Free Body Diagrams - Fundamental Physics - Solved Past Paper, Exams of Physics Fundamentals

Download Free Body Diagrams - Fundamental Physics - Solved Past Paper and more Exams Physics Fundamentals in PDF only on Docsity! Solutions Semester 1, 2008 Page 1 THE UNIVERSITY OF SYDNEY FACULTIES OF ARTS, EDUCATION & SOCIAL WORK, ENGINEERING AND SCIENCE PHYS 1001 PHYSICS 1 (REGULAR) SOLUTIONS JUNE 2008 Time allowed: THREE Hours MARKS FOR QUESTIONS ARE AS INDICATED TOTAL: 90 MARKS INSTRUCTIONS • All questions are to be answered. • Use a separate answer book for section A and section B. • All answers should include explanations in terms of physical principles. DATA Density of fresh water ρ = 1.00 × 103 kg.m-3 Free fall acceleration at earth's surface g = 9.81 m.s-2 Gravitational constant G = 6.67 × 10-11 N.m2.kg-2 Speed of light in a vacuum c = 3.00 × 108 m.s-1 Speed of sound in air v = 344 m.s-1 Avogadro constant NA = 6.023 × 10 23 mol-1 Universal gas constant R = 8.314 J.mol-1.K-1 Boltzmann constant k = 1.380 × 10-23 J.K-1 Stefan–Boltzmann constant σ = 5.67 × 10 -8 W.m- 2 K -4 REG Question 1 = ADV Question 1 A mass of 2m is connected to two masses of mass m and 3m as shown below. The strings connecting the masses are light. The pulleys and the slopes are frictionless. (a) Draw separate free body diagrams for each of the masses, taking care to identify all forces acting. (b) If the angle of the slope θ = 30°, and the system is released from rest, does the hanging mass 2m remain where it is, accelerate upwards or accelerate downwards? Justify your answer. (5 marks) Solution (a) (1 mark for left diagram; ½ mark each for similar right diagrams) (b) Take vertically upwards as the positive direction for the 2m mass and downwards along the slope as the positive direction for the m and 3m masses Apply Newton’s Second Law to the 2m mass as follows: (mr2) ω1 = m r2 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ω2 ω2 = 4ω1 (1 mark) change in kinetic energy = 1 2 I2ω2 2 − 1 2 I1ω1 2 = 1 2 m r 2 4 .16 ω1 2 − 1 2 m r2ω1 2 = 1 2 m r2 4ω1 2 −ω1 2[ ] = 3 x ORIGINAL KE (1 mark) (e) When the stone travels from one circle to a smaller circle it does so in a spiral. The tension from the string now has a component along the direction of motion and can now do work on the stone to increase its kinetic energy. (1 mark) REG Question 3 A student sits on a stool, which is at rest initially but is free to rotate around a vertical axis. She is handed a bicycle wheel spinning with angular momentum, u , pointing vertically upwards (not shown in this orientation in the picture below). wheelL r (a) Seen from above, what is the direction of rotation of the wheel about the vertical axis? The student now changes the direction of the axis of the bicycle wheel so as to reverse the direction of the angular momentum of the wheel (i.e. the axis is rotated by 180 ). o (b) Describe what now happens. (c) Apply the conservation of angular momentum to the system consisting of student, stool and wheel. Justify its use and account for the motions in part (b) in both magnitude and direction. In rotating the axle of the wheel, the student had to exert a torque at right angles to the direction of the wheel's angular momentum. (d) What is the name for the motion that results from a torque acting at right angles to the direction of the angular momentum associated with a spin? (5 marks) Solution (a) Right Hand rule, as in the diagram below, gives the direction as anti-clockwise as viewed from above. (1 mark) (b) As the student changes the axis of rotation of the bicycle wheel the stool (and student) begins to rotate. (½ mark) (c) There are no external torques acting on the system and so angular momentum is conserved. So the final angular momentum of the system is equal to its initial angular momentum. (1 mark) In the following we will define the angular momentum upwards as a positive value. Initially, the bicycle wheel has angular momentum 1L+ and the student/stool has zero angular momentum. So 1initL L= . Finally, the axis of the bicycle wheel has been reversed and it has angular