2020 EJC J2 H3 Prelim SS

Page 1 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution EJC H3 Physics 2020 Preliminary Examination Marks Scheme and Solution Question Solution Marks 1 (a) Change in displacement of B relative to A = ቚቀ30000sin30°+20000sin10° 30000 cos 30° − 20000 cos 10°ቁቚ = 19512 𝑚 OR Using cosine rule, 𝐵ଵ𝐵ଶ =ඥ30ଶ+ 20ଶ− 2(20)(30)cos (100° − 60°)= 19.512 𝑘𝑚 𝑠𝑝𝑒𝑒𝑑=𝑠 𝑡 =19512 3600=5.42 𝑚 𝑠ିଵ M1 (M1) A1 (b) Let F be the point of closest approach for A and B. Using sine rule: sin 𝜃 20 =sin 40° 19.51 𝜃 = 41.2° 𝐵ଵ𝐹 = 30 cos 41.2° = 22.5685 𝑘𝑚 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 =22568.5 5.42 = 4163.9𝑠 = 69 min 24 𝑠 𝑓𝑟𝑜𝑚 12 𝑝𝑚 Time they are the closest is 1309HR C1 C1 A1

Page 2 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution (c) ∠𝐵ଵ𝐴𝐹=180°−90°−41.2°=48.8° 𝐵𝑒𝑎𝑟𝑖𝑛𝑔=60°+48.8°=108.8° 𝑠𝑝𝑒𝑒𝑑 = 𝐴𝐹 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛=30000 sin 41.2° 9 × 60 = 36.6 𝑚 𝑠ିଵ Also accept if student use time taken as 9×60+24=564 𝑠 and gets speed as 35.0 m s-1. A1 A1 Question Solution Marks 2 (a) (i) The expression for the moment of inertia of a sphere can be developed by summing the moment of infinitely thin disks about the z axis. 𝑑𝐼 =1 2𝑦ଶ𝑑𝑚 =1 2𝑦ଶ𝜌𝑑𝑉 =1 2𝑦ଶ𝜌𝜋𝑦ଶ𝑑𝑧 𝐼஼ெ=1 2𝜌𝜋න 𝑦ସ𝑑𝑧 ோ ିோ =1 2𝜌𝜋න (𝑅ଶ− 𝑧ଶ)ଶ𝑑𝑧 ோ ିோ = 8 15𝜌𝜋𝑅ହ =2 5𝑀𝑅ଶ Accept summing up spherical hollow shells. Clearly defined variables Correct definition of dI Correct algebra leading to final expression B1 B1 B1 (ii) Using parallel axis theorem 𝐼௉ =𝐼஼ெ+𝑀𝑑ଶ 𝐼௉ =2 5𝑀𝑅ଶ+𝑀(𝐿+𝑅)ଶ M1 A1 (b) (i) 𝜏=−𝑀𝑔(𝐿+𝑅)sin𝜃 (no marks if negative sign is omitted.) A1 (ii) Using Newton’s 2nd law for rotation: 𝜏=𝐼𝛼 −𝑀𝑔(𝐿+𝑅)sin𝜃=൤2 5𝑀𝑅ଶ+𝑀(𝐿ଶ+2𝐿𝑅+𝑅ଶ൨𝛼 𝛼 = 𝑀𝑔(𝐿 + 𝑅)sin 𝜃 𝑀ቀ2 5𝑅ଶ+ 𝑅ଶ+ 2𝐿𝑅 + 𝐿ଶቁ = − 𝑔(𝐿 + 𝑅)sin 𝜃 ቀ7 5𝑅ଶ+ 𝐿ଶ+ 2𝐿𝑅ቁ B1 A1

Page 3 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution (iii) θ must be small such at sin𝜃≈𝜃 𝛼=− 𝑔(𝐿+𝑅) ቀ7 5𝑅ଶ+ 𝐿ଶ+ 2𝐿𝑅ቁ sin𝜃 𝛼 = − 𝑔(𝐿 + 𝑅) ቀ7 5𝑅ଶ+ 𝐿ଶ+ 2𝐿𝑅ቁ 𝜃 𝜔ଶ = 𝑔(𝐿 + 𝑅) ቀ7 5𝑅ଶ+ 𝐿ଶ+ 2𝐿𝑅ቁ 𝑇 = 2𝜋ඩ 7 5𝑅ଶ+ 𝐿ଶ+ 2𝐿𝑅 𝑔(𝐿 + 𝑅) =2𝜋ඨ 7 5(0.30)ଶ+ 1ଶ+ 2(0.30)(1) 9.81(1+0.30) =2.31 𝑠 M1 M1 A1

