2023 A Level H2 Physics P2 Soln

2023 A Levels H2 Physics 9749/02 Paper 2 suggested solutions 1a When submerged, there is a higher pressure on the base of the block compared to the pressure on the top of the block. This difference in pressure produces a net upward force that causes an upthrust on the block. 1bi Explain: Upthrust equals to the weight of fluid displaced. 3 3 , of fluid displaced 27.81.0 10 9.81(100) 0.27272 0.27 N upthrust U mg Vg = = =    = = *Note: Do not assume that stating the quantities meant explanation. 1bii Taking moments about the pivot Sum of clockwise moments = sum of anticlockwise moments For Fig. 1.1 (18) (8.3) (1 )FW = −−−−− For Fig. 1.2 ( )(19) (7.8) (2)F U W− = −−−−− Solving equation (1) and (2) F = 2.49 N Note: For Fig. 1.2, there are two forces acting on the block, namely F (weight) and U (upthrust) 2a At the highest point of the path, KE of the object is 65 J. 2 1 1 1 ( cos(42 )) 652 20.69 m s vertical component of 20.69sin(42 ) 13.8 14 m s mv v v − − = = = = 2b 0 14 9.81 1.43 s yyv u at t t =+ =− = Time taken for object to land 2 2.86 st== Horizontal displacement ( )20.69cos 42 2.86 44.0 m=   = 2c Time taken to rise reach maximum height is shorter than the time it takes to return to the ground as the deceleration of the object rising upwards is larger than the acceleration of the object as it is falling. 3ai 2 GMmF r= 11 24 32 6 63 10 6 0 10 6800 6400 36 000 10 ( . )( . )( ) (( ) ) −= + = 1500 N 3aii 2 2 23 2 24GMm mr T rr T GM =  = period T is independent of m (mass of satellite) so all satellites in geostationary orbit need not have mass of 6800 kg, i.e. can be any mass. 3bi 2mv mvBqv r r Bq=  = 23 3 19 3 4 10 1 8 10 1 60 10 . .. − −− =    = 0.12 m

3bii radius r depends on the momentum of the ion, ions with different momentum would have different radius of path (there could be different isotopes of different mass and momentum) . 4ai A stationary wave is formed when the sound wave from the loudspeaker is incident on the metal plate and reflects. The incident wave and reflected waves have the same wavelength and speed but are travelling in opposite directions. The incident and reflected waves overlap and are superposed, and the resulting wave is a stationary wave with points of maximum and minimum amplitude. 4aii (showing construction lines on the graph) Distance between two successive troughs = 10.0 cm Time for two periods = 10.0 x 0.50 = 5.0 ms frequency = 3 1 4005.0 10 2− = Hz 4aiii distance between adjacent nodes is half a wavelength. 340 0.85400 0.4252 v f  = = = = distance = 0.425 m 4bi The path difference is 0.4m which is 2 wavelengths. Since the waves from X and Y are emitted in phase and the path difference is an integer number of wavelengths, the waves meet in phase at point P. Constructive interference occurs and the amplitude of the sound at P is a maximum, thus the intensity of sound at P is a maximum. 4bii 2 2 2