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 T