【文档说明】广东省汕头市潮阳区2022-2023学年高一上学期期末教学质量监测物理答案及评分标准.docx，共(4)页，43.136 KB，由小赞的店铺上传

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潮阳区2022-2023学年度第一学期高一级教学质量监测试卷物理一、单项选择题:本题共7小题每小题4分，共28分。二、多项选择题:本题共4小题，每小题5分，共20分。全部选对得5分，选对但不全的得3分，有选错的得0分。题号1234567891011答案CBCBDDCABDBCCDAC三、实验：本大

题共2小题，共16分。12.（6分，每空2分）（1）10；（2）1000；（3）F=1000(L-0.1)13．(10分)（1）重物会落在桌面上(或“纸带打点过短”等类似的答案)；应先接通电火花计时器电源，再释放纸带。(4分)（2）b（2分）；（3）

左（2分）；9.75（2分）四、计算题(本题共3小题，共36分)14（8分）.(1)冰车和小孩受力如图所示竖直方向的合力为零，则有FN＋Fsinθ＝mg···································

···········（2分）解得支持力FN＝188N··············································································

·········（1分）(2)在水平方向，根据牛顿第二定律得Fcosθ-Ff＝ma·····························（2分）滑动摩擦力Ff＝μ𝐹压由牛顿第三定律得FN＝𝐹压·······

··································（2分）解得加速度a＝0.33m/s2······································································

·······（1分）15.（12分）解：据题意小轿车初速度𝑣0=30m/s，加速度a=—0.5m/s2；驾驶员的视距d=60m，反应时间𝑡1=0.6𝑠全科免费下载公众号-《高中僧课堂》(1)从刹车到停止时间为t2，则：0206sv

ta−==··········································（3分）(2)反应时间内做匀速运动，则：10118mxvt==··················································（2分）从刹车到停止汽车做匀

减速运动直到速度为0，则：202090m2vxa−==········（3分）小轿车从发现物体到停止的全部距离为：12108mxxx=+=····················（2分）三角警示牌距车的最小距离为：48mxxd=−=···

···································（2分）16.(16分)解：（1）（6分）对小物块和小车分别使用牛顿第二定律得：μmg=mam·····································

·············（2分）F-μmg=MaM····················································（2分）代入数据解得：am=2m/s2·····································

············（1分）aM=0.5m/s2··················································（1分）（2）（4分）对小物块和小车分别使用运动学公式得：v共=amt1，v共=v0+aMt1·····（2分）解得：t1=1s·····

·····································································（2分）（3）（6分）当两者达到相同的速度后，假设两者保持相对静止，以共同的加速度a

做匀加速运动对小物块和小车整体，由牛顿第二定律有：F=(M+m)a解得：a=0.8m/s2此时小物块和小车之间的静摩擦力f=ma=1.6N而小物块和小车之间的最大静摩擦力fm=μmg=4Nf<fm，所以两者达到相同的速度后保持相对静止。························（2分）在

开始的t1=1s内，小物块的位移sm=12amt21=1m末速度v=amt1=2m/s·····（2分）在接下来的t-t1=0.5s内，小物块与小车相对静止，以共同的加速度a=0.8m/s2做匀加速运动，这0.5s内小物块

通过的位移s=v(t-t1)+12a(t-t1)2代入数据解得：s=1.1m从小物块放上小车开始，经过t=1.5s它通过的位移大小为s总=sm+s=2.1m·····（2分）评卷说明：判断两者达到相同的速度后，两者保持相对静止，共给2分。缺少判断的应扣相应的分数。