2022～2023学年第二学期高一年级期中质量监测
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
选项 B B D A C A B C C D
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
选项 BD BC BD AD BC
三、实验题：共 14分。
16.（6分）
（2）相同 （1分） B （1分）
（3）1:2 （2分） 1:4 （2分）
17.（8分）
（1）匀速直线运动 （2分） 自由落体运动（或 匀加速直线运动） （2分）
（2） 0.050（2分） 1.0 （2分）
四、计算题：共 41分。
18.（7分）
航天器在地、月间飞行时，受到的引力大小与地球、月球半径无关。
设地、月间距为d,航天器质量为m受到地球、月球的引力大小相等时，
它到月心距离为 ，与地心距离为 。
地 月
= ···········································································（3分）
地 = 月 ············································································（1分）
+ = ··············································································（1分）
=3.8× m·············································································（2分）
19.（7分）
（1）当B绳刚伸直，不受到拉力作用时，两绳与杆构成直角三角形，
设此时角速度为ω0，绳A与竖直方向夹角为θ
mgtanθ =m ··········································································· （1分）
tanθ = ······················································································（1分）
r = 5L
ω 0= ··················································································（1分）
（2）设此时，绳A对小球的拉力为FA,B绳对小球的拉力FB,
FAsinθ +F = m B ······································································（2分）
FAcosθ =mg ················································································（1分）
F B= ··················································································（1分）
20.（7分）
（1）运动员在圆弧最低点B，应用牛顿运动定律：
=
F1=1800N··················································································· （1分）
由牛顿第三定律得F1’= - F1
所以运动员给B点的压力为 1800N.··················································· （1分）
（2）从B点到C点,由牛顿运动定律F合 =ma = 得
A =0.5m/s2···················································································（1分）
由运动学公式 = ···························································（1分）
由牛顿运动定律 = ······················································（1分）
得F2=1746N················································································（1分）
由牛顿第三定律得F2’= - F2 所以运动员给C点的压力为 1746N···············（1分）
21A.（10分）
1 2
（1）竖直方向： y gt ·························································（2分）
2
水平方向： x v0t ···································································· （1分）
y x2结合 6
t =1s···························································································（2分）
（2）竖直方向速度大小 vy gt ······················································（2分）
根据速度的合成法则，则小物块刚到 p 点时速度的大小为
v v20 v
2
y ···············································································（2分）
v= 101m/s =10.05m/s····································································（1分）
21B.（10分）
(1)设斜面顶端与轨道最高点的水平距离为x,小球在轨道最高点时速度为v0,
到达斜面时的竖直分速度为
= 2gh ··················································································· （1分）
= ·················································································· （2分）
h = ·······················································································（2分）
v = x = 2h0 ····································································（1分） tanθ
(2)设小球在轨道最高点时受到的支持力为F，
+ = ·············································································（2分）
小球对轨道的压力为F`, F` = F ·····················································（1分）
