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 [1] Spacecraft from Earth turns through 180 degree Angle between Earth-Venus at start = 233.7 -180 = 53.7o [1] 4 (a) 𝑉𝑐 = 𝐸 (1 − exp (− 𝑡 𝑅𝐶)) 𝐸 2 = 𝐸 (1 − exp (− 𝑡 𝑅𝐶)) Simplifies to t = RC ln(2) [2] (b) (i) Effective inductance = 2L + 2M [1] 𝑓 = 1 2𝜋√(2𝐿+2𝑀)𝐶 [1] (ii) (2𝐿 + 2𝑀) ( 𝑑𝐼 𝑑𝑡) + 𝐼𝑅 + 𝑞 𝐶 = 0 [1] (2𝐿 + 2𝑀) ( 𝑑2𝑞 𝑑𝑡2) + 𝑅 ( 𝑑𝑞 𝑑𝑡) + 𝑞 𝐶 = 0

Substitute 𝑞 = 𝑄𝑜 exp(−𝛾𝑡) sin (𝜔𝑡 + 𝜙) into above equation: [1] Group terms in cos: 𝛾 = 𝑅 2(2𝐿+2𝑀) [1] Group terms in sin: 𝛾2 − 𝜔2 − 𝑅 2𝐿+2𝑀 𝛾 + 1 (2𝐿+2𝑀)𝐶 = 0 𝜔 = √ 1 (2𝐿+2𝑀)𝐶 − 𝑅2 4(2𝐿+2𝑀)2 [1] 5(a) E=½ CV2= 1 2 𝜖𝑜𝐴 𝑑 𝑉2 [1] = 6.91 x 10-7 J [1] (b) 𝐸 = 1 2 𝜖𝑜𝐴 𝑑 𝑉2 ∝ 1 𝑑 , new E = 6.91 x 10-7/2 = 3.45 x 10-7J [1] Less energy stored. Work done by hand and energy lost by capacitor because charge flows back to charge battery. [1] (c) Energy = ½ Q2/C = ½ (CoV)/( Co/2) = 1.38 x 10-6J [1] The increase in energy is accounted for by the work done in pulling the capacitor plates. [1] (d) Energy stored decrease because of higher capacitance [1] Energy stored reduces because work is done by the capacitor plates to attract the dielectric into the plates. [1] (Work has to be done to pull out the dielectric, to restored back the energy stored in the capacitor) (6)(a)(i) The wavelength of sound sent out by siren = c/f The speed of sound relative to the observer = u + c Frequency heard by observer = (u+c)/(c/f) = (u+c)f/c [2] (ii)As the motorcyclist if far away from the siren, its velocity is almost parallel to the velocity of sound and add on to it. So the observed frequency is higher than f by (c+u)f/c [1] As the motorcyclist approaches the siren, the component of velocity in the direction of the wave becomes smaller 𝑢𝑐𝑜𝑠(𝜃), 𝑤ℎ𝑒𝑟𝑒 𝜃 is the angle between the original direction and the current position. So the frequency is now (c+ 𝑢𝑐𝑜𝑠(𝜃)𝑓/𝑐, which decreases as 𝜃 increase [1] As 𝜃 becomes 90, the observed frequency drops to f. [1] As 𝜃 increase beyond 90, the relative speed decrease to c-| 𝑢𝑐𝑜𝑠(𝜃)|, so the frequency decrease till (c-u)f/c [1]

