2023 RI H2 Physics Prelims P3 Answers

1 © Raffles Institution [Turn over 2023 Preliminary Examinations H2 Physics Paper 3 Solutions 1 (a) (i) Gravitational force provides the centripetal force 2 2 2 3 3 2 GMm mr r GM r GM r    − = = = 3 2k =− = – 1.5 (ii) 3 22 2 11 3 2 2 () 2 6700( )( )24 3600 42000 r r    − − = =  2 = 1.14  10−3 rad s−1 (b) (i) 2 2 2 2 22 k GMm mv GMm mvrrr mv GMmE r =  = == 2 2 T K PE E E GMm GMm rr GMm r =+ = + −  =− (ii) 21 12 11 24 7 5 7 7 ()22 11()2 (6.67 10 )(6.0 10 )(700) 1 1 ()2 4.2 10 (3.0 10 4.2 10 ) 2.37 10 J T GMm GMmE rr GMm rr −  =− − − =− =−   +  = (c) ET = EP + EK (ET and EP are negative values, EK is positive.) Moon’s orbit Earth x

2 © Raffles Institution [Turn over ET is constant, when Moon is nearest to the Earth, its EP is smallest (more negative), hence EK and speed is the largest. OR By conservation of energy, when Moon is nearest to the Earth, its EP is smallest (more negative), hence EK and speed is the largest. 2 (a) (i) By constantpV T = , CC BB CB pV pV TT = ( ) 5 5 7.0 10 0.0015 960 1260 0.0040 2.00 10 Pa CBB C BC TpVp TV=  = = (ii) ( ) ( ) 53 Area under the graph 1 7.0 2.0 10 4.0 1.5 102 1130 J W − = = +   −  = (b) 1. The process is carried out slowly. 2. The walls of the engine are good conductors of heat. 3. The engine is surrounded by a heat reservoir at constant temperature. Maximum 2 marks (c) change work done on gas / J heat supplied to gas / J increase in internal energy / J A → B −1000 2625 1625 B → C −1125 500 −625 C → D 0 −1000 −1000 D → A 555 −555 0 (d) During an explosion, the pressure at different points of the gas is not uniform. Hence, pressure is undefined and cannot be represented as a point or a curve on the p−V diagram. 3 (a) For a to be minimum, the minimum intensity detected directly below it must be the 1st minimum from the central bright fringe. i.e., 0.5n = Path difference = 0.5  222.00 2.00 0.5a + − = ( ) ( ) 222 22 2.00 2.00 0.5 4 4 2 0.25 a a   + = + + = + + ( ) ( )( ) 2 9 9 6 33 4 0.25 750 10 2 0.25 750 10 1.50 10 1.22474 10 1.22 10 a  − − − −− =+ =  +  =  =  = 

3 © Raffles Institution [Turn over  has to be substituted in at the last step and not earlier as calculator has insufficient number of digits. OR Since the interference pattern is symmetrical about perpendicular bisector between S1 and S2, 0.5 0.5a x x x=  +  = OR Distance (on the screen) between the central maximum and the 1 st minimum is 22 xa= ( )( ) 2 9 6 33 750 10 2.00 1.50 10 1.22474 10 1.22 10 m DDa xa aD a   −− −− == = =  =  =  =  (b) (i) ( ) ( )( ) 9 3 31 3 1.0 3 3