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.

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1 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 epressions for vector dot and cross products: u.v u v u v u v u v Section A: Answer ALL Questions u v u v, u v u v, u v A. State, giving reasons, whether the following quantities are vectors or scalars: (,, 3) Answer: Vector, because three comma-separated scalars [] a) f f f Answer: Scalar, because scalar quantities added together. d d d [] b) 0ˆi ˆj kˆ Answer: Vector, because three unit vectors added (or because i, j and k bold with hats on) [] [3] A. State Newton s Third Law of Motion, and eplain what it sas about momentum. Answer: Action equals Reaction. [] It sas that momentum is conserved [] [3] A3. A ramp is at 30º to the horiontal, and a mass m slides down it with no friction. a) Indicate on a sketch the force eerted b gravit on the mass, and resolve it into vertical and horiontal components. Give their values. u v mg (mg) vert = mg (mg) hori = 0 [] b) On a new sketch, indicate the components parallel and perpendicular to the ramp of the force eerted b gravit on the mass. Give their values.

2 mg par = mg/ mg perp = m 3 [] mg par mg perp c) On a new sketch, indicate the force eerted b the ramp on the mass. Give its value. mg perp mg perp = m 3 [] d) Calculate the acceleration of the mass. Answer: a = g perp = ½g = 5 m s [] [7] A4. A light stick of length m has kg point masses attached at the centre and at one end (end A). Find the moments of inertia about a) An ais perpendicular to the stick and passing through the centre. Answer: I = mr [] = 0 + = kg m [] b) An ais perpendicular to the stick and passing through the other end (end B) Answer: I = + = 5 kg m [] [3] A5. A biccle and its rider weigh 80kg. a) The ride km up a road with gradient in 0, to an altitude of 00m. Calculate the gravitational potential energ. Answer: U = mgh = = J [] b) The descend without using the brakes. Neglecting friction and air resistance, calculate the kinetic energ and velocit at the end of the descent. Answer: KE at end = PE at start = mgh, [] so ½mv = mgh, v gh [] 00 = 80 m s [] [4]

3 A6. A planet mass m in a circular orbit around a star of mass M has an orbital speed of v. Give epressions for, a) The radius R of its orbit, mv GMm Answer: Using centripetal force = centrifugal force,, [] then r r GMm GM r [] mv v mv GMm b) Its kinetic energ, Answer: Using where m is the mass of r r the star, the kinetic energ is ½Mv = ½ GMm/r. [] c) Its gravitational potential energ. du GMm Answer: The gravitational potential U is such that dr, [] so that r GMm U [] r [5] Paper continues

4 Section B: Answer ONE Question B. a) With the technolog available to the Ancient Greeks, how could the have found (i) The diameter of the Earth, Answer: Eratosthenes method (a well with the Sun directl overhead at noon, and a flagpole at a known distance awa to the North with a known length of shadow at the same moment (assume Sun is at infinit). OR distance to ship dropping below horion. [3] (ii) The distance from the Earth to the Moon, Answer: Sie and shape of Earth s shadow on Moon during a lunar eclipse. OR duration of lunar eclipse epressed as fraction of month. [] (iii) The distance from the Earth to the Sun? Answer: Triangulate at midmonth, when the moon is an eact semicircle E M S is then eactl 90 degrees. Observation will give the angle M E S as a little under 90 degrees, from which and knowing the distance to the moon, the distance of the Sun is readil obtained. [3] [8] b) A flwheel with a moment of inertia I = 5 kg m is mounted on a light ale. It is initiall stationar. A torque of 0 Nm is applied to the ale for 00 s. (i) Calculate the final angular velocit of the flwheel. Answer: = /I = rad s. [] = T = 00 rad s. [] (ii) Calculate the final angular momentum of the flwheel. Answer: = dl/dt, [] so L = T = 000 Nms OR L = I = 000 kg m s. [] (iii) Calculate the final rotational energ of the flwheel. Answer: ½ I []= 0 5 J. [] (iv) If a torque of 0 N m is now applied about an ais perpendicular to the ale of the flwheel, what will the motion be? Eplain our answer. Answer: Groscope, therefore precession about the third ais. [] More precisel, if the ale is along, the torque is applied about, L initiall along gains 5 dt in the direction in a short time dt. I.e. the ale moves round b the angle d = 5/000 dt, i.e. there is a precession angular velocit of p = rad s about the -ais. [] [8]

5 c) The potential energ of a nitrogen atom in an ammonia (NH 3 ) molecule 4 varies with position as U. (i) Sketch U(). Give an epression for the force F on the nitrogen atom and sketch F(). Answer: F = du/d = (ii) At what positions will the nitrogen atom be in stable equilibrium? Eplain our answer. Answer: The force is ero at the maima and minima of the first sketch, [] therefore the molecule is in equilibrium at these points. [] However, onl at the two minima is the force at small displacements a restoring force, so these are points of stable equilibrium. [] [9] B. a) A pendulum consists of a kg bob attached to a high ceiling b a 0 m light rope. It is pulled to one side and released to swing back and forth. As the rope swings through the vertical, the speed of the bob is 4.4 m.s -. (i) What is the acceleration of the bob, in magnitude and direction, at this instant? Answer: The bob is following a curvilinear path, of radius of curvature 0m at speed 4.4 m s at this instant, [] so its centripetal acceleration is v /r = 00/0 = 0 m s. [] The tangential force is ero at

6 this instant, so the tangential acceleration is 0. [] So the acceleration is a vector pointing straight upwards and of magnitude 0 m s. [] (ii) What is the tension in the rope at this instant? Answer: T = mg + ma = = 60 N. (iii) How close to the ceiling will the ball be at the end-points of the motion? Answer: Energ = PE + KE = ½ mv = 00 J at the bottom. [] It rises to a height at which KE = 0 and PE = mgh = 00 J. [] So h = 0 m. [] That is, the bob just touches the ceiling. [] [9] b) A steel hoop weighing 0.5kg rolls without slipping down a ramp inclined at 30 to the horiontal. Calculate its acceleration. Answer: When the hoop rolls along at a speed v, the KE for its centre of mass (its translational KE) is ½ mv. [] At this speed, it is rotating at = v/r and it has moment of inertia I = mr (as all its mass is at radius r) []. So its rotational KE is ½ I = ½ mr (v/r) = ½ mv. Total KE = mv []. Suppose it has started off stationar and rolled down through a height h; then KE gained = PE lost = mgh []. For a ramp at 30, h = sin 30 d = ½ d where d is the distance moved, so PE = ½ mgd []. Now KE = F d, so F = ½ mg and v = gh = ½ gd []. But for linear acceleration we have v = u +as [], with here u = 0 and s = d. So v = ad = ½gd. Then a = ¼ g =.5 m s [] [8] c) Given the planet of Question A5, if its speed v = 0 kms -, to what speed must it be accelerated if it is to leave its circular orbit and escape to infinit? Answer: In a circular orbit, centripetal acceleration v /r = gravitational acceleration GM/r, so kinetic energ ½ mv = ½ GMm/r []. Gravitational potential energ is GMm/r, so total energ is ½ GMm/r []. To escape to infinit requires that total energ = 0 [], so it needs kinetic energ of GMm/r twice as much []. So v esc must be v orb = 4.4 km s []. Does it matter in what direction it is accelerated? Give reasons. This discussion makes no mention of direction [], so direction cannot matter [] (as long as it is not in the direction that leaves it travelling along the radial direction directl into the sun) [8] End of Paper Prof. D.J. Dunstan

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.

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. 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