From Newton to Einstein. Mid-Term Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.

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1 Fom Newton to Einstein Mid-Tem Test, a.m. Thu. 3 th Nov. 008 Duation: 50 minutes. Thee ae 0 maks in Section A and 30 in Section B. Use g = 0 ms in numeical calculations. You ma use the following epessions fo vecto dot and coss poducts: u.v = u v + u v + u v u v = Section A: Answe ALL Questions ( u v u v, u v u v, u v u v ) A. State, giving easons, whethe the following quantities ae vectos o scalas: a) (,, 3) Answe: Vecto, because thee comma-sepaated scalas b) + + Answe: Scala, because scala quantities added d d d togethe. c) 0 ˆi + ˆj + kˆ Answe: Vecto, because thee unit vectos (o because i, j and k bold with hats on) [3] A. State Newton s Thid Law of Motion, and eplain what it sas about momentum. Answe: Action equals Reaction. It sas that momentum is conseved. [3] A3. A amp is at 45º to the hoiontal, and a mass m slides down at constant speed. a) Indicate on a sketch the foce eeted b gavit on the mass, and esolve it into vetical and hoiontal components. Give thei values. mg (mg) vet = mg (mg) hoi = 0 b) On a new sketch, indicate the components paallel and pependicula to the amp of the foce eeted b gavit on the mass. Give thei values. mg 7 = mg/ mg = mg/ mg mg 7

2 c) On a new sketch, indicate the foce eeted b the amp on the mass. Give its value. mg d) Calculate the coefficient of fiction between the mass and the amp. [7] Answe: Fictional foce F = mf nomal = mmg/. But fo eo acceleation, this must be equal and opposite to the paallel component of foce, mg 7 = mg/. Theefoe m = A4. A light stick of length m has kg point masses attached at the cente and at one end (end A). Find the moments of inetia about a) An ais pependicula to the stick and passing though the cente. Answe: I = m = 0 + = kg m b) An ais pependicula to the stick and passing though the othe end (end B) Answe: I = + = 5 kg m [3] A5. A planet mass M in a cicula obit aound a sta has an obital speed of v. Give epessions fo its kinetic eneg and gavitational potential eneg. [4] Mv GMm Answe: Using = whee m is the mass of the sta, the kinetic eneg is ½Mv = ½ GMm/. The gavitational potential U is GMm GMm such that du =, so that U = d

3 Section B: Answe ONE Question B. a) With the technolog available to the Ancient Geeks, how could the have found (i) The diamete of the Eath, Answe: Eatosthenes method (a well with the Sun diectl ovehead at noon, and a flagpole at a known distance awa to the Noth with a known length of shadow at the same moment (assume Sun is at infinit). OR distance to ship dopping below hoion. 3 (ii) The distance fom the Eath to the Moon, Answe: Sie and shape of Eath s shadow on Moon duing a luna eclipse. OR duation of luna eclipse epessed as faction of month. 3 (iii) The distance fom the Eath to the Sun? [9] Answe: Eatosthenes method (a well with the Sun diectl ovehead at noon, and a flagpole at a known distance awa to the Noth with a known length of shadow at the same moment (assume Sun is at infinit). OR distance to ship dopping below hoion. 3 b) A flwheel with a moment of inetia I = 0 kg m is mounted on a light ale. It is initiall stationa. A toque of 0 Nm is applied to the ale fo 00 s. (i) Calculate the final angula velocit of the flwheel. Answe: α = τ/i = ad s. ω = αt = 00 ad s. (ii) Calculate the final angula momentum of the flwheel. Answe: τ = dl/dt, so L = τt = 000 Nms OR L = Iω = 000 kg m s (iii) Calculate the final otational eneg of the flwheel. Answe: ½ Iω = J. (iv) If a toque of 5 N m is now applied about an ais pependicula to the ale of the flwheel, what will the motion be? Eplain ou answe.[] Answe: Goscope, theefoe pecession about the thid ais. Moe pecisel, if the ale is along, the toque is applied about, L initiall along gains 5 dt in the diection in a shot time dt. I.e. the ale moves ound b the angle dϕ = 5/000 dt, i.e. thee is a pecession angula velocit of ω p = ad s about the -ais.

