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