(b) In unit time, if the source is moving with a speed vs, the f waves occupy a distance of (c- vs), so the wavelength is now shorter in value, given by (c- vs)/f. [1] The speed of sound remains the same. So the observed frequency = c/[(c- vs)/f] or 𝑐 𝑐−𝑣𝑠 𝑓 [1] (c) Answer: 𝑓𝑜 = 𝑐+𝑢 𝑐−𝑣𝑠 f (d) The velocity of air relative to ground = vw Velocity of sound sound relative to air = c Therefore velocity of sound wave relative to ground = vw+ c [1] So have to replace c by vw+ c in the expression in part (c) [1] (e) 𝚫𝒇 𝒇 = 𝒇𝒐−𝒇 𝒇 = [( 𝒄 𝒄−𝒗𝒔 )𝒇−𝒇] 𝒇 = 𝟏 𝟏−𝒗𝒔 𝒄 − 𝟏 = (𝟏 + 𝒗𝒔 𝒄 + 𝒉𝒊𝒈𝒉𝒆𝒓 𝒕𝒆𝒓𝒎𝒔) − 𝟏 = 𝒗𝒔 𝒄 [1] (ii) 𝛿𝑓 𝑓 = 𝑣 𝑐 = 2𝜋𝑅 𝑇𝑐 [1] Or 𝑓 = 2𝜋𝑅 𝑇𝑐 𝑓 . The side moving towards Earth will have f+ 𝛿𝑓, and the side moving away from the Earth has 𝑓 − 𝛿𝑓 [1] (iii) Volume = 4𝜋𝑅3 3 = 2.0 𝑥 1030 1.4 𝑥 103 , R = 6.99 x 108 [1] δf/f = 6.91 x 10-6 [1] (iv) Line width = 2Δ𝜆, 2Δλ 𝜆 = 2Δ𝑓 𝑓 = 2𝑣 𝑐 = 8 𝑥 10−5, [1] v = 12000 m/s =√3𝑘𝑇 𝑚 , T = 5800 K [1] 7(a) Take pivot at contact point: 𝑎𝑐𝑚 𝑎 = 2𝑟 3𝑟 , 𝑎𝑐𝑚 = 2𝑎/3 [1] (b) Take contact as pivot, 𝛼 = 𝑎𝑐𝑚 2𝑟 = 2𝑎 3 2𝑟 = 𝑎 3𝑟 = 13.3a [2] (if take CM as pivot, 𝛼 = 𝑎−2𝑎 3 𝑟 = 𝑎 3𝑟) (c) 2Mg -T = 2Ma force equation for M [1] T + f – Mg sin(30) = M(2a/3) force equation for yoyo [1]

(d) Net torque about centre of mass = I x angular acceleration [1] Tr – 2fr = Mr2(a/(3r)) [1] (e) Solving 3 equations T =16 Mg/23 = 0.696 Mg a = 15g/23 = 0.652 g f = 51Mg/46 = 0.239 Mg [3] (f) f = 𝜇𝑚𝑔𝑐𝑜𝑠(30) [1] 𝜇 = 11 23√3 = 0.276 [1] (g) 𝛼 = 𝑎 3𝑟 = 0.652𝑔 3𝑟 = 85.3 [1] 𝜃 = 1 2 𝛼𝑡2 = 10.7 𝑟𝑎𝑑 [1] (h) (i) Translational KE + rotational KE = mgh Consider centre of mass as reference: v = 2x2.5/3, 𝜔 = 2.5/3𝑟 1 2 𝑀 ( 2𝑥2.5 3 ) 2 + 1 2 𝑀𝑟2 ( 2.5 3𝑟) 2 = 𝑀𝑔ℎ [1] Solving gives h = 0.177 m [1] (ii) Distance travelled by centre of mass along slope = 0.177/sin(30) = 0.354 m [1] 𝑆 = 𝑣+𝑢 2 𝑡 = 0+2𝑥2.5 3 2 𝑡 , t = 0.425 s [1] (iii) Linear retardation of centre of mass = (2𝑥2.5 3 ) 2 2𝑥0.354 = 3.92 [1] 𝛼 = 3.92 2𝑟 = 3.92 0.050 = 78.4 𝑟𝑎𝑑𝑠−2 [1] Or use 𝛼 = Δ𝜔 𝑡 = 0− 2.5 3𝑥0.025 0.425 =78.4 8(a)(i)Treat the cylinder as a summation of I flowing out of page in the outer cylinder superimposed on a current flowing into the page in the hole. The B-field due to outer cylinder inside the hole circulates in anticlockwise with magnitude that