Page 4 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution Question Solution Marks 3 (a) Using Gauss Law, ර𝐸ሬ⃑∙𝑑𝐴⃑ =𝑞௘௡௖௟௢௦௘ௗ 𝜀଴ Using a cylindrical Gaussian surface, with 𝑟 < 𝑅 𝐸(2𝜋𝑟𝐿)=𝜌𝜋𝑟ଶ𝐿 𝜀଴ 𝐸= 𝜌𝑟 2𝜀଴ M0 A1 (b) Using Gauss Law, ර𝐸ሬ⃑∙𝑑𝐴⃑ =𝑞௘௡௖௟௢௦௘ௗ 𝜀଴ Using a cylindrical Gaussian surface, with 𝑅 < 𝑟 𝐸(2𝜋𝑟𝐿)=𝜌𝜋𝑅ଶ𝐿 𝜀଴ 𝐸= 𝜌𝑅ଶ 2𝜀଴𝑟 M1 A1 (c) Linear graph from 0 < r < R and 1/r graph for R < r Labelling of coordinates of max point B1 B1 (d) Let 𝑟⃑ locate a point within the hole, relative to the axis of the cylinder and let 𝑟⃑′ locate this point relative to the axis of the hole. Let 𝑏ሬ⃑ locate the axis of the hole relative to the axis of the cylinder. As shown in diagram, 𝑟′ሬሬ⃑ = 𝑟 ⃑ − 𝑏ሬ⃑ 𝐸ሬ⃑௢௙௙ି௔௫௜௦=𝜌 𝑟′ሬሬ⃑ 2𝜀଴ =𝜌(𝑟 ⃑ − 𝑏ሬ⃑) 2𝜀଴ 𝐸ሬ⃑௛௢௟௘= 𝐸ሬ⃑௖௬௟௜௡ௗ௘௥− 𝐸ሬ⃑௢௙௙ି௔௫௜௦ = 𝜌𝑟 ⃑ 2𝜀଴ −𝜌൫𝑟 ⃑ − 𝑏ሬ⃑൯ 2𝜀଴ = 𝜌𝑏ሬ⃑ 2𝜀଴ Direction: Along the direction from centre of cylinder to centre of hole. B1 B1 (e) Cylindrical Gaussian surface about central axis encloses no charge (based on equation in (d), as 𝑏ሬ⃑ =0ሬ⃑) Electric field strength is zero. M1 (M1) A1

Page 5 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution Question Solution Marks 4 Using 𝑣=𝐿ௗ௜ ௗ௧ Instantaneous power supplied to initiate the current in the inductor is 𝑃 = 𝑖𝑣 = 𝐿𝑖𝑑𝑖 𝑑𝑡 𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 =න 𝑃 𝑑𝑡 ௧೑ ଴ =න 𝐿𝑖𝑑𝑖 𝑑𝑡 𝑑𝑡 ௧೑ ଴ =න 𝐿𝑖 𝑑𝑖 ூ೑ ଴ =൤1 2𝐿𝑖ଶ൨ ଴ ூ೑ =1 2𝐿𝐼௙ଶ B1 M1 C1 A1

Page 6 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution Question Solution Marks 5 (a) The total energy of a moving satellite m under the influence of the gravitational field due to the Earth of mass M is given by: 𝑇𝐸 = 𝐾𝐸 + 𝐺𝑃𝐸 =1 2𝑚𝑣ଶ−𝐺𝑀𝑚 𝑟 Since the satellite have an ellipsoidal orbit, 𝑣ଶ = 𝑣௧௔௡௚௘௡௧ଶ + 𝑣௥௔ௗ௜௔௟ଶ Since satellite is in a central force field 𝜏 ⃑ = 𝑟 ⃑ × 𝐹⃑ = 0ሬ⃑ and resulted in 𝐿 = 𝑚𝑟𝑣௧ Therefore, sub in the about equations and simplifying, 𝑇𝐸 =1 2𝑚𝑣௧ଶ+1 2𝑚𝑣௥ଶ−𝐺𝑀𝑚 𝑟 =1 2𝑚𝑣௥ଶ+ 𝐿ଶ 2𝑚𝑟ଶ−𝐺𝑀𝑚 𝑟 B1 B1 A0 (b) (i) Kepler’s first law states that all planets move in elliptical orbits with the Sun at one focus of the ellipse. B1 (ii) At turning points, 𝑣௥ =0 𝐸 = 𝐿ଶ 2𝑚𝑟ଶ−𝐺𝑀𝑚 𝑟 𝐸𝑟ଶ+ 𝐺𝑀𝑚𝑟 −𝐿ଶ 2𝑚= 0 𝑟ଶ+𝐺𝑀𝑚 𝐸 𝑟 −𝐿ଶ 2𝑚𝐸= 0 By comparing coefficients, 𝐴 =𝐺𝑀𝑚 𝐸 𝐵=− 𝐿ଶ 2𝑚𝐸 M1 A1 A1 (iii) Since the roots of the equation in (b)(ii) are 𝑟௔ and 𝑟௣, ൫𝑟−𝑟௣൯(𝑟−𝑟௔)=0⇒𝑟ଶ−൫𝑟௣+𝑟௔൯𝑟+𝑟௔𝑟௣ =0 Comparing coefficients, −ீெ௠ ா = (𝑟௔+ 𝑟௣) 𝐸 = −𝐺𝑀𝑚 ൫𝑟௔+ 𝑟௣൯ = −(6.67 × 10ିଵଵ)(5.97 × 10ଶସ)(10.0) (25 + 35)× 10଺ = −6.64 × 10଻𝐽 (accept if algebraic method gives correct final answer) C1 C1 C1 A1

