2016 H2 Physics MS EJC Suggested

©EJC 2020 H2 PHYSICS SUGGESTED MARK SCHEME 9646 October/November 2016 Paper 1 Multiple Choice Qns Key Qns Key Qns Key Qns Key 1 B 11 A 21 C 31 A 2 C 12 D 22 D 32 C 3 B 13 D 23 D 33 D 4 D 14 C 24 C 34 B 5 D 15 C 25 C 35 B 6 C 16 A 26 B 36 B 7 D 17 D 27 C 37 D 8 D 18 A 28 A 38 C 9 B 19 B 29 B 39 C 10 C 20 D 30 B 40 C Notes: Q5: Ball must fall again to reach ground after rising, so need a multiplier of 2.

2 ©EJC 2020 Paper 2 Structured Questions Note: Need to read questions, diagrams and graphs carefully. Definitions need to be precise. Qns Marks 1(a) SI unit for electrical resistance is when a potential difference of one volt per ampere of current flows through a conductor Note: do not make mistake of defining resistance or mixing up units/quantities. B1 1(b)(i) ( ) 1 50 0 32 4 6875 VR R V. . . = = = = Ω I I ( ) ( ) 2 2 23 2 7 2 4 0 23 10 1 4 0 4 87 10 40 R dR RA R LL L R L A d . . ρ π πρ π − − − =     = = = × = × = ×  Ω Note: be careful of the powers of ten for mm and cm C1 C1 M1 A1

3 ©EJC 2020 Qns Marks 1(b)(ii) ( ) 22 7 8 4 0 01 0 01 1 4 2 2 00 1 014 87 10 2 6 2 10 50 0 23 0 3 40 0 m 2 RV LL Vd L Vd L .. . .. . dd Vd L Vd L ... . ππρ ρ ρ ρρ − − = =   ∆∆ ∆∆∆= + ++ ∆ ∆∆∆∆= + + + =× + ++  =   ×Ω I I I I I Note: since d is squared, there needs to be a multiplier of 2 to its associated fractional uncertainty. OR ( ) ( ) ( ) ( ) 22 max min 2 2 max max min min min min max max 2233 22 4 1 2 24 0 24 10 4 1 51 1 49 03 1 3 99 03 0 22 10 81 3 0410 0 1 RV LL V V LL .. .. . dd d d .. . ππρ ρρρ π π −− −− = =   ∆=  =       −  −           ××  = −      ××    I II 86 2 10 m. −    = ×Ω Note: actual uncertainty should be quoted to 2 s.f. It is only 1 s.f. when paired with the actual quantity, as per the next part. M1 A1 M1 A1 1(b)(iii) ( ) 749 06 1 0 m..ρ −=±×Ω Note: need to check that the power of tens are the same before quoting the quantity to the 1 s.f. version of the actual uncertainty. A1 1(c) Accuracy is how close a measurement is to the true value. Accepted value lies within the range in the answer to (b)(iii) of ( ) 749 06 1 0 m..ρ −=±×Ω and so is accurate. Precision is the degree to which repeated measurements agree with each other. The range of values result in a percentage uncertainty 06 4100 100 129 . .% %%ρ ρ ∆ ×= ×= and so precision is poor. Note: answer needs to make reference to the value in part (b)(iii). B1 B1

4 ©EJC 2020 Qns Marks 2(a) When t = 0.70 s, F = 6.4 N 2 64 40 1 6 ms F ma F.a . . m −= = = = M1 (read graph) A1 2(b) B1: 21 6 ms. a −= until t = 1.4 s, B1: straight line until (4.2, 0) 2(c) Area under F – t graph is impulse (change in momentum) area from t = 0 to t = 1.4s: 6.4 × 1.4 = 8.96 N s Note: the area required is the initial rectangle only and not of the triangle later, past t = 1.4 s. M1 A1

5 ©EJC 2020 Qns Marks 2(d) B1: object starts from rest so zero momentum at t = 0 s use answer to part (c) to plot (1.4, 8.96) connected to origin via straight line B1: calculate impulse from t = 1.4 s to t = 4.2 s via area of triangle in Fig 2.1 ( )( ) 4.2 s 1 4 s 8 96 8 96 1 42 14 642 8 96 Ns Ns17 92 t t. p . p ... . . p . p= == + = + ∆ = ∆= − = B1: smoothly curving line from (1.4, 8.96) to (4.2, 17.92) Note: the momentum is still generally increasing as the force applied remains positive in direction. mark out the points accurate to half a small square