momentum of . If the angular momentum of the student/stool is 1L− sL then 1final sL L L= − + . But 1 1 12 final init s s L L L L L L L = ⇒ − + = ⇒ = The angular momentum of the student/stool is twice that of the angular momentum of the bicycle wheel. (1 mark) This means that the student/stool has positive angular momentum and so is rotating anti-clockwise as viewed from above. (1 mark) (d) The name for the motion that results from a torque acting at right angles to the direction of the angular momentum associated with a spin is precession. (½ mark) REG Question 5 = ADV Question 5 A loudspeaker emits a pure sinusoidal sound of frequency f , which can be varied. The loudspeaker is placed in front of a wine glass and the sound wave from the loudspeaker makes the glass oscillate. If the frequency is set at a particular value, 1f , there is a high chance that the glass will break. (a) Explain in just a few words why the glass could break when the frequency of the emitted sound wave is “just right”. Assume that you have two wine glasses that “ring” (i.e. they vibrate with a ringing sound when you tap them gently with your finger) with the same frequency. However, one of them “rings” for much longer than the other. (b) The glasses are put next to the loudspeaker under the same conditions. Is the glass that “rings” for longer more likely or less likely to break than the other glass? Briefly explain your answer. (5 marks) Solution (a) If the frequency coming from the loudspeaker is a natural frequency of the glass the energy in the oscillating glass bowl will build up (amplitude increase) due to resonance, and hence stress the glass more. The glass is therefore likely to break. (natural frequency or resonance + large amplitude response 2 marks) b) The glass that rings for longer has less damping (students might answer “lower damping rate” or “longer damping time” which are both correct and equivalent answers), and will therefore accumulate more energy (obtain larger amplitude) in the oscillation of the glass, and hence stress the glass more which would make it more likely to break. (more likely to break 1 mark; less damping in glass that rings longer 1 mark; larger amplitude response because of less damping 1 mark) FND Question 6 = REG Question 6 = ADV Question 6 A string is attached to a wall. A lecturer holding the unattached end pulls with a constant force (parallel to the string) while moving his hand up and down (perpendicular to the string) to create a pulse travelling towards the wall. The lecturer now wants to produce a pulse that takes a longer time to reach the wall. Four ways of doing this are considered below. (a) Choose the best answer from the following and give a brief justification. i. He should move his hand up and down more quickly. ii. He should move his hand up and down more slowly. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall. (b) Choose the best answer from the following and give a brief justification. i. He should use a heavier string of the same length, under the same tension. ii. He should use a lighter string of the same length, under the same tension. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall. (c) Choose the best answer from the following and give a brief justification. i. He should displace his hand a greater distance up and down but at the same speed. ii. He should displace his hand a lesser distance up and down but at the same speed. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall. (d) Choose the best answer from the following and give a brief justification. i. He should pull the string harder to increase the tension in the string. ii. He should pull the string less hard to decrease the tension in the string. iii. Neither (i) or (ii) will change the time for the pulse to each the wall. The lecturer now detaches the string from the wall and invites one of the students to hold that end of the string and keep it fixed. The lecturer and student both pull the string with the same force as the lecturer did before. The lecturer then moves his hands up and down in the same way as he did