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

3 © Raffles Institution [Turn over Markers’ Comments (a) This part proves to be mathematically c hallenging for many students who left this part blank. (b) Several students forgot to consider the initial kinetic energy of the carriage. (c) At A, the resultant force is towards the centre of the circle which is downwards. Hence the length of arrow for weight should be longer than the normal contact force. When you are required to draw and label forces acting on a body, it is important to spell out the forces instead of giving your own notations. (f) Generally well done. 3 (a) A transverse wave is one in which its particl es oscillate in a direction perpendicular to the direction of energy transfer. A longitudinal wave is one in which its particles oscillate in a direction parallel to the direction of energy transfer. B1 B1 (b) polarisation B1 (c) (i) 2  k r 1.2  105  1.52 = I6  6.02 I6 = 7.50  107 W m2 M1 A1 (ii) P = I  A = 7.50  107  1.3  103 = 9.75  1010 W B1 (iii) I =kA2 57 22 0 1.2 10 7.50 10 84 x   x0 = 21 m M1 A1 Markers’ Comments This question is generally well done. 4 (a) The electric field strength at the point is the electric force exerted per unit positive charge placed at that point. B1 B1 (b) (i) - smooth lines with correct field direction (at least 8 for each sphere), lines originate from spheres perpendicularly. - Lines not touching and not crossi ng, symmetrical and null space in centre of both spheres. B1 B1

4 © Raffles Institution Zero credit when field lines between spheres act through direction of P. (ii) Distance of X from centre of sphere A = (0.08 + 14.0) = 14.08 m Taking rightwards as positive, 99 12 2 12 2 34 31 31 electric field strength 0.0400 10 0.0400 10 4 (8.85 10 )(14.08) 4 (8.85 10 )(60.0 14.08) 1.814268 10 ( 1.70570 10 ) 1.64 10 N C 1.64 10 N C                   Towards sphere A M1 A1 B1 (c) (i) AB 00 99 12 12 Potential at P 44 (0.040 10 ) (0.040 10 ) 4 (8.85 10 )(30.0) 4 (8.85 10 )(30.0) 0.023978 0.0240 V              QQ rr A1 (ii) - symmetrical curve about P, value higher than potential within spheres. (values must match convention of (c)(i). - Potential is a constant value within both spheres B1 B1 Markers’ Comments (a) Many student miss out the terms ‘electric’ force, ‘per unit’ and ‘positive’ in their definition. Statements often read as the electrical forc e, (N) as the subject and acting on (unit charges) which is not the same as electric force per unit positive charge (NC1). sphere A sphere B V/V P x -4.50 0 0.024

5 © Raffles Institution [Turn over A handful mention ‘masses’ and even moving charges from infinity to the point. (b) (i) Very badly done. Sketching of field lines has been tested before, but most students totally ignore feedback. It is hard to sketch lines going perpendicularly into the surface of the sphere. Little right angles can easily settle this issue, but only a handful adopted that technique. Many are totally clueless of how field lines ought to look like. (ii) Very badly done. Many overlooked the fact that the distance should be taken from the centre of the spheres, and that the spheres have diameter of 0.16 m. Many probably forgot that that electric field strength is a vector, giving us a scalar sum instead, this arises from students not sketching diagrams to guide their thought processes. (c) (i) Horribly done. Many assumed that at the null point, no field lines are sketched and hence potential is zero. This is NOT true! Potential is a scalar and they add up. Many students get confused (with gravitational potential) and put in the negative sign for the equation. For electric fields, the polar ities will take care of themselves when you substitute the proper (positive or negative) charges into the equation!!!! Note that potential gradient = 0 does not mean that potential is also 0. (Potential at infinity is taken as zero, and the point P is nowhere near infinity) Equation for electric potential is given in the data sheet!! Many students did not read the question and though that we were asking for the potential at point X. (ii) Well done (except for the flipping of the graph which stemmed from the error in (c)(i). Note that the sign convention for this graph has to follow that of part (i). No credit awarded if the correctly shaped gr aph extends into both the positive and negative axes. 5 (a) Faraday's law of electromagnetic induction states that the induced e.m.f. is proportional to the rate of change of magnetic fl