2018 RI H2 Physics P1 solutions

© Raffles Institution 9749/01   2018 Raffles Institution Preliminary Examinations – H2 Physics Paper 1 Answers 1 C 6 B 11 C 16 D 21 B 26 D 2 D 7 D 12 B 17 C 22 B 27 B 3 C 8 C 13 C 18 D 23 C 28 A 4 A 9 A 14 A 19 D 24 B 29 D 5 D 10 A 15 B 20 A 25 C 30 A Paper 1 Suggested Solutions 1 C ker ker 70 20 50 g 112 g Liq Total Bea Liq Total Bea mm m mm m           3 20 . 6 50 10.0 20 . 6 5.0 0.5 g cm (1 s.f.)50 10.0 Liq Liq Liq Liq Liq Liq m V mV mV             2 D After first bounce, 22 2 00 0 79 3 1 1 31 16 16 4 2 2 4 where is the speed just before the ball hits the ground left initial initialKE KE KE mv m v v                 If the height of the second triangle is now 0 3 4 v , time duration of the second triangle is 3 0.754 tt (due to similar triangles). Since air resistance is negligible, the time taken by the ball to move up after the first bounce is the same as the time taken to move down to the ground again. OR Since air resistance is negligible, the gradient of the velocit y-time graphs will all be the same with the value of g. 0 0 22 2 00 10 1 0 79 3 1 1 31 16 16 4 2 2 4 3speed after bounce, 4 31 31 3 0.7544 4 left initial initial v gv g tt KE KE KE mv m v vv vtv g tt tgg g                               

2 © Raffles Institution 9749/01   3 C Horizontal motion: 4.0 x x sv t st vv   Vertical motion: 2 2 1 1 2 14 . 01.5 1.1 2 14 m s ysg t g v v      4 A This is a completely inelastic collision. Hence, total mechanical energy is not conserved. 5 D By Newton’s second law, net force, 3 5 41 54 5 FFm m a FaF mm       Consider the forces acting on Y: 355 43 55 5 XY XY FFFm m FF FFm m     6 B There cannot be any normal reaction when the sphere is in equil ibrium because it will be the only force with a horizontal component that is not balanced. 7 D Taking moments about the point on the base of the block through which R acts,       225.0cos30 5.0sin30 10 2 10 1.0 8.0 10 0.0473 m 4.7 cm x x       8 C Work done to stretch it 10 cm:  2 1 1 0.10 4.02 800 N m k k    Work done to stretch it 20 cm:    21 800 0.20 162  Additional work required = 16 – 4 = 12 J 9 A 2 2 1 cos 40 --- (1) sin40 --- (2)1.2sin40 (2) tan 40 2.5 m s(1) 1.2 sin 40 Tm g mvT v vg       

3 © Raffles Institution 9749/01   10 A The equations for potential energy and kinetic energy are given by : p GMmE R and 2 k GMmE R As R increases, pE increases (becomes less negative) and kE decreases. 11 C 2 3 3 2 , , 12 0 2 4and since = 3 42 83 3 For a body to escape a planet’s gravitational influence from its surface: escape escape escape escape X xx xx escape Y y Y YY GMm GMmv vRR MR GR GRv R vR v R R vR R               39 9 4.50142   12 B The components of the particle’s motion in the horizontal x-direction is simple harmonic. Hence ax . 13 C Between 2.0 s and 2.5 s, the block is in contact with the sprin g and its motion is simple harmonic i.e. the graph is ¼ of a cosine graph. Period when block is in SHM = 4  0.5 s = 2.0 s   = 2/T = 3.14 rad s1 0.30 0.095 mooxv   14 A Using Malus’ law, 22 21 0 1cos 10 cos 102  II I 2 2 cos where is the angle between the axes of polarisation of the second and third filters   II 22 2 20 22 00 1 2 1cos = cos 10 cos2 11 cos 10 cos32 2cos 33.99 343cos 10           II I II       I0 I1 I2 I

4 © Raffles Institution 9749/01   15 B distance travelled in 1 s = 110 2 time taken to travel 11 s21 0  distance = speed time 11 2.021 0 12 2.0 0.40 m10       At initial position M, detector detects maximum intensity (antinode). Minimum intensity detected will be ¼ wavelength away (node). 11distance moved 0.40 0.10 m44  16 D 2 2 0.40 1.5 10tan 22 1.5 10 2.1 m0.402t an 2 D D         9 4 sin 2 2 600 10 3.44 10 0.34 mm0.40sin 2 a a            OR  9 2 600 10 2.1 0.34 mm 1.5 10 4 Dx a Da x         17 C   11 2 2 12 11 22 2 1 1 11 11 1.25 0.75 0.9375 PV PV TT TTTP V P V TPV PV         2 2 21 2 11 13 22 0.9375 0.97 KEm ck T cT c T TTcc c c TT     18 D For a fixed mass of gas in the pump and basketball, UWQ    Compression implies W > 0. Well-insulated pump/basketball implies Q  0. Hence, ∆U > 0 (increases) and the temperature of the gas increases.

5 © Raffles Institution 9749/01   19 D Electric force acting on ion P is eE (in direction of the E field), and that acting on ion Q is also eE (but in the opposite direction to the E field). Resultant force on the molecule is therefore zero.    19 9 26 Torque of couple 1.60 10 4200 0.12 10 8.1 10 Nm Fd qEd        20 A When oil drop is at equilibrium, qE mg Vqm gd   When p.d. is 2V, net forc 2 2 (upwards) e Vqm gd mg mg mg    Upward acceleration is thus g. 21 B 2 6 10 120 JEV t   I 22 B Effective external resistance for max. power delivered = 5  111 10 5 10 R R    Potential difference across the 10 resistor 5 (12) 6 V10   22 6 3.6 W10 VP R 23 C At d, the net field due to both wires is zero. Hence the magnetic flux density of each wire is equal in magnit ude and opposite in directions. This means IX and IY are in opposite directions. 22 4.00 3 XY oX o Y X Y BB Ld d Ld d Ld       II I I 24 B Since both X and Y have the same mass to charge ratio, 22 2 1 0.252 XX X YY Y KE v R KE v R    

6 © Raffles Institution 9749/01   25 C Since dsEB L v B L dt , the magnitude of E can be deduced from the gradient of the s-t graph. 26 D Resistance of heater 2 2110 15.125 800 ACVR P    Power dissipated when d.c. supply is used 2 2156 1609 W15.125 DCV R  27 B hc E   and 2hc E    2 E E      Since 1 122 E E    2  28 A Shortest wavelength min hc eV   minlg lg lg hc Ve   Graph of minlg against lgV is a straight line with negative gradient and positive intercept. 29 D The sample consists of the parent as well as the daughter nuclei.  particles are electrons that have mass much smaller than that of the parent nuclei. Hence the mass of the sample is only slightly smaller than its initial mass, differing only by the small mass of  particles emitted. 30 A energy released = (39.25) + (28.48) – (64.94) = 2.79 MeV  -ray energy = 2.79  2.31 = 0.48 MeV