Sound Wave Pulse - Physics - Solved Paper, Exams of Physics

Download Sound Wave Pulse - Physics - Solved Paper and more Exams Physics in PDF only on Docsity! 5 Phys 1121 T1 2004 Question 1. (16 marks) A scientist is standing at ground level, next to a very deep well (a well is a vertical hole in the ground, with water at the bottom). She drops a stone and measures the time between releasing the stone and hearing the sound it makes when it reaches the bottom. i) Draw a clear displacement-time graph for the position of the falling stone (you may neglect air resistance). On the diagram, indicate the depth h of the well and the time T1 taken for the stone to fall to the bottom. ii) Showing your working, relate the depth h to T1 and to other relevant constants. iii) The well is in fact 78 m deep. Take g = 9.8 ms-2 and calculate T1. iv) On the same displacement-time graph, show the displacement of the sound wave pulse that travels from the bottom to the top of the well. Your graph need not be to scale. v) Taking the speed of sound to be 344 ms-1, calculate T2, the time taken for the sound to travel from the bottom of the well to reach the scientist at the top. Show T2 on your graph. vi) State the time T between release of the stone and arrival of the sound. Think carefully about the number of significant figures. The scientist, as it happens, doesn't have a stop watch and can only estimate the time to the nearest second. Further, because of this imprecision and because she is solving the problem in her head, she neglects the time taken for the sound signal to reach her. For the same reason, she uses g ≅ 10 ms-2. vii) What value does the scientist get for the depth of the well? viii) Comment on the relative importance of the errors involved in (a) neglecting the time of travel of sound, (b) approximating the value of g and (c) measurement error. 6 Question 1. y t h T 1 T 2 T falling stone rising sound 1 2 (T + T ) T 2 y t -h T 1 T 2 T falling stone rising sound 1 2 (T + T ) T 2 (5 marks for diagram) i) Depending on the choice of origin, the graph might look like these. The algebra below is for the upper case. ii) y = yo + vyot + 1 2 ayt 2 = h + 0 − 12 gt 2 hits the bottom when y = 0, so 0 = h − 12 gT1 2 ∴ T12 = 2h g T1 = √ 2hg (4 marks) iii) h = 78 m → T1 = 4.0 s. (1mark) v) speed = distance travelled/time taken, so T2 = h/vs = 0.23 s. (1 mark) vi) T = T1 + T2 = 3.99 s + 0.227 s → Τ = 4.2 s. (2 significant figures) (2 marks, incl sig figs) vii) She says 0 = h − 12 gT1 2 ∴ h = 12 gT1 2 ≅ 12 gT 2. So she calculates h ≅ 12 (10ms -2)(4 s)2 = 80 m. (3 marks) viii) Her time estimate is 4.0 ± 0.5 s, an error of 13%. Neglecting the time for the sound to travel (6%) and taking 9.8 ≅ 10 (2%) are small errors by comparison. <She is lucky to have worked out an answer so close to the precise one.> (Any reasonable comment about the accuracy earns two marks.) (2 marks) 9 Question 3 (10 marks) R h car A toy racing car is placed on a track, which has the shape shown in the diagram. It includes a loop, which is approximately circular with radius R. The wheels of the car have negligible mass, and turn without friction on their axle. You may also neglect air resistance. The dimensions of the car are much smaller than R. i) Showing all working, determine the minimum height h from which the car may be released so that it maintains contact with the track throughout the trip. Question 3 R h v car i) v must be sufficiently great that the centripital force at the top of the loop at least equals the weight of the car. (Faster than this, a downwards normal force is required.) No non-conservative forces do work, so conservation of mechanical energy applies: Ui + Ki = Uf + Kf mgh + 0 = mg.2R + 1 2 mv 2 v2 = 2g(h − 2R) (4 marks) It loses contact when normal force = 0. N + mg = Fcentrip = mv2/R i.e. falls when mg = mv2/R v2 = gR (3 marks) Therefore it just falls off if h satisfies gR = v2 = 2g(h − 2R) 5gR = 2 gh ∴ hmin = 5R/2 (3 marks) 10 Question 4. (17 marks) i) An apple, attached to a tree a distance of 6370 km from the centre of the Earth, falls to the ground, and appears to accelerate at 9.80 ms-2. The average Earth-moon distance is 3.84×108 m. Making the approximation that the Earth is an inertial frame, using these two data and the inverse square law of gravitation, butwithout using a value for the gravitational constant G or the mass of the Earth, determine the period of the moon's orbit around the Earth. Express your answer in days. Give at least one reason why your answer might differ from a lunar month (29.5 days). i) ag = F m = const r2 When r = 6.37 106 m, ag ≅ g = 9.80 ms-2, so const = agr2 ≅ 3.98 1014 m3s-2. amoon = rmoonω2 = const rmoon2 Τ = 2π ω = 2π √rmoon3const = ... ≅ 27.4 days. (7 marks) This is approximately equal to the lunar month. However, ag is the acceleration in a frame of reference that accelerates around the Earth's axis of rotation. The real acceleration of the apple is greater than than this by rearthωearth2. Furthermore, during a month, the Earth moves ~360°/13 around the sun, so the lunar month is longer than T by about (14/13). (2 marks for either. Plus a bonus mark for anyone who gets both.) ii) The International Space Station has an orbital period of 91.8 minutes. The mass of the Earth is 5.98×1024 kg and its radius is 6.37×106 m. G = 6.673×10-11N m2 kg-2. From these data and the law of universal gravitation, determine the elevation of the station above the Earth and its speed. |F| = GMm r2 = macentrip = mrω2 r3 = GM ω2 = GMT2 22π2 r = √ 3 GMT2 22π2 = 6.74 106 m (6 marks) h = r − rEarth = 370 km. (1 mark) v = 2πr T = 7.7 km.s -1. (1 mark) 11 Question 5. (12 marks) a) M mv In a circus performance, a clown lies on his back with a brick, mass M, on his chest. An assistant uses a hammer with a mass m = 1.0 kg, to crack the brick. The head of the hammer is travelling vertically down at v = 20 ms-1. The mass of the handle is negligible. The collision between hammer and brick is of extremely short duration. However, because the brick cracks at the surface, the collision is completely inelastic. i) Derive an expression for the velocity V of the brick plus hammer immediately after the collision with the brick. ii) In an earlier part of the performance, a selection of audience members with different weights has stood on the clown's chest. The deformation of the chest is proportional to the weight of the person standing, and a 100 kg man produces a depression of 30 mm in his chest. Derive an expression for the spring constant of the clown's chest. iii) The Occupational Health and Safety Officer for the circus decides that the breaking brick trick should not depress the clown's chest more than 30 mm beyond the resting position of the brick before the collision. Derive a value for the required mass M of the brick. You may neglect the gravitational potential energy associated with deformation of the clown's chest. iv) Express your answer to part (iii) as an inequality. Describe the reason for the direction of the inequality. Caution. Do not try this exercise at home.