2013 ap physics c 力学 scoring guidelines答案及解析
vlsi-2013
vlsi-2013UNIVERSITY OF GLASGOWDegree of MSc in EngineeringVLSI DESIGN AND CAD (ENG5092)Friday 13 December 201309:30-11:30Answer FOUR questionsAnswer only TWO questions from each of sections A and B Each question is worth 25 marksThe numbers in square brackets in the right-hand margin indicate the marks allotted to the part of the question against which the mark is shown. These marks are for guidance only.An electronic calculator may be used provided that it does not have a facility for either textual storage or display, or for graphical display.Continued overleafPage 1 of 6Continued overleaf Page 2 of 6Section A : Attempt any TWO questions [50 marks] Q1 (a)A pipelined system architecture must be able to arbitrarily shift data one bit to the left, one bit to the right, or not at all, in a single clock cycle. Sketch a circuit that will do this using pass-transistor logic. You may assume that there is an input and an output register associated with the device. [8] (b)A simple digital multiplier relies on a process of successive shifting of data to the left, and addition. (i) Show how you can express a m -bit unsigned binary number using radix-2 notation.[4] (ii) By expressing two unsigned binary numbers X and Y, oflength m and n respectively, in radix-2 notation, derive a formula for the product Z = XY . [8] (c)Using your answer for part (b(ii)) of this question, write down the Boolean expression for the partial products that would appear in a logical implementation of the multiplier, and sketch the logic circuit required to calculate a partial product. [5] Q2 (a) Sketch the circuit diagram for a CMOS circuit with the function: D C B A Z ).(++= [8] (b)A layout is required for the circuit. (i) In the layout for the circuit, what is the minimum number of regions of active required to make all the transistors? [2] (ii) Draw a stick diagram for the circuit, including features such as merged transistor active layers and bulk connections. Clearly label each part of your diagram. Coloured pens or pencils may be used, but are not essential. [9] (c)What is the advantage of using techniques such as placing more than one transistor on a single region of active. [3] (d) How are the standard cells designed in practice to achieve regular layouts for blocks? Illustrate your answer with a sketch if necessary. [3]Q3 (a) State the key dimensions th constructed us (b) The Elmore d between cells also be used, compared to u (c) Figure Q3 sho of three furthe the inputs layers of met connected usi interconnect, assume that al(d) Cell placemen how cell pla minimize delaTable Q3. InterconneM1 area capacitance = 0.2 fF/μm M2 edge capacitance = 0.05 fF/μM2 area capacitance = 0.1 fF/μm M2 edge cap acitance = 0.05 fF/μ Co Page 3 of 6e key MOSFET properties and their relationshi ons that determine the propagation delay a ted using them. more delay model can be used to estimate the int cells in a layout. Identify and describe two other m used, and say what advantages or disadvantages ed to using the Elmore method. Q3 shows the routing connecting the output of a cel further cells, labeled A, B and C. The load capaci uts is labeled in Figure Q3. The routing is requ f metal (M1 (horizontal) and M2 (vertical)) th ed using vias. Using Table Q3 for the electrical p nect, calculate the Elmore delay from Z to input A that all tracks are laid out to be of minimum width,cement strongly influences the overall delay of a c l placement affects delay, and what methods e delay by means of judicious cell placement. Figure Q3. Interconnect routing layout.nnect electrical properties.fF/μm 2 5 fF/μm fF/μm 2 5 fF/μm Via resistance = 0.5 ?M1 resistivity = 0.1 ?/square M2 resistivity = 0.1 ?/squareContinued overleafonship to the deviceof a digital circuit[5]he interconnect delayother methods that canages they have when[8]f a cell, Z, to the inputcapacitance at each ofrequired to use two) that are electricallyrical properties of thenput A only. You maywidth, which is 1 μm.[6] of a circuit. Describehods can be used to[6]Continued overleafPage 4 of 6Section B : Attempt any TWO questions [50 marks]Q4 (a)Draw a clearly labeled diagram showing the cross-section of a n-channel MOSFET in a p-type substrate. [4] (b)Explain what is meant by the term inversion in the operation of a MOSFET. [3] (c) Consider a MOSFET of gate length L and gate width W .(i) Write down an expression for the gate charge in terms of the gatecapacitance per unit area C ox , the applied gate-to-source voltageV GS and threshold voltage V th . [2](ii) Show that at very low drain-to-source voltage V DS (near 0 V) thedrain current I D is given by()DS th GS ox n D V V V L W C I ?=μwhere μn is the mobility of electrons near the si licon surface.[8](d) Sketch a transistor layout indicating the active region, the polysilicon gate, and the contact areas for the smallest transistor that can be realized in a CMOS process. Using this, list and explain at least three basic design rules for laying out transistorsassuming that each mask has a worst case misalignment of 0.75λ, where λ is half the gate length. [8]Q5 (a) Draw a labelled block diagram of a typical sampled data system. [6](b) Complex MOS ICs, such as microcontrollers, require on-chip dataconversion capabilities using only MOSFETs and capacitors.A weightedcapacitor digital-to-analog converter (DAC) is a good example of such aconverter.