TJC 2023 H2 Prelim Paper 3 Solutions

2023 JC2 H2 Prelim Paper 3 Solutions 1 (a) The electric field lines, along which the electric forces exerted by the charged sphere on other charges act, are always perpendicular to the surface of the sphere (i.e. the electric field lines are radial). Hence the field lines appear to originate from the centre of the sphere. M1 A0 (b) (i) arrow pointing to the left labelled ‘electric force’ (FE) and arrow pointing downwards labelled ‘weight’(W) C1 (ii) 1. Electric force on the particle FE = QE FE = VQ / d = (2.0 × 102 × 8.0 × 10-19) / 4.0 × 10-2 = 4.0 × 10-15 N resultant force ( ) ( ) 2215 15 153.9 10 4.0 10 5.6 10 N− − −=  +  =  A1 C1 A1 2. let  be the angle to horizontal tan = weight /electric force  = tan-1 (3.9 × 10-15 / 4.0 × 10-15) = 44° A1 (c) (i) downward sloping line from (0, 2.0) M1 magnitude of gradient of line increases with time and line ends at (T, 0) A1 (ii) Horizontally: s = ½ at2 2.0 x 10-2 = ½ (4.0 x 10-15 /(3.9 x 10-15 / 9.81) T2 T = 0.063 s C1 A1 2 (a) (i) gravitational field strength is equal to the negative gravitational potential gradient Or state equation g = - d/dr B1 (b) (i) G = GM/r2 gA = G(3M)/(2R)2 = ¾ GM/R2 gB = G(M)/(R)2 = GM/R2 so conclude gA< gB alternative soln: by calculating the gradient at surface and using gradient = field strength deduce gradient at surface of A < gradient at surface of B C1 A1 (ii) M is the point/position in which net gravitational field strength is zero B1 (iii) sketch: one curve, starting with gradient of decreasing magnitude at 2R and finishing with gradient of increasing magnitude at D – R field strength shown as zero near the point of maximum potential (position of M correct), with gravitation field strength at surface of A< B B1 B1 Substitution must be shown

(c) (i) The gravitational force of attraction between the stars are of the same magnitude. This force provides the centripetal force for each star to orbit about their common centre of mass. B1 (ii) There is no external force acting on the binary stars, so the c.m. should not change position. B1 (iii) centripetal force = MAdω2 = MB(2.8 × 108 –d)ω2 Or MAdA = MBdB MA / MB = 3.0 = (2.8 × 108 – d) / d d = 7.0 × 107 km C1 A1 3 (a) (i) Molecules collide with one another in a haphazard manner hence they possess different kinetic energies at any time. (word “random” does not get marks in answer) B1 (ii) No intermolecular forces in ideal gas, hence no potential energy (Do not accept specific forces eg. “attraction”; vague description eg. “interaction” B1 (iii) p = 1/3 ρ <c2> => pV = 1/3 M <c2> since ρ = M/V For ideal gas, pV = n RT Hence, 1/3 M <c2> = n RT