2024学年第一学期丽水五校高中发展共同体期中联考高一年级物理学科参考答案命题1：庆元中学吴锋命题2：松阳一中王潜龙命题3：青田中学林燕平一．选择题1234567891011121314CDCBABCCABDAAA15161718ADABBCCD二．非选择题19.（1）0.05（2）BD（3）0.4751.2520.（1）3.70（3.68，3.69，3.70，3.71，3.72均给分）（2）B（3）CD（4）丙121.（1）根据匀变速直线运动公式，可得xat2·············································（2分）2解得a=8m/s2·························································································（1分）（2）根据匀变速直线运动公式，可得v=at······················································（2分）解得v=24m/s························································································（1分）vv（3）由匀变速直线运动平均速度公式可得：v0······································（2分）2解得v12m/s··················································································（1分）（）对结点受力分析可知（分）22.1OF弹TOBtan·················································2联立解得（分）又F=kx，TOB=mBgx=0.32m····························································1mg（2）对结点O受力分析可知TB·························································（1分）OCcos对受力分析可知（分）ATOCmAgsinθ································································1联立解得mA=5kg·························································································（1分）（）再次对受力分析可知（分）3ATOCmAgsinθf············································2解得f=8N，方向沿斜面向下···········································································（2分）高一物理答案第1页(共2页)v2v2（）由匀变速规律可知mm（分）23.1xm······················································32a12a222又a1=2a2解得a1=1.6m/s，a2=0.8m/s·························································（1分）2v1（2）由匀变速规律可知x0.2m解得v1=vm=0.8m/s，·······························（1分）2a1v所以t10.5s························································································（1分）a1又感应距离l=v人t························································································（1分）解得l=0.75m·······························································································（1分）（3）搬运物体到感应区后，门先匀加速移动0.2m，若搬运物体成功，门至少还要匀减速移动的1xDs10.3m距离为2···············································································（1分）12xvmt1a2t1由2可得门做匀减速运动用时t10.5s或t11.5s（舍去）ttt1s则开门用时01················································································（1分）Lv物0.75m/sv可得t所以物体的移动速度物不能超过0.75m/s·····························（1分）高一物理答案第2页(共2页)