(i) Give the circuit diagram of a 3-bit weighted capacitor DAC andexplain its operation. [8] (ii) Comment on the drawbacks of this DAC architecture. [2](iii) What is the output voltage of a 3-bit weighted capacitor DACwhen the input word is 110 and the reference voltage V ref = 5V?[2](c) Sketch the schematic diagram of a potentiometric DAC using a 2-bit DACas an example. What is the main advantage of this DAC implementation?[7]Continued overleafPage 5 of 6Q6 (a) Flash analog-to-digital converters (ADC) are used in high-speed applications such as video and radar signal processing.(i) Sketch the schematic diagram of a flash ADC using a 3-bit ADC toillustrate your answer. Specify the relative reference resistor valuesof the ADC and explain how high conversion speed is achieved. [7] (ii) If the reference voltage for the ADC is 3 V, specify the actualreference voltage levels used in the conversion process. What willbe the digital output for an input voltage of 1 V? What range ofinput voltages would give the same digital output in this case?[6](b) Sigma-delta analog-to-digital converters (Σ-? ADC) are very popular forvery high resolution (≥16 bit) low-to-medium speed applications such asdigital audio.(i) Explain what is meant by the term quantization noise. [3](ii) State and briefly explain the two techniques employed in Σ-?ADCs to improve the signal-to-noise ratio. [6] (iii) The signal-to-noise (SNR) for a first order Σ-? ADC is given bySNR = 6.02(n + 1.5m) – 3.41 dB, where the basic ADC is n-bit andthe oversampling ratio (OSR) is given by 2m. What sample rate isrequired to obtain 16-bit resolution if the system uses a 1-bit ADCand the Nyquist sampling rate is 44 kHz? [3]End of question paperPage 6 of 6。
2013国际物理奥林匹克竞赛理论试卷(2)
IntroductionIn this problem we study an efficient process of steam production that has been demonstrated to work experimentally. An aqueous solution of spherical nanometer-sized silver spheres (nanoparticles) with only about particles per liter is illuminated by a focused light beam. A fraction of the light is absorbed by the nanoparticles, which are heated up and generate steam locally around them without heating up the entire water solution. The steam is released from the system in the form of escaping steam bubbles. Not all details of the process are well understood at present, but the core process is known to be absorption of light through the so-called collective electron oscillations of the metallic nanoparticles. The device is known as a plasmonic steam generator.Figure 2.1(a)A spherical charge-neutral nanoparticle of radius R placed at the center of the coordinate system. (b) A sphere with a positive homogeneous charge density (red), and containing a smaller spherical charge-neutral region (0, yellow) of radius , with its center displaced by. (c) The sphere with positive charge density of the nanoparticle silver ions is fixed in the center of the coordinate system. The center of the spherical region with negative spherical charge density –(blue) of the electron cloud is displaced by , where . (d)An external homogeneous electric field . For time-dependent , the electron cloud moves with velocity . (e) The rectangular vessel () containing the aqueous solution of nanoparticles illuminated by monochromatic light propagating along the -axis with angular frequency and intensity .A single spherical silver nanoparticleThroughout this problem we consider a spherical silver nanoparticle of radius and with its center fixed at the origin of the coordinate system, see Fig. 2.1(a). All motions, forces and driving fields are parallel to the horizontal -axis (with unit vector ). The nanoparticle contains free (conduction) electrons moving within the whole nanoparticle volume without being bound to any silver atom. Each silver atom is a positive ion that has donated one such free electron.The electric field in a charge-neutral region inside a charged sphereFor the rest of the problem assume that the relative dielectric permittivity of all materials is . Inside a charged sphere of homogeneous charge density and radius R is created a small spherical charge-neutral region of radius by adding the opposite charge density , with its center displa-ced by from the center of the R-sphere, see Fig. 2.1(b).The restoring force on the displaced electron cloudIn the following, we study the collective motion of the free electrons, and therefore model them as a single negatively charged sphere of homogeneous charge density with a center position , which can move along the x-axis relative to the center of the positively charged sphere (silver ions) fixed at the origin of the coordinate system, see Fig. 2.1(c). Assume that an external force displaces the electron cloud to a new equilibrium position with . Except for tiny net charges at opposite ends of the nanoparticle, most of its interior remains charge-neutral.The spherical silver nanoparticle in an external constant electric fieldA nanoparticle is placed in vacuum and influenced by an external force due to an applied static homogeneous electric field , which displaces the electron cloud the small distance , where .The equivalent capacitance