RVHS H2 Physics P2 Soln

River Valley High School Pg 1 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 RVHS JC2 H2 Physics Prelims Paper 2 Mark Scheme 1 (a) From the measured width of the cubes, take average. Compare the average value to the expected value of 2.0 cm; the closer it is to 2.0 cm, the more accurate is the dimension of the cubes. If the average width is between 1.95 cm and 2.04 cm, it can be considered to be accurate. B1 B1 (b) From the measured width of the cubes, compare the widths of each cube to one another. The closer the widths of each cube, the more precise is the dimensions of his printed cubes. M1 A1 2 (a) The Principle of Conservation of Linear Momentum states that the total momentum of a system remains constant provided no external resultant force acts on the system. B1 B1 (b) An elastic collision between two or more objects is one in which kinetic energy is conserved. B1 (c) initial momentum = (0.400)(5.0) = 2.0 kg m s−1 momentum of P after collision = (0.400)(–0.40) = –0.16 kg m s−1 momentum of Q after collision = 2.0 – (–0.16) = 2.16 kg m s−1 M1 A1 (d) For elastic collision, relative speed of approach = relative speed of separation uP – uQ = vQ – vP 5.0 – 0 = vQ – (– 0.40) vQ = 4.6 m s−1 mQvQ = 2.16 kg m s−1 mQ = (2.16) / (4.6) = 0.47 kg M1 A1 (e) F = Δp / Δt = (2.16 – 0) / (0.060) = 36 N A1 3 (a) (i) The positive direction was defined to be the direction from A to B. ( ) 22 11 24 24 82 31 6.67 10 5.07 10 3.23 10 (0.5 3.85 10 ) 3.31 10 ABGM GMgravitational field strength rr N kg − −− =− + = −  +   =−  Accept 3.31 x10−3 N kg−1 M1 A1

River Valley High School Pg 2 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (ii) General shape correct. Gravitational potential at surface of A is lower than that at B Marker’s comments: • Many students left a gap between the line to the dotted lines (the surface of both planets). This is not acceptable since the instruction stated specifically to draw from the surface of one planet to the other. Marks cannot be awarded for incomplete diagrams. • Some students drew the curve as if the dotted lines are the asymptotes. This is incorrect, since it suggests that the magnitude of the gravitational potential at the surface of the planet approaches infinity. Incorrect example: B1 (iii) General shape B1 B1

River Valley High School Pg 3 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 Magnitude of g at surface of A larger than that at B and neutral point closer to planet B Marker’s comments: • Similar to previous question, some students drew the curve as if the dotted lines are the asymptotes. This is incorrect, since it suggests that the magnitude of the gravitational field strength at the surface of the planet approaches infinity. (b) (i) • By Newton’s 3rd law of motion, Force on C by D = force on D by C • The same magnitude of force provides for the centripetal force on each star about P. • The angular velocity ω is the same for both stars. Hence, 22 1 2 C C D D CD CD C D M r M r rM Mr r r = = = M1 (ii) Since 1 2 C D r r = and 2C D M M = , 2 22 2 ( 2 ) 18 C C C C C C MGM GMgravitational force r r r == + M1 (iii) Gravitational force on C by D provides for centripetal force on C 2 2 2 23 212 3 11 9 30 18 18 2 18(2.40 10 ) 2 6.67 10 3.84 10 9.99 10 C CC C C C GM Mr r rM GT kg    − = =   =   = M1 M1 A1 4 (a) (i) 𝑣 = 𝑓𝜆 0.90 = 𝑓(0.30) 𝑓 = 3.0 Hz M1 (ii) 𝑓 = 1 2𝜋 √28 𝑚 m = 0.0788 kg A1 (iii) 1. Driving force is larger due to the larger amplitude of wave, OR more energy transfer. M1

