Displacement Time Graph - Higher Physics - Solved Past Paper, Exams of Physics

Download Displacement Time Graph - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 6 Phys 1131 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. 7 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) 10 Question 3 (13 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. ii) If a spherical ball is placed at the height calculated in part (i), will it maintain contact with the track? Explain your answer in a few clear sentences. 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) ii) For the marble, some of the initial potential energy is converted into rotational kinetic energy, so there is proportionlly less translational kinetic energy at the top of the loop, so its speed is less. So the centripital force required is less than the weight, so it falls off the track. (3 marks) 11 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). 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. 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) |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) 12 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. 15 i) ω v h h Rolling: point of application of friction stationary ∴ non-conservative forces do no work ∴ Uf + Kf = Ui + Ki 0 +    1 2 M v 2 + 1 2 Iω 2 = Mgh + 0 rolling ∴ ω = vR and write I = Mk 2 1 2 Mv 2 + 1 2 Mk 2 v 2 R2 = Mgh 1 2 v 2    1 + k 2 R2 = gh v = √2gh1 + k2/R2 (7 marks) ii) from (i), v decreases increasing ratio k/R khoop R = 1 > kdisc R = √ 12 > ksphereR = √ 25 ∴ vhoop < vdisc < vsphere ∴ sphere arrives before disc, which arrives before hoop. (3 marks) iii) During the descent, gravitational potential energy is converted into kinetic energy of two types: translational and rotational. The equation above shows that the ratio of rotational to translational kinetic energy is (k/R)2, so objects with large k/R ratios have proportionally less translational kinetic energy and so lower (translational) speed, all else equal. (2 marks) iv) the answer to part (i) does not depend on mass m or radius R, so the large and small spheres should travel equally quickly. (1 mark) v) The wheels of the car are the only part of it that rotates, and they are small, so only a small fraction of the initial gravitational potential energy is converted into rotational kinetic energy, and nearly all is converted into translational kinetic energy. This v will be larger for the car than for an object with a substantial fraction of its kinetic energy in rotation. (2 marks) vi) The soup wins. The spaghetti behaves nearly as a solid, so it rotates at the same speed as the can. Therefore the spaghetti has rotational kinetic energy. The soup, at least at first, does not rotate with the can. For the spaghetti, the initial gravitational potential energy is converted into translational kinetic energy plus rotational kinetic energy. For the soup, the same quantity of potential energy is converted into translational kinetic energy plus a much smaller quantity of rotational kinetic energy. So the soup has more translational kinetic energy and therefore travels faster. (3 marks) 16 L d ii) Having run out of fuel, a cosmonaut is initially a distance d = 100 m away from a rocket ship. The rocket ship has mass M, length L = 80 m and is approximately cylindrical. The astronaut is on the plane of symmetry, as shown. They are in deep space, and both are initally stationary with respect to an inertial frame. Due to gravitational attraction, the cosmonaut (whose mass and dimensions are negligible compared with those of the ship) moves towards the ship. Derive an expression for the time taken for the cosmonaut to reach the ship m M θy r x ii) dF = − Gm.dM r2 cos θ = − Gm.dM r2 cos θ iii) 17 ii) Australia Post is investigating the possibility of delivering mail to distant countries by drilling a hole through the Earth and simply dropping the mail through. Consider a straight hole dug through the Earth and passing through the centre. Showing your working, calculate how long it would take for an object to pass from one end of the hole to the other under the influence of gravity alone. 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. For your calculation, assume that the Earth's density is uniform. (Hint: you may use without proof the shell theorem.) r R <Shell theorem: the gravitational force due to a thin, uniform spherical shell on another mass is zero if the other mass is inside the shell. If the other mass is outside, it equals the force exerted by a point mass equal to the mass of the shell, but concentrated at its centre.> When the delivered mail (mass m) is at radius r, the mass of all the shells with smaller radius is Mr = ρ. 4 3 πr3 so the force exerted on m at r is Fr = − G mρ.43 π r 3 r2 = − Kr where K = Gmρ.43 π ∴ motion is SHM with ω = √ Km Τ = 2π ω = 2π √Gρ.43 π = 2π √GM/R3 = 3π Gρ = .... = 84 minutes ∴ falls through (one half cycle) in 42 minutes (actually faster for real density profile) b)