and inductance of the silver nanoparticleFor both a constant and a time-dependent field , the nanoparticle can be modeled as an equivalent electric circuit.The equivalent capacitance can be found by relating the work , done on the separation of charges , to the energy of a capacitor, carrying charge . The charge separation will cause a certain equivalent voltage across the equivalent capacitor.For a time-dependent field ,the electron cloud moves with velocity , Fig. 2.1(d). It has the kinetic energy and forms an electric current flowing through the fixed yz-plane. The kinetic energy of the electron cloud can be attributed to the energy of an equivalent inductor of inductance carrying the current .The plasmon resonance of the silver nanoparticleFrom the above analysis it follows that the motion, arising from displacing the electron cloud from its equilibrium position and then releasing it, can be modeled by an ideal LC-circuit oscillating at resonance. This dynamical mode of the electron cloud is known as the plasmon resonance, which oscillates at the so-called angular plasmon frequency .The silver nanoparticle illuminated with light at the plasmon frequencyIn the rest of the problem, the nanoparticle is illuminated by monochromatic light at the angular plasmon frequency with the incident intensity . As the wavelength is large, , the nanoparticle can be considered as being placed in a homogeneous harmonical-ly oscillating field ( ) . Driven by , the center ( )of the electron cloud oscillates at the same frequency with velocity and constant amplitude . This oscillating electron motion leads to absorption of light. The energy captured by the particle is either converted into Joule heating inside the particle or re-emitted by the particle as scattered light.Joule heating is caused by random inelastic collisions, where any given free electron once in a while hits a silver ion and loses its total kinetic energy, which is converted into vibrations of the silver ions (heat). The average time between the collisions is , where for silver nanoparticle we use .The incident light beam loses some time-averaged power by scattering on the oscillating elec-tron cloud (re-emission). depends on the scattering source amplitude , charge , angular fre-quency and properties of the light (the speed of light and permittivity in vacuum). In terms of these four variables, is given by .The above equivalent circuit elements are combined into an LCR series circuit model of the silver nanoparticle, which is driven by a harmonically oscillating equivalent voltage ( ) determined by the electric field of the incident light.Steam generation by lightAn aqueous solution of silver nanoparticles is prepared with a concentration. It is placed inside a rectangular transparent vessel of size andilluminated by light at the plasmon frequency with the same intensity at normal incidence as above, see Fig. 2.1(e). The temperature of the water is and we assume, in fair agreement with observations, that in steady state all Joule heating of the nanoparticle goes tothe production of steam of temperature , without raising the temperature of the water. The thermodynamic efficiency of the plasmonic steam generator is defined by the power ratio , where is the power going into the production of steam in the entire vessel, while is the total power of the incoming light that enters the vessel.Most of the time any given nanoparticle is surrounded by steam instead of water, and it can thus be described as being in vacuum.。
APIO2013中文版
2013亚太地区信息学奥林匹克竞赛APIO 2013竞赛时间：2013年5月11日9:00-14:00编译器版本及编译选项见考试注意事项。
机器人【问题描述】VRI（V oltron机器人学会）的工程师建造了n个机器人。
任意两个兼容的机器人站在同一个格子时可以合并为一个复合机器人。
我们把机器人用1至n编号（n≤ 9）。
如果两个机器人的编号是连续的，那么它们是兼容的，可以合并成一个复合机器人。
最初这n个机器人各自都只有唯一的编号。
而一个由两个或以上的机器人合并构成的复合机器人拥有两个编号，分别是构成它的所有机器人中最小和最大的编号。
例如，2号机器人只可以与1号或3号机器人合并。
若2号机器人与3号机器人合并，可构成编号为2-3的复合机器人。
如果编号为2-3的复合机器人与编号为4-6的复合机器人合并，可构成编号为2-6的复合机器人。
当所有机器人合并以后则构成1-n复合机器人。
工程师把这n个机器人放在了一个封闭的房间中，房间四周均是墙。
该房间被划分成w ×h个方格。
有些方格有障碍物，机器人不可经过或停留；其余方格允许多个机器人停留，同时允许机器人经过。
任何时候一个机器人只占用一个方格。
初始时刻，所有机器人均在不同的方格中。
这些原始的机器人不会自发地移动。
它们只有被工程师沿x轴或y轴推动后，才会沿推动的方向不断向前直线移动，直至碰到障碍物或墙停止移动。
停止移动后，它会扫描当前的格子是否存在可以与它合并的机器人，如果有，则合并并继续检查，直至不能再合并为止。
工程师只能沿水平向左、水平向右、竖直向上、竖直向下四个方向推动机器人，并且，在机器人尚未停止移动时，不允许推动其它机器人，因此任何时刻，房间中都只能有一个机器人移动。
为了帮助机器人转向，工程师在一些格子中放置了转向器。
具体地说，转向器分为顺时针转向器（右转器）和逆时针转向器（左转器），顺时针转向器可以使到达该格子的机器人沿顺时针方向转向90°；逆时针转向器可以使到达该格子的机器人沿逆时针方向转向90°。
国外比较好的损伤力学教材
国外比较好的损伤力学教材以下是一些国外较为流行和高质量的损伤力学教材：1. "Mechanics of Composite Materials" by Robert M. Jones - 这本书是关于复合材料力学的经典教材，其中包括了对于损伤分析和断裂力学的深入讲解。
2. "Damage and Fracture Mechanics: Failure Analysis of Engineering Materials and Structures" by Taoufik Boukharouba -这本书提供了广泛的损伤和断裂力学理论的详细介绍，并通过案例研究来说明其在工程材料和结构分析中的应用。
3. "Applied Mechanics of Solids" by Allan F. Bower - 这本书讲解了固体力学中的基本原理和方法，包括损伤力学和断裂力学的内容，并给出了实际应用的例子。
4. "Fracture Mechanics: Fundamentals and Applications" by T.L. Anderson - 这本书详细介绍了断裂力学的基础理论和实际应用，包括对于裂纹扩展和损伤分析的深入探讨。
5. "Damage Mechanics with Finite Elements: Practical Applications with Python and MATLAB Programs" by Christian Linder - 这本书以Python和MATLAB程序为基础，介绍了有限元法在损伤力学中的应用，帮助读者理解和解决实际问题。
请注意，这些教材的选择还要根据你的具体需求和程度来决定。
对于初学者，推荐选择较为综合的教材以了解基本概念和原理；对于有一定背景的学者，可以选择更加专业和深入的教材来进一步研究和应用相关内容。
2013 ap physics c 力学 scoring guidelines答案及解析
© 2013 The College Board. Visit the College Board on the Web: .
