2020 RI Prelims H2 Phy Paper 2 Soln

© Raffles Institution 9749/02   2020 Raffles Institution Preliminary Examinations – H2 Physics Paper 2 – Suggested Solutions 1 (a) The projectile undergoes a trajectory sy mmetrical about the c entre (or parabolic trajectory), with vertical displacement sy = 0. Taking upwards as positive and considering vertical motion,  2 2 1 2 10s i n 2 2s i n (shown) yy o o oo o su t a t ut g t ut g      Alternative method: Since the projectile undergoes a parabolic trajectory, Taking upwards as positive and considering vertical motion, Vertical component of initial velocity, sinyuu  Vertical component of final velocity, sinyvu  sin sin 2s i n (shown) yy o o o vua t uug t ut g      Alternative method: Taking upwards as positive and considering vertical motion of the projectile, At the highest point of it’s trajectory, vertical component of velocity, 0yv  0s i n sin where is the time taken for it to reach the peak of its trajectory yyvua t ug t utt g      Since the projectile undergoes a parabolic trajectory, its total flight time is twice the time taken to reach the peak of its trajectory. 2 sin2 (shown)o utt g  (b) (i) 1. horizontal displacement of projectile = horizontal displacement of cart + 45 m  cos 45ut v t   1 45cos cos 45 2s i n 9.8135cos23 45 23 5 s i n 2 3 16.08 16.1 m s vu t gu u            

2    © Raffles Institution 9749/02 2. *Shape of graph, symmetrical about minimum turning point *Position of graph (graph above horizontal axis, correct values of t at minimum turning point and end of graph) (ii) Launch the projectile with a larger speed. OR Lauch the projectile at a larger angle to the horizontal. (but less than 45 ) Comments (a)  Students need to be very mindful when they see a question which states ‘Explain your working’. While the question may seem st raightforward, the need to explain the working will demand more attention and effort.  Many students simply stated the symbols and equations, without really explaining how certain conclusions were made.  Many who used the 21 2su t a t approached often did not give a good explanation why total displacement of the projectile upon landing is zero. Stating the reason to be that it lands at the same level of launch did not earn credit. The main point here is that the path is symmetrical about the centre. A projectile not undergoing a symmetrical path about its centre, would also have zero total displacement if it lands on the same level of launch.  Those who used the other 2 alternat ive approaches did not explain that vy = 0 at the peak of the path, and why vy =  uy.  Students must remember that this is an ‘explain’ question and they should explicitly mention these points in words, and not expect the examiner