6 ©EJC 2020 Qns Marks 3(a) take pivot at A, by Principle of Moments , sum of clockwise moments = sum of anticlockwise moments    ( ) ( ) ( ) ( )36 cos 40 45 cos 0 12 8 0 8 6 N X. X. . ° = °   = M1 M1 A0 3(b)(i) vector sum of forces is zero along all directions; horizontally, weight of bar has no rightward component of force to cancel leftward component of X no resultant moment about any point; the 3 lines of action of (i) force at A, (ii) weight of bar and (iii) X must meet at a common point the perpendicular distance of all 3 forces from the common point, and therefore the sum of moments at this common point, is zero Note: the explanation has to relate how the specific situation satisfies the two conditions for equilibrium B1 B1

7 ©EJC 2020 Qns Marks 3(b)(ii) Let F be at an angle of θ anticlockwise to the horizontal axis. By cosine rule, ( )( ) ( )( ) 22 2 22 36 2 36 cos20 8 8 36 2 36 8 8 cos20 27 9 N FX X F. . . =+− ° = +− ° = Note: since part (a) is a “Show …”, it is safe to quote X as the printed value of 8.8 N. Note that F has both vertical and horizontal components. Be careful of the angles used; it is easy to mix up when applying cosine rule. 3(b)(iii) 4(a)1. vast majority of alpha particles pass straight through thing metal foil or are deviated by small angles most of the target atom is empty space and the volume of the nucleus is very small compared to the atom B1 4(a)2. a very small minority of about 1 in 8000 alpha particles are scattered through angles greater than 90° the mass of the positively charged nucleus makes up the majority of the mass of the atom, and is concentrated in a very small nucleus region. B1 F X 36 N

8 ©EJC 2020 Qns Marks 4(b)(i) ( )( ) ( ) 2 K KK 6 19 27 71 1 2 22 4 2 4 8 10 1 6 10 4 1 66 10 1 52 10 ms E mv EEv mu .. . . αα α α − − − = = = ×× = × = × M1 M1 A0 4(b)(ii) work done per unit positive charge in moving a small test charge from infinity to that point 4(b)(iii) By Principle of Conservation of Energy, ( )( ) ( ) ( )( ) KP Au K 0 Au 0K 0 K 19 12 loss in gain in 79 2 79 1 6 10 1 2 10 4 11 44 1 4 8 85 E Q d E QE eeQd E e . . Q E α α πε πε πε π − − = = × = × = = ( ) ( ) 64 8 10. e× ( ) 144 74 10 m. −= × Note: common mistakes include • using the field strength equation, ending up with 2 1 d • omitting electronic charge for the gold nucleus/alpha • using 4e instead of 2e for the alpha charge • using mass numbers instead of proton numbers for charge. M1 M1 A1 5(a) For sinusoidal alternating voltage, 0 r.m.s. 2 VV = 2 r.m.s. 2 0 max mean po 1 1 wr 2 2 e V RP V R P = = = M1 5(b)(i) 3 1 10 349 rad 2 2 1 s 8 T πω π − − = = × = A1

9 ©EJC 2020 Qns Marks 5(b)(ii) 0 r.m.s. 2 170 2 120 V V V = = = A1 5(c) ( ) 22 2 m.s. 2 2 r.m.s. 170 170 9 18 170 2 170 2 120 V70 2 1 V V V = = = = = = 5(d)(i)1. alternating p.d. in the primary coil results in alternating current in the primary coil sets up a changing magnetic flux there is rate of change of magnetic flux linkage with secondary coil Note: need to specific actual location of “source” of changing magnetic flux and location of “receiver” of changing magnetic flux linkage. Answers should not omit terms such as “flux linkage” and “rate of change” of flux linkage. B1 B1 5(d)(i)2. ideal transformer so same magnetic flux at both secondary and primary coil less coils at secondary coil so proportionally less magnetic flux linkage induced e.m.f. is directly proportional to the rate of magnetic flux linkage, so the e.m.f. induced across secondary coil is proportionally lower Note: “use Faraday’s law … to explain” is required so cannot quote ss pp NV NV= . The transformer is described as ideal and the secondary circuit is an open circuit so cannot attribute the lower p.d. to eddy current losses or heat losses. B1 B1

10 ©EJC 2020 Qns Marks 5(d)(ii) ( ) sec sec pri pr