4 c) The potential eneg of a nitogen atom in an ammonia (NH 3 ) molecule 4 vaies with position as U = 4. (i) Sketch U(). Give an epession fo the foce F on the nitogen atom and sketch F(). Answe: F = du/d = (ii) At what positions will the nitogen atom be in stable equilibium? Eplain ou answe. [9] Answe: The foce is eo at the maima and minima of the fist sketch, theefoe the molecule is in equilibium at these points. Howeve, onl at the two minima is the foce at small displacements a estoing foce, so these ae points of stable equilibium.

5 B. a) A pendulum consists of a kg bob attached to a high ceiling b a 0 m light ope. It is pulled to one side and eleased to swing back and foth. As the ope swings though the vetical, the speed of the bob is 4.4 m.s -. (i) What is the acceleation of the bob, in magnitude and diection, at this instant? Answe: The bob is following a cuvilinea path, of adius of cuvatue 0m at speed 4.4 m s at this instant, so its centipetal acceleation is v / = 00/0 = 0 m s. The tangential foce is eo at this instant, so the tangential acceleation is 0. So the acceleation is a vecto pointing staight upwads and of magnitude 0 m s. (ii) What is the tension in the ope at this instant? Answe: T = mg + ma = = 60 N. (iii) How close to the ceiling will the ball be at the end-points of the motion? [9] Answe: Eneg = PE + KE = ½ mv = 00 J at the bottom. It ises to a height at which KE = 0 and PE = mgh = 00 J. So h = 0 m. That is, the bob just touches the ceiling. b) A steel hoop weighing 5kg olls without slipping down a amp inclined at 30 to the hoiontal. Calculate its acceleation. [9] Answe: When the hoop olls along at a speed v, the KE fo its cente of mass (its tanslational KE) is ½ mv. At this speed, it is otating at ω = v/ and it has moment of inetia I = m (as all its mass is at adius ). So its otational KE is ½ Iω = ½ m (v/) = ½ mv. Total KE = mv. Suppose it has stated off stationa and olled down though a height h; then KE gained = PE lost = mgh. Fo a amp at 30, h = sin 30 d = ½ d whee d is the distance moved, so PE = ½ mgd. Now KE = F d, so F = ½ mg and v = gh = ½ gd. But fo linea acceleation we have v = u +as, with hee u = 0 and s = d. So v = ad = ½gd. Then a = ¼ g =.5 m s c) Given the planet of Question A5, if its speed v = 0 kms, to what speed must it be acceleated if it is to leave its cicula obit and escape to infinit? Does it matte in what diection it is acceleated? Give easons. [] Answe: In a cicula obit, centipetal acceleation v / = gavitational acceleation GM/, so kinetic eneg ½ mv = ½ GMm/. Gavitational potential eneg is GMm/, so total eneg is ½ GMm/. To escape to infinit equies that total eneg = 0, so it needs kinetic eneg of GMm/ twice as much. So v esc must be v ob = 4.4 km s. This discussion makes no mention of diection, so diection cannot matte (as long as it is not in the diection that leaves it tavelling along the adial diection diectl into the sun). End of Pape Pof. D.J. Dunstan

PHY4116 From Newton to Einstein. Mid-Term Test, 10a.m. Wed. 16 th Nov Duration: 50 minutes. There are 25 marks in Section A and 25 in Section B.

PHY4116 From Newton to Einstein. Mid-Term Test, 10a.m. Wed. 16 th Nov Duration: 50 minutes. There are 25 marks in Section A and 25 in Section B. PHY46 From Newton to Einstein Mid-Term Test, 0a.m. Wed. 6 th Nov. 06 Duration: 50 minutes. There are 5 marks in Section A and 5 in Section B. Use g = 0 ms in numerical calculations. You ma use the following