山西省太原市2023-2024学年高三上学期期末学业诊断物理答案

2023～2024学年第一学期高三年级期末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BBBBCBBD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ACBDABCD三、实验题：共16分。13.（6分）红（1分）1.5（1分）变大（2分）1.4（2分）14.（10分）（1）AC（2分）（2）（2分）（3）（图像为一条直线）（2分）1.36~1.50（2分）3.41~4.65（2分）四、计算题：共36分。15.（8分）（1）物块P第一次在传送带上向右运动的最远点为D点······························（3分）�????�−��????????�????°∙????�????°−��????��=�−�m·····················································································（1分）��=���<�????（2）物块P从D点运动到B点··································································（1分）����????��=����−�6m/s��=���/�<物块P从B点运动到斜面最高点······································（2分）����−????��−????????�????°∙????�????°=�−����m····················································································（1分）���=�16.（8分）(1)粒子在磁场中做圆周运动的半径为rθ由几何关系得rtan＝R·············································································（1分）22v0qv0B＝m··················································································（1分）rB＝��????�????粒子从O到Q做类平抛运动，运动时间为t22L＝v0t2·····································································································································（1分）t2＝��????1qE2L＝··t2··················································································（1分）2m＝E��????���···················································································（1分）����=��????(2)粒子在磁场中做匀速圆周运动的周期为Tv0＝························································································（1分）????��粒子在磁场中运动时间为t1t1＝·······················································································（1分）�????�t1＝????��????粒子从P运动Q的时间为t＝t1＋t2＝·····································(1分)��+????��????17.（10分）(1)P物块下摆2mv10·········································································（1分）�????�=�−P物块上摆2mv2····································································（1分）��−????��=�−�以右为正方向v1=v2=????�????�−�mv1=mv2+3mv3···········································································（1分）222mv1=mv2+3mv3··································································（1分）������等式成立，P与Q碰撞为弹性碰撞（2）Q物块从B到C223mg=3mvc3mv3··················································（1分）�����−��−�vc=v3=方向向上????��物块在从C点飞出后水平方向做匀加速直线运动，竖直方向做竖直上抛运动=g,=g···························································································（1分）����=����在竖直方向上，以下为正22vD-vC=2L··········································································································（1分）��vD=vC=-????��????���t=···············································································································（1分）��−����在水平方向上，以右为正x=t2=L·····························································································（1分）��+�����x总=x+L=L···································································································（1分）�+��18.（10分）（1）撤去外力瞬间，甲棒的速度为v，此时回路的电流为BLvI2R甲、乙受到的安培力大小相同，都为B2L2vFBIL·········································································（1分）A2RF=maB2L2vB2L2va甲，a乙·····························································（2分）4mR2mR（2）在t时间内，甲、乙两棒的电流相同，流过的甲乙的电荷量也相同甲棒FtBLIt2mv·······························································（1分）qIt·························································································（1分）Ft2mvq··············································································（1分）BL（3）撤去外力F后，甲做加速度减小的减速运动，乙棒做加速度减小的加速运动，最后两棒速度相等，末速度为v，甲乙组成系统动量守恒，以右为正2mv3mv1212Q2mv3mv···································································（1分）221Qmv230Q甲1Q212Q甲mv················································································（1分）60在甲乙两棒达到共速之前，设两棒的速度分别是v甲和v乙，22BLv甲-v乙乙棒tmv0··················································（1分）2R−两棒的位移差xv甲-v乙t4mvRx·················································································（1分）3B2L2