Weighing Scales - Fundamental Physics - Solved Past Paper, Exams of Physics Fundamentals

Download Weighing Scales - Fundamental Physics - Solved Past Paper and more Exams Physics Fundamentals in PDF only on Docsity! 93/06(a) Semester 1, 2009 Page 1 of 1 THE UNIVERSITY OF SYDNEY PHYS 1002 PHYSICS 1 (FUNDAMENTALS) JUNE 2010 Solutions 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 at 20 °C and 1 atm ρ = 3 31.000 10 kg.m−× Density of sea water at 20 °C and 1 atm ρ = 3 31.024 10 kg.m−× Atmospheric air pressure P = 51.013 10 Pa× Free fall acceleration at earth's surface g = 29.80m.s− Speed of light in vacuum c = 8 13.00 10 m.s−× FND_Q01 Question 1 A large beaker full of water is placed on weighing scales and the weight of beaker plus water is noted. (a) A rock suspended from a cord is then lowered into the water, and held fully immersed but not touching the bottom. (The water which overflows is drained on to the floor away from the weighing scales.) Explain briefly, why the reading on the weighing scales does not change when the rock is lowered into the water. (b) The rock is now lowered so that it rests on the bottom of the beaker. Will the reading on the weighing scales increase, decrease, or stay the same? Explain your answer. (5 marks) FND_Q02 Question 2 A cabbage of mass 1.0 kg is attached to a weighing scale and is fired as a projectile as shown in the diagram above. Answer the following questions about the reading on the weighing scale at various stages of the motion of the projectile composed of the cabbage and weighing scale. What is the reading on the weighing scale: (a) as the projectile is rising; (b) at the top of the motion of the projectile; (c) as the projectile is falling? In each case, briefly explain the reasons for the answers that you have given. (5 marks) Solution (a) Reading on scale is zero (1 mark) (b) Reading on scale is zero (1 mark) (c) Reading on scale is zero (1 mark) • the cabbage and the scale are both under free fall • the cabbage and scale are really falling with 29.8m.sg −= throughout the motion • relative to one another they have the same acceleration so they do not exert a contact force or tension or pull force on one another (2 marks if a coherent explanation based around the above) FND_Q03 Question 3 Christopher and Jayne were sitting on laboratory chairs, facing each other, initially at rest on a horizontal floor. Christopher had a larger mass than Jayne. They pushed against each other. (Assume that the chairs can move across the floor without friction) (a) While they were pushing each other, who experienced the greater magnitude of pushing force? Briefly explain your answer. (b) At some time t after they finished pushing each other, who will have travelled further from their initial position? Briefly explain your answer. (c) After they finished pushing each other, whose momentum was larger in magnitude? Briefly explain your answer. (5 marks) Solution (a) The free-body diagram for the stationary beetle is N is the normal force, W is the weight force. The net force on the beetle is zero (1 mark) (b) Now the turntable is spinning, the free-body diagram is (1 mark) (c) The centripetal force is provided by the friction f between the beetle’s feet and the turntable. (1 mark) (d) The centripetal force on the beetle while moving in circular motion is given by: 2 cF m rω= If the turntable is spinning faster, a greater force between the beetle and the turntable is needed to provide the centripetal force and prevent the beetle from sliding outwards. Friction provides this force; once the force required is greater than the maximum force that can be provided by static friction, the beetle will slip. As ω increases, cF increases and so to keep cF less than the maximum frictional force, the beetle should decrease r , i.e. by moving towards the centre of the turntable. (1 mark for direction, 1 mark for reasoning) FND_Q05 Question 5 The picture above shows a metal plate after it has been sprinkled with sand grains and forced to vibrate. (a) Explain why the sand grains are present in some regions but not in others. (b) Explain how to obtain a different