience a magnetic force that is always perpendicular to velocity of the electrons. In this case, the magnetic force provides for the centripetal force that causes the electrons to move in a circular path. 8cii Magnetic force provides for the centripetal force 2 31 7 3 19 9.11 10 2.40 10 1.43 1.8 10 1.6 10 0.0531 m mvBqv r mvr Bq              8d The component of the velocity (of the electrons) perpendicular to the field will result in the electron to move in circular path. The component of the velocity (of the electrons) parallel to the field will cause the electron to move (at constant speed) in the direction of the field. Hence, the net effect is that the electron will move in a helical path.

9a 1. This observation supports the particulate nature because radiation arrives as discrete bundles of energy (photons) with energy, E = hf. Each photon interacts with an electron in the metal surface on a one-to-one basis as each photon can transfer its energy to at most one electron. A higher intensity beam of radiation cont