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www.ks5u.com二、选择题：本题共8小题，每小题6分。在每小题给出的四个选项中，第14~18题只有一项符合题目要求，第19~21题有多项符合题目要求。全部选对的得6分，选对但不全的得3分，有选错的得0分。14.由玻尔理论可知氢原子核外电子A.能自发的从低能级跃迁到高能级B.激发态的能量满足En=n2E1(n=2，3，…)C.从基态电离所需的能量一定恰好等于基态的能量D.从第4能级向第1能级跃迁时发出光的频率一定比向第2能级跃迁时发出光的频率高15.如图所示，装有弹簧发射器的小车放在水平地面上，现将弹簧压缩锁定后放人小球，再解锁将小球从静止斜向上弹射出去，不计空气阻力和一切摩擦。从静止弹射到小球落地前的过程中，下列判断正确的是A.小球的机械能守恒，动量守恒B.小球的机械能守恒，动量不守但C.小球、弹簧和小车组成的系统机械能守恒，动量不守恒D.小球、弹簧和小车组成的系统机械能守恒，动量守恒16.如图所示为一含有理想变压器的电路，U为正弦交流电源，输出电压的有效值恒定，开关S闭合前后理想电流表的示数比为1：3，则电阻R1、R2的比值为A.1:1B.2：1C.3：1D.4:117.2018年1月12日，我国以“一箭双星”
1方式成功发射第26、第27颗北斗导航卫星，拉开2018年将发射16颗北斗卫星的序幕。北斗导航卫星的轨道有三种：地球静止轨道（高度35809km）、倾斜地球同步轨道（高度35809km)、中圆地球轨道（高度21607km），如图所示。下列说法正确的是A.中圆地球轨道卫星的周期一定比静止轨道卫星的周期长B.中圆地球轨道卫星受到的万有引力一定比静止轨道卫星受到的万有引力大C.倾斜同步轨道卫星始终位于地球表面某点的正上方D.倾斜同步轨道卫星每天在固定的时间经过同一地区的正上方18.如图所示，直角坐标系中（a，0）处固定有点电荷+Q，（-a，0）处固定有点电荷-nQ，各自产生的电场强度大小分别为E1、E2，已知圆心在（3a，0）处半径为2a的圆周上E1=E2，则n等于A.1B.2C.3D.419.如图所示，“旋转秋千”中座椅（可视为质点）通过轻质缆绳悬挂在旋转圆盘上。当旋转圆盘以角速度ω匀速转动时，不计空气阻力，缆绳延长线与竖直中心轴相交于O点，夹角为θ，O点到座椅的竖直高度为h，则当ω增大时
2A.h不变B.θ增大C.ω2h不变D.ω2h增大20.如图所示，C、D、E、F、G是圆上的等分点，O为圆心，五角星顶角为36°。圆形区域内有垂直纸面向里的匀强磁场，磁感应强度大小为B。质量为m，电荷量为-q的粒子（重力不计）以不同速率沿DG方向从D点射入磁场，则下列说法正确的是A.从F点射出的粒子一定经过O点B.从F点射出的粒子不一定经过O点C.从下射出的粒子在磁场中运动的时间为D.从F射出的粒子在磁场中运动的时间为21.如图所示，质量分别为3m和m的1、2两物块叠放在水平桌面上，物块2与桌面间的动摩擦因数为μ，物块1与物块2间的动摩擦因数为2μ。物块1和物块2的加速度大小分别用a1、a2表示，物块l与物块2间的摩擦力大小用f1表示，物块2与桌面间的摩擦力大小用f2表示，最大静摩擦力等于滑动摩擦力。当水平力F作用在物块1上，下列反映a和f变化的图线正确的是
3第Ⅱ卷三、非选择题：包括必考题和选考题两部分。第22题~第32题为必考题，每个试题考生都必须作答。第33题~第38题为选考题，考生根据要求作答。（一）必考题（共129分）22.（6分）用如图甲所示装置探究小车加速度与其质量的关系，图乙是其俯视图，实验操作步骤如下：（1）取两个相同的小车A、B，放在光滑的水平桌面上，小车的左端各系一条细绳，绳的另一端跨过定滑轮各挂一个相同的重物，小车的右端各系一条细绳，细绳另一端用夹子固定；（2）A车质量为mA，通过在B车上加砝码改变B车质量，B车和车上砝码的总质量为mB，并使A、B两车的质量均______重物质量（选填“远大于”、“约等于”、“远小于”）；（3）打开夹子，重物牵引小车运动，合上夹子，两小车同时停止。用刻度尺分别测出A、B两小车通过的位移为xA、xB。则两车的加速度之比aA:aB=__________；（4）某同学猜想小车加速度与质量成反比，则xA、xB，与mA、mB，应满足的关系式为_________；（5）改变B车上砝码的质量重复实验，验证该同学的猜想并得出结论。23.（9分）利用如图甲所示电路，可以测量电源电动势和内阻，所用的实验器材有：待测电源，电阻箱R（最大阻值999.9Ω)，表头G（量程为200μA，内阻为900Ω），定值电阻R0、R1，开关S，导线若干。
4实验步骤如下：（1）先利用R0和表头G构成2mA的电流表，再将2mA的电流表改装成量程为6V的电压表，根据题设条件定值电阻的阻值应为：R0=________Ω，R1=________Ω；（2）将电阻箱阻值调到________（选填“最大“或“最小”），闭合开关S；（3）多次调节电阻箱，记下表头G的示数I和电阻箱相应的阻值R；（4）以为纵坐标，为横坐标，作-图线如图乙所示；（5）若把流过电阻箱的电流视为干路电流，根据-是图线求得电源的电动势E=_______V.内阻r=________Ω；（6）由于步骤（5）的近似处理会导致测量的电动势_______，内阻_______（此两空均选填“偏小”、偏大”、“不变”）。24.（14分）如图甲所示，ac、bd是两条间距为L=2m的粗糙平行金属轨道，轨道平面与水平面夹角为θ=30°，轨道足够长且电阻不计。a、b之间接一定值电阻R=3Ω，与ab相距x=1m处有一质量为m=0.6kg的金属杆ef垂直轨道放置。且恰能处于静止状态，金属杆ef接入电路的电阻值为r=1Ω，总是与轨道良好接触，若某时刻在导轨所在空间加一个磁扬，方向垂直于轨道平面向上，磁感应强度B随时间t的变化关系如图乙所示。重力加速度g=10m/s2，求：（1）加磁场后，金属杆ef两端的电压；