pattern of lines by vibrating the same metal plate. (5 marks) Solution (a) Electromagnetic waves such as X-rays and infrared can be modeled as a stream of particles called photons when they interact with matter. The energy E of an individual photon is given by E h f= where h is Planck’s constant and f is the frequency of the radiation. The frequency of X-rays is considerably larger than infrared radiation. This means that the individual X-rays have much more energy than infrared photons. When X-ray photons interacts with our bodies they have sufficient energy to ionize atoms and molecules which interferes with the chemistry of cells and can lead to destruction of the cells and if a person is exposed to large doses than it can lead to the malfunction of organs and death. Infrared radiation is non-ionizing and therefore is not as dangerous to our bodies. (1 mark for photon concept; 1 mark for greater energy more damaging) (b) When light enters the glass from the vacuum, the electromagnetic field of the light interacts with the electrons and consequently the light slows down. The speed of the light is characterized by its refractive index n cn v = where, c is the speed of light in vacuum, and, v is the speed of light in glass. Hence 1n v c> ⇒ < . (1 mark) For all waves, the frequency f is the number of oscillations per second of the wave. This does not change as it moves from one medium to another. Therefore, the frequency of the light is the same in the vacuum and the glass. (1 mark) The speed of a wave is given by v f λ= where λ is the wavelength of the light. Therefore, as the speed of the wave v is reduced and the frequency f is the same, the wavelength λ of the light in the glass must be smaller than in the vacuum. (1 mark) FND_Q07 Question 7 A geologist has a rock sample and wishes to identify it. The rock has a mass of 0.870 kg. The rock is suspended from a spring balance and a cord and fully immersed in pure water with the rock not touching the bottom. The spring balance shows a tension T in the cord of 6.62 N. (a) Draw a diagram showing the forces on the rock while it is fully immersed in the water. (b) Write an equation relating (in symbols) the forces acting on the rock. (c) Find the volume of the rock. Show your reasoning. (d) Referring to the densities listed below, determine what the rock is made from. Show your reasoning. Densities: Pure water: 1.00 × 103 kg.m-3 Gypsum: 2.30 × 103 kg.m-3 Quartz: 2.65 × 103 kg.m-3 Siderite: 3.90 × 103 kg.m-3 Barite: 4.48 × 103 kg.m-3 (10 marks) Solution (a) (2 marks) (b) For static equilibrium, 0Bw F T+ + = or Bw F T= + . (1 mark) (c) Weight of rock (in air) (0.870)(9.80) 8.53 Nm g= = = . When immersed in water, scale reads 6.62 NT = . Therefore, 8.53 6.62 1.91N.BF = − = By Archimedes’s Principle, this is the weight of the displaced water. Therefore the mass of the displaced water / 1.91/ 9.80 0.194 kgw g= = = And the volume of the displaced water 3 4 3/ 0.194 /1.00 10 1.94 10 m .m ρ −= = × = × (5 marks) (d) Density of rock 3 34 0.870/ 4.47 10 kg.m . 1.94 10 m V −−= = = ×× The mineral is therefore barite. (2 marks) ( ) 2 1 2 1 2 10 2 2 (2)(2.5) 0.714 s. 9.80 h h gt h t g − = + ⇒ = = = (1 mark) In this time the ball travels a horizontal distance 2d of 2 2 (26.3)(0.714) 18.8m.d v t= = = (1 mark) This is 18.8 15.0 3.8 m− = beyond the net and is therefore within the target area. (1 mark) FND_Q09=REG_Q07 Question 9 or 7 A ballistic pendulum can be used to determine the speed of a bullet fired into it. The above diagram shows an example of such a pendulum. A bullet of mass 0.10 kg is fired horizontally at a speed v into the block of wood of mass 1.0 kg, which is suspended motionless from the ceiling by a string. The distance from the ceiling to the point of impact at the centre of the block is 0.50 m . The bullet stops in the block. (a) Write an expression for the following quantities in terms of the initial speed of the bullet v which is not yet known: (i) the kinetic energy of the bullet just before the impact; (ii) the momentum of the bullet just before the impact; (iii) the speed of the block (with the bullet embedded in it) just after the