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高2025届【物理答案】·第1页(共2页)重庆市名校联盟2023-2024学年度第一期期中联合考试物理试题参考答案(高2025届)一、单项选择题(本大题共7小题,共28分)1234567BABCDCA二、多项选择题(本大题共3小题,共15分)8910BDADBC三、
实验题(本大题共2小题,共15分)11题(1)电势差(2分)(2)减小(2分)减小(2分)12题(1)C(2分)(2)D(2分)(3)130220mxxmxx(2分)m1(x1-x0)=130220(())mxxmxx(3分)四
、解答题(解答应写出必要的文字说明、方程式和重要演算步骤,只写出最后答案的不能得分,有数值计算的题,答案中必须明确写出数值和单位,若解答过程合理且结论正确可给相应分值)13题(12分)答案:(1)60V(2)110V(3)-2.5×10-6J解析:(1)A、B间的
电势差:..........................................................................................................................................
..........(2分)UAB=60V......................................................................................................
........................................................(2分)(2)因为UAB=φA-φB........................................................................
..................................................................(2分)则:φA=φB+UAB=50V+60V=110V.....................
...........................................................................................(2分)(3)点电荷放在B点时它的电势能:EP=qφB..............................
.....................................................................................................
.....................(2分)EP=-5×10-8×50J=-2.5×10-6J.........................................................(2分)14题(12分)答案:(1)30N(2)1.28m
解析:(1)根据题意:由机械能守恒定律有:212BmvmgR解得:24m/sBvgR.................................................(2分)小物块在B点时,根据牛顿第二定律
有:2NBvFmgmR...........................................................(2分)解得:N30NF......................................
..............................................................................................(1分)根据牛顿第三定律得:NN30NFF..................
.............................................................................(1分)(2)根据题意,设小物块到达木板右端时恰好和长木板达到共同速度v,根据动量守恒定律有:BmvMmv
解得:0.8m/sv.......................................................................................................................
..........(2分)设此时长木板的长度为L,根据功能关系有:2211()22BmgLmvMmv...........................................(2分)解得:1.28mL(分数表示为m2532L)(其它解法合理也可).
...........................................(2分)qWUABAB{#{QQABJYSAggiIAAAAAAgCQw1ACkAQkAEACAoOBBAMsAABQAFABAA=}#}高2025届【物理答案】·第2页(共2页)15题(18分)答案:(1)
0Pv3m/sQv(2)1.5m(3)为保证绳子不松懈需满足为:53.410V/mE;当42.510V/mE时,Q在AB段滑行的路程为12ms;当452.510V/m3.410V/mE时,Q在
AB段滑行的路程为24.5ms解析:(1)小球P运动至与Q碰撞前,有:22101122mgLmvmv....................................................................
...........................................(1分)解得:13m/sv....................................................
.................................................................................(1分)P与Q发生弹性碰撞,有:1PQmvmvmv....
...................................................................................(1分)2221PQ111222mvmvmv.........................
.....................................................................(1分)解得:P0v...................................................................
......................................................................(1分)Q3m/sv...........................................................
...................................................................................(1分)(2)设碰撞后,Q没有滑离A
B段,当Q减速至0,有:211Q102qExmgxmv........................................................................
...........................................(2分)解得:11.5mxd假设成立,即P与Q第一次碰撞后,滑块Q离A点的最远距离为1.5m.............
...........................(2分)(3)当P与Q第一次碰撞后,Q恰好能滑到B点,则:21Q102qEdmgdmv解得:412.510V/mE..................................................
...........................................................................(1分)当42.510V/mE时,Q从B点离开电场,则Q在AB段滑行的路程为:1
2msd.................................................................................................(1分)当42.510V/mE时,Q没有从B离开
,因为:qEmg则Q返回再次与P相碰。设Q第一次在AB段减速至0的位移为2x时,返回与P碰撞后,P恰好能回到O点等高位置,则:2222Q102qExmgxmv...............
......................................................................(1分)设Q回到A点时速度为2v,Q从A开始运动到又回到A点时,有:2222Q11222mgxmv
mv................................................................................................................(1分)P与Q再次碰撞,质量相等弹性碰撞,交换速度,则
碰后P的速度也为v2,此后P运动至圆心等高,有:22102mgLmv解得:523.410V/mE...................................................................................(1分)
当53.410V/mE时,P超过O点等高位置,因为2012mvmgL,P无法到达最高点,绳子松懈,不满足条件............................................
..............................................................................................................(1分)当452.
510V/m3.410V/mE时,因为qEmg,Q无法停在AB段,最终停在A点,全过程,对系统,由能量守恒,有:22012mgsmvmgL.....................................................
.............................(1分)解得:24.5ms,则Q在AB段滑行的路程为:24.5ms.....................................................
................(1分)综上:为保证绳子不松懈需满足为:53.410V/mE当42.510V/mE时,Q在AB段滑行的路程为12ms当452.510V/m3.410V/mE时,Q在AB段滑行的路程为24.5ms(其它解法合理也可){#{QQ
ABJYSAggiIAAAAAAgCQw1ACkAQkAEACAoOBBAMsAABQAFABAA=}#}获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com