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 V1) 00(1.3 A ABVI R=  ∆   ==  M1

©EJC 2019 9646/N15/Suggested_Solution Question Marks ( )( )AD (1.6 (4.8) 7.7) V XXV IR = ∆ = = M1 CD 7.7 3.9 3.8 VV∆ −== Notes Need to be clear between p.d. across resistor vs potential at a point. A1 Notes The explanations for Q3 were challenging to some candidates. A learning point here is that explanations in terms of energy is important. 3(a) a single discrete packet of energy of electromagnetic radiation is absorbed by an electron at the surface of M. B1 if the energy absorbed is at least the work function of M, electron will be emitted with minimum amount of kinetic energy Notes Examiners were on the look-out for explanations that discuss energy changes. It was insufficient to refer to EM radiation or threshold frequency. B1 3(b)(i) electrons with the most kinetic energy loses all their kinetic energy as electric potential energy in the electric field between M and C. Zero current as no electrons reach C. Notes Insufficient to explain the meaning of stopping potential in genera. Need to apply to the situation and discuss the energy changes from max KE to EPE. B1 3(b)(ii) Positive plate is collecting all emitted electrons giving rise to maximum current value. Rate of emission of electrons at M is limited by the number of photons reaching M. Notes Need to answer question in terms of why current is constant despite p.d. becoming positive. B1 3(c) ( ) max max 19 34 34 14 42.2 1.8 6.63 1.6 10 10 10 9.65 10 6 H .6 z 3 s KE hf K hf E eV h ee − −− Φ ΦΦ= = × = = = × + ×× + = − + M1 A1

©EJC 2019 9646/N15/Suggested_Solution Question Marks 3(d) graph stretched vertically upwards by factor of 2 same sV B1 B1 4(a) ( ) 2 2 2 53 5 2 4.22 2 210 0. 10 1.53 10 m12 s 7 car r T πω π −   ××  = = ×  =  = M1 A1 4(b)(i) Only force acting on Amalthea is the gravitational force by Jupiter and acts in same direction as the centripetal force. Gravitational field strength g is gravitational force per unit pass so the net force netF mg= B1 B1 By N2L, the gravitational force results in the rate of change of momentum of Amalthea that takes place in the direction of the gravitational force. Amalthea’s mass is constant so net centripetal acceleration ca is net force per unit mass: net c dp m madt dvF dt= == 4(b)(ii) gravitational field strength provides centripetal acceleration Jupiter 2 Jupiter Io Am althea Amalthea Io 3.8 0.712 7 2.33 c c GM gar r ra ra GM a = = = = = = M1 A1 5(a)(i) ( ) 19 5 14 2 (2)(1.60 10 )(1.5 10 ) 4.8 10 N EF qE e E − − = = = ×× = × A1

©EJC 2019 9646/N15/Suggested_Solution Question Marks 5(a)(ii) 22 00 19 19 12 12 2 5 (2) (2)11 44 1 (2)(1.60 10 )(2)(1.60 10 ) 4 (8.85 10 ) (4.0 10 ) 5.8 10 N AB E QQ eeF rrπε πε π −− −− − +−= = ××= ×× = × Notes Common mistake was to leave out the square for the separation. M1 M1 A1 5(b) [from (a)(i)] A and B have same amount of charge in opposite polarities. They each experience an electric force that is equal in magnitude but opposite in direction when subject to the same external uniform electric field. [from (a)(ii)] By N3L, the electrostatic forces between A and B are equal and opposite. The forces are internal within the molecule. vector sum of forces acting within and on molecule AB is zero Notes Need to take the hint from the earlier part of the question and discuss forces within (internal to the) molecule as well as external forces acting on the moledule. M1 M1 A0 5(c) ( ) 14 12 25 sin60 (4.8 10 ) (4.0 10 )(sin60 ) 1.7 10 N m Fd F dτ ⊥ −− − = = ° = ×× °  = × M1 A1 6(a) I128 0 128 53 1 54 D−→ β+ 6(b) Assume no background count, ( ) 0 (2000) (6000) half half ln 4000 1 4000 l 75 28 28 ln2 175 4000 ln2 28 17n 5 5 1 10 s tC Ce e e t t λ λ λ λ − − −=   = −= −   = =    −= M1 M1 A1

©EJC 2019 9646/N15/Suggested_Solution Question Marks 7(a) [ ] ( ) 1 11 22 11 22 3 3 1 2 ln 5.00 10ln ln(1.2)6.00 10 1.2 ln ln ln rs s bdy kMg d d ds d s y y s y s y y y − − − =    =        ×= = −  ×  − = = l s is -3 to 1 s.f. 7(a)(ii) ln (y1 / m) = – 3.170 ln (l / m) = – 0.693 7(a)(iii) s -3.0269 -3.0004 -3.0231 -3.0319 -3.0693 -3.0693

©EJC 2019 9646/N15/Suggested_Solution Question Marks 7(a)(iv) 12 12 1.4 ( 3. gradient 3.0 8) 0.1 ( 0 9) 0 . yy xx −= − = − − = −− −− M1 A1 7(a)(v) k, M, g, d, and b are constant, 1 1ln ln ln rs s y bdkMg kMgdyr b −= = +  l l plot of 1y against l is a straight line with gradient r, y-intercept ln skMgd b   graph supports the expression. Notes Answers need to go beyond what is to be expected after taking log on both side of the given equation, and to compare with the graph that drawn.