RI 2016 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9646 H2 Physics 2016 GCE A-Level Suggested Solutions Paper 1 1 B 2 1 32 3 3 02 6 9 32 12 6 18 2 A: (10 m)(10 m) (10 m ) B: (10 m)(10 m) (10 m ) C: (10 m)(10 m) (10 m ) D: (10 m)(10 m) (10 m ) −− − − −− −− − = = = = 2 C 34 8 19 9 of green light ~ 550 nm (6.63 10 )(3.00 10 ) 3.62 10 J550 10 hcE λ λ − − − ××= = = ×× 3 B 1 a 1 w 22 2 2 R aw 11 700 km h 250 km h 700 250 653.8 km h 650 km h v v v vv − − −− = = = −= − = = 4 D A: velocity-distance graph B: displacement-time graph C: acceleration-time graph 5 D Taking upwards as positive: 2 2 2 2 0 ( sin ) 2( )(210) sin 420 0 sin 420 420 yy yyy yyu as ug ug u at u gt g gt gt g v v θ θ θ + = +− = = + = − = − = = 420)(2)( ) 250cos 420 sin 420 cos 250 420tan 250 6 500 ( 0 cos xx g g gu g ug u s ut u g θ θ θ θ θ θ = = = = = = ° Vw Va VR 500 m θ 210 m u

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 6 C 1 Momentum after collision Momentum before collision 2(12) 4(0) 24 kg m s− = = + = 7 D Constant force ⇒ constant acceleration No force ⇒ constant speed 8 D (1020)(3)(4)(5)(9.81) 600372 weight of concrete (2300)(3)(4)(5)(9.81) 13 upthru 5 st on concret 378 e 0 sw sw c c Vg N m m Vg g g N ρ ρ = = = = = = = = 5 6 force required to lift concrete when fully submerged 1353780 600372 753408 7.53 10 N force required to lift concrete when out of the water weight of concrete 1.35 10 N = − = = × = = × 9 B Assuming there is friction, total reaction force from the slope on skier is the resultant of the normal reaction force and friction. 10 C Torque of a couple = force × perpendicular distance between forces Since force and distance does not change, torque of the couple remains as M. 11 A 6316 (48 10 ) 400(1 10 )100 0.0521 kg 52 g x x ×=× = = 12 D -5 122 (24)(36)(36 7.27 10) radT πω π −= = ×= Normal reaction force friction

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 13 D ( ) ( ) 22 2 1 1 2 On a dry surface, friction provides for centripetal fo 16 On a wet surface: ) 16 ) rc ) 16 e. 1 2 1 ( 2 1( )(2 2 mmv rr mv r mv r m r f f r f m = = = = = = 14 C GM rφ = − , scalar, algebraic summation. 15 C ( ) 22 22 2 2 2 23 1/ 2312 2 11 (Keplar's Law) 1.43 10 11.9 29.7 years7.78 10 2 ss JJ s GMm mr M r Tr Tr Tr T Tr πω = =  ∝ =  ×= = ×     16 A Maximum KE at equilibrium 22 0 K.E. P.E. constant -v xxω± += = 17 D Decrease in GPE = Increase in KE 2 2 1 1(cos0.030 cos0.050) 0 1(cos0.030 cos0.05 2 ()2 0.125 0.13 r 0) ad mv r mg g ω ω − −= − −= = = ORω: angular speed, Ω: angular frequency = 2 T π ( ) ( ) ( ) 2 22 2 0 2 22 2 0 2 2 22 1 Similar to 2 0.050 0.030 0.126 0.13 rad v xx T ω ω θθ πω ω − = − = Ω− = − = = 0.050 rad 0.030 rad

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 18 A 11 2 2 12 5 22 1 1 5 21 3 (1.0 10 )(0.025) 600 300 5.0 10 0.010 m PP TT PT TP VV VV =  × =  × =    = 19 B is the same for both processes UQW U ∆= + ∆ 8 (3 ) (1 ) 6 J into gas Q Q +− = +− = 20 D 23 21 3 2 28 3 (1.38 10 )(28 273.15)2 6.23 10 J k k E kT TC E − − ×+ = × = ≈° = 21 B Note: there is inconsistency in the information given in this question Approach using intensity: Let XX AI , be intensity and amplitude after passing through 1st polariser YY AI , be intensity and amplitude after passing through 2nd polariser ZZ AI , be intensity and amplitude after passing through 3rd polariser ( ) ( ) X YX Zy II III IIII = = °= = °= = 2 2 cos 45 4 1cos 45 ( )( ) 42 2 8 However, if use information given in the second part of the question: ( ) y yy zy AA AAkA k k Ak Ak I II I = = = =  = °=   = = 2 2 2 2 2 2 1cos 45 ( )2 2 2 4 22 4 22 D 23 D Consider vector sum of the two waveforms at every point.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 24 C 12 2 0 22 00 1 4 (3 )(3 ) 4 (3 ) 4 QQ r Q Q QQ F FrF r πε πε πε = = == 25 C 2 0 1 2 0 1 2 0 22 22 00 1 ' 32.0 ( : side of 1 square)4 1 (32.0) 3.6 N C94 (3 ) 1 16.0 N C24 ( 2) 4 ( 5 )4 (4 ) 1 6.4 N C5 s R Q rr Q r Q E r QQ rrr E E E E E πε πε πε πεπε − − − = = = = = = = = = = = + = 26 B Terminal p.d. = p.d. across variable resistor Resistance of variable resistor ↓, p.d. ↓ Effective resistance of circuit ↓ I ↑ ⇒ power loss in internal resistance = I2R ↑ 27 C 28 3 21 41 Copper and semiconductor in series (8.6 10 )(0.586 10 ) 4.3 10 1.16 10 m s cs s sc c cc s s neAv n eAv n eAv nvv n − − = = = = ××= = × = × I III 28 A LDR: Brightness ↑, R ↓ Thermistor: Temperature ↑, R ↓ Smallest voltmeter reading when total resistance is the smallest. 29 B 111 11 1Effective resistance 4.0 6.04.0 4.0 8.0 8.0 10 −−−= ++ ++ + = Ω 8.0 20 2.0 A10 terminal p.d. 20 (2.0)(6.0) 8.0 V 8.0 1.0 A8.0 Ω = = = −= = = I I