5（2）加磁场开始为记时起点，经过多长时间金属杆ef可开始运动。25.（18分）如图甲所示，一轻质弹簧一端固定在倾角为37°的固定直轨道的底端，另一端位于直轨道上O点，弹簧处于自然状态，距O点为l0的C处有质量为m(可视为质点）的小物块A，以初速度v0沿斜面向下作匀加速直线运动，加速度大小为0.4g，小物块A将弹簧压缩至最低点O1，随后沿斜面被弹回，小物块A离开弹簧后，恰好回到C点，已知重力加速度大小为g，sin37°=0.6，cos37°=0.8。求：（1）物块A与直轨道间的动摩擦因数μ；（2）小物块A运动到O1点时弹簧的弹性势能；（3）若将小物块A与弹簧上端栓接，另一个质量为2m的物块B（可视为质点）与A叠放在一起（不粘连），A、B与直导轨之间的动摩擦因数相同，外力推动B使弹簧压缩到O1点位置静止（图乙），然后撤去外力.A、B一起向上运动一段距离后分离，求小物块B向上滑行距O1点的最大距离。（二）选考题（共45分）33.[物理一选修3-3]（1）（5分）下列说法正确的是_________（选对1个得2分，选对2个得4分，选对3个得5分。每选错1个扣3分，最低得分为0分）A.分子间距离增大分子引力增大、斥力减小B.温度与分子热运动的平均动能成正比C.气体的压强是由大量气体分子频繁地碰撞器壁而产生的D.液体表面张力的方向与液面垂直并指向液体内部E.同种元素可以构成不同的晶（2）（10分）如图所示，缸壁光滑的柱形气缸固定在水平面上，中间开口与大气相通，大气压强P0=l×105Pa，一劲度系数k=1×103N/m的弹簧与两活塞连接，处于原长，活塞截面积S=1×10-3m2。A、B两室有同种理想气体，压强均为P0，两活塞距左、右两壁均为l0
6=0.10m，除B室右壁导热外，其他缸壁和活塞均绝热。现对左室缓慢加热，当A室气体吸收1000J的热量时它对活塞做了200J的功，B室活塞向右移动了0.02m，该过程中设外界温度保持不变，A、B气室均不漏气。求：①A中气体内能的变化量；②A中活塞移动的距离。34.[物理--选修3-4]（1）（5分）由波源S形成的简谐横被在均匀介质中向左、右传播。已知介质中P、Q两质点位于波源S的左右两侧如图甲所示，P、Q的平衡位置到S的平衡位置之间的距离分别为5.1m、8.2m.图乙中实线和虚线分别表示P、Q的振动图像，下列判断正确的是____。（选对1个得2分，选对2个得4分，选对3个得5分。每选错1个扣3分，最低得分为0分）A.P、Q的起振方向相同B.P、Q的运动方向始终相同C.当P在波峰时.Q向下运动D.这列波的频率为10HzE.这列波的波速可能为10m/s（2）（10分）如图所示，一个三棱镜的截面为等边三角形ΔABC。此截平面内的光线以入射角i1从AB边射入，进入棱镜后直接射到AC边，并以出射角i2射出，当i1=i2时出射光线相对入射光线偏转角为30°。求：①此时入射角i1；②三棱镜的折射率n。
7
8泸州市高2015级第二次教学质量诊断性考试物理部分参考答案及评分意见第I卷选择题（共48分）选择题（本题包括8小题，每小题6分，共48分。每个小题所给出的四个选项中，有一个或多个是符合题目要求的。全部选对的得6分，选不全的得3分，有选错或不答的得0分）。14D15C16B17D18B19BC20AD21AC第II卷非选择题（共62分）22.（6分）（2）远大于（3）xA:xB（4）=(凡是满足这种关系式都可以)··········每空2分23.（9分）（1）R0=100WR1=2910ΩW····································································每空1分（2）最大············································································································1分（5）E=6.0Vr=r=0.2Ω············································································每空2分（6）偏小偏小························································································每空1分24.（14分）解：(1)设感应电动势为E，由法拉第电磁感应定律得
9············································································································2分由于则·········································································································1分设流过电阻R的电流为I，由闭合电路的欧姆定律可知···········································································································2分设金属杆ef两端的电压为Uef，由部分电路的欧姆定律可知Uef=IR·············································································································1分解得：Uef=1.5V·································································································1分（2）设金属杆ef与导轨之间的最大静摩擦力为fm，由平衡得fm=mgsinθ······································································································2分设金属杆ef刚好运动时的磁感应强度为B，由题意可知Mgsinθ+fm
10=BIL······························································································2分由数学关系可知B=kt···············································································································2分代入数据t=6s·······································································································1分25.（18分）解：（1）设物块A与直轨道间的动摩擦因素为μ，由牛顿第二定律可知mgsinθ-μmgcosθ=ma····················································································2分解得：μ=0.25····································································································2分（2）设小物块A将弹簧压缩到最低点时，弹簧的压缩量为x，小物块从初位置到再次回到起始、位置整个过程，由动能定理知2μmgcosθ(l0+x)=0-mv02·············································································2分解得：
