(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.

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