2023届山东省青岛地区西海岸、平度、胶州、城阳四区高三上学期期中考试物理试题 PDF版

这是一份2023届山东省青岛地区西海岸、平度、胶州、城阳四区高三上学期期中考试物理试题 PDF版，文件包含山东省青岛地区西海岸平度胶州城阳四区2022-2023学年高三上学期期中考试物理试题pdf、202211高三物理答案docx等2份试卷配套教学资源，其中试卷共12页， 欢迎下载使用。
2022－2023学年度第一学期期中学业水平检测       高三物理答案及评分标准    2022.11一、单项选择题：本大题共8小题，每小题3分，共24分。1．B  2．B  3．A  4．D   5．C   6．A  7． D   8． C 二、多项选择题：本大题共4小题，每小题4分，共16分，选不全得2分，有选错得0分。  9．BC  10．AC  11．BCD  12．BD 三、非选择题13．（6分）（1）F－v2（或F－vn，都可以）··················（2分）（2）成反比…（2分）、当圆周运动的速率不变而半径改变时，F与r的乘积相等。（2分）14.（9分）（1）m<<M … （1分）； （2）…（2分）； （3）丙…（2分）；（4）mgs=（2分）空气阻力和滑轮与细线间的摩擦（2分）15．（8分）（1）设每块砖厚度为d上升过程：8d－4d＝a1T2····················（1分）mg＋f＝ma1······························（1分）下降过程：5d－3d＝a2T2····················（1分）mg－f＝ma2······························（1分）解得：f＝mg·····························（1分）（2）设上升的最大高度为H上升过程：v02＝2a1H······················（1分）下降过程：v2＝2a2H·······················（1分）解得：v＝v0·····························（1分）评分标准：第1问，5 分；第2问，3 分。共8 分。    16．(8分) （1）设水桶做匀速运动时受到细绳的拉力为F，则有F＝(M＋m)g·······································(1分)P＝Fvm··········································(1分)解得vm＝·········································(1分) （2）设水桶在水中受到的浮力为F浮，桶口运动到井口的过程中，由动能定理得W－(M＋m)gd＋d＝0·································(2分)F浮＝mg··········································(1分)解得W＝gd········································(2分)评分标准：第1问，3 分；第2问，5 分。共8分。17．(13分) （1）由动能定理：m0gR＝m0vB2 ······················（1分）     得：vB＝3m/s    由牛顿第二定律：FN+m0g＝m ······················（1分） 得：FN＝2.5N·································（1分）由牛顿第三定律可得：小球对轨道的作用力为2.5N，方向竖直向上。（1分）（2）因为小球与滑块碰撞时交换速度，所以轨道间在运动的其实可以只考虑一个物体对滑块分析：因为mgsin53°>μmgcos53°，所以小球不能停在轨道上（1分）则小球做往复运动后，最后停在D点··················（1分）由mg(R+2r+2rsin30°)＝μmgscos53°·····················（1分）得s＝5m··········································（1分）     （3）当mgsinθ＝μmgcosθ时，θ＝37°0≤θ≤37°时，小滑块能停在倾斜轨道上················（1分）由mg(R+2r+2rsin30°)－mgxsinθ－μmgxcosθ＝0···········（1分）    得x＝则Wf＝μmg xcosθ＝······························（1分）37°＜θ≤60°时，小滑块最后停在D点Wf＝mg(R+2r+2rsin30°)···························（1分） 则Wf＝11.25J··································（1分）评分标准：第1问，4 分；第2问，4 分；第3问，5 分。共 13 分。  18．(16分)（1）在地球表面：·····································（1分）     由：得：v＝········································（1分）（2）设地球密度为ρg＝＝···········································（1分）距地心为h时：g1＝＝······························（1分）得：g1＝········································（1分）（3）设卫星从B向O′运动的过程中距离O′为x卫星在不同位置时的重力加速度g2＝···················（1分）卫星受到的地球引力：F引＝mg2＝mg··················（1分）卫星受到的地球引力沿通道方向的分力：F＝F引＝（1分）即：F随x均匀变化，且F与物体偏离O′位置的位移始终相反，所以物体作的是以O′为平衡位置的简谐运动。              （1分） （4）卫星从B点运动到O′位置过程由动能定理：W＝＝··········································（1分）解得：v0＝······································（1分）m、M在O′发生弹性碰撞：Mv0－m v0＝Mv1+m v2······························（1分）···············································（1分）解得：v2＝，由于M>>m可知：v2＝3 v0····································（1分）令卫星回到B点时的速度为第一宇宙速度v由动能定理：·····································（1分）解得：h＝·······································（1分） 评分标准：第1问，2 分；第2问，3 分；第3问，4分；第4问，7 分。共16分。