RI 9749 2017

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT ( )12 1 1 1 1 6 32ratio 31 6 2 vv v vv +× = = = ×× GCE A−level 2017 H2 PHYSICS 9749/1 answers 1 A 11 D 21 A 2 B 12 B 22 D 3 C 13 D 23 D 4 D 14 D 24 C 5 C 15 B 25 A 6 A 16 C 26 B 7 D 17 D 27 D 8 C 18 C 28 C 9 D 19 B 29 B 10 A 20 C 30 C Suggested detailed solutions: 1 A 2 B 2 Let , then 2 2 2 1 2 6xy A x y zA % %%%z Ax yz ∆∆ ∆∆= = + + = +× + = . 3 C Distance travelled is given by the area under the speed-time graph. OR By similar triangles, from the graph: 3 31Ratio = = 4 D The maximum velocity (terminal velocity Tv ) happens when air resistance is equal to weight: T0 60. v mg= . The acceleration a at any speed is found using Newton’s 2nd Law: ( ) 2net 3 0 9 81 0 60 120 60 7 4 ms30 . . .( )F mg . va. mm . −−−= = = = 12 21 00 2 6 0 12 0 vv vv−− = ⇒= −− 1 T 30 98 1 49 ms0 60 max ..vv . −×= = =

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT Acceleration when v = 12 ms-1 : 5 C Since the momentum reverses direction in the collision, the change in momentum is greater than p. Since the collision is inelastic, the final momentum must be less than p in magnitude (so that the final KE is less than the initial KE). So the change in momentum must be some value in between p and 2p. 6 A Action-reaction pairs satisfy the (A exerts on B) and (B exerts on A) relations. 7 D KE is converted into GPE on the way up, so it decreases; it increases on the downward trip since GPE is converted into KE. 8 C For a satellite in orbit, Total energy of satellite, Change in total energy of satellite, OR 11 24 2 66 9 11E= 2 6.67 10 6.0 10 6.9 10 1 1 2 6.5 10 7.2 10 2.1 10 J T fi GMm RR − ∆− −   × ×× ×× =−−  ×× = −× 9 D Angular velocity ω is common to every point on the disc. 2 2 24 Since and is a constant, a r, 1 8 ms2 rr ar aa ωω − = = = ∝ = = 10 A 2 30 98 1 06 0 1 2 30 7 4 ms . . . .a a. − × − ×= × ∴= 2 2 GMm mv RR = 2GMm mvR⇒= 2 22 T GMm mvE R= −= − 2 22 96 9 10 7900 7500 2 1 10 J2 T .E. × ∆= − − = −× 1GM RRφ = −∝ SX SX R R φ φ = ( ) 6 7 71 6 6 371 10 6 257 10 6 208 10 J kg6 371 10 50000 X . ...φ −×⇒ = × −× = −××+

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 11 D Note that 273 15T/K T/ C .= +  . 12 B The system is insulated from heat supply, therefore Q = 0. Since temperature increases, internal energy increase. From the first law of thermodynamics: doneonsystemUQW∆=+ Since Q =0, 0doneonsystemWU = ∆> By stirring, work is done on the system. 13 D For a good car suspension system, the oscillation should stop as quickly as possible so that the system may return to and stay at equilibrium position. A and C are light damping. B is over damping. D is critical damping. 14 D The dotted wave is lagging the solid wave by 2/π (quarter of a period), or leading it by 32 /π