collision; (iv) the kinetic energy of the block (with the bullet embedded in it) just after the collision. (b) Derive an expression for the maximum height h above the test position that the block (with bullet embedded in it) reaches after the collision. Suppose that a bullet is fired into the block with a speed such that the block rises until the string is horizontal. (c) Calculate the value of the initial speed of the bullet. (10 marks) Solution Deduct ½ mark one time for missing units. No units are required when no specific values have been substituted. (a) (i) 2 21 0.050 J 2 K m v v= = (1 mark) (ii) 10.10 kg.m.sp mv v −= = (1 mark) (iii) All forces are internal to the system, momentum is therefore conserved, and so after beforep p= Therefore ( ) 10.10 0.0909 m.s 1.10 after after after p m M v mv mv v m M v − = + = ⇒ = + = = (1 mark for method, 1 mark for correct result) (iv) The kinetic energy after the collision is ( ) 2 2 2 1 2 (0.5)(1.10)(0.0909 ) 0.004545 J afterK m M v v v = + = = (1 mark for method, 1 mark for correct result) (b) The block rises to a height h until all the kinetic energy is transformed into potential energy: ( ) ( )1.10 (9.80) 10.78 J K m M g h h h = + = = (2 marks for energy transformation: numerical value not required) (c) If the string is horizontal, then 50cm 0.50 mh = = so 2 2 1 0.004545 (10.78)(0.50) 5.39 5.39 / 0.004545 1185.8 34.4 m.s v v v − = = ⇒ = = ⇒ = (2 marks) 1 2 p Xτ τ τ τ τ= + + + ( ) ( ) ( ) ( )4 0.8 9 13.5 2 2.0 0 15 16.3 m x x x x = − + − + − + = − (2 marks) (c) For the pole to balance, the net torque must be zero, so 15 16.3 0 16.3 /15.0 1.09 m x x − = ⇒ = = (1 mark for balance condition; 1 mark for answer) FND_Q11 Question 11 If a diver bobs up and down on the end of a diving board, the motion can be modelled as that of a mass oscillating on a spring. A light and flexible diving board deflects by 0. from the horizontal when Robert, a diver, stands on its end as shown in the diagram. 225m 68.0 kg (a) Calculate the effective spring constant for the diving board. Robert then jumps a little and lands back on the end of the board, depressing it by an extra after which the board moves up and down so the end of the board and Robert moves as a simple harmonic oscillator. 0.105 m (b) Calculate the period of the oscillation. (c) Sketch a graph of the position of the end of the diving board against time for two complete cycles of oscillation. Take upwards as the positive vertical direction. On your graph also show the position of the undeflected board and the position of maximum deflection downwards. Show numerical values on the time and position axes. The amplitude of the oscillation of the diving board increases if Robert drives the motion of the board with his legs at the right frequency. (d) Calculate the frequency of the driving force exerted by Robert’s legs to give the largest amplitude of vibration. Briefly justify your answer. (10 marks) Solution (a) We have an equilibrium deflection with a mass 0.225meqy = 68.0 kgm = At the equilibrium position, the restoring force from the spring board is equal to the gravity force and so k x . m g= This gives 3 1(68.0)(9.80) 2.96 10 N.m (0.225)eq m gk y −= = = × . (2 marks) (b) 3 68.02 2 0.952 s 2.96 10 mT k π π= = = × (2 marks) (c) (3 marks) Solution (a) One possible shape of the standing wave A series of snapshots of the wave It is the third harmonic (2 marks) (b) For the 3rd harmonic 3 1366 3f Hz f= = Therefore the fundamental (1st harmonic) is 1 366 / 3 122 Hzf = = (1 mark) (c) For 1.50 mL = the wavelength of the wave must be 1.00 m.λ = The velocity is given by ( )( ) 1 3 3 366 1.00 m 366m.sv f λ −= = = or equivalently . ( )( ) 1 1 1 122 3.00 m 366m.sv f λ −= = = (2 marks) (d) -1 6 -1 2 2 0.154 kg.m 1.15 10 kg.m 366 T T Fv F v μ μ − = = = = × (2 marks) (e) TFv μ = If tension TF doubles and the linear density μ remains constant then: • the speed v increases by a factor of 2 • the wavelength is determined by boundary conditions and so there is no change. • the frequency increases by a factor of 2 (because v f λ= and the speed increases by a factor of 2 and the wavelength remains the same). (3 marks)