Constant Acceleration - Higher Physics - Solved Past Paper, Exams of Physics

Download Constant Acceleration - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! PHYS 1131 T1, 2007 UNSW Question 1 a) Your car is stopped on the side of a straight highway. The street is busy: many cars are going past, all of them at 100 km per hour. It's a sunny day and the solar cells are clean, so assume that your car accelerates at a constant rate a = 1.9 m.s−2 from zero until it reaches a final speed vf = 100 km per hour. You need a long gap between cars to be able to accelerate to 100 km per hour to join the stream of the traffic. In this question you will work out how long. i) Sketch a displacement - time or x(t) graph showing the position of a car accelerating from rest to vf at constant acceleration a, and then continuing at constant speed vf. ii) On the same graph, show the displacement of a following car. This is a car, which travels at a constant speed vf at all times and which, when your car has finished accelerating, is a safe distance L behind yours. Show L clearly on the graph. iii) Below or above your displacement graph, and using the same scale for the time axis, sketch a velocity - time (v(t)) graph for the two cars. Show vf on the graph. iv) Some authorities judge that, in good conditions, the safe distance L between cars on a highway is the distance travelled by a car in 2 seconds*. What is L for this case? v) Assume that you start accelerating when the car ahead is a distance L in front of you and finish accelerating when the car behind (which travels at constant speed vF) is a distance L behind you. (It is not required, but it may help to draw the x(t) for the car ahead of yours, as well.) Showing your working, calculate the minimum necessary distance between the car in front of you and the car behind you. b) With respect to the ground, the wind is blowing from the North East at speed vw = 15 km per hour. You are bicycling South at speed v (with respect to the ground). i) Relative to you, the wind is coming directly from the East. Determine your speed v. ii) Relative to a second cyclist, also travelling South, the wind is coming directly from the SouthEast. Determine her speed (with respect to the ground). * A comment for street safety but not for marks. The safe distance is a minimum for good conditions. In poor visibility, leave larger gaps. The timing requires judgment and the acceleration 1.9 m.s−2 over this distance is not always achievable. Consider this calculation as an underestimate. Question 1 i-iii) t t x t1 L x you v you v x behind x ahead L v f d1 d2 (not required) v behind v behind v you iv) vf = 100 km/hr = 105 m 3600 s = 28 m.s-1. L = (2s).vf = 56 m. v) Distance d1 travelled by you while accelerating: 2ad1 = vf2 − 02 d1 = vf2/2a Time taken to reach vf: t1 = vf/a Distance travelled by following car in that time d2 = t1vf = vf2/a Minimum initial separation between you and following car = d2 − d1 + L Add L for car in front: minimum gap = d2 − d1 + 2L = vf2/2a + 2L = 310 m. (which is why there are speed matching lanes on freeway entrances) Attention marker: some candidates may just write somethink like gap = d1 + 2L. This gives the correct numerical value but loses marks as indicated. b) v w v'w v 45°N v w v'w v 45°N 45° Let the relative wind velocity be v'w vw = v + v'w as shown v = vw sin 45° = 15 km per hour * sin 45° = 11 km per hour. As before, vw = v + v'w v = 2 vw sin 45° = 21 km per hour. Question 3 H x pop! spring counter balance cable lift As a safety precaution, you decide to install a large spring in the bottom of a lift shaft. (A lift is the same thing as an elevator. The shaft is the volume in which it travels.) The mass of the spring is negligible compared to the mass M of the lift (and, happily, there are no passengers in the lift for this problem). Assume that the spring constant is k, and the spring obeys Hooke's law for the range considered in this problem. Suppose that bottom of the lift is a distance H above the spring when the cable breaks, while the lift is travelling downwards at speed vi. It then falls and hits the spring. Air resistance and friction are negligible. i) Explaining your reasoning, derive an expression for the maximum compression xm of the spring. To simplify the algebra, you may assume that H >> xm and that d > xm. ii) Briefly explain why the dimensions (or units) in your equation are correct. iii) When will the acceleration of the lift be greatest? Explain your