2024 ACJC J2 H1 Physics Prelim P1 AS

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H1 (8867) 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 D Density 3 3 where m V r dV    Δ Δ Δ 100% 3 100% 0.1 0.043 100%32.5 1.87 6.7% m d m d                           3 C o o o o 15cos20 6.0 sin40 17.95 18.0 N 15sin20 6.0cos 40 0.5340 0.534 N x y F F         4 B At highest point, vertical component of the velocity is zero. Hence the ratio is sin sinv v   . 5 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 6 D 2 2 2 2 1 2 2 2 1 1 2 2 ( ) Using sin , 2 ( )sin s ut at ha t t g a hg t t          7 B When the bucket is moving at uniform speed, 0T mg T mg    When the bucket is decelerating uniformly,   ' ' mg T ma T m g a vm g t          Difference in tension = T’  T vm g mgt mv t       

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H1 (8867) Physics JC2 2024 Page 2 of 5 8 A 2 2 2 2 2 2 2 ( )Rate of change of momentum ( vol of water expelled) ( ) 1000 0.012 10 45 N rel rel rel mv t mv t v t r hv t hv r t v r v r v                    9 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 p p 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 p p v v        10 D A collision is elastic if relative speed of approach equals relative speed of separation. 1 2 2 1 1 2 2 1 ( )u u v v u u v v        Options A and B are wrong. An elastic collision is also one in which K.E. is conserved. total initial K.E. of system = total final K.E. of system 2 2 2 2 1 2 1 2 2 2 2 2 1 2 1 2 1 1 1 1 2 2 2 2mu mu mv mv u u v v       11 C cos18 520 550 Ncos18 T W T       sin18 520 sin18 170 Ncos18 T R R     

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H1 (8867) Physics JC2 2024 Page 3 of 5 12 C Taking moments about pivot,    30 70 7 3 P Q P Q   13 B Action-reaction pair must act on different bodies. 14 A 240 1000(60)(9.81)( ) 39 kW3600 TE U t t TE mgh mgvt t         15 C P Fv With constant force, velocity increases linearly with time, P also increases linearly. 2 2 ( 2 ) (2 ) P F as P F as   With constant force, power increases non-linearly with distance. 16 C 1000 0.2 9.81 10 65.4 W5 60 65.4 required 163.5 W0.4 mg h Vg hP t t P              17 A -4 -1 Time taken for hour hand to complete one revolution, 12 60 60 2Angular speed, 1.5 10 rad s T T       18 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 r r v r v a r a               19 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, then A would be the answer. 20 B The circuit is equivalent to 2 sets of parallel resistors in series with each other. 1 1 1Effective resistance 2 R R R        

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H1 (8867) Physics JC2 2024 Page 4 of 5 21 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 22 B Ω ( ) (1.00 3.0) 1.00 (0.40 12.0) 0.40 Hence, 3.0 4.8 0.40 3.0 E R r E r E r r r r              23 A Ω1 1 1 2.50 3.0 (6.0 4.0 5.0) PQ PQ RR       Ω1 1 1 4.44 (5.0 3.0) (6.0 4.0) PR PR RR      Ω1 1 1 4.00 6.0 (5.0 4.0 3.0) PS PS RR       Ω1 1 1 4.50 (6.0 3.0) (5.0 4.0) QS QS RR      Largest current is when resistance is lowest, i.e. across PQ. 24 C By potential divider principle, 10( ) LDR LDR LDR RV R R   When RLDR = 5.3 , VLDR = 4.5 V 5.34.5 10(5.3 ) 6.478 Ω R R     When RLDR = 3.1 , 3.1 10(3.1 6.478) 3.23 V LDRV     25 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. 26 C Determine the force on wire Z due to X and Y and find the resultant force.

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 1 Guide H1 (8867) Physics JC2 2024 Page 5 of 5 27 B 2 1 1 2 1 2 2 2 1 magnetic force provides centripetal force. Since , 2 1 2 mvBqv r mvr Bq mvr B q r r mvr B q B B       28 A Majority of the particles will still pass through without deviation. Although the charge of the gold nucleus is higher, the probability of head on interaction remains approximately the same and atom consists mainly of space. 29 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. 30 A At t = 4T, Activity of A = 4 0 1 2 A     Activity of B = 2 0 1 2 A     2 activity of A 1 0.25activity of B 2     