2021 A Level H2 Physics P3 Soln

2021 A Levels H2 Physics Paper 3 suggested solutions 1 (a) (i) 2 sin 9.81sin40 6.31 m s ag  − = = = 1 (a) (ii) ( )( ) 22 2 1 2 2 6.3057 0.56 2.66 m s v u as v v − =+ = = 1 (b) (i) ( ) 22 c 1.50.072 1.35 N0.12 vF ma m r= = =  = 1 (b) (ii) ( )( ) centripetal 1.35 0.072 9.81 0.644 N N mg F N += = − = The force is directed vertically downwards. 2 (a) Since gravitational force is attractive in nature, in order to separate two masses, positive work must be done by an external force since the exernal force and displacement are in the same direction. Hence, gravitational potential energy and hence potential increases with separation of the masses. Since the maximum gravitational potential at infinite separation is defined to be zero, thus at any separation less than infinite, the gravitational potential is less than zero, hence it will be negative in value. 2 (b) (i) ( )( ) 11 23 71 6 6.67 10 6.2 10 1.22 10 J kg3.4 10 GM r − − = − = − = −   2 (b) (ii) Kinetic energy: ( )( ) 22 3 7 K 11 2.8 3.8 10 2.02 10 J22E mv= =  =  Gain in Gravitational potential energy to travel out into space: ( ) ( ) 772.8 0 1.22 10 3.42 10 JUm   =  = − −  =  Since the kinetic energy is not sufficient to overcome the gravitational potential required to travel out into space, the rock will return to the surface of the planet. 3 (a) Internal energy of an ideal gas is a function of its state and is the sum of the microscopic kinetic energy due to the random motion of the gas particles. There is no potential energy related to the relative position of the gas particles due to negligible forces of attraction between particles in an ideal gas. Internal energy is equivalent to 3 2 nRT 3 (b) (i) Using ideal gas equation PV = nRT (where T is in kelvin K): Given that n remains the same and that R is a constant,

( )( ) ( )( ) ( ) 1 1 2 2 12 5 3 5 31 10 3.2 10 1 10 3.6 10 12 273 273 3.6 12 273 273 47.625 47.6°C3.2 PV P V TT   −− =     =++ = + − = = 3 (b) (ii) ( ) 5 3 31.0 10 3.6 10 3.2 10 40 J W p V −−=  =    −  = 3 (c) (i) Using 1st Law of Thermodynamics:  = +  = + − =101 ( 40) 61 J U Q W U (Won gas = - Wby gas against atmosphere) 3 (c) (ii) ( ) ( )( ) ( )( ) 23 53 22 61 1.38 10 12 273 Increase in KE per molecule 1 10 3.2 10 7.50 10 J PV NkT U kTU N PV − − − =   += = =  = 4 (a) (i) 0.50 cm 4 (a) (ii) 122 7.85 rad s0.8T  −= = = 4 (a) (iii) ( )( ) 21 00 0.50 10 7.8540 0.0393 m svx  −−= =  = 4 (b) Same starting amplitude, with decreasing amplitude a time passes due to larger damping effect due to air resistance from the increased surface area of the card. To draw the decrease in amplitude clearly, students can consider using exponential decay dashed lines to guide succeeding amplitudes. Due to damping, the period may increase slightly (not shown here, but if increases slightly, must make sure that the new period is consistent over the 3 cycles). Note that -0.75 cm is