2015 H2 Physics MS EJC Suggested

©EJC 2019 9646/N15/Suggested_Solution H2 PHYSICS SUGGESTED SOLUTIONS 9646 November 2015 Paper 1 Multiple Choice Question Key Question Key 1 A 21 C 2 D 22 D 3 C 23 B 4 A 24 B 5 C 25 A 6 A 26 B 7 D 27 B 8 C 28 B 9 A 29 D 10 C 30 B 11 B 31 B 12 A 32 D 13 D 33 B 14 A 34 C 15 A 35 C 16 A 36 A 17 C 37 C 18 C 38 A 19 C 39 D 20 B 40 B Notes Q4: mass of tanker and water is decreasing. Q11: the minute hand completes a revolution in 1 hour, not 1 minute *(that’s the second hand) Q18: there has to be a 180 degree phase change after reflection. Q20: RS wave will always have a component in that direction and will not be stationary. Q28: ammeter is only reading half the current output from cell. Q36: need to determine initial uncertainties.

©EJC 2019 9646/N15/Suggested_Solution Paper 2 Structured Questions Question Marks Notes Qns is not straightforward because force acting on mass is not same as time progresses. Must be reminded that force is directly proportional to extension and not the full length of spring. The types of energy involved are GPE, KE and EPE. 1(a) natural length 0 40 cmx = when M = 300 g, equilibrium length x = 60 cm ( ) ( )( ) ( ) 0 0 1 2 0.300 9.81 14.715 Nm60 40 10 Fk Mk x xM x g g x − − = −= =    − = =− × M1 elastic potential energy = ( ) ( ) ( ) 22 2 0 11 14.715 70 4 102 0 2 0.662 J kx x − = ×−= − Notes Cannot assume constant force from spring, and to work in terms of extension. A1 1(b)(i) on release, before passing through equilibrium position GPE is converted to KE + Elastic PE B1 After passing through equilibrium position as mass reaches lowest point, GPE + KE is converted to Elastic PE Notes Markers noted confusion between zero G PE vs minimum GPE at lower point. B1 1(b)(ii) By PCE, decrease in GPE is gain in KE and gain in Elastic PE ( ) ( )( ) h mv mg hv mg m − ∆= + ∆−= − = = 2 1 1 Elastic PE2 2 Elastic PE 2 0.3 (9.81)(0.3) 0.662 0.3 1.21 ms Notes Do not ignore lose in GPE. KE must be positive. B1 M1 C1 A1

©EJC 2019 9646/N15/Suggested_Solution Question Marks 1(b)(iii) By PCE, decrease in GPE is gain in Elastic PE 2) 2 2(0.300)(9. 1 (2 8 1) 14.715 0.400 m hh mgh gk k m ∆= ∆ ∆= = = Notes: Do not use kinematics equations (they only work for constant acceleration). B1 M1 A1 2(a) 12 12 1 gradient 0.208 V 2.5 1.25 12 6 yy xx − − −= − − = = I M1 to find y-intercept c: ( ) 1 1 0 1. 2.5 1.25 12 6 2.5 1.25 2 0 5 12 66 yc x c − =− = − − −  −  = − A1 plot of I against V is a straight line with uniform gradient 0.208 and passes through origin, I is proportional to V A0 2(b)(i) resistance 12 4.8 2.5 R V = = Ω = I A1 2(b)(ii)1. A 5 12 1.3 A45 A R V R= = = + + I A1 2(b)(ii)2. 12 1.6 A4.8 2.7 Y X XR R V= = = + + I A1 2(b)(iii) ( )AC 75 100 75 (4) 3.9