2020 9749 H2 Physics P1 MS EJC Suggested ver Aug2022

©EJC 2020 9749/H2 PHYSICS 2020 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2020 Paper 1 Multiple Choice Question Key Question Key Question Key 1 B 6 C 11 B 2 D 7 B 12 A 3 D 8 A 13 A 4 A 9 D 14 B 5 C 10 B 15 B 16 A 21 A 26 D 17 C 22 B 27 C 18 C 23 A 28 D 19 D 24 B 29 A 20 C 25 B 30 D Notes: Candidates found Questions 8, 16, 19, 22 and 24 more challenging. Question 8 distractor B – did not account for two ropes distractor C – confused the air densities distractor D – did not account for the weight of air inside the balloon Question 11 Takes 1 hour (3600 s) for a minute hand to complete one revolution. Question 14 Note that line MN is NOT an isothermal change. Using PV = nRT, temperature is decreasing. Question 16 Note that the energy of the oscillations reduced by E / 4 rather than to E / 4. Question 19 Single slit equation accounts for only one side of the central maximum. The width of the central maximum is double of this. Question 24 Note that the question asks for potential difference, not potential.

2 ©EJC 2020 9749/H2 PHYSICS 2020 1 let radius be r 34volume of sphere 3 3 Vr Vr Vr π= ∆∆ =  so y quantity is directly proportional to x quantity 2 for N3L, the forces involved should • be of same type • be of same magnitude • be of different directions • act on different interacting bodies Option D fails first bullet point 3 let wire be a cylinder length L and cross- sectional radius r 2 2 3 stress strain Nmunits of area under curve mm J m F r L L π − = ∆=   =     = 4 at t = 0, both car are at same position at t = T, both cars same position again so equal displacement from 0 to T since area under v-t graph is displacement P + (Q + R) = (Q + R) + S 5 relative velocity taking car as reference crb v vv = −       so rv   has a rightwards component and a downwards component 6 equilibrium so no resultant force in any direction no resultant torque about any point C is wrong because while the torque have same magnitude, one must act clockwise and the other anti-clockwise (so not “equal”) 7 moment of a couple can be calculated by either of the following: • (magnitude of one of the pair of equal and opposite forces)(perpendicular distance between forces) • 2(magnitude of one of the pair of equal and opposite forces)( (perpendicular distance between 1 force and the pivot) [useful tip for angles] check what it means by 0 and 90θθ= °= ° . in this case when 90θ = ° we should get max force so the function should be sinθ 8 there are two ropes. consider free body d iagram of balloon+basket: ( ) ( )( ) ( )( )( )( ) cool deflated hot air cool hot def lated upthrust weight of displaced air = total weight W equilibrium so 1 2 1 0 8982 N 2 1 204 2800 700 9 81 769 U Vg mm g UW g T V . Tm . . ρ ρρ − = = − + − = = + − = = 2T mg upthrust

3 ©EJC 2020 9749/H2 PHYSICS 2020 9 some thermal energy is not converted into useful work ( )( ) ( )( ) useful useful 6 produced useful produced 6 10 efficiency 100 1600 22 33 60 1600 22 33 6 1001 0 0 64 PF P %P % % v . P . × = × = ×× = = = = 10 given period, ( ) 2 c 2 2 8 2 2 2 210 27 3 24 63 85 0 0 m s00273 ar r . . T . ω π π −   ×  ×× = = = = 11 minute hand takes 1 hour to go around 2 1 2 2 60 0 00175 rad s T . πω π − = = = 12 let increase in GPE be U ( )final initial 11 2 12 2 2 Gm Um Gmm rr mGm r φφ −−=  = = − − 13 gravitational field strength ( )( ) ( ) 2 11 31 263 1 66 7 20 1 0 5 1 0 10 10 10 0 0593 N kg GMg . . r .− − ×× = ×× = = 14 ( )pV nR T= for L  M, process is constant pressure V is directly proportional to T, eliminate A and D for M  N, ( ) ( ) ( ) ( ) 6 MM 3 6 NN 3 10 Pa m 10 Pa 2 0 0 003 6000 0 8 0 005 40 m00 p. . p V V .. × = = = × = temperature drops so eliminate C (as a second layer of confirmation, direct proportionality above should suffice) 15 by first law of thermodynamics, UQW∆=+ for expt 1, 0UQ∆=+ for expt 2, ( )UQ W∆ = +− due to expansion (negative work done on gas) so 12UU∆ >∆ 16 energy is transferred to and fro between max ke and max pe every quarter cycle ( ) 2 max max 0 2 22 2 00 new 0 1E KE 2 1 2 0.8674 mv m E xx E xE x x ω±− − = = = = = ∝ Change in amplitude = (1– 0.867)x = 0.134x 17 phase difference between a sine and negative cosine wave is 90° 18 wavelength is 8 m 12 1 5 Hz8 vf .λ = == displacement

4 ©EJC 2020 9749/H2 PHYSICS 2020 19 angle of first minima involves half of x: for small angles i 2 sn 2 2 x b D / D x/ f x Dcb λ λ θ = = =   20 taking ratios: new 0 old 0 2 2 4 4 Q V d QV d πε πε = = 21 electric field strength of point charge: ( )( ) 2 0 0 0 1 gradient gradient 4 4 4 QE Q r Q πε πε πε    = = = 22 same p.d. across both elements: 2 XXY YX 2 YX Y XY 2 X 2 X 2 YY XX YY 0 816 2 2 1 15 VL PRR AA LVPR A AR d d dd dP dP . . ρ ρ π π = = =     = = =   = = =  23 effective external resistance of circuit 2 is R / 2 2 3 2 3 2 3 2 2 2 2 2 4 23 E Rr E r EE r r E r R R R r R r R R r R = =  =  + = + + + +  = + + = I I 24 to have zero p.d., R XY has to have ability to be zero, eliminate A, C and D 25 let time of flight within plates be t ( ) ( ) ( )( ) 2 7 e 2 2 e 2 e 219 2 731 2 5 1 97 0 5 1 979 11 10 10 10 1 2 1 2 1 2 1 6 10 30001 10 2 1010 10 0 017 m x x y y x x st v. d q md q md sq md VF ma qE q Va v . h ut . t V V . . a t − − − −− ×= × ∆= = =  ∆ ∆= +  ∆     = = = + = ∆     ×  ×  ×××  = = x θ

5 ©EJC 2020 9749/H2 PHYSICS 2020 26 let vertical component of Earth’s flux density be B ( ) ( )( ) 5 2 210 2 1 20 0 10 T 08 m 4 NBA . . − = ×= 27 2 rms rms PR P R = = I I 28 12 12 hcf h E E E E h c λ λ − = − = = 29 ( ) ( )( ) ( ) ( )( ) rxt pdt 2 rxt pdt 2 227 8 11 1 66 10 10 235 04 1 01 140 91 91 91 3 1 01 1 02 02 3 29 0 9 J m . .u m . . .u E mc . uc . . m . − − − = + = ++ = = = × = ×× 30 12 0 12 12 12 0 0 0 0 8 7 3 00 10 5 00 10 3 00 10 2 00 1 15 11 0 ln ln ln10 7 97 10 s 52 t t N N Ne N eN N N N N . N. . N. N. t t t . λ λ λ λ − − − −= × ×= × = × =  =−   =  =  ×  = × − =  