Page 7 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution Question Solution Marks 6 (a) v = f 340 = (20 × 103) or 340 = (80 × 103)  = 17 mm or  = 4.3 mm 4.3 mm to 17 mm. (also accept 12.8 mm for the range in between) M1 A1 (b) (i) Displacement at L = 1.5 – 0.7= 0.8 (units) Displacement at M = 0 (units) since the two curves intersect Displacement at N = 0 – 1.0 = –1.0 (units) Two answers correct All three answers correct B1 B1 (ii) Correct line through points L, M and N Complete waveform drawn (symmetrical about the vertical line through N) B1 B1 (c) (i) Value of Δf = 51.25 - 50.80 = 0.45 kHz and f = 50.80 kHz c v f f 2 . v . 0 45 2 50 80 340 Insect’s speed, v = 1.51 m s–1 C1 A1 (ii) Increase in frequency due to more frequent reflection of wave. (sound wave that is being reflected off the insect is being compressed) Towards bat. B1 (B1) B1 (iii) Any reasonable factor that interfere with the reflected waves - Echo from surrounding objects - Presence of multiple bats using the same frequency sound waves B1 wave P

Page 8 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution - Evolved insects that produces sonic clicks to jam echolocation. (iv) Higher frequency wavelength of wave is smaller. More sensitive to small change in frequency due to Doppler’s effect. B1 B1 (d) (i) From Fig 6.4: Graph of ln I against x is a straight line with negative gradient, equation of the straight line is : ln I = mx + ln Io where m is the gradient and ln Io is the y-intercept x o mln       I I  xme oI I or xm oeII This equation is of the form of x oe II where m = -  B1 B1 B1 (ii) Using Fig. 6.4, gradient = 7.0657.000.175.2 15.200.1   Hence α = 0.7 m-1 B1 (iii) Density of air. Denser air leads to smaller attenuation coefficient. B1 B1

Page 9 of 15 9814(2020) H3 Physics Preliminary Examination (MS) and solution Question Solution Marks 7 (a) (i) Consider a hollow cylindrical shell with radius 𝑟, and height L 𝑑𝑚 = 2𝜋𝜌𝐿𝑟 𝐼 =න𝑟ଶ 𝑑𝑚 =න 2𝜋𝜌𝐿𝑟ଷ𝑑𝑟 ோ೚ೠ೟ ோ೔೙ =൤𝜋𝜌𝐿 2 𝑟ସ൨ ோ೔೙ ோ೚ೠ೟ =𝜋𝐿 2 ቆ 𝑀 𝜋൫𝑅௢௨௧ଶ − 𝑅௜௡ଶ ൯𝐿ቇ ൫𝑅௢௨௧ସ − 𝑅௜௡ସ ൯ =𝜋𝐿 2 ൬𝑀 𝜋𝐿൰൫𝑅௢௨௧ଶ + 𝑅௜௡ଶ ൯ =1 2𝑀(𝑅௢௨௧ଶ +𝑅௜௡ଶ ) M1 M1 A1 (ii) For a solid cylinder, Rin = 0 𝐼௦௢௟௜ௗ=1 2𝑀𝑅௢௨௧ଶ A1 (b) (i) All force correctly labelled Relative length of vector arrows are correct (W all same and W > T2) T2 >T1 Horizontal component of reaction force by axel equals T1. B1 B1 B1 B1 (ii) T2 >T1 so that there is resultant clockwise torque acting on the pulley This ensures that the string does not slip on the pulley. B1 B1 (iii) From the block: 𝑀𝑔−𝑇ଶ =𝑀𝑎 --- (1) From the cylinder: 𝑇ଶ𝑅−𝑇ଵ𝑅=ଵ ଶ𝑀𝑅ଶ𝛼 --- (2)

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