2024 HCI H2 Physics Paper 2 Solutions

1 2024 HCI Preliminary Examination Paper 2 Suggested Solutions Q1 (a) During the collision, there are no external forces acting on the photon and electron or system, hence linear momentum is conserved. B1 (b)(i) 1. 22 -17.30 10 kg m sxp − =  B1 2. -10 kg m sYp= B1 (b)(ii) By principle of conservation of linear momentum, (→) ( )( ) ( )( ) 227.3 10 cos 60 cos 25 oo pepp− = + …(1) () ( )( ) ( )( )sin 25 sin 60oo eppp = …(2) Solving (1) and (2) gives, ( )( ) ( )( ) ( )( ) ( ) ( )( ) 22 sin60 sin25tan60 7.3 1o 0 coc s60 s25 p e p e p p p p−  = =  −  22 -16.35 10 kg m sep −= M1, M1 – 2 equations showing application of COLM A1 – final answer after solving two equations. M1 M1 A1

2 Q2 (a) Every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. B1 B1 (b)(i) The vertical components of the two forces due to B and C cancel each other, hence, Resultant force in the y-direction = 0 Hence, Resultant force F = Resultant force in the x-direction ( ) ( ) 211 242 21 22 21 9 6.67 10 6.20 10 2 cos30 2 cos30 2.5487 10 N 1.32 10 =2.55 10 N xGM d − =    =   =     Alternative methods accepted. B1 M1 (b)(ii) The resultant gravitational force provides for the centripetal force required for the rotation of planet A. B1 2F m R= 21 7 -1 24 8 2.5487 10 7.3546 10 rad s6.20 10 7.60 10 F mR −= = =     6 7 22 8.54 10 s7.3546 10T   −= = =   M1 A1 (b)(iii) The gravitational potential is set to be zero at infinity. Gravitational force is attractive in nature and the force exerted on a test mass by the external agent will be in opposite direction to the displacement of the mass. Thus negative work is done by the external force (to bring a test mass from infinity to the point and hence potential is negative). OR The gravitation potential is set to be zero at infinity. B1 B1 OR B1 mass A 30° 30° Fby B Fby C

3 Since gravitational force is attractive in nature, (positive) work is done by an external agent to bring a (test) mass from the point (in the field if these 3 planets) to infinity, hence the (initial) potential is therefore (lower than at infinity and hence) negative. B1 (b)(iv) ( ) 2 2 2 211 24 3 9 0 6.67 10 6.20 10 3 1.32 10 5.83 10 J AB AC BCU U U U Gm Gm Gm d d d x− = + +    =− + − + −         =  −  =−  M1 A1

4 Q3 (a) A polarised wave is one in which the vibrations/oscillations of the wave are restricted to only one direction in the plane normal/perpendicular to the direction of energy transfer. B1 B1 (b)(i) As long as polarising filter B is perpendicular to either polarising filter A or filter C, the emergent light from filter C will be zero. Hence, the other angle that occurs will be when polarising axis of B is 90 of C, i.e. θ = 135° A1 (b)(ii) Using Malus’ Law, Intensity of light emergent from polarising filter B = I cos2(60°) Intensity of light emergent from polarising filter C = I cos2(60°) cos2(60°- 45°) = 0.233 I M1 A1

5 Q4 (a)(i) (Use FLHR to get the direction of force – the force will provide for centripetal acceleration for circular motion as it is always perpendicular to the velocity and hence points towards centre of circle. The center of circle should be below point A) Complete circle with centre of circle vertically below point A and the arrow tangential to the circle. B1 (a)(ii) The magnetic force provides for the centripetal force required for circular motion, B1 2 31 7 1 3 19 (9.11 0.133 m (1.2 = 13 cm (shown) 10 kg)(2.8 10 m s ) 10 T)(1.60 10 C) mv mvBqv R R Bq −− −−=  = =   = M1 (b)(i)1 Direction of electric field is downward in the plane of the paper. A1 (b)(i) 2 For the electron to be undeflected, the net force on it must be zero. 73 41 10 )(1.2 10 ) 3.36 1 (2 N .8 0 C qvB qE E vB − − = = = = M1 A1 (b)(ii) Helix (out of the plane of the paper) with an increasing pitch Remarks : • Spiral not accepted as answer. Spiral is not helix – a spiral has a changing radius. • “Increasing pitch” can be marked from the diagram if it is included. But at least four turns needs to be drawn for any credit. B1 B1 A • FB circle •

