物理答案_2024-2025高一（7-7月题库）_2024年11月试卷_1121山西省太原市2024-2025学年高一上学期11月期中考试

2024 ~ 2025 学年第一学期高一年级期中学业诊断 物理参考答案及评分建议 一、单项选择题：本题包含10小题，每小题3分，共30分。 题号 1 2 3 4 5 6 7 8 9 10 答案 A D C C D B A A C C 二、多项选择题：本题包含5小题，每小题3分，共15分。 题号 11 12 13 14 15 答案 AC BD BC AC ACD 三、实验题：本题包含2小题，共14分。 16.（9分） （1）B（2分） （2）1.0或0.99(3分） （3）1.9(4分） 17.（5分） 1 实验原理：查阅本地重力加速度的值 自由落体位移时间关系h  gt 2 (2分） 2 实验器材：毫米刻度尺（1分） 实验步骤：一名同学竖直握刻度尺上端，刻度尺下端零刻度线对齐被测试同学的虎口。静 止释放刻度尺，被测试同学握住刻度尺，被测试同学的虎口对应位置即反应时间内刻度尺 下落的高度。代入公式，计算被测试同学的反应时间。（2分） （其余合理答案均可得分） 四、计算题：本题包含5小题，共41分。 18.（7分） 汽车通过该位置的瞬时速度可近似用平均速度表示 高一物理答案 第 1 页 {#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}v= ···························································································（4分） ∆ v= =25m/s=90km/h ································································（2分） . 90km/h>80km/h 汽车超速·······························································（1分） 19.（7分） （1）运动员下落过程做自由落体运动 v2 -02gh···················································································（2分） 0 v  2gh 8m/s··········································································（1分） 0 （2）以竖直向下为正方向，加速度为 v-v a  0 ······················································································· （2分） Δt a 18m/s2 ··················································································（1分） 加速度大小为18m/s2，方向竖直向上···················································（1分） 20.（8分） （1）玩具小车做匀加速直线运动 1 x  at 2 ·····················································································（2分） 1 2 11 t 3s···························································································（1分） 1 （2）玩具小车匀加速至B点，在B点速度 v at 3m/s··············································································· （2分） 1 11 玩具小车接着保持加速后的最大速率3m/s沿BC转弯，最后匀减速至D点 0v 2 x  1 ···················································································（2分） 2 2a 2 x 2.25m··················································································· （1分） 2 21.（9分） （1）计时开始后第1s内运动员位移 高一物理答案 第 2 页 {#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}1 x  at2 0.5m········································································（1分） 1 2 计时开始后第1s运动员运动的时间 t 1t ······················································································（1分） 0 计时开始后前两秒内运动员位移 1 x a  1t 2 ··············································································（2分） 2 计时开始后运动员第2s内位移 x 4m 2 计时开始后前两秒内运动员位移 x x x 4.5m··········································································（1分） 1 2 t 0.5s························································································（1分） 0 （2）计时开始后第1s内，运动员的位移 1 x  at2 0.5m 1 2 t 1t 0.5s·············································································（1分） 0 运动员起跑的加速度的大小 a 4m/s2······················································································（2分） 22.（10分） （1）根据速度时间关系 ··············································································（1分） m = + A车加速到最大速度的时间为 = A车行驶2.5s后的位移为 + m = + m − ······················································································（2分） = 高一物理答案 第 3 页 {#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}这段时间内B车的位移 ··············································································（1分） = = 此时A、B两车相距 ·········································································（1分） = + − ······················································································（1分） = （2）当A车减速到两车速度相等时恰好不相撞 ··············································································（1分） = m− ··········································································（2分） m− − = A车减速时的加速度的大小为 ················································································（1分） 2 = . m/s 高一物理答案 第 4 页 {#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}