2019 RI Prelims Paper 2 Soln

© Raffles Institution [Turn over 2019 Preliminary Examinations H2 Physics Paper 2 Solutions 1 (a) Speed is a quantity and cannot be defined by a unit (seconds). Speed is distance travelled per unit time. B1 B1 (b) (i) 7.5 15cos 60      B1 (ii) 2 2 15sin60 12.99 1 2 12.0 12.99 ( 9.81)2 2.79 or 0.146 (rej) y yy y u su t a t tt t       C1 M1 A1 (c) (i) (ii) (i) Straight line (negative area larger than positive area) (ii) Correct shape B1 negative area larger than positive area B1 Gradient at x-intercept is g B1 Markers’ Comments (a) Most candidates were able to define speed co rrectly. A common mistake was to define speed as “the rate of change of distance travelled per unit time”. The next part of the question was poorly attempted. Most explanations stated that the incorrect definition referred to an average value rather than an instantaneous value, or that there were alternative units to seconds but failed to mention that units are generally not stated in definitions. Many candidates stat ed that “seconds” are not the only units of time, which implies that speed can be defined in terms of other units of time, which is incorrect. (b) (i) Most students answered this question correctly. (ii) Most candidates who did not get full credi t assumed the ball returned to its height above the ground after the motion or failed to include their sign conventions. (c) (i) Most students answered this question correctly. (ii) Some graphs had kinks in them. Mathematical ly, the gradient at a kink is undefined and cannot be the case on a v-t graph. v / m s1 t / s tf R Q

2 © Raffles Institution 2 (a) 2 15 15cos40 7.0 mh   M1 (b) By Principle of the Conservation of Energy,        22 2 -1 Loss in GPE Gain in KE 117.0 3.5 22 2 9.81 3.5 1 8.35 m s B B mg mg mv mu v      M1 A1 (c) Correct direction of arrows and correct label. Arrow of weight should start from the centre of the body and arrow of the normal contact force should start from the ground. Arrow of the weight is longer than the arrow of the normal contact force. One mark is deducted for any additional force or mistake. B1 B1 (d)   2 2 2 450 8.35450 9.81 20 2850 N mvmg N r mvNm g r     M1 A1 (e) (i) B1 (ii) At ‘X’, (normal force – weight) provi des the centripetal force. Hence the normal force is equals to the weight + centripetal force which is the largest during the entire journey. The speed of the carria ge is the highest at the bottom of the circular arc. B1 B1 (f) 2mvmg N r N = 0 when the carriage just loses contact,    2 -1 20 9.81 14.0 m s mvmg r vr g     M1 A1 Normal contact force Weight of the carriage X

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