RI 9646 2014

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2014 GCE A-Level 9646 H2 Physics Suggested Solutions Paper 1 Solutions 1 C Absolute error presented to 1 sf = 3 m s−1 Quantity presented to same precision as absolute error = 348 m s−1 Speed = (348 ± 3) m s−1 2 A 5 µm = 5000 nm (not visible ⇒ not B, C) 0.5 µm = 500 nm (visible ⇒ A or D) Radio waves are in the range of cm to m (not D) 3 A Gravitational field strength = initial acceleration = gradient of graph when object is just released from rest = 6 m s−2 4 D Let total distance travelled be x. Use s = ut + ½ at2. For initial 0.25 x: 0.25 x = 0 + ½ (9.81) t2 For final 0.75 x: x = 0 + ½ (9.81) (t + 1)2 ⇒ 2 (9.81) t2 = ½ (9.81)(t2 + 2t + 1)2 ⇒ 3t2 – 2t – 1 = 0 ⇒ t = 1 s Total = 1 + 1 = 2 s 5 D Since system total momentum is non -zero, both spheres cannot be at rest simultaneously in order to conserve momentum. 6 C ( ) ( ) ( ) 4 -2 Let the required tension be . Isolate th e wagons 3, 4, 5 and 6 and apply N2L. 4 4 4000 N 4 6.0 10 kg 0.15 m s 52 kN T T f ma T T −= −× = × × × = 7 C Work done in stretching the fibre is given by the area under the force -extension graph. When force is increased from 0 to F, the extension increases, work is done on the fibre, and this work is the area P. When the force is reduced from F to 0, the extension decreases, work is done by the fibre, and this work is the area Q. The net work done on the fibre is therefore (P – Q). Note: if the axes are swapped, work done will be area between graph and vertical axis. 8 D When the ball is falling vertically with constant speed v, the result ant force acting on it is zero. Therefore, weight W is balanced by the sum of upthrust U and viscous force kv: W U kv= + Note: if velocity is high, the viscous is kv2.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 9 C ( ) ( )( ) 45 6 Let the power developed by the locomotiv e be . If its acceleration is when travelling at speed , the equation of motion is pulling force frictional force mass 5.0 10 3.0 10 0.5010 2.0 10 P av P av P P −= ⇒−× = × ⇒= × max 6 4 max 6 -1 max 4 W The speed reaches a maximum when the acceleration decreases to zero: frictional force 0 2.0 10 5.0 10 2.0 10 40 m s5.0 10 P v v v −= ×⇒= × ×∴= = × 10 D The direction of electric field V/x points from left to right within the parallel plates. The electric force (q)(V/x) acts to the right on the charge having displacement x to the right. The work done by this electric force on the charge is +[( q)(V/x)] ( x) = +qV, causing its kinetic energy to increase by qV. Therefore, its electric potential energy decreases by qV, due to conservation of mechanical energy. 11 B A & D are equivalent statements and both are wrong. If the centripetal acceleration of the astronaut and capsule is the same, their centripetal force may not be the same because they may have different mass. 12 B The car tends to lose cont