2025 RI H2 Physics Prelims P2 Answers

2025 Preliminary Examination H2 Physics Paper 2 Solutions 1 (a) ( )( )( ) 2 2 2 3 0.234 5.13 10 11.38 10 1.72 10 2330 kg m m abc − − − − = =    = (b) ( ) 3 0.002 0.01 0.01 0.01 23300.234 5.13 11.38 1.72 40 kg m m abc m a b c m a b c     − =     = + + +  = + + + = ( ) 32330 40 kg m −= (c) zero error / incorrect calibration of calipers / balance 2 (a) Let n be number of 5.0 g masses. By principle of moments, take moments about the edge of the table, Total anti-clockwise moments = total clockwise moments ( )( ) ( )( )( ) ( )( )( ) − = − +  = 30.101.5 9.81 0.11 9.81 0.50 0.10 5.0 10 9.81 0.802 7.75 n n Therefore, the maximum number of masses is 7. (b) Let tension in string be T. Consider forces acting on the metre rule and taking moments about the edge of the table, string holding the 1.5 kg mass will cause a reduction of anti-clockwise moments 0.10 0.0502TT==  string attached to ruler at midpoint will cause a reduction of clockwise moments ( )( )sin60 0.40 0.346TT=  = Since the reduction of clockwise moments is greater, t here will be a net anticlockwise moment acting on the metre rule. Hence more 5.0 g masses can be placed (compared to (a)) before the cantilever topples. 3 (a) The vertical displacement of the yoke is r sin(t). Hence, the acceleration is ( ) 22 sinr t x  − =− , which is the defining equation of simple harmonic motion.

2 © Raffles Institution (b) (i) 1. ( ) 00 1 2 0.0800.40 1.3 m s vy r    − = = = = 2. ( ) 2 00 2 2 2 0.0800.40 20 m s ay   − = = = or ( ) 00 2 2 1.30.40 20 m s av   − = = = (ii) 4 (a) (i) When the path difference between the waves that meet at the screen is integer multiple of wavelength of the light, they meet in phase and there is constructive interference, forming bright fringe. When the path difference between the waves that meet at the screen is odd integer multiple of half wavelength of the light, they meet in antiphase and there is destructive interference, forming dark fringe. correct path difference meeting in phase / antiphase constructive / destructive interference (ii) The single slit diffraction pattern from each slit forms the envelope of the double slit interference pattern. Where the minima of the single slit diffraction pattern coincide with the bright fringes of the double slit interference pattern, these bright fringes cannot be observed and will be seen missing from the interference pattern. (b) (i) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9  3.00 m 17.7 mm t / s F / N 5.9 −5.9 0 | | | | | | | | | |

3 © Raffles Institution [Turn over ( ) 9 3 4 sin sin sin tan as is smalltan 590 10 35.4 10 1 2 3.00 1.00 10 m 0.100 mm b b         − − − = = =   = 