RI 9646 2013

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2013 A-Level H2 Physics Suggested Solutions Paper 1 1 A The scattering of data points around the best fit line suggests that the results are imprecise. Using 2212 t2s ut at sg= + ⇒= and value of gradient, g = 9.804 m s−2 (close to 9.81 m s−2) and is accurate. 2 C Normal A4 paper is 80 grams per square meter (gsm). Area of A4 paper is 21.0 cm by 29.7 cm. (Use ruler to measure exam question paper). Mass of a single sheet = 80 × 0.210 × 0.297 = 5 g 3 A 22 22 2 22 3 to 4 11 22 1112 ( 2) (1)22 8.00 m s 11 (4) (3) 28 m22 s ut at s at aa a saa − = + ⇒= = − = =−= 4 D Velocity of stone passing edge of cliff on its way down is 10 m s−1 downwards 2211 (10)(1.2) (9.81 )(1.2) 19.1 m22s ut at= += + = 5 C For elastic collision, relative speed of approach = relative speed of separation, u1 − u2 = v2 − v1. Only option C is correct. 6 B 2 m Ax Avtt p mvF Avtt ρ ρ ρ ∆∆ = =∆∆ ∆∆= = =∆∆ 7 A pressure due to fluid (600)(9.81)(0.200) 1177 Paghρ= = = 8 D For equilibrium, the third force must form a closed vector triangle with the other two forces. Hence, it must act in the Z direction. 9 C 1001400 9.81 1.6 109872 W0.20 20 FvP = = × ×× = 10 A The sum of the three types of energies must be 120 J. Option B is wrong because kinetic energy at bottom must be 0. Option C is wrong because the total energy in middle is not conserved. Option D is wrong because kinetic en ergy throughout the motion cannot be constantly at 0. 11 B There are only two forces acting on the ball, the weight of the ball and the tension in the thread. The resultant of these two forces provides the required centripetal force for the ball to move in a circle.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 12 D 2 13 7 2 72 14 2 1 5.0 102 2.447 10 (2.447 10 ) 9.98 10 m s0.6 c mv v va r − − = × = × ×= = = × 13 A 2 2 2 64when 2 , 16 2 64when 4 , 4 4 GMg r rg rg = = = = = = = 14 A Centripetal force is provided by gravitational force 2 22 At P, 11At Q, 99(3 ) Q GMa r GM GMaa rr = = = = 15 D GPE is linear as it is proportional to height. KE is an inverted parabola with a value of zero at the equilibrium position. EPE is a curve with increasing values as extension increases (displacement downwards). 16 C 1 max 0 2 (75)(0.01) 4.71 m svx ωπ −= = = 17 C ' 5 and ' 2 2 0.45 0.4 0.6 final escape PVn RT TTPP VPnn RT n nnn = →→ = =  = −= 18 C 11 22 12 1 2 12 12 Pm h Pm h PP m m PP mm = + = + −= − −= − l l ll l 19 B 2 0 max 0 kx vx ω= I= Thus maximum speed of air molecules increases. Speed of wave is independent of intensity and is dependent on temperature and mass of a molecule in the medium.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 20 A The crests are locations of nodes of the stationary wave, which have no amplitude and has no effect on the powder. 2 0.2 m 330 3300 Hz0.1 vf f λ λ = = = = 21 B For the first order maxima to be symmetrical about the zeroth order, the incident beam must be normal to the grating. Option A will result in two sets of diffraction patterns. Option B will result in greater separation between the first bright fringe and central fringe on the screen. Option D will result in fringes that are not sharp. 22 B With no change in frequency and wavelength of the wave, the two waves undergo destructive interference at P and constructive interference at Q. The destructive interference at P is less complete and results in an increase in loudness. The constructive inte rference at Q will result in a lower resultant amplitude and decrease in loudness. 23 C ( ) ( ) ( ) 22 13 2 31 By conservation of energy, initial KE + initial EPE final KE + final EPE 11 + 0 ( )22 1 2 mv mv e V m v v eV = = +− −= 24 B Direction of gravitational field always point towards the mass, while the direction of the electric field lines can be towards or away from the charge, depending on whether it is a negative or positive charge. 25 A LDR, metal wire and thermistors have I-V graphs that pass through origin and only a diode has nearly linear portion that is far from the origin. 26 A ( ) terminal pd thus, since is larger, terminal pd of P is larger terminal pd thus power output of P is also larger Q output r r P ε= − = I I 27 D ( ) let be the thickness of the metal sheet s, 4.0 2total resistance 4.0 4.0 4 20 2 R A xt xt x x x xtt ρ ρ ρ ρ = = = += + =   l