originally. (e) Compared with the original travel time for the pulse, choose the best answer from the following options and give a brief justification. i. The travel time does not change because the tension in the rope is the same. ii. The travel time gets longer because the tension in the rope is greater. iii. The travel time gets shorter because the tension in the rope is greater. (5 marks) Solution (a) Option (iii) (½ mark) The velocity of the pulse along the string does not depend on frequency. (½ mark) (b) Option (i). (½ mark) The velocity depends on 1/μ1/2 where μ is the mass per unit length of the string. Hence greater μ means smaller velocity (and hence longer travel time). (½ mark) (c) Option (iii)(½ mark) The size of the vertical displacement does not chance the velocity. He also changes the period of the wave but this also does not affect the velocity. (½ mark) (d) Option (ii). (½ mark) The velocity depends on F1/2 where F is the tension in the string. Hence smaller F means smaller velocity (and hence longer travel time). (½ mark) (e) Option (i). (½ mark) The tension in the string will be the same. The action-reaction pair “lecturer-wall” is the same as “lecturer-student”. Flecturer= -Fwall= -Fstudent. Hence the travel time stays the same. (½ mark) Total (5 marks) 21 cos 2 sin 2 (sin cos ) sin k k hmv m g h m g g hv μ α α α μ α α = + + ⇒ = (1 mark for energy conservation; 2 marks for correct answer) FND Question 9 = REG Question 8 A 0.060 kg arrow is moving vertically upward at a speed of 50 m.s–1 just as it hits a block of wood of mass 5.0 kg. The arrow stops in the block and the block is free to move vertically. (a) What is the kinetic energy of the arrow just before the collision? (b) What is the momentum of the arrow just before the collision? (c) What is the momentum of the block (with the arrow embedded in it) just after the collision? (d) What is the velocity of the block just after the collision? (e) How much energy was lost in the collision? (f) How high does the block rise above its initial position? (10 marks) Solution (a) Just before the collision, 2 21 (0.5)(0.06)(50 ) 75 J 2 K m v= = = (2 marks) (b) Just before the collision -1(0.06)(50) 3.0 kg.m.sp mv= = = (2 marks) (c) Since momentum is conserved, the combined momentum of the block + arrow after the collision is ( ) -1 -1 3.0 kg.m.s 3.0 0.59 m.s (5.0 0.06) i f f p p M m v v = = + = ⇒ = = + (1 mark for method; 1 mark for correct result) (d) The kinetic energy just after collision is 2 21 ( ) (0.5)(5.06)(0.59 ) 0.88 J 2 f K m M v= + = = Or 0.89 if they kept all the sig figures for fv . So the energy lost is 75 0.88 74 JEΔ = − = . (i.e. nearly all of the energy) (1 mark for method; 1 mark for correct result) (e) The block will rise until all of the kinetic energy is transformed into potential energy so: ( )0.88 (5.06)(9.81) 0.88 0.018m 1.8cm (5.06)(9.81) K J m M g h h = = + = ⇒ = = = h (1 mark for method; 1 mark for correct result) Energy required to convert water at C to steam at 1 C is the sum of 23.0 o 12 o ΔQ = [water from C to 100 C] + [liquid to gas change of state] + [steam from 10 to 112 C] 23.0 o o 0 o o (1 mark) 3 6 6 6 (100 23.0) (112 100) (175)(4.19 10 )(77) (175)(2.26 10 ) (175)(2.12 10 )(12) (56.5 395.5 4.5) 10 J 456.5 10 J = 456 MJ water v steamQ m c m L m cΔ = − + + − = × + × + × = + + × = × 3 (1 mark) Power absorbed by the collector is: 5(757)(255) 1.93 10 WP I A= = = × Energy absorbed in a time T (seconds) is 51.93 10 Joules = 0.193 MJ absE T T = × (1 mark) Time required to heat water as above is 456 2360 (2363 or 2365 depending on calculation precision) seconds 0.193 39.4 minutes = 0.66 hours T = = = (1 mark) In part (b) award ½ mark in each subpart for correct method; do not penalize more than once for a calculation error. REG Question 10 = ADV Question 10 A 0.12 mol sample of a diatomic ideal gas is permanently in contact with a thermal reservoir that holds the temperature of the sample fixed at a temperature of . The initial volume of the sample is o9.80 C 3 31.30 10 m−× . The gas expands and does 14.0 J of work on the surroundings. (a) What