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 to infer from the equations stated.  It is also important to state the sign convent ions, and be consistent with the convention for all parts of the question. (b) (i) 1.  This part was generally well done.  No credit was given for only finding th e value of time time using the equation described in part (a).  When calculating the displacement of the cart, many students rounded off the value to 2 s.f. (90 m) rather prematurely. Student s should leave the answers in intermediate steps to more s.f. if possible. t / s EK 0 2.8 1.4

3    © Raffles Institution 9749/02 2. The kinetic energy of a projectile is not zero at any point along its flight. Many students mistakenly drew their graphs with Ek = 0 in the middle. (ii) There were several answers that suggested to decrease the angle of launch. Whie mathematically this increases the horizont al component of velocity, if the angle is too small, the object may not stay in the air long enough.     2 (a) (i) By Newton’s second law,   300 1.0 300 9.81 3243 3240 N AB AB Tm g m a T    (ii) Consider the forces on the jib. Since the jib is in equilibrium, taking moments about D, sum of anticlockwise moments = sum of clockwise moments         sin25 10.0 2400 9.81 4.0 3240 8.0 28417 28400 N AC AC T T    (b) (i) (ii) When the crane is just about to topple about G, the normal contact force on the left wheel just goes to zero. Considering the forces on the crane and taking moments about G, sum of anticlockwise moments = sum of clockwise moments         max max 16000 9.81 2.5 2400 9.81 4.0 8.0 37278 37300 N W W   (c) The force by the wind on the slab will cause it to swing/ sway/ oscillate. This results in additional clockwise moment from the increased tension in cable AB as the slab is displaced rightwards by the wind, OR increased perpendicular distance from G when wind is blowing such that the slab sways away from the crane, hence it will topple clockwise about G. A supporting cable 5.0 m F E jib 8.0 m 25 cab load B C D 10.0 m G

4    © Raffles Institution 9749/02 Comments (a) (i) Did not accept “ ABTm a m g ” as this does not show application of Newton’s 2nd law and understanding that ma is the resultant of TAB and mg. Some students erroneously equated a to be 10.81 m s –2. Also, the directions of T and ma were opposite for quite a number of students from this group. (ii) Common mistake was to use 300 x g x 8.0 when it should be TAB x 8.0. Students should have directly used perpendicular distances 4.0 m and 8.0 m given in Fig. 2.1 for 2400g and T AB respectively, instead of wasting ti me to further trigonometry in determining perpendicular distances. The weight of the cab is irrelevant in this part, as this part deals with the jib itself being in equilibrium. Some students failed to see this part as a rotational equilibrium problem. (b) (i) Mostly well-done. (ii) A number of students erroneously included TAC here. This is wrong, as this part deals with rotational equilibrium of the entire crane system. This means TAC, the force exerted by the cab on the jib (and vice -versa), is an internal force in this system. There is no moment due to internal forces. Only external forces such as gravitational forces, should be consider ed. Note that by taking moments about G, the normal force at G yields zero moment. Students should recognize question parts l abelled (a), (b), though related to the same question, are usually independent parts , whereas (i) and (ii) under the same part are inter-dependent (i.e. answer for (ii) is usually related to (i)). (c) Mostly well-done. Answers need to be spec ific in indicating concrete sways rightward / away from the cab, otherwise if it is leftwar d/ towards the cab, the weight of the slab actually contributes to additional anticlockwise moments about G, hence stabilising rather than toppling the crane.     3 (a) (i) C 65 cm string tangential path parabolic path

5    © Raffles Institution 9749/02 Path of the ball is such that it leaves the vertical circle horizontally to the left along the tangent of the circle (linear speed of the ball at any point along the circular path is at a tangent), followed by a parabolic path (due to gravitational force). (ii) The resultant of the tension and weight on the ball provides the centripetal force.   2 1 1 1 16 0.30 9.81 0.30 0.65 8.183 8.18 rad s Tm gmr Tm g mr        (b) (i) When the angular speed increases, a larger centripetal force is required. Since the horizontal component of the tension in the string provides for this centripetal force, tension a