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 28 A when rheostat is set to 0 , 10 40 20 V20 when rheostat is set to 20 , 10 40 10 V40 V V Ω =×= Ω = ×= 29 A 5 71 2 for charges passing through undeflected, 3.0 10 2.0 10 m s1.5 10 BEFF Bqv qE Ev B − − = = ×= = = × × 30 B ( )( )( ) ( )( ) 45 7 45 8 maximum magnetic flux linkage, cos 35 2.1 10 7.0 10 5.145 10 maximum magnetic flux, cos 2.1 10 7.0 10 1.47 10 NBA BA θ φθ −− − −− − Φ= = ×× = × = = ×× = × 31 A Magnetic flux density due to wire XY is into the page on the right side of XY. Magnetic flux density is higher near the wire. Induced e.m.f. in option C and D is zero. 32 A Q = It. Area under I – t graph is the amount of charge 33 B ( ) amplitude, 0.5 V period, 0.2 s 2equation is sin 0.5sin 10 A T VA tT π π = = = =

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 34 B st 1 1 nd 2 2 10turns ratio for 1 transformer 10 10 30turns ratio for 2 transformer 30 30 pp s s sp s ps s p s s s ps s p s s NV NV N VNV N N VNV N = = = = = = = = I I II II 35 B max max 34 8 19 9 18 KE KE 6.63 10 3.0 10 1.6 10 4.380 10 1.798 10 J 11.23 eV hc hc φλ φλ − − − − = + = − × ××= −× ×× = × = 36 C Not in 9749 syllabus 37 B Not in 9749 syllabus 38 B Not in 9749 syllabus

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 39 C 0 0 1/ 2 0 0 00 let be the initial number of Antimony-1 24 nuclei, be the number of Antimony-124 after tim e , be the number of Tellurium-124 after ti me ln2 An Te t An Te An Te An An An N Nt Nt N Ne t N NN N NN NN e NN λ λ λ − − = = = − −−= = 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 ln2 () ln2 ()0 ln2 ln2() () ln2 () 1 6 61 71 ln 2 1( ) ln60 7 168.4 days tt t t tt tttt tt e Ne e ee e t t λ − − − −− − −= = = − = −=  = 40 D In b eta decays , the nucleon number is unchanged and the proton number increases by 1. In alpha decay, the nucleon number decreases by 4 and the proton number decrease by 2.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 Paper 2 Structured Questions 1 (a) ( )( ) 2211Distance moved along cable, 1.5 2.022 3.0 m s at= = = Vertical height 3.0sin40 1.93 m = ° = M1 M1 A1 (b) (i) At constant speed, N = W = (95)(9.81) = 932 N A1 (ii) Since the cable car is at constant speed, there is no resultant force acting on it. Hence, the vector sum of forces in all directions must be zero. Since N and W are the only forces in the vertical direction, their magnitudes must be equal. B1 (c) (i) Weight of the man, normal contact force from the floor of the cable car and friction between the man and the floor of the cable car. B1 (ii) Normal reaction is larger than weight, giving rise to acceleration in the upward direction. Friction causes a horizontal acceleration. The resultant acceleration is in the direction of the rope. B1 B1 (d) ( )( )Distance moved along cable for 120 s at constant speed 3.0 120 360 m= = ( )( )1Distance moved along cable for last 3.0 s 3.0 0 3.0 4.5 m2= += Total vertical height =1.93 360 sin 40 4.5 si n 40 236 m + °+ ° = Gain in GPE (95)(9.81 )(236) 220000 J mgh= = = M1 M1 A1 (e) (i) Note: • At t = 0 gradient is 0 • Positive increasing gradient • Curve must reach t = 2.0 s A1 h t/ s 2.0 0 0

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 8 (ii)