TJC 2023 H2 Prelim Paper 1 Solutions

2023 H2 Prelim Paper 1 Solutions: 1 2 3 4 5 6 7 8 9 10 D D B B A A C D D C 11 12 13 14 15 16 17 18 19 20 D D B B D B A A D B 21 22 23 24 25 26 27 28 29 30 B A A A C C B C A D 1 D ( ) ( ) ( ) ( ) 1 1 385 115 270 1353.50 1.50 2.00 1 1 0.02 0.02 135 270 2.00 3.7 4 sv mm st v s t v s t v v mm s − − −= = = = −   =+ ++ =+  = = 2 D Magnitude of 2nd displacement = √1502 + 1402 = 205 𝑐𝑚 At a direction  = 90 – 30 – tan-1 (140/150) = 17o below x axis or 343o anticlockwise to +x axis 3 B At t = 0 s, 1.2 x 105 = 2(1.0 x 105) a a = 0.60 m s-2 At t = 20 s, v = at = 0.60 x 20 = 12 m s-1 back carriage moves with constant speed of 12 m s -1, so in another 20 s, distance moved = 12 x 20 = 240 m front carriage now has acceleration of 1.2 m s-2. In another 20 s, s = ut + ½ at2 s = 12 (20) + ½ (1.2)(20)2 = 480 m distance between front and back carriage = 480 - 240 = 240 m 150 cm 140 cm

4 B Ball A: h = ½ gt2 → time of flight t = h g 2 xA = (2v)t = 22 hv g Ball B: 2h = ½ gt2 → time of flight t = h g 22 xB = vt = hv g 22 A B x x ===22 1.4112 5 A Consider the whole system, 𝐹 − 6𝑓 = 6𝑀𝑎 𝑎 = 𝐹 − 6𝑓 6𝑀 Let the force that 3M acts on 2M be F1. Consider Newton’s 2nd law on 3M, 𝐹1 − 3𝑓 = 3𝑀𝑎 𝐹1 = 3𝑀 (𝐹 − 6𝑓 6𝑀 ) + 3𝑓 = 𝐹 2 6 A Apply COM to the collision: (5.0)(200) = ( 5.0 + 95) u u = 10 m s-1 Apply COE after collision: ½ (100) (102) = (100)(9.81) H H = 5.1 m 7 C The 3 lines of action of the forces should coincide at a point. 8 D For each mass, Tension + Upthrust = Weight  Tension = Weight - Upthrust Since rod is horizontal, tension is the same for both P and Q Upthrust = Vg is greater for P since liquid X is denser. So weight of P is greater to get the same tension as Q. 9 D Let T be the tension in the string. For vertical equilibrium of mass, Tcos30.00 = mg (1) For horizontal circular motion of mass, Tsin30.00 = mr2 (2) )1( )2( → tan30.00 = r2/g tan30.00 = (3.0 + 5.0sin300)2/9.81  = 1.01 rad s-1 Time for one revolution T =  2 = 6.2 s. 3M 2M F M central axle m 5.0 m 3.0 m

10 C Apply Newton 2nd law: Along tangent of circular path, F – mg sin  = 0 Along the radial direction, T – mg cos  = mr2 Thus T = mr2 + mg cos , so T varies with cos  Note: Tension is largest at lowest point and smallest at highest point in a vertical circle. 11 D At point P, )r/ GM()r/ GM(Φ B P 323 −+−= = )M.M(r G B513 +− At point Q, )r/ GM()r/ GM(Φ B Q 332 −+−= = )MM.(r G B351 +− )M.M(.)MM.( so,Φ.Φ BB PQ 513251351 251 +=+ =  MB = 2M 12 D Consider the forces on the satellite: 22 23 1 =  GMm mrrr So when r decreases, both gravitational force on the satellite and the its angular velocity increase. So A and B are wrong statements. Consider the various energies associated with an orbiting satellite: r GMmKE 2+= , r GMmGPE −= So when radiu s r decreases, KE increases while GPE decrease. So only D is correct. 13 B 10 = mcΔθ t + h –(1) where h is rate of heat loss to surroundings 18 = 3mcΔθ t + h---(2) Solving (1) and (2) gives h = 6.0 W

14 B At constant P, V  T. When cooled, T decreases, so V decreases. Since gas undergoes compression, work is done on the gas. W = +ve When T decreases, U = -ve, according to 1st law of thermodynamics U = q + W -ve = q + (+ve)  q must be -ve 15 D Given xo = 0.030 m, from graph → period T = 2.0 s Max v = xo  = 0.030 x (2/2.0) = 0.094 ms-1 16 B At X, the two neighbouring particles are moving towards the particle at X, so it is a compression region . At Y the two neighbouring particles are moving away from the particle at Y. 17 A Rayleigh’s criteria, minimum angle of resolution  = /b. size of the pupil ≈ slit width b. If b increases,  decreases. (If  is small, images are easily resolved.) 18 A When the two polarisers are parallel( = 0), the intensity of emergent beam is I0 When Q is rotated at angle  , the intensity of emergent beam is reduced by 30%, which means final intensity should be 0.70I0. Appy Malus Law, II = = = + =  2 000.70 cos In the first quadrant, 33 In the third quadrant, 180 33 213    19 D The electric field within the conductor will be zero(since no net charge/equipotential inside a conductor) while the external regions will still have a uniform electric field pointing from positive to negative plate. displacement distance along the straight line from source

20 B R = ρL / A , R = V / I V/I = ρL / A L = VA / Iρ = 240 [π(3.0x10-4)2] / (2.0 x 1.4 x 10-6) = 24 m 21 B A: IA = V/R B: IA = V/3R C: IA = V/R D: IA = V/2R 22 A V = E – Ir When current I = 0, V = E = 2.5V 0.9 = 2.5 – 0.80 r r = 2.0 Ω or use gradient of graph = -r 23 A Apply RH Grip Rule to determine the direction of B-field at O due to the currents flowing in a straight wire at P, Q, R and S respectively. Vector sum of the B fields gives direction A. R S P Q BQ O BS BP BR Bresultant

24 A electron passes undeflected so electric force FE = magnetic force FB qE B q= v conditions for passing through undeflected does not depends on mass or charge so outcome is the same. 25 C Similar to Faraday’s homopolar generator. The induced current I =  /R = Br2f/R, which is a constant value when r and f are kept constant. However the direction of current changes clockwise or anticlockwise, depending on which radii is cutting the flux(use Fleming RH rule). 26 C de Broglie wavelength  = h/mv, so   1/m for same speed v. Since mass of electron is much smaller than proton, e > p For single slit diffraction pattern, bsin  = , so sin   , → diffraction angle for electron is much larger than protons. 27 B Electrons: dNe/dt = I / e = 2.010−6 / 1.610−19 Photons: dNp/dt = P / E = 0.3110−3 / 3.111.610−19 Ratio = 0.020 28 C p = mv = (9.11 × 10-31)(1.50× 106)=1.37 × 10-24 kg m s-1 p = 0.2% x p = 0.002 p = 0.002(1.37 × 10-24) = 2.73 × 10-27 kg m s-1 x = h/(p) =6.63 × 10-34 / (2.73 × 10-27)= 2.4 × 10-7 m FE FB FE FB I I  

29 A Energy released = BE of products BE of reactants = (8.32×136) + (8.58×98) (7.60× 235) = 186 MeV − − 30 D Initial true count rate = 90 -10 = 80 No of half lives = 12/24 = 0.5 C = 80(1/2)0.5 = 57 min-1 Final count = 57+10 = 67 min-1