type of thermal process occurs during the expansion? Explain. (b) Calculate the initial pressure of the gas. (c) Calculate the final volume of the gas. (d) Calculate the final pressure of the gas. (e) Calculate the change in internal energy of the gas. (f) Calculate the heat transfer. Is heat transferred to the gas or from the gas? Explain. (g) If the gas had expanded in an adiabatic process from its initial volume to the final volume calculated in part (c), would the temperature of the gas be lower, the same as, or higher than ? Briefly explain your answer. o9.80 C (h) If the gas had expanded in an adiabatic process from its initial volume to the final volume calculated in part (c), would the work done by the gas on the surroundings be less than, equal to, or greater than 14.0 J? Briefly explain your answer. (10 marks) Solution n = 0.12 mol γ = 1.4 (diatomic gas) T1 = 9.8 oC = (9.8 + 273.15) K = 282.95 K T = T2 = T1 held constant by thermal reservoir V1 = 1.30×10-3 m3 W = 14.0 J R = 8.314 J.mol-1.K-1 (a) Since the temperature is held constant, it is an isothermal change. (1 mark) (b) Ideal gas equation 1 1 3 1 5 1 (0.12)(8.314)(282.95) Pa 1.3 10 2.17 10 Pa pV n RT nRTp V nRTp V p − = = = = × = × (1 mark) (c) The work done in an isothermal change is given by 2 1 ln VW n RT V ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ The final volume is 2 1 2 1 2 1 3 3 2 3 3 2 ln exp exp 14(1.3 10 )exp m (0.12)(8.314)(282.95) 1.37 10 m V W V n RT V W V n RT WV V n RT V V − − ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = × ⎜ ⎟ ⎝ ⎠ = × (2 marks) (d) For an isothermal change 2 2 1 1 1 1 2 2 5 3 2 3 5 2 (2.17 10 )(1.3 10 ) Pa (1.37 10 ) 2.06 (or 2.07) 10 Pa p V p V p Vp V p p − − = = × × = × = × (2 marks) (e) Isothermal change 0TΔ = ⇒ Δ = 0U 4 J 0 zero change in the internal energy (1 mark) (f) 1st law 0 0 0 1 T U U Q W Q W Δ = ⇒ Δ = Δ = − = ⇒ = = + Q is positive – heat is added to the gas (1 mark) (g) In an adiabatic process, Q = 0 0 0 since 0U Q W W U W TΔ = − = − ⇒ Δ < > ⇒ Δ < The temperature is less than 9.80 oC. Also can show adiabat on pV diagram, as below, which will show temperature decreases. Solution (a) -14.91 9.82 3.13 rad.s 0.500 k m ω = = = = . (2 marks) (b) 2 2 22 3.13 f T T 2.01sπ π πω π ω = = ⇒ = = = . (1 mark) (c) (3 marks consisting of 1 mark for numerically correct x-axis; 1 mark for numerically correct y-axis;1 marks for correct labels and units. Do not penalise if part of the graph is on the t<0 side. Do not penalise twice if T is not correct from Part (b)). (d) Locations of maximum speed are shown as an * and locations of zero speed shown as an O. (1 mark for maximum speed; 1 mark for points of zero speed) (e) Locations of maximum KE are shown as an * and locations of Maximum Potential Energy shown as an O. (1 mark for maximum KE; 1 mark for points of Maximum Potential Energy) REG Question 12 (a) The frequency of the fundamental mode (f1) for a certain cylindrical pipe, open at both ends, is 100 Hz. (i) Draw a diagram of the pipe to illustrate where along the pipe the displacement nodes and antinodes occur for this standing wave. (ii) Calculate the frequency of the first overtone of the pipe. (iii) For the first overtone, draw a diagram of the pipe to illustrate where along the pipe the displacement nodes and antinodes occur for this standing wave. On a separate diagram mark where the pressure nodes and antinodes occur. (b) Consider a wave travelling in the x direction described by the equation: y = 0.1 sin(10x – 100 t ), where x and y are in metres and t is in seconds. (i) In which direction, positive or negative x, does this wave travel? (ii) Calculate the wavelength, period, velocity and amplitude of this wave. (iii) Describe qualitatively the result from combining the above wave with another wave of the form: y = 0.1 sin(10x + 100 t ). Hint: consider in which direction this wave travels. (10 marks) Solution (a)(i) A N A (1 mark (full mark also if only A,N,A are indicated or only displacement function is indicated)) (a)(ii) 2 12 (2)(100) 200 Hzf f= = = (1 mark) (a)(iii) Displacement A N A N A (1 mark)