(ACJC) 2024 J2 H2 Physics Prelim P1 Solution

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2024 Page 1 of 5 Qn Ans Discussion 1 C base units of P = base units of ( A v n) (kg m s−2)(m s−1) = (kg m−3)(m2)( m s−1)n kg m2 s−3 = kg m−3+2+n s−n Comparing the indices for s, n = 3 2 A Option A – Non-linear gradient indicates largest change in velocity with time Option B – Almost constant gradient indicates almost constant velocity with no acceleration. Option C – Constant gradient indicates constant velocity with no acceleration Option D – Zero velocity 3 D 2 22 21 22 21 1 2 2 () Using sin , 2 ( )sin s ut at ha tt ga hg tt   =+ = − = = − 4 A 2 2 2 22 22 ()Rate of change of momentum ( vol of water expelled) () 1000 0.012 10 45 N rel rel rel mv t mv t v t rhv t hvr t v r v rv       = = = = = = = =    =

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2024 Page 2 of 5 5 C Assume Charlie throws the ball to the right and take rightwards as positive. Consider Charlie and the ball as 1 system, Before throwing the ball, ( ) before throw after throw 1 0 0 (1.5)(2.5) (60)( ) 0.0625 m s ball ball Charlie Charlie Charlie Charlie pp m u m u u u − = = + − =+ =− Charlie is moving to the left with speed 0.0625 m s−1 after throwing the ball. Consider the returning ball and Charlie as 1 system now, before catch after catch after catch 1 after catch (60)( 0.0626) (1.5)( 2.5) (61.5)( ) 0.122 m s pp v v − = − + − = =− 6 C cos18 520 550 Ncos18 TW T = ==  ( ) sin18 520 sin18 170 Ncos18 TR R = =  =  7 C Taking moments about pivot, ( ) ( )30 70 7 3 PQ P Q = = 8 C 1000 0.2 9.81 10 65.4 W5 60 65.4 required 163.5 W0.4 mg h Vg hP tt P     = = = =   ==  9 A -4 -1 Time taken for hour hand to complete one revolution, 12 60 60 2Angular speed, 1.5 10 rad s T T =   = =  10 D 2 2 Since , 1When reduces to and increases to 2 ,2 12 ( ) and 2 1( )(2 ) 22 new new vv r and a r r rr v r v a r a = = = == ==       11 B Gravitational field strength due to mass M is directly proportional to the mass M and inversely proportional to the (square of the) distance. Considering the system of the 3 masses, Planet 1 has the smallest mass, hence the neutral point is B (closest point to the Planet 1). A is wrong because it is along the straight line connecting the centres of Planet 1 and Planet 2. If we to consider only the neutral point between only Planet 1 & Planet 2 excluding Planet 3, A would be the answer.

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2024 Page 3 of 5 12 A 1 When and are decreased by half, gradie nt increases by 4. pV nRT V p nRT nT = = 13 D 3 ( ) ( ) 2.1 (2108)(4.5) 2.1 (334 10 ) 2.1 (4190)(96) 1.57 MJ ice f waterQ mc ml mc=  + +  = +  + = 14 C The volume of the molecules is much smaller in comparison to the volume of container. 15 D There are no oscillations, and the amplitude will return to equilibrium in a short time. 16 B ( ) 22 2 2 2 2 2 3 24 o o o v x x Av A v   = −   = − =   kinetic energy 3 total energy 4= 17 B Wavelength is from compression to compre ssion. 100 m spans across 5 wavelengths. 100Therefore, 20 m 5 Using , 350 (20) Therefore, frequency 18 Hz v f f == == =   18 C 36 400 2 (2 ) 0.18 (2 ) 36 400 4 phase difference of 2 is equivalent to a phase difference of 4 . v f x x A         == = = = = 19 B Consider Rayleigh’s criterion, -2 0.5 Therefore, 2.0 10 rad new old old new new b b b      == =

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2024 Page 4 of 5 20 B Since potential is a scalar quantity and distances of A and C to B are the same, potential due to A cancels the potential due to C. Thus, the potential at B is due to charge at D alone. 6 12 2 1 100 10() 4 (8.85 10 ) 20 10 4.5 MV V x − −− =  =+ 21 B Direction of electric field line is directed away from a positive charge that sets it up in the first place. Whether the field strength increases or not depends on how closely together the electric field lines are to one another. However, the further the d istance from the electric field line, the lesser is the electric potential due to the positive charge. 22 B The circuit is equivalent to 2 sets of parallel resistors in series with each other. 1 11Effective resistance 2 RR R − = +  = 23 D Total potential drop from J to L = 3 IR where I is the current through series resistors R and 2R [Path JML] Potential VL = 0, potential VJ = 3IR Potential drop across JM = IR and potential drop across ML = 2IR 24 C The path cannot be due to a magnetic field because the magnetic force acting on the charged particle will be into the page (positive charge) or out of the page (negative charge). Hence, the field must be an electric field and since the charged particle is deflected downwards, the charge is positive. 25 D Let the velocity of the rod down the slope be v. e.m.f. induced = B L v cos θ current induced = e.m.f. induced / R Magnetic force = B I L to the left When terminal velocity is achieved, Net F along slope = 0 Resolving forces along the slope, 22 sin magnetic forcecos cos cos tan cos mg BLvBL R mgRv BL      = =   =

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H2 (9749) Physics JC2 2024 Page 5 of 5 26 B Note that as the value of V is a straight line with a constant gradient, the square of this function will be a curve. Option D is wrong because it is obtained from an alternating function of constant V0. Option C is wrong because it is obtained from an alternating sinusoidal function. Option A is wrong because it is the root-mean-square voltage is not the average voltage. Note: 2 22 2 2 2 2 00 20 0 0 22 33 2 00 33 0 2 00 41 2 2 () 22 88 3 3 8 3 3 T T T rms T VVV V t dt t dt t dtTT TT T VV tT TT VV  = = =  ==   ==    27 A Photoelectric equation: hf =  – eVs where  = hfm s hVf e =− Smaller work function means threshold frequency fm is lower. Gradient of Vs – f graph h e = is constant. Hence the line for the new metal should be parallel to the original line but shifted to the left. 28 A When operating voltage is decreased, the electrons have less energy and the cut -off wavelength 1 is longer. Since the target material has not changed, there is no difference in the characteristic spectrum and hence there is no change in 2. 29 A Majority of the particles will still pass through without deviation. Although the charge of the gold nucleus is the same , the probability of head on interaction remains approximately the same and atom consists mainly of empty space. 30 A Y and Z are more stable than X. Hence, the binding energy per nucleon of X should be lower than that of Y and of Z.