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 cy