2022 H2 CJC Physics Promo Paper 2 Suggested Solutions

[Turn over CANDIDATE NAME CLASS 1T PHYSICS 9749/2 Paper 2: Structured Questions 30 September 2022 2 hours Candidates answer on the Question Paper No Additional Materials are required READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. MARK SCHEME This document consists of 28 printed pages and 0 blank page. FOR EXAMINER’S USE DIFFICULTY L1 L2 L3 Q1 / 11 Q2 / 10 Q3 / 10 Q4 / 12 Q5 / 9 Q6 / 15 Q7 / 13 PAPER 2 / 80 Catholic Junior College JC1 Promotional Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 [Turn over Formulae uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p V hydrostatic pressure p = gh gravitational potential  = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin t magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-t) decay constant λ = 1 2 ln2 t

4 Answer all the questions in the spaces provided. 1 (a) A ball of mass 0.50 kg leaves the edge of a table with a horizontal velocity v, as shown in Fig. 1.1. Fig. 1.1 The height of the table is 1.25 m. The ball travels a distance of 1.50 m horizontally before hitting the floor. Air resistance is negligible. For the ball, (i) Show that the horizontal velocity v is 3.0 m s-1, [2] L2 Solution: Assume downwards and rightwards directions as positive. Consider horizontal motion, 1.51.5 (1) xxs u t vt t v = =  = −−−− M1 ball table path of ball v 1.25 m 1.5 m floor part (c)

5 [Turn over Consider vertical motion, ( ) 2 2 1 2 11.25 0 9.81 (2)2 y y ys u t a t t =+ = + −−−−− Sub (1) into (2), ( ) 2 1 1 1.51.25 9.812 2.97 or 3.0 m s v v − =  = M1 A0 Examiner’s Comments Mostly well done. (ii) Calculate the velocity just as it hits the floor, magnitude of velocity = ………...……..……………. m s-1 direction of velocity = ………………………………………. [3] L2 Solution: Assume downwards and rightwards directions as positive. Let v1 be the magnitude of velocity just as it hits the floor. Consider vertical motion, ( )( ) 22 2 1 2 0 2 9.81 1.25 4.95 m s y y y y y y v u a s v v − =+ =+ = 22 1 1 3.0 4.95 5.788 5.79 or 5.8 m s v −  = + == Let θ be the angle below the horizontal, M1 A1 vx Vy θ

6 y x v 4.95tan θ = =v3 oθ = 58.8 below the horizontal A1 Examiner’s Comments Generally well done. Many students did not obtain credit for the answer on direction of velocity. Most students did not give the answer on the direction of velocity accurately and clearly. For example answers such as “from the horizontal” or “to horizontal” are not accepted. The accepted answers are “below the horizontal” or “clockwise below the horizontal”. The context of the question is not related to bearing or compass, answers stating “south of east” or “east of south” are not accepted. (iii) Using the floor as reference where the potential energy of the ball is zero, calculate the kinetic energy and potential energy of the ball at the top of the table. kinetic energy = ………...….…..……………. …J potential energy = ………...….…..……………. …J [2] L1 Solution: Kinetic energy ( )( ) 2211 0.50 3 2.25J22mv= = = Potential energy ( )( )( )0.50 9.81 1.25 6.13Jmgh= = = A1 A1 Examiner’s Comments Mostly well done. However, a handful of students thought that K.E. at top of table is zero even though the ball has velocity when projected horizontally at the top of the table. (b) The horizontal distance, along the floor, from the bottom of the table is x. Fig. 1.2 shows the variation with x of the potential energy Ep of the ball.

7 [Turn over On Fig 1.2, sketch the variation with x of the kinetic energy Ek of the ball. Fig. 1.2 [2] L3 1 mark – award mark for correct shape of graph 1 mark – award only if the sketched graph has all the following 2 features: B1 x / m 1.5 0 Energy / J EP 0.75 x / m 1.5 0 Energy / J 0.75 EP Ek

8 • Ek graph start below Ep but not 0. • Ek graph ends at x = 1.5 m with a value higher than the initial Ep value at x = 0. Additional Notes to determine the expressions for EP and EK (for student learning): ( ) ( )Py 1.25 ------ 1E mg h mg s=  = − Taking downwards as positive, ( ) 22 y y y 2 y 11 0 22 1 ------- 22 s u t a t gt s gt = + = + = ( )xxSince 2.97 or ---- 3 2.97 xs u t x t t=  = = Subs (2) & (3) into Eqn (1) ( )( ) ( ) 2 2 P 2 P 111.25 = 0.5 9.81 1.25 9.81 2 2.97 2 8.82 6.13 2.73 xxE mg g Ex  = − −    =− ( ) ( ) ( ) K P P K P 2 K 2 K Total Energy Initial Initial 2.25 6.13 6.13 2.73 2.25 2.73 E E E E E Ex Ex = − = + − = + − − =+ B1 Examiner’s Comments Averagely well done. The common mistake is that many students sketch the K.E. graph starting at 0 J and ending at the same value as the Initial EP. This in incorrect. Assuming no loss of energy to air resistance, the loss in G.P.E will be entirely converted to gain in K.E. Since the ball has an initial K.E., the final K.E. will be higher than the value of the Initial EP. (c) On Fig 1.1, draw the path of the ball if air resistance was not negligible. [2] L2 Solution: Shorter range Strike ground at a steeper angle B1 B1 Examiner’s Comments Mostly well done.

9 [Turn over 2 (a) State the principle of moments. …………………………………………………………………………………………….. …………………………………………………………………………………………….. [1] L1 Solution: For a body in equilibrium, sum/total of clockwise moments about a