River Valley High School Pg 4 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 Amplitude of the vertical oscillations will also increase. A1 2. Wavelength of wave increases. Since the speed remains the same, by 𝑣 = 𝑓𝜆, driving frequency of wave will decrease. Since driving frequency is not equal to natural frequency, amplitude will decrease (no more resonance) M1 A1 3. Damping force increases. Therefore, amplitude decreases. M1 A1 (b) (i) Light in which the oscillations of the electromagnetic fields are all in a single plane. A1 (ii) The displacements due to the two waves are in perpendicular axes, thus their vector sum will not be able to produce distinct maxima and minima, since they can never be in the same or in opposite directions (OR there will not be complete cancellation). Hence, the contrast between bright and dark fringes is not observable. M1 A1 (iii) For X, after the initial unpolarised light passed through the polariser, the light became plane polarised with 1 2=XII with amplitude XAA= (i.e. unchanged for any  value of X). Hence, 22(cos 45°)(cos 45°) 11 2 2 2 8 =   =      = ZXII I I M1 M1 A1 5 (a) (i) Correct directions with line of action of force vector passing through charges. (Arrow for negative charge pointing to the left and positive charge to the right) A1 (ii) F qE= ( )( ) 15 31 2 10 2 0 10.. −=   122 4 10 N. −= A1 (iii) Fd sin= ( )( ) 12 32 4 10 2 5 10 35. . sin −−=   M1 ( ) 153 4 4 10 N m. −= A1 (iv) either rotates to align with the field or oscillates (about a position) clockwise / with the positive charge on the right of the centre of dust particle B1

River Valley High School Pg 5 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (b) (i) electric force is negative of potential energy gradient B1 (ii) The two charges have opposite signs since the force between the charges is attractive / negative. B1 (iii) gradient = ( ) ( ) 10 10 9 1 00 3 00 9 6 10 4 4 10 3 846 10 .. .. . −− − − −=  −  = M1 9 193 846 10 1 6 10F . . −=    C1 106 2 10.N −= A1 6 (a) Let J be the contact point between the jockey and the wire XY 0.48 (20.0) 8.01.2 8.0(4.0) (4.0) 1.391323.0 XJ XJ XY XY XJ XJ XY LRR L RVV R = = =  = = = By considering the secondary circuit, p.d. across resistor R = 1.3913 V p.d. across 1.0 Ω resistor = 0.50 V By principle of potential divider, 11 1.3913 (1.0) 2.80.50 resistor R resistor R resistor R RV RV R  = = =  C1 C1 A1 (b) (i) I = nAvq v = I / nAq = 2.8 / ( (8.49 × 1028)(0.20 × 10−6)(1.6 × 10−19) = 1.03 × 10−3 m s−1 M1 A1 (ii) 6 8 0.72 2.8 0.72 (0.20 10 )2.8 1.7 10 m3.0 VR R A RA   − − == =  = = =   I l l C1 A1 (iii) As the current increases, the temperature increases since the rate at which electrons collide into the metal lattice increases. As temperature increases, lattice ions vibrate more vigorously, hindering the flow of electrons. Hence, its resistance increases. B1 B1 7 (a) A discrete unit/quantum/packet of electromagnetic radiation energy. A1 (b) An emission line spectra is due to emission of photons during transition between energy levels. Since spectrum consists of discrete lines [rather than a continuous spectrum], the energy levels must be discrete. B1 B1

River Valley High School Pg 6 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (c) Excited level = –13.6 + 13.2 = –0.4 eV, nearest level is 5 Highest frequency transition from 5 to 1. Energy = hf = –0.54 –(–13.6) ev = 13.06 eV f = 3.15 x 1015 Hz M1 M1 A1 (d) (i) Negative potential difference will prevent the most energetic electrons from reaching the collector plate. B1 (ii) photon energy = work function + KE of electrons (3.4–0.54) x 1.6 x10–19 = work function + 0.72 x 1.6 x 10–19 Work function = 3.42 x 10–19 J M1 A1 (iii) No of electrons per sec = 0.21 × 10–6 /1.6 × 10–19 No of photons p