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国际课程IGCSE剑桥物理chap13
Application of Reflection: Mirrors •The mirror is a reflecting surface and using the Law of Reflection we
•Incoming parallel light rays are converged to into the focal point of the lens.
•Light rays that enter the lens through the centre are unaltered and proceed with their original path.
•Converging lenses: lenses that converge incoming parallel rays of light
•Diverging lenses: lenses that diverge incoming parallel rays of light
•In this chapter we focus on converging lenses. These lenses bend light in the following way:
The majority of light refracts and a minority will reflect if i < c
The majority of light refracts with r = 90° and a minority will reflect if i = c.
ASS-2013-1
Applied Surface Science 264 (2013) 633–635Contents lists available at SciVerse ScienceDirectApplied SurfaceSciencej o u r n a l h o m e p a g e :w w w.e l s e v i e r.c o m /l o c a t e /a p s u scInfluence of pulsed electron beam treatment on microstructure and properties of TA15titanium alloyYu-kui Gao ∗College of Aerospace Engineering and Applied Mechanics,Tongji University,No.100of Zhangwu Road,Shanghai 200092,Chinaa r t i c l ei n f oArticle history:Received 8June 2012Received in revised form 16July 2012Accepted 15October 2012Available online 23 October 2012Keywords:Pulsed electron beam treatment Grade characteristics Nanoindenta b s t r a c tThe surface of TA15titanium alloy was modified by pulsed electron beam and the hardness distribution along the treated surface layer was investigated by nanoindent technology.The grade characteristics were therefore analyzed by studying the distribution of hardness along surface layer of specimens.Moreover,the microstructure was investigated by OM,XRD and TEM techniques.Furthermore,the correlation of hardness to microstructure was analyzed.The results show that the grade fine grain microstructure is formed in the upper surface layer and the temperature grade or heat effect caused by pulsed electron beam treatment is the main reason to form grade fine grain microstructure in the surface layer.© 2012 Elsevier B.V. All rights reserved.1.IntroductionTitanium alloys are employed in aerospace industry because of their remarkable characteristics:the high ratio of strength to weight,excellent corrosion resistance and good fatigue perfor-mance.TA15titanium alloy is widely used in Chinese aircraft to make parts,which have lighter weight,longer life and good corro-sion properties [1].Surface determines performance and many components’prop-erties are affected by the surface.Wear,fatigue,and corrosion are the main properties of many metallic materials.The effect of sur-face integrity on fatigue was investigated by many researchers.Surface integrity is a comprehensive concept which includes the change of chemical and mechanical properties of materials at the surface layer.It has four main categories and microstructure is the critical one [2].The wear of TA15titanium alloy is bad due to its lower hardness and worse surface resistance to fretting.How to improve the wear property of TA15titanium alloy by increasing surface hardness is an important problem in its applications.Pulsed electron beam treatment is a new surface modifica-tion process and can be applied in many fields [3–7].There are many investigations on the effects of pulsed electron beam for measurements of properties and microstructures of the titanium alloys [8–12].Therefore,the wear property may be improved with increasing hardness in surface layer by finer grain modified by pulsed electron beam.The hardness is greater when the grain is∗Corresponding author.Tel.:+862165981290;fax:+862165983267.E-mail address:yukuigao@finer,therefore the wear property could be modified with the hard-ness increase of surface layer.However,the effect of surface fine grain microstructure by electron beam treatment on hardness has been less investigated and therefore it is interesting and impor-tant to study the change of hardness along surface layer by electron beam treatment and to obtain the characteristics of grade fine grain microstructure in surface layer with the nanoindent technology.2.Experimental material and methods2.1.Experimental materialTA15titanium alloy was employed in this investigation.Its chemical composition is listed in Table 1.The metal was annealed at 800◦C for 1h then had an air cooling.Its microstructure is mainly ␣phase and a little phase.The tensile property of TA15titanium alloy is listed in Table 2.2.2.Experimental methodsThe surfaces of specimens were pulsed electron beam treated by the Solo type pulsed electron beam machine.The parameters of pulsed electron beam treatment are the pulsed duration time,t =15s,pulsed times,N =3pulses,kinetic energy,E =10–40keV,and pulsed energy density,E S =15J/cm 2.The microstructure of pulsed electron beam treated specimens was analyzed by OM.The phases and their contents were deter-mined by the D/max-2500/PCX type XRD tester.Moreover,the fine grain characteristics were also analyzed by TEM method.0169-4332/$–see front matter © 2012 Elsevier B.V. All rights reserved./10.1016/j.apsusc.2012.10.083634Y.-k.Gao /Applied Surface Science 264 (2013) 633–635Table 1Chemical composition of TA15titanium alloy (wt.