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 current through first 4.0 2.0 1.0 1.0 A p.d. across first 4.0 (1.0)(4.0) 4.0 V terminal p.d. 4.0 8.0 4.0 4.0 V xyV Ω= − = Ω= = = − =−= 30 B Use Fleming’s left hand rule: 31 A Force perpendicularly out of page. 16 19 5 51 sin23 7.3 10 (0.084)(1.60 10 )( sin23 ) 1.39 10 1.4 10 m s F Bqv v v −− − = ° ×= × ° = ×= × The component of the velocity parallel to the magnetic field will cause the electron to move in a helix. 32 C 2000(0 0.34) 28330.24 2800 V fi dE dt E tt Φ= − Φ −Φ∆Φ= =∆∆ −= = = 33 D 22 32 , (10.0 10 )(8.0 10 ) 8.0 10 m BA N A φφ −− − = Φ= = ×× = × 34 B , 80 80 80 90 V80 40 2 22.5 W 2 1peak ( )(180) 90 V2 90 V 3 W80 2 s s rms mP V V V V Ω Ω  +  = = = = = = = 35 B All other options demonstrate the wave-nature of electromagnetic radiation. F I B 23° v

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 36 B 1 1 Gradient 1 y-intercept 1 RT RT y mx c += = −+ = + = − = 37 D N-type semiconductor 38 C Higher concentration of electrons in n-type → during formation of depletion regions, electrons diffuse from n to p type (i.e. in the y-direction) During easy flow of charge →i.e. forward bias Holes move from p to n type (i.e. in the x-direction) 39 C 1/ 2 120 01 2 1 ln2() ,2 1ln( ) ln 2 t tN tN Nt Nt λ= = = 12 00 21 23 8 8 8 ln 2ln( ) ln 11ln ln22 8.6 10 ln2ln 17.7 10 (2.6 10 )ln 2 1.73 10 s 1.7 10 s tNNt NN n − = ×= × ×= ×× × = × = × 40 C ( ) 00 0 00 ln2 , when the curves intersect, 1 21 ln 2 1 ln 2 1ln ln1 ln 22 ln 2 ln 2 t x yx TT xy TT T N Ne N N N N N Ne N Ne ee e T T λ λλ λλ τ τ ττ τ − −− −− − = = − = ⇒= − = − = −=    = −= − −=      

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 8 Paper 2 1 (a) One ohm is the resistance of a conductor when a potential difference of one volt across it causes a current of one ampere to flow through it. (b) (i) ( ) ( ) ( )( ) ( )( ) 2 2 23 2 7 7 2 4 1 50 0 23 10 4 0 32 40 0 10 4 869 10 4 87 10 m LR A VI dRA Vd L L IL .. .. . . ρ π πρ π − − − − = = = = × = × = × =×Ω (ii) ( ) 7 88 2 00 1 00 1 00 1 012 4 869 101 50 0 23 0 32 40 0 6 202 10 6 10 (1 s.f.) V dIL V dI . .. . .. L .. . . ρ ρ ρ − −− ∆ ∆ ∆ ∆∆= + ++ ∆= + + + × = ×= × (iii) ( ) 774 9 10 0 6 10 m..ρ −−= × ±× Ω (c) Accuracy refers to the closeness of the calculated value in (b)(iii) to the true value. Precision refers to the size of the uncertainty (0.6 in this case) relative to the calculated value in (b)(iii). 2 (a) 264 1 6 m s40 F.a. m m. Fa −= = = = (b)

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 9 (c) 1 1 impulse = area under the F-t graph 64 14 8 96 kg m s 9 0 kg m s .. . . − − = × = = (d) 3 (a) Taking moments about A: By Principle of Moments, Sum of clockwise moments = Sum of anti-clockwise moments ( ) ( )( ) ( )( ) ( ) ( ) ( ) 36 0.45cos 60 sin70 1.2cos 60 cos70 1.2sin60 36 0.45cos 60