2021 SAJC H3 Physics Solution

Q1 (a) 4u(2000) = (4u+12u)Vcm Vcm= (4/16)2000 = 500 m/s [1] (b) [1] 𝑉𝛼,𝑐𝑚 = 2000 − 500 = 1500 𝑚𝑠−1 [1] sin(𝛼) 500 = sin(30) 1500 , 𝛼 = 9.59𝑜, total = 39.59 from horizontal 𝑉𝛼,𝐿𝑎𝑏 = √15002 + 5002 − 2 𝑥 1500𝑥500𝑥𝑐𝑜𝑠(180 − 9.59 − 30) = 1912 𝑚𝑠−1 [2] (c) 𝑉𝐶,𝑐𝑚 = 0 − 500 = −500 𝑚𝑠−1 [1] Angle between C and Vcm= 180-39.6-(180-39.6)/2 =70.2o (opposite side of alpha) 𝑉𝐶,𝑙𝑎𝑏 = √5002 + 5002 − 2𝑥500𝑥500 cos(39.59) = 338.7 𝑚𝑠−1 [2] Q2 (a) Let mass = M = 𝜌𝜋(4𝑅2 − 𝑅2) = 3𝜌𝜋𝑅2 Circular sheet can be taken as a complete disc with density of 𝜌 and radius 2R (mass of 4M/3), superimposed with a disc of R with density of – 𝜌 or mass of -M/3 [1] 30 1500 500 1500 39.59 30 𝑉𝛼,𝐿𝑎𝑏 =1912 500 VC, lab 500 500

X-position of CM = at vertical mid-line of circular disc Y-position of CM: Y = [(4𝑀 3 )2𝑅−(𝑀 3 )3𝑅] 𝑀 = 5𝑅 3 from bottom of circular sheet. [1] (b) (i) Path shows normal projectile/parabolic path of the CM with loops in the forward direction. [1] The separation of the loops in the x-directions are equal. [1] The separation of the loops in the y-direction are closer nearer the top. [1] (ii) The horizontal position x= 22.6cos(45)t+(7R/3)sin(wt). The vertical position y= 22.6sin(45)t +(7R/3)cos(wt) + 5R/3- ½ .9.81 t2 [1] Time of flight can be estimated by ignoring the 2nd and 3rd term y which are much smaller in magnitude Estimated time of flight = 2x 22.6sin(45) /9.81 = 3.3 s or 1.65 s to the highest point [1] The horizontal velocity is 22.6 cos(45) + 7𝑅(5𝜋) cos(5𝜋𝑡) /3 = 16+21.98cos(5𝜋𝑡) [1] Move backwards means horizontal velocity < 0 Solving for 16+21.98cos(5𝜋𝑡) < 0, for t values less than 1.65 s, T = 0.4 gives t = 0.152s, 0.152+0.4=0.552, 0.552+0.4=0.952, 0.952+0.4=1.35, 1.35+0.4=1.75 (exceed) So the outer most edge moves back 4 times before reaching the top [1] (iii) At t = 0.152 s, vertical velocity = 22.6sin(45) -(7Rw/3)sin(wt) -9.81 t = 16 -15.08 -2.43 m/s = -1.51 m/s [1] Q3 (a)(i) Speed at start = 2x3.14x1.496x1011/3.256x107=29768 m/s Speed of venus orbit = 2x3.14 x 0.7233x1.496x1011/1.941x107= 35009 m/s [1] Semi-major axis of transfer orbit = (1+0.7233)/2 AU = 1.289 x 1011 m [1]

𝐺𝑀𝑚 𝑟2 = 𝑚𝑟 ( 4𝜋2 𝑇2 ) , 𝐺𝑀 = 4𝜋2𝑟3 𝑇2 = 1.326 𝑥1020 At start of transfer 1 2 𝑚𝑣𝐸 2 − 𝐺𝑀𝑚 1.496𝑥 1011 = − 𝐺𝑀𝑚 2𝑥1.289𝑥1011 𝑣𝐸 = √2𝐺𝑀 ( 1 1.497𝑥1011 − 1 1.289𝑥1011) = 27276 m/s Reduction of speed at LEO = 29768 – 27276 = 2490 m/s [1] At destination: 1 2 𝑚𝑣𝑉 2 − 𝐺𝑀𝑚 0.7233𝑥1.496𝑥 1011 = − 𝐺𝑀𝑚 2𝑥1.289𝑥1011 VV = 37711 m/s, Reduction in speed at venus orbit = 37711 -35009 = 2700 m/s [1] (b) Transfer time = 1 2 √4𝜋2𝑎3 𝐺𝑀 = 1 2 √4𝑥3.142𝑥(1.289𝑥1011)3 √(𝐺𝑀) =1.26x107s = 146 days [1] (c) Venus turn through angle of 360x(1.26x107/1.941 x 107) = 233.7