%).AlZrMoVFeSiONCTi5.9 2.1 1.71.90.210.110.080.020.09BalTable 2Tensile property of TA15titanium alloy.Tensile strengthYield strengthElongation 985MPa881MPa14%The hardness distribution along surface layer was determined by the Tribolndente nanoindent tester.The maximum applied load is 1N,and the grade characteristics were analyzed based on the hardness distribution along surface layer.3.Results and discussion3.1.Microstructures and phases of TA15titanium alloyThe microstructure of pulsed electron beam treated specimens is shown in Fig.1and there are three zones in Fig.1.The three zones are melted layer,heat affected layer and substrate [3]from the upper or top surface layer to matrix illustrated in Fig.1.Fig.1also shows two special zones,the melt layer and the affected layer along the cross section surface of pulse electron beam treated spec-imens.In the melt layer,fine grain microstructure can be found and the phase cannot be clearly seen because of great resistance of sin-gle ␣phase corrosion in fine grain microstructure layer,as shown in Fig.1.The fine grain microstructure is determined by XRD peak width illustrated in Fig.2and the size of the diffraction grain is about 168nm for the pulsed electron beam treated specimens.The size of the diffraction grain for the unmodified specimens is also deter-mined by XRD peak width as shown in Fig.3.The size of the diffraction grain is about 310nm for the unmodified annealed spec-imens.The finer grains or diffraction domains can increase the hardness and make surface have a good wear property.3.2.Nanoindent hardness and grade characteristicsTo determine the grade characteristics of fine grain microstruc-ture in surface layer and the effect of fine grain microstructure on hardness in surface layer of the pulsed electron beam treated specimens,the distribution of hardness along surface layer was determined by the nanoindent hardnessmeasurement.Fig.1.Microstructure of TA15titanium alloy treated by pulsed electronbeam.Fig.2.XRD of pulsed electron beam treated specimens.The hardness along the cross section surface of the pulsed elec-tron beam specimens is illustrated in Fig.4.The matrix hardness is about 4.62GPa and the surface hardness of pulsed electron beam specimens is about 7.11GPa,the increment of hardness is about 2.49GPa and the increase percentage is 53%.The grade characteristics of fine grain microstructure in the sur-face layer and the effect of fine grain microstructure on hardness in surface layer can be determined by nanoindentmeasurementFig.3.XRD of referenced annealedspecimens.Fig.4.Hardness along cross section surface of pulsed electron beam specimens.Y.-k.Gao/Applied Surface Science264 (2013) 633–635635Fig.5.TEM images of pulsed electron treated specimens.(a)BF image(b)SAED pattern(c)indexed diffraction pattern.technology.It can be seen that the gradient of hardness in surface layer about25m as shown in Fig.4is greater and this depth is equal to the depth of the melt layer,which is illustrated in Fig.1. The gradient of hardness beneath the melt layer between25m and50m is smaller and the gradient of hardness for the distance of50m from the surface to the distance of80m is about zero.The great increase in hardness of the surface melt layer should be related to the formation offine grain microstructure during pulsed electron beam treatment.The grain or domain size is gettingfiner induced by pulsed electron beam treatment;the hardness will be higher because of the higher yield strength.The smaller width of martensite lentil microstructure can be found in the melt layer,as shown in Fig.5.The width of some martensite lathes is about90nm and the SAED pattern in Fig.5shows the lattice distort and the finer grain which make the diffraction dots become the extended diffraction dots or even some short diffraction continue lines.The hardness is affected mainly by the factors offine grain microstructure and the heat effect during pulsed electron beam treating process.Thefine grain or small domain size can improve the hardness,and at the same time the surface melted layer has the less effect of heat after pulsed electron beam treatment due to short time,therefore,the gradient of hardness from surface to50m is great and mainly determined by thefine grain effect,althoughfine grains are formed through being re-melted with a high temperature induced by pulsed electron beam treatment.The gradient of hardness beneath the melt layer where from the surface between50m and80m is smaller and near to zero just because of a balance of thefine grain size effect and the heat effect induce by pulsed electron beam treatment.Thefine grain effect makes an increase in hardness,but the heat effect decreases hardness.The increase in hardness may also be related to the phase trans-formation,but the quantitative analysis of phases content by XRD illustrates that no any new phase occurs in the pulsed electron beam treated specimens and the change of phase content of both␣andis less than5%,therefore,the increase of hardness by pulsed electron modification should be mainly due to thefine grain size effect.4.Conclusions(1)Thefine grain microstructure can be formed in melt layerinduced by pulsed electron beam treatment.(2)The grade characteristics offine grain microstructure in surfacelayer and the effect offine grain microstructure on hardness can be determined by nanoindent technology.(3)The great increment of hardness in the surface melt layer shouldbe related to the formation offine grain microstructure during pulsed electron beam treatment.(4)The gradient of hardness beneath the melt layer is greater dueto thefine grain size effect.AcknowledgementThe author is grateful for the foundation support from Pro-gram of Ministry of Science and Technology of China(contract nos. 200473-3641and08-039577).References[1]J.Y.Wang,Z.M.Ge,B.Y.Zhou,Titanium Alloys in Aeronautical Application,Shanghai Science Press,Shanghai,1985,p.68.[2]Y.K.Gao,X.B.Li,Q.X.Yang,M.Yao,Influence of surface integrity on fatiguestrength of42CrNi2Si2MoVA steel,Mater.Lett.61(2007)466–469.[3]A.D.Pogrebnyak, D.I.Proskurovskii,Modification of metal surface layerproperties using pulsed electron beams,Phys.Status Solidi A145(1994) 9–49.[4]D.I.Proskurovsky,V.P.Rotshtein,G.E.Ozur,A.B.Markov,D.S.Nazarov,Pulsedelectron-beam technology for surface modification of metallic materials,J.Vac.Sci.Technol.A16(1998)2480–2488.