11································································································1分设小物块从最低点运动到弹簧处于原长时，弹簧所做的功为w，从最低点到C点的过程中，由动能定理得：······································································2分弹簧所做的功为w=-ΔEP······················································································1分解得：···································································································1分（3）设小物块A、B一起到达弹簧原长处的速度为v1由动能定理可知···································································2分····························································································1分设小物块B从弹簧原长处继续上滑的距离为x1，根据动能定理·····························································
122分设小物块B向上滑行的最大距离为l'l'=x+x1············································································································1分·············································································································1分33题（1）BCE····································································································5分(2)解：①由ΔU=Q+W··················································································2分可得：DΔU=1000-200=800J······························································2分即内能增加800J②因B室气体作等温变化P0Sl0=PSl·························································································1分l=l0-0.02
13··························································································1分B中活塞受力平衡P0S+kx=PS·······················································································1分····································································1分x=0.025m·······························································································1分所以A中活塞向右移动：x+0.02m=0.045m·············································1分34.（1）ADE·······························································································5分(2)解：①令AB界面的折射角为r1，AC界面的入射角为r2，则偏转角········································2分i1=i2·······················································································1分r1=r
142······················································································1分由几何关系得：······································································1分·············································································1分②由折射定律可知···················································2分n=·····················································2分