answer briefly. You may assume that kxm > 2 Mg. iv) Derive an expression for the greatest acceleration amax in terms of the parameters of the problem. v) Consider the case of a lift falling from rest from height H above the spring. How long must d be so that amax ≤ 5g? (Hint: use your results for (i) and (iv)) Question 3 i) During both the fall and during the compression of the spring, non-conservative forces are negligible, so mechanical energy is conserved. Ki + Ui = Kf + Uf Let's take the point of maximum compression of the spring as the zero for gravitational potential energy. Maximum compression of the spring occurs when the lift is instantaneously stationary, so 1 2 Mvi2 + Mg(H+xm) = 0 + 1 2 kxm2 but H >> xm, so 1 2 Mvi2 + MgH ≅ 0 + 1 2 kxm2 kxm2 = Mvi2 + 2MgH xm = √Mk (vi2 + 2gH) ii) Mv2 and MgH both have dimensions of energy (or units of Joules), which is force times distance (Newtons times metres). The spring constant k has dimensions of force per unit length, (units of Newtons per metre), so the argument of the square root is Force x distance Force / distance = distance 2    has units of N*m N.m-1 = m2 , so the RHS has dimensions of distance (units of m), which is the same as the LHS. iii) When the spring is compressed, the total upwards force on the lift will be Σ F = Fspring − W = kx − Mg. At maximum compression, the force is ΣF = kxm − Mg. (In practice, Mg << kxm. so only penalise 1 mark lost if −Mg omitted.) (FYI but not required: We are given kxm > 2 Mg. Unless H is extremely small or unless the spring is very weak and also very long, kxm will be > 2 Mg. We are told that H >> xmax, so the acceleration is greatest when the spring is most compressed.) iv) amax = Fm M = k√Mk (vi2 + 2gH) − Mg M = √kM(vi2 + 2gH) − g. v) For vi = 0, amax ≤ 5g implies that √2kgHM ≤ 6g so rearranging gives k ≤ 18 MgH For maximum compression xm, using parts (i) and (iv) and vi = 0, xm = √Mk 2gH = √MH18Mg 2gH = H3 By the way, and not for marks, most lifts have emergency brakes, that are more practical than springs. However, the lift in the physics building has a pair of springs under it. (Two similar springs in parallel act like one spring of half the k, so the analysis above applies.) Just looking at it, I estimate that xm ~ 40 cm Question 4 A traffic accident investigator finds a scene represented by the following schematic. She makes the following observations and deductions: A B B A φ θ dA dB dC v 0B v 0A side road main road φ North gravitational acceleration = g mB mA Car A has skidded a distance dA along a side street before colliding with car B. Vehicle B has skidded a distance dB along a main street before colliding with car A. Squashed together during the collision, the two wrecked cars have skidded together, sideways but without rotating. They have skidded a distance dC, at an angle θ to the main street and have come to rest together as shown. From the black marks left on the street, we know that all four wheels on both vehicles skidded both before and after the collision. Neither street has any slope. The investigator assumes that the initial velocities v 0A and v 0B, before either vehicle started skidding, are in the directions shown. For legal reasons, she wishes to calculate their magnitudes, v 0A and v 0B. The masses of the cars A and B are mA and mB, respectively. The coefficients of static and kinetic friction between the rubber and the street are µs and µk, respectively. i) Explaining any assumptions and reasoning, derive an expression for the momentum p_ c of the two cars together, after the collision, in terms of mA, mB, dC, g and the appropriate µ. ii) Between the time when either of the cars begins to skid and the time when they come to rest, is there any stage where conservation of mechanical energy is an appropriate approximation? If so, explain when and why. If not, explain why not. iii) Between the time of the cars beginning to skid and the time when they come to rest, is there any stage where conservation of momentum is an appropriate approximation? If so, explain when and why. If not, explain why not. iv) Explaining any assumptions and reasoning, derive an expression for the speed v 0A of car A, before it started skidding, in terms of parameters given in the sketch. v) Explain in one clear sentence why there are skid marks before the collision? (In practice, the wreckage usually rotates about a vertical axis, giving rise to loops and sometimes discontinuities in the skid marks, but the principles are similar.)