F` = ( - 1) ····································································· （1分）
22A.（10分）
(1)小球从B处沿圆弧切线方向进入圆管,将小球运动到B处时的速度
沿水平方向与竖直方向分解,如图
v = 可得 B = 5 m/s···································································（2分）
A点的初速度v0= 3m/s····································································（1分）
(2)小球运动到B处时,竖直方向的分速度
vy= vB sin α = 5× . m/s = 4 m/s·······················································（1分）
从A→B,小球在竖直方向做自由落体运动,有vy= gt································（1分）
解得下落时间t = 0.4 s
A B h = 、 间的高度差 g t2······························································（1分）
解得h =0.8 m··············································································· （1分）
(3)因为小球恰好能到达D点，所以到达D点的速度为 0··························（1分）
对于小球有：FN= mg = 2N·····························································（1分）
再由牛顿第三定律可知小球对轨道的弹力FN′ = FN=2N，方向竖直向下····（1分）
22B.（10分）
(1)小球运动到 B处时,竖直方向的分速度
vy= vB sin α = 5× . m/s = 4 m/s······················································ （1分）
从 A到 B,小球在竖直方向做自由落体运动,有 vy= gt·····························（1分）
解得下落时间 t = 0.4 s
A、B 间的高度差 h = gt2 = 0.8 m···················································（1分）
(2)设小球在 D处时受到管壁对其的作用力方向竖直向下,大小为 FN
根据牛顿第二定律有 mg + F = m N ··················································（2分）
解得 FN=0.4 N
根据牛顿第三定律,小球在 D处时对管壁的作用力
FN′ =FN=0.4N，方向竖直向上·························································（1分）
(3)因为小球沿切线方向从 B点进入圆管
且有 tanα = = ，·····································································（2分）
再由平均速度关系可知： = ······················································（1分）
所以，弹射装置出口到 B点的高度 H与支架距离 B点的水平位移 x应满足的
关系式为：H = ······································································（1分）
4.马刀锯是一种木匠常用的电动工具。其内部安装了特殊的传动装置，简化后如图所示，可
2022~2023学年第二学期高一年级期中质量监测
以将电动机的转动转化为锯条的往复运动。电动机带动圆盘O转动，圆盘0上的凸起P在
转动的过程中带动锯条左右往复运动。已知电动机正在顺时针转动，转速n=2400r/min,
物理试卷
0P=2cm。当0P与锯条运动方向的夹角0=37°时，
锯条运动的速度大小约为
(考试时间：上午7：30一9：00)
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
A.3m/s
B.4m/s
C.5m/s
D.6m/s
题
号
四
总分
5.2022年夏天，我国南方地区遭遇多轮暴雨袭击，抗洪救灾的战士们利用直升飞机参与救援
得分
物资的运送。已知直升飞机在距离地面某高度以2.5m/s的速度沿水平方向匀速飞行并进
行物资投送，投出的物资恰好落在了正前方距离投放位置水平距离5m的地方。若物资可
一、单项选择题：本题包含10小题，每小题3分，共30分。请将正确选项前的字母填入表内相
应位置。
看成质点，不计空气阻力，重力加速度g取10m/s2。下列说法正确的是
题号
1
4
5
6
7
8
9
9
A.从直升飞机上拍摄的视频看到物资做平抛运动
答案
B.直升飞机的飞行高度为15m
C.物资在投出后第1秒内和第2秒内的速度变化量相同
1.关于曲线运动的叙述正确的是
D.物资在投出后第1秒内和第2秒内的位移之比为1：3
A.做曲线运动的物体某时刻有可能处于平衡状态
6.2023年3月22日CBA常规赛战罢第38轮，山西男篮在这场关键的“卡位”赛中以119比94
B.做曲线运动的物体，速度一定变化
大胜吉林队，比赛中原帅手起刀落命中三分球。如图所示，比赛中原帅将篮球从地面上方
C.做曲线运动的物体加速度一定改变
B点以速度斜向上抛出，恰好垂直击中篮板上A点。若该运动员向前运动到等高的C点
D.做曲线运动的物体，速度方向时刻变化，故曲线运动不可能是匀变速运动
投篮，还要求垂直击中篮板上A点，运动员需
2.在某次足球比赛中，一运动员踢出“香蕉球”，足球划出一道曲线后射门得分。在足球加速
A.减小抛出速度o,同时增大抛射角日
向下射门时，关于足球所受的合力F和速度：的关系可能正确的是
B.增大抛出速度o,同时增大抛射角0
Q
0
C.减小抛射角0，同时减小抛射速度。
9
A.
C.F
D.
D.减小抛射角0，同时增大抛射速度
3.2023年3月山西省文化和旅游厅为了宣传山西省的人文美景，以“走进表里山河，晋享美好
7.随着2022年北京冬奥会的成功举办，冰雪项目在我国也越来越受欢迎。如图所示，某运动
生活”为主题，利用无人机拍摄了宣传片“粤来悦晋”。其中一台无人机，上升向前飞行，竖
员训练时从助滑道上不同位置下滑，都从斜坡顶点B水平飞出。若以速度，飞出，用时：，落
直向上的速度：，及水平方向速度u,与飞行时间的关系图像如图甲、乙所示。则下列说法
在斜坡上的C点；若以速度，飞出，用时，落在斜坡上的D点。已知，=V2t,不计空气阻
正确的是
力。关于运动员从B点飞出，落回斜面的过程，下列判断正确的是
A.无人机在t,时刻处于失重状态
A.两次位移关系有BD=V2BC
B.无人机在0~，这段时间内沿直线飞行
B.两次初速度大小之比=1V2
C.无人机在，时刻上升至最高点
C.两次落在斜面上时速度与斜面夹角之比0，：0，=1：V2
D.无人机在2~时间内做匀变速运动
D.两次落在斜面上时速度大小之比v:v2=1:2
料
高一物理第1页（共8页）
高一物理第2页（共8页）