6 Q5 (a)(i) 22 628 1.00 10 sTT  −= =  =  A1 (a)(ii) 0 15 10.6 V = 11 V 22 rms VV === A1 (a)(iii) 1 1 1 7 12.0 3.0 6.0 36effR = + = +  Reff = 5.1429  M1 0 0 15 2.92 A = 2.9 A5.1429 V R= = =I A1 (a)(iv) Vrms across 6.0  resistor = 0 6.0(15 ) 10.0 V9.0 7.071 V 2 2 2 V  = = = M1 Mean power = 22 7.071 8.33 W = 8.3 W6.0 V R == A1 (b) 𝑃0 = 8.33 × 2 = 16.7 𝑊 B1 : Correct shape (sin2 graph) B1: at least 2 cycles shown with period labelled correctly ( t  2T i.e. 4 maximum power in graph) B1: correct peak value labelled on graph. B3 20 0.0100 /W /s 10 0.0200 16.7

7 Q6 (a)(i) Gain in kinetic energy of electron = Loss in EPE of system 21 02 mv eV−= ( )( ) 3 1 19 6 1 2 1.6 10 1002 5.93 109.1 10 s1 meVv m − − −  = = =   M1 A1 (a)(ii) ( ) ( )( ) 34 31 6 10 6.63 10 9.11 10 5.93 10 1.23 10 m h p − − −    == = M1 A1 (b)(i) 2d sin  B1 (b)(ii)1 . The electrons emerge with a larger speed/kinetic energy and hence momentum. B1 By de Broglie relationship ( h p = ), the wavelength of the electrons decreases. B1 (b)(ii) The path difference of the electron waves (from the different atomic planes) arriving at the detector remains constant, however the wavelength of the electrons decreases continually. B1 When the path difference is integer multiple of the de Broglie wavelength of the electrons (0, , 2, ….. ), constructive interference occurs/ the electron waves meet in phase, B1 the likelihood/chance/probability of the electrons arriving at the detect is large and a maximum value of I is detected. B1 B1 - path difference is constant B1 - CI /maxima occurs when path difference is integer multiple of the wavelength of the electrons. B1 – maxima corresponds to high chance probability of electron arriving there

8 Q7 (a) The half-life of a radioactive nuclide is the average time taken for half of the original number of nuclei in a sample of the radioactive nuclide to decay. Or the activity of a sample of the radioactive nuclide to halve. B1 (b) Activity is the number of disintegrations per unit time. B1 (c)(i) Half-life = 12.5 h = 12.5 × 60 × 60 = 45000 s Decay constant = ln2 ln2 half-life 45000= = 1.54 × 10-5 s-1 B1 M1 A1 (c)(ii)  − ==  = =  00 1/2 6 0 ln2 ln2(0.22 ) (0.22 )(12.5 60 60) 3.39 10 Bq N N NT N A M1 A1 (d)(i) and (ii) For P, the gradient is the negative of the decay constant. Same nuclide same decay constant and hence same gradient. For Q with a very much shorter half-life compared to K-42, it will approach the original graph quite quickly. ( 12 10 20 ttA A e A e−−=+ . Cannot linearise to give a straight line.) B1 B1 t lg A tadd P Q

9 Q8 (a)(i) From graph, when v = 25 m s-1, P = 500 W B1 (a)(ii) 500 20 N25 P Fv PF v = = = = M1 A1 (a)(iii) At constant velocity, the net force on the rider is zero. Furthermore, Fslope = 0 since the ground is level. Hence, the propulsive force = the drag force Fair  = 1.0 x 10-3 g cm-3 = 3 6 3 3 31.0 kg 1.0 kg m10 m 10 10−− − −  = 2 2 2 1 2 120 (1.0)( )(25 )2 Effective drag area, 0.064m air D D D F C Av CA CA = = = C1 – appreciate that Fair = F allow mark as long as 20 N is substituted for M1 – for correct conversion of density to kg m-3 A1 – correct calculation of drag area. C1 M1