山西省太原市2023-2024学年高三上学期期中学业诊断 物理答案

2023～2024学年第一学期高三年级期中学业诊断物理参考答案及评分建议一、单项选择题：本题包含8小题，每小题4分，共32分。题号12345678选项CCDADCCC二、多项选择题：本题包含4小题，每小题4分，共16分。题号9101112选项CDCDACBD三、实验题：共16分。13.（8分）（1）4.80（2分）（）（分）（分）22�2�����????（3）减小遮光片宽度（其它合理建议也可得分）（2分）14.（8分）（3）小于（2分）（4）（2分）�����（2分）����=����−����（2分）��=��+��四、计算题：共36分。15.（7分）（1）赤道处·····································································（1分）两极处·········································································（1分）·············································································（1分）（2）两极处·········································································（1分）········································································（1分）···················································································（1分）·········································································（1分）16.（8分）（1）根据题意，由平衡条件有··········································································（2分）��+���=���由玻意耳定律有pVpV0011···················································································（1分）联立解得·············································································（1分）�����=��+�����带入数据得3V1=432cm·················································································（1分）（2）根据题意，由理想气体状态方程有pVpV000xTT02·················································································（1分）解得TV2VxT00····················································································（1分）则漏出气体与剩余气体质量之比VV1x0V300············································································（1分）17.（10分）（1）小球B质量为m1，下落h时速度为v,以下为正方向v2gh················································································（1分），与小球A碰后反弹速度大小为，小球A碰后瞬间速度为v2，在碰撞的前���=�后瞬间mvm(v)mv1112·····································································（2分）121212m1vm1(v1)mv2222·························································（2分）1v2v2·····················································································（1分）····················································································（1分）���=��（2）碰后的小球A再次压缩弹簧x0的过程，设弹簧弹性势能的增加量为△Ep12mv2mgx0Ep2···································································（2分）1Epmghmgx04····································································（1分）18.(11分)（1）以A为研究对象，由动量定理···············································（1分）��????−��????=����=���/�物块A以10m/s滑入传送带，物块在传送带上的加速度为a=μg=2m/s2··················································································（1分）匀减速阶段所用时间t1··········································································（1分）���=�????�−�????�1s��=此时A速度=8m/s，A以v2=8m/s撞向B−��=��????�····················································（1分）����+����=����+����·············································（1分）�������������+�����=�����+�����解得v4=-1m/s,v5=5m/s·····························································（1分）物块A以1m/s返回进入传送带，经过t2时间速度减为零，返回时从传送带离开仍以v6=1m/s离开,之后从平台飞出=0.5s�−����=�在传送带上总时间t=t1+2t2=2s··················································（1分）(2)AB下落时间为t3，=0.6s················································（1分）????��=�=v6t3=0.6m············································································（1分）��=v5t3=3m·············································································（1分）��xB-xA=2.4m···············································································（1分）