2018 RI H2 Physics P2 solutions

© Raffles Institution 9749/02   2018 Raffles Institution Preliminary Examinations – H2 Physics Paper 2 – Solutions 1 (a) (i) The rate of change of (total) momentum of a system of bodies is d i r e c t l y proportional to the resultant external force acting on the syst em and the direction of the change is in the direction of the force. (ii) If the system is isolated from all external forces, then the re sultant external force acting on it is zero. By Newton’s second law, the total momentum of the system will not change, which means that it is conserved. (b) (i) One quarter of SHM sinusoidal function (ii) By conservation of energy, 22 2111 222 AA BBmv mv k x  2220.100 0.050 80 0.060 0.288ABvv    2225 . 7 6ABvv  ----- (1) By conservation of momentum, 0AA BBmv mv 2BAvv  ----- (2) Substitute (2) into (1), 22245 . 7 6AAvv  10.98 m sAv  Hence, 11.96 m sBv  (c) (i) Taking moments about O,   sin55 0.15 0.050 9.81 0.25 4.0 0.20 7.509 7.51 N T T      (ii) Resolving the forces horizontally, cos55 7.509 cos55 4.307 4.31 NxFT    Horizontal component is 4.31 N (Direction is leftward.) force t 0

2    © Raffles Institution 9749/02 Resolving the forces vertically,     sin55 0.050 9.81 4.0 0.050 9.81 4.0 7.509 sin55 1.661 1.66 N y y FT F             Vertical component is 1.66 N (Direction is downward.) (iii) 1.661tan 4.307 21.09 21.1 y x F F        2 (a) (i) 1. 1 c y c l e  3 1.90 ms 11 526.3 526 Hz 1.90 10 T f T       (Students are allowed to use one or more cycles.) 2 . uncertainty in the time scale is the smallest division: 0.1 ms If used one period in (a)(i)1., 0.1 msT 0.1 526.31.9 27.7 30 Hz (1 s.f.) fT fT Tff T       Absolute uncertainty given to 1 s.f. OR If used two periods in (a)(i)1., 20 . 1 m sT 1 0.1 0.05 ms2T  B rope 0.15 m 0.10 m O 0.15 m table surface wall centre line Fy Fx F

3    © Raffles Institution 9749/02 0.05 526.31.90 13.85 10 Hz (1 s.f.) fT fT Tff T       Absolute uncertainty given to 1 s.f. (ii)  2 4 25 42 1.90 0.76 ms52 5 t T t T Tt             1.50 0.76 2.26 ms (iii)  22 2 11 2 1 1 22 1 1 1 4 for 0.25 0.25 120 60 cm PPII rr IP r IP r II Prr P Prr P        (b) (i) All the particles in a progressive wave oscillate with the same amplitude. The particles in a stationary wave oscillate with amplitudes that range from zero at the nodes to a maximum at the antinodes. (ii) All the particles within a wavelength of a progressive wave have different phases. All the particles between two adjacent nodes of a stationary wa ve have the same phase. Particles in adjacent segments have a phase difference of  radians.   3 (a) (i) At equilibrium, pressures in both chambers are the same. Using pV = nRT XX Y Y XY nR T nR T VV     300 2.51.2 450 4.0 0.50 mol YX XY XY TVnn TV              (ii) nRTp V   1.2 8.31 300 0.50 8.31 450OR4.0 2.5 XYpp  

4    © Raffles Institution 9749/02 747.9 750 PaXYpp      (iii) With some gas removed, there are fewer gas molecules in Y which results in a smaller overall rate of change of momentum of molecules as they collide with the walls of the chamber. Hence the average force on the walls decr eases which implies gas pressure is reduced. ( b ) ( i ) Since process A to B is isothermal  TA = TB pA AB BpV V       5 55 2.9 10 0.015 0.040 1.09 10 1.1 10 Pa Ap        (ii) Using 21 3 Nmp cV     5 2 3 1.09 10 0.0403 0.060 pVc Nm     21 466.9 470 m srmscc    (iii) 33 ()22Un R T p V      55 3 2 3 0.015 5.2 10 2.9 102 5175 5200 J CC BBUp V p V          Using the First Law of Thermodynamics, 0 5200 J UWQ Q Q        ( i v ) By ensuring that the process from C to A takes place rapidly su ch that there is insufficient time for any heat transfer to take place between t he system and its surroundings.

5    © Raffles Institution 9749/02 4 (a) Electric field strength is numerically equal to the potential gradient at that point. OR Electric field strength is the negative of the potential gradient. (b) (i) 6.0 m (ii) 9 12 0 7.2 10224 4 8.85 10 6.0 21.6 22 V Q r        (iii) ( i v ) 1 . No it will not reach the surface of Q. It does not have sufficient kinetic energy to reach Q, as the electric potential at the surface of Q is higher than the electric potential at the surface of P. 2 . Being positively charged, it will be repelled by sphere P’s ele ctric field and accelerates with increasing kinetic energy as the electric potential decreases towards x = 6.0 m / mid-point. After passing x = 6.0 m, it decelerates towards sphere Q with decreasing kinetic energy as the electric potential increases, but it does not have sufficient energy to reach Q. Its kinetic energy drops to zero somewhere before 11.92 m and i t returns to sphere P with its kinetic energy increasing towards x = 6.0 m then decreasing after x = 6.0 m, and the cycle repeats again. OR Being positively charged, it will be repelled by sphere P’s ele ctric field and accelerates towards the point x = 6.0 m as the resultant electric field is in the positive x-direction, as shown by the negative of the potential gradient. After passing x = 6.0 m, it decelerates towards sphere Q as the resultant electric field direction is in the negative x-direction, and its velocity reaches zero before reaching Q. It will then return to sphere P, accelerating towards x = 6 . 0 m , t h e n decelerating after the mid-point, and the cycle repeats again.   V / V x / m 0 0.16 11.92 6.00 12.00

6    © Raffles Institution 9749/02 5 (a) (i) Resistance of a conductor is defined as the ratio of the potent ial difference across it to the current flowing through it i.e. VR I Resistivity is the constant of proportionality for the relation ship between a conductor’s resistance and its length and cross-sectional area i.e. R A l    OR Resistance of a conductor is dependent on the length and cross-sectional area (i.e. R A l  ) of the conductor whereas resistivity is a characteristic of the conductor’s material which is independent of length and cross-sectional area. (ii)   23 6 1.0 10(2.0) 2 1.5 10 1.047 1.0 m LR A RAL            (b) (i) 6.0 2.0 4.0 1.0 A VR V R      I I (ii) number density of conduction electrons,  3 23 28 28 3 no. of mol volume density molar mass 8.96 10 6 . 0 2 1 00.064 8.428 10 8.4 10 m A A Nn N            28 3 2 19 55 1 1.0 8.43 10 (1.0 10 / 2) 1.60 10 9.440 10 9.4 10 ms d d nAv q v nAq             I= I

7    © Raffles Institution 9749/02 (iii) 5 33 t 0.20 9.44 10 2.119 10 2.1 10 s d d v        ( i v ) When the switch is closed, the electric field is established in the circuit almost instantaneously. Hence, all free electrons including those in the lamp filament present in the circuit will start to drift at the same time. (c) (i) XJ XYLL 3.0 ( 1 . 0 )6.0 0 . 5 0 m XJ XY V V   (ii)  XJ XYLL 1.0 3.01.0 0.5 ( 1 . 0 )6.0 0.333 0.30 m XJ XY V V     

8    © Raffles Institution 9749/02 6 (a) max 373 300 373 0.196 e   (b) (i)