TJC H1 PHY P1 Solutions

2016 Preliminary Examination H1 Paper 1 Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 D A C A D B D B B A C C D A D 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C C C A B C C D A A D B D C A 1 D Vm = volume per unit mol  unit of b = unit of Vm = m3 mol-1 Unit of [a/V m2] = unit of P = Pa  Unit of a = Pa (m3 mol-1)2 = Pa m6 mol-2 2 A Average = (7.6 + 7.5 + 7.8 + 7.4 + 7.6) /5 = 7.6 cm. True value given = 7.5 cm , so average value of 7.6 cm is accurate to within 0.1 cm. The 5 values are not precise to within 0.1 cm as there is a spread of values from 7.4 cm to 7.8 cm. 3 C A: In the measurement of the diameter of a sphere, take more readings and finding the average value of these readings will help to reduce the fractional uncertainty of the diameter. (random error) B: Plotting a graph of voltage and current readings for an ohmic conductor and using its gradient to find resistance will help to eliminate (reduce/minimize) random error. D: Checking for zero error on a voltmeter before measuring voltage will help to reduce random error (eliminate systematic error). 4 A sf) 1 (to s m 0.3 0.295 .7899x42.1 02.02500.0 001.0 2 s m 789.9)42.1( )500.0(44 2- 2- 2 2 2 2            g T T l l g g T lg  g = (9.8 ± 0.3) m s-2 5 D Taking upwards as positive, since net force is weight downwards, by N2L, acceleration is constant g value, downwards (negative constant value) At rebound, there is a sharp positive peak as there is sudden impulsive force acting upwards on the ball within a short time of impact As the ball is moving upwards after rebound, the net force is still weight downwards  acceleration is still negative constant value. 6 B When body moving upwards, net force is downwards = W+ drag force = ma, deceleration a > g When body is at maximum height, velocity = 0, net force is downwards = W; a = g When body is moving downwards, net force is downwards = W – drag force = ma such that acceleration downwards a < g So, value of acceleration downwards < value of deceleration upwards  time taken downwards > time taken upwards

2 7 D For X: applying v = u + a t in the vertical direction: 0 = u sin  - g t Hence g ut sin Similarly for Y: Time to reach max height = g u sin)2( = 2 t Since horizontal velocity for Y is 2 u cos  (or 2 times that of X) and time taken for Y to reach max height is twice that for X, horizontal displacement of Y at max height is 4 times that of X. 8 B When joined in parallel, both springs would have the same extension, but the applied force would be shared between them. So if the extension is e, spring 1 would be supporting F 1 and spring 2 would be supporting F1. The applied force is thus (F1 + F2). 9 B The weight of the block has to be included. Since there are three forces on the block, the forces should be concurrent. 10 A The net force is -mgsin  is a constant. So the net force = rate of change of momentum = slope = constant 11 C Use relative speed of approach = relative speed of separation [6-1 = 2-(-3)], as well as conservation of ke. The incorrect answer A also fulfils the relative speed equation, but because M is heavier, the system seems to have an increase in ke after the collision. spring 1 spring 2 F2 B e extension 0 F1 F1 + F2

3 12 C Initially, when v=0, viscous force is zero. Since mg = ma, the initial acceleration is g for both masses. Viscous force is proportional to speed v. At terminal speed, mg = kv, if mass is large, terminal velocity is large . 13 D Pressure P = hg, so dP/dh = g = 1000x9.81 = 104 Pa m--1 14 A at initial angle 45o, the initial horizontal velocity = vertical velocity, i.e. ux = uy. at any instant G PE = mgh. Since h = uyt -1/2 gt2, so PE =mg(uyt -1/2 gt2), at any instant KE = ½ mv2 = ½ m(vx2 + vy2) and vy = uy – gt, so KE decrease, then increase quadratically with t. At maximum height, both KE and PE are equal at ½ mu x2 15 D By definition, work done by force F = force F x displacement in the direction of force z 16 C Power P = Force x distance/time = force x velocity 17 C Taking to the right to be +ve displacement, At the next instant of time, particle 1 has –ve displacement ( which means it is moving to the left ) while particle 7 has +ve displacement ( which means it is moving to the right) 18 C Period T of the wave = Since the wave is moving to the right, Q will be moving upwards in the next instant of time. Hence, shortest time taken for Q to move to zero position = 19 A 20 B Using pythagoras theorem = path length L2 D = (402+92)½ = 41 m Path difference between L1 D and L2 D = 41 – 40 =1 m v = fλ For 1st maximum, path difference = 1λ   = 1 m 330 = fλ = f(1) f= 330 Hz 21 C Point X is a node. At one particular instant of time, particles on the left of X will move to the right while particles on the right of X will move to the left, producing a region of compression at X . At another instant of time, particles on the left of X will move to the left while particles on the right of X will move to the right, producing a region of rarefaction at X. 22 C Power = Energy/time = 120/2.0 = 60 W A is wrong since current I=P/V = 60/120 = 0.5 A = 0.5 Cs -1 B is wrong since 120 V means 120 J delivered for each Coulomb of charge D is wrong. It should be 60 J per second when current is 2.0 A 71 Direction of motion Displacement Distance

4 23 D The resistance at 1.2 V is 0.4 x 103 Ω, not 0.4 Ω The trend of the graph shows that as V increases, current I increases sharply, so ratio V/I decreases 24 A When pd is applied across BD, combined resistance across BD, R BD = 20/3  When pd is applied across AB, combined resistance in branches BD and BCD = 20/2 = 10  Add to branch AD which is in series = 10 + 10 = 20  Combined resistance across AB = (1/10 + 1/20)-1 = 20/3  This is same as RBD in the previous case. Since power to circuit, P = V2/R  PBD = PAB 25 A Original distribution of resistance across the two loops in the circuit is R/2 to R/2 i.e. 1:1. The new distribution is R to R/2 i.e. 2:1. Due to the new distribution of resistance, the voltage across X will increase (2/3 V) and the voltage across Y and Z will decrease (1/3 V). Since current is V / R, and R is constant, the increase in voltage across X will cause an increase in current through X while the decrease in voltage across Y and Z will cause a decrease in current through Y & Z. 26 D Using LHR, for the magnetic force on the wire to be downwards, the magnetic field must be towards the right. 27 B Use the Right Hand Grip for current in wire X to find the magnetic field at P and Q, then use the LHR to find the forces at P and Q. 28 D hf =  + K ……(1) h(2f) =  + K’ ……(2) 2( + K) =  + K’ K’ =  + 2K > 2K 29 C E3 – E1 = h f3 ; E3 – E2 = h f1 ; E2 – E1 = h f2 E3 – E1 = ( E3 – E2 ) + (E2 – E1) h f3 = h f1 + h f2 30 A Decrease in electron’s KE is transferred to the atom, exciting it. This is subsequently released as a photon when the atom de-excites. Thus ½ m(u 2 – v2) = hf. E3 E2 E1 f3 f2 f1