[5]A.D.Pogrebnjak,Metastable states and structural phase change in metals andalloys exposed to high power pulsed ion beams,Phys.Stat.Sol.117A(1990) 17–35.[6]V.P.Rotshtein,D.I.Proskurovsky,G.E.Ozur,Yu.F.Ivanov,A.B.Markov,Surfacemodification and alloying of metallic materials with low-energy high-current electron beams,Surf.Coat.Technol.180–181(2004)377–381.[7]Y.K.Gao,Surface modification of TA2pure titanium by low energy highcurrent pulsed electron beam treatments,Appl.Surf.Sci.257(2011) 7455–7460.[8]A.D.Pogrebnjak,A.P.Kobzev,B.P.Gritsenko,Effect of Fe and Zr ion implantationand high-current electron irradiation treatment on chemical and mechanical properties of Ti–V–Al Alloy,J.Appl.Phys.87(2000)2142–2148.[9]A.D.Pogrebnjak,E.A.Bazyl,Modification of wear and fatigue characteristics ofTi–V–Al alloy by Cu and Ni ion implantation and high-current electron beam treatment,Vacuum64(2002)1–7.[10]A.D.Pogrebnjak,S.Bratushka,V.I.Boyko,et al.,A review of mixing processes inTa/Fe and Mo/Fe systems treated by high current electron beams,Nucl.Instrum.Meth.B B145(1998)373–390.[11]T.Grosdidier,J.X.Zou,B.Bolle,S.Z.Hao,C.Dong,Grain refinement,hardeningand metastable phase formation by high current pulsed electron beam(HCPEB) treatment under heating and melting modes,J.Alloys Compd.504S(2010) S508–S511.[12]X.D.Zhang,S.Z.Hao,X.N.Li,C.Dong,T.Grosdidier,Surface modification of puretitanium by pulsed electron beam,Appl.Surf.Sci.257(2011)5899–5902.。
AP物理C力学模拟卷MultipleChoiceQuestions
AP物理C⼒学模拟卷MultipleChoiceQuestionsMultiple Choice QuestionsTime: 45 minutes. You may refer to the Constants sheet. However, you may not use the Equations sheet, and you may not use a calculator on this portion of the exam.1. A cannon is mounted on a truck that moves forward at a speed of 5m/s. The operator wants to launch a ball from a cannon so the ball goes as far as possible before hitting the level surface. The muzzle velocity of the cannon is 50 m/s. What angle from the horizontal should the operator point the cannon?A.5°B.41°C.45°D.49°E.85°2. A car moving with speed v reaches the foot of an incline of angleθ. The car coasts up the incline without using the engine.Neglecting friction and air resistance, which of the following is correct about the magnitude of the car's horizontal acceleration aand vertical acceleration a y?xA.a x = 0; a y < gB.a x = 0; a y = gC.a x < g ; a y < gD.a x < g ; a y = gE.a x < g ; a y > g3. A bicycle slows down with an acceleration whose magnitude increaseslinearly with time. Which of the following velocity–time graphs could represent the motion of the bicycle?4. A cart is sliding down a low friction incline. A device on the cartlaunches a ball, forcing the ball perpendicular to the incline, as shown above. Air resistance is negligible. Where will the ball land relative to the cart, and why?A.The ball will land in front of the cart, because the ball'sacceleration component parallel to the plane is greater thanthe cart's acceleration component parallel to the plane.B.The ball will land in front of the cart, because the ball hasa greater magnitude of acceleration than the cart.C.The ball will land in the cart, because both the ball and thecart have the same component of acceleration parallel to theplane.D.The ball will land in the cart, because both the ball and thecart have the same magnitude of acceleration.E.The ball will land behind the cart, because the ball slowsdown in the horizontal direction after it leaves the cart.5.The quantity "jerk," j, is defined as the time derivative of anobject's acceleration,What is the physical meaning of the area under a graph of jerk vs.time?A.The area represents the object's acceleration.B.The area represents the object's change in acceleration.C.The area represents the object's change in velocity.D.The area represents the object's velocity.E.The area represents the object's change in position.6. A particle moves along the x-axis with a position given by theequation x(t) = 5 + 3t, where x is in meters, and t is in seconds.The positive direction is east. Which of the following statements about the particle is FALSE.0.The particle is east of the origin at t = 0.1.The particle is at rest at t = 0.2.The particle's velocity is constant.3.The particle's acceleration is constant.4.The particle will never be west of position x = 0.7. A mass hangs from two ropes at unequal angles, as shown above. Whichof the following makes correct comparisons of the horizontal and vertical components of the tension in each rope?8.The force of air resistance F on a mass is found to obey the equationF = bv2, where v is the speed of the mass, for the range of speedsinvestigated in an experiment. A graph of F N vs. v2 is shown above.What is the value of b?.0.83 kg/mA. 1.7 kg/mB. 3.0 kg/mC. 5.0 kg/mD. 1.0 kg/mE.zero9. A box sits on an inclined plane without sliding. As the angle ofthe plane (measured from the horizontal) increases, the normal force.increases linearlyA.decreases linearlyB.does not changeC.decreases nonlinearlyD.increases nonlinearly10.Which of the following conditions are necessary for an object tobe in static equilibrium?.The vector sum of all torques on the object must equal zero.I.The vector sum of all forces on the object must equal zero.II.The sum of the object's potential and kinetic energies must be zero.C.I onlyD.II onlyE.III onlyF.I and II onlyG.I, II, and III11.A student pushes a big 16-kg box across the floor at constant speed.He pushes with a for ce of 50 N angled 35° from the horizontal, as shown in the diagram above. If the student pulls rather than pushes the box at the same angle, while maintaining a constant speed, what will happen to the force of friction?.It must increase.A.It must decrease.B.It must remain the same.C.It will increase only if the speed is greater than 3.1 m/s.D.It will increase only