2022一2023学年第二学期期末数学质量调研
高一物理试题
注意辜项：
1.答卷前，考生务必将自己的学校、班级、姓名、考生号等填写在答题卡指定位置。
2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用稼
皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在等题卡上，写在本试卷上无效。
一、单项选择题：本题共8小题，每小题3分，共24分。每小题只有一个选项符合题目要求。
1,2023年5月30日0931，搭载景海鹏、朱杨柱、桂海潮3名航天员的神舟十六号载人飞船由长征二号F
遥十六运载火箭在酒泉发射场发射升空，正式开启空间站新阶段首次载人飞行任务。载人飞船在远离地
球的过程中，所受地球引力大小F随它距地面的高度h变化的关系图像可能正确的是
B
D
2,在如图所示的四种电场中，分别标记有4、b两点。其中a、b两点电场强度相同的是
⊙-
…
●力
印
Z
丙
A.甲图中与点电荷等距离的a、b两点
B.乙图中两等量异种电荷连线的中垂线上与连线等距离的Q、b两点
C.丙图中两等量同种电荷连线的中垂线上与连线等距离的α、b两点
D.丁图中，与无穷大金属板等距离的关于垂线对称的4、b两点
3.如图所示，原来不带电的半径为r的金属球放在绝缘支架上，右侧放一个电荷量为十Q的点电荷，点电
荷到金属球表面的最近距离为2。设静电力常量为k,下列说法正确的是
A,金属球右端感应出正电荷
B.点电荷Q激发的电场在金属球内任意位置电场强度均为零
物理试颗第1而（共7而）
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C.感应电荷在金属球球心处激发的电场强度为
9-3
D.感应电荷在金属球球心处激发的电场强度为9
4.在长约1m的一端封闭的玻璃管中注满清水，水中放一个大小适当的圆柱形红蜡块，玻璃管的开口端用
胶塞塞紧，保证将其迅速竖直倒置后，红蜡块能沿玻璃管由管口匀速上升到管底。
现将此玻璃管倒置安装在置于粗糙桌面上的小车上，如图所示。小车从A位置以
初速度开始向右先做匀减速直线运动，再做匀加速直线运动，同时红蜡块沿玻
璃管匀速上升。按照题图建立的坐标系，在这一过程中红蜡块实际运动的轨迹可
能是下列选项中的
B
5.长征七号A运载火箭于2023年1月9日在中国文昌航天发射场点火升空，托举实践二十三号卫星直冲
云臂，随后卫星进入预定轨道，发射取得圆满成功。已知地球表面的重力加速度为g,地球的半径为R,
实践二十三号卫星距地面的高度为h(小于地球同步卫星距地面的高度)，入轨后绕地球做匀速圆周运
动，则该卫星
A.线速度大于7.9m/s
B.周期可能超过24小时
C.向心加速度大于地面重力加速度g
D.角速度大于地球同步卫星的角速度
6.如图所示，一条不可伸长的轻质软绳跨过定滑轮，绳的两端各系一个质量分别为m、3m的小球a和，
用手按住α球静止于地面时，b球离地面的高度为。两物体均可视为质点，定滑轮的质量及一切阻力
均不计，重力加速度为g。释放a球后，b球刚要落地前，下列说法正确的是
A.a球的机械能守恒
B,b球的机械能增加
C.b球刚要落地时的速度为√gh
D.b球刚要落地时的速度为√2gh
物理试题第2页（共7页）
00000002022—2023学年第二学期期末教学质量调研
高一物理参考答案及评分标准
一、单项选择题：本题共 8小题，每小题 3分，共 24分。每小题只有一个选项符合题目要求。
1. B 2. B 3. C 4. A 5. D 6. C 7. D 8.D
二、多项选择题：本题共 4小题，每小题 4分，共 16分。每小题有多个选项符合题目要求。全部
选对得 4分，选对但不全的得 2分，有选错的得 0分。
9. BD 10. AC 11.CD 12. ABD
三、非选择题：本题共 6小题，共 60分。
2
13.（1）C (2分) 2 m hC hA （ ） (2分) （3）略大于 (2分)
8T 2
14.（1） 8.0 10-3 或7.5 10-3 8.5 10-3 (2分)； 1.3 10-3 或1.2 10-3 1.4 10-3 (2分)
（2）变大 (2分)； 不变 (2分)
a
15．（8分）解: （1）设 AC的长度为 L，由几何关系： L 0 ·······································①sin30
Q
A点点电荷在 C点的电场强度大小为 EA： EA k 2 ······················································②L
kQ
解得： EA 2 ····································································································③4a
（2）由 C点电场强度方向可知，B点的点电荷带正电 ···················································④
Q
设 B点点电荷在 C点的电场强度大小为 E BB：EB k 2 ··················································⑤a
EAsin30
0 = EB ····································································································⑥
1
解得：QB Q ·································································································⑦8
评分标准：①②③④⑤⑦各 1分，⑥式 2分
16 v
2
．（8分）解：（1）设质点通过 A点时受到轨道的支持力大小为 FN：F mg FN m ········①R
由牛顿第三定律得，质点对轨道的压力大小： F N FN ·················································· ②
解得： F N 7mg ····································································································· ③
v 2
（2）质点恰好通过 B点时： F mg m B ··································································④
R
解得： vB 6gR ·····································································································⑤
评分标准：①④⑤式各 2分，②③各 1分
17．（14分）解：（1）对工件受力分析，根据牛顿第二定律得：
mgcos mgsin ma ···························································································①
a mgcos mgsin 2.5m/s2解得：
m
经过 t v1时间工件与传送带速度相等，则匀加速运动时间为： t1 0.8s ······························②a
v
（2）工件加速至与传送带速度相等时运动的距离为： x1 t1 0.8m ·································③2
传送带运动的位移： x2 vt1 ·······················································································④
工件相对于传送带运动的位移： x x2 x1 ···································································⑤
工件与传送带摩擦产生的热量：Q mgcos x ·····························································⑥
解得：Q 6J ··········································································································⑦
（3）传送带持续传送工件时每经过 1s掉落一个物块，故以 1s为周期，传送带多做的功为：
W 1 Q mv2 mglsin ····························································································⑧
2
W
故电机增加的平均功率： P ················································································⑨
t
P 68w ···············································································································⑩
评分标准：①②式各 2分，③④⑤⑥⑦⑨⑩式各 1分，⑧式 3分
qE
18.（16分）解：（1）粒子从 A点运动到 B点，由牛顿第二定律得： a 01 ··························①m
1
粒子在 x方向： l0 a t
2
1 1 ··························································································②2
粒子在 y方向： l0 vBt1 ···························································································· ③
qE0l0
联立可得粒子运动到 B点时的速率： vB ···························································④2m
1 2 1 2
（2）粒子从 B点运动到 D点，由动能定理得： qE04l0 mvD mvB ···························· ⑤2 2
v 17qE0l解得： D 0 ································································································⑥2m
（3）粒子从 B点到 D点： t
2l
2
0 2 2ml0 ·································································⑦
vB qE0
2qE l
粒子在 D点时的水平方向的速度： v 0 0Dx a1t2 2 ··················································⑧m
粒子到达 D点后，在 0-T内粒子所受电场力水平向左，沿 x轴方向先做匀减速直线运动，
a qE 12 ·················································································································⑨m
粒子沿 x轴方向减速为零：0 vDx a2t3 ······································································· ⑩
t 9ml0 T3 ····································································································· 2qE0 2
T
再经 -T 粒子在水平方向上反向加速至 vDx，由对称性可知粒子在 x方向的坐标： x 3l0 ··· 2
粒子在 y方向上匀速运动: y vBT 3l0 ········································································
粒子在 y方向的坐标： y 2l0 y ···············································································
故粒子 T时刻的坐标（3l0 ,5l0）·················································································
评分标准：①②③④⑥⑦⑧⑨⑩ 式各 1分，⑤式 2分