if the speed is less than 3.1 m/s.12.Consider a system consisting only of the Earth and a bowling ball,which moves upward in a parabola above Earth's surface. The downward force of Earth's gravity on the ball, and the upward force of the ball's gravity on the Earth, form a Newton's third law force pair.Which of the following statements about the ball is correct?.The ball must be in equilibrium since the upward forces must cancel downward forces.A.The ball accelerates toward the Earth because the force ofgravity on the ball is greater than the force of the ball onthe Earth.B.The ball accelerates toward the Earth because the force ofgravity on the ball is the only force acting on the ball.C.The ball accelerates away from Earth because the forcecausing the ball to move upward is greater than the force ofgravity on the ball.D.The ball accelerates away from Earth because the forcecausing the ball to move upward plus the force of the ballon the Earth are together greater than the force of gravityon the ball.13.A mass m is attached to a mass 3m by a rigid bar of negligible massand length L. Initially, the smaller mass is located directly above the larger mass, as shown above. How much work is necessary to flip the rod 180° so that the lar ger mass is directly above the smaller mass?.4mgLA.2mgLB.mgLC.4pmgLD.2pmgL14.A ball rolls horizontally with speed v off of a table a height habove the ground. Just before the ball hits the ground, what is its speed?.A.B.C.vD.15.A pendulum is launched into simple harmonic motion in two differentways, as shown above, from a point that is a height h above its lowest point. During both launches, the bob is given an initial speed of3.0 m/s. On the first launch, the initial velocity of the bob isdirected upward along the pendulum's path, and on the second launch it is directed downward along the pendulum's path. Which launch will cause the pendulum to swing with the larger amplitude?.the first launchA.the second launchB.Both launches produce the same amplitude.C.The answer depends on the initial height h.D.The answer depends on the length of the supporting rope.16.The mass M is moving to the right with velocity v0 at position x= x0. Neglect friction. The spring has force constant k. What is the total mechanical energy of the block at this position?17.A sphere, a cube, and a cylinder, all of equal mass, are releasedfrom rest from the top of a short incline. The surface of the incline is extremely slick, so much so that the objects do not rotate when released, but rather slide with negligible friction. Which reaches the base of the incline first?.the sphereA.the cubeB.the cylinderC.All reach the base at the same time.D.The answer depends on the relative sizes of the objects.18.Block B is at rest on a smooth tabletop. It is attached to a longspring, which is in turn anchored to the wall. Block A slides toward and collides with block B. Consider two possible collisions: Collision I: Block A bounces back off of block B.Collision II: Block A sticks to block B.Which of the following is correct about the speed of block Bimmediately after the collision?.It is faster in case II than in case I ONLY if block B is heavier.A.It is faster in case I than in case II ONLY if block B isheavier.B.It is faster in case II than in case I regardless of the massof each block.C.It is faster in case I than in case II regardless of the massof each block.D.It is the same in either case regardless of the mass of eachblock.19.A 0.30-kg bird is flying from right to left at 30 m/s. The birdcollides with and sticks to a 0.50-kg ball which is moving straight up with speed 6.0 m/s. What is the magnitude of the momentum of the ball/bird combination immediately after collision?.12.0 N?sA.9.5 N?sB.9.0 N?sC. 6.0 N?sD. 3.0 N?s20.The force F on a mass is shown above as a function of time t. Whichof the following methods can be used to determine the impulse experienced by the mass?.multiplying the average force by t maxI.calculating the area under the line on the graphII.taking the integralC.II onlyD.III onlyE.II and III onlyF.I and II onlyG.I, II, and III21.A projectile is launched on level ground in a parabolic path so thatits range would normally be 500 m. When the projectile is at the peak of its flight, the projectile breaks into two pieces of equal mass. One of these pieces falls straight down, with no further horizontal motion. How far away from the launch point does the other piece land?.250 mA.375 mB.500 mC.750 mD.1000 mQuestions 22 and 23A rigid rod of length L and mass M is floating at rest in space farfrom a gravitational field. A small blob of putty of mass m < M is moving to the right, as shown above. The putty hits and sticks to the rod a distance 2L/3 from the top end.22.How will the rod/putty contraption move after the collision?.The contraption will have no translational motion, but will rotate about the rod's center of mass.A.The contraption will have no translational motion, but willrotate about the center of mass of the rod and putty combined.B.The contraption will move to the right and rotate about theposition of the putty.C.The contraption will move to the right and rotate about thecenter of mass of the rod and putty combined.D.The contraption will move to the right and rotate about therod's center of mass.23.What quantities are conserved in this collision?.linear and angular momentum, but not kinetic energyA.linear momentum onlyB.angular momentum onlyC.linear and angular momentum, and linear but not rotationalkinetic energyD.linear and angular momentum, and linear and rotationalkinetic energy24.A car rounds a banked curve of uniform radius. Three forces act onthe car: a friction force between the tires and the road, the normal force from the road, and the weight of the car. Which provides the centripetal force which keeps the car in circular motion?.the friction force aloneA.the normal force aloneB.the weight aloneC. a combination of the normal force and the friction forceD. a combination of the friction force and the weight25.A ball of mass m anchored to a string swings back and forth to amaximum position A, as shown above. Point C is partway back to the vertical position. What is the direction of the mass's acceleration at point C?.along the mass's path toward point BA.toward the anchorB.away from the anchorC.between a line toward the anchor and a line along the mass'spathD.along the mass's path toward point A26.In a carnival ride, people of mass m are whirled in a horizontalcircle by a floorless cylindrical room of radius r, as shown in the diagram above. If the coefficient of friction between the people and the tube surface is µ, what minimum speed is necessary to keep the people from sliding down the walls?Questions 27 and 28The uniform, rigid rod of mass m, length L, and rotational inertiaI shown above is pivoted at its left-hand end. The rod is released from rest from a horizontal position.27.What is the linear acceleration of the rod's center of mass the moment after the rod is released?28.What is the linear speed of the rod's center of mass when the mass passes through a vertical position?29.The 1.0-m-long non-uniform plank, shown above, has weight 1000 N.It is to be supported by two rods, A and B, as shown above. The center of mass of the plank is 30 cm from the right edge. Each support bears half the weight of the plank. If support B is 10 cm from the right-hand edge, how far from the left-hand edge should support A be?.0 cmA.10 cmB.30 cmC.50 cmD.70 cm30.A mass m on a spring oscillates on a horizontal surface with periodT. The total mechanical energy contained in this oscillation is E.Imagine that instead a new mass 4m oscillates on the same springwith the same amplitude. What is the new period and total mechanical energy?31.A mass m is attached to a horizontal spring of spring constant k.The spring oscillates in simple harmonic motion with amplitude A.What is the maximum speed of this simple harmonic oscillator?32.An empty bottle goes up and down on the surface of the ocean, obeyingthe position function x= Acos(t). How much time does this bottle take to travel once from its lowest position to its highestposition?33.The Space Shuttle orbits 300 km above the Earth's surface; theEarth's radius is 6400 km. What is the acceleration due to Earth's gravity experienced by the Space Shuttle?. 4.9 m/s2A.8.9 m/s2B.9.8 m/s2C.0.8 m/s2D.zero34.An artificial satellite orbits Earth just above the atmosphere ina circle with constant speed. A small meteor collides with thesatellite at point P in its orbit, increasing its speed by 1%, but not changing the instantaneous direction of the satellite's velocity. Which of the following describes the satellite's new orbit?.The satellite now orbits in an ellipse, with P as the farthest approach to Earth.A.The satellite now orbits in an ellipse, with P as the closestapproach to Earth.B.The satellite now orbits in a circle of larger radius.C.The satellite now orbits in a circle of smaller radius.D.The satellite cannot maintain an orbit, so it flies off intospace.35.Mercury orbits the sun in about one-fifth of an Earth year. If 1AU is defined as the distance from the Earth to the sun, what is the approximate distance between Mercury and the sun? .(1/25) AUA.(1/9) AUB.(1/5) AUC.(1/3) AUD.(1/2) AU。
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Question 1 (continued)
Distribution of points ii. 3 points
For a smooth curve that begins with increasing x and is concave up for t 0.79 s , ending between t 0.79 s and 1.0 s For a straight line with a positive slope, beginning between t 0.79 s and 1.0 s For a smooth transition of the curve from non-linear to linear in the region between t 0.79 s and 1.0 s (c) 3 points For using the graph to determine the distance traveled during the first part of the motion, beginning at t 0 and ending somewhere between 0.79 s (when the glider and spring lose contact) and 2 s (the maximum time shown on the velocity graph For calculating using the graph between 0 and 1.0 s, d1 (2.9 large grid squares)( 0.125 m square) 0.36 m (1 square = 0.25 m/s × 0.5 s = 0.125 m) For a correct expression indicating constant velocity during the last part of the motion d2 u Dt u(t 1.0 s) For adding the two distances and solving for the time at which the glider hits the bumper d1 d 2 2.0 m 0.36 m 0.50 m s (t 1.0 s) 2.0 m (2.0 0.36) m t 1.0 s 4.3 s 0.50 m s
AP® PHYSICS C: MECHANICS 2013 SCORING GUIDELINES
Question 1
15 points total Distribution of points
(a)
3 points
For labeling the axes with appropriate values For a smooth curve that begins with increasing u and is concave down For a horizontal line near u 0.50 m s , beginning between t = 0.79 and 1.0 s (b) i. 1 point For a correct method of plotting position x as a function of time t Examples Plot the area under the velocity curve from part (a) as a function of time. x udt The slope of x as a function of t would yield the u versus t graph in part (a).
1 2 1 2 kx1 mu2 2 2
k
2 mu2 2 x1
1 point
1 point
1 point
2 x1 For a correct answer, with correct units k 1.6 N m
k
2 mu2
0.40 kg 0.50 m s 2 0.25 m 2
1 point 1 point 1 point
1 point
1 point
1 point
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AP® PHYSICS C: MECHANICS 2013 SCORING GUIDELINES
1 point
Alternate Points 1 point 1 point
For recognizing that the 0.79 s of contact time is one quarter of a period For giving the period as four times the contact time T 4 0.79 s 3.2 s
AP® Physics C: Mechanics 2013 Scoring Guidelines
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1 point
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AP® PHYSICS C: MECHANICS 2013 SCORING